RESERVOIR ENGINEERING
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INTRODUCTION TO RESERVOIR ENGINEERING
RESERVOIR PRESSURES AND TEMPERATURES
RESERVOIR FLUIDS COMPOSITION
PHASE BEHAVIOUR OF HYDROCARBON SYSTEMS
BEHAVIOUR OF GASES
PROPERTIES OF RESERVOIR LIQUIDS
FUNDAMENTAL PROPERTIES OF RESERVOIR ROCKS
ROCK PROPERTIES MEASUREMENT
PERMEABILITY-ITS VARIATIONS
FLUID FLOW IN POROUS MEDIA
DRIVE MECHANISMS
VAPOUR LIQUID EQILIBRIA
EQUILIBRIUM RATIO PREDICTION AND CALCULATION
PVT ANALYSIS
MATERIAL BALANCE EQUATION
MATERIAL BALANCE EQUATION APPLICATION
WATER INFLUX
IMMISCIBLE DISPLACEMENT
EXAMINATION AND MODEL SOLUTIONS
RESERVOIR ENGINEERING
RE
This Reservoir Engineering module covers material presented in a range of reservoir engineering texts and a
number of the figures and examples are based on these texts and copyright is currently being sought. The student
may find the more detailed analysis in these texts supportive when going through these notes. The following
books are considered useful in building up a reservoir engineering library.
1.Fundamentals of Reservoir Engineering.
L.P.Dake. Elsevier. 1978
ISBN:0-444-41667-6
2.The Practise of Reservoir Engineering.
L.P.Dake. Elsevier. 1994.
ISBN: 0-444-82094-9
3.Principles of Petroleum Reservoir Engineering.
G.H.Chierici. Springer-Verlag 1994.
ISBN:3-540-56037-8
4.Fundamental Principles of Petroleum Reservoir
Engineering
B.F. Towler. Society of Petroleum Engineers Inc
ISBN:55563-092-8
5.Applied Reservoir Engineering
B.C.Craft & M.F.Hawkins. Prentice Hall.
1959.
6.The Properties of Petroleum Fluids 2nd Ed
W.D.McCain Pennwell Books . 1990
ISBN:0-87814-335-1
7.Petroleum Engineering Principles and Practise.
J.S.Archer & C.Wall.Graham & Trotman.
1986. ISBN:0-86910-715-9
8.Petroleum Reservoir Engineering.
J.W.Amyx,D.M.Bass & R.L.Whiting.
McGraw-Hill. 1960. ISBN:07-001600-3
9.PVT and Phase Behaviour of Petroleum Reservoirs A. Danesh. Elsevier. ISBN: 0-444-82196-1
Adrian C Todd
All rights reserved no part of this publication may be reproduced, stored in a retrieval system or
transmitted in any form or by any means, electronic, mechanical, photocopying, recording or
otherwise without the prior permission of the Copyright owner.
Reservoir Engineering notes cover an extensive amount of material. They are support
material for the examination in this topic but are also considered to be useful material
in subsequent career use. Not all the material in the text can be covered in a limited
time examination.
In the context of the examination a student should consider the learning objectives at
the front of each section which should help in the level of detail and analysis which
is required in relation to an examination covering the various topics.
Detailed below is a graded analysis of each section which should help the candidate
in examination preparation. These should be considered alongside the learning
objectives.
Grading structure:
5
4
3
2
1
-
Core material for examination purposes
Core material less analytical than 5 - examinable.
Between 4 & 2
General awareness. Not so examinable with respect to analysis of detail.
Other information not examinable.
OM- Material covered in another module not for examination purposes in Reservoir
Engineering.
Equations – It is not necessary to memorise complicated equations. Equations unless
asked to be derived will be given.
Clearly some basic equations one should know and would not be given e.g.
Darcy’s Law,
PV = nzRT
STOOIP equation
Equilibrium Ratio K=y/x
Insitute of Petroleum Engineering, Heriot-Watt University
3
4
Chapter 3 Reservoir Composition
1
5
2
5
3
4
4
5
5.1
5
5.2
2
5.3
2
Chapter 2 Reservoir Pressures
1
5
2
5
3
5
4
5
5
5
6
4
Chapter 1 Introduction
Section – grading
1.1
4
1.2
4
1.3
4
2
4
3
–
3.1
4
3.2
4
3.3
4
3.4
4
4
OM
5
OM
6
4
7
4
8
4
9
4
Chapter 7 Reservoir Rocks
1
3
2
4
3
3
4.1
5
4.2
5
4.3
3
4.4
3
4.5
5
4.6
5
4.7
4
4.8
5
4.9
2
4.10
2
5
3
6
5
7.1
5
7.2
5
7.3
5
8.1
5
8.2
5
8.3
5
Chapter 6 Liquids
1
5
2
5
3
5
4
5
5
5
6
5
7
3 note there is an error in some texts with another 7 heading
8.1
5
8.2
5
9
5
10
3
11
1
12
5
Chapter 5 Gases
1.1
5
1.2
5
1.3
5
1.4
5
1.5
5
1.6
5
1.7
5
1.8
5
1.9
5
2.1
5
2.2
5
2.3
5
2.4
1
2.5
5
3
5
4
3
5
3
6
2
Chapter 4 Phase Behaviour
All material 5
Chapter 10 Fluid Flow
1
3
2
3
3.1
3
3.2
3
3.3.1
3
3.3.2
3
3.3.3
3
3.3.3.1
5
3.3.4
5
3.4
5
3.4.1
3
3.5
5
4
1
5
5
5.2
5
5.3
5
6
5
Chapter 9 - Permeability Variations
1
3
2
5
3
5
Chapter 8 Rock Measurement
1.1
2
2.1
2
2.2
2
3.1
2
3.2
2
4.1
3
4.2
3
4.3
3
4.4
3
5
2
6.1
5
6.2
3
6.3
5
6.4
5
7
2
Insitute of Petroleum Engineering, Heriot-Watt University
5
Vapour Liquid Equilibrium
2
2 Eq 11 – 5
5
5
5
5
3
3
3
3
Chapter 14 PVT
1
4
2
2
3.1
5
3.2
5
3.3
5
3.4
2
3.5
2
4
5
5
2
6
5
7
3
8.1
2
8.2
2
9
5
10
5
11
5
12
5
13
3
14
3
15
1
Chapter 13 Equilibrium Ratio
1
3
2
3
3
2
4
2
Chapter 12
1
2
3
4.1
4.2
4.3
5.1
5.2
5.3
5.4
Chapter 11 Drive Mechanisms
All sections 5
MB Application
5
5
5
4
5 (5.1.2.2 Eq46 -1 )
4
4
2
5
5
2
2
1
3
1
Chapter 17 Water Influx
1
5
2.1
3
2.2
3
2.3
3
2.4
3
2.5
5
3
5
4
4
5
3
6
2
7
2
Chapter 16
1
2
3
4
5.1
5.2
5.3.1
5.3.2
5.3.3
5.3.4
5.3.5
5.4
5.5
5.6
6
Chapter 15 Material Balance
1
5
2
3
3
5
4
5
5
5
6
5
7
5
8
5
9
5
10
5
11
5
12
5
Chapter 18 Immiscible Displacement
1
5
2
5
3.1
3
3.2
5
3.3
3 ( Eqn 1 – 5 -should be expected to know )
3.4
4 ( post equation 14 – 5 )
4
5
5.1
2
5.2
5 ( from equation 72+ - 2)
6.1
3
6.2
3
6.3
3
6.4
5
6.5
5
6.6
1
6.7
5
7
5
8.1
2
8.2
2
8.3
2
8.4
1
8.5
1
Introduction To Reservoir Engineering
CONTENTS
1 INTRODUCTION
8. PRODUCTION OPERATIONS OPTIMISATION
8.1
Development Phase
8.2
History Matching
8.3
Phases of Development
2 RESERVOIR ENGINEERING TECHNIQUES
9. THE UNIQUENESS OF THE RESERVOIR
3 RESERVE ESTIMATING
3.1
Definitions
3.2
Proven Reserves
3.2.1 Exercises - Reserve Definitions
3.3
Unproved Reserves
3.3.1 Probable Reserves
3.3.2 Possiible Reserves
3.4
Reserve Status Categories
3.4.1 Developed:
3.4.1.1 Producing
3.4.1.2 Non-producing:
3.4.2 Undeveloped Reserves:
10. CONCLUSIONS
1.1
1.2
1.3
Reserves Estimation
Development Planning
Production Operations Optimsation
4 PROBABILISTIC REPRESENTATION OF RESERVES
5 VOLUME IN - PLACE CALCULATIONS
5.1
Volume of Oil and Gas in-Place
5.2
Evolution of Reserve Estimate
5.3
Reservoir Area
5.4
Reservoir Thickness
5.5
Reservoir Porosity
5.6
Water Saturation
5.7
Formation Volume Factors
5.8
Recovery Factors
5.9
Production Capacity
5.10 Hydrocarbon Pore Volume Map
6 OTHER APPRAISAL ROLES
7 DEVELOPMENT PLANNING
7.1
Reservoir Modelling
7.2
Technoconomics
7.3
Coping with Uncertainty
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•Show using a block diagram the integration of reservoir engineering with other
petroleum engineering and other subjects.
•
Define the SPE definitions of reserves; proven reserves, unproved reserves;
probable reserves and possible reserves.
•
Calculate given the prerequisite data proved, probable and possible reserves.
•Describe in general terms reserve estimation.
•
Sketch a diagram showing the probability versus recoverable reserves indicating,
proven, proven + probable and proven + probable + possible reserves.
•Present a simple equation for volumes of oil and gas in-place.
•Describe in general terms the evolution of reserves through successive exploration wells.
•
Describe briefly with the aid of a sketch the various maps used to represent
reservoir; area, thickness porosity, saturation.
•
Describe briefly the use of the production (well0 test to determine reservoir
flowability and properties.
•
Describe briefly the various elements of development planning: reservoir
modeling technoeconomics and uncertainty.
•Illustrate with a sketch the impact of different technical parameters on the associated uncertainties on a project.
•
Describe in general terms in the context of production operations, optimization
in history matching.
•Draw a sketch showing the various phases of production from build up to economic limit.
•Draw a sketch illustrating the various recovery scenarios from primary to tertiary recovery.
Introduction To Reservoir Engineering
1 INTRODUCTION
With the petroleum industry’s desire to conserve and produce oil and gas more efficiently
a field of specialisation has developed called Petroleum Reservoir Engineering. This
new science which can be traced back only to the mid 1930’s has been built up on a
wealth of scientific and practical experience from field and laboratory. In the 1959
text of Craft & Hawkins1 on Applied Reservoir Engineering it is commented that “as
early as 1928 petroleum engineers were giving serious consideration to gas-energy
relationships and recognised the need for more precise information concerning
physical conditions as they exist in wells and underground reservoirs. Early progress
in oil recovery methods made it obvious that computations made from wellhead or
surface data were generally misleading.” Dake2, in his text "The Practise of Reservoir
Engineering", comments that “Reservoir Engineering shares the distinction with
geology in being one of the ‘underground sciences’ of the oil industry, attempting
to describe what occurs in the wide open spaces of the reservoir between the sparse
points of observation - the wells”
The reservoir engineer in the multi-disciplinary perspective of modern oil and gas
field management is located at the heart of many of the activities acting as a central
co-ordinating role in relation to receiving information processing it and passing it on
to others. This perspective presented by Dake2 is shown in the figure below.
Exploration
Geophysics/
Geology
Petrophysics
Reservoir Engineering
Economics
(Project viability)
Production
Process Egineering
General Engineering
Platform Topsides Design
2
Figure 1 Reservoir Engineering in Relation to Other Activities (adapted Dake )
Dake2 has usefully specified the distinct technical responsibilities of reservoir
engineers as:
• Contributing, with the geologists and petrophysicists , to the estimation of
hydrocarbons in place.
• Determining the fraction of discovered hydrocarbons that can be recovered.
• Attaching a time scale to the recovery.
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• Day-to-day operational reservoir engineering throughout the project lifetime.
The responsibility of the first is shared with other disciplines whereas the second is
primarily the responsibility of the reservoir engineer. Attaching a time scale to recovery
is the development of a production profile and again is not an exclusive activity. The
day-to-day operational role is on going through the duration of the project.
A project can be conveniently divided into two stages and within these the above
activities take place, the appraisal stage and the development phase. The appraisal
phase is essentially a data collection and processing phase with the one objective of
determining the ‘viability’ of a project. The development phase covers the remaining
period if the project is considered viable from the time continuous production commences to the time the field is abandoned. Reservoir engineering activity in various
forms takes place during both of these stages.
The activities of reservoir engineering fall into the following three general categories:
(i) Reserves Estimation
(ii) Development Planning
(iii) Production Operations Optimisation
1.1 Reserves Estimation
The underground reserves of oil and gas form the oil company’s main assets. Quantifying such reserves forms therefore a very important objective of the practising
reservoir engineer but it is also a very complex problem, for the basic data is usually
subject to widely varying interpretations and on top of that, reserves may be affected
significantly by the field development plan and operating practice. It is an on-going activity during, exploration, development planning and during production. It is
clearly a key task of the appraisal phase for it is at the heart of determining project
viability.
Before any production has been obtained, the so-called ‘volumetric estimate of
reserves’ is usually made. Geological and geophysical data are combined to obtain
a range of contour maps with the help of a planimeter and other tools the hydrocarbon bearing rock volumes can be estimated. From well log petrophysical analysis,
estimates of an average porosity and water saturation can be made and when applied
to the hydrocarbon rock volume yield an estimate of oil in place (STOIIP). Since
it is well known that only a fraction of this oil may in fact be ‘recoverable’, laboratory tests on cores may be carried out to estimate movable oil. The reserve estimate
finally arrived at is little more than an educated guess but a very important one for
it determines company policy.
In 1987 the Society of Petroleum Engineers in collaboration with the World Petroleum
Congress published definitions with respect to reserves and these are now accepted
world-wide 3. These definitions have been used in the summary of reserve definitions which follow.
Introduction To Reservoir Engineering
1.2 Development Planning
Oilfield development, particularly in the offshore environment, is a ‘front loaded’
investment. Finance has to be committed far in advance not only of income guaranteed by the investment, but frequently also of good definitive data on the character
of the field. Much of the responsibility for this type of activity falls on the reservoir
engineers because of their appreciation for the complex character of sub-surface fluid
behaviour under various proposed development schemes.
1.3 Production Operations Optimisation
Producing fields will seldom behave as anticipated and, of course, by the very nature
of this sort of activity, the balance of forces in the reservoir rock gets severely upset by
oil and gas production. The reservoir engineer is frequently called upon to ‘explain’
a certain aspect of well performance, such as increasing gas-oil ratio, sand and/or
water production and more importantly will be asked to propose a remedy. The actual
performance of the reservoir as compared to the various model predictions is another
ongoing perspective during this phase.
2 RESERVOIR ENGINEERING TECHNIQUES
In the past the traditionally available reservoir engineering tools were mainly
designed to give satisfactory results for a slide rule and graph paper approach. For
many problems encountered by reservoir engineers today this remains a perfectly
valid approach where the slide rule has been replaced by the calculator. Increasingly,
however, the advance of computing capability is enabling reservoir engineering
modelling methods (‘simulations’) to be carried out at the engineers desk, previously
considered impossible.
The basis of the development of the 'model' of the reservoir are the various data
sources. As the appraisal develops the uncertainty reduces in relation to the quality
of the forecasts predicted by the model. Building up this ‘geological’ model of the
reservoir progresses from the early interpretation of the geophysical surveys, through
various well derived data sets, which include drilling information, indirect wireline
measurements, recovered core data, recovered fluid analysis, pressure depth surveys,
to information generated during production.
3. RESERVE ESTIMATING
The Society of Petroleum Engineers SPE and World Petroleum Congress WPO1987
agreed classification of reserves3 provides a valuable standard by which to define
reserves, the section below is based on this classification document.
3.1 Definitions
Reserves are those quantities of petroleum which are anticipated to be commercially
recovered from known accumulations from a given date forward.
All reserve estimates involve some degree of uncertainty. The uncertainty depends
chiefly on the amount of reliable geologic and engineering data available at the time
Insitute of Petroleum Engineering, Heriot-Watt University
of the estimate and the interpretation of these data. The relative degree of uncertainty
may be conveyed by placing reserves into one of two principal classifications, either
proved or unproved.
Unproved reserves are less certain to be recovered than proved reserves and may
be further sub-classified as probable and possible reserves to denote progressively
increasing uncertainty in their recoverability.
Estimation of reserves is carried out under conditions of uncertainty. The method of
estimation is called deterministic if a single best estimate of reserves is made based
on known geological, engineering, and economic data. The method of estimation is
called probabilistic when the known geological, engineering, and economic data are
used to generate a range of estimates and their associated probabilities. Identifying
reserves as proved, probable, and possible has been the most frequent classification
method and gives an indication of the probability of recovery. Because of potential
differences in uncertainty, caution should be exercised when aggregating reserves
of different classifications.
Reserves estimates will generally be revised as additional geologic or engineering
data becomes available or as economic conditions change. Reserves do not include
quantities of petroleum being held in an inventory, and may be reduced for usage or
processing losses if required for financial reporting.
Reserves may be attributed to either natural energy or improved recovery methods.
Improved recovery methods include all methods for supplementing natural energy
or altering natural forces in the reservoir to increase ultimate recovery. Examples of
such methods are pressure maintenance, gas cycling, waterflooding, thermal methods,
chemical flooding, and the use of miscible and immiscible displacement fluids. Other
improved recovery methods may be developed in the future as petroleum technology
continues to evolve.
3.2 Proven Reserves
Proven reserves are those quantities of petroleum which, by analysis of geological
and engineering data, can be estimated with reasonable certainty to be commercially
recoverable, from a given date forward, from known reservoirs and under current
economic conditions, operating methods, and government regulations.
Proved reserves can be categorised as developed or undeveloped.
If deterministic methods are used, the term reasonable certainty is intended to express
a high degree of confidence that the quantities will be recovered. If probabilistic
methods are used, there should be at least a 90% probability that the quantities actually recovered will equal or exceed the estimate.
Establishment of current economic conditions should include relevant historical
petroleum prices and associated costs and may involve an averaging period that is
consistent with the purpose of the reserve estimate, appropriate contract obligations,
corporate procedures, and government regulations involved in reporting these
reserves. In general, reserves are considered proved if the commercial producibility
of the reservoir is supported by actual production or formation tests. In this context,
Introduction To Reservoir Engineering
the term proved refers to the actual quantities of petroleum reserves and not just
the productivity of the well or reservoir. In certain cases, proved reserves may
be assigned on the basis of well logs and/or core analysis that indicate the subject
reservoir is hydrocarbon bearing and is analogous to reservoirs in the same area that
are producing or have demonstrated the ability to produce on formation tests.
The area of the reservoir considered as proved includes (1) the area delineated by
drilling and defined by fluid contacts, if any, and (2) the undrilled portions of the
reservoir that can reasonably be judged as commercially productive on the basis of
available geological and engineering data. In the absence of data on fluid contacts, the
lowest known occurrence of hydrocarbons controls the proved limit unless otherwise
indicated by definitive geological, engineering or performance data. Reserves may be
classified as proved if facilities to process and transport those reserves to market are
operational at the time of the estimate or there is a reasonable expectation that such
facilities will be installed. Reserves in undeveloped locations may be classified as
proved undeveloped provided (1) the locations are direct offsets to wells that have
indicated commercial production in the objective formation, (2) it is reasonably
certain such locations are within the known proved productive limits of the objective
formation, (3) the locations conform to existing well spacing regulations where
applicable, and (4) it is reasonably certain the locations will be developed. Reserves
from other locations are categorised as proved undeveloped only where interpretations
of geological and engineering data from wells indicate with reasonable certainty that
the objective formation is laterally continuous and contains commercially recoverable
petroleum at locations beyond direct offsets.
Before looking at further detail we will carry out some tests to help emphasise the
above definition.
3.2.1 Exercises - Reserve Definitions
The section on Reserve Definitions as put together by the SPE and the World Petroleum Congress, defines the various aspects of reserve definitions. These definitions,
are important both to companies and countries, and they can have very significant
commercial impact. The following tests are presented to help understand the working of these earlier definitions.
Test 1
There are 950 MM stb ( million stock tank barrels) of oil initially in place in a reservoir. It is estimated that 500 MM stb can be produced. Already 100 MM stb have
been produced. In the boxes below, identify the correct answer.
950
500
400
MM stb
The Reserves are: 450
400
500
MM stb
STOIIP is:
Turn to page 9 for answers
Insitute of Petroleum Engineering, Heriot-Watt University
Test 2
Before starting production it was estimated that there was a 90% chance of producing at least 100 MM stb, 50% chance of producing 500 MM stb and 10% chance of
producing 700MM stb. That is we are sure we can produce at least 100MM stb, and
we will probably produce as much as 500 MM stb, and we will possibly produce as
much as 700 MM stb.
Tick the correct answers.
Proved reserves (MM stb):
400
500
600
700
200
400
500
600
700
Possible reserves
200
100
400
500
600
700
100
200
Probable reserves
100
Turn to page 9 for answers
Test 3
What is wrong with the following definitions?
1. Reserves are those quantities of petroleum which are anticipated to be recovered
from a petroleum accumulation.
Test 4
1. We have a structure in our licence area which we intend to explore. We anticipate
it to contain a STO IIP of 2000 MM stb, and recovery factor of 65% using primary
methods (30%), secondary (25%) and tertiary (10%) recovery methods. What are
the reserves?
Test 5
A reservoir has been discovered by drilling a successful exploration well, and drilling
a number of producing wells. We have even produced some 200 MM stb of oil.
STOIIP = 2000MM stb
What are the reserves?
Recovery factor = 35%
Introduction To Reservoir Engineering
Test 1 answer
There are 950 MM stock tank boards in place. It is estimated that 500 MM stb can
be produced and 100 MM stb have been produced then 400 recoverable reserves
remain.
950
√
500 X
400
X
MM stb
The Reserves are: 450
X
400 √
500
X
MM stb
STOIIP is:
Test 2 answer
Proved :
Probable :
Possible : Proved :
Proved & Possible
Proved & Probable & Possible :
100 MM stb
500 - 100 = 400 MM stb
700 - 500 = 200 MM stb
100 MM stb
500 MM stb
700 MM stb
Test 3 answer
Reserves are those quantities of petroleum which are anticipated to be commercially
recovered from a petroleum accumulation.
Clearly economics is a very important aspect of the definition.
Economic Variables
What economic factors are used in the calculations? What oil and gas price do we
use for proved reserve estimates? Is inflation taken into account? Do we predict
future price trends? Do we apply discount factors to calculate present value of the
project? Are all these used in proved reserve calculations? The current economic
conditions are used for the calculations, with respect to prices, costs, contracts and
government regulations.
Test 4 answer
1. Answer is zero by SPC/WPC definition.
2. Intentions and anticipations are not the basis for reserves. In this case no well
has yet been drilled.
Note: Some companies allocate potential reserves for internal use but these cannot
be used for public and government figures. Reserves are those quantities of petroleum which are anticipated to be commercially
recovered from a known accumulation.
Requirements for “Proved” include
The following sources are required for proved reserves. Maps (from seismic and/
geological data). Petrophysical logs. Well test results and rock properties from core
analysis tests on recovered core.
Insitute of Petroleum Engineering, Heriot-Watt University
Facilities
An important perspective which might be forgotten by the reservoir engineer, is that
for reserves to be classified as “proven”, all the necessary facilities for processing
and the infrastructure for transport must either be in place or that such facilities will
be installed in the future, as backed up by a formal commitment.
Contribution to the Proved Reservoir Area
This comes from drilled and produced hydrocarbons, the definition of the gas and oil
and water contacts or the highest and lowest observed level of hydrocarbons. Also
the undrilled area adjacent to the drilled can be used.
Test 5 answer
Ultimate recovery = 2 000 x 0.35 = 700 MM stb
Minus production to date = 200
Reserves = 500 MM stb
Reserves are those quantities of petroleum which are anticipated to be commercially
recovered from known accumulations from a given date forward.
i.e. Reserves refer to what can be produced in the future.
RESERVE CATEGORIES
Probability Levels
Figure 2 gives a schematic of reserves showing the progression with time.
P10
Potential
SPE / WPC Definitions
Possible
Possible
Probable
Probable
P50
Provan
P90
Cumulative Production
Provan
Time
Seismic Discovery of Start of Dev Start of
Data
Well
Planning Production
PERIOD
TYPE OF
DATA
METHOD
Before Drilling
Exploration Well
Geophysical
and Geological
Prior and During Delineation, Evaluation,
Appraisal
Development
Geophysical,
Geological,
Petrophysical
and Well Test Data
Mostly Probabilistic
Abandonment
Production
Geophysical,
Reservoir Performance
Geological,
and Production Data
Petrophysical
and Well Tests and Production Data
Deterministic and Probabilistic
Figure 2 Variations of Reserves During Field Life
What are the amounts termed that are not recoverable? The quantity of hydrocarbons that remains in the reservoir are called remaining hydrocarbons in place, NOT
remaining reserves!
Reserves which are to be produced through the application of established improved
recovery methods are included in the proved classification when :
10
Introduction To Reservoir Engineering
(i) Successful testing by a pilot project or favourable response of an installed
program in the same or an analogous reservoir with similar rock and fluid
properties provides support for the analysis on which the project was based,
and,
(ii) It is reasonably certain that the project will proceed. Reserves to be recovered
by improved recovery methods that have yet to be established through
commercially successful applications are included in the proved classification
only:
(i) After a favourable production response from the subject reservoir from either
(a) A representative pilot or
(b) An installed program where the response provides support for the analysis
on which the project is based and
(ii) It is reasonably certain the project will proceed.
3.3 Unproved Reserves
Unproved reserves are based on geologic and/or engineering data similar to that
used in estimates of proved reserves; but technical, contractual, economic, or
regulatory uncertainties preclude such reserves being classified as proved.
Unproved reserves may be further classified as probable reserves and possible reserves. Unproved reserves may be estimated assuming future economic conditions
different from those prevailing at the time of the estimate. The effect of possible
future improvements in economic conditions and technological developments can
be expressed by allocating appropriate quantities of reserves to the probable and
possible classifications.
3.3.1. Probable Reserves
Probable reserves are those unproved reserves which analysis of geological and
engineering data suggests are more likely than not to be recoverable. In this context,
when probabilistic methods are used, there should be at least a 50% probability that
the quantities actually recovered will equal or exceed the sum of estimated proved
plus probable reserves. In general, probable reserves may include :
(1) Reserves anticipated to be proved by normal step-out drilling where subsurface
control is inadequate to classify these reserves as proved,
(2) Reserves in formations that appear to be productive based on well log
characteristics but lack core data or definitive tests and which are not analogous
to producing or proved reservoirs in the area,
(3) Incremental reserves attributable to infill drilling that could have been classified
as proved if closer statutory spacing had been approved at the time of the
estimate,
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11
(4) Reserves attributable to improved recovery methods that have been established
by repeated commercially successful applications when;
(a) a project or pilot is planned but not in operation and
(b) rock, fluid, and reservoir characteristics appear favourable for commercial
application,
(5) Reserves in an area of the formation that appears to be separated from the
proved area by faulting and the geologic interpretation indicates the subject
area is structurally higher than the proved area,
(6) Reserves attributable to a future workover, treatment, re-treatment, change of
equipment, or other mechanical procedures, where such procedure has not been
proved successful in wells which exhibit similar behaviour in analogous
reservoirs, and
(7) Incremental reserves in proved reservoirs where an alternative interpretation of
performance or volumetric data indicates more reserves than can be classified
as proved.
3.3.2. Possible Reserves
Possible reserves are those unproved reserves which analysis of geological and engineering data suggests are less likely to be recoverable than probable reserves.
In this context, when probabilistic methods are used, there should be at least a 10%
probability that the quantities actually recovered will equal or exceed the sum of
estimated proved plus probable plus possible reserves. In general, possible reserves
may include:
(1) reserves which, based on geological interpretations, could possibly exist
beyond areas classified as probable,
(2) reserves in formations that appear to be petroleum bearing based on log and
core analysis but may not be productive at commercial rates,
(3) incremental reserves attributed to infill drilling that are subject to technical
uncertainty,
(4) reserves attributed to improved recovery methods when
(a) a project or pilot is planned but not in operation and
(b) rock, fluid, and reservoir characteristics are such that a reasonable doubt
exists that the project will be commercial, and
(5) reserves in an area of the formation that appears to be separated from the
proved area by faulting and geological interpretation indicates the subject area
is structurally lower than the proved area.
12
Introduction To Reservoir Engineering
3.4 Reserve Status Categories
Reserve status categories define the development and producing status of wells and
reservoirs.
3.4.1. Developed:
Developed reserves are expected to be recovered from existing wells including reserves
behind pipe. Improved recovery reserves are considered developed only after the
necessary equipment has been installed, or when the costs to do so are relatively minor.
Developed reserves may be sub-categorised as producing or non-producing.
3.4.1.1 Producing:
Reserves subcategorised as producing are expected to be recovered from completion intervals which are open and producing at the time of the estimate. Improved
recovery reserves are considered producing only after the improved recovery project
is in operation.
3.4.1.2. Non-producing:
Reserves subcategorised as non-producing include shut-in and behind-pipe reserves.
Shut-in reserves are expected to be recovered from (1) completion intervals which
are open at the time of the estimate but which have not started producing, (2) wells
which were shut-in for market conditions or pipeline connections, or (3) wells not
capable of production for mechanical reasons. Behind-pipe reserves are expected to
be recovered from zones in existing wells, which will require additional completion
work or future recompletion prior to the start of production.
3.4.2. Undeveloped Reserves:
Undeveloped reserves are expected to be recovered:
(1) From new wells on undrilled acreage,
(2) From deepening existing wells to a different reservoir, or
(3) Where a relatively large expenditure is required to
(a) Recomplete an existing well or
(b) Install production or transportation facilities for primary or improved
recovery projects.
4. PROBABILISTIC REPRESENTATION OF RESERVES
Whereas in the deterministic approach the volumes are determined by the calculation
of values determined for the various parameters, with the probalistic statistical analysis
is used, using tools like Monte Carlo methods. The curve as shown in the figure 3
below presents the probability that the reserves will have a volume greater or equal
to the chosen value.
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13
Probability that the reserve is at least
as large as indicated.
1.0
'Proven'
0.9
0.5
'Proven + Probable'
'Proven + Proable
+ Possible'
0.1
0
Recoverable Reserve
Figure 3 Probabilistic Representation of Recoverable Reserves.
On this curve:
The proven reserves represent the reserves volume corresponding to 90% probability
on the distribution curve.
The probable reserves represent the reserves volume corresponding to the difference
between 50 and 90% probability on the distribution curve.
The possible reserves represent the reserves volume corresponding to the difference
between 10 and 50% probability on the distribution curve.
As with the deterministic approach there is also some measure of subjectivity in the
probalistic approach. For each of the elements in the following equation, there is a
probability function expression in low, medium and high probabilities for the particular
values. A schematic of a possible distribution scenario for each of the elements and
the final result is given below in the figure 4.
Net rock
volume.
Net rock
average
porosity
Connate Formation Estimated
water
volume
recovery
saturation factor
factor
[ Vnr x φ x (1 - Swc) / Bo ] x RF
Uniform
P
Triangular
= Reserves
Gaussian Uniform
=
Figure 4 Probablistic Reserve Estimates.
14
p90
p50
p10
Introduction To Reservoir Engineering
The resulting calculations result in a probability function for a field as shown in
the figure 5 below, where the values for the three elements are shown
Proven = 500 MM stb the P90 figure.
Probable = 240 MM stb which together with the proven makes up the P50 figure.
of 740MMstb
Possible = 120 MM stb which together with the proven and probable makes up the
P10 value of 860MMstb
Reserves distribution for a new field.
100
P90
90
P10 = 860 MMstb
P50 = 740 MMstb
P90 = 500 MMstb
Probability / %
80
70
60
Proven 500 MMstb
50
Probable 240 M
40
30
P+P+P = 860 MMstb
20
120
10
0
P50
Proven
0
200
Probable
400
600
Reserves / MMstb
P10
Possible
800
1000
Figure 5 Reserves Cummulative Probability Distribution.
As a field is developed and the fluids are produced the shape of the probability curve
changes. Probability figures for reserves are gradually converted into recovery leaving less uncertainty with respect to the reserves. This is illustrated in figure 6.
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15
100
P90
90
Probability / %
80
70
60
P50
50
40
Proved ultimate recovery.
30
20
10
0
Production
0
200
P10
Proved reserves
400
600
Reserves / MMstb
800
1000
Figure 6 Ultimate Recovery and Reserves Distribution For a Mature Field.
5. VOLUME IN-PLACE CALCULATIONS
5.1 The volume of oil and gas in-place depends on a number of parameters :
The aerial coverage of the reservoir. A
The thickness of the reservoir rock contributing to the hydrocarbon volume. hn
The pore volume, as expressed by the porosity ,φ , the reservoir quality rock.
The proportion of pore space occupied by the hydrocarbon ( the saturation ).
1-Sw
The simple equation used in calculation of the volume of fluids in the reservoir, V,
is
V=Ahnφ(1-Sw):
(1)
where:
A= average area
hn = nett thickness. nett thickness = gross thickness x nett: gross ratio
φ = average porosity
Sw = average water saturation.
When expressed as stock tank or standard gas volumes, equation above is divided
by the formation volume factor Bo or Bg.
V = Ahnφ (1 − Sw ) / Bo
(2)
To convert volumes at reservoir conditions to stock tank conditions formation volume
factors are required where Bo and Bg are the oil and gas formation volume factors.
These are defined in subsequent chapters. The expression of original oil in place is
termed the STOIIP.
16
Introduction To Reservoir Engineering
The recovery factor, RF, indicates the proportion of the in-place hydrocarbons expected to be recovered. To convert in place volumes to reserves we need to multiply
the STOIIP by the recovery factor so that:
Reserves = STOIIP x RF
(3)
The line over the various terms indicates the average value for these spatial
parameters.
The reservoir area A, will vary according to the category; proven, probable or possible, that is being used to define the reserves.
Before examining the contributions of the various parameters it is worthwhile to
give consideration of the evolution of the reserve estimate during the exploration
and development stage.
5.2 Evolution of the Reserve Estimate
Figure 7 gives a cross section view of a reservoir structure as suggested from seismic
and geological data.
Oil
Suggested 0il and water contact
Figure 7 Cross Section Interpretation From Seismic and Geological Data.
Using this data and possible suggested structure we can carry out some oil in place
calculations and estimate reserves. These figures however are not admissible in public
reserve estimates. They are useful inside the company to justify project expenditure!
The question is where do we locate the first exploration well and get involved in large
exploration expenditure costs. Figure 8 suggest three alternatives
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17
Suggest this location.
Oil
Suggested oil and water contact
Figure 8 Alternative locations of Exploration Wells
In figure 9 an exploration well has been drilled and a core recovered and the structure of the field with respect to formations and contacts redefined. The redefined
structure can now be used to provide an estimate of reserves according to the
three, proven, probable and possible perspectives. Figure 10
Oil
Oil and water contact
Cored interval
Figure 9 Interpretation After Exploration Well Drilled and Cored.
18
P
Oil
e
bl
si
os
ba
ble
Proved
Pr
o
Pr
ob
ab
l
e
Introduction To Reservoir Engineering
ible
s
Pos
Figure 10 After The Exploration Well Was Drilled.
Subsequent appraisal wells are now drilled to give better definition of the reserves
of the field. Well 2 aimed at defining the field to the left identifies some additional
isolated hydrocarbon structure with its own oil water contact. Figure 11. The well, as
well as increasing the proven reserves, further identifies previous unknown reserves. The next appraisal well is aimed at defining the reserves in the other direction. During well testing on wells 1or 2 indications of faulting are also helping to define the
flowing nature of the accumulation. Figure 12 for the further appraisal well confirms
the accumulation to the right and also identifies the impact of the fault with a new
oil water contact. Subsequent appraisal wells and early development give greater
definition to the field description. Figure 13
Well 2.
ven
Pro
Well 1.
Proven
Proposed
delineation
well 3.
Oil
Initial appraisal stage.
Figure 11 Further Delineation Well.
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19
Well 2.
Well 1.
Well 3.
Gas
ven
Pro
Proven
Oil
New oil water contact.
Figure 12 After Further Appraisal.
Well 1. Well 4.
Well 2.
Well 3.
Gas
ven
Pro
Proven
Oil
New oil water contact.
Figure 13 Final Appraisal Well.
From a deterministic perspective the various reserve estimates, that is, proven,
probable and possible can be further determined. The indication of the various
elements based on the top structure map are shown. Figure 14
20
Introduction To Reservoir Engineering
Probable
1
Proved
2
3
4
Possible
Figure 14 Reserves Uncertainties by Deterministic Method.
5.3 Reservoir Area
The reservoir area can be obtained by separately evaluating the individual units
making up the reservoir as obtained from various reservoir maps. These maps are
derived from the evidence given from seismic and subsequent drilled wells. The maps
generally indicate the upper and lower extent of the reservoir section or sections and
the aerial extent as defined by faults or hydrocarbon contacts. Figure 15 shows an
aerial section with the defined limits. The contour lines are lines of constant subsea
depths. Figure 16 gives a cross section of a reservoir unit. The combination of the
two representations of the unit(s) can be used to calculate the gross rock volume.
Fault B
ounda
ry
Porosity
Boundary
Fluid
Contact
ry
unda
Bo
Fault
Figure 15 Structure Contour Map.
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7
21
Heighest Elevation
on Top Structure
To
p
Contour Elevation
(units ss)
Heighest Elevation
on Base Structure
St
R
r
Ro ese uctu
re
ck rvo
Ba
Vo ir
se
lu
St
m
ru
e
c
tu
re
Hydrocarbon Water
Contact Elevation
o
Area Contained by Contour
Figure 16 Reservoir cross section.
7
Figures 17 & 18 show an example of a top structure map and cross section of the
Rough Gas field in the North Sea.
47/2 47/3
Completed Producers
Gw
C
Proposed Well Locations
C
C.I. = 50ft.
960
47/7 47/8
G
0
w
9
95500
00
x Abandoned Wells
955000
8
95 50
8
94 00
94 50
93 00 8
8
8 93
8
B 250
9 A2 00
8
92
8
A
47/8-1x
8
A
8
8
A
A5
A
A4
A
Platform A
A3
9100
A6
93
00
93 50
92
9200
50
91
50
x 47/8-2
Figure 17 Top Sand Structure Map Rough Gas Field.
22
5
Introduction To Reservoir Engineering
Depth (ft)
subsea
9000
A5
Unc
Rot
onf
orm
ity
es
onfo
rmit
y
lieg
Fault
9400
A2
A4
Unc
end
Fault
9200
A1
A3
Tentative
hydrocarbon/
water contact
9600
9800
Carboniferous
Sands
Figure 18 Schematic Cross Section of The Rough Field.
5
5.4 Reservoir Thickness
Another representation of the reservoir formations is the reservoir thickness map.
Where the areal contour maps show the thickness normal to the plane of the reservoir
the contours are called isopachs. When the thickness is mapped as a vertical thickness
then the contour is called an isochore. Not all the reservoir thickness will contribute to fluid recovery and will include non-productive strata. Those contours which
include these non-productive material are called gross reservoir isopach and those
where non-productive material is excluded are called net reservoir isopach maps.
Those intervals contributing to flow are termed pay. The ratio of net to gross, hn/ht ,
is an important aspect in reservoir evaluation. Figure 19 shows a net pay thickness
isopach and the isopach map for the Rough field is shown in figure 20 75
0
150
125
Isopach C I
25 Units
100
7
Figure 19 Net Pay Thickness Isopach.
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23
47/2 47/3
GwC
140
130
Gw
C
120
0
11
0
10
A2
47/7 47/8
47/8-1 x
A5
A4
80
A1
0
11
100
90
A3
6
11
70
A6
x 47/8-2
Figure 20 Rough Field Isopach.
5
The isopach map can also be used to calculate reservoir volume. For example in figure
21 the area under a plot of net pay thickness vs. area contained within the contour
provides a net pay volume. These plots can be generated for each section or rock
type. The thickness plots for each section are called isoliths.
0
Net Pay Isopach Value
40
Area Enclosed = Net Rock Volume
80
120
140
180
OWC
Area Contained by Contour
7
Figure 21 Hydrocarbon Volume From Net Pay Isopach.
5.5 Reservoir Porosity
The variation of porosity can also be represented . The average porosity, φ, in a well
can be calculated from the thickness-weighted mean of the porosities 4 .
24
Introduction To Reservoir Engineering
m
φw =
∑φ h
k =1
k n, k
hn
(4) where φk is the average porosity derived from the log over a small thickness hn,k
within the net pay thickness, hn.
These values of porosity can then be plotted to generate an isoporosity map as illustrated in figure 22. The example of an isoporosity map for the Rough Field is
shown in figure 23.
5
10
25
20
15
Porosity C I
5%
Figure 22 Iso Porosity Map.
7
47/2 47/3
G
w
C
C
Gw
x
A5
A
A1
A6
6%
A3
A4
10
%
8%
47/8-1
12%
47/7 47/8
14%
A2
47/8-2
x
7
Figure 23 Rough Field Iso Porosity Map.
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25
5.6 Water Saturation, Sw
The water saturation in a reservoir is influenced by the characteristics of the reservoir
rock and the location with respect to the position above the free water level near
the oil-water or gas-oil contact (see section Reservoir Rock Properties Chapter 7).
The average water saturation Sw,w , can be calculated in a similar way to porosity by
calculating the volume weighted mean across the producing elements of the formation, the pay.
m
Sw,w =
∑S
k =1
w, k
φ k h n,k
φwh n
(5)
The values of Sw,w can be plotted and contours of constant saturation (isosaturation) presented. Figure 24.
Shale
15
20
25
30 35
40
WOC
4
Figure 24 Iso Saturation (sw) Map.
A more detailed description together with exercises are given in the mapping section
of the geology module.
5.7 Formation Volume Factors Oil, Bo and Gas, Bg
These properties of the oil and gas which convert reservoir volumes to surface volumes, are generated from measurements made on fluid samples from the reservoir. They do
not vary significantly across the reservoir when compared to the other rock related
parameters. These parameters are covered in the gas properties and oil properties
chapters. In some reservoirs where the formations are thick there is a compositional
gradient over the depth. This variation in composition from heavier (less volatile
components) to lighter components at the top results in a variation of the oil formation volume factor, Bo over the thickness. In such cases an average value based on
values measured or calculated at depth would be a preferred value.
26
Introduction To Reservoir Engineering
5.8 The Recovery Factor, ER
The proportion of hydrocarbons recovered is called the recovery factor. This factor is influenced by a whole range of factors including the rock and fluid properties
and the drive mechanisms. The variability of the formation characteristics, the heterogeneity can have a large influence on recovery. The development process being
implemented and the geometries and location of wells again will also have a large
influence. Calculating recovery therefore in the early stages is not feasible and many
assumptions have to be included in such calculations. It is in this area that reservoir
simulation can give indications but the quality of the calculated figure is limited by
the sparse amount of quality data on which the simulation is based.
The American Petroleum Institute6 has analysed the recoveries of different fields and
correlations have been presented for different reservoir types and drive mechanisms.
Figures 25 and 26 give the residual saturations and oil recovery efficiences for different drive mechanisms. The API also presents correlations for recoveries,ER,
For sandstone and carbonate reservoirs with solution gas drive
ER, o
 φ (1 − Sw ) 
= 0.4185

 Bob 
0.1611
 k 


 µob 
0.0979
(Sw )
0.3722
 pb 
 
 pa 
0.1741
(6)
For sandstone reservoirs with water drive
ER, o
 φ (1 − Sw ) 
= 0.54898

 Boi 
0.0422
 k µ wi 


 µoi 
0.0770
p
 a
(Sw )− o.1903  pi  − 0.2159
(7)
b refers to bubble point conditions, i is the initial condition and a, refers to abandonment
pressure.
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27
2
5
10
20
30
40
50
60
70
80
95
98
RESIDUAL SATURATIONS
1.00
Sor In Water Drive
Reservoirs
0.50
0.50
Sgr In Solution Gas Drive
Reservoirs
0.10
0.10
0.05
0.05
−σ
0
+σ
MEDIAN
Sor (OR Sgr) as Fraction of Total Pore Space
1.00
2
5
10
20
30
40
50
60
70
80
95
98
0
PERCENTAGE OF CASES LARGER THAN
Figure 25 Log - Probability Residual Oil Saturation For Water Drive and Solution Gas
6
Drive Reservoirs. (API )
5
OIL RECOVERY EFFICIENCY AT FIELD ABANDONMENT
IN PERCENT OF OIL PLACE
2
10
20
30
40
50
60
70
80
95
98
RESIDUAL SATURATIONS
1.00
1.00
Water Drive
Gas Cap Drive
0.50
0.50
Gas Cap Drive +
Water Injection
0.10
0.10
Solution Gas Drive
0.05
0.05
0
+σ
MEDIAN
−σ
2
5
10
20
30
40
50
60
70
80
95
98
0
PERCENTAGE OF CASES LARGER THAN
6
Figure 26 Log - Probability of Oil Recovery For Various Drive Mechanisms. (API )
28
Introduction To Reservoir Engineering
5.9 Production Capability
Another concept, isocapacity, is used to signify production capability. Isocapacity
denotes equal values of permeability-net thickness product. This product can be
mapped instead of permeability. The figure 27 shows an isocapacity map where the
absolute permeability has been obtained as an arithmetic average in the zone.
4
0.5
5
4
3
2
1
123
0.25
7
Figure 27 Isocapacity Map.
The permeability map for the Rough Field is given in figure 28
G
w
C
C
Gw
47/2
47/7 47/8
120 A2
100
80
47/8-1 x
A4
A5
Platform B
60
40
A3
0
A6
x 47/8-2
Contour Intervals 20 millidarcies
5
Figure 28 Rough Field Permeability Map.
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29
5.10 The Hydrocarbon Pore Volume Map
The hydrocarbon pore volume can be obtained by combining the net rock volume with
a mean porosity and a mean hydrocarbon saturation. An alternative is the mapping of
hydrocarbon thickness (HPT) at each well. HPT at a well in a given zone is:
_
_
HPT = φ .hn . Sh
(8)
where:
_
_
Sh = 1 − Sw
Figure 29 gives an HPT map and the Rough Field HPT map is given in figure 30
12
11
0
10
14
15
14
13
13
12
10
0
11
9
7
Figure 29 Hydrocarbon Pore Thickness Map.
0
10
A2
9
8
A5
A4
7
A1
6
A3
5
4
A6
5
Figure 30 Rough Field Hydrocarbon Pore Thickness.
30
Introduction To Reservoir Engineering
6. OTHER APPRAISAL ROLES
In building up the ‘picture’ to enable the reserves estimates and recoveries to be
determined the reservoir engineer will be involved in an number of aspects. One of
the most powerful tools is the production test.
In a well test an exploration or appraisal well is converted to a short term producing
well, with all the associated facilities put in place to handle the produced fluids and
monitor fluid rates. A downhole pressure monitoring device is also located in the
well. Figure 31. The well is flowed at a constant rate , and sometimes two rates as
illustrated in figure 32a, a two rate test. The downhole pressure device responds to
the production and pressure declines. After a short or longer time period depending
on the nature of the test, the well is “shut in”, i.e. the flow is stopped. In the well the
pressure builds up and eventually as monitored by the downhole pressure device,
recovers to the original pressure. Figure 32b. It is in the analysis of the pressure
drawn down and build up curves and the rates that the reservoir engineer is able to
determine the flowability of the reservoir. If the flowing interval thickness is known,
the permeability can be calculated. The presence of faults can also be detected.
A considerable amount of reservoir data can be obtained from these well tests
sometimes called DST’s ( drill stem tests). It has been the practise over recent years
for the produced fluids to be flared since there is unlikely to be an infrastructure to
collect these fluids. Now that companies are moving to a zero or reduced hydrocarbon
emission policy the nature and facilities required for these tests are changing. A
feature of the flaring approach is a public demonstration of the productivity of the
well being tested.
Insitute of Petroleum Engineering, Heriot-Watt University
31
Surface casing
Production casing
Production tubing
Cement
Packer
Perforations
Down hole
pressure monitor
Figure 31 Production Test Assembly.
32
q bbls / day
Introduction To Reservoir Engineering
Flow 1
Flow 2
Well shut in
t
Pf. psig
Pi
Pressure draw down
Pressure build up
t
Figure 32 Production Test Analysis. Two Rate Test.
Well test analysis is a powerful reservoir engineering tool and is treated in depth in
a subsequent module of the Petroleum Engineering course.
The nature of the fluids is key to reservoir behaviour and also subsequent processing
in any development. The collection and analysis of these fluids is an important
role and is at the focus of PVT analysis. This topic is covered in Chapter 14 PVT
Analysis. The pressure profile in a well is another important aspect of reservoir
characterisation and can be used to identify fluid contacts. When used during the
early stages of production it can be a powerful means of refining the structure and
hydrodynamic continuity characteristics of the reservoir. This is covered in the next
chapter. Like PVT analysis where the information is based on samples removed
from the reservoir, core analysis is based on recovered core from the formation.
Various tests on this material and its reaction to various fluids provides many of the
reservoir engineering parameters important in determining the viability of a project.
Core analysis also provides a cross check for indirect measurements made downhole.
These core analysis perspectives are covered in chapters 7 and 8.
It is clear from what we have discussed that reservoir engineering is an important
function in the appraisal of the reservoir. The focus for this appraisal so far has concentrated on determining the characteristics and potential flow behaviour of a reservoir
under development. Clearly there could be a whole range of possibilities with respect
to the plan that could be used to develop the field. This development planning perspective is an important part of the reservoir engineers role. Again it is a team effort
Insitute of Petroleum Engineering, Heriot-Watt University
33
involving the geological community who understand the ‘reservoir’ and the various
engineers who have the responsibilities of designing and operating the hardware to
enable production. An important part of any future development are the facilities that
would be required for sustained production and its is therefore an important part of
the appraisal stage to provide data for those who would have responsibility for good
quality data predictions which will enable optimised facility design.
In any project new data is always being generated. Indeed for a reservoir, its
characteristics are unlocked over the whole lifetime of the project. The duration of
the appraisal stage clearly is a techno economic decision related to the confidence
to go ahead based on a good foundation of quality data and forecasts. Fine tuning
can always be carried out but this is costly if this delays the development stage. It
is important to identify and fill the gaps for the largest uncertainties, and having
sufficient information to design a system which is safe and cost effective. The
difficulty is making the decision on the data under which a line is drawn which
defines the basis for field development design. In reservoir development the reservoir
is always revealing its properties, indeed it is in the production phase that the true
characteristics are revealed.
7 DEVELOPMENT PLANNING
7.1 Reservoir Modelling
Given appraisal well data, and test results the reservoir engineer can consider some
alternative development plans, relying heavily on experience and insight. Since the
80’s computer based reservoir simulation has played a major role.
The starting point will invariably be a reservoir map used to calculate reserves, but
in addition use will be made of the material balance equation (chapter 15), together
with some drive concepts (chapter 11), to predict reservoir behaviour. One of the
problems faced in making predictions is to adequately take into account knowledge
about geological trends and, although individual well models can be adjusted to reflect
local conditions, there is no practical ‘desk calculator’ technique for using say, the
material balance equation and well models to come up with a predictive reservoir
performance. Displacement models such as those derived by Buckley and Leverett
(chapter 18), mainly from observations in the laboratory, give some insight into
reservoir behaviour but again do not significantly assist in allowing the engineer to
study the effect of alternative development plans on a heterogeneous reservoir.
With insight and ingenuity, the reservoir can be divided into a number of simple
units that can be analysed by the traditionally available techniques but such an
approach remains unsatisfactory. Over recent years the integration of geological and
geophysical perspectives is contributing considerably to the ‘confidence’ in reservoir
modelling.
7.2 Technoeconomics
For hydrocarbon accumulations found on dry land the traditional reservoir engineering
techniques available for field development planning were, in fact, quite adequate. This
is mainly so because land development operations offer a high degree of planning
34
Introduction To Reservoir Engineering
flexibility to oil companies and hence allow them to make optimal use of the latest
information. In an offshore environment this is not the case; once platforms have
been ordered most development options are closed. It is with respect to offshore field
development planning that reservoir simulation models have found their greatest
application potential.
7.3 Coping with Uncertainty
The challenge to the exploration & production business of the oil & gas industry is
considerable. The looking for the “needle in the haystack” scenario is not too far from
the truth, when compared to other industrial sectors. With the challenge of reserves
being found in technically challenging areas and the oil price moving in response
to political as well as demand scenarios, there is the need to define more accurately
forecasts of production and recovery. Reducing uncertainty is the message of the
current decade and not least in reservoir engineering and its related disciplines.
It is clear from what we have overviewed in this chapter and the topics which will be
covered in the subsequent chapters that there are many parameters which contribute
to the viability of the various aspects of successful oil and gas production. It is also
clear that the various forms of data required, the confidence in the absolute values
vary according to the type, and therefore the final impact on the final result will vary
according to the particular parameter.
The following list summarises some of the principal uncertainties associated with
the performance of the overall reservoir model. The type of data can for example
be subdivided into two aspects “static” and “dynamic” data .
Static Properties
• Reservoir structure
• Reservoir properties
• Reservoir sand connectivity
• Impact of faults
• “thief” sands
Dynamic Properties
• Relative permeability etc
• Fluid properties
• Aquifer behaviour
• Well productivity (fractures, welltype, condensate drop out etc.)
The impact of each of these parameters will vary according to the particular field but
it is important that the company is not ignorant of the magnitude of the contributing
uncertainties, so that resources can be directed at cost effectively reducing specific
uncertainties. Figure 33 illustrates an outcome which might arise from an analysis
of various uncertainties for a particular field. It demonstrates for this particular field
and at the time of analysis the impact of the various data has on the final project cost. Clearly in this case the aquifer behaviour uncertainties has the least impact whereas
reservoir structure and well productivity uncertainties had the most significant. Another field would result in different impact perspectives, and therefore a different
strategy to reduce overall project uncertainty would be required.
Insitute of Petroleum Engineering, Heriot-Watt University
35
Q
Reservoir
area
P
Well
production
Project
Cost
Reservoir
structure
Sand
conectives
Aquifer
behaviour
Fluid properties
Relative
permeabilities etc.
Thief zones
Faults
-
Changes
+
Figure 33 Impact on a Project of Different Uncertainties
8 PRODUCTION OPERATIONS OPTIMISATION
8.1 The Development Phase
The development phase covers the period from the time continuous production
starts until the production from the field stops i.e. abandonment. The decision when
to stop production clearly is a techno-economic decision based to a large extent on
the costs of the development. Low volume producers can be allowed to continue in
an onshore development where well operating costs might be low but the high costs
associated with for example in an expensive offshore operation sets a much higher
economic limit for the decision to abandon a field.
During the development phase Dake2 has identified a number of roles for the Reservoir
Engineering which are targeted at optimising production. It is an irony that some
of the best data is generated during the production phase. Through production the
reservoir unveils more of its secrets. Some of these may cause modifications to the
development, perhaps in defining new well locations. The nature of the hydrodynamic
continuity of the reservoir is mainly revealed through pressure surveys run after a period
of production. This may define zones not being drained and therefore modifications
to the well completions might result.
As production progresses fluid contacts rise and therefore these contacts need to
be monitored and the results used to decide, for example, to recomplete a well as a
result of, for example excessive water production. As is pointed out in the chapter on
reservoir pressure, development wells before they are completed provide a valuable
resource to the reservoir engineer to enable surveys of pressure to be run to provide
a dynamic pressure-depth profile.
36
Introduction To Reservoir Engineering
8.2 History Matching
Throughout the production phase the comparison of the actual performance with that
predicted during the appraisal stage and more recent predictions is made. It is during
this stage that the quality of the reservoir simulation model comes under examination. The production pressure decline is compared to that predicted and the reservoir
simulation model adjusted to match. This process is called history matching. Clearly
if the simulation cannot ‘predict’ what has happened over the recent past it cannot
be used with much confidence to forecast the future!
More simple approaches not requiring the resources of a complex simulator can also
be used to up date early predictions, for example material balance studies.
Once production has been obtained, the additional data becomes available and makes
an important contribution to the refining of the initial reserves estimates. Two techniques historically used are decline curve analysis and material balance studies.
In material balance studies, the pressure-volume behaviour of the entire field is
studied assuming an infinite permeability for the reservoir. By assuming an initial
oil-in-place from volumetric calculations, the pressure is allowed to decline following
fluid withdrawal. This decline is matched against the observed pressure behaviour
and, if necessary, the original oil-in-place figure is modified until a match is obtained. In the presence of a water drive, additional variables are included by allowing water
influx into the ‘tank’. Water influx is governed by mathematical relationships such
as van Everdingen and Hurst (These concepts are covered in Chapters 11, 12, and
13 MB/MB Applications and Water Influx). Decline curves are plots of rate of withdrawal versus time or cumulative withdrawal
on a variety of co-ordinate scales. Usually a straight line is sought through these observations and extrapolated to give ultimate recovery and rates of recovery. Decline
curves only use rates of withdrawal and pay relatively little attention to the reservoir
and flowing pressures. A change in the mode of operation of the field could change the
slope of the decline curve; hence, this is one of the weaknesses of this technique.
A noteworthy feature of these two approaches is that the engineer in fact ‘fits’ a simple model to observe data and uses this model to predict the future by extrapolation.
As more data becomes available the model gets ‘updated’ and predicted results are
adjusted. Decline curve analysis has not been used to the same extent as in the 60’s
and 70’s. With the power of computing and the efforts made to integrate geological
understanding , the physics of the flow and behaviour of rock and fluid systems into
reservoir simulation, the ‘fitting” and the uncertainty of earlier methods are being
superseded by integrated reservoir simulation modelling.
The routine company function will generate the need for on going production profile updates. The generation of these is generally the responsibility of the reservoir
engineer, who might chose simple analytical approaches to the more costly reservoir
simulation methods.
8.3 Phases of Development
During the development there are a number of phases. Not all of these phases may
be part of the plan. There is the initial production build up to the capacity of the facilInsitute of Petroleum Engineering, Heriot-Watt University
37
ity as wells are brought on stream. There is the plateau phase where the reservoir is
produced at a capacity limited by the associated production and processing facilities.
Different companies work with different lengths of the plateau phase and each project
will have its own duration. There comes a point when the reservoir is no longer able
to deliver fluids at this capacity and the reservoir goes into the decline phase. The
decline phase can be delayed by assisting the reservoir to produce the fluids by the
use of for example ‘lifting’ techniques such as down-hole pumps and gas lift. The
decline phase is often a difficult period to model and yet it can represent a significant
amount of the reserves. These phases are illustrated in figure 34
Production rate
Plateau phase
Artificial lift
Decline phase
Build up phase
Economic limit
Time - years
Figure 34 Phases of Production.
The challenge facing the industry is the issue of the proportion of hydrocarbons left
behind. The ability to extract a greater proportion of the in-place fluids is obviously
a target to be aimed at and over recent years recoveries have increased through the
application of innovative technology. Historically there have been three phases of
recovery considered. Primary recovery, which is that recovery obtained through the
natural energy of the reservoir.
Secondary recovery is considered when the energy is supplemented by injection of
fluids, for example gas or water, to maintain the pressure or partially maintain the
pressure. The injected fluid also acts as a displacing fluid sweeping the oil to the
producing wells. After sweeping the reservoir with water or gas there will still be
remaining oil; oil at a high saturation where the water for a range of reasons, for
example; well spacing, viscosity, reservoir characteristics to name just a few, has
by-passed the oil. The oil which has been contacted by the injected fluid will not be
completely displaced from the porous media. Because of characteristics of the rock
and the fluids a residual saturation of fluid is held within the rock. Both of these
unrecovered amounts, the by-passed oil and the residual oil are a target for enhanced
recovery methods, EOR.
Much effort was put into enhanced oil recovery (EOR) research up until the mid
seventies. Sometimes it is termed tertiary recovery. When the oil price has dropped
the economics of many of the proposed methods are not viable. Many are based on
38
Introduction To Reservoir Engineering
the injection of chemicals which are often oil based. The subject of EOR has not been
forgotten and innovative methods are being investigated within the more volatile
oil price arena. Figure 35 gives a schematic representation of the various phases of
development and includes the various improved recovery methods. More recently
a new term has been introduced called Improved Oil Recovery (IOR). IOR is more
loosely defined and covers all approaches which might be used to improve the recovery of hydrocarbons in place. Clearly it is not as specific as EOR but provides more
of an achievable target than perhaps some of the more sophisticated EOR methods.
As we have entered into the next millennium it is interesting to note that a number
of major improved recovery initiatives are being considered particularly with respect
to gas injection. One perspective which make a project more viable is that of the
disposal of gas for example which is an environmental challenge in one field can
be the source of gas for another field requiring gas for a gas injection improved oil
recovery process.
Primary
Recovery
Natural
Flow
Artifical Lift
Pump gas lift etc.
Secondary
Recovery
Pressure
Maintenance
Natural
Flow
C
O
N
V
E
N
T
I
O
N
A
L
Water, gas injection
Tertiary
Recovery
E
O
R
Thermal
Gas
Steam In-situ
combustion.
Hydrocarbon
miscible, CO2
N2 immiscible
gas
Chemical
Microbial
Polymer
surfactant/
polymer
Figure 35 Oil Recovery Mechanisms.
9. THE UNIQUENESS OF THE RESERVOIR
As we have discussed the role of the reservoir engineer in combination with other
disciplines is to predict the behaviour of the reservoir. Whereas in the early years of
oil exploration little attention was paid to understanding the detailed characteristics
of the reservoir, it is now recognized that detailed reservoir properties associated with
often complex physical and chemical laws determine field behaviour. The unlocking
of these characteristics and understanding the laws enable engineering plans to be
put in place to ensure optimised developments are implemented. This is schematically illustrated in figure 36.
Insitute of Petroleum Engineering, Heriot-Watt University
39
Reservoir
Behaviour
Development
Plan
Reservoir Description
Unique
Dynamic and Static
Figure 36 Relationship between Reservoir Description, and Reservoir Behaviour.
At one extreme for example in a blow - out situation, a reservoir produces in an uncontrolled manner only restricted by the size of the well through which is producing.
Optmised development however based on a thorough understanding of the reservoir
enables the reservoir to be produced in a controlled, optimised manner.
In many other industries the effort expended on one project can be utilised in engineering a duplicate or a similar size unit elsewhere. Such opportunities are not possible in the engineering of a reservoir. Reservoirs are unique in many aspects. The
composition of the fluids are unique, the rock characteristics and related properties
are unique, the size and shape are unique and so on. From our perspective this reservoir description is dynamic as the reservoir over a period of time gives up its secrets.
From the reservoir’s perspective however the description is static, except with the
changes resulting from the impact of fluid production or injection. The challenge
to those involved is reducing the time it takes for our dynamic description to match,
our static description known only to the reservoir or whoever was responsible for
its formation! The answer perhaps is more of a philosophical nature. The reality is
shown in figure 37 where the top structure map for a North Sea gas field with a ten
year gap shows the impact of knowledge gained from a number of wells as against
that interpreted from the one well. Considerable faulting is shown not as a result of
major geological a activity over the ten years but knowledge gained from the data
associated with the new wells.
40
Introduction To Reservoir Engineering
2°00
2°10
2°20
100
0
00
200
0
21
53°10
SHELL/ESSO 49/26
2200
20
21
00
00
Gas /water contact
Depths in metres
scale 1 100,000
21
00
49/26.1
53°05
00
12
53°05
53°10
AMOCO 49/27
0
80
100
0
00
20
10
00
00
00
20
20
0
210
20 100
00 0
10
00
100
0
2000
2°00
2°20
Present interpretation of Leman Gas-field, showing contours on top of Rotliegendes in feet below sea-level
Figure 37 (a) The Leman Field as it Appeared to be When The Exploration Well Was
The Leman field as it appeared to be when the exploration well was drilled
Drilled.
2°00
2°10
53°10
2°20
2°30
53°10
SHELL/ESSO 49/26 AMOCO 49/27
Depth in feet
Miles
0 1
0 1 2 KMS
70
00
Gas /water contact
A permanent platform
63
00
00 63
53°05
53°05
6400
6300
620
0
69
00
610
0
6900
690
00
69
70
53°00
2°00
69
00
6300 6
90
0
0 64
0
00
6300
6400
2°10
0
53°00
2°20
2°30
Present interpretation of Leman Gas-field, showing contours on top of Rotliegendes in feet below sea level.
Leman
field ten
years
discovery
Figure 37b
Leman
Field
Tenafter
Years
After Discovery
The coverage of the reservoir has also changed effecting the equity associated with
the blocks. This illustrates the early benefits to be gained from drilling a number of
exploration wells. These equity agreements, are called unitisation agreements and such
agreements are shortened when good quality and comprehensive reservoir description data is available. Clearly there can never be sufficient description, however the
Insitute of Petroleum Engineering, Heriot-Watt University
41
economics of project management will determine when decisions have to be taken
based on description to date. The value of extra information has to be balanced by
the cost of delay in going ahead with a project.
10. CONCLUSION
In order to accomplish these objectives the Petroleum Reservoir Engineer should
have a broad fundamental background both theoretically and practically in the basic
sciences and engineering. The basic areas are:
(i)
(ii)
(iii)
(iv)
The properties of petroleum reservoir rocks
The properties of petroleum reservoir fluids
The flow of reservoir fluids through reservoir rock
Petroleum reservoir drive mechanisms
It is also important that the Petroleum Reservoir Engineer has a thorough basic
understanding in general, historical and petroleum geology. The influence of geological
history on the structural conditions existing in a reservoir should be known and
considered in making a reservoir engineering study. Such a study may also help to
identify and characterise the reservoir as to its aerial extent, thickness and stratification
and the chemical composition, size distribution and texture of the rock materials.
In his latest text, Dake2 comments on some of the philosophy of approach to reservoir engineering, and identifies the importance of pinning down interpretation and
prediction of reservoir behaviour to well grounded laws of physics.
Reservoir forecasting has moved on considerably since wells were drilled with little
interest and concern into the production and forecasting of what was happening in
the reservoirs thousands of feet below. The approach to coping with uncertainty as
jokingly reflected in the cartoon below, (Figure 38) is no longer the case as sophisticated computational tools enable predictions to be made with confidence and where
uncertainty exists the degree of uncertainty can be defined.
42
Introduction To Reservoir Engineering
"We feed the geological data for the area, the computer produces a schematic topological
overview designating high probability key points, then we stick the printout on the wall and
Lever throws darts at it."
Figure 38 A Past Approach to Uncertainty!
REFERENCES
1. Craft, B.C. and Hawkins, M.F. Applied Reservoir Engineering, Prentice-Hall
Inc. 1959
2. Dake, L.P., The Practise of Reservoir Engineering. Elsevier. 1994
3. Society Of Petroleum Engineers. Reserves Definitions 1995.
4. Chierici,G.L. Principles of Petroleum Reservoir Engineering. Vol 1 Springer
Verlag 1994
5. Hollois,A.P. Some petroleum engineering considerations in the change over of
the Rough Gas field to the storage mode. Paper EUR 295 Proc Europec. 1982,
pg 175
6. API. A Statistical Study of the Recovery Efficiency. American Petroleum Institute.
Bull D14, 1st Edition ,1967
7. Archer,J.S. and Wall,C.G. Petroleum Engineering Principles and Practise, Graham
and Trotman ,1986.
Insitute of Petroleum Engineering, Heriot-Watt University
43
Reservoir Pressures and Temperatures
CONTENTS
1 INTRODUCTION
2 ABNORMAL PRESSURES
3 FLUID PRESSURES IN HYDROCARBON
SYSTEMS
4 PRESSURE GRADIENTS AROUND WATEROIL CONTACT
5. TECHNIQUES FOR PRESSURE
MEASUREMENT
6. RESERVOIR TEMPERATURE
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•
Having worked through this chapter the student will be able to:
•
Define the terms; lithostatic pressure, hydrostatic pressure and hydrodynamic
pressure.
•
Draw the normal hydrostatic pressure gradient for water systems.
•
Define normal pressured reservoirs, overpressured reservoirs and underpressured
reservoirs
•
Describe briefly and sketch the pressure gradients associated with overpressured
and underpressured reservoirs.
•
Describe briefly , sketch and present equations for the pressures in a water
supported oil and gas bearing formation.
•
Illustrate how a downhole formation pressure device can be used to discriminate
permeability layers after production has commenced.
•
Comment briefly what geothermal gradient is in a reservoir where flow
processes occur at constant reservoir temperature.
Reservoir Pressures and Temperatures
1. INTRODUCTION
Determining the magnitude and variation of pressures in a reservoir is an important
aspect in understanding various aspects of the reservoir, both during the exploration
phase but also once production has commenced.
Oil and gas accumulations are found at a range of sub-surface depths. At these depths
pressure exists as a result of the depositional process and from the fluids contained
within the prous media. These pressures are called lithostatic pressures and fluid
pressures. These pressures are illustrated in figure 1.
The lithostatic pressure is caused by the pressure of rock which is transmitted through
the sub-surface by grain-to grain contacts. This lithostatic or sometimes called geostatic
or overburden pressure is of the order of 1 psi/ft. The lithostatic pressure gradient
varies according to depth, the density of the overburden, and the extent to which the
rocks are supported by water pressure. If we use this geostatic pressure gradient of
1 psi/ft. then the geostatic pressure Pov, in psig at a depth of D feet is
pov = 1.0D
(1)
The geostatic pressure is balanced in part by the pressure of the fluid within the pore
space, the pore pressure, and also by the grains of rock under compaction. In unconsolidated sands, loose sands, the overburden pressure is totally supported by the
fluid and the fluid pressure Pf is equal to the overburden pressure Pov . In deposited
formations like reservoir rocks the fluid pressure is not supporting the rocks above
but arises from the continuity of the aqueous phase from the surface to the depth D in
the reservoir. This fluid pressure is called the hydrostatic pressure. The hydrostatic
pressure is imposed by a column of fluid at rest. Its value depends on the density of
the water ρw, which is affected by salinity. In a sedimentary basin, where sediment
has settled in a region of water and hydrocarbons have been generated and trapped,
we can expect a hydrostatic pressure. For a column of fresh water the hydrostatic
pressure is 0.433 psi/ft. For water with 55,000 ppm of dissolved salts the gradient is
0.45 psi/ft; for 88,000 ppm of dissolved salts the gradient is about 0.465 psi/ft.
Its variation with depth is given by the equation.
Pf = ρwDg
(2)
where g is the acceleration due to gravity.
There is another fluid pressure which arises as a result of fluid movement and that
is called the hydrodynamic pressure. This is the fluid potential pressure gradient
which is caused by fluid flow. This however does not contribute to in-situ pressures
at rest. Institute of Petroleum Engineering, Heriot-Watt University
Depth (Ft.)
14.7
0
Pressure (psia)
FP
GP
Overpressure
Underpressure
Overburden
Pressure (OP)
Normal
(FP = Fluid Pressure, GP = Grain Pressure)
Figure 1 Gives the relationship between the lithostatic pressure and the hydrostatic
1
pressure.
Fluid pressure in hydrocarbon accumulations are dictated by the prevailing water
pressure in the vicinity of the reservoir. In a normal situation the water pressure at
any depth is:
dP
Pw =  
x D + 14.7psia
 dD  water
(3)
where dP/dD is the hydrostatic pressure gradient
This equation assumes continuity of water pressure from the surface and constant
salinity. In most cases even though the water bearing sands are divided between
impermeable shales, any break of such sealing systems will lead to hydrostatic pressure continuity, but the salinity can vary with depth.
Reservoirs whose water pressure gradient when extrapolated to zero depth give an
absolute pressure equivalent to atmospheric pressure are called normal pressured
reservoirs.
EXERCISE 1
If the average pressure gradient in a region is 0.47 psi/ft, calculate the pore
pressure in a normally pressurised formation at 7400ft. Convert the pressure from
psi to KPa, then express the pressure in MPa. What is the pressure gradient in
KPa/m?
Reservoir Pressures and Temperatures
2. ABNORMAL PRESSURE
Under certain conditions, fluid pressures may depart substantially from the normal
pressure. Overpressured reservoirs are those where the hydrostatic pressure is greater
than the normal pressure and underpressured reservoirs are below normal pressure.
Figure 1. They are called abnormal pressured reservoirs and can be defined by the
equation:
dP
Pw =  
x D + 14.7 psia + C
 dD  water
(4)
where C is a constant, being positive for overpressured and negative for an underpressured system.
For abnormally pressured reservoirs, the sand is sealed off from the surrounding strata
so that there is not hydrostatic pressure continuity to the surface.
Conditions which cause abnormal fluid pressure in water bearing sands have been
identified by Bradley 2 and include (Figure 2):
FP-Too High
Upthrust
(a)
(b)
Original Deposition
Dense Shale
Reservoir
Shale deposited too
quickly to allow
fluid equilbrium
North Sea
(c)
Glacier
Normal Surface
Greenland 3 km thick
1300 psi/1000 m ice
Figure 2 Causes of overpressurring
•
Thermal effects, causing expansion or contraction of water which is unable
to escape ; an increase in temperature of 1˚F can cause an increase of 125 psi
in a sealed fresh water system.
Institute of Petroleum Engineering, Heriot-Watt University
•
Rapid burial of sediments consisting of layers of sand and clay. Speed of burial
does not allow fluids to escape from pore space.
•
Geological changes such as uplifting of the reservoir, or surface erosion both
of which result in the water pressure being too high for the depth of the burial. The opposite occurs in a down thrown reservoir.
•
Osmosis between waters having different salinity, the sealing shale acting as a
semi-permeable membrane. If the water within the seal is more saline than the
surrounding water, the osmosis will cause a high pressure and vice versa.
Overpressured reservoirs are common in Tertiary deltaic deposits such as the North
Sea, Niger delta and the Gulf Coast of Texas. In the North Sea one mechanism for
overpressure is the inability to expel water from a system of rapidly compacted
shales.
With abnormally pressured reservoirs a permeability barrier must exist, which inhibit
pressure release. These may be lithological or structural. Common lithological
barriers are evaporates and shales. Less common are the impermeable carbonates
and sandstones. Structure permeability barriers may result from faults which, in
some cases, seal. The subject on of abnormal pressures is covered more fully in
the Geology Module
If reservoirs are all normal pressured systems then the pressure gradient for these
reservoirs would be virtually all the same, other than from the influence of salinity.
The figure below shows the water pressure gradients for a number of reservoirs in
the North Sea and indicates the significant overpressuring in this region. Often these
overpressuring show regional trends. For example the fields depicted in figure 3
show an increase in abnormal pressure in the south east direction. Clearly if all these
reservoirs were normally pressured then the pressure depths values would lie on the
same gradient line with a zero depth pressure value of atmospheric pressure.
Reservoir Pressures and Temperatures
8,000
Statfjord OWC
Brent OWC
9,000
Thistle OWC
Cormorant
OWC
4
Subsea Depth (Feet)
10,000
11,000
1
2
Heather
OWC
Ninian
OWC
3
Lyell
5
12,000
Alwyn
N.W. Alwyn
S.W> Ninian
13,000
Note:
Water gradient lines drawn
through known or projected
oil/water contacts
5000
6000
7000
8000
9000
10,000
Pressure, psig
3
Figure 3 Examples of overpressured reservoirs in the North Sea
3. FLUID PRESSURES IN HYDROCARBON SYSTEMS
Pressure gradients in hydrocarbon systems are different from those of water systems
and are determined by the oil and gas phase in-situ specific gravities, ρo and ρg of
each fluid.
The pressure gradients are a function of gas and oil composition but typically are:
 dP 
= (0.45 psi / ft)
 dD  water
(5)
 dP  = (0.35 psi / ft)
 dD  oil
(6)
 dP 
= (0.08 psi / ft)
 dD  gas
Institute of Petroleum Engineering, Heriot-Watt University
(7)
For a reservoir containing both oil and a free gas cap a pressure distribution results,
as in the Figure 4 As can be seen, the composition of the respective fluids gives rise
to different pressure gradients indicated above. These gradients will be determined by
the density of the fluids which result from the specific composition of the fluids.
Depth (Ft.)
13
8500
12
Depth (Ft.)
8600
11
10
9
Gas-Oil Contact
0.17 psi/ft
ρf = 0.39 gm/cc
8
7
6
0.29 psi/ft
ρf = 0.67 gm/cc
5
8700
Oil-Water Contact
4
0.47 psi/ft
ρf = 1.09 gm/cc
3
8800
4000
2
1
4050
4100
4150
Formation Pressure (PSI)
Figure 4 Pressure distribution for an oil reservoir with a gas cap and an oil-water contact.
The nature of the pressure regime and the position and recognition of fluid contacts
are very important to the reservoir engineer in evaluating reserves, and determining
depletion policy.
The data used for these fluid contacts comes from:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Pressure surveys
Equilibrium pressures from well tests
Flow of fluid from particular minimum and maximum depth
Fluid densities from reservoir samples
Saturation data from wireline logs
Capillary pressure data from cores
Fluid saturation from cores
EXERCISE 2
If the pressure in a reservoir at the OWC is 3625 psi, calculate the pressure at the top
if there is a 600ft continuous oil column. If a normal pressure gradient exists outwith
the reservoir, calculate the pressure differential at the top of the reservoir. Redo the
calculations for a similar field, but this time containing gas.
Reservoir Pressures and Temperatures
4. PRESSURE GRADIENTS AROUND THE WATER-OIL CONTACT
Water is always present in reservoir rocks and the pressure in the water phase Pw
and the pressure in the hyrocarbon phase Po are different . If P is the pressure at the
oil/water contact where the water saturation is 100%, then the pressure above this
contact for the hydrocarbon and water are :
Po = P - ρogh
(8)
Pw = P - ρwgh
(9)
The difference between these two pressures is the capillary pressure Pc: see Chapter
8. In a homogenous water-wet reservoir with an oil-water contact the variation of
saturation and phase pressure from the water zone through the capillary transition
zone into the oil is shown in Figure 5). In the transition zone the phase pressure
difference is given by the capillary pressure which is a function of the wetting phase
saturation. (Chapter 8).
Oil Zone
Sw
h=
Vertical
Depth
D
Oil Phase Pressure
po = pFWL - ρogh
Oil Gradient
Capilliary
Transition
Zone
pc
pc (Sw)
∆ρg
WOC
Water Gradient
FWL
(pc = o)
Water Phase Pressure
pw = pFWL - ρwgh
Water Zone
0
Swc
1
Water Saturation, Sw
pFWL
Pressure, P
Figure 5 Pressure Gradients around the Water-Oil Contact
Pc = Po - Pw
(10)
at hydrostatic equilibrium
Pc(Sw) = ∆ρgh
∆ρ = ρw-ρo
h = height above free water level
Institute of Petroleum Engineering, Heriot-Watt University
The free water level, FWL, is not coincident with the oil-water contact OWC. The
water contact corresponds to the depth at which the oil saturation starts to increase
from water zone. The free water level is the depth at which the capillary pressure
is zero.
The difference in depth between the oil-water contact and the free water level depends
on the capillary pressure which in turn is a function of permeability, grain size etc.
Providing the phase is continuous the pressures in the respective phases are:
Po = PFWL - ρogh
(11)
Pw = PFWL - ρwgh
(12)
On the depth-pressure diagram the intersection of the continuous phase pressure line
occurs at the free water level.
5. TECHNIQUES FOR PRESSURE MEASUREMENT
Earlier tests for vertical pressure logging have been replaced by open-hole testing
devices that measure the vertical pressure distribution in the well, and recover formation samples.
One such device which was introduced in the mid seventies which has established
itself in reservoir evaluation is the repeat formation tester RFT (Schlumberger trade
name). It was initially developed as a device to take samples. Over the years however
its main application is to provide pressure -depth profiles over reservoir intervals. The
device places a probe through the well mud cake and allows small volumes of fluid
to be taken and pressure measurements to be made (Figure 6). It can only be operated
therefore in an open hole environment. The unit can be set at different locations in
the well and the pressure gradient thereby obtained. This device has been superseded
by different tools provided by a number of wireline service providers. The principle
is the same of measuring with a probe in open hole the pressure depth profile.
10
Reservoir Pressures and Temperatures
Packer
Mud Cake
Packer
Filter
Flow Line
Equalising Valve
(To Mud Column)
Piston
Pressure Guage
Formation
Flow Line
Chamber 1
Probe Closed
Chamber 2
Seal Valve
to Upper Chamber
Seal Valve
to Upper Chamber
Probe Open and
Sampling
Figure 6 Original Schematic of the RFT Tool
These open hole pressure measurements have proved valuable at both the appraisal
stage and can be used to establish fluid contacts. It has also proved particularly valuable during the development stage in accessing some of the dynamic characteristics
of the reservoir. The pressure changes in different reservoir layers resulting from
production reveal the amount of interlayer communication and these pressure measurements can be a powerful tool in understanding the characteristics of the reservoir
formation. By comparing current pressure information with those obtained prior to production,
important reservoir description can be obtained which will aid reservoir depletion,
completion decisions and reservoir simulation.
In 1980 Amoco3 published a paper with respect to the Montrose Field in The North
Sea which illustrates the application of pressure-depth surveys. Figure 7 shows the
pressure depth survey in 1978 of a well after production since mid 1976. Only the
top 45ft of the 75ft oil column had been perforated. The initial pressure gradient indicates the oil and water gradients at the condition of hydrostatic equilibrium. The
second survey shows a survey after a period of high production rate, and reveals the
reservoir behaviour under dynamic conditions. The various changes in slope in the
pressure profile reveal the partial restricted flow in certain layers. Similar surveys
in each new development wells (Figure 8) show the similar profiles and enable the
detailed layered structure of the reservoir to be characterised which is important for
reservoir simulation purposes.
Institute of Petroleum Engineering, Heriot-Watt University
11
Gr%
0 100
Sw%
100 0
Reservoir pressure - psig
θ%
0 50
2500
3000
3500
4000
Top paleocene
Layer 1
Layer 2
True vertical subsea depth - metres
2500
8100
Original
pressure
gradient
8200
Layer 3
8300
2550
8400
Layer 4
8500
2600
8600
Layer 5
2650
8700
True vertical subsea depth - feet
Perforations
8800
14
26
24
18
22
16
20
Reservoir pressure - MPa
Figure 7 RFT Pressure Survey in Development Well of Montrose Field 3.
Reservoir pressure - psig
3000
3400
3200
A15 A11
A17
A18
2500
A6
A8
Original
pressure
gradient
8100
8200
8300
2550
8400
8500
2600
8600
2650
symbol
2700
8000
18
?Well number
22/17-A6
A8
A11
A15
A17
A18
20
Date
05/04/77
27/01/78
20/12/77
15/08/78
02/11/78
28/03/79
26
22
24
Reservoir pressure - MPa
8700
8800
True vertical subsea depth - feet
True vertical subsea depth - metres
2450
8900
28
9000
Figure 8 RFT Pressure Syrveys on a number of Montrose Wells3.
12
Reservoir Pressures and Temperatures
6. RESERVOIR TEMPERATURE
The temperature of the earth increases from the surface to centre. The heat flow outwards through the Earth’s crust generates a geothermal gradient, gc. This temperature
variation conforms to both a local and regional geothermal gradient, resulting from
the thermal characteristics of the lithology and more massive phenomenon associated
with the thickness of the earth’s crust along ridges, rifts and plate boundaries.
In most petroleum basins the geothermal gradient is of the order of 1.6˚F/100 ft.
(0.029 K/m) The thermal characteristics of the reservoir rock and overburden give
rise to large thermal capacity and with a large surface area in the porous reservoir
one can assume that flow processes in a reservoir occur at constant reservoir temperature. The local geothermal gradient will be influenced by associated geological
features like volcanic intrusions etc. The local geothermal gradient can be deduced
from wellbore temperature surveys . However they have to be made under stabilised
conditions since they can be influenced by transient cooling effects of circulating
and injected fluids.
During drilling the local thermal gradient can be disturbed and by analysis of the
variation of temperature with time using a bottom hole temperature (BHT) gauge
the local undisturbed temperature can be obtained.
Without temperature surveys the temperature at a vertical depth can be estimated
using a surface temperature of 15 oC (60 oF) at a depth D.
T(D) = 288.2 + gcD (K)
Solutions to Exercises
EXERCISE 1
If the average pressure gradient in a region is 0.47 psi/ft, calculate the pore pressure
in a normally pressurised formation at 7400ft. Convert the pressure from psi to KPa,
then express the pressure in MPa. What is the pressure gradient in KPa/m?
Multiply KPa by 0.145 to get psi.
1 US foot = 0.3048m.
SOLUTION
Pressure in formation = 0.47 * 7400 = 3478 psi
Converting to KPa = 3478 / 0.145 = 23986 Kpa
Converting to MPa = 23986 / 1000 = 23.99 MPa
Pressure gradient
= 0.47 psi/ft = (0.47 / 0.145) KPa/ft = 3.2414 KPa/ft
= (3.2414 /0.3048) KPa/m
= 10.63 KPa/M
Institute of Petroleum Engineering, Heriot-Watt University
13
EXERCISE 2
If the pressure in a reservoir at the OWC is 3625 psi, calculate the pressure at the top
if there is a 600ft continuous oil column. If a normal pressure gradient exists outwith
the reservoir, calculate the pressure differential at the top of the reservoir. Redo the
calculations for a similar field, but this time containing gas.
SOLUTION
Typical pressure gradients are (psi/ft):
Water – 0.45
Oil
– 0.35
Gas
– 0.08
Pressure at seal = 3625 - (600*0.35) = 3415 psi
To calculate the pressure differential across seal, look at fluid gradient differential
from OWC to seal 600ft above…
Differential = (0.45-0.35) * 600 = 60 psi
If the reservoir is gas then the differential becomes…
(0.45 – 0.08) * 600 = 222 psi higher in the reservoir than surrounding area
REFERENCES
1.
Dake,L.P. Fundamentals of Reservoir Engineering. Elsevier 1986
2. Bradley,J.S. Abnormal Formation Pressure. The American Association of
Petroleum Geologists Bulletin. Vol 59, No6, June 1975
3. Bishlawi,M and Moore,RL: Montrose Field Reservoir Management. SPE Europec Conference, London,(EUR166) Oct.1980
14
Reservoir Fluids Composition
CONTENTS
1 INTRODUCTION
2 HYDROCARBONS
2.1 Chemistry of Hydrocarbons
2.2 Alkanes or Paraffinic Hydrocarbons
2.3 Isomerism
2.4 Unsaturated Hydrocarbons
2.5 Napthene Series
2.6 Aromatics
2.7 Asphalts
3 NON-HYDROCARBON COMPOUNDS
4 COMPOSITIONAL DESCRIPTION FOR
RESERVOIR ENGINEERING
4.1 Definitions of Composition in Reservoir
Engineering
5 GENERAL ANALYSIS
5.1 Surface Condition Characterisation
5.2 Refractive Index
5.3 Fluorescence of Oil
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•
Describe briefly the origin, nature and appearance of petroleum fluids
•
Be aware that the principal components of petroleum fluids to be
hydrocarbons.
•
Draw a diagram illustrating the classification of hydrocarbons and to identify;
paraffin’s (alkanes ), aromatics and cyclic aliphatics ( napthas).
•
List the non- hydrocarbon compounds which might be present in small qualities
in reservoir fluids.
•
Define the black oil model description of the composition of a reservoir fluid.
•
Explain briefly what PNA analysis is and its application.
•
Describe briefly the concept of pseudo components in fluid composition
characterization.
•
Be aware of general analysis descriptors for petroleum fluids e.g. oAPI, refractive
index and flourescence.
•
Be able to calculate the API gravity given the specific gravity
•
Calculate given the prerequisite data proved, probable and possible reserves.
•
Describe in general terms reserve estimation.
Reservoir Fluids Composition
1 INTRODUCTION
Petroleum deposits vary widely in chemical composition and depending on location
have entirely different physical and chemical properties. The very complex
characteristics are evident from the many products which can be produced from oil
and gas.
What is petroleum? Petroleum is a mixture of naturally occurring hydrocarbons
which may exist in the solid, liquid or gaseous states, depending on the conditions
of temperature and pressure to which it is subjected.1
Petroleum deposits occurring as a gaseous state are termed natural gas, in the liquid
state as petroleum oil or crude oil and in the solid state as tars, asphalts and waxes.
For a mixture with small molecules it will be a gas at normal temperature and pressure
(NTP). Mixtures containing larger molecules will be a liquid at NTP and larger
molecules as a solid state, for example, tars and asphalts.
The exact origin of these deposits is not clear but is considered to be from plant,
animal and marine life through thermal and bacterial breakdown.
The composition of crude oil consists mainly of organic compounds, principally
hydrocarbons with small percentages of inorganic non-hydrocarbon compounds. such
as carbon dioxide, sulphur, nitrogen and metal compounds. The hydrocarbons may
include the lightest (C1 methane ) to napthenes and polycyclics with high molecular
weights.
The appearance varies from gases, through very clear liquids, yellow liquids to a
dark, often black, highly viscous material, the variety obviously being a function of
composition. Although the principal elements are carbon (84-87%), and hydrogen
(11-14%), crude oil can vary from a very light brown liquid with a viscosity similar
to water to a very viscous tar like material .
Water is always present in the pore space of a reservoir, since the original depositional
environment for the rocks was water. This water has subsequently been displaced
by the influx of hydrocarbons but not totally since surface tension forces acting in
the rock pore space cause some of the water to be retained.
For reservoir engineering purposes the description of the composition is an important
characterisation parameter for the determination of a range of physical parameters
important in various reservoir volumetric and flow calculations. It is not the concern
of the reservoir engineer to determine the composition with respect to understanding
the potential to separate the material to a range of saleable products. For this reason
therefore simplistic characterisation approaches are used. The two compositional characterisation approaches used are the compositional model
and the black oil model. The basis of the compositional model is a multicomponent
description in terms of hydrocarbons and the black oil model is a two component
description in terms of produced oil, stock tank oil and produced gas, solution gas.
The compositional model is the topic covered in this chapter and the black oil model
is covered in the liquid properties chapter.
Institute of Petroleum Engineering, Heriot-Watt University
2 HYDROCARBONS
2.1 Chemistry of Hydrocarbons
The compositional model uses hydrocarbons as the descriptor since hydrocarbons
represent the largest proportion in petroleum fluids. It is important to review briefly
the chemistry of hydrocarbons.
The hydrocarbon series is represented in figure 1 below
Hydrocarbons
Aliphatic
Alkanes
Alkenes
Aromatics
Alkynes
(Paraffins)
Cyclic Aliphatics
(Napthenes)
Figure 1 Classification of Hydrocarbon.
The hydrocarbons divide into two groupings with respect to the arrangement of the
carbon molecules and the bonds between the carbon molecules. The arrangement of
the molecules are open chain or cyclic and the bonds between the carbon are saturated
(single) bonds or unsaturated or (multiple) bonds.
2.2 Alkanes or Paraffinic Hydrocarbons
The largest series is the alkanes or paraffins which are open chain molecules with
saturated bonds. Carbon has a valance of four and therefore the formula for these
compounds is CnH2n+2. These saturated hydrocarbons include all the paraffins in
which the valence of the carbon atoms is satisfied by single covalent bonds. This type
of structure is very stable. Unsaturated hydrocarbons are those where the valence
of some of the carbon atoms is not satisfied with single covalent bonds so they are
connected by two or more bonds which make them less stable and more prone to
chemical change.
The paraffin series begins with methane (CH4), and its basic formula is CnH2n+2. Pentane to pentadecane are liquids and the chief constituents of uncracked gasoline. Its higher members are waxy solids. In a given bore hole the wax may clog the pore
space next to the hole as gas expands and cools.
The paraffins are the largest constituent of crude oil and are characterised by their
chemical inertness. Clearly they would not have remained as they are if this were not
so.
2.3 Isomerism
From methane to propane there is only one way to arrange the branched chains however
above propane there are alternative arrangements and these are called isomers.
Reservoir Fluids Composition
Structural formulae do not represent the actual structure of the molecules. Isomers
are substances of the same composition that have different molecular structure and
therefore different properties, for example, normal butane and isobutane.
normal butane
CH3CH2CH2CH3 - B.Pt. 31.1˚F
isobutane
CH3CH CH3
- B.Pt. 10.9˚F
CH3
Pentane has three structures (isomers). Clearly the number of isomers increase as the
number of carbon atoms increases. Hexane has 5 isomers and heptane 9.
Table 1 below gives some of the basic physical properties of the more common
hydrocarbons of the paraffin series and Table 2 lists the state of the various pure
components demonstrating that components which might be solid on their own
contribute to liquid states when part of a mixture. Figure 2 gives some structural
formula for three paraffin compounds.
Name
Chemical
Formula
Molecular
Weight
Boiling Point
(°C) at normal
Critical
Temp °C
Density
Gas
Liquid
(air = 1)
(water = 1)
conditions
sp.gr.
Methane
CH4
16.04
-161.4
-82.4
0.554
0.415 (-614°)
Ethane
C 2H6
30.07
-89.0
32.3
1.038
0.54 (-88°)
Propane
C 3H8
44.09
-42.1
96.8
1.522
0.585 (-44.5°)
n-butane
C4H10
58.12
0.55
153.1
2.006
0.601 (0°)
Isobutane
C4H10
58.12
-11.72
134.0
2.006
0.557
n-pentane
C5H12
72.15
36.0
197.2
2.491
0.626
Isopentane
C5H12
72.15
27.89
187.8
2.491
0.6197
n-hexane
C6H14
86.17
60.30
228.0
2.975
0.6536
Table 1 Physical properties of common hydrocarbons.
Institute of Petroleum Engineering, Heriot-Watt University
ALKANES or PARAFFIN HYDROCARBONS
Cn H 2n+2
No of carbon
atoms
1
Name
State (ntp)
Methane
Gas
2
Ethane
Gas
3
4
Propane
Butane
Gas
Gas
5
6
Pentane
Hexane
Liquid
Liquid
7
Heptane
Liquid
8
Octane
Liquid
9
10
C5-C17
Nonane
Decane
Liquid
Liquid
Liquid
C18+
Solid
Table 2 Alkanes or Paraffin Hydrocarbons Cn H 2n + 2
H
PARAFFINS
H
H
C
H
H
Methane
H
H
H
C
H
H
C
C
C
H
H
H
H
Iso-butane
H
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
n-octane
Figure 2 Gives some standard formula for saturated hydrocarbons
2.4 Unsaturated Hydrocarbons
These are hydrocarbons which have double or triple bonds between carbon atoms. They have the potential to add more hydrogen or other elements and are therefore
termed unsaturated. There are termed the olefins, and there are two types, alkenes,
for example ethylene, CH2=CH2, which have a carbon-carbon double bond and
alkynes, for example acetylene,CH=CH which have a carbon carbon triple bond. Both compound types being unsaturated are generally very reactive and hence are
not found in reservoir fluids.
2.5 Napthene Series
The napthene series (CnH2n) sometimes called cycloparaffins or alicyclic hydrocarbons
are identified by having single covalent bonds but the carbon chain is closed and is
saturated. They are very stable and are important constituents of crude oil. Their
chemical properties are similar to those of the paraffins. A crude oil with a high napthene
content is referred to as an napthenic based crude oil. An example is cyclohexane
C6H12. Figure 3 gives the structural formula for two napthenic compounds.
Reservoir Fluids Composition
NAPHTHENES
H
H
H
C
H
C
H
H
C
H
C
H
H
H
C
H
H
C
C
C
H
H
C
C
H
H
H
C
C
H
H
H
H
H
H
Methyl
Cyclopentane
H
Cyclohexane
Figure 3 Structural formula for two naphenic compounds.
2.6 Aromatics
The aromatic series (CnH2n-6) is an unsaturated closed-ring series, based on the benzene
compound and the compounds are characterised by a strong aromatic odour. Various
aromatic compounds are found in crude oils. The closed ring structure gives them
a greater stability than open compounds where double or triple bonds occur. Figure
4 gives the structural formula for two aromatic compounds.
AROMATICS
H
H
H
C
C
C
H
C
C
H
H
C
C
C
H
H
C
C
H
H
C
C
C
H
C
H
Benzene
C
C
H
H
Naphthalene
Figure 4 Structural formula for two aromtic compounds.
The aromatic-napthene based crudes are usually associated with limestone and dolomite
reservoirs such as those found in Iran, the Arabian Gulf and Borneo.
Some crude oils used to be described, more from a refining perspective, according
to the relative amount of these non paraffin compounds. Crude oils would be called
paraffinic, napthenic or aromatic. It is not a classification of value in reservoir
engineering.
Institute of Petroleum Engineering, Heriot-Watt University
Physical Properties of some Common Petroleum
Reservoir Fluid Constituents
Component
Paraffins
Methane
Ethane
Propane
n-Butane
Iso-Butane
n-Pentane
n-Hexane
Iso-octane
n-Decane
Naphthenes
Cyclopentane
Methyl cyclo-pentane
Cyclohexane
Aromatics
Benzene
Toluene
Xylene
Naphthalene
Formula
Melting Point
(˚C)
Normal Boiling Point
(˚C)
Density (g/cm3)
at 1 atm and 15˚C
CH4
C2H6
C3H8
C4H10
C4H10
C5H12
C6H14
C8H18
C10H22
-184
-172
-189.9
-135
-145
-131.5
-94.3
-107.4
030
-161.5
-88.3
-42.2
-0.6
-10.2
36.2
69.0
99.3
174.0
0.626
0.659
0.692
0.730
C5H10
C6H12
C6H12
-93.3
-142.4
6.5
49.5
71.8
81.4
0.745
0.754
0.779
C6H6
C7H8
C8H10
C10H8
5.51
-95
-29
80.2
80.1
110.6
144.4
217.9
0.885
0.867
0.880
0.971
Table 3 Physical properties of some common petroleum reservoir fluid constituents
2.7 Asphalts
Asphalt is not a series by itself. Asphalts are highly viscous to semi-solid, brownblack hydrocarbons of high molecular weight usually containing a lot of sulphur
and nitrogen, which are undesirable components, and oxygen. Asphalts are closely
related to the napthene series and because of their high nitrogen and oxygen content
they may be considered juvenile oil, not fully developed.
3 NON-HYDROCARBON COMPOUNDS
Although small in volume, generally less than 1%, non-hydrocarbon compounds have
a significant influence on the nature of the produced fluids with respect to processing
and the quality of the products.
The more common non-hydrocarbon constituents which may occur are:
sulphur, oxygen, nitrogen compounds, carbon dioxide and water.
Sulphur and its associated compounds represent 0.04% - 5% by weight. These
corrosive compounds include sulphur, hydrogen sulphide (H2S ),which is very toxic,
and mercaptans of low molecular weight ( these are produced during distillation and
require special metals to avoid corrosion). Non-corrosive sulphur materials include
sulphides. Sulphur compounds have a bad smell and both the corrosive and noncorrosive forms are undesirable. On combustion these products produce S02 and S03
which are undesirable from an environmental perspective.
Reservoir Fluids Composition
Oxygen compounds, up to 0.5% wt., are present in some crudes and decompose to
form napthenic acids on distillation, which may be very corrosive.
Nitrogen content is generally less than 0.1% wt., but can be as much as 2%. Nitrogen
compounds are complex . Gaseous nitrogen reduces the thermal quality of natural
gas and needs to be blended with high quality natural gas if present at the higher
levels.
Carbon Dioxide is a very common constituent of reservoir fluids, especially in gases
and gas condensates. Like oxygen it is a source of corrosion. It reacts with water to
form carbonic acid and iron to form iron carbonate. Carbon dioxide like methane has
a significant impact on the physical properties of the reservoir fluids.
Other compounds. Metals may be found in crude oils at low concentration and are
of little significance. Metals such as copper, iron, nickel, vanadium and zinc may be
present. Produced natural gas may contain helium, hydrogen and mercury.
Inorganic compounds The non-oil produced fluids like water will clearly contain
compounds arising from the minerals present in the rock, their concentration will
therefore vary according to the reservoir. Their composition however can have a
very significant effect on the reservoir behaviour with respect to their compatibility
with injected fluids. The precipitation of salts, scale, is a serious issue in reservoir
management.
Many of these salts need to be removed on refining as some generate HC1 when
heated with water.
4. COMPOSITIONAL DESCRIPTION FOR RESERVOIR ENGINEERING
4.1 Definitions of Composition in Reservoir Engineering
In petroleum engineering, and specifically in reservoir engineering, the main issue
is one of the physical behaviour and characteristics of the petroleum fluids. The
composition of the fluid clearly has a significant impact on the behaviour and
properties. In petroleum engineering therefore the description of the composition is
a key to determine the physical properties and behaviour. For the oil refiner or chemical manufacturer the composition of the fluid is the key to
determine what chemical products can be extracted or processed from the material. The petroleum engineer is not concerned with the fact that the oil might contain, albeit
in small concentrations, hundreds of different components. The petroleum engineer
wants as simple a description as possible which still enables the determination of the
physical properties and behaviour under different temperature and pressure conditions. Two models are used in this industry to describe the composition for physical property
prediction purposes, the black-oil model and the compositional model.
The black-oil model is a 2 component description of the fluid where the two components
are, the fluids produced at surface, stock tank oil and solution gas. Associated with
this model are black-oil parameters like solution gas-oil ratio and the oil formation
volume factor. These parameters are discussed in the chapter on liquid properties.
Institute of Petroleum Engineering, Heriot-Watt University
The compositional model is a compositional description based on the paraffin series
CnH2n+2. The fluid is described with individual compositions of normal paraffins up
to a limiting C number. Historically C6, more common now to go up to C9, or even
higher. Components greater than the limiting C number are lumped together and
defined as a C+ component.
Isomers, normal and iso are usually identified up to pentane. Non paraffinic
compounds are assigned to the next higher paraffin according to its volatility. The
material representing all compounds above the limiting carbon number are called the
C+ fraction , so C7+ for a limiting value of C6 and C10+ for a limiting value of C9.
The physical properties of paraffins up to the limiting C number are well known
and documented. The C+ component is however unique to the fluid and therefore
two properties are used to characterise it, apparent molecular weight and specific
gravity.
The behaviour of some fluids are complex and the paraffin based description may
have difficulty in predicting properties under certain conditions. Consideration may
be required to also identify napthenic and aromatic compounds, (PNA analysis),
which could be contributing to complex behaviour. This is particularly the case for
gas condensates existing at high pressures and high temperatures.
Figure 4 illustrates the compositional model and its application as reservoir fluids are
produced to surface. Although the individual components contribute to a single liquid reservoir phase for an oil, when the fluids are produced to surface they produce a gas
phase, solution gas, and a liquid phase, stock tank oil. The distribution characteristics
of the individual components is complex and not just a function of temperature and
pressure. For reservoir fluids the composition is also an influence on the distribution. This makes it a difficult task to predict this distribution perspective since reservoir
fluid compositions are unique. This topic is further dealt with in the chapter on
vapour liquid equilibrium. Improved methods of chemical analysis make it possible
to describe the oil up to a C value of C29. Although such definitions provide a very
accurate description, the associated computer effort in using such a comprehensive
description does lead to the use of pseudo components. Pseudo components are
obtained by grouping the various C number compositions, thereby reducing the
description to 4 or 5 "pseudo components". A number of methods exist to group the
various C values and other components.
10
Reservoir Fluids Composition
Reservoir Fluid
Gas at Surface Conditions
Oil at Surface Conditions
C1
C2
C3
C4
C5
C6
C7+
The relative amounts of C1 - C7+ are a
function of :
Temperature, Pressure, Composition (particularly at high temperature)
Figure 5 Compositional Model
5. GENERAL ANALYSIS
5.1 Surface condition characterisation
Reservoirs as well as having unique compositions also exist at specific pressures and
temperatures. It is important therefore to provide a common basis for describing the
quantities of fluids in the reservoir and throughout the production process.
The basis chosen is the fluids at surface conditions, the surface conditions being 14.7
psia or 101.3 kPa and 60oF or 298K. These conditions are called standard conditions. For gas therefore this yields standard cubic feet SCF or standard cubic meters SCM. It is useful to consider these expression not as volumes but as mass, the volume of
which will vary according to density. For liquids we express surface conditions as
stock tank volumes either stock tank barrels STB or stock tank cubic meters STM3. The relative amount of gas to oil is expressed by the gas-oil ratio GOR SCF/STB.
Since there are so many types of oil, each with a wide range of specific gravity, an
arbitrary non-linear relationship was developed by the American Petroleum Institute
(API) to classify crude oils by weight on a linear-scaled hydrometer. The observed
readings are always corrected for temperature to 60oF, by using a prepared table of
standard values.
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11
Degrees API =
141.5
-131.5
Sp.Gr.at 60ºF
(1)
Sp.Gr = specific gravity relative to water ar 60oF.
The API gravity of water is 10º. A light crude oil would have an API gravity of 40º,
while a heavy crude would have an API gravity of less than 20º. In the field, the API
gravity is readily measured using a calibrated hydrometer.
There are no definitions for categorising reservoir fluids, but the following table 5
indicates typical GOR, API and gas and oil gravities for the five main types. The
compositions show that the dry gases contain mostly paraffins, with the fraction
of longer chain components increasing as the GOR and API gravity of the fluids
decrease.
In chapter 4 we give a classification for the various reservoir fluid types in the context
of phase behaviour. Type
Dry Gas
Appearance Colourless
at surface
Gas
Initial GOR
(scf/stb)
WetGas
Gas Condensate
Volatile Oil
Black Oil
Colourless
Gas +
clear liquid
Colourless
+ significant
clear/straw
Colour
Brown liquid
Some
Red/Green
Liquid
Black
Viscous
Liquid
No Liquids
>15000
3000-15000
2500-3000
100-2500
-
60-70
50-70
40-50
<40
0.60-0.65
0.65-0.85
0.65-0.85
0.65-0.85
0.65-0.85
96.3
88.7
72.7
66.7
52.6
C2
3.0
6.0
10.0
9.0
5.0
C3
0.4
3.0
6.0
6.0
3.5
C4
0.17
1.3
2.5
3.3
1.8
C5
0.04
0.6
1.8
2.0
0.8
C6
0.02
0.2
2.0
2.0
0.9
C7+
0.0
0.2
5.0
11.0
27.9
ºAPI
Gas S.G.
(air=1)
Composition (mol %)
C1
Table 5 Typical values for different reservoir fluids
5.2 Refractive index
The refractive index provides another indicator of the density of produced oils. The
general refractive index range for oil is 1.39 to 1.49. The heavier the crude, the
higher the refractive index and the lower the API gravity. This can be measured with
a refractometer or by the same methods used in optical mineralogy with reference gravity oils.
12
Reservoir Fluids Composition
5.3 Fluorescence of oil
The fluorescence of oil which is measured by its colour under ultraviolet light provides
another indicator, and is often used by those analysing the cuttings as the well is
drilled. The rock sample should be placed as quickly as possible under ultraviolet
light since fluorescence of oil subsides with evaporation and the activity of ‘live’ oil
decreases. If whole core is being examined then the whole core should be passed
under UV light to determine the fluorescent colour and the pattern of oil-in-place in
the cored interval.
When possible, pictures should be taken of the core showing the fluorescence. These
are very useful when accompanying reports to the head office which may be hundreds
if not a few thousand miles away.
The degree of fluorescence is indicated below for different compositions as reflected
in the API gravity.
2˚
10˚
18˚
45˚
- 10˚
- 18˚
- 45˚
- above API
API
non-fluorescent to dull brown
API
yellow brown to gold
API
gold to pale yellow
blue-white to white
It should be pointed out that most oils increase in API gravity with depth in a given
lithologic column with the reason being that younger juvenile oils, heavier with a
lower API gravity, have not yet been transformed from the initial formation conditions
to higher petroleum members. Two well-known exceptions to this pattern are found
in the Burgan sands of Kuwait and the shallow sands of the Bibi Eibat field in the
USSR where the high-gravity members are found higher up in the stratified column
than the low-gravity members.
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EXERCISE 1
Calculate the Specific Gravity (SG) of a 38o API oil. What is its density in lbs/cu.ft?
(62.32 lbs/cu.ft equals an SG of 1.0 and 43.28 API)
Now convert an oil with an SG of 0.744 to Degrees API.
EXERCISE 2
A reservoir oil is quoted as having a Gas Oil Ratio (GOR) of 604 scf/bbl. Convert
this to Standard Cubic Meters (SCM)gas per Stock Tank Cubic Meters (SM3)
1 Foot = 0.3048m
1 barrel = 5.615 cu ft.
1 barrel = 0.159 M3
EXERCISE 3
A reservoir is said to contain an ‘initial GOR’ of 11,000scf/bbl. What type of
reservoir is described, and what API oil could be typically expected from such a
field?
EXERCISE 4
Define the ‘Black Oil Model’ and the ‘Compositional Model’
14
Reservoir Fluids Composition
Solutions to Exercises
EXERCISE 1
Calculate the Specific Gravity (SG) of a 38o API oil. What is its density in lbs/
cu.ft?
(62.32 lbs/cu.ft equals an SG of 1.0 and 43.28 API)
Now convert an oil with an SG of 0.744 to Degrees API.
SOLUTION
Convert using the equation 1:
API = (141.5 / SG) -131.5
38 = (141.5 / SG) -131.5
Sg= 141.5 / (131.5 + 38)
SG = 0.835
Similarly, to convert SG into API:
API = (141.5 / 0.744) -131.5
API = 58.7o
EXERCISE 2
A reservoir oil is quoted as having a Gas Oil Ratio (GOR) of 604 scf/bbl. Convert
this to Standard Cubic Meters (SCM)gas per Stock Tank Cubic Meters (SM3)
1 Foot = 0.3048m
1 barrel = 5.615 cu ft.
1 barrel = 0.159 M3
SOLUTION
604 scf/bbl = 604 * 0.30483 STM/bbl = 17.09 SCM/bbl= 107.48 SCM/STM3
EXERCISE 3
A reservoir is said to contain an ‘initial GOR’ of 11,000scf/bbl. What type of reservoir
is described, and what API oil could be typically expected from such a field?
SOLUTION
A reservoir with a GOR of 11,000 scf/bbl would be typically termed a ‘Gas Condensate
Reservoir’. The API gravity would probably be in the low 50’s.
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EXERCISE 4
Define the ‘Black Oil Model’ and the ‘Compositional Model’
SOLUTION
Black Oil Model.
Two component description of the reservoir fluid consisting of stock tank oil and
solution gas. Compositional changes with varying pressure and temperature are
ignored. Terms such as ‘Gas Oil Ratio’ and ‘Formation Volume Factor’ are black
oil model terms.
Compositional Model.
The compositional model is based on the paraffin series CnH2n+2. To keep the number
of components in the model manageable, long chain members are grouped together
and given an average property. These compounds are termed collectively as the ‘C+
fraction’. Typically this covers the hydrocarbons above Heptane and therefore is
called the C7+ fraction, which is characterised using the terms Apparent Molecular
Weight and Specific Gravity.
REFERENCES.
1. Amyx, J.W., Bass, D.M., and Whiting, R.L."Petroleum Reservoir
Engineering", McGraw-Hill Book Company, New York 1960
16
Phase Behaviour of Hydrocarbon Systems
CONTENTS
1 DEFINITIONS
2 PHASE BEHAVIOUR OF PURE SUBSTANCES
2.1 The Phase Diagram
3 TWO COMPONENT SYSTEMS
3.1 Pressure - Temperature Diagrams
3.2 Pressure Volume Diagram
4 MULTI-COMPONENT HYDROCARBON
4.1 Pressure Volume Diagram
4.2 Pressure Temperature Diagram
4.3 Critical Point
4.4 Retrograde Condensation
5 MULTI-COMPONENT HYDROCARBON
5.1 Oil Systems (Black Oils and Volatile Oils)
5.2 Retrograde Condensate Gas
5.3 Wet Gas
5.4 Dry Gas
6 COMPARISON OF THE PHASE DIAGRAMS OF
RESERVOIR FLUIDS
7 RESERVOIRS WITH A GAS CAP
8 CRITICAL POINT DRYING
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
General
• Define; system, components, phases, equilibrium, intensive and extensive
properties.
Pure Components
• Sketch a pressure-temperature (PT) diagram for a pure component and illustrate
on it; the vapour-pressure line, critical point, triple point, sublimation-pressure
line, the melting point line, the liquid, gas and solid phase zones.
• Define the critical pressure and critical temperature for a pure component.
• Describe briefly with the aid of a PT diagram the behavior of a pure component
system below( left|) and above ( right) of the critical point.
• Sketch the pressure- volume (PV) diagram for a pure component illustrating the
behavior above the bubble point, between the bubble and dewpoint and below
the dewpoint.
• Sketch a series of PV lines for a pure component with a temperature below, at
and above the critical temperature.
• Sketch the three dimensional phase diagram for pure component systems.
Two Components
• Plot a PV diagram for a 2 component system and identify key parameters.
• Plot a PV diagram for a 2 component system and identify key parameters and
the relationship to the vapour pressure lines for the two pure components.
• Sketch the critical point loci for a series of binary mixtures including methane
and indicate how a mixture a mixture of methane and another component can
exist as 2 phases at pressures much greater than the 2 phase limit for the two
contributing components.
• Draw a PT diagram for a two component system, to illustrate the cricondentherm,
cricondenbar and the region of retrograde condensation.
• Define the terms cricondentherm and cricindenbar. • Explain briefly what retrograde condensation is.
Multicomponent Systems
• Sketch a PT and PV diagrams to illustrate the behaviour at constant temperature
for a fluid in a PVT cell. Identify key features.
• Draw a PT diagram for a heavy oil, volatile oil, retrograde condensate gas,
wet gas and dry gas. Illustrate and explain the behaviour of depletion from the
undersaturated condition to the condition within the phase diagram.
• Describe briefly with the aid of a sketch, the reasons for and the process of gas
cycling, for retrograde gas condensate reservoirs.
• Plot a PT diagram for a reservoir with a gas cap to illustrate the gas at dew point
and oil at bubble point.
Miscellaneous
• With the aid of sketch explain the process of critical point drying.
Phase Behaviour of Hydrocarbon Systems
Oil and gas reservoir fluids are mixtures of a large number of components which when
subjected to different pressure and temperatures environments may exist in different
forms, which we call phases. Phase behaviour is a key aspect in understanding the
nature and behaviour of these fluids both in relation to their state in the reservoir and
the changes which they experience during various aspects of the production process.
In this chapter we will review the qualitative aspects of the behaviour of reservoir
fluids when subjected to changes in pressure and temperature.
1 DEFINITIONS
Before we consider the effect of temperature and pressure on hydrocarbon systems
we will define some terms:
• System - amount of substances within given boundaries under specific conditions
composed of a number of components. Everything within these boundaries are
part of the system and that existing outside of the boundaries are not part of the
system. If anything moves across these boundaries then the system will have
changed.
•Components - those pure substances which produce the system under all
conditions.
For example, in the context of reservoir engineering, methane, ethane, carbon dioxide
and water are examples of pure components.
•Phases - This term describes separate, physically homogenous parts which are
separated by definite boundaries.1 Examples in the context of water are the three
phases, ice, liquid water and water vapour.
•Equilibrium - When a system is in equilibrium then no changes take place with
respect to time in the measurable physical properties of the separate phases.
•Intensive and extensive properties - physical properties are termed either
intensive or extensive. Intensive properties are independent of the quantity
of material present. For example density, specific volume and compressibility
factor are intensive properties whereas properties such as volume and mass are
termed extensive properties; their values being determined by the total quantity
of matter present.
The physical behaviour of hydrocarbons when pressure and temperature changes
can be explained in relation to the behaviour of the individual molecules making up
the system. Temperature, pressure and intermolecular forces are important aspects
of physical behaviour.
The temperature is an indication of the kinetic energy of the molecules. It is a physical
measure of the average kinetic energy of the molecules. The kinetic energy increases
as heat is added. This increase in kinetic energy causes an increase in the motion of
the molecules which also results in the molecules moving further apart.
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The pressure reflects the frequency of the collision of the molecules on the walls of
its container. As more molecules are forced closer together the pressure increases.
Intramolecular forces are the attractive and repulsive forces between molecules. They
are affected by the distance between the molecules. The attractive forces increases
as the distance between the molecules decreases until however the electronic field of
the molecules overlap and then further decrease in distance causes a repulsive force,
which increases as the molecules are forced closer together.
The molecules in gases are widely spaced and attractive forces exist between the
molecules whereas for liquids where the molecules are closer together there is a
repelling force which causes the liquid to resist further compression.
The hydrocarbon fluids of interest in reservoir systems are composed of many components however in understanding the phase behaviour of these systems it is convenient
to reflect on the behaviour of single and two component systems.
2 PHASE BEHAVIOUR OF PURE SUBSTANCES
2.1 The Phase Diagram
It is beneficial to study the behaviour of a pure hydrocarbon under varying pressure
and temperature to gain an insight into the behaviour of more complex hydrocarbon
systems.
Phase diagrams are useful ways of presenting the behaviour of systems. They are
generally plots of pressure versus temperature and show the phases that exist under
these varying conditions.
Figure 1 gives a pressure - temperature phase diagram for a single-component system
on a pressure temperature diagram and the following points are to be noted.
Phase Behaviour of Hydrocarbon Systems
1
2
Melting P
Pressure
Solid
oint
C
Liquid
u
po
Va
blim
Su
a
Critical Point
r
s
es
Pr
e
ur
3
Vapour
Gas
tion
Triple Point
Temperature
Figure 1 Pressure temperature diagram for a single component system
•
Define the black oil model description of the composition of a reservoir fluid.
•
Explain briefly what PNA analysis is and its application.
Vapour Pressure Line
The vapour pressure line divides regions where the substance is a liquid, 2, from
regions where it is a gas, 3. Above the line indicates conditions for which a substance
is a liquid, whereas below the line represent conditions under which it is a gas. Conditions on the line indicate where both liquid and gas phases coexist.
Critical Point
The critical point C. is the limit of the vapour pressure line and defines the critical
temperature, Tc and critical pressure, Pc of the pure substance. For a pure substance
the critical temperature and critical pressure represents the limiting state for liquid and
gas to coexist. A more general definition of the critical point which is both applicable
to multi component as well as single component systems is; the critical point is the
point at which all the intensive properties of the gas and liquid are equal.
Triple Point
The triple point represents the pressure and temperature at which solid, liquid and
vapour co-exist under equilibrium conditions. Petroleum engineers seldom deal
with hydrocarbons in the solid state, however, more recently solid state issues are a
concern with respect to wax, asphaltenes and hydrates.
Sublimitation-Pressure Line
The extension of the vapour-pressure line below the triple point represents the conditions which divides the area where solid exists from the area where vapour exists
and is also called the sublimation - pressure line.
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Melting Point Line
The melting line divides solid from liquid. For pure hydrocarbons the melting point
generally increases with pressure so the slope of the line is positive. (Water is exceptional in that its melting point decreases with pressure).
3 USE OF PHASE DIAGRAMS
3.1 Pressure -Temperature Diagrams (PT)
Consider the behaviour of a cell charged with a pure substance and the volume varied
by the frictionless displacement of a piston as shown in figure 2, below.
P1
Pb
P
Pd
P2
Liquid
Gas
Figure 2 Phase Changes With Pressure at Constant Temperature
For example, following the path 1 - 2 in figure 3 on the pressure-temperature diagram,
ie holding temperature constant and varying pressure by expansion of the cylinder.
Phase Behaviour of Hydrocarbon Systems
3
E
Pc
c
oint Line
1
A
B
Liquid
Melting P
Solid
Pressure
F
r-
u
po
Va
ne
4
e li
sur
s
pre
2
G
Gas
T
Temperature
Tc
Figure 3 Pressure-Temperature Diagram for a Single-Component System
As the pressure is reduced, the pressure falls rapidly until a pressure is reached lying
on the vapour pressure line. A gas phase will begin to form and molecules leave the
liquid. At further attempts to reduce the pressure the volume of gas phase increases,
while liquid phase volume decreases but the pressure remains constant. Once the
liquid phase disappears further attempts to reduce pressure will be successful as the
gas expands.
Above the critical temperature, following the path 3 - 4, a decrease in pressure will
cause a steady change in the physical properties, for example a decrease in density but
there will not be an abrupt density change as the vapour pressure line is not crossed. No phase change takes place.
Consider the behaviour of the system around the critical point. If we go from point
A to point B, by increasing the temperature, we go though a distinctive phase change
on the vapour pressure line where two phases, liquid and gas co-exist. If we now go
a different route to B, starting with the liquid state at ‘A’ increase the pressure isothermally (constant temperature ) to a value greater than Pc at E. Then keeping the
pressure constant increase the temperature to a value greater than Tc at point F. Now
decrease the pressure to its original value at G. Finally, decrease the temperature
keeping the pressure constant until B is reached. The system is now in the vapour
state and this state has been achieved without an abrupt phase change. The vapour
states are only meaningful in the two phase regions. In areas far removed from the
two phase region particularly where pressure and temperature are above the critical
values, definition of the liquid or gaseous state is impossible and the system is best
described as in the fluid state.
The pressure-temperature diagram for ethane is given in Figure 4.
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800
c
Pressure - PSIA
700
Liquid
600
Vapor
500
400
40
60
80
100
120
Temperature - º F
Figure 4 Pressure-Temperature diagram of Ethane
3.2 Pressure Volume Diagram (PV)
The process just described in 3.1 can also be represented on a pressure-volume diagram at constant temperature (Figure 5). As the pressure is reduced from 1, a large
change in pressure occurs with small change in volume due to the relatively low
compressibility of the liquid. When the vapour pressure is reached gas begins to
form. This point is called the bubble point, ie the point at which the first few molecules leave the liquid and form small bubbles of gas. As the system expands more
liquid is vaporised at constant pressure. The point at which only a minute drop of
liquid remains is called the dew point. Sharp breaks in the line denote the bubble
point and dew point.
Phase Behaviour of Hydrocarbon Systems
PVT CELL
PV DIAGRAM
All Liquid
T > Tc
SINGLE PHASE
1
Liquid state-rapid change of
pressure with small volume change
First Gas Bubble
Pressure
Last Drop of Liquid
T < Tc
Pressure remains constant while
both gas and liquid are present
4
Dew Point
Gas
Bubble Point
T2 > Tc
2
TWO PHASE REGION
All Gas
Volume
Figure 5 Pressure-Volume diagram for a Single-Component System
For a pure substance vapour pressures at bubble point and dew point are equal to the
vapour pressure of the substance at that temperature. Above the critical point, ie 3
- 4 , the PV behaviour line shows no abrupt change and simply shows an expansion
of the substance and no phase change. This fluid is called a super critical fluid. A series of expansions can be performed at various constant temperatures and a
pressure volume diagram built up and the locus of the bubble point and dew point
values gives the bubble point and dew point lines which meet at the critical point. Conditions under the bubble point and dew point lines represent the conditions
where two phases coexist whereas those above these curves represent the conditions
where only one phase exists. At the critical temperature the P,T curve goes through
the critical point. Figure 6
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3
T = Tc
Liquid state rapid
change of temperature
with small volume change
T > Tc
1
SINGLE PHASE
T < Tc
4
De
Curve
Bubble
Point
Pressure
Critical Point
w
Po
in
t
Pressure remains constant while Cur
ve
both gas and liquid are present
2
TWO PHASE REGION
Volume
Figure 6 Series of PV lines for a pure component
The pressure volume curve for pure component ethane is given in figure 7
The locus of the bubble points and dew points form a three-dimensional diagram
when projected in to a P-T diagram give the vapour pressure line (Figure 8).
900
Pressure - PSIA
800
C
700
90
600
A
400
0
ºF
Two Phase Region
Liquid
500
11
0
B
D
0.05
0.10
0.15
ºF
Vapor
60 º
F
0.20
0.25
Specific Volume - Cu. Ft. per lb.
Figure 7 Pressure-Volume Diagram of Ethane
10
Phase Behaviour of Hydrocarbon Systems
Bubble Point Line
uid
Liq
Critical Point
G
as
an
d
Vo
lu
s
Ga
me
id
e
tur
ra
pe
m
Te
Critical Point
u
Liq
Pressure
Pressure
Li
qu
id
Dew Point Line
Vapor Pressure Curve
s
Ga ure
rat
pe
m
Te
Figure 8 Three Dimensional Phase Diagram for a Pure Component System
4 TWO COMPONENT SYSTEMS
Reservoir fluids contain many components but we will first consider a system containing two components, such a system is called a binary.
4.1 Pressure Volume Diagram
The behaviour of a mixture of two components is not as simple as for a pure substance. Figure 9 shows the P-V diagram of a two-component mixture for a constant
temperature system.
Pressure
Liquid
Bubble Point
Liquid
and
Gas
Dew Point
Ga
s
Volume
Figure 9 Pressure-Volume Line for a Two-Component System at Constant Temperature
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11
The isotherm is very similar to the pure component but the pressure increases as the
system passes from the dew point to the bubble point. This is because the composition of the liquid and vapour changes as it passes through the two-phase region. At
the bubble point the composition of the liquid is essentially equal to the composition of the mixture but the infinitesimal amount of gas is richer in the more volatile
component. At the dew point the composition of vapour is essentially the mixture
composition whereas the infinitesimal amount of liquid is richer in the less volatile
component. Breaks in the line are not as sharp as for pure substances.
The pressure-volume diagram for a specific n-pentane and n-heptane mixture is given
in Figure 10. Clearly a different composition of the two components would result
in a different shape of the diagram.
600
500
45
400
300
200
100
0
Bubble Point Line
Pressure - PSIA
0º
Critical point
45
4º
F
425
º
400
º
350
º
300
º
0.1
Dew Point Line
0.2
0.3
0.4
0.5
Specific Volume - Cu. Ft. per lb.
Figure 10 Pressure-Volume Diagram for N-Pentane and N-Heptane (52.4 mole %
Heptane) ref. 4
4.2 Pressure Temperature Diagram
Compared to the single line representing the vapour pressure curve for pure substances
there is a broad region in which the two phases co-exist. The two-phase region of
the diagram is bounded by the bubble point line and the dew point line, and the two
lines meet at the critical point. Points within a loop represent two-phase systems
(Figure 11).
Consider the constant temperature expansion of a particular mixture composition. At
1 the substance is liquid and as pressure is reduced liquid expands until the bubble
point is reached. The pressure at which the first bubbles of gas appear is termed the
bubble point pressure. As pressure is decreased liquid and gas co-exist until a minute
amount of liquid remains at the dew point pressure. Further reduction of pressure
causes expansion of the gas.
12
Phase Behaviour of Hydrocarbon Systems
By carrying out a series of constant temperature expansions the phase envelope is
defined and within the envelope contours of liquid to gas ratios obtained. These are
called quality lines and describe the pressure and temperature conditions for equal
volumes of liquid. The quality lines converge at the critical point.
4.3 Critical Point
In the same way as pure components, when more than one component is present liquid and gases cannot coexist, at pressures and temperatures higher than the critical point. The critical point for a more than one component mixture is defined as a
point at which the bubble point line and dew point line join, ie. it is also the point
at which all the intensive properties of the liquid are identical. This aspect is a very
severe test for physical property prediction methods.
If the vapour pressure lines for the pure components are drawn on the P-T diagram
then the two-phase region for the mixture lies between the vapour pressure lines. In the figure 11 the critical temperature of the mixture TcAB lies between TcA and TcB
whereas the critical pressure PcAB lies above PcA and PcB. It is important to note that
the PcAB and TcAB of the mixture does not necessarily lie between the Pc & Tc of the
two pure components.
1
Critical Point
PCAB
PCA
% Liq.
Liquid
CA
100
75
50
PCB
Pressure
b
Bu
ble
P
t
o in
e
Li n
0
t
Poin
Dew
Temperature
CB
25
TCA
2
Gas
TCAB
TCB
Figure 11 Pressure-Temperature Diagram for a Two Component System
A specific mixture composition will give a specific phase envelope lying between the
vapour pressure lines. A mixture with different proportions of the same components
will give a different phase diagram. The locus of the critical point of different mixture compositions is shown in Figure 12 for the ethane and n-heptane system, and in
Figure 13 for a series of binary hydrocarbon mixtures. Figure 13 demonstrates that
for binary mixture e.g. Methane and n-decane two phases can coexist at conditions
of pressure considerably greater than the two phase limit, critical conditions for the
separate pure components. Methane is a significant component of reservoir fluids.
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13
1400
C2
Composition
No Wt % Ethane
C 100.00
C1 90.22
C2 50.25
C3 9.78
C7 N-Heptane
1200
C1
800
C
C3
A1
an e
600
A
le
bb
Bu
i
Po
e
in
L
nt
C7
A3
0
i
100
B3
B2
B1
De
w
Po
an
e
nt
A2
200
0
li n
e
400
E th
Pressure, lbs./Sq. In. ABS
1000
N-
He
pt
B
200
300
400
500
600
Temperature º F
Figure 12 Pressure-Temperature Diagram for the Ethane-Heptane System 2
14
Phase Behaviour of Hydrocarbon Systems
6000
Single Phase
5000
Pressure Lbs. (psia)
4000
Two Phases
3000
2000
et
ha
ne
1000
M
0
0
-100
Eth
e
an
0
pa
Pro
ne
100
e
an xane ptane
ne
ut a
ent
ane
N-B N- P N-He N-He
N-Dec
200
300
400
500
600
700
Temperature º F
Figure 13 Critical Point Loci for a Series of Binary Hydrocarbon Mixtures 2
4.4 Retrograde Condensation
Within the two phase region our two component system there can be temperatures and
pressures higher than the critical temperature where two phases exist and similarly
pressures. These limiting temperatures and pressures are the cricondentherm and
cricondenbar . The cricondentherm can be defined as the temperature above which
liquid cannot be formed regardless of pressure, or expressed differently, as the maximum temperature at which two phases can exist in equilibrium. The cricondenbar can be defined as the pressure above which no gas can be formed regardless of temperature or as the maximum pressure at which two phases can exist in equilibrium.
(Figure 14).
These limits are of particular significance in relation to the shape of the diagram in
figure 14.
Consider a single isotherm on Figure 14. For a pure substance a decrease in pressure
causes a change of phase from liquid to gas. For a two-component system below Tc
a decrease in pressure causes a change from liquid to gas.
We now consider the constant temperature decrease in pressure, 1-2-3 , in figure 14 at
a temperature between the critical temperature and the cricondentherm. As pressure
is decreased from 1 the dew point is reached and liquid forms, i.e., at 2 the system is
such that 5% liquid and 95% vapour exists, i.e. a decrease in pressure has caused a
change from gas to liquid, opposite to the behaviour one would expect. The phenomena is termed Retrograde Condensation. From 2 - 3, the amount of liquid decreases
Institute of Petroleum Engineering, Heriot-Watt University
15
and vaporisation occurs and the dew point is again reached where the system is gas. Retrograde condensation occurs at temperatures between the critical temperature and
cricondentherm. The retrograde region is shown shaded in the figure.
Region of retrograde condensation
Cricondenbar
Liquid
1
% Liq.
Pressure
100
10
5
0
2
e
Dew Point Lin
3
Gas
Cricondentherm
25
Po
e
50
Bu
bb
l
in
t
Li
ne
75
Temperature
Figure 14 Phase Diagram Showing Conditions for Retrograde Considerations
5. MULTI-COMPONENT HYDROCARBON
Using two component systems we have examined various aspects of phase behaviour. Reservoir fluids contain hundreds of components and therefore are multicomponent
systems. The phase behaviour of multicomponent hydrocarbon systems in the liquid-vapour region however is very similar to that of binary systems however the
mathematical and experimental analysis of the phase behaviour is more complex.
Figure 15 gives a schematic PT & PV diagram for a reservoir fluid system. Systems
which include crude oils also contain appreciable amounts of relatively non-volatile
constituents such that dew points are practically unattainable.
16
Phase Behaviour of Hydrocarbon Systems
PVT CELL
PHASE DIAGRAM
All Liquid
Liqu
id
"a"
Critical Point
First Gas Bubble
Bubble Point
uid
Liq
%
%
40
%
20
%
w
De
int
Po
Lin
Pressure
60
e
Bu
bb
le
Pressure
Last Drop of Liquid
Po
in
Gas / 40% Liquid
80
tL
i ne
Bubble Point
Dew Point
Dew Point
All Gas
Temperature
Volume
Figure 15 Phase Diagrams for Multicomponent Systems
We will consider the behaviour of several examples of typical crude oils and natural
gases:
Low-shrinkage oil (heavy oil - black oil)
High-shrinkage oil (volatile oil)
Retrograde condensate gas
Wet gas
Dry Gas
Figure 16 is a useful diagram to illustrate the behaviour of the respective fluid types
above. However it should be emphasised that for each fluid type there will be different
scales. The vertical lines help to distinguish the different reservoir fluid types.
Isothermal behaviour below the critical point designates the behaviour of oil systems
and the fluid is liquid in the reservoir, whereas behaviour to the right of the critical
point illustrates the behaviour of systems which are gas in the reservoir.
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Single Phase Region (Gas)
Single Phase Region (Liquid)
Black
Oil
P
Pressure
Bu
% Liquid
100
75
b
int
Po
ble
Volatile
Gas
Oil
Condensate
P
m
b
Gas
CP
e
Lin
2
Two Phase Region
TM
Where:
P = Bubble point pressure
b
at indicated temperature
P = Maximum pressure at which
m
two phases can coexist
50
25
20
15
10
5
0
Dew
Poin
T = Maximum temperature at
m
which two phases can coexist
e
t Li n
Single Phase Region
C = Critical conditions
Gas
X5
X = Cricondentherm
5
Temperature
Figure 16 Phase diagram for reservoir fluids
5.1 Oil Systems ( Black Oils and Volatile Oils)
Figures 17&18 illustrate the PT phase diagrams for black and volatile oils.
The two-phase region covers a wide range of pressure and temperature. Tc is higher
than the reservoir temperature. In figure 17 the line 1-2-3 represents the constant
reservoir temperature pressure reduction that occurs in the reservoir as crude oil is
produced for a black oil. These oils are a common oil type. The dotted line shows
the conditions encountered as the fluid leaves the reservoir and flows through the
tubing to the separator.
If the initial reservoir pressure and temperature are at 2, the oil is at its reservoir
bubble point and is said to be saturated, that is, the oil contains as much dissolved
gas as it can and a further reduction in pressure will cause formation of gas. If the
initial reservoir pressure and temperature are at 1, the oil is said to be undersaturated,
i.e. The pressure in the reservoir can be reduced to Pb before gas is released into the
formation. For an oil system the saturation pressure is the bubble point pressure.
18
Phase Behaviour of Hydrocarbon Systems
1 Undersaturated
Mole % Liq.
100
Lin
e
2 Saturated
Critical Point
Pb
3
75
De
50
w
Po
int
Sep.
line
Po
int
Bu
bb
le
Pressure
Liquid
Gas
25
0
Temperature
Figure 17 Phase Diagram for a Black Oil
As the pressure is dropped from the initial condition as a result of production of fluids, the fluids remain in single phase in the reservoir until the bubble point pressure
corresponding to the reservoir temperature is reached. At this point the first bubbles
of gas are released and their composition will be different from the oil being more
concentrated in the lighter ( more volatile) components. When the fluids are brought
to the surface they come into the separator and as shown on the diagram, the separator conditions lie well within the two phase region and therefore the fluid presents
itself as both liquid and gas. The pressure and temperature conditions existing in the
separator indicate that around 85% liquid is produced, that is a high percentage and as
a result the volume of liquid at the surface has not reduced a great amount compared
to its volume at reservoir conditions. Hence the term low-shrinkage oil.
As the pressure is further reduced as oil is removed from the reservoir, point 3 will
be reached and 75% liquid and 25% gas will be existing in the reservoir. Strictly
speaking once the reservoir pressure has dropped to the bubble point, beyond that the
phase diagram does not truly represent the behaviour of the reservoir fluid. As we will
see in the chapter on drive mechanisms, below the bubble point gas produced flows
more readily than the associated oil and therefore the composition of the reservoir
fluid does not remain constant. The system is continually changing in the reservoir
and therefore the related phase diagram changes.
The summary characteristics for a black oil sometimes termed a heavy oil or low
shrinkage oil are as follows.
Broad-phase envelope
High percentage of liquid
High proportion of heavier hydrocarbons
GOR < 500 SCF/STB
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Oil gravity 30˚ API or heavier
Liquid - black or deep colour
Volatile oil contains a much higher proportion of lighter and intermediate hydocarbons than heavier black oil and therefore they liberate relatively large volumes of gas
leaving smaller amounts of liquid compared to black oils. For this reason they used
to be called high shrinkage oils. The diagram in figure 18 shows similar behaviour to
the black oil except that the lines of constant liquid to gas are more closely spaced.
Points 1 and 2 have the same meaning as for the black oil. As the pressure is reduced
below 2 a large amount of gas is produced such that at 3 the reservoir contains 40%
liquid and 60% gas.
At separator conditions 65% of the fluid is liquid, i.e. less than previous mixture.
The summary characteristics for a volatile sometimes termed a heavy oil or high
shrinkage oil when compared to black oils are as follows.
Not so broad phase envelope as black oil
Fewer heavier hydrocarbons
Deep coloured
API < 50˚
GOR < 8000 SCF/STB
1
2
Liquid
Critical Point
Mole % Liq.
100
3
50 40
e
Sep.
Gas
w
po
in
t
lin
Bu
b
ble
po
int
lin
e
Pressure
75
De
25
0
Temperature
Figure 18 Phase Diagram for a Volatile Oil
Clearly, for these fluids, it is the composition of the fluid that determines the nature
of the phase behaviour and the relative position of the saturation lines, (bubble point
and dew point lines), the lines of constant proportion of gas/liquid and the critical
point.
20
Phase Behaviour of Hydrocarbon Systems
For both of these fluids types one can prevent the reservoir fluid going two phase
by maintaining the reservoir pressure above its saturation pressure by injecting fluids into the reservoir. The most common practise is the use of water as a pressure
maintenance fluid.
5.2 Retrograde Condensate Gas
If the reservoir temperature lies between the critical point and the cricondentherm a
retrograde gas condensate field exists and Figure 19 gives the PT diagram for such
a fluid. Above the phase envelope a single phase fluid exists. As the pressure declines to 2 a dew point occurs and liquid begins to form in the reservoir. The liquid
is richer in heavier components than the associated gas. As the pressure reduces to
3 the amount of liquid increases. Further pressure reduction causes the reduction of
liquid in the reservoir by re-vaporisation. It is important to recognise that the phase
diagram below for a retrograde condensate fluid represents the diagram for a constant
composition system.
Before production the fluid in the reservoir exists as a single phase and is generally
called a gas. It is probably more accurate to call it a dense phase fluid. If the reservoir
drops below the saturation pressure the dew point, then retrograde condensation occurs within the formation. The nature of this condensing fluid is only in recent years
being understood. It was previously considered that the condensing fluid would be
immobile since its maximum proportion was below the value for it to have mobility. It was considered therefore that such valuable condensed fluids would be lost to
production and the viability of the project would be that from the ‘wet’ gas.
1
Mole % Liq.
B
Pressure
Liquid
e
bl
ub
P
tL
oin
ine
Critical Point
2
3
100
75
Sep.
50
25
10
5
0
De
t
oin
wP
e
Lin
Gas
Temperature
Figure 19 Phase Diagram for a Retrograde Condensate Gas
One of the development options for such a field therefore is to set in place a pressure
maintenance procedure whereby the reservoir pressure does not fall below the
saturation pressure. Water could be used as for oils but gas might be trapped behind
the water as the water advances through the reservoir. Gas injection, called gas
Institute of Petroleum Engineering, Heriot-Watt University
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cycling ( Figure 20 ), is the preferred yet very expensive option. In this process the
produced fluids are separated at the surface and the liquid condensates, high value
product relative to heavy oil, are sent for export, in an offshore situation probably by
tanker. The ‘dry’ gas is then compressed and reinjected into the reservoir to maintain
the pressure above the dew point. Clearly with this process the pressure will still
decline because the volume occupied by the gas volume of the exported liquid is
not being replaced. Full pressure maintenance is obtained by importing dry gas
equivalent to this exported volume from a nearby source. Eventually the injected dry
gas displaces the ‘wet’ gas and then the field can be blown down as a conventional
dry gas reservoir, if a suitable export route for the gas is then in place. The process
described is very costly and carries with it a number of risks not least the possibility
of early dry gas breakthrough.
Imported Gas
Gas
Dry Gas Reinjection
Surface Separation
Condensate Sales
Injection Well
Production Well
Gas Water Contact
Figure 20 Gas cycling process
Recent research has shown that the nature of oil forming in porous media by this retrograde process may not be as first considered. The isolation of condensing liquids in
porous rock is dependant on the relative strength of the interfacial tension and viscous
forces working in the rock. If the relative magnitude of these is high then the fluid
will be trapped however if they are low as a result of low interfacial tension, which
is the case nearer the critical point, then the condensing liquids may be mobile and
move as a result of viscous and gravity forces. Condensate liquids have been able
to flow at saturations well below the previously considered irreducible saturation
proportion. Established relative permeability thinking is having to be reconsidered in
the context of gas condensates. The phenomena just described may give explanation
to the observation sometimes made of an oil rim below a gas condensate field.
Looking at the PT phase diagram one might consider that "blowing the reservoir down"
22
Phase Behaviour of Hydrocarbon Systems
quickly might be an option and as a result vaporise the condensed liquids in the formation. This is not a serious option since once the reservoir pressure falls below the
dew point the impact of the increasing liquid proportion remaining in the reservoir
causes the phase diagram to move to the right relative to reservoir conditions, and any
vaporising will be of the lightest components which are likely to be in good supply
and therefore not of significant value.
The summary characteristics for a retrograde gas condensate fluid are as follows.
Contains more lighter HC’s and fewer heavier HC’s than high-shrinkage oil
API up to 60˚ API
GOR up to 70,000 SCF/STB
Stock tank oil is water-white or slightly coloured
5.3 Wet Gas
The phase diagram for a mixture containing smaller hydrocarbon molecules lies
well below the reservoir temperature. Figure 21. The reservoir conditions always
remain outside the two-phase envelope going from 1 to 2 and therefore the fluid exists as a gas throughout the reduction in reservoir pressure. For a wet gas system,
the separator conditions lie within the two-phase region, therefore at surface heavy
components present in the reservoir fluid condense under separator conditions and this
liquid is normally called condensate. These liquid condensates have a high proportion of light ends and sell at a premium. The proportion of condensates depend on
the compositional mix of the reservoir fluid as represented by the iso-volume lines
on the PT diagram.
Liquid
1
Pressure
Critical Point
Mole % Liq.
100
75
50
25
5
0
2
Sep.
Gas
Temperature
Figure 21 Phase Diagram for a Wet Gas
The reference wet gas, clearly does not refer to the system being wet due to the presence of water but due to the production condensate liquids.
Institute of Petroleum Engineering, Heriot-Watt University
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In some locations where there are natural petroleum leakages at the surface, when
condensates are produced they are sometimes called white oil.
The summary characteristics for wet gas are as follows.
GOR < 100,000 SCF/STB
Condensate liquid > 50˚ API
5.5 Dry Gas
The phase envelope of the dry gas, which contains a smaller fraction of the C2-C6
components, is similar to the wet gas system but with the distinction that the separator
also lies outside the envelope in the gas region (Figure 22). The term dry indicates
therefore that the fluid does not contain enough heavier HC’s to form a liquid at
surface conditions.
The summary characteristics for a dry gas are as follows.
GOR > 100,000 SCF/STB
Pressure
1
Critical Point
Liquid
75
50
25
2
Sep.
Gas
Temperature
Figure 22 Phase Diagram for a Dry Gas
6 COMPARISON OF THE PHASE DIAGRAMS OF RESERVOIR FLUIDS
Figure 16 gave a rather simplistic representation of the various types of fluids with
respect to the relative position of reservoir temperature with respect to the phase
diagram. In reality it is the phase diagram which changes according to composition
and the relative position of the reservoir temperature and separator conditions, and
these determine the character of the fluid behaviour. Figure 23 gives a better indication of the various reservoir types with respect to a specific pressure and temperature
24
Phase Behaviour of Hydrocarbon Systems
Pressure
scales. As the proportion of heavier components in the respective fluids increases
the phase envelope moves to the right.
Separator
Dry Gas
Gas
Wet Gas Condensate
Volatile
Oil
Black
Oil
Temperature (ºC)
Critical Point
Figure 23 Relative positions of phases envelopes
7 RESERVOIRS WITH A GAS CAP
Figure 24 illustrates a simplification of the phase diagrams associated with an oil
reservoir with a gas cap. The phase diagram for the gas cap fluid, the oil reservoir
fluid and for a fluid representing the combination fluid of a mixture of gas and liquid
in the same proportions as they exist in the reservoir are presented.
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Reservoir Temperature
Reservoir Gas
Total Reservoir Fluid
CG
C
Reservoir
Liquid
Pd=Pb
Pressure
Initial
Reservoir
Pressure
CL
Separator
Temperature
Figure 24 Phase Diagram for an Oil Reservoir with a Gas Cap
The diagram illustrates that at the gas-oil contact the gas is at its dew pressure, the oil
is at its bubble point pressure and the combination fluid lies on the constant proportion quality line representing the ratio of the gas and oil as they exist in the reservoir
system. The gas cap may be dry, wet or condensate depending on the composition
and phase diagram of the gas.
8 CRITICAL POINT DRYING
Although not part of the topic of phase behaviour in the context of reservoir fluids it
is useful to illustrate the application in a very practical application in the context of
the evaluation of rock properties. Critical point drying has been used by a number
of sciences to prepare specimens of delicate materials for subsequent micro visual
analysis where conventional preparation techniques will destroy delicate fabric. Critical point drying takes advantage of the behaviour of fluids around the critical
point where one can go from one phase type, like liquid to gas without a visually
observed phase change. In the 1980’s it was observed in a UK offshore field that the interpreted permeability
for a well sand in the zone where water injection was proposed was different from
well injectivity tests when compared to the core analysis value where the value was
many times more. The extent of this difference was such that permeabilities from
the well test gave values which would prevent injection to take place whereas those
from the core tests would result in practical injectivities. Clearly the difference was
important.
26
Phase Behaviour of Hydrocarbon Systems
The company concerned embarked on a more sophisticated core recovery and analysis process suspicious that perhaps the fabric of the rock was being affected by core
preparation methods. They resorted to critical point drying.
The core recovered from the water zone of the reservoir from a subsequent new well
was immersed and transferred to the test laboratory submerged in ‘formation water’.
At the laboratory a core plug sample was extracted, cut to size and loaded into a
core holder still submerged in the water. The core was then mounted in a flow rig
(figure 25) and an alcohol which is miscible with water displaced the water in the
core. Carbon dioxide at a pressure and temperature where it is in the liquid state was
then introduced which miscible displaced the alcohol. The temperature and pressure
was then adjusted taking them around the critical point rather than across the vapour
pressure line of the PT phase diagram (figure 26) ending up with a temperature and
pressure below the vapour pressure line with the fluid now in a gaseous state. After
this process the permeability was measured to be of the same order as that interpreted from the well injectivity test.
The reason for this difference was subsequently demonstrated to be a very fragile
clay which during conventional core recovery and cleaning was damaged to an extent
that its pore blocking structure was destroyed.
T
P
Core In Holder
Figure 25 Critical point drying system
Pressure
Critical Point Drying Route
Critical Point
LIQUID
Vapour
Pressure Line
GAS
Temperature
Figure 26 Critical point drying
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REFERENCES
1. Fig 1 Daniels, F Farrington: “Outlines of Physical Chemistry,” John Wiley
& Sons,Inc New York, 1948
2. Fig 2 Brown,GG et al. “ Natural Gasoline and Volatile Hydrocarbons,”
Natural Gasoline Association of America, Tulsa, Okl., 1948.
Fig 10 Sage, S.G.,Lacy,W.N. Volumetric and Phase Behaviour of Hydrocarbons,
Gulf Publishing Co.Houston 1949
28
Behaviour of Gases
CONTENTS
1 IDEAL GASES
1.1 Boyle's Law
1.2 Charles' Law
1.3 Avogadro's Law
1.4 The Equation of State For an Ideal Gas
1.5 The Density of an Ideal Gas
1.6 Standard Conditions
1.7 Mixtures of Ideal Gases
1.7.1 Dalton's Law of Partial Pressures
1.7.2 Amagat's Law
1.8 Apparent Molecular Weight
1.9 Specific Gravity of a Gas
2 BEHAVIOUR OF REAL GASES
2.1 Compressibility Factor For Natural Gases
2.2 Law of Corresponding States
2.3 Pseudocritical Properties of Natural Gases
2.4 Impact of Nonhydrocarbon Components on
z Value
2.5 Standard Conditions For Real Reservoir
Gases
3 GAS FORMATION VOLUME FACTOR
4 COEFFICIENT OF ISOTHERMAL
COMPRESSIBILITY OF GASES
5 VISCOSITY OF GASES
5.1 Viscosity
5.2 Viscosity of Mixtures
6 EQUATIONS OF STATE
6.1 Other Equations-of-State
6.2 Van de Waals Equation
6.3 Benedict - Webb - Rubin Equation (BWR)
6.4 Redlich - Kwong Equation
6.5 Soave, Redlich Kwong Equation
6.6 Peng Robinson Equation of State
6.7 Application to Mixtures
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•
Present the ideal equation of state, PV=nRT.
•
Calculate the mass of an ideal gas given PV 7T values.
•
Derive an equation to calculate the density of an ideal gas.
•
Convert a mixture composition between weight and mole fraction.
•
Present an equation and calculate the apparent molecular weight of a mixture.
•
Define and calculate the specific gravity of a gas.
•
Present the equation of state, EOS, for a ‘real gas’ and explain what ‘Z’ is,
PV=ZnRT.
•
Define the pseudocritical pressure and psuedocritical temperature and be able
to use them to determine the ‘Z’ value for a gas mixture.
•
Express and calculate reservoir gas volumes in terms of standard cubic
volumes.
•
Define the gas formation volume factor and derive an equation fore it using the
EOS.
•
Calculate the volume of gas in a reservoir in terms of standard cubic volumes
given prerequisite data.
•
Calculate the viscosity of a gas of a specific composition given perquisite
equations and figures.
•
Be aware of the development of EOS’s to predict reservoir fluid properties.
Behaviour of Gases
INTRODUCTION
A gas is a homogenous fluid that has no definite volume but fills completely the vessel
in which it is placed. The system behaviour of gases is vital to petroleum engineers
and the laws governing their behaviour should be understood. For simple gases these
laws are straightforward but the behaviour of actual hydrocarbon gases particularly
at the conditions occurring in the reservoir are more complicated.
We will review the laws that relate to the pressure, volume and temperatures of gases
and the associated equations. These relationships were previously termed gas laws;
it is now more common to describe them as equations of state.
1 IDEAL GASES
The laws relating to gases are straightforward in that the relationships of pressure,
temperature and pressure are covered by one equation. First consider an ideal gas.
An ideal gas is one where the following assumptions hold:
•
•
•
Volume of the molecules i.e. insignificant with respect to the total volume of
the gas.
There are no attractive or repulsive forces between molecules or between
molecules and container walls.
There is no internal energy loss when molecules collide.
Out of these assumptions come the following equations.
1.1 Boyle’s Law
At constant temperature the pressure of a given weight of a gas is inversely proportional
to the volume of a gas.
i.e.
Vα
1
or PV = constant, T is constant
P
(1)
P = pressure, V = volume, T = temperature.
1.2 Charles’ Law
At constant pressure, the volume of a given weight of gas varies directly with the
temperature:
i.e. V α T or
V
= constant, P is constant
T
(2)
The pressure and temperature in both laws are in absolute units.
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1.3 Avogadro’s Law
Avogadro’s Law can be stated as: under the same conditions of temperature and
pressure equal volumes of all ideal gases contain the same number of molecules. That
is, one molecular weight of any ideal gas occupies the same volume as the molecular
weight of another ideal gas at a given temperature and pressure.
Specifically, these are:
(i) 2.73 x 1026 molecules/lb mole of ideal gas. (ii) One molecular weight (in lbs) of any ideal gas at 60˚F and 14.7 psia
occupies a volume of 379.4 cu ft.
One mole of a material is a quantity of that material whose mass in the unit system
selected is numerically equal to the molecular weight.
eg.
one lb mole of methane CH4 = 16 lb
one kg mole of methane CH4 = 16kg
1.4 The Equation of State for an Ideal Gas
By combining the above laws an equation of state relating pressure, temperature and
volume of a gas is obtained.
PV
= constant
T
(3)
R is the constant when the quantity of gas is equal to one mole.
It is termed the Universal Gas Constant and has different values depending on the
unit system used, so that;
R in oilfield units = 10.732
cu ft psia
lb mole R
Table 1 gives the values for different unit systems.
p
psia
atm
atm
atm
atm
mm Hg
in.Hg
V
T
n
cu ft
cu ft
cc
litre
cu ft
litre
cu ft
R
K
K
K
R
K
R
lb - mole
lb - mole
gm - mole
gm - mole
lb - mole
gm - mole
lb - mole
R
10.73
1.3145
82.06
0.08206
0.730
62.37
21.85
Table 1 Values of R for different unit systems
Behaviour of Gases
For n moles the equation becomes:
PV = nRT
(4)
T= absolute temperature oK or oR where
ºK=273 +oC and oR=460 +oF
To find the volume occupied by a quantity of gas when the conditions of temperature
and pressure are changed from state 1 to state 2 we note that:
n =
PV
PV
PV
is a constant so that 1 1 = 2 2
RT
T1
T2
EXERCISE 1.
A gas cylinder contains methane at 1000 psia and 70°F. If the cylinder has a volume of 3 cu.ft assuming methane is an ideal gas calculate the mass of methane in
the cylinder.
1.5 The Density of an Ideal Gas
Since density is defined as the weight per unit volume, the ideal gas law can be used
to calculate densities.
ρg = weight / volume =
where ρg is the gas density
For 1 mole m = MW
V =
m
V
MW = Molecular weight
RT
P
∴ ρg =
MW.P
RT
(5)
EXERCISE 2.
Calculate the density of the gas in the cylinder in exercise 1.
Institute of Petroleum Engineering, Heriot-Watt University
1.6 Standard Conditions
Oil and gas at reservoir conditions clearly occur under a whole range of temperatures
and pressures.
It is common practice to relate volumes to conditions at surface, ie 14.7 psia and
60˚F.
ie
Pres Vres
P V
= sc sc
Tres
Tsc
(6)
sc - standard conditions res - reservoir conditions
This relationship assumes that reservoir properties behave as ideal. This is NOT the
case as will be discussed later.
EXERCISE 3.
Assuming methane is at the conditions of exercise 1, calculate the volume the gas
would occupy at standard conditions.
1.7 Mixtures of Ideal Gases
Petroleum engineering is concerned not with single component gases but mixtures
of a number of gases.
Laws established over early years governing ideal gas mixtures include Dalton’s
Law and Amagat’s Law.
1.7.1 Dalton’s Law of Partial Pressures
The total pressure exerted by a mixture of gases is equal to the sum of the pressures
exerted by its components. The partial pressure is the contribution to pressure of
the individual component.
Consider a gas made up of components A, B, C etc
The total pressure of the system is the sum of the partial pressures
ie
P = PA + PB + PC + .....
where A, B and C are components.
therefore
(7)
Behaviour of Gases
P = nA
i.e.
∴
RT
RT
RT
+ nB
+ nC
V
V
V
P =
RT
Σn j
V
Pj
n
= j = yj
P
n
(8)
where yj = mole fraction of jth component.
The pressure contribution of a component, its partial pressure, is the total pressure
times the mole fraction.
1.7.2 Amagat’s Law
Amagat’s Law states that the volume occupied by an ideal gas mixture is equal to the
sum of the volumes that the pure components would occupy at the same temperature
and pressure. Sometimes called the law of additive volumes.
i.e.
V = VA + VB + VC
V = nA
V =
i.e.
(9)
RT
RT
RT
+ nB
+ nC
P
P
P
RT
Σn j
P
Vj
n
= j = yj
V
n
(10)
i.e, for an ideal gas the volume fraction is equal to the mole fraction.
It is conventional to describe the compositions of hydrocarbon fluids in mole terms.
This is because of the above laws. In some circumstances however weight compositions
might be used as the basis and it is straight forward to convert between the two.
EXERCISE 4.
A gas is made up of the following components; 25lb of methane, 3 lb of ethane and
1.5 lb of propane. Express the composition of the gas in weight and mole fractions.
Institute of Petroleum Engineering, Heriot-Watt University
1.8 Apparent Molecular Weight
A mixture does not have a molecular weight although it behaves as though it had a
molecular weight. This is called the apparent molecular weight. AMW
If yj represents the mole fraction of the jth component:
(
AMW = Σ y j × MWj
)
AMW for air = 28.97, a value of 29.0 is usually sufficiently accurate.
EXERCISE 5.
What is the apparent molecular weight of the gas in exercise 4
1.9 Specific Gravity of a Gas
The specific gravity of a gas, γg is the ratio of the density of the gas relative to that of
dry air at the same conditions.
γg =
ρg
ρair
(11)
Assuming that the gases and air are ideal.
γg
MgP
Mg
Mg
= RT =
=
M air P
M air
29
RT
Mg = AMW of mixture, Mair = AMW of air.
EXERCISE 6.
What is the gas gravity of the gas in exercise 4 ?
2 BEHAVIOUR OF REAL GASES
The equations so far listed apply basically to ideal systems. In reality, however,
particularly at high pressures and low temperatures the volume of the molecules are
no longer negligible and attractive forces on the molecules are significant.
Behaviour of Gases
The ideal gas law, therefore, is not too applicable to light hydrocarbons and their
associated fluids and it is necessary to use a more refined equation.
There are two general methods of correcting the ideal gas law equation:
(1) By using a correction factor in the equation PV = nRT
(2) By using another equation-of-state
2.1 Compressibility Factor for Natural Gases
The correction factor ‘z’ which is a function of the gas composition, pressure and
temperature is used to modify the ideal gas law to:
PV = znRT
(12)
where the factor ‘z’ is known as the compressibility factor and the equation is known
as the compressibility equation-of-state or the compressibility equation.
The compressibility factor is not a constant but varies with changes in gas composition,
temperature and pressure and must be determined experimentally (Figure 1).
To compare two states the law now takes the form:
P1V1
PV
= 2 2
z1T1
z 2 T2
(13)
z is an expression of the actual volume to what the ideal volume would be.
Vactual
Videal
co
ns
ta
nt
(14)
at
ur
e
=
1.0
pe
r
=
Compressibility factor, Z
z
Te
m
i.e.
0.5
0
0
PRESSURE, P
Figure 1 Typical plot of the compressibility factor as a function of pressure at constant
temperature.
Institute of Petroleum Engineering, Heriot-Watt University
Although all gases have similar shapes with respect to z the actual values are component
specific. However through the law of corresponding states all pure gases are shown
to have common values.
2.2 Law of Corresponding States
The law of corresponding states shows that the properties of many pure liquids and
gases have the same value at the same reduced temperature (Tr) and pressure (Pr)
where:
Tr =
T
P
and Pr =
Tc
Pc
(15)
Where, Tc and Pc are the pure component critical temperature and pressure.
The compressibility factor ‘z’ follows this law. It is usually presented vs Tr and Pr.
Although in many cases pure gases follow the Law of Corresponding States, the gases
associated with hydrocarbon reservoirs do not. The Law has however been used to
apply to mixtures by defining parameters called pseudo critical temperature and
pseudocritical pressure .
For mixtures a pseudocritical temperature and pressure, Tpc and Ppc is used such
that:
Tpc = Σy jTcj and Ppc = Σ y j Pcj
(16)
where y is the mole fraction of component j and Tcj and Pcj are the critical temperature
and pressure of component j.
It should be emphasised that these pseudo critical temperature and pseudocritical
pressures are not the same as the real critical temperature and pressure. By definition
the pseudo values must lie between the extreme critical values of the pure components
whereas the actual critical values for mixtures can be outside these limits, as was
observed in the Phase Behaviour chapter.
EXERCISE 7.
Calculate the pseudo critical temperature and pseudocritical pressure of the mixture
in exercise 4 .
For mixtures the compressibility factor (z) has been generated with respect to natural
gases 1, where ‘z’ is plotted as a function of pseudo reduced temperature, Tpr and
pseudo reduced pressure Ppr where
10
Behaviour of Gases
Compressibility Factors for Natural Gases as a
Function of Pseudoreduced Pressure and Temperature.
1.1
0
Pseudo Reduced Pressure, Pr
1
2
3
4
5
6
8
Pseudo Reduced Temperature
3.0
2.8
2.6
2.4
2.2
2.0
1.9
1.8
1.0
0.9
5
1.
4
1.0
1.05
1.2
0.95
1.
1.5
05
1.6
1.
1.7
1
0.8
1.3
1.1
1.1
1.
1.7
1.45
0.7
0.6
1.
3
1.2
1.6
1.8
1.15
0.4
2.0
3.0
2.8
1.1
1.3
2.6
1.2
3.0
2.2
2.0
1.8
1.7
1.6
0.9
7
1.9
1.1
Compressibility of
Natural Gases
(Jan. 1, 1941)
2.6 2.4
1.2
1.0
1.1
1.4
1.3
8
1.4
2.2
1.05
0.25
1.0
1.7
1.9
2.4
1.1
0.3
1.5
4
1.
1.5
1.25
0.5
1.6
2
1.4
1.35
1.3
1.
Compressibility Factor, z
7
1.05
0.9
9
10
11
12
13
Pseudo Reduced Pressure, Pr
14
15
Figure 2 Compressibility factors for natural gas1 (Standing & Katz, Trans AIME, 1942)
Institute of Petroleum Engineering, Heriot-Watt University
11
Tpr =
T
P
and Ppr =
Tpc
Ppc
(17)
The use of this chart , figure 2 ,has become common practise to generate z values for
natural gases. Poettmann and Carpenter 2 have also converted the chart to a table.
Various equations have also been generated based on the tables.
EXERCISE 8.
For the gas of exercise 4 determine the compressibility factor at a temperature of
150°F and a pressure of 3500psia.
2.3 Pseudocritical Properties of Natural Gases
Pseudocritical Pressure, psia
The pseudocritical properties of gases can be computed from the basic composition
but can also be estimated from the gas gravity using the correlation presented in Figure 3.
Pseudocritical Properties of Natural Gases
700
Condens
650
Miscellaneous
ate Wel
l Fluid
Gases
s
600
550
Pseudocritical Temperature, R
500
450
400
n
Co
s
se
Ga
e
an
ell
sc
Mi
s
ou
ids
Flu
ell
W
e
sat
den
350
300
0.5
0.6
0.7
0.8
0.9
1.0
1.1
Gas Gravity (air = 1)
Figure 3 Pseudocritical properties of natural gases 3
12
1.2
Behaviour of Gases
2.4 Impact of Nonhydrocarbon Components on z value.
Components like hydrogen sulphide, and carbon dioxide have a significant impact
on the value of z. If the method previously applied is used large errors in z result.
Wichert and Aziz 4 have produced an equation which enables the impact of these
two gases to be calculated.
T'pc = Tpc - e
(18)
and
p′pc =
p pc Tpc′
(
)
Tpc + yH 2 S 1 − yH 2 S e
(19)
T'pc and p'pc are used to calculate Tpr and ppr. The value for ε is obtained from
the figure 4 from the Wichert and Aziz paper
80
15
70
PER CENT C02
60
50
20
40
E
25
30
30
20
30
25
10
20
15
0
5
0
10
10
20
30
40
34.5
50
60
70
80
PER CENT H2S
Figure 4 Adjustment factors for pseudocritiacl properties for non hydrocarbon
gases(Wichert & Aziz)
Institute of Petroleum Engineering, Heriot-Watt University
13
EXERCISE 9.
Calculate the pseudo critical properties of the gas in exercise 4 if it also contained 3
lb of hydrogen sulphide, 10lb of carbon dioxide and 2.5lb of nitrogen
Gas
Components
1
2
3
4
5
6
Methane
Ethane
Propane
Hydrogen
sulphide
Carbon
Dioxide
Nitrigen
Total
Weight
Wgt
Mol
fraction weight
lb moles
Mole
fraction
pc-psi
Tc °R
ppc
psia
Tpc
255.70
26.17
10.81
28.25
25
3
1.5
3
0.56
0.07
0.03
0.07
16.04
30.07
44.09
34.08
0.035
0.002
0.001
0.002
0.743
0.048
0.016
0.042
667.00
708.00
616.00
1306
344
550
666
673
495.8
33.7
10.0
54.8
10
0.22
44.01
0.005
0.108
1071
548
116.1 59.38
2.5
45
0.06
1.00
28.02
0.002
0.0466
0.043
1.000
493
227
21.0
731
9.66
390
From Wichert & Azis chart for compositions of H2S and CO2 ε = 19
Tpc′ = Tpc - e = 371o R
p′pc =
p pc Tpc′
(
)
Tpc + yH 2 S 1 − yH 2 S e
Ppc′ = 694.3
2.5 Standard Conditions for Real Reservoir Gases
As indicated in section 1.6 for ideal gases it is convenient to describe the quantity of
gas to a common basis and this is termed the standard conditions, giving rise to the
standard cubic foot and the standard cubic metre. The petroleum engineer is primarily
interested in volume calculations for gaseous mixtures. Throughout the industry gas
volumes are measured at a standard temperature of 60˚F (15.6˚C) and at a pressure of
14.7 psia (one atmosphere). These conditions are referred to as standard temperature
and pressure STP. Standard Cubic Feet, the unit of volume measured under these
conditions is sometimes abbreviated SCF or scf (SCM is Standard Cubic Metres). It
is helpful to consider these expressions not as volumes but as an alternate expression
of the quantity of material. For example a mass of gas can be expressed as so many
standard cubic feet or metres.
EXERCISE 10.
Express the quantity of 1 lb mole of a gas as standard cubic feet.
14
Behaviour of Gases
EXERCISE 11.
Express the mass of gas in exercise 4 as standard cubic feet.
3 GAS FORMATION VOLUME FACTOR
The petroleum industry expresses its reservoir quantities at a common basis of surface
conditions which for gases is standard cubic volumes. To convert reservoir volumes
to surface volumes the industry uses formation volume factors. For gases we have Bg, the gas formation volume factor, which is the ratio of the volume occupied at
reservoir temperature and pressure by a certain weight of gas to the volume occupied
by the same weight of gas at standard conditions. The shape of Bg as a function of
pressure is shown in figure 5.
Bg =
volume occupied at reservoir temperature and pressure
volume occupied at STP
The gas formation volume factor can be obtained from PVT measurements on a gas
sample or it may be calculated from the equations-of-state discussed previously. One definition of the gas formation volume factor is: it is the volume in barrels
that one standard cubic foot of gas will occupy as free gas in the reservoir at the
prevailing reservoir pressure and temperature.
Depending on the definition the units will change and the units will be; rb free
gas/scf gas or rm3 free gas/scm gas
.008
.006
Bg
rb/scf
.004
.002
1000
2000
3000
PRESSURE (psig)
Figure 5 Gas Formation Volume Factor, Bg
Institute of Petroleum Engineering, Heriot-Watt University
15
For example Bg for a reservoir at condition 2 is;
Bg =
V2
P Tz
= sc 2 2
Vsc P2 Tsc zsc
(20)
‘sc’ refers to standard conditions. z at standard conditions is taken as 1.0
The reciprocal of Bg is often used to calculate volumes at surface so as to reduce the
possibility of misplacing the decimal point associated with the values of Bg being
less than 0.01, ie:
volume at surface
1
=
=E
volume in formation Bg
E is sometimes referred to as the expansion factor.
Usually the units of Bg are barrels of gas at reservoir conditions per standard cubic
foot of gas, ie bbl/SCF or cubic metres per standard cubic metre.
Bg =
VR
Vsc
(21)
R and sc are reservoir and standard conditions respectively.
VR =
znRT
P
(22)
T and P at reservoir conditions:
Vsc =
zsc nRTsc
Psc
(23)
z = 1 for standard conditions
∴ Bg = z
T Psc cu. ft
. .
Tsc P SCF
Since Tsc = 520˚Rm Psc = 14.7 psia for most cases
Bg = 0.0283
Bg = 0.0283
or
zT cu. ft
P SCF
zT cu. ft
bbl
×
P SCF 5.615 cu ft
Bg = 0.00504
16
zT res bbl
P SCF
(24)
Behaviour of Gases
Bg = 0.0283
or
zT cu. ft
bbl
×
P SCF 5.615 cu ft
Bg = 0.00504
zT res bbl
P SCF
(25)
EXERCISE 12.
Calculate the gas formation factor for a gas with the composition of exercise 4
existing at the reservoir conditions given in exercise 8.
EXERCISE 13.
A reservoir exists at a temperature of 150°F (as for exercise 8) suitable for storing
gas. It has an areal size of 5 miles by 2 miles and is 200ft thick. The average porosity
is 20% and there is no water present. How much gas of the composition of exercise
4 can be stored at a pressure the same as in exercise 8 i.e. 3500 psia ? (1 mile=
5280 ft.)
4 Coefficient of Isothermal Compressibility of Gases
The compressibility factor, z, must not be confused with the compressibility which is
defined as the change in volume per unit volume for a unit change in pressure, or
cg = −
1  ∂V 
1  ∂Vm 
  or = −


V  ∂P 
Vm  ∂P 
(26)
Vm is the specific volume or volume per mole.
cg is not the same as z, the compressibility factor.
For an ideal gas:
PV = nRT or:
 dV  = − nRT
 dP 
P2
1 − nRT  1
cg =   
=
 V   P2  P
(27)
For real gases:
V =
znRT
P
Institute of Petroleum Engineering, Heriot-Watt University
17
dz
P
−z
 ∂V 
  = nRT dP2
 ∂P  T
P
cg = −
cg =
P  nRT  ∂z

P
− z 
2 


nRTz  P  ∂P
1 1 ∂z
− .
P z ∂P
(28)
dz/dP can be obtained from the slope of the z vs P curve.
The Law of Corresponding states can be used to express the above equation in
another form
P = Ppc Ppr
∂z  ∂Ppr   ∂z 
=

∂P  ∂P   ∂Ppr 
∂Ppr
1
=
∂P Ppc
∂z  1   ∂z 
=
∂P  Ppc   ∂Ppr 
Combining this equation with eqn 28 above yields
cg =
1
1  ∂z 
−
Ppc Ppr zPpc  ∂Ppr 
Tpr
c g Ppc =
1 1  ∂z 
−
Ppr z  ∂Ppr 
Tpr
Units of cg = P-1, and cgPc is dimensionless
cpPpc is called pseudo reduced compressibility, cpr
18
(29)
Behaviour of Gases
Since the pseudo reduced compressibility is a function of ‘z’ and pseudo reduced
pressure, the graph of Figure 2 can be used with Equation 29 to calculate values of
cpr.
5 VISCOSITY OF GASES
5.1 Viscosity
Viscosity is a measure of the resistance to flow. It is given in units of centipoise. A centipoise is a gm/100 sec.cm. The viscosity term is called dynamic viscosity
whereas kinematic viscosity is the dynamic viscosity divided by the density.
kinematic vis cos ity =
dynamic viscosity
density
Kinematic viscosity has units of cm2/100 sec and the term is called centistoke.
Viscosoty, micropoises
Gas viscosity reduces as the pressure is decreased. At low pressures an increase in
temperature increases gas viscosity whereas at high pressures gas viscosity decreases
as the temperature increases. Figure 6 gives the values for pure component ethane.
1000
900
800
700
600
500
Viscosity of ethane
Pressure, psia
5000
400
3000
300
4000
2000
15000
200
750
1000
600
100
90
80
70
50
14.7
100
150
200
250
300
350
400
Temperature, deg F
Figure 6 Viscosity of ethane
The viscosity of gases at low pressures can be obtained from correlations presented
by different workers. Institute of Petroleum Engineering, Heriot-Watt University
19
0.024
m
liu
He
0.022
Air
0.020
e
x id
Dio
n
rbo
Ca
0.018
0.016
Viscosity, cp
en
rog
Nit
nS
ge
dro
y
H
0.014
id
ulf
e
han
M et
e
ylen
Eth
0.012
e
ane
Eth
e
pan
pr o
n
ta e
i-Bu
t ane
n-Bu
0.010
0.008
e
ntan
n-pe
a
x ne
n-He
tane
p
e
n-H
ane
n-Oct
ane
n
o
N
nne
n-Deca
0.006
0.004
50
100
150
200
250
300
350
400
Temperature, ?ºF
Figure 7 Viscosity of paraffin hydrocarbon gases at one atmosphere
Figure 7 and Figure 8 give the viscosities of individual components and paraffin
hydrocarbons at one atmosphere. For systems greater than 1 atmos the viscosities
can be obtained from the literature. Another way is by calculating the reduced
temperature and reduced pressure and use the chart developed by Carr6 which gives
a ratio of µ at reservoir conditions. This is given in Figure 9 in terms of pseudo
reduced conditions.
20
Behaviour of Gases
1.0
Gas Gravity (Air = 1)
2.0
2.5
1.5
Correction added to
Viscosity, c.p.
0.015
0.013
0.012
0.011
0.010
0.009
0.008
0.007
0.006
0.005
Correction added to
Viscosity, c.p.
Viscosity, at 1 atm, µ1, centipoise
0.014
0.0010
G = 20
1.5
1.0
G = 20
0.0010
G = 06
0.0005
0
0
400
ºF
300
ºF
200
ºF
100
ºF
5
10
15
Mole per cent N2
3.5
CO2
0.0015
0.0005
0
0
1.5
1.0
G = 20
0.0010
G = 06
5
10
15
Mole per cent CO2
1.5
1.0
0.0005
G = 06
5
10
15
Mole per cent H2S
0
0
0.004
10
N2
0.0015
H2S
0.0015
3.0
Correction added to
Viscosity, c.p.
0.016
0.5
20
30
40
50
60
70
Molecular Weight
80
90
100
Figure 8 Viscosity of gases at atmospheric pressure6
6.0
5.0
µ =
Viscosity at operating temperature
and pressure, centipoises
µA =
Viscosity at 14.7 psia (1atm) and
operating temperatures, centipoises
4.0
Viscosity, µ / µA
3.5
3.0
20
15
2.5
ps
eu
do
red
10
uc
ed
8
2.0
pre
s
su
re
6
,p
R
4
3
1.5
2
1
1.0
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
Pseudoreduced Temperature, TR
Figure 9 Viscosity ratio vs pseudo reduced temperature and pseudo pressure.
Institute of Petroleum Engineering, Heriot-Watt University
21
5.2 Viscosity of Mixtures
Another formula that is used for mixtures is:
µ mix =
Σµ j y j M j
Σy j M j
(30)
j = 1, n
where:
y j = mole fraction of jth component
M j = molecular weight of component
µ j = the viscosity of jth component
n = number of components
The presence of other gases can also make a significant difference on the viscosity
(Figure 7).
EXERCISE 14.
Calculate the viscosity of the gas mixture in exercise 4 at 200°F and a pressure of
one atmosphere.
EXERCISE 15.
Use the gas gravity method to calculate the viscosity of the gas in exercise 4
EXERCISE 16.
Determine the viscosity of the gas in exercise 4 at 150°F and 3500 psia (ref ex 4, 7,
&8)
22
Behaviour of Gases
6 EQUATIONS OF STATE
6.1 Other Equations-of-State
As indicated at the start of section 2 the compressibility factor evolved out of the
need to use an equation derived out of ideal gas behaviour and incorporating it
into it a correction factor to suit real gas behaviour. One of the difficulties of the
compressibility equation:
PV = ZnRT
to describe the behaviour of gases is that the compressibility factor is not constant and
therefore mathematical manipulations cannot be made directly but must be carried out
through graphical or numerical techniques. Rather than use this modified equation
of state many have developed equations specifically to represent the behaviour of
real gases. It is an irony however that because of the long use of the equation above
incorporating z many of the real gas equation of states have been worked to calculate
z for use in the above equation.
6.2 Van de Waals Equation 1873
The well known van der Waal’s equation was one of the earliest equations to represent
the behaviour of real gases. This most basic EOS, which corrects for the volume of the
molecules and attractive and collision forces using empirical constraints a and b.
(P + a/V2) (V-b) = RT
(31)
The two corrective terms to overcome the limiting assumptions of the ideal gas
equation are:
(i) The internal pressure or cohesion term , which accounts for the cohesion forces,
is a/V2.
(ii) The co-volume b, which represents the volume occupied by one mole at infinite
pressure and results from the repulsion forces which occur when the molecules
move close together.
The equation can also be written as:
V3 - (+ b) V2 + (a/P)V - ab/P = 0
Such equations are therefore called cubic equations of state.
The equation written to solve for z, the compressibility factor , becomes:
Z3 - Z2 (1 + B) + Z A - AB = 0
(32)
where A=
aP
bP
and B =
2
( RT )
RT
Institute of Petroleum Engineering, Heriot-Watt University
(33)
23
Values of a and b are positive constants for a particular fluid and when they are
zero the ideal gas equation is recovered. One can calculate P as a function of V for
various values of T. Figure 10 is a figure of 3 isotherms. Also drawn is the curve
for saturated liquid and saturated vapour.
Isotherm T1 is the single phase isotherm, Tc is the critical isotherm and T2 gives the
isotherm below the critical temperature.
c
T1>Tc
P
Tc
Psat
T2<Tc
Vsat (liq)
Vsat (vap)
V
Figure 10 PV behaviours of pure components predicted by EOS.
At the critical point , for a pure substance , the equation of state should be such
that:
 ∂ 2P
 ∂P 
=  2
=0
 
 ∂V  T = Tc  ∂V  T = T
c
That is the critical isotherm exhibits a horizontal inflection point at the critical
point.
24
Behaviour of Gases
The application of these conditions to the van de Waals equation yields:
a=
RT
27 R2 Tc2
and b =
64 Pc
8 Pc
(34)
EXERCISE 17.
Calculate the critical constants for n- heptane.
For the curve, T2<Tc, the pressure decreases rapidly in the liquid region with increasing
V; after crossing the liquid saturated line a minimum occurs, rises to a maximum
and then decreases at the saturated vapour line. Real behaviour does not follow this
behaviour. They contain a horizontal segment where saturated liquid and saturated
vapour coexist in varying proportions.
This equation is not able to represent gas properties over a wide rage of temperatures
and pressures and over subsequent years many equations have been developed. A
number are given including those which are finding favour in their application in
this industry.
6.3 Benedict-Webb - Rubin Equation (BWR) 1940
This equation developed for pure light hydrocarbons found considerable application
in predicting thermodynamic properties of natural gases, since natural gases are
essentially mixtures of light hydrocarbons and it can be written in a form similar to
Van der Waals equation.
PT Bo RT − Ao − Co / T 2 bRT − a
+
+
+
V
V2
V3
aα
C
γ
−γ
+ 3 o 2 1 + 2  exp 2 
6
V 
V
V T  V 
P=
(35)
where a, b, c, Ao, Bo and Co are constants for a given gas.
These equations are derived for pure components for which the empirical parameters
need to be obtained. For mixtures mixing rules are required to obtain these
constants.
6.4 Redlich-Kwong Equation 1949
Numerous equations were developed with increasing numbers of constants specific to
pure components. More recently there has been a move back to the cubic equations
like van der Waals. We will describe briefly those which have found favour in the
oil and gas sector.
This modern development of cubic equations of state started in 1949 with the Redlich
and Kwong equation which involves only two empirical constants.
Institute of Petroleum Engineering, Heriot-Watt University
25
RT
a( T )
−
V - b V (V + b )
P =
(36)
where a and b are functions of temperature.
The term a(T) depends on the temperature and Redlich Kwong expressed this as a
function of the reduced temperature Tr using
a(T ) =
ac
TR
By applying the limiting condition at the critical points yields values of ac and b related to critical constants. Such that ;
ac = 0.42748
R2 Tc2
RT
and b = 0.08664 c
Pc
Pc
(37)
6.5 Soave, Redlich Kwong equation
Soave, in 1972, modified the Redlick-Kwong (RK) equation and replaced the a/T0.5
term with a temperature dependent term aT where aT = acα. .
The Soave, Redlich-Kwong (SRK) equation is therefore:
P=
RT
acα
−
(V − b) [V (V + b)]
(38)
where
α is a non dimensionless temperature dependent term which has a value of 1.0 at the
critical temperature.
α is obtained from
[
(
α = 1 + m 1 − Tr
)]
2
where m = 0.480 + 1.574ω - 0.176ω 2
where ω is the Pitzer accentric factor 8 .
6.6 Peng Robinson Equation of State 1975
Peng and Robinson modified previous equations in relation to the attractive term.
They introduced it to improve the predictions of the Soave modification in particular
for the calculation of liquid densities.
26
Behaviour of Gases
P=
RT
acα
−
V − b [V (V + b) + b(V − b)]
(39)
R2 Tc2
RT
ac = 0.457235
and b = 0.0778 c
Pc
Pc
(40)
and α is the same function as for the Soave equation except the ω function is
different;
where
m = 0.37464 + 1.54226w - 0.26992w2
These equations, in particular the SRK and PR equation are widely used in simulation
software used to predict behaviour in reservoirs, wells and processing. There are
other equations of state which are as competent at predicting physical properties
which have been developed mainly focusing on the need to improve the accuracy of
liquid volumes predictions. There is, however, great reluctance to change from those
presently used because of the investment in their associated parameters. An excellent
review of these equations and application is given by Danesh 9.
6.7 Application to Mixtures
When properties of mixtures are required mixing rules are required to combine the
data from pure components.
For both the SRK and PR equation
(
b = ∑ y j b j and a = ∑ ∑ yi y j ai aj 1 − kij
j
i
j
)
(41)
where the term kij is termed the binary interaction coefficients which are independent
of pressure and temperature. Values of binary interaction coefficients are obtained
by fitting equation of state (EOS) predictions to gas-liquid data for binary mixtures. They have NO physical property significance. Each equation has its own binary
interaction coefficient.
Effort is underway and methods exist to not use binary interaction parameters but to
use physical property related parameters to enable good quality predictions.
Institute of Petroleum Engineering, Heriot-Watt University
27
EXERCISE 18.
A PVT cell contains 0.01 cu ft ( 300cc) of gas with at composition of ; methane 0.67
mol.frac, ethane 0.235 and n-butane 0.05. The temperature is increased to 300°C.
Use the SRK equation to calculate the pressure at this increased temperature. Use
binary interaction coefficients of C1-nC4 0.02, C2-nC4 0.01 and C1-C2 0.0
Solutions to Exercises
EXERCISE 1.
A gas cylinder contains methane at 1000 psia and 70 oF. If the cylinder has a volume
of 3 cu.ft assuming methane is an ideal gas calculate the mass of methane in the
cylinder.
SOLUTION
PV
n
where n
m
M
m
m=
= nRT
= m/M
= number of moles
= mass
= molecular weight
= PMV/RT
lb 
(3cuft )
lbmole 
10.73 psia.cuft  530 o R
(
)

lbmole.o R 
(1000 psia)16.04
Mass of methane, m = 8.46 lb
28
Behaviour of Gases
EXERCISE 2.
Calculate the density of the gas in the cylinder in exercise 1.
SOLUTION
ρg =
ρg =
MW.P
RT
(1000 psia)16.04
lb 
lbmole 
10.73 psia.cuft  530 0 R
(
)

lbmole.oR 
Density of gas, ρg = 2.82
lb
cu. ft.
EXERCISE 3.
Assuming methane is at the conditions of exercise 1, calculate the volume the gas
would occupy at standard conditions.
SOLUTION
P1V1
PV
P V
= 2 2 = sc sc
T1
T2
Tsc
Vsc =
P1 Tsc V
Psc T1
1000 psia 520 o R
Vsc =
x3ft 3
o
14.7 psia 530 R
Vsc = 200.23 scf
EXERCISE 4.
A gas is made up of the following components; 25lb of methane, 3 lb of ethane and 1.5
lb of propane. Express the composition of the gas in weight and mole fractions.
Institute of Petroleum Engineering, Heriot-Watt University
29
SOLUTION
Gas
Components
1 Methane
2 Ethane
3 Propane
Totals
A
Weight
25
3
1.5
29.05
B
Mol weight
C
lb moles
16.04
30.07
44.09
D
Mole fraction
1.559
0.100
0.034
0.921
0.059
0.020
1
EXERCISE 5.
What is the apparent molecular weight of the gas in exercise 4
SOLUTION
Gas
Components
1
2
3
Methane
Ethane
Propane
A
Mol weight
mw
16.04
30.07
44.09
B
Mol fraction
yi
0.921
0.059
0.020
1.000
C
A*B
14.77
1.77
0.89
17.43
Apparent Molecular weight= 17.43
EXERCISE 6.
What is the gas gravity of the gas in exercise 4 ?
SOLUTION
γg =
Mg
M
= g
Mair 29
Μg = AMW = 17.43
Gas gravity = 0.6
EXERCISE 7.
Calculate the pseudo critical temperature and pseudocritical pressure of the mixture
in exercise 4 .
30
Behaviour of Gases
SOLUTION
Gas
Components
1
2
3
Methane
Ethane
Propane
Total
A
B
Mol weight Mole fraction
mw
yi
16.04
30.07
44.09
0.921
0.059
0.020
1.0
C
pc-psi
667.00
708.00
616.00
D.
Tc °R
ppc
344
550
666
Tpc
614.3
41.7
12.4
668.4
316.81
32.42
13.39
362.6
Pseudocritical pressure = 668.4 psia
Pseudocritical temperature = 362 oR
EXERCISE 8.
For the gas of exercise 4 determine the compressibility factor at a temperature of 150
o
F and a pressure of 3500psia.
SOLUTION
Ppr = P/Ppc, Tpr = T/Tpc
From exercise 6 Ppc = 668 psia, Tpc = 362.6°R
P = 3500 psia, and T = 150°C ie. 610°R
Ppr = 5.24, and Tpr = 1.68
From standing Katz chart, figure 2
Compressibility factor, z = 0.88
EXERCISE 9.
Calculate the pseudo critical properties of the gas in exercise 4 if it also contained 3
lb of hydrogen sulphide, 10lb of carbon dioxide and 2.5lb of nitrogen
Gas
Components
1
2
3
4
5
6
Methane
Ethane
Propane
Hydrogen
sulphide
Carbon
Dioxide
Nitrigen
Total
Weight
Wgt
Mol
fraction weight
lb moles
Mole
fraction
pc-psi
Tc °R
ppc
psia
Tpc
255.70
26.17
10.81
28.25
25
3
1.5
3
0.56
0.07
0.03
0.07
16.04
30.07
44.09
34.08
0.035
0.002
0.001
0.002
0.743
0.048
0.016
0.042
667.00
708.00
616.00
1306
344
550
666
673
495.8
33.7
10.0
54.8
10
0.22
44.01
0.005
0.108
1071
548
116.1 59.38
2.5
45
0.06
1.00
28.02
0.002
0.0466
0.043
1.000
493
227
21.0
731
9.66
390
From Wichert & Azis chart for compositions of H2S and CO2 ε = 19
Institute of Petroleum Engineering, Heriot-Watt University
31
Tpc′ = Tpc - e = 371o R
p′pc =
p pc Tpc′
(
)
Tpc + yH 2 S 1 − yH 2 S e
Ppc′ = 694.3
EXERCISE 10.
Express the quantity of 1 lb mole of a gas as standard cubic feet.
SOLUTION
Equation of state PV = RT for 1 mole
R = 10.732 psia. cu.ft/lb.mole °R T = 60+460 = 520 °R, P = 14.65 psia
or V for 1 lb.mole = RT/P = 380.9 scf/lb.mole.
EXERCISE 11.
Express the mass of gas in exercise 4 as standard cubic feet.
SOLUTION
Total mass of gas = 29.5 lb.
Apparent mol.wgt of gas exercise 5 = 17.43 lb./lb.mole
lb.moles of gas = 1.6924
Standard cubic feet of gas = 380.9 x 1.6924
= 644.68 scf
EXERCISE 12.
Calculate the gas formation factor for a gas with the composition of exercise 4 existing
at the reservoir conditions given in exercise 8.
SOLUTION T = 150 oF ie 610 oR and P = 3500 psia
Compressibility factor at these conditions from exercise 8 = 0.88
Bg using equation above = 0.0008 res bbl/scf
32
Behaviour of Gases
EXERCISE 13.
A reservoir exists at a temperature of 150oF (as for exercise 8) suitable for storing
gas. It has an areal size of 5 miles by 2 miles and is 200ft thick. The average porosity
is 20% and there is no water present. How much gas of the composition of exercise
4 can be stored at a pressure the same as in exercise 8 i.e. 3500 psia. ? (1 mile=
5280 ft.)
SOLUTION
Volume of reservoir pore space = 5x2 x (5280)2 x 200 x 0.2
= 11,151,360,000 cu. ft.
=1,985,994,657 bbls
Bg , exercise 11 =0.00077299 res. bbls/SCF
Volume of gas =2.56923E+12 scf
EXERCISE 14.
Calculate the viscosity of the gas mixture in exercise 4 at 200°F and a pressure of
one atmosphere.
SOLUTION Gas
Components
Mol Weight Mole fraction
yj
16.04
30.07
44.09
Methane
Ethane
Propane
µ mix =
0.921
0.059
0.020
1.000
Viscosity
from fig 7
µj
0.013
0.0112
0.0098
√Mj
yj√Mj
µjyj√Mj
4.0050
5.4836
6.6400
SUM
3.6884
0.3233
0.1335
4.1451
0.0470
0.0036
0.0013
0.529
Σµ j y j M j
Σy j M j
µmix = 0.0529/4.1451
µmix =0.01275 cp
EXERCISE 15.
Use the gas gravity method to calculate the viscosity of the gas in exercise 4
SOLUTION Gas
Components
Methane
Ethane
Propane
Mol Weight Mole fraction
mw
yj
16.040
30.070
44.090
0.000
0.921
0.059
0.020
1.000
Institute of Petroleum Engineering, Heriot-Watt University
14.7720
1.773
0.886
17.431
33
γg=AMW/Mair
Mol weight air =
AMW of gas =
Gas Gravity =
µg =
γg=AMW/29 Temperature = 150°F
29.000
17.431
0.601
0.01265 from fig 8
EXERCISE 16.
Determine the viscosity of the gas in exercise 4 at 150oF and 3500 psia (ref ex 4, 7,
&8)
SOLUTION From exercise 7
Ppc = 668.4
Tpc = 362.6
3500
Pr = P P =
= 5.24
c
668.4
610
Tr = T T =
= 1.68
c
362.6
From Lee correlation
µ / µatmos = 1.75
Viscosity at atmospheric pressure
From exercise 13 and 14 = 0.01275 cp
Viscosity at conditions = 0.0223 cp
EXERCISE 17.
Calculate the critical constants for n- heptane.
SOLUTION R = 10.732. Tc for heptane = 973 oR and Pc = 397 psia
Using equations above a = 115,872 cu ft 2/lb mole
and b = 3.2878 cu ft./lb mole
34
Behaviour of Gases
EXERCISE 18.
A PVT cell of volume 0.01 cu ft ( 300cc) contains 0.008 lb mole. of gas with
a composition of; methane 0.67 mol.frac, ethane 0.235 and n-butane 0.05. The
temperature is increased to 300°C. Use the SRK equation to calculate the pressure
at this increased temperature. Use binary interaction coefficients of C1-nC4 0.02,
C2-nC4 0.01 and C1-C2 0.0
SOLUTION Calculate the constants a and b for each component
R2 Tc2
RT
ac = 0.42748
and b = 0.08664 c
Pc
Pc
where m = 0.480 + 1.574ω - 0.176ω 2
[
(
α = 1 + m 1 − Tr
Components
)]
2
pc
bj
ac
ω
y
Tc°R)
m
a
a=a*α
Methane
0.67
344
667
0.4759
8735
0.0104 0.49635 0.57546
ethane
0.235
550
708
0.7223
21036
0.0979 0.63241 0.79033 16625
n-butane
0.05
766
551
1.2926
52429
0.1995 0.78701 1.00619 52753
5027
Now calculate the mixture values.
(
b = ∑ y j b j and a = ∑ ∑ yi y j ai aj 1 − kij
j
i
j
)
where a ij = (1- k ij )(a i a j )0.5
Components
yi
b
Methane
0.67
0.312
ethane
0.235
0.181
n-butane
0.05
0.129
1
kij
Methane
0.00
kij
ethane
kij
n-butane
aij
Methane
0.00
0.02
2123.7
1485.5 1037.29 4646.52
0.00
0.01
1485.5
1039.1 732.969 3257.56
0.00
1037.3
732.97 527.535 2297.8
0.622
aij
ethane
aij
n-butane
sum
sum
10201.9
Now use SRK to calculate pressure.
Institute of Petroleum Engineering, Heriot-Watt University
35
P=
RT
acα
−
(V − b) [V (V + b)]
Vm = 1.25 cu ft / lb mole
b = 0.622
a cα = 10201.9
P = 8617.6 psia
REFERENCES
1. Standing MB and Katz DL Density of Natural Gases. Trans AIME, 146(1942).
p140
2. Poettmann FH and Carpenter PG The Multiphase Flow of Gas and Water through
Vertical Flow Strings with Application to the Design of Gas Lift Installations.
API Drilling and Production Practise. 1952, pp 279-91
3. Brown GG et al. Natural Gasoline and Volatile Hydrocarbons” National Gasoline
Assoc. of America, Tulsa, Okl. 1948
4. Wichert, E and Aziz,K “ Calculate Z’s for sour gases” Hyd Proc.(May 1972)
51, 119-122
5. Katz, D.L., Handbook of Natural Gas Engineering, McGraw Hill, NY, 1959
6. Carr N et al. Viscosity of natural gases under pressure. Trans AIME 201, 264,
(1954)
7. Lee et al “The viscosity of natural gases.” Trans AIME 1966 237, 997-1000
8. Pitzer K S et al The Volumetric and Thermodynamic Properties of Fluids II.
Compressibility Factor, Vapour Pressure and Entropy of Vaporisation. J .Am.
Chem. Soc. (1955) 77, No 13,3433-3440
9. Danesh, A PVT and Phase Behaviour of Petroleum Reservoir Fluids. 1998
Elsevier ISBN:0 444 82196 1 p129-162
36
Properties of Reservoir Liquids
CONTENTS
1
COMPOSITION BLACK OIL MODELS
2
GAS SOLUBILITY, Rs
3
OIL FORMATION VOLUME FACTOR, Bo
4
TOTAL FORMATION VOLUME FACTOR, BT
5
BELOW THE BUBBLE POINT
6
OIL COMPRESSIBILITY
7
BLACK OIL CORRELATIONS
8
FLUID DENSITY
8.1 Specific Gravity of a Liquid
8.2 Density Based on Ideal Solution Principles
9
FORMATION VOLUME FACTOR OF GAS
CONDENSATE, Bgc
10 VISCOSITY OF OIL
11 INTERFACIAL TENSION
12 COMPARISON OF RESERVOIR FLUID
MODELS
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• Define gas solubility, Rs and plot vs. P for a reservoir fluid.
• Define undersaturated and saturated oil.
• Explain briefly flash and differential liberation
•
Define the oil formation volume factor Bo, and plot Bo vs. P for a reservoir
fluid.
•
Define the Total Formation Volume factor Bt, and plot Bt vs. P alongside a Bo
vs. P plot.
•
Present an equation to express Bt in terms of Bo, Rs and Bg.
•
Express oil compressibility in terms of oil formation volume factor.
•
Use black oil correlations and their graphical form to calculate fluid
properties.
•
Calculate the density of a reservoir fluid mixture, using ideal solution principles,
at reservoir pressure and temperature, using density correction chart for C1 &
C2 and other prerequisite data.
•
Define the formation volume factor of a gas condensate
•
Calculate the reserves and production of gas and condensate operating above
the dewpoint, given prerequisite data.
•
Use viscosity equations and correlations to calculate viscosity of fluid at reservoir
conditions.
•
Calculate the interfacial tension of equilibrium gas-oil systems given prerequisite
equations and data.
•
List the comparisons of the black oil and compositional model in predicting
liquid properties
Properties of Reservoir Liquids
1 COMPOSITION - BLACK OIL MODEL
As introduced in the chapter on Composition, petroleum engineers are requiring
a compositional description tool to use as a basis for predicting reservoir and well
fluid behaviour. The two approaches that are commonly used are the multicomponent
compositional model described in the earlier chapter and the two component black oil
model. The latter simplistic approach has been used for many years to describe the
composition and behaviour of reservoir fluids. It is called the “Black Oil Model”.
The black oil model considers the fluid being made up of two components - gas dissolved
in oil and stock tank oil. The compositional changes in the gas when changing pressure
and temperature are ignored. To those appreciating thermodynamics this simplistic
two component model is difficult to cope with. The Black Oil Model, illustrated in
Figure 1, is at the core of many petroleum engineering calculations, and associated
procedures and reports.
Associated with the black oil model are Black Oil model definitions in relation to
Gas Solubility and Formation Volume Factors.
Reservoir Fluid
Solution Gas
/ = Rs
Stock Tank Oil
/ = Bo
Bo = Oil Formation Volume Factor
Rs = Solution Gas to Oil Ratio
Figure 1 "Black Oil Model"
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2 GAS SOLUBILITY
Although the gas associated with oil and the oil itself are multicomponent mixtures
it is convenient to refer to the solubility of gas in crude oil as if we were dealing with
a two-component system.
The amount of gas forming molecules in the liquid phase is limited only by the
reservoir conditions of temperature and pressure and the quantity of light components
present.
The solubility is referred to some basis and it is customary to use the stock tank
barrel.
Solubility
=
f (pressure, temperature, composition of gas
composition of crude oil)
For a fixed gas and crude, at constant T, the quantity of solution gas increases with
p, and at constant p, the quantity of solution gas decreases with T
Rather than determine the amount of gas which will dissolve in a certain amount of
oil it is customary to measure the amount of gas which will come out of solution as
the pressure decreases. Figure 2 illustrates the behaviour of an oil operating outside
the PT phase diagram in its single phase state when the reservoir pressure is above
its reservoir bubble point at 1. Fluid behaviour in the reservoir is single phase and
the oil is said to be undersaturated . In this case a slight reduction of pressure causes
the fluid to remain single phase. If the oil was on the boundary bubble point pressure
line at 2 then a further reduction in pressure would cause two phases to be produced,
gas and liquid. This saturated fluid is one that upon a slight reduction of pressure
some gas is released. The concept of gas being produced or coming out of solution
gives rise to this gas solubility perspective. Clearly when the fluids are produced to
the surface as shown by the undersaturated oil in figure 2 the surface conditions lie
within the two phase area and gas and oil are produced. The gas produced is termed
solution gas and the oil at surface conditions stock tank oil. These are the two components making up the reservoir fluid, clearly a very simplistic concept.
The gas solubility Rs is defined as the number of cubic feet (cubic metre) of gas
measured at standard conditions, which will dissolve in one barrel (cubic metre)
of stock tank oil when subjected to reservoir pressure and temperature.
In metric units the volumes are expressed as cubic metre of gas at standard conditions
which will dissolve in one cubic metre of stock tank oil.
Properties of Reservoir Liquids
Solution Gas
Surface
Rsi scf/stb
Pi 1
Pressure
2
+
P
Stock Oil Tank
1 st b. oil
Temperature
Phase Diagram
Oil Reservoir
Bo rb.oil
Oil and Dissolved Gas
Figure 2 Production of reservoir hydrocarbons above bubble point
Figure 3 gives a typical shape of gas solubility as a function of pressure for a reservoir fluid at reservoir temperature. When the reservoir pressure is above the bubble
point pressure then the oil is undersaturated, i.e. capable of containing more gas. As
the reservoir pressure drops gas does not come out of solution until the bubble point
is reached, over this pressure range therefore the gas in solution is constant. At the
bubble point pressure, corresponding to the reservoir temperature, two phases are
produced, gas and oil. The gas remaining in solution therefore decreases.
The nature of the liberation of the gas is not straight forward. Within the reservoir
when gas is released then its transport and that of the liquid is influenced by the relative
permeability of the rock ( discussed in Chapter 10). The gas does not remain with its
associated oil i.e. the system changes. In the production tubing and in the separator
it is considered that the gas and associated liquid remain together i.e. the system is
constant. The amount of gas liberated from a sample of reservoir oil depends on the
conditions of the liberation. There are two basic liberation mechanisms:
Institute of Petroleum Engineering, Heriot-Watt University
600
Rs scf/stb
Rsi
400
200
Pb
1000
2000
3000
Pressure (psig)
Figure 3 Solution Gas - Oil Ratio as a Function of Pressure.
Flash liberation
the gas is evolved during a definite reduction in
pressure and the gas is kept in contact with the liquid
until equilibrium has been established.
Differential liberation
the gas being evolved is being continuously
removed from contact with the liquid and the liquid is in
equilibrium with the gas being evolved over a finite
pressure range.
The two methods of liberation give different results for Rs. This topic is covered in
more detail in the PVT analysis chapter.
Production of a crude oil at reservoir pressures below the bubble point pressure occurs
by a process which is neither flash or differential vaporisation. Once enough gas is
present for the gas to move toward the wellbore the gas tends to move faster than the
oil. The gas formed in a particular pore tends to leave the liquid from which it was
formed thus approximating differential vaporisation, however, the gas is in contact with
liquid throughout the path through the reservoir. The gas will also migrate vertically
as a result of its lower density than the oil and could form a secondary gas cap.
Fluid produced from reservoir to the surface is considered to undergo a flash process
where the system remains constant.
3 OIL FORMATION VOLUME FACTOR, B o
The volume occupied by the oil between surface conditions and reservoir or other
operating changes is that of the total system; the ‘stock tank oil’ plus its associated
or dissolved ‘solution gas’. The effect of pressure on the complex stock tank liquid
and the solution gas is to induce solution of the gas in the liquid until equilibrium is
reached. A unit volume of stock tank oil brought to equilibrium with its associated
Properties of Reservoir Liquids
gas at reservoir pressure and temperature will occupy a volume greater than unity
(unless the oil has very little dissolved gas at very high pressure).
The relationship between the volume of the oil and its dissolved gas at reservoir
condition to the volume at stock tank conditions is called the Oil Formation Volume
Factor Bo. The shape of the Bo vs. pressure curve is shown in Figure 4. It shows
that above the bubble point pressure the reduction in pressure from the initial pressure causes the fluid to expand as a result of its compressibility. This relates to the
chapter on Phase Behaviour where for an oil the PV diagram shows a large decline
in pressure for a small increase in volume, being again an indication of the compressibility of the liquid. Below the bubble point pressure this expansion due to
compressibility of the liquid is small compared to the ‘shrinkage’ of the oil as gas is
released from solution.
The oil formation volume factor, is the volume in barrels (cubic metres) occupied in
the reservoir, at the prevailing pressure and temperature, by one stock tank barrel
(one stock tank cubic metre) of oil plus its dissolved gas.
Units - rb (oil and dissolved gas)
Bo rb./stb
1.2
1.1
1.0
Pb
1000
2000
3000
Pressure (psig)
Figure 4 Oil formation volume factor
These black oil parameters, Bo and Rs are illustrated in Figure 5 a,b,&c from Craft
and Hawkins 1 reservoir engineering text., where they present the Rs and Bo curve for
the Big Sandy field in the USA. The visual concept of the changes during pressure
and temperature decrease is also presented.
Institute of Petroleum Engineering, Heriot-Watt University
PA
PA
P
PB
P01
(a)
Free Gas
2.990 Cu. Ft.
Free Gas
676 Cu. Ft.
1,310 BBL
1,333 BBL
1,210 BBL
1,040 BBL
1,000 BBL
P01 = 3500 PSIA
T01 = 160º F
PB = 2500 PSIA
T01 = 160º F
P = 1200 PSIA
T01 = 160º F
PA = 14.7 PSIA
T01 = 160º F
PA = 14.7 PSIA
T01 = 60º F
A
B
C
D
E
567SCF/STB
500
AT 1200 PSIA
RS = 337
400
300
200
100
0
0
500
1000
1500
2000
2500
INITIAL PRESSURE
BUBBLE POINT PRESSURE
Solution Gas, SCF/STB
600
(b)
Free Gas
567 Cu. Ft.
3000
3500
Pressure, PSIA
Figure 5 Gas to oil ratio and oil formation volume factor for Big Sandy Field reservoir
oil 1.
2500 PSIA
BOB = 1.333
1200 PSIA
BO = 1.210
1.20
14.7 PSIA & 160º F
BO = 1.040
1.10
1.00
14.7 PSIA & 60º F
BO = 1.000
0
500
1000
1500
2000
Pressure, PSIA
Figure 5b
2500
INITIAL PRESSURE
1.30
3500 PSIA
BOI = 1.310
BUBBLE POINT PRESSURE
(b)
Formation Volume Factor, BBL/STB
1.40
3000
3500
Properties of Reservoir Liquids
The reciprocal of the oil formation volume factor is called the ‘shrinkage factor bo
bo =
1
Bo
The formation factor Bo may be multiplied by the volume of stock tank oil to find
the volume of reservoir required to produce that volume of stock tank oil. The
shrinkage factor can be multiplied by the volume of reservoir oil to find the stock
tank volume.
It is important to note that the method of processing the fluids will have an effect
on the amount of gas released and therefore both the values of the solution gas-oil
ratio and the formation volume factor. A reservoir fluid does not have single Bo or Rs
values. Bo & Rs are dependant on the surface processing conditions. This simplistic
reservoir model (Figure 6) demonstrates that the black oil model description of the
reservoir fluids is an after the event, processing, description in terms of the produced
fluids. This simplistic approach to modelling reservoir fluids becomes more difficult
to consider when one is involved in reservoirs which become part of a total reservoir
system (Figure 7).
Rs
BO
Figure 6 Black oil description of reservoir fluid
Institute of Petroleum Engineering, Heriot-Watt University
Multi Reservoir System
Bo
Rs
?
Rs 1
Bo 1
Rs 3
Bo 3
Rs 2
Bo 2
Rs 4
Bo 4
Figure 7 Integrated system of reservoir common pipeline and final collection system.
4 TOTAL FORMATION VOLUME FACTOR, Bt
In reservoir engineering it is sometimes convenient to know the volume occupied
in the reservoir by one stock tank barrel of oil plus the free gas that was originally
dissolved in it. A factor is used called the total formation-volume factor Bt, or
the two-phase volume-factor and is defined as the volume in barrels that 1.0 STB
and its initial complement of dissolved gas occupies at reservoir temperature and
pressure, i.e. it includes the volume of the gas which has evolved from the liquid
and is represented by:
Bg (Rsb - Rs)
i.e.
Bt = Bo + Bg (Rsb - Rs)
Rsb = the solution gas to oil ratio at the bubble point
10
(1)
Properties of Reservoir Liquids
B0b
Gas
Oil
Bg(Rsb-Rs)
Bt
Oil
B0
Hg
Figure 8a Total formation volume factor or two phase volume factor
Its application comes from the Material Balance equation (Chapter 15) where it is
sometimes used to express the volume of oil and associated gas as a function of pressure. It is important to note that Bt does not have volume significance in reservoir
terms since the assumption in Bt is that the system remains constant. As mentioned
earlier if the pressure drops below the bubble point in the reservoir then the gas
coming out of solution moves away from its associated oil because of its favourable
relative permeability characteristics.
Figure 8b gives a comparison of the total formation-volume factor with the oil formation-volume factor. Clearly above Pb the two values are identical since no free
gas is released. Below Pb the difference between the values represents the volume
occupied by free gas.
Bt
Bo
Pressure
Pb
Figure 8b Total and oil formation volume factor
The value of BT can be estimated by combining estimates of BO and calculation of
Bg and known solubility values for the pressures concerned.
Institute of Petroleum Engineering, Heriot-Watt University
11
5 BELOW THE BUBBLE POINT
Figure 9 depicts the behaviour below the bubble point when produced gas at the
surface comes from two sources, the solution gas associated with the oil entering the
wellbore plus free gas which has come out of solution in the reservoir and migrated to
the wellbore. The total producing gas to oil ratio is made up of the two components
solution gas Rs and the free gas which is the difference. The diagram illustrates the
volumes occupied by these two in the reservoir, the solution gas being part of Bo and
the free gas volume through Bg.
Free Gas
& Solution Gas
Pressure
Surface
R= Rs + (R - Rs)
Pi
+
Stock Oil Tank
P
1 st b. oil
Temperature
Oil Reservoir
Reservoir
Gas
Oil
Bo rb (oil and dissolved gas) /stb
(R - Rs) Bg
rb (free gas) /stb
Figure 9 Production of reservoir hydrocarbons below bubble point
6 OIL COMPRESSIBILITY
The volume changes of oil above the bubble point are very significant in the context
of recovery of undersaturated oil. The oil formation volume factor variations above
the bubble point reflect these changes but they are more fundamentally embodied
in the coefficient of compressibility of the oil, or oil compressibility.
The equation for oil compressibility is
co = −
1  ∂V 
 
V  ∂P  T
in terms of formation volume factors this equation yields
12
Properties of Reservoir Liquids
co = −
1  ∂Bo 


Bo  ∂P  T
Assuming that the compressibility does not change with pressure the above equation
can be integrated to yield ;
co ( P2 − P1 ) = − ln
V2
V1
where P1 & P2, and V1 & V2 represent the pressure and volume at conditions 1 & 2.
7 BLACK OIL CORRELATIONS
Over the years there have been many correlations generated based on the two component based black oil model characterisation of oil. The correlations are based
on data measured on the oils of interest. These empirical correlations relate black
oil parameters, the variables of Bo and Rs to; reservoir temperature, and oil and gas
surface density. It is important to appreciate that these correlations are empirical and
are obtained by taking a group of data for a particular set of oils and finding a best fit
correlation. Using the correlation for fluids whose properties do not fall within those
for the correlation can result in significant errors. Danesh 2 has given an excellent
review of many of these correlations
A number of empirical correlations, based on largely US crude oils, and other locations across the world have been presented to estimate black oil parameters of gas
solubility and oil formation volume factor. The most commonly used is Standing’s
3
correlation. Other correlations include, Lasater 4, and recently Glaso 6
Pb = f (Rs, γg, po, T)
where Pb = bubble point pressure at ToF
Rs = solution gas-oil ratio (cu ft/ bbl)
γg= gravity of dissolved gas
ρo = density of stock-tank oil .(specific gravity)
Standing’s correlation for the calculation of Pb, bubble point pressure is:
 R  0.83

s

Pb = 18.2   10 (0.00091T − 0.0125( API )) − 1.4 
 γ g 

His correlation for the oil formation volume factor is;
(2)
1.2
  γ g  0.5

Bo = 0.9759 + 0.000120  Rs   + 1.25T 
  ρo 

Institute of Petroleum Engineering, Heriot-Watt University
(3)
13
Standing's correlations have been presented as nomographs enabling quick look
predictions to be made. Figures 10 & 11 give the nomogram forms of these correlations
for gas solubility and oil formation volume factor. Standing’s correlation is based on
a set of 22 California crudes.
Other correlations have been presented by Lasater 4 based on 137 Canadian ,USA
and South American crudes, Vasquez and Beggs5 using 6000 data points, Glaso 6 using 45 North Sea crude samples, and Mahoun 7 who used 69 Middle Eastern crudes.
Danesh2 gives a very useful table showing the ranges covered by the respective black
oil correlations
14
Properties of Reservoir Liquids
2000
Tank oil gravity, ºAPI
50
1000
900
800
700
600
bl
r b 400
,
tio
10
500
pe
1.
20
ft
30
1.
10
1500
300
cu
1.90
1.80
1.
30
ra
il
-o 200
as
1.70
G
1.60
50
1.
40
150
1.50
1.
100
90
80
70
1.40
60
50
Temperature, ºF
il
1.20
1.02
1.03
1.04
1.05
1.06
1.08 1.10
1.07 1.09
bb
lp
er
bb
lo
ft
0.50
0.60
0.70
0.80
0.90
1.00
20
an
30
Gas gravity Air=1
1.30
100
120
140
160
180
200
220
240
260
ko
40
Formation volume of bubble-point liquid
Figure 10 Oil-formation volume factor as a function of gas solubility, temperature, gas
gravity and oil gravity (Standing)
Institute of Petroleum Engineering, Heriot-Watt University
15
16
600
700
800
900
1000
Ta
nk
20
pe
ra
Bubble-point
500
m
Te
400
1.50
t
e
ur
,º
1500
F
2000
50
3000
ºA
4000
PI
70
e
Pr
ss
u
,
re
ps
ia
Figure 11 Gas solubility as a function of pressure. Temperature, gas gravity and oil
gravity
6000
5000
1
80
90
100
Ai
r=
60
gr
av
ity
ty,
vi
ra
lg
oi
40
G
as
30
150
200
180
240
220
260
200
ti
160
300
ft
500
600
700
800
900
1000
120
140
60
80
100
400
l
300
1.30
1.40
as
ra
u
bb
200
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
52
54
56
58
60
1.20
1.10
1.00
0.90
0.80
0.70
0.60
0.50
G
l
oi
c
o,
r
pe
(STANDING)
1500
2000
Properties of Reservoir Liquids
Correlation
Ref
Bubble - point pressure (psia)
Temperature, °F
Bo
Gas - oil ratio (scf/stb)
Oil Gravity, oAPI
Gas Gravity
Separator Pressure
Searator Temperature °F
Standing
3
130-7000
100-258
1.024-2.15
20-1425
16.5-63.8
0.59-0.95
265-465
100
Lasater
4
45-5780
82-272
Vasquez-Beggs
5
15-6055
162-180
1.028-2.226
3-2905
0-2199
17.9-51.1 15.3-59.5
0.574-1.22 0.511-1.651
15-605
60-565
36-106
76-150
Glaso
6
165-7142
80-280
1.025-2.588
90-2637
22.3-48.1
0.65-1.276
415
125
Marhoun
7
130-3573
74-240
1.032-1.997
26-1602
19.4-44.6
0.752-1.367
Table 1 Black oil correlation and their ranges at application 2 8 FLUID DENSITY
Liquids have a much greater density and viscosity than gases, and the density is affected
much less by changes in temperature and pressure. For petroleum engineers it is
important that they are able to estimate the density of a reservoir liquid at reservoir
conditions.
8.1 Specific Gravity of a Liquid
γo =
ρo
ρw
(4)
The specific gravity of a liquid is the ratio of its density to that of water both at the
same T & P. It is sometimes given as 60˚/60˚, i.e. both liquid and water are measured
at 60˚ and 1 atmos.
The petroleum industry uses another term called ˚API gravity where
° API =
141.5
− 131.5
γo
(5)
where γo is specific gravity at 60˚/60˚.
There are several methods of estimating the density of a petroleum liquid at reservoir
conditions. The methods used depend on the availability and nature of the data of
data. When there is compositional information on the reservoir fluid then the density
can be determined using the ideal solution principle. When the information we have
is that of the produced oil and gas then empirical methods can be used to calculate
the density of the reservoir fluid.
8.2 Density based on Ideal Solution Principles
Mixtures of liquid hydrocarbons at atmospheric conditions behave as ideal solutions.
An ideal solution is a hypothetical liquid where no change in the character of the
liquids is caused by mixing and the properties of the mixture are strictly additive.
Institute of Petroleum Engineering, Heriot-Watt University
17
Petroleum liquid mixtures are such that ideal-solution principles can be applied for the
calculation of densities and this enables the volume of a mixture from the composition and the density of the individual components. The principle is illustrated using
the following exercise. Data for the specific components are given in the tables at
the end of the chapter
Exercise 1.
Calculate the density at 14.7psia and 60 ºF of the hydrocarbon liquid mixture with
the composition given below:
Component
nC4
nC5
nC6
Mol.
fract.
1b mol.
0.25
0.32
0.43
1.00
Solution Exercise 1
Solution
Component
nC4
nC5
nC6
Mol.
Mol.
Weight
Liquid
density
Liquid
fract.
1b mol.
weight
1b/1b
mol.
1b
at 60˚F
Density at volume
and 14.7 cu ft
psia
1b/cu ft
0.25
0.32
0.43
____
1
58.1
72.2
86.2
14.525
23.104
37.066
_____
74.695
36.45
39.36
41.43
0.3985
0.5870
0.8947
_____
1.8801
Liquids at their bubble point or saturation pressure contain large quantities of dissolved gas which at surface conditions are gases and therefore some consideration
for these must be given in the additive volume technique. This physical limitation
does not impair the mathematical use of a “pseudo liquid density “ for methane and
ethane since it is only a step in its application to determine a reservoir condition
density. This is achieved by obtaining apparent liquid densities for these gases and
determining a pseudoliquid density for the mixture at standard conditions which can
then be adjusted to reservoir conditions.
Standing & Katz 8 carried out experiments on mixtures containing methane plus other
compounds and ethane plus other compounds and from this were able to determine
a pseudo-liquid (fictitious) density for methane and ethane
18
Properties of Reservoir Liquids
Apparrent density of Methane, g/cc
Apparrent density of of Ethane, g/cc
Correlations have been obtained by experiment giving apparent liquid densities of
methane and ethane versus the pseudoliquid density (Figure 12).
0.6
0.5
0.4
0.3
0.4
0.3
0.2
Ethane - N - Butane
Ethane - Heptane
Ethane - Crystal oil
Methane - Cyclo Hexane
Methane - Benzene
Methane - Pentane
Methane - Hexane
Methane - Heptane
Methane - Propane
Methane - Crystal oil
Methane - Crude oil
0.1
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Density of system, 60ºF B atm. pressure
Figure 12 Variation of apparent density of methane and ethane with density of the system 8.
To use the correlations a trial and error technique is required whereby the density of
the system is assumed and the apparent liquid densities can be determined. These
liquid densities are then used to compute the density of the mixture by additive volumes and the value checked against the initial assumption. The procedure continues
until the two values are the same.
When non hydrocarbons are present, the procedure is to add the mole fractions of
the nitrogen to methane, the mole fraction of carbon dioxide to ethane and the mole
fraction of hydrogen sulphide to propane.
Institute of Petroleum Engineering, Heriot-Watt University
19
Exercise 2:
Calculate the “surface pseudo liquid density” of the following reservoir
composition.
Component
Mole percent
Methane
Ethane
Propane
Butane
Pentane
Hexane
Heptane +
44.04
4.32
4.05
2.84
1.74
2.9
40.11
Properties of
heptane +
API gravities = 34.2
SG = 0.854
Mol wt = 164
Solution Exercise 2
Estimate ρο
From fig 12
Component
44.65 lb/cu ft.
Mole
fraction
z
Methane
0.4404
Ethane
0.0432
Propane
0.0405
Butane
0.0284
Pentane(n&i) 0.0174
Hexane(n&i) 0.029
Heptane+
0.4011
Total
1
Density =
=
0.716 gm/cc
Density 0.326
C1
Density 0.47
C2
Mol
Weight
Weight
lb/lb
lb
mole
M
16
30.1
44.1
58.1
72.2
86.2
164
zM
7.0464
1.30032
1.78605
1.65004
1.25628
2.4998
65.7804
81.31929
81.32 lb /
44.65 lb/cu.ft
lb/cuft
20.3424
29.328
Liq
Density
at 60°F &
14.7 psia
lb/cu.ft
ρo
20.3424
29.328
31.66
35.78
38.51
41.43
53.26
Liquid
Volume
cu ft.
zM/ρo
0.34639
0.04434
0.05641
0.04612
0.03262
0.06034
1.23508
1.8213
1.82 cu ft
This trial and error method is very tedious so Standing and Katz devised a chart which
removes the trail and error required in the calculation. The densities have been converted into the density of the heavier components, C3+, and the weight percent of the
two light components, methane and ethane in the C1+ and C2+ mixtures. Figure 13.
20
Properties of Reservoir Liquids
30
20
0
10
ne p
lus m
a
70
%e
than
e in
etha
50
0
60
30
t%
W
ne
ha
et
m
in
e
tir
en
st
sy
em
10
50
20
40
30
30
20
10
Density of system including methane and ethane, lb/cu ft
60
40
40
terial
50
Wt
Density of propane plus, lb/cu ft
70
Figure 13 Pseudo-liquid density of systems containing methane and ethane 10.
We shall examine through examples various ways of calculating downhole reservoir
fluids densities dependant on the data available.
The three considered are:
1. The composition of the reservoir fluid is known.
2. The gas solubility , the gas composition and the surface oil gravity is known
3. The gas solubility, and gas and liquid gravities are known.
1. The composition of the reservoir fluid is known.
The procedure is illustrated using the following two exercises .
Institute of Petroleum Engineering, Heriot-Watt University
21
Exercise 3.
Calculate the surface density of the mixture in exercise 2 using the chart of figure 13
The pseudodensity is converted to reservoir conditions firstly by taking the effect
of pressure and secondly accounting for the effect of temperature. The variation of
density with respect to pressure and temperature has been investigated and it has
been demonstrated that thermal expansion is not affected by pressure. Standing &
Katz took National Petroleum Standards data and with supplementary data produced
correction factors for pressure and temperature to convert atmospheric density to
reservoir density.
The compressibility and thermal expansion effects have been expressed graphically
in Figures 14 and 15.
10
8
,0
15
00
7
6
4
1
0
25
su
re
,
ps
ia
00
2
Pr
es
0
3,
3
00
00 00
6,0 5,0 4,0
5
0
,00 8,000
10
Density of pressure minus density at 60ºF β 14.7 psia lb/cu ft
9
2,
00
0
1,0
00
30
35
40
45
50
55
60
Density at 60ºF and 14.7 psia, lb/cu ft
Figure 14 Density correction for compressibility of liquids 8.
22
65
Properties of Reservoir Liquids
10
8
7
6
24
0
5
0
22
0
4
20
Density at 60ºF minus density at temperature, lb/cu ft
9
3
16
0
18
0
Te
mp
e ra
ture
ºF
140
120
2
100
1
80
0
25
60
30
35
40
45
50
55
60
65
Density at 60ºF and pressure P, lb/cu ft
Figure 15 Density correction for thermal expansion of liquids 10.
Exercise 4.
Calculate the density of the reservoir liquid of exercise 3 at a reservoir temperature
of 5,500 psia and 180 oF
Full compositional data may not always be available and the characterisation of the
produced fluids will vary from full compositional analysis to a description of the
fluids in terms of gas and oil gravity. The procedure just described is for the situation where the composition of the reservoir fluid is known. The procedures which
follow cover the situation where a less comprehensive analysis is available. These
methods make use of empirical correlations.
Institute of Petroleum Engineering, Heriot-Watt University
23
2. Reservoir Density when the Gas Solubility , the gas composition and the surface
oil gravity are known
By considering surface liquid as a single component and knowing the composition
of the collected gas the techniques previously discussed can be used to determine
reservoir liquid density. Again we will illustrate the procedure with an example
Exercise 5.
A reservoir at a pressure of 4,000 psia and a temperature of 200oF has a producing
gas to oil ratio of 600 scf/STB. The oil produced has a gravity of 42 oAPI. Calculate
the density of the reservoir liquid. The produced gas has the following composition
Component
Methane
Ethane
Propane
Butane
Pentane
Hextane
Mole Fraction
0.71
0.13
0.08
0.05
0.02
0.01
3. The Gas Solubility, and Gas and Liquid gravities are known.
Katz has produced a correlation (figure 16) to enable densities to be determined when
the only information on the gas is its solubility and its gravity. The figure gives apparent liquid densities of gases against gravity for different API crudes
Apparent Liquid density of Dissolved Gas at
60 F and 14.7 psia, lb/cu. ft.
45
40
35
20 API Crude
30
30
40
50
60
25
20
15
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
Gas Gravity
Figure 16 Apparent liquid densities of natural gases
24
Properties of Reservoir Liquids
Exercise 6.
Use the correlation of Katz to calculate the reservoir fluid density of a field with
a GOR of 500scf/STB with a gas gravity of 0.8 and a 35oAPI oil for reservoir
conditions of 4,000psia and a temperature of 180oF.
Katz method
9 FORMATION VOLUME FACTOR OF GAS CONDENSATE
The situation for a wet gas or gas condensate is different for a conventional oil when
one is considering the volume changes taking place upon release to surface conditions. For a wet gas or condensate system liquid at surface is gas in the formation.
The comparison therefore with respect to conditions in the reservoir to those at the
surface is distinctly different from an oil system, where an oil in the reservoir produces
gas and liquids at the surface. For a wet gas or condensate, a gas in the reservoir
produces gas and liquids at the surface.
The formation-volume factor therefore for a condensate, Bgc is defined as the
volume of gas in the reservoir required to produce 1.0 STB of condensate at the
surface. The units are generally barrels of gas at res. conditions per barrel of stock
tank oil. There are a number of methods of estimating Bgc. To calculate the properties of the reservoir fluid from the information on the produced
fluids requires a combination of the quantities and characteristics of these fluids. The
methods used depends on the level of detail of the characteristics of the produced
fluids. A number of methods are presented using examples which vary according
to the level of detail.
Exercise 7.
A gas condensate produces gas and liquids with the compositions detailed below,
with a producing GOR of 30,000 SCF/STB. Determine the composition of the
reservoir gas.
Component
Methane
Ethane
Propane
Butane
Pentane
Hexane
Heptane +
Composition
Gas
0.84
0.08
0.04
0.03
0.01
1.00
Institute of Petroleum Engineering, Heriot-Watt University
Liquid
0.15
0.36
0.28
0.12
0.09
1.00
25
Exercise 8.
The gas condensate reservoir above is contained in reservoir sands with an
average pay thickness of 100ft, with a porosity of 0.18 and a connate water
saturation of 0.16. The aerial extent of the field is 5 sq. miles. The initial reservoir
pressure is 5,000 psia and the reservoir temperature is 180 oF. Determine the initial
reserves of the field in terms of condensate and gas.
Exercise 9.
Calculate the gas condensate formation factor for the example in exercise 8.
10 VISCOSITY OF OIL
The viscosity of oil at reservoir temperature and pressure is less than the viscosity
of the dead oil because of the dissolved gases and the higher temperature. Correlations are available which enable the dissolved gas and pressure effect on the dead
oil viscosity to be determined. Danesh 2 has given a good review of many of the
empirical approaches. The favoured correlations are those of Beggs and Robinson
11
Egbogah and Ng 12 ,Vazquez and Beggs13 , and Labedi14 . Figure 17 gives plots,
,
presented by McCain 17 , of the correlation of dead oil viscosity from Egbogah and
Ng 12 , and figure 18s the impact of dissolved gas from the Beggs and Robinson 11
correlation.
1000
800
700
600
500
400
300
200
Viscosity of Gas-Free Oil, µoD, cp
100
80
70
60
50
40
30
20
100º
150º
200º
250º
300º
R
es
er
10
8
7
6
5
4
3
vo
irT
em
pe
ra
tur
e,
ºF
2
1
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
10
20
30
40
Stock - Tank Oil Gravity, ºAPI
26
Figure 17 Dead oil viscosities 17.
50
Properties of Reservoir Liquids
0
200
0
10
0
20
at
io
20
0
50
G
as
-O
il R
10
8
7
6
5
4
3
io
n
00
10
0
150
0
0
20
So
lu
t
Viscosity of Gas-Saturated Oil, µoD, cp
100
80
70
60
50
40
30
2
1
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.4 0.6 0.81
2
3 4 5 6 7 8 10
20 30 40 60 80100
200 300
Viscosity of Gas-Free Oil, µoD, cp
Figure 18 Viscosities of saturated black oils 11.
Beggs and Robinson 11 examined 600 oil samples over a wide range of pressure and
temperature and came up with the following correlation.
µod = 10A - 1
(6)
where, log A = 3.0324 - 0.0202oAPI -1.163 log T
µod is the dead oil viscosity in cp and T is in oF.
Egbogah and Ng 12, had a different expression for A
log A = 1.8653 - 0.025086oAPI -0.56441 log T
Examination of these correlations has shown that they are not very reliable with
errors of the order of 25% (DeGetto15)
Beggs and Robinson 11 gave a correlation to give the impact of dissolved gas.
µob = CµodB
(7)
where
C
= 10.715 (Rs + 100)-0.515
and
B
= 5.44 (Rs + 150)-0.338
µob is the saturated oil viscosity
Vazquez and Beggs13 presented an equation to take into account pressure on viscosity
above the saturation pressure.
Institute of Petroleum Engineering, Heriot-Watt University
27
µo = µob (P/Pb)D
where
D
(8)
= 2.6 P1.187 e-11.513-8.98 x10-5P
This is presented in figure 19 from McCain 17.
10,000
9,000
8,000
7,000
6,000
4,000
3,000
Pre
2.000
1.000
900
800
700
600
500
60
00
psi
a
00
200
100
60
40
100
90
80
70
60
50
40
20
30
20
10
6
4
f Oil At
yo
Viscosit
Bubble
p
Point, c
2
10
9
8
7
6
5
4
3
1
2
0.6
0.4
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.2
0.1
0.3
Viscosity of Oil Above Bubble Point, µo, cp
200
30
00
re
00
300
40
ssu
00
20
400
50
0
100
500
Bubble Point pressure, Pb, psia
5,000
0.2
0.1
Figure 19 Viscosities of undersaturated black oils 17.
Labedi (ref 14) also produced an empirical correlation to determine viscosity at pressures above the bubble point
µo= µob + (P/Pb-1)(10-2.488µob0.9036 Pb0.6151 /100.0197oAPI )
(9)
Danesh 2 in his text compared the various correlations from a published experimental
viscosity value in a well known PVT report, using the following exercise.
28
Properties of Reservoir Liquids
Exercise. 10
Calculate the viscosity of oil in the PVT report of chapter 12 at a pressure of
5,000psig and 220°F. The °API of the oil is 40.1 and the GOR, Rs is 795 scf/ST
Beggs and Robinson
µod = 10A -1
Log A = 3.0324 - 0.0202°API - 1.163 log Tx
µod = dead oil viscosity cp.
(Beggs
3.0324 0.0202
1.163)
(Egbogah 1.8653 0.025086 0.56441)
Beggs
Egbolgah
API =40.1
T
=220
Rs =795
P
=5,000 psig
Pb =2,635 psig
log A =
-0.5031
-0.46
A
=
0.3140
0.34
Viscosity
dead oil = 1.06 cp
1.21 cp
Measured value = 1.29 cp
Viscosity at bubble point
Beggs
µob = CµobB
µob = oil viscosity at bubble point pressure
C = 10.715 (Rs + 100) -0.515
B = 5.44 (Rs + 150) -0.338
C = 0.3234
B = 0.5369
µob = 0.3584 cp
Measured value = 0.355 cp
Viscosity at pressure of 5015 psig
Vazquez - Beggs
µo = µob (P/Pb)D
-5
D = 2.6p 1.187 e -11.513 - 8.98x 10 p
e function = -11.9633
D = 0.4663 cp
Labed, correlation
o
µo= µob + (P/Pb-1)(10 -2.488µob0.9036 Pb0.6151 /10 0.0197 API )
µo = 0.4304 cp
Measured value = 0.45 cp
Institute of Petroleum Engineering, Heriot-Watt University
29
11 INTERFACIAL TENSION
In recent years interfacial tension has become to be realised as an important physical
property in the context of the recovery of reservoir held hydrocarbons, in particular
for gas condensates. Interfacial tension, arises from the imbalance of molecular
forces at the interface between two phases. For many years it has been neglected but
more recently it has been realised that in gas injection and condensation processes
the magnitude of the various forces; surface, gravitational and viscous forces can
have a significant impact on the mobility of the various phases. A major advance in
knowledge has been that in the context of gas condensates where it was considered
that in the tradition of relative permeability knowledge liquid formation by retrograde
condensation would be immobile. Recent research has shown that such fluids are
mobile because of the associated low interfacial tension 16. Danesh 2 in his text covers
the topic of interfacial tension extensively. Mentioned briefly below are some of the
techniques which are currently used in predicting IT for reservoir fluids.
Interfacial tension decreases as temperature and pressure increases as shown for the
effect of temperature for pure components in figure 20 from McCain’s text 17 adapted
from Katz19 data.
35
Surface Tension, dynes per cm
30
25
20
Mol wt.
240
220
200
180
160
15
140
10
5
0
-200
-200
0
100
200
300
400
500
600
Temperature, ºF
Methane
Ethane
Propane
n - Pentane
l - Butane
n - Heptane
n - Hexane
n - Octane
n - Butane
Figure 20 Interfacial tensions of hydrocarbons. (Adapted from Katz, et al., J. Pet. Tech.,
Sept. 1943.)
30
Properties of Reservoir Liquids
There are several methods for predicting IFT, and they require experimentally determined
parameters. Work on pure compounds have shown that IFT can be related to density,
compressibility and latent heat of vaporisation. The multicomponent perspective of reservoir fluid properties has made use of the IFT/density relationships.
The Parachor method of McLeod 18 has gained acceptance where the IFT between
vapour and liquid is related to the density difference of the respective phases.
 ρ − ρg 
σ =  Pσ L


M 
4
(10)
where ρL and ρg are the density of the liquid and gas phases, and M is the molecular
weight. σ is the IFT . Pσ is called the parachor.
Katz19 has provided the parachors for pure components as shown in the table below and
they are also presented in the figure 21 prepared by MaCain using Katz’s19 data.
Parachors, Ps, for IFT
Component
Methane
Ethane
Propane
i-Butane
n-Butane
i-Pentane
n-Pentane
n-Hexane
n-Heptane
n-Octane
Hydrogen
Nitrogen
Carbon dioxide
Parachor
77
108
150.3
181.5
189.9
225
231.5
271
312.5
351.5
34
41
78
Parachors have been shown to have a linear relationship with molecular weight according to the relationship;
Pσ = 21.99 + 2.892M
(11)
and also to the critical properties
Institute of Petroleum Engineering, Heriot-Watt University
31
600
500
Parachor, P
400
300
i - C5
200
i - C4
100
0
50
100
150
200
Molecular Weight
Figure 21 Parachors for computing interfacial tension of normal paraffin hydrocarbons 19.
Pσ = 0.324Tc1/4vc7/8
where Tc is in K and the critical volume vc is in cm3/gmol.
To apply the parachor approach to mixtures the molar averaging approach of Weinaug
and Katz20 can be used.

 ρ
ρ 
σ = ∑ Pσ  x j L − yj g  
Mg  
 ML
 j
4
(12)
xj and yj are the mole fractions of the components in the liquid and gas phases.
Firoozabadi 21 has provided parachors to enable calculations to be made for heavy
components using the following equation.
Ps= -11.4 + 3.23M-0.0022M2
where M is the molecular weight of the heavy component. Figure 22.
32
(13)
Properties of Reservoir Liquids
1400
1200
Parachor. P
1000
800
600
400
200
0
100
200
300
Molecular weight
400
500
Figure 22 Parachors of heavy fractions for computing interfacial tension of reservoir
liquids. McCain17
This method is illustrated using an example from McCain 17.
Exercise 11.
Calculate the IFT of the following volatile oil mixture at 2315 psia and 190°F for
the oil with the following composition.
12 COMPARISON OF RESERVOIR FLUID MODELS
It is useful to summarise the differences between the Black Oil Model approach
compared to the Compositional Model in predicted fluid properties.
The suitability of the two approaches is largely related to the nature of the fluid. For
heavier oils where there are low GOR’s as compared to volatile oils with high GOR’s,
black oil models are likely to be suitable. For the more volatile systems where there
are more significant compositional variations in a reservoir as pressure is depleted,
compositional models are considered more capable of predicting fluid behaviour.
The computational requirements of compositional models used to be a restriction when
carrying out large reservoir simulation. The continued development of computing
and associated equations of state modelling reduces these former restrictions.
Institute of Petroleum Engineering, Heriot-Watt University
33
Companies are now focusing their attention on being capable of modelling the total
process from fluid extraction from the reservoir, through well production and facility treatment. At present separate modelling occurs, and many of these models are
not compatible. The black oil approach is certainly considered by many to be from
a former era.
The list below gives a summary comparison of the two approaches.
Black Oil Models
• 2 components - solution gas and stock tank oil
• Bo, Rg, etc.
• Empirical correlations
• After the event description of fluid properties
Compositional Models
• N components based on paraffin series
• Equation of state based calculations
• Feed forward calculation of fluid properties
In a subsequent chapter on vapour liquid equilibria we will describe how the volumes
and compositions of vapour and liquid equilibrium mixtures can be calculated when
a mixture is processed at a particular pressure and temperature. These calculations
although calculation intensive can be considered feed forward calculations and enable the effects of temperature and pressure changes to be determined on a particular
feed mixture.
The black oil approach which has been a major theme of this chapter uses the characteristics of the produced fluids to determine the composition and properties of the
feed reservoir mixture, i.e. a back calculation. As will be seen in the section on PVT
reports, the quantities and characteristics of the produced fluids are dependant on the
pressure and temperature conditions used to separate the fluid.
At the back of this chapter are tables of physical properties which are useful in many
of the procedures described.
34
Properties of Reservoir Liquids
Institute of Petroleum Engineering, Heriot-Watt University
35
36
Properties of Reservoir Liquids
Institute of Petroleum Engineering, Heriot-Watt University
37
38
Properties of Reservoir Liquids
Solutions to Exercises
Exercise 1.
Calculate the density at 14.7psia and 60 ºF of the hydrocarbon liquid mixture with
the composition given below:
Component
Mol.
fract.
1b mol.
nC4
nC5
nC6
0.25
0.32
0.43
1.00
Mol.
Mol.
Weight
Liquid
density
fract.
1b mol.
weight
1b/1b
mol.
1b
at 60˚F
Density at volume
and 14.7 cu ft
psia
1b/cu ft
0.25
0.32
0.43
____
1
58.1
72.2
86.2
14.525
23.104
37.066
_____
74.695
36.45
39.36
41.43
Solution Exercise 1
Solution
Component
nC4
nC5
nC6
Liquid
0.3985
0.5870
0.8947
_____
1.8801
Exercise 2:
Calculate the “surface pseudo liquid density” of the following reservoir
composition.
Component
Mole percent
Methane
Ethane
Propane
Butane
Pentane
Hexane
Heptane +
44.04
4.32
4.05
2.84
1.74
2.9
40.11
Institute of Petroleum Engineering, Heriot-Watt University
Properties of
heptane +
API gravities = 34.2
SG = 0.854
Mol wt = 164
39
Solution Exercise 2
Estimate ρο
From fig 12
Component
44.65 lb/cu ft.
Mole
fraction
Mol
Weight
Weight
lb/lb
lb
mole
z
Methane
0.4404
Ethane
0.0432
Propane
0.0405
Butane
0.0284
Pentane(n&i) 0.0174
Hexane(n&i) 0.029
Heptane+
0.4011
Total
1
M
16
30.1
44.1
58.1
72.2
86.2
164
Density =
=
0.716 gm/cc
Density 0.326
C1
Density 0.47
C2
zM
7.0464
1.30032
1.78605
1.65004
1.25628
2.4998
65.7804
81.31929
lb/cuft
20.3424
29.328
Liq
Density
at 60°F &
14.7 psia
lb/cu.ft
ρo
20.3424
29.328
31.66
35.78
38.51
41.43
53.26
81.32 lb /
44.65 lb/cu.ft
Liquid
Volume
cu ft.
zM/ρo
0.34639
0.04434
0.05641
0.04612
0.03262
0.06034
1.23508
1.8213
1.82 cu ft
Exercise 3.
Calculate the surface density of the mixture in exercise 2 using the chart of figure
13
Solution Exercise 3
Component
Mole
fraction
z
Methane
0.4404
Ethane
0.0432
Propane
0.0405
Butane
0.0284
Pentane(n&i) 0.0174
Hexane(n&i) 0.029
Heptane+
0.4011
40
Mol
Weight
Weight
lb/lb
lb
mole
M
16
30.1
44.1
58.1
72.2
86.2
164
zM
7.0464
1.30032
1.78605
1.65004
1.25628
2.4998
65.7804
1
Weight of propane +
lbs.
=
Density of propane +
=
Weight per cent ethane in
ethane +
Weight per cent methane in
methane +
From figure 13 pseudo liquid
density =
Liq
Density
at 60°F &
14.7 psia
lb/cu.ft
ρo
Liquid
Volume
31.66
35.78
38.51
41.43
53.26
0.05641
0.04612
0.03262
0.06034
1.23508
cu ft.
zM/ρo
72.97
Volume =
1.43
51.01 lb cu ft
1.75
8.67
45 lb/cu ft
Properties of Reservoir Liquids
Exercise 4.
Calculate the density of the reservoir liquid of exercise 3 at a reservoir temperature
of 5,500 psia and 180 oF
Solution Exercise 4
Density of following reservoir liquid at 6,000 psia and 180˚F.
Step 1
Pseudoliquid density at standard conditions
from exercise 3 ρo = 45 lb/cu ft
Step 2
Adjust to 60˚F and 5,500 psia
i.e. correction = +1.9 lb/cu ft
(Figure 14)
i.e. ρo = 45 + 1.9 = 46.9 lb/cu ft at 60˚F 6,000 psi
Step 3
Adjust to 180˚F.
(Figure 15)
i.e. thermal correction = -3.18 lb/cu ft
ρo = 46.9 - 3.18 = 42.32 lb/cu ft at 180˚ and 6,000 psia
ρo = 42.32 lb/cu ft @ 180˚F and 6,000 psia
Exercise 5. A reservoir at a pressure of 4,000 psia and a temperature of 200oF has a producing gas
to oil ratio of 600 scf/STB. The oil produced has a gravity of 42 oAPI. Calculate the
density of the reservoir liquid. The produced gas has the following composition
Component
Methane
Ethane
Propane
Butane
Pentane
Hextane
Mole Fraction
0.71
0.13
0.08
0.05
0.02
0.01
Institute of Petroleum Engineering, Heriot-Watt University
41
Calculation of pseudo density of gas.
From PV=znRT, Solubility of gas, Rs = 600 scf/STB
1 lb mole =
379 scf
Oil =
42 API
Density of crude = 50.87 lb/cuft
285.62 lb/STB
Density of water = 62.37 lb./cuft
Component
Methane
Ethane
Propane
Butane
Pentane(n&i)
Hexane(n&i)
Oil 42 API
Totals
Mole
fraction
volume
fraction
Solubility Mol
Weight
Weight
scf
lb/lb mole lb/STB
gas/STB
z
0.71
0.13
0.08
0.05
0.02
0.01
zRs
426
78
48
30
12
6
M
16
30.1
44.1
58.1
72.2
86.2
600
Liq Density Liquid Volume
zRsM/379
17.98
6.19
5.59
4.60
2.29
1.36
285.62
323.63 lb
at 60°F
& 14.7 psia
lb/cu.ft
cu ft/STB.
ρo
zm/ρo
31.66
35.78
38.51
41.43
0.176
0.129
0.059
0.033
5.615
6.01 cu ft
Density of propane + =
323/6.01/lb cuft = 49.81 lb/ cu ft
Weight % C2+ =
2.315
Weight% C1+ =
5.557
From Figure 13 Pseudoliquid density of reservoir fluid at 60°F & 14.7 psia = 46.5 lb / cu ft
Correction for pressure Fig 14
Correction for temperature Fig 15
=
1.23 +
3.55 -
= 47.73
= 44.18
Density of Reservoir Fluid = 44.18 lb/cu ft
Exercise 6. Use the correlation of Katz to calculate the reservoir fluid density of a field with a
GOR of 500scf/STB with a gas gravity of 0.8 and a 35oAPI oil for reservoir conditions of 4,000psia and a temperature of 180oF.
Katz method
Solution Exercise 6. Mass o f gas per STB.
Molecular weight of gas = molecular weight air x 0.8 = 29.2 x 0.8 = 23.2
Mas og gas / STB = 500
42
scf
lb. mole
23.2 lb
x
x
= 30.61 lb / STB
stb
379 scf
lb mole
Properties of Reservoir Liquids
Component
Weight
lb/STB
Gas
Oil
30.61
297.62
328.23
Liq Density
at 60ºF
& 14.7 psia
lb/cu.ft
26.3
from chart
Liquid Volume
cu ft/STB.
1.164
5.615
6.779
Pseudodensity
of reservoir fluid= 328.23 / 6.779 =
48.42
Correction for pressure at Fig 14
+1.13 =
49.55
Correction for pressure at Fig 15
-2.9
46.65
Reservoir density=
=
46.65 lb/cu ft
Exercise 7. A gas condensate produces gas and liquids with the compositions detailed below,
with a producing GOR of 30,000 SCF/STB. Determine the composition of the
reservoir gas.
Component
Methane
Ethane
Propane
Butane
Pentane
Hexane
Heptane +
Composition
Gas
0.84
0.08
0.04
0.03
0.01
1.00
Institute of Petroleum Engineering, Heriot-Watt University
Liquid
0.15
0.36
0.28
0.12
0.09
1.00
43
Solution Exercise 7
Liquid
Component
Mol. Fractn
lb mole
C3
0.15
C4
0.36
C5
0.28
C6
0.12
C7+*
0.09
* C8 used for C7+
Mol.Wgt.
lb/lb mol
Wgt.
lb/lb mole
44.1
58.1
72.2
86.2
114.2
6.615
20.916
20.216
10.344
10.278
68.369
68.369
Mol.Wgt.
liq.
Density of liquid=
GOR=
30000 scf/STB
=
79.16 lb mole gas/STB
Note: 1 lb mole = 379 SCF
GOR =
25.36 lb mole gas/lb mole liquid
Liquid
density
lb/cu ft
31.66
35.78
38.51
41.3
43.68
Liquid
volume
cu ft
0.223
0.585
0.506
0.25
0.235
1.799
38.00 lb/cu ft
213.39 lb/STB
3.12 lb mole /STB
2. Recombination according to the above GOR of 25.36 lb mole gas / lb moleliquid
Component
Methane
Ethane
Propane
Butane
Pentane
Hexane
Heptane +
Composition
Gas
lb mole
y
0.84
0.08
0.04
0.03
0.01
1
Liquid
lb mole
x
0.15
0.36
0.28
0.12
0.09
1
lb mole gas/
lb mole oil
25.36
25.36y
21.30
2.03
1.01
0.76
0.25
25.36
lb moles
Res fluid
Composition
Res Fluid
25.36y + x
21.30
2.03
1.16
1.12
0.53
0.12
0.09
26.36
0.808
0.077
0.044
0.043
0.020
0.005
0.003
1.000
Exercise 8.
The gas condensate reservoir above is contained in reservoir sands with an average
pay thickness of 100ft, with a porosity of 0.18 and a connate water saturation of 0.16.
The aerial extent of the field is 5 sq. miles. The initial reservoir pressure is 5,000 psia
and the reservoir temperature is 180 oF. Determine the initial reserves of the field in
terms of condensate and gas.
44
Properties of Reservoir Liquids
Solution Exercise 8
Component
Mol. Fract.
Critical Temperature
Critical Pressure
C1
C2
lb mole yj
0.808
0.077
R
Tcj
344
551
R
yjTcj
278.00
42.41
psia
Pcj
667
708
psia
yjPcj
539.026
54.491
C3
C4
0.044
0.043
666
750
29.42
31.89
616
540
27.210
22.960
C5
C6
C7+
Totals
0.020
0.005
0.003
1
838
914
1025
16.96
4.16
3.50
406.34
489
437
360
9.899
1.989
1.229
656.80
Tpc=
406.34
Ppc =
656.80
Reservoir pressure
=
5000 psia
Reservoir temperature
=
180 F
Pseudo reduced pressure
=
640 R
= 7.61
Pseudo reduced temperature = 1.58
Compressibility factor from Standing & Katz chart figure 2 Gas properties chapter
z=
R=
0.98
10.73 cu ft. psi/lb.mol R
Volume of the reservoir = 5 square miles x 100 feet
Volume of the reservoir = 2.1076 x 109 cu ft
(1 mole = 5280 ft)
PV=znRT
V/n=zRT/P
Specific volume at reservoir conditions = 1.3460 cu ft/lb.mol
No of lb moles in reservoir= 1.5658 x 109 lb moles
No. of standard cubic feet of gas in reservoir = 5.9345 x 1011 SCF (1 lb mole 379 scf)
Reserves in reservoir in terms of produced fluids
From previous exercise GOR of = 30,000 SCF/STB
= 25.36 lb mole gas/lb mole condensate
For each 26.36 lb mole of reservoir fluid 25.36 lb mol is produced gas
and 1 lb mole is condensate
Reserves in terms of produced fluids
Gas 1.506428 x 109 lb moles = 5.70936 x 1011 SCF
Condensate 1.9643E+09 lb moles = 6.2935E+08 STB
Exercise 9.
Calculate the gas condensate formation factor for the example in exercise 8.
Solution Exercise 9.
Bgc = bbls of gas in reservoir/STB condensate
Volume of gas in reservoir = 6.9696 x 1010 cu ft = 1.2412 x 1010 bbls
Condensate = 6.2935 x 106 STB
Bgc = 1972.2
bbls res gas/STB condensate
In some cases full compositional information may not be available but only black
oil descriptions of the oil and gas gravity for the gas. In this case correlations can be
used to provide the necessary data to calculate the same data as for exercise 8 & 9.
Institute of Petroleum Engineering, Heriot-Watt University
45
Exercise 10
Calculate the viscosity of oil in the PVT report of chapter 12 at a pressure of 5,000psig
and 220°F. The °API of the oil is 40.1 and the GOR, Rs is 795 scf/ST
Beggs and Robinson
µod = 10A -1
Log A = 3.0324 - 0.0202°API - 1.163 log Tx
µod = dead oil viscosity cp.
(Beggs 3.0324 0.0202 1.163)
(Egbogah 1.8653 0.025086 0.56441)
Beggs Egbolgah
API = 40.1
T = 220
Rs = 795
P = 5,000 psig
Pb = 2,635 psig
log A = -0.5031 -0.46
A = 0.3140 0.34
Viscosity
dead oil =
1.06 cp 1.21 cp
Measured value = 1.29 cp
Viscosity at bubble point
Beggs
µob = CmobB
µob = oil viscosity at bubble point pressure
C = 10.715 (Rs + 100) -0.515
B = 5.44 (Rs + 150) -0.338
C = 0.3234
B = 0.5369
µob = 0.3584 cp
Measured value = 0.355 cp
Viscosity at pressure of 5015 psig
Vazquez - Beggs
µo = µob (P/Pb)D
-5
D = 2.6p 1.187 e -11.513 - 8.98x 10 p
e function = -11.9633
D = 0.4663 cp
Labed, correlation
o
µo= µob + (P/Pb-1)(10 -2.488µob0.9036 Pb0.6151 /10 0.0197 API )
µo = 0.4304 cp
Measured value = 0.45 cp
46
Properties of Reservoir Liquids
Exercise 11
Calculate the IFT of the following volatile oil mixture at 2315 psia and 190°F for
the oil with the following composition.
Solution Exercise 11
Component
Carbon dioxide
Nitrogen
Methane
Ethane
Propane
i - Butane
n - Butane
i - Pentane
n - Pentane
Hexane
Heptanes plus
Liquid Composition
Mole fraction
0.0159
0.0000
0.3428
0.0752
0.0564
0.0097
0.0249
0.0110
0.0140
0.0197
0.4303
Gas Composition
Mole fraction
0.0259
0.0022
0.8050
0.0910
0.0402
0.0059
0.0126
0.0039
0.0044
0.0040
0.0049
Properties of heptanes plus of liquid
Specific gravity = 0.868
Molecular weight = 217 lb/lb mole
Density of liquids and gas from previous methods
PL = 0.719 g/cc
Pg = 0.137 g/cc
Molecular weight
Component
Co2
N2
C1
C2
C3
i-C4
n-C4
i-C5
i-C5
C6
C7+*
ML = 110.1 g/s mole
Mg = 21.1 g/g mole
xj
0.0159
0.0000
0.3428
0.0752
0.0564
0.0097
0.0249
0.0110
0.0141
0.0197
0.4303
1.000
yi
0.0259
0.0022
0.8050
0.0910
0.0402
0.0059
0.0126
0.0039
0.0044
0.0040
0.0049
1.000
Pσ
78.0
41.0
77.0
108.0
150.3
181.5
189.9
225.0
231.5
271.0
*586.6
Equation 12
-0.0050
-0.0006
-0.2301
-0.0108
0.0161
0.0046
0.0154
0.0105
0.0147
0.0278
1.6297
1.4723
from figure 23
Institute of Petroleum Engineering, Heriot-Watt University
47
REFERENCES
1. Craft,BC & Hawkins, MF. Applied Reservoir Engineering” 1959 Prentice Hall,
NY
2. Danesh, A, PVT and Phase Behaviour of Petroleum Reservoir Fluids. 1998
Elsevier. pp 66-77
3. Standing MB “A pressure-Volume-Temperature Correlation for Mixtures of
Californian Oils and Gases”, Drill & Prod, Proc.275-287 (1947)
4. Lasater, J.A. “ Bubble Point Correlation “, Trans AIME, 213,379-381 (1958).
5. Vasquez,M and Beggs,HD “Correlations for Fluid Physical Property Prediction
“ JPT,968-970, (June 1980)
6. Glaso, O “Generalised Pressure Volume Temperature Correlations” JPT,785
795 (May 1980)
7. Marhoun,MA, “PVT Correlations for Middle East Crude Oils” JPT, 650-665
(May 1988)
8. Standing, M.B. and Katz,D.L. “ Density of Crude Oils Saturated with Natural
Gas” Trans AIME 146, 159 (1942)
9. Kessler, MG and Lee,BI,: “Improved Prediction of Enthalpy of Fractions,” Hyd
Proc.(Mar.1976) 55,153-158.
10. Standing,M “Volumetric and Phase Behaviour of Oil Field Hydrocarbon Systems”
SPE Dallas 1951
11. Beggs,HD. and Robinson,JR: Estimating the Viscosity of Crude Oil Systems”
JPT,27,1140-1141 (1975)
12. Egboghah,EO and Ng,JT: ‘An improved Temperature Viscosity Correlations
for Crude Oil Systems”, J.Pet Sci and Eng.,5,197-200 (1990)
13. Vasquez,M. and Beggs,HD :” Correlations for Fluid Physical Property Predictions”.
JPT,968 (June 1980)
14. Labedi,R: “Use of Production Data to Estimate Volume Factor, Density and
Compressibility of Reservoir Fluids”, J.of Pet.Sci and Eng. 4.375-90,(1990)
15. DeGhetto,G.,Paone,F. and Villa,M.: “Reliability Analysis of PVT Correlations
“,SPE 28904, Proc of Euro.Pet Conf. Lndn, 375-393 (Oct.,1994)
16. Danesh,A.,Krinis,D.,Henderson G.D., and Peden,J>M> “Visual Investigation
of Retrograde Phenomena and Gas Condensate Flow in Porous Media” 5th
European Symposium on Improved Oil Recovery ,Budapest (1988)
17. McCain,WD., “The Properties of Petroleum Fluids” Pennwell Books ,Tulsa,
Ok 1990. ISBN 0-87814-335-1
18. Macleod, DB., “On a Relation Between Surface Tension and Density,” Trans.,
Faraday Soc. (1923) 19,38-42.
19. Katz,DL.,”Handbook of Natural Gas Engineering”, McGraw Hill Book Co
Inc., New Yk,(1959)
20. Weinaug,KG and Katz,DL,: “Surface Tension of Methane-Propane Mixtures”.
I&EC,239-246 (1943)
21. Firoozabadi,A , Katz,D.L., Soroosh,H.M and Sajjadian,V.A.: “Surface Tension
of Reservoir Crude-Oil/Gas Systems Recognising the Asphalt in the Heavy
Fraction,” SPE Res Eng.(Feb) 1988,3,No 1, 265-272.
48
fundamental Properties of Reservoir Rocks
CONTENTS
INTRODUCTION
1. CHARACTERISTICS OF RESERVOIR ROCKS
2. PHYSICAL CHARACTERISTICS OF
RESERVOIR ROCKS
3. POROSITY
3.1 Range of Values
3.2 Factors Which Affect Porosity
3.2.1 Packing and Size of Grains
3.2.2 Particle Size Distribution
3.2.3 Particle Shape
3.2.4 Cement Material
3.3 Subsurface Measurement of Porosity
3.3.1 Density Log
3.3.2 Sonic Log
3.3.3 Neutron Log
3.4 Average Porosity
4. PERMEABILITY
4.1 Darcy's Law
4.2 Factors Affecting Permeability
4.3 Generalised Form of Darcy's Law
4.4 Dimensions of Permeability
4.5 Assumptions For Use of Darcy's Law
4.6 Applications of Darcy's Law
4.7 Field Units
4.8 Klinkenberg Effect
4.9 Reactive Fluids
4.10 Average Reservoir Permeability
5. STRESS EFFECTS ON CORE MEASUREMENTS
5.1 Stress Regimes
5.2 Compressibility of Poros Rock
5.3 Types of Compressiblilty
5.4 Measurements of Pore Volume Compressiblity
5.5 Effect of Stress on Permeability
6. POROSITY - PERMEABILITY RELATIONSHIPS
7. SURFACE KINETICS
7.1 Capillary Pressure Theory
7.2 Fluid Distribution in Reservoir Rocks
7.3 Impact of Layered Reservoirs
8. EFFECTIVE PERMEABILITY
8.1 Definition
8.2 Water Displacement of Oil
8.2.1 Water - Oil Relative Permeability
8.3 Gas Displacement of Oil and Gas - Oil Relative Permeability
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•
Define porosity and express it as an equation in terms of pore, bulk and grain
volume.
• Explain the difference between total and effective porosity.
• Define permeability and present an equation, Darcy’s Law, relating flow rate to
permeability in porous media.
• List the assumptions for the applicability of Darcy’s Law.
• Derive an equation based on Darcy’s Law relating flow of gas in a core plug
and the upstream and downstream pressures.
• Derive an equation relating flow rate to permeability for a radial incompressible
system.
• Comment on the difference between gas and liquid permeability (Klinkenberg
effect ).
• Sketch a figure relating liquid permeability to gas permeabilities plotted as a
function of reciprocal mean pressure.
• Briefly describe the impact of reservoir stresses on permeability and porosity
• Draw a sketch demonstrating the result of interfacial tension between oil, water
and a solid, and locate the contact angle and define its values for wetting and
non-wetting phases.
• Express the capillary pressure Pc as two equations, one in terms of interfacial
tension, contact angle and pore radius, and the other in terms of height and
density of fluids.
• Define the free water level.
•Draw the Pc or height vs. saturation capillary pressure curve and identify
significant features.
• Sketch and explain the impact of saturation, history, density difference and
interfacial tension in capillary pressure curves.
•Sketch the impact of capillary pressure effects on the saturation distribution of
stratified formations
• Define effective and relative permeability and plot typical shapes.
• Define imbibition and drainage in the context of capillary pressure and relative
permeability curves.
• Sketch the pore doublet model and use it to explain the retention of trapped oil
in large pores and briefly relate it to the principle behind some enhanced oil
recovery methods.
• Define mobility ratio.
•Sketch a shape for gas- oil relative permeability curves.
fundamental Properties of Reservoir Rocks
Introduction
The properties of reservoir rocks with respect to the fluids they contain and with respect
to the fluids which will be injected into them are important when characterising a
reservoir in terms of its reserves and the mobility of the fluids. This next section gives
a brief over view of these properties, and is followed by chapters on their measurement
and variation. In relation to the detailed description of rock characteristics the reader
is referred to the Geology module of this Petroleum Engineering course.
The reservoir engineer is concerned with the quantities of fluids contained within the
rocks, the transmissivity of fluids through the rock and other related properties.
1. Characteristics of Reservoir Rocks
The specifications of a reservoir rock are such that there must be a large enough
capacity to store economically viable amounts of hydrocarbon and the hydrocarbon
must flow at economical rates when penetrated by a well. The factors which may
affect the capacity and the flow properties are the porosity, permeability, capillary
pressure, compressibility and fluid saturation. In the case of a reservoir rock, these are
not standard characteristics determined before formation of the rock, but are closely
linked to the geological processes that brought the sediments together and deposited
them in the sequences and under the chemical and physical changes inherent in the
system.
In order to contain enough oil or gas to make production economically viable, a
reservoir rock must exceed: a minimum porosity, a minimum thickness, a minimum
permeability, and a minimum area.
In order to extract the fluids the rock must be permeable which requires that there be
sufficiently large, interconnecting pores.
Although a permeable rock must also be porous, a porous rock is not necessarily
permeable. Certain volcanic rocks are porous but not permeable because the voids are
not interconnecting; shale may be quite porous but impermeable because the pores are
extremely small, thereby preventing free movement of the fluids contained within.
2. Physical Characteristics of Reservoir Rocks
Considering a common reservoir rock, sandstone, the grains making up this rock
are all irregular in shape. The degree of irregularity, or lack of roundness reflects
the source sediments and the physical and chemical processes to which they were
subsequently exposed. Violent crushing or grinding action between rocks causes
grains to be very irregular and sharp-edged. The tumbling action of grains along the
bottom of streams or seabeds smoothes sand grains. Wind-blown sand, as occurs in
moving dunes in deserts, results in sand grains that are even more rounded. Sand
grains that make up sandstone beds and fragments of carbonate materials that make
up limestone beds do not fit together congruently: the void space between the grains
forms the porosity.
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The pore spaces (or interstices) in reservoir rock provide the container for the
accumulation of oil and gas and these give the rock its characteristic ability to absorb
and hold fluids. Most commercial reservoirs of oil and gas occur in sandstone,
limestone or dolomite rocks, however, some reservoirs occur in fractured shale and
even in basement rocks such as in Vietnam. Knowledge of the physical characteristics
of the pore space and of the rock itself (which controls the characteristics of the pore
space) is of vital importance in understanding the nature of a given reservoir.
For the reservoir engineer, porosity is one of the most important rock properties as
a measure of the space available for accumulation of hydrocarbon fluids.
3. Porosity
The first step in forming a sandstone, for example, is to have a source of material
which is eroded and transported to low lying depressions and basins such as would
be found off the coasts of a landmass. The material would consist of a mixture of
minerals, but for a sandstone, the majority would be made of quartz in the form of
grains. When these were deposited, they would be surrounded by sea water or brine,
and as the sediment thickness increased, the weight or the pressure produced by the
overlying sediments would force the grains together. Where they contacted each other
large stresses would be produced and a phenomenon called pressure solution would
occur which dissolved the quartz at the points of contact between the grains until the
stresses reduced to a level which was sustainable by the grains. The dissolved material
would be free to precipitate in other regions of the sediment. In this way the initially
loose material would be solidified with discrete connections between the grains.
Initially, if subsea, the pore spaces would be filled with brine, and as the lithification
process occurred, some pore spaces would be isolated with the brine trapped inside.
If the vast majority were interconnected then the initial pore fluid would be free to be
swept through the rock by other fluids such as hydrocarbons. In this way the geometry
of the grains produces an assembly of solids with voids in between them. The grains
vary in diameter but may be from a few microns to several hundred microns. The
geometry of the pore spaces is such that they have narrow entrances (pore throats)
where the edges of the grains touch each other and larger internal dimensions (between
the grains). The complicated nature of these interconnected pores is illustrated in
figure 1 which is a metal cast of the pores in a sandstone rock.
fundamental Properties of Reservoir Rocks
Figure 1 Metallic Cast of Pore Spaces in a Consolidated Sand
One method of classifying reservoir rocks, therefore, is based on whether pore spaces
( in which the oil and gas is found) originated when the formation was laid down
or whether they were formed through subsequent earth stresses or ground water
action.
The first type of porosity is termed original porosity and the latter, secondary or
induced porosity. This is illustrated in figure 2.
Cementing material
Sand grain
Effective porosity 25%
Isolated porosity 5%
Total porosity 30%
Figure 2 Effective, isolated and total porosity
Secondary porosity in limestone beds occurred as a result of fracturing, jointing,
dissolution, recrystallisation or a combination of these processes.
Where water is present in a carbonate formation, there is a continuous process of
solution and deposition or recrystallization. If solution is greater than deposition in
any zone, porosity will be developed between the crystal grains. An important type
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of porosity of this kind is found in dolomite zones which occur in conjunction with
large limestone deposits. Dolomite may be deposited originally as a sedimentary
rock, or it may be formed by replacing the calcium carbonate in limestone rock with
magnesium.
The impact of isolated pore space clearly cannot contribute to recoverable reserves
of fluid nor contribute to permeable pore space as illustrated in figure 3.
Total Pore Space
Isolated Pore Space
Effective Pore Space
Permeable Pore Space
dead End
Pore
Figure 3 Total, effective, isolated permeable and dead end pore space
Porosity is defined as the ratio of the void space or pore space (Vp) in a rock to the
bulk volume (Vb) of that rock and it is normally expressed as a percentage of total
rock volume. The porosity is usually given the symbol φ. The matrix volume is the
volume of the solid grains, Vm.
Porosity =
Porosity =
Bulk volume − Grain volume
x 100
Bulk volume
Porosity =
Void volume
x 100
Bulk volume
pore volume
× 100
void volume + grain volume
fundamental Properties of Reservoir Rocks
Bulk Volume
Representation
Grain Volume
Representation
Pore Volume
Representation
Figure 4 Representation of bulk, grain and pore volumes
These components are illustrated in figure 4 for monosize spheres.
Total porosity is defined as the ratio of the volumes of all the pores to the bulk of a
material, regardless of whether or not all of the pores are interconnected. Effective
porosity is defined as the ratio of the interconnected pore volume of a material.
If the grains are represented by spheres stacked together as in figure 4, then the pore
space can be seen between the solid grains.
Total Porosity =
Total Void Space
Vb
Effective porosity =
Interconnected Void Space
Vb
Induced or Secondary Porosity = porosity from fractures or vugs (large chambers formed
in certain carbonates and limestones caused by groundwater flow and dissolution).
3.1 Range Of Values
The maximum porosity of porous media can be considered in relation to an assembly
of spheres arranged as a cubic packing of spheres. If the sides of a cube are assumed
to be formed by the lines drawn from the centre of each sphere to the adjacent spheres,
the cube in figure 5 would be produced.
Institute of Petroleum Engineering, Heriot-Watt University
Figure 5 Cube defined by the centres of each adjacent sphere
The length of each side would be 2x radius, giving the bulk volume as:
Vb = (2r)3 = 8r3
The grain volume would be the equivalent of the volume of one sphere
Vm =
4πr 3
3
and the porosity (given the symbol φ) would be
φ=
Vb − Vm
=
Vb
4πr 3
3 = 1 - π = 0.476
3
8r
6
8r 3 −
If the spheres fit in the cusps generated by the lower layer then a porosity of 26%
occurs. For a size distribution of spheres the ultimate minimum porosity would be
zero which would be the case if sufficient grains were available to completely fill the
pore spaces as shown in figure 6 for part filling of the void.
Figure 6 Minimum porosity when all pore spaces are filled
fundamental Properties of Reservoir Rocks
Several factors may combine to affect the porosity of a rock, but the main distinction
to be made is as follows based on the amount of connected pore volume, and whether
the pore space has been altered by dissolution or by fracturing after deposition and
lithification.
3.2 Factors Which Affect Porosity
The porosity (and permeability) of sandstone depend upon many factors, among
which are the packing, size and shape of the grains, variations in size of grains,
arrangement in which grains were laid down and compacted, and amount of clay
and other materials which cement the sand grains together.
3.2.1 Packing And Size Of Grains
The absolute sizes of the sand grains which make up a rock do not influence the
amount of porosity occurring in the rock. However variations in the range of sand
grains sizes do influence considerably the porosity.
3.2.2 Particle Size Distribution
If spheres of varying sizes are packed together, porosity may be any amount from 48
per cent to a very small amount approaching 0 per cent as shown in figure 7.
3.2.3 Particle Shape
If the sand grains are elongated or flat and are packed with their flat surfaces together,
porosity and permeability may both be low we will discuss further in the context of
permeability.
Pore Space
Figure 7 Reduction in porosity due to a range of particle sizes
3.2.4 Cement Material
Sandstones are compacted and usually cemented together with clays and minerals. The
porosity and permeability of a sandstone are both influenced to a marked degree by
the amount of cementing material present in the pore space and the way this material
occupies the pore space between the sand grains. The cementing material may be
uniformly located along the pore channels to reduce both porosity and permeability
Institute of Petroleum Engineering, Heriot-Watt University
or the cementing material may be located at the pore throats which reduces the ability
of fluid to enter the pore, but may not reduce the overall porosity of the rock by a
significant amount.
Limestone formations may have intergranular porosity. However, the pore openings
are more often inter-crystalline, that is spaces between microscopic crystals. They also
may take the form of pits or vugs caused by solution and weathering, or by shrinkage
of the matrix. These forms of porosity are called secondary porosity. Another type of
secondary porosity is that caused by fracturing and is very important in that it permits
many limestone rocks of otherwise low porosity to become excellent reservoirs.
Porosity may range from 50% to 1.5% and actual average values are listed below:
Recent sands (loosely packed)
Sandstones (more consolidated)
Tight/well cemented sandstones
Limestones (e.g. Middle East)
Dolomites (e.g. Middle East)
Chalk (e.g. North Sea)
35 - 45%
20 - 35%
15 - 20%
5 - 20%
10 - 30%
5 - 40%
A point that needs to be emphasised is that the concept of ‘porosity’ is complex and
therefore difficult to define and determine. It may refer to spaces between sand grains
or it may refer to limestone caves: it may even exclude a fraction of the free water
(water not bound chemically) present in the rock. Sometimes good estimates, (i.e.
relevant to reservoir development problems) may be obtained from laboratory studies,
or core samples, and sometimes such measurements are irrelevant.
In summary, the amount of porosity is principally determined by shape and arrangement
of sand grains and the amount of cementing material present, whereas permeability
depends largely on the size of the pore openings and the degree and type of cementation
between the sand grains. Although many formations show a correlation between
porosity and permeability, the factors influencing these characteristics may differ widely
in effect, producing rock having no correlation between porosity and permeability.
3.3. Subsurface Measurement Of Porosity
Porosity is measured directly from recovered rock samples as part of core analysis
and also downhole by special tools which indirectly measure a property which can
be related to the formation porosity. These downhole measurement techniques are
very sophisticated in both their engineering and in their practice. For example, the
porosity of a formation can be logged while the hole is being drilled, giving almost
real time indications of the nature of the reservoir. Core analysis procedures will be
reviewed later.
In general the downhole porosity may be related to the acoustic and radioactive
properties of the rock.
3.3.1 Density Log
The density log is derived from the response of the atoms in the minerals in the
rock to bombardment with gamma radiation. The atoms accept energy of a specific
frequency and emit energy of a different frequency; this energy is detected. The
10
fundamental Properties of Reservoir Rocks
energy density is related to the number of atoms and therefore to the density of the
rock being bombarded. If the formation under test is known, for instance a sandstone,
then changes in the density measured within the sandstone result from a change in
the porosity of the formation rather than a change in the mineralogical nature of the
sandstone. This obviously relies on a good description of the geology of the formation.
In a porous formation, the pore fluid will also affect the response of the tool in that
the atoms of the fluid will also react to the bombardment and affect the energy
detected. With reference to calibration samples of different rock types, the effect of
both mineralogy and pore fluid content can be accounted for. Empirical relationships
have been developed to relate the porosity to the values of density which have been
logged. In the following relationship, the logged density, ρL, matrix density, ρm , and
the fluid density, ρf , are related to the porosity, φ
ρL = ρm (1- φ ) + ρf φ
φ =
ρ L − ρm
ρ f − ρm
The contribution of the matrix and the pore fluid are in relation to the relative amounts
of each, and these are related to the porosity. Typically, matrix densities and fresh
water density are as follows
ρQuartz ρLimestone ρWater = 2.65 gcm-3
= 2.71 gcm-3
= 1.00 gcm-3
3.3.2 Sonic Log
This log is similar in concept to the density log, however, it is acoustic energy which is
radiated into the formation from sonic transducers in the logging tool. These produce
compression waves which travel along the side of the borehole in the formation. The
time taken for the wave to travel from the transmitter to the receiver (travel time) is
related to the acoustic properties of the formation. As for the case of the density log,
if the formation is known and its mineralogy is not changing, then variations in the
travel time must result from the changes in the formation acoustic properties, the most
significant of which is the density which is related to the porosity. As with the density
tool, the density of the formation fluids in the pore spaces will affect the travel time
and this must be accounted for. Calibration samples of different rock types have lead
to an empirical relationship between the logged travel time, ∆TL , matrix travel time,
∆Tm , the fluid travel time, ∆Tf , and the porosity, φ .
∆TL = ∆Tm (1- φ ) + ∆Tf φ
φ =
∆TL − ∆Tm
∆Tf − ∆Tm
The contribution of the matrix and the pore fluid are in relation to the relative amounts
of each, and these are related to the porosity. Typically, matrix travel times and fresh
Institute of Petroleum Engineering, Heriot-Watt University
11
water travel time are as follows
∆TQuartz
∆TLimestone
∆TWater
=
=
=
55µs ft-1
47µs ft-1
190µs ft-1
3.3.3 Neutron Log
This is another radioactive logging technique which measures the response of the
hydrogen atoms in the formation and can give an indication of the porosity. Neutrons
of a specific energy are fired into the formation and they disrupt the steady state
activity of hydrogen atoms. They then radiate energy which is detected by the tool:
the energy returned is related to the number of hydrogen atoms which is related to
the hydrocarbon and water in the pore spaces. By calibration, the porosity can be
determined.
3.4 Average Porosity
Porosity is normally distributed and an arithmetic mean can be used for averaging.
For unclassified data,
n
φa =
∑φ
i =1
i
n
(1)
where φa is the mean porosity, φi is the porosity of the ith core measurement and n is
the number of measurements.
4 PERMEABILITY
4.1 Darcy's Law
The permeability of a rock is the description of the ease with which fluid can pass
through the pore structure.
At one extreme, the permeability of many rocks is so low as to be considered zero
even though they may be porous. Such rocks may constitute the cap rock above a
porous and permeable reservoir and they include in their members clays, shales,
chalk, anhydrite and some highly cemented sandstones.
The permeability is a term used to link the flowrate through and pressure difference
across a section of porous rock. The problem is complicated in that the number of
pore spaces, their size and the interconnections is not standard. Thus the application
of the general energy equation, for example as in the case of flow through pipes,
becomes very difficult for flow through porous media.
In petroleum engineering the unit of permeability is the Darcy, derived from the
empirical equation known as Darcy’s Law named after a French scientist who
investigated the flow of water through filter beds in 1856. His work provided the
basis of the study of fluid flow through porous rock. 12
fundamental Properties of Reservoir Rocks
Q=
k∆P. A
µL
(2)
where:
Q =
A =
∆P =
µ =
L =
k =
flow rate in cm3/sec
cross sectional area of sample in cm2
pressure different across sample, atm
viscosity in centipoise
length of sample in cm
permeability in Darcy
Darcy’s law of fluid flow states that rate of flow through a given rock varies directly
with the pressure applied, the area open to flow and varies inversely with the viscosity
of the fluid flowing and the length of the porous rock. In terms of equating the
parameters, the constant of proportionality in the equation is termed the permeability.
The unit of permeability is the Darcy which is defined as the permeability which
will permit a fluid of one centipoise viscosity (= viscosity of water at 68°F) to flow
at a linear velocity of one centimetre per second under a pressure gradient of one
atmosphere per centimetre. Permeability has the units Darcys. Figure 8 illustrates
the concept and the units of permeability
∆p = 1 atmos
Q = 1 cm 3
sec 1cm 2
µ = 1 cp
L = 1 cm
k = 1 darcy
Figure 8 Concept of permeable rocks
Darcy’s Law experiment consisted of a sandpack through which water flowed at a
constant rate (figure 9).
Institute of Petroleum Engineering, Heriot-Watt University
13
Manometric
heads of water
Length, L
Sand
Flowrate, Q
h1
h2
Flowrate, Q
Area of the end of the sandpack
Figure 9 Schematic of Darcy’s experiment
His results showed that the flowrate was directly proportional to the area open to flow,
the difference in pressure and inversely proportionate to the length of the sandpack,
i.e.
Q ∝ A, ∆h,
or
Q=k
1
L
A(h1 − h2 )
L
where Q is the flow rate, A is the area of the end of the core, h1 and h2 are the static
heads of water at the inlet and outlet of the core (the equivalent of the static pressure),
L is the length of the core. K is the constant of proportionality. It is constant for a
particular sand pack. When other workers replicated the experiment, the results were
different to those of Darcy. This was accounted for by inclusion of the viscosity of
the flowing fluid and the equation becomes:
Q=
kA(h1 − h2 )
µL
where the original terms have the same meaning and µ is the viscosity of the fluid
in centipoise.
On a more theoretical basis, Poiseuille formulated the relationship between flow rate
and pressure drop for fluid flowing in a pipe. The form of the relationship is
πr 4 ∆P
Q=
8µL
(3)
where Q is the flowrate, r is the radius of the tube, µ is the viscosity of the fluid and
L is the length of the tube. In this case the dependence of the flowrate / pressure drop
relationship can be seen to be dependent on the radius of the tube. In a similar manner,
the radius of the pores in a rock dictate the nature of the relationship, specifically, the
14
fundamental Properties of Reservoir Rocks
radius of the pore throats is of most significance, since these are the smallest radii
and therefore affect the flowrate/ pressure drop relationship most.
The practical unit is the millidarcy (mD) which is 10-3 Darcy. Formation permeabilities
vary from a fraction to more than 10000 milli-Darcies. At the low end of the range,
clays and shales have permeabilities of 10-2 to 10-6 mD. These very low permeabilities
make them act as seals between more permeable layers.
4.2 Factors Affecting Permeability
Permeability along the flat surfaces will be higher, than the permeability in a direction
perpendicular to the flat surfaces of the grains. In a reservoir, the permeability
horizontally along the bed is usually higher than the permeability vertically across the
bed because the process of sedimentation causes the grains to be laid down with their
flattest sides in a horizontal position (minimising the area exposed to the prevailing
currents during deposition). Figure 10 illustrates the concept.
If sand grains of generally flat proportions are laid down with the flat sides nonuniformly positioned and located in indiscriminate directions, both porosity and
permeability may be very high. To illustrate, if bricks are stacked properly, the space
between the bricks is very small; if the same bricks are simply dumped in a pile, the
space between the bricks might be quite large.
Horizontal permeability 400mD
Vertical permeability 200mD
Horizontal permeability 900mD
Vertical permeability 500mD
Porosity 16%
Porosity 32%
Figure 10 Directional Permeability
The shape and size of sand grains are important features that determine the size of the
openings between the sand grains. If the grains are elongated, large and uniformly
arranged with the longest dimension horizontal, permeability to fluid flow through the
pore channels will be quite large horizontally and medium-to-large vertically. If the
grains are more uniformly rounded, permeability will be quite large in both directions
and more nearly the same. Permeability is found generally to be lower with smaller
grain size if other factors (such as surface tension effects) are not influential. This
occurs because the pore channels become smaller as the size of the grains is reduced,
and it is more difficult for fluid to flow through the smaller channels.
Institute of Petroleum Engineering, Heriot-Watt University
15
This directional perspective to any property is termed anisotropy. As shown above
permeability is a directional property and gives rise to different permeabilities
depending on the shape and depositional characteristics. Very dramatic anisotropy
is generated if a rock is fractured. These anisotropic perspectives are illustrated in figure
11. Porosity is a non directional property and therefore is isoptropic.
Sandstone
Fractured Core
Figure 11 Directional permeability.
4.3 Generalised Form Of Darcy’s Law
A three dimensional rock can be defined within the co-ordinate system illustrated
in figure 12.
-Z
Vs
s
0
+x
+y
+Z
Figure 12 Co-ordinate system for rock permeability
The x and y co-ordinates increase from zero to the left and out from the page; the
z co-ordinate increases downwards. The flow velocity in a particular direction can
be defined as the flowrate in that direction divided by the area open to flow. In any
direction, s, the flow velocity is termed Vs and is equated to the static pressure gradient
in that direction (i.e. the change in pressure, dP, over a small element of length, ds in
that particular direction) minus a contribution from the difference in head (because
of the difference in elevation) of the fluid across the section ds. Therefore,
Vs = -
16
k dp
ρg
dz
( −
)
6
µ ds 1.0133 x10 ds
(4)
Q=k
L
Q=
kA(h1 − h 2 )
µL
Q=
πr 4 ∆P
8µL
fundamental Properties of Reservoir Rocks
the change in elevation head is equal to the sine of the angle to the horizontal
A(hand
1 − h2 )
Q
=
k
k dp
ρg
dzL
Vs = - ( −
)
6
µ ds 1.0133x10 ds
= sine θ, where θ is in degrees.
kA(h1 − h 2 )
Q=
The Darcy units are:
dz
µL
= sin θ , where θ is in degrees.
ds
Vs
πr ∆P k
Qρg=
K dP
dz µ
8µL
Vs = - ( −
)
6
µ ds 1.0133x10 ds ρ
g
=
=
=
=
=
4
L
Vs =
T
velocity along path s - cms-1
permeability - Darcys
viscosity - centipoise
density of fluid - gcm-3
acceleration due to gravity - 980 cms-2
k dp
ρg
dz
−
)
MVs = - µ ( M
6
ds
µ =
ρ = ds3 1.0133x10
= pressure gradient along s - atm cm-1
LT
L
6
−2
1.0133 x 10 converts from dynes cm to atmospheres
dz
, where θ is in degrees.
M
L dP = sinMθ4.4
P =
ds = 2 2 Dimensions Of Permeability
2 g =
2
k dp
ρg
dz
LT
T ds
LT
Vs = - ( −
)
6
µ ds 1.0133 x10 ds the dimensions of each
From
Darcy’s
equation,
K dP
ρg
dz
Vs = - term
( −can be deduced
)in terms of length, L, mass, M and time, T
L
kLT M
ML
6
=
( 2 2 − 3 2 ) µ ds 1.0133x10 ds
T
M LT
LT
L
M
M
L
K
Vs =
µ =
ρ = 3
=
L
L
MT
M LT
T
LT
Vs =
µ =
ρ = 3
T
LT
L
K = L2
M
L dP
M
P =
g = 2
= 2 2
2
LT
LT
M
L dP
M T ds
K dP
ρg
P ρg
=
g = 2
= 2 2 dz
2 dz
Vs = ( −
), T ds
= zero
LT
L T6
µ ds 1.0113x106 Therefore,
ds
1.0113x10
ds in terms of the dimensions (and keeping permeability as k)
the equation
is
L
kLT M
ML
=
( 2 2 − 3 2)
Q
kLT
M
ML
T
M L LT L
T
Vs = Vx =
=
( 2 2 − 3 2)
A
L
K
T
M LT
LT
=
kA dP
T
LT
Q = µ dx
K = L2
L
K
=
T
LT
L
kA P2
ρg
dz
ρg
dz
Q∫ dx = dP V = - K ( dP −
),
= zero
∫
s
6
6
µ P1
µ ds
ds
1.0113x10 ds
0
K =1.0113x10
L2
(5)
kA
Q(L - 0) = (P − P )
µ 2 1 It can
Q be seen that the dimensions reflect the nature of the constant of proportionality
Vs = Vx and
= it should not be confused with, for example, the area open to flow, A, of the end
kA(P1 − P2 )
Q =
(6)units, since 1 atm = 14.73 psi = 1.013
of aAcore or a sand pack. In terms of metric
µL
kAbar
dPand 1 cp = 10-3 Pas it follows that
Q = µ dx
= 9.87 xdz
10 m2 ~ 1 x 10 m
dP
ρg 1Ddz
ρg
2
−
),
zero
1mD
= 9.876 x 10=-16m
~ 1 x 10-15m2
6
P2ds
ds L1.0113x10
1.0113x10
ds
kA
Q∫ dx = dP
µ P∫1
0
-13
Vs = Vx = - k(
Q
A
kA dP
Q = µ dx
Vs =
-12
2
kAof Petroleum Engineering, Heriot-Watt University
Institute
Q(L - 0) = -
µ
(P2 − P1 )
17
Other units of inches2 or cm2 could be used but they are all too large for porous media
and they would also require conversion to relate to permeabilities quoted in other
units. Darcys and milliDarcys are most commonly used.
4.5 Assumptions For Use Of Darcys Law
The simple Darcy Law, as used to determine permeability, only applies when the
following conditions exist:
(i)
(ii)
(iii)
(iv)
(v)
Steady state flow
Laminar flow;
One phase present at 100% pore space saturation.
No reaction between fluid and rock;
Rock is homogenous
1. Steady state flow, i.e. no transient flow regimes. This becomes unrealistic in
terms of flow in a reservoir where the nature of the fluids and the dimensions of
the reservoir may produce transient flow conditions for months or even years. For
laboratory based tests, the cores are small enough that transient conditions usually
last only a few minutes.
2. Laminar flow, i.e. no turbulent flow. For most reservoir applications this is valid
however near to the well bore when velocities are high for example in gas production
turbulent flow occurs. Sometimes it is termed non- darcy flow. Figure 13
Laminar Flow
Turbulent Flow
Q
A
Q k . ∆P
=
µ
A
L
∴K =
Q . L .
µ
A ∆P
∆P
L
Figure 13 Effect of Turbulent Flow on Measured Permeability
18
fundamental Properties of Reservoir Rocks
3. Rock 100% saturated with one fluid, i.e. only one fluid flowing.
In the laboratory this can be achieved by cleaning cores, however, there will be a
certain connate water saturation in the reservoir, and there may be gas, oil and mobile
water flowing through the same pore space. The concept of relative permeability can
be used to describe this more complex reservoir flow regime. Relative permeability
is discussed later.
4. Fluid does not react with the rock, i.e. it is inert and there is no change to the pore
structure through time.
There are cases when this may not happen, for example when a well is stimulated
during an hydraulic fracturing workover. The fluids used may react with the minerals
of the rock and reduce the permeability. In such cases, tests on the rock to determine
the compatibility of the treating fluids must be conducted before the workover.
5. Rock is homogeneous and isotropic, i.e. the pore structure and the material
properties should be the same in all directions and not vary. In reality, the layered
nature and large areal extent of a reservoir rock will produce variations in the vertical
and horizontal permeability.
4.6 Applications of Darcys Law
To examine the applicability of this simple relationship, approximations to the type
of flow encountered in a reservoir can be made: linear flow along a reservoir section
and radial flow into a wellbore. More complex geometries cannot be analysed using
this simple analytical equation and forms of approximating the geometry and flow
are required.
In the following expressions, the nomenclature is identical to that used above.
(i) Horizontal, linear, incompressible liquid system (figure 14)
A
Q
L
P2
P1
Figure 14 Linear flow regime
From the basic Darcy equation
Vs = -
K dP
ρg
dz
(
−
),
6
µ ds 1.0113 x10 ds
ρg
dz
= zero
6
1.0113 x10 ds
Institute of Petroleum Engineering, Heriot-Watt University
19
The flow rate and area open to flow is substituted for the flow velocity. The variables
are separated and integrated over the length (for the flow rate) and the pressures P1
to P2 for the change in pressure. The pressure drop P2 minus P1 is negative and is
corrected by the negative sign on the left hand side of the equation.
Q
A
kA dP
Q = µ dx
Vs = Vx =
L
P
kA 2
Q ∫ dx = dP
µ P∫1
0
Q(L - 0) = Q =
kA
( P2 − P1 )
µ
kA( P1 − P2 )
µL
(6)
The final form is as formulated by Darcy and the permeability will have the units of
Darcys if the other units are:
flow rate, Q - cm3s-1
area open to flow, A - cm2
viscosity, µ - centipoise
pressure, P - atm
length, L - cm
(ii) Horizontal, linear, compressible ideal gas system
The flow regime is the same as for the linear liquid system and from the basic Darcy
equation:
Vs = Vx = - k (
Q
A
kA dP
Q = µ dx
dP
ρg
dz
ρg
dz
−
),
= zero
6
6
ds 1.0113 x10 ds 1.0113 x10 ds
Vs =
In this case, the laboratory measurement of the gas flow would usually be conducted
downstream from the core at almost atmospheric conditions (i.e. there would not
be a large pressure drop across the flow meter). It is assumed that the gas used is
ideal, however, there needs to be a correction to the volumetric flow rate measured
to account for the higher pressure in the core. Figure 15.
20
fundamental Properties of Reservoir Rocks
P1
P
L
A
Core
Pb
P2
Valve
Qb
Flow
measurement
Figure 15 Configuration for gas permeability measurements.
The flowrate measured, Qb at ambient pressure, Pb is related to the flowrate, Q in the
core at the pressure in the core, P via the ideal gas law. If the assumption is made
that the temperature is constant, then
QP = Q b Pb
Q =
Q b Pb
P
and substituting into the equation, separating the variables and integrating
produces
Q b Pb
kA dP
=P
µ dx
L
P
kA 2
Q b Pb ∫ dx = PdP
µ P∫1
0
Q b Pb (L - 0 = -
Qb =
k=
kA ( P22 − P12 )
µ
2
kA( P12 − P22 )
2 µLPb
2 µQb Pb L
A( P12 − P2 2 )
(7)
(8)
Comparing the two expressions equations 6 and 7, it is seen that the gas flow rate is
proportional to the difference in the pressure squared, whereas the liquid flowrate
is proportional to the difference in the pressure. In well testing, the flow rates are
measured at the surface and for gas wells one of the diagnostic plots is the flowrate
versus difference in pressure squared plot. Neglecting the fact that the gas is real, it
gives an indication of the ability of the reservoir to produce gas.
Institute of Petroleum Engineering, Heriot-Watt University
21
Gas Q b =
kA( P12 − P22 )
kA( P1 − P2 )
Liquid Q =
2 µLPb
µL
In certain circumstances, the mean flow rate,Q is measured at a mean pressure,P
which, in the case of a laboratory experiment on a core, is the mean of the upstream
and downstream pressure, i.e.
P=
P1 + P2
2
and
Q = Volume flow rate at P
P Q = PbQb
substituting this into the above gas equation 7.
kA( P12 − P22 )
Pb Q b = PQ =
2 µL
and since
1
1 kA
(P1 − P2 )(P1 + P2 )
(P1 + P2 )Q =
2
2 µL
Q=
kA( P1 − P2 )
µL
(9)
The ideal gas permeability can be calculated from the liquid equation using mean
flowrate, Q measured at mean pressure.
(iii) Horizontal, radial, incompressible liquid system (figure 16)
Radial flow
re
Pe
rw
re
Pw
h
rw
Well
Plan
Elevation
Figure 16 Radial geometry with radial flow from the outer boundary to the wellbore
22
fundamental Properties of Reservoir Rocks
re
rw
Pe
Pw
is the outer boundary radius
is the inner boundary radius (well)
is the pressure at the external boundary
is the pressure at the inner boundary
Starting from the basic Darcy expression again,
Vs = -
k dP
ρg
dz
ρg
dz
(
−
),
= zero
6
6
µ ds 1.0113 x10 ds 1.0113 x10 ds
Substituting for flow velocity, Vs = Vr =
Q
A
In this case the direction of flow is in the opposite sense to the co-ordinate system,
therefore
ds = -dr
For radial geometry, the area, A, is now radius dependent therefore
A = 2πrh
Substitution into the basic expression gives
Q
k dP
= 2πrh
µ − dr
(10)
separating the variables and integrating
r
P
Q e dr k e
= ∫ dP
2πh r∫w r
µ Pw
Q
k
(ln re − ln rw ) =
( Pe − Pw )
2πh
µ
which gives the final form
Q =
2πkh( Pe − Pw )
r
µ ln e
rw
(11)
(iv) Horizontal, radial, compressible real gas system
In this case the geometry is identical to that of the radial flow of incompressible
fluid with the modifications for the compressibility of a gas as per the linear gas
flow system.
Institute of Petroleum Engineering, Heriot-Watt University
23
Vs = -
k dP
ρg
dz
ρg
dz
(
−
),
= zero
6
6
µ ds 1.0113 x10 ds 1.0113 x10 ds
Q
k dP
= 2πrh
µ − dr
If the assumption is made that the temperature is constant, then
QP = Q b Pb
Q =
Qb Pb
P
and substituting into the equation, 10
Q b Pb
k dP
= 2πrh
P
µ dr
separating the variables
re
P
dr 2πkh e
Q b Pb ∫
=
PdP
r
µ P∫w
rw
and integrating produces
 r  2πkh  Pe 2 − Pw 2 
Q bPb ln e  =

µ 
2

 rw 
Qb =
πkh
Pe 2 − Pw 2 )
(
r 
µPb ln  e 
 rw 
4.7 Field Units
(10)
Measurements made in the field are often quoted in field units and to ensure
compatibility with the Darcy equation, a conversion is required. The field units are
usually as follows:
Flow rate, Q - bbl/day or ft3/day
Permeability, k - Darcy
Thickness or height of reservoir, h - feet
Pressure, P - psia
Viscosity, m - centipoise
Radius, r - feet
Length, L - feet
24
fundamental Properties of Reservoir Rocks
In order to convert the Darcy equation for liquid flow, Q =
KA( P1 − P2 )
µL
bbl 5.615 ft 3 1728in3 16.39cm 3 day
hr
Q
(
)(
)(
)(
)(
)
3
3
day
bbl
ft
in
24hr 3600 s
=
( K )( Aft 2 )(
929cm 2
atm
)( ∆Ppsia)(
)
2
ft
14.696 psia
30.48cm
( µ )( Lft )(
)
ft
to oil field units, the following conversion factors are used:
Q
bbl
KA( P1 − P2 )
= 1.1271
day
µL
and these produce the following version of Darcy’s equation in field units:
Q
bbl
KA( P1 − P2 )
= 1.1271
day
µL
(11)
4.8 Klinkenberg Effect
Darcy’s Law would indicate that the permeability should be the same irrespective
of the fluid transmitted, since viscosity is included in the equation. Measurements
made on gas as against liquid for some conditions give higher permeabilities than
the liquid. This phenomenon is attributed to Klinkenberg, who attributed the behaviour to the
effect of the slippage of gas molecules along the solid grain surfaces. This occurs
when the diameter of the capillary opening (pore throat diameter) approaches the
mean free path of the gas (i.e. there is in effect only one gas molecule per capillary).
Darcys Law assumes laminar flow and viscous theory specifies zero velocity at the
boundary of the flow channel. This is not valid when the mean free path of the gas
approaches the diameter of the capillary and the result is that low pressure permeability
measurements are unrealistically high because there is insufficient gas molecules to
form a zero velocity boundary layer at the edges of the pores and to form a mass of
flowing gas within the pores. In this case, too many gas molecules flow through the
pores and the permeability appears to be higher than it actually is: the effect reported
by Klinkenberg.
Since the mean free path is a function of the size of the molecule, the permeability
is a function of the type of gas used in the permeability measurement. This gas
permeability is corrected for the Klinkenberg effect by plotting the gas permeability
at each reciprocal mean pressure. This is illustrated for hydrogen, nitrogen and carbon
dioxide in figure 17:
Institute of Petroleum Engineering, Heriot-Watt University
25
100
Gas Permeability: Millidarcies
80
60
40
Hydrogen
Liquid permeability
Nitrogen
20
Carbon Dioxide
0
0
1
2
3
4
5
Reciprocal Mean Pressure: (Atm.)
Figure 17 Variation in gas permeability with reciprocal mean pressure.
Pm is the mean pressure of the gas (the mean of the upstream and downstream pressures
either end of the core orp in figure 15). In effect, if the gas pressure is raised infinitely
high, the gas will perform as an incompressible liquid would, therefore if several
measurements of permeability are made at different mean pressures, the relationship
between mean pressure and permeability can be extrapolated to the equivalent
pressure conditions of a liquid. In reality, extrapolation to infinity is impossible, so
the reciprocal mean pressure is used and the results are extrapolated to zero reciprocal
mean pressure (i.e. 1/infinitely high mean pressure). This point corresponds to the
liquid permeability. The different gasses have different slopes, but they all extrapolate
to the same equivalent liquid permeability.
The form of the equation developed by Klinkenberg is of the form
kL =
kG
b
l+
Pm
where
kL = equivalent liquid permeability
kG = permeability to gas
Pm = mean flowing pressure
26
(12)
fundamental Properties of Reservoir Rocks
b = Klinkenberg constant for a particular gas and rock (slope of the gas permeability,
inverse mean pressure relationship).
The Klinkenberg effect is greatest for low permeability rocks and low mean
pressures.
4.9 Reactive Fluids
Darcys Law assumes that the fluid does not react with the formation. Many
formation
waters react with clays in the rock to produce a lower permeability to liquid than
would be obtained with gas. Therefore the permeability to water in the formation
may be much lower than would be determined to gas in the laboratory. Any water
injected into the formation may severely reduce the permeability due to clay swelling.
The change in permeability may be substantial, for example from several hundred
millidarcys to less than one millidarcy.
4.10 Average Reservoir Permeability
Permeability is not normally distributed but has an exponential distribution, therefore
a geometric mean is used to obtain an average reservoir permeability.
The Geometric Mean of n numbers is the nth root of their product:
5 Stress Effects on Core Measurements
5.1 Stress Regimes
In reservoir engineering the impact of reservoir stresses on reservoir flow and capacity
parameters has been considered for a number of years but, increasingly, the interest
in stress related measurement has grown. The effect of removing a core from the
formation is to remove all the confining forces on the sample, allowing the rock
matrix to expand in all directions, partially changing the shapes of the fluid-flow
paths inside the core.
It is worth considering the stresses associated with reservoir rock parameters. Figure
18 illustrates the likely configuration of a core extracted from a vertical well, and the
orientation of the core plug extracted for permeability and porosity measurements.
Institute of Petroleum Engineering, Heriot-Watt University
27
Core plug
for horizontal
k measurement
Core plug
for vertical
k measurement
Whole core
4 Inch
Formation
Figure 18 Trends in Reservoir Rock Characterisation
Within a reservoir the stresses in the formation can be expressed in three directions,
the major and two minor principal stresses. Figure 19a. The major principal stress
acting mainly in the vertical direction. Clearly the depositional environment and
formation structure will result in slight changes to these orientations.
28
fundamental Properties of Reservoir Rocks
Major Principal Stress
Minor Principal Stress
(a)
Minor Principal Stress
Equal Stresses
Kh
(b)
(c)
Equal Stresses
Figure 19 Stress States in Reservoirs and Cores
In core analysis, service companies have been asked to measure porosity and
permeability under reservoir stress conditions. They have done this by applying
different stresses for the axial and radial stresses. As can be seen in Figure 19b for a
conventional plug the radial stress would be a combination of the major and a minor
principal stress. To enable the true stress field to be represented, a varying radial
stress distribution would be required. If a vertical plug was used, Figure 19c, then a
constant radial stress could be an acceptable value for the average minor stresses. In
this case, however, the permeability value would be Kv, the vertical permeability.
The effect of the overburden and the pore pressure on the matrix is to produce a net
force between the grains of the matrix (which, when the area over which the force
acts is accounted for produces a net stress). If the matrix is considered to be elastic,
that is, there is a unique relationship between the stress and the strain within the
matrix, then the matrix will strain as the stress is altered. If the stress increases, the
Institute of Petroleum Engineering, Heriot-Watt University
29
strain reduces the radius of the pore throats and reduces the volume of the pore space.
This effect may be different for different rock types and even within the same rock
type if the amount of cementing material is altered. The significant aspects of this
phenomenon are when cores are removed from subsurface to the laboratory (since
the overburden and pore pressure will change) and when the pore pressure in the
reservoir changes due to local pressure conditions around the wells (drawdown) and
within the reservoir as a whole as it is depressurised, for example. The impact of
the net overburden stress which increases as the reservoir pressure ( pore pressure )
decreases is illustrated in figure 20.
1.0
?Well Cemented
Permeability: Fraction of Original
.8
.6
?Friable
?Unconsolidated
.4
.2
0
0
2000
4000
6000
8000
10000
Net Overburden Pressure: PSI
Figure 20 Permeability Reduction with Net Overburden Pressure
In general, the stress regime subsurface is considered to be hydrostatic (as in the case
of the pore fluid) and that the stresses can be resolved into one vertical stress, and
two horizontal stresses. For hydrostatic conditions, all of these are the same. In core
analysis, therefore, the porosity at equivalent subsurface conditions may be determined
by applying an external pressure to the core. This is usually done by inserting the core
into a cell rated for pressures up to 10000 psi (68.9MPa) and applying a stress to the
ends of the core and to the sides. The nature of these tests are such that usually the
stress applied to the sides of the core represents the horizontal stress and the stress
applied to the ends represents the vertical stress. Once trapped inside the cell, the
pore pressure may be increased to a representative level and measurements of pore
volume and permeability made under these stress conditions.
30
fundamental Properties of Reservoir Rocks
More recently, the effect of non-hydrostatic stress conditions has been shown to
be important in certain reservoir conditions, such as in tectonically active areas
(Columbia, South America where the formation of the Andes mountains is associated
with large horizontal stresses) or in areas associated with faults or very compressible
reservoir rocks such as some chalks. In this case the conventional test cells are not
appropriate and special true triaxial cells are required. In these cells the ends of the
core are subjected to the vertical stress as per the conventional cells, but the sides of
the core are wrapped in a cage of individual tubes which can be pressurised in banks
around the core to represent the different horizontal stresses.
In summary, when the properties of the cores are measured in the laboratory, they
can be subjected to
Zero stresses
No effect of the stress on the property
Hydrostatic stresses
The effect of the magnitude of the stresses are measured
Triaxial stresses
The effect of stresses resolved in the three principal
directions are measured
Real stress behaviour
The effect of the magnitude and direction of the stresses are measured
This topic is covered in more detail in the subsequent chapter.
5.2 Compressibility Of Porous Rock
As the rock matrix is subjected to a stress, it will deform and alter the pore space
volume as the rock is compressed. For simplicity, the overburden will be considered
to produce hydrostatic stress (called the compacting stress) on the reservoir, i.e. a
grain-to-grain stress in the rock. Within the pores, fluid pressure acts on the surface
of the grains and reduces the grain-to-grain (or compacting) stress. Therefore in a
real reservoir there is a balance between the effect of the overburden stress and the
pore pressure. This can be described by the relationship
Pcompacting = Poverburden - Ppore pressure
where Pcompacting is the grain-to-grain stress, Poverburden is the stress produced by the
weight of the overburden at a particular depth and Ppore pressure is the pressure of the
fluids in the pores. The expression shows the balance between the overburden and
the pore pressure in compacting the rock matrix: if the pore pressure declines, the
compacting stress increases and the pore volume declines. This assumes that the
overburden remains constant which is logical over the time period of a producing
reservoir. The balance can be represented by figure 21:
Institute of Petroleum Engineering, Heriot-Watt University
31
Surface
Po
Cap Rock
Depth
Pf and Pc
Reservoir
Pc
Pc
Pc
Pf
Pc
Pc
Enlarged view of the pore space
Pc
Grains
Pore space filled with fluid
Figure 21 The balance between overburden & rock stress and fluid pressure
P o = Pf + Pc
Po = overburden pressure
Pf = fluid pressure
Pc = compacting stress
The effect of the change in the balance between the overburden stress and the pore
pressure is to change the compacting stress. If there is an increase in pore pressure,
then the pore volume will increase, however, this is rare and in the main, pore
pressure declines during production and the pore spaces compact under the increasing
compact stress. Two issues are significant: the initial porosity in the reservoir (i.e.
to correctly define the volume of oil in place) and the reduction in that porosity (or
pore volume) as the pressure declines (for material balance and simulation studies).
Figure 22 shows the relationship between porosity and depth (or stress). As the depth
(and stress) increases, the porosity declines. Care needs to be taken when assessing
porosity values: were they measured under overburden or at ambient conditions?
The shale sample shows a large change in porosity as the plate-like clay minerals
are compacted and fit together in a more congruent manner.
32
fundamental Properties of Reservoir Rocks
50
Porosity, φ
40
Sandstone
30
20
10
0
Shale
0
3000
6000
Depth of burial (ft) or stress (psi)
Figure 22 Alteration in porosity with depth of burial (or stress)
The rate of change of pore volume with pressure change can be represented by an
isothermal compressibility (assuming temperature is constant):
Cf = -
1 dv
v dP
(15)
where Cf is the isothermal compressibility, v is the volume, dv is the change in volume
and dP is the change in pressure (the negative sign accounts for the co-ordinate system:
as the pressure increases, the volume decreases).
5.3. Types Of Compressibility
An issue with regard to the compressibility is: which part of the reservoir is
being compressed and which part is significant in calculating the response of the
reservoir.
Three types of compressibility can be considered:
(i) Matrix volume compressibility - the change in volume of the rock grains. This is
very small and usually not of interest in sandstones since it is a purely mechanical
change in volume of the very stiff grains.
(ii) Bulk volume compressibility - the change in the unit volume of the rock. This
is of interest in reservoirs near the surface because of the problem of subsidence;
Changes in volume of the reservoir around faults which may cause the fault to slip
and alter the conductivity both through the fault and across it;
Reservoirs composed of unconsolidated or very weakly consolidated material where
the changes in porosity can be significant. The changes in the volume of the reservoir
both in a vertical sense leading to subsidence and in a horizontal sense leading to
shearing of the wellbore and the associated loss in integrity.
(iii) Pore volume compressibility - change in pore volume. This is of greatest interest
since the pore volume affects the porosity which affects reservoir performance.
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For completeness, all aspects of the reservoir compressibility should be considered,
however, in many problems only specific aspects of the compressibility may be
required such as in a well cemented sandstone reservoir where the bulk volume
change is very small and the subsidence is negligible, but the pore compressibility
is an important feature of the drive mechanism.
5.4 Measurement Of Pore Volume Compressibility
The measurement of pore compressibility is usually conducted in a coreholder which
applies an equal compacting pressure around the core. An inner liner ensures the power
fluid (usually hydraulic oil) does not contaminate the pores of the sample. The pore
pressure is usually kept at ambient, i.e. the compacting pressure mimics the net effect
of the overburden and the pore pressure in the reservoir. This makes the test simpler,
however, there may be conditions where the compressibility of the grains themselves
plays a significant role in the system and the test may require to be conducted at
true overburden and pore pressure conditions. For the test at ambient pore pressure
conditions, an outlet is connected to the core holder and this is lead to a pipette or a
balance to measure the amount of pore fluid expelled. The pressure of the hydraulic
oil is increased in stages and for each stage the amount of fluid expelled is measured
after the rock has come to equilibrium. The data can then be analysed to indicate the
change in porosity or pore compressibility. Figure 23 shows the concept.
Pipette
Sealed core
Pump
Pressure vessel
Figure 23 Measurement of the reduction inpore volume as the external stress (or compacting pressure) is increased
The results show the change in pore volume relative to the original pore volume,
for a given change in the compacting pressure (this assumes that changes in the
compacting pressure have the same effects as changes in the pore pressure) which
can be substituted in to the isothermal compressibility as
Cp = -
1 dv p
v p dPc
where:
Cp = pore volume compressibility
vp = initial pore volume
dvp = change in pore volume (amount of fluid expelled)
dPc = change in compacting pressure
34
fundamental Properties of Reservoir Rocks
Typical values of pore compressibility are in the range 3 x10-6 psi-1 to 10 \x10-6 psi-1,
however, soft sediments can have compressibilities in the range 10 \x10-6 psi-1 to 20x10-6
psi-1 or 30 *10-6 psi-1. Figure 24 illustrates the values determined for some limestones
and sandstones.
Pore compressibility 10-6 psi-1
10
9
Sandstone
Limestone
8
7
6
5
4
3
0
10
Porosity %
20
Figure 24 Compressibility of Sandstones and Limestones
5.5 Effect of stress on permeability
As the effect of a stress on the rock matrix affects the pore volume, it also affects the
pore throat radii and the permeability of the rock. In general, an increase in stress
reduces the pore throat radii and the permeability declines. For most rocks subjected
to an hydrostatic stress, this is the case as the stress is equal in all directions. Figure
25 shows typical permeability declines for increase in stress for sandstone.
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Permeability stress sensitivity for various sandstones
1000
Permeability (mD)
100
10
1
0
20
40
60
Hydrostatic stress (MPa)
80
Figure 25 The reduction in permeability for a range of sandstone samples (the porosity is
in the range 15% to 22%)
Unconsolidated material has larger absolute changes in permeability as the total
strain is greater.
In true triaxial stress regimes, the stresses are not identical and the strain (and therefore
pore throat radii) may cause the sample to dilate in one direction and increase the
pore throat radii therefore enhancing the permeability. This can be illustrated better
by considering a fractured core (figure 26).
36
fundamental Properties of Reservoir Rocks
Fracture
σ
v
Permeability
σh maximum
Fracture closing under stress
Core
σh minimum
σh maximum
σh maximum perpendicular to fracture
Fracture
σv
Permeability
σh maximum
Fracture opening under stress
Core
σh minimum
σ
h maximum
σh maximum parallel to fracture
Figure 26 Triaxial stresses applied to a fractured core
If the largest horizontal stress acts across the fracture (i.e. perpendicular to the faces of
the fracture) then it will be clamped shut; if the largest horizontal stress acts parallel
to the fracture, then it may split open. In this way the anisotropy (or difference in the
properties) may lead to different permeabilities and porosities from the same sample
if the stresses are applied in different ways around the core.
6. Porosity-Permeability Relationships
Whereas for porosity there are a number of downhole indirect measurement methods,
the same is not the case for permeability. The downhole determination of permeability
is more illusive. Down hole permeability is mainly obtained by flow and pressure
determination and requires other characteristics for example the flowing interval. There has been a continued interest in porosity-permeability correlations, on the
basis if one has a good correlation of laboratory measured porosity and permeability
then down hole measurements of porosity could unlock permeability values for
those formations where recovered core has not been practical. Although porosity is
an absolute property and dimensionless, permeability is not and is an expression of
flow which is influenced by a range of properties of the porous media, including the
shape and dimensions of the grains and the porosity. Since porosity is an important
parameter in permeability it is not surprising for those rocks which have similar particle
characteristics that a relationship exists between porosity and permeability. Figure
27 below gives examples of permeability correlations for different rock types.
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1000
?Oolitic Limestone
Sucrosic Dolomite
Permeability: Millidarcies
Reef Limestone
Well Cemented
Hard Sand
100
Chalky
Limestone
10
Intercrystalline
Limestone and
Dolomite
Fine Grained
Friable Sand
1.0
0
5
10
15
20
25
30
35
Porosity: Percent
Figure 27 Permeability and Porosity Trends for Various Rock Types
(Core Laboratories Inc)
7 Surface Kinetics
If core for a particular section cannot be recovered, or for example is formed as a pile
of sand on the rig floor, then correlations like these in figure 27 are used. Porosity
measurements obtained indirectly from wireline methods can be used to obtain the
laboratory porosity vs down hole porosity cross plot. Using this laboration porosity
value the associated permeability value can be determined from an appropriate
correlation as in figure 27.
The simultaneous existence of two or more phases in a porous medium needs terms
such as the capillary pressure, relative permeability and wettability to be defined.
With one fluid only one set of forces needs to be considered: the attraction between
the fluid and the rock. When more than one fluid is present there are three sets of
active forces affecting capillary pressure and wettability.
Surface free energy exists on all surfaces between states of matter and between
immiscible liquids. This energy is the result of electrical forces. These forces cause
molecular attraction between molecules of the same substance (cohesion) and between
molecules of unlike substances (adhesion).
38
fundamental Properties of Reservoir Rocks
Surface tension (or interfacial tension) results from molecular forces that cause the
surface of a liquid to assume the smallest possible size and to act like a membrane
under tension.
7.1 Capillary pressure theory
The rise or depression of fluids in fine bore tubes is a result of the surface tension and
wetting preference and is called capillarity. Capillary pressure exists whenever two
immiscible phases are present, for example, in a fine bore tube and is defined as the
pressure drop across the curved liquid interface. The equilibrium in force between the
molecules of a single phase is disrupted at an interface between two dissimilar fluids.
The difference in masses and the difference in the distances between the molecules
of the different phases produces an initially unbalanced force across the interface. Figure
28 shows the interface between oil and water molecules.
W
W
O
W
O
O
W
Different mass.
Different space
between molecules.
O
W: water molecule
O: oil molecule
distance between molecules
Figure 28 Representation of an oil water boundary
Interfacial tension deforms the outer surface of immiscible liquids to produce droplets.
If the two liquids are present on a surface, the interfacial tension deforms the liquids
to produce a characteristic contact angle as shown in Figure 29.
A wetting phase is one which spreads over the solid surface and preferentially wets
the solid. The contact angle approaches zero (and will always be less than 90˚). A non-wetting phase has little or no affinity for a solid and the contact angle will
be greater than 90˚
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Interfacial tension, s, defined as force / unit length
σwo
Oil
Water
θ
Contact angle, θ
σso
σsw
Solid
σwo Interfacial tension between the water and oil
σ
sw Interfacial tension between the solid and water
σ
so
Interfacial tension between the solid and oil
Figure 29 Interfacial tension between oil, water and a solid
The contact angle describes the nature of the interaction of the fluids on the surface:
for the oil-water system shown above: an angle less than 90˚ indicates that the surface
is water wet. If the angle were greater than 90˚ then the surface would be oil wet. The composition of the surface also affects the interfacial tension. Figure 30 shows the
effect of octane and napthenic acid on a water droplet on silica and calcite surfaces.
The water is not affected by the change in surface in the water/octane system, however,
the napthenic acid causes the water to wet the silica surface, but to be non-wetting
on the calcite surface.
Napthenic acid
Octane
35°
30°
Silica
Octane
Napthenic acid
106°
30°
Calcite
Figure 30 The effect of a change in the surface on wetting properties
The Adhesion tension, At is defined as the difference between the solid water and
solid oil interfacial tension. This is equal to the interfacial tension between the water
and oil multiplied by the cosine of the contact angle,
At = σsw - σso = σwo Cos θwo
40
fundamental Properties of Reservoir Rocks
If a container of oil and water is considered as in figure 31, the denser water lies
below the oil.
σcosθ
θ
σ
OIL
h
radius, r
.c
Water
Figure 31 Capillary rise in an oil/water system
If a glass capillary tube of radius, r is inserted such that it pierces the interface between
the oil and water, the geometry of the tube and the imbalance in forces produced
between the glass, oil and water cause the interface to be pulled upwards into the tube.
If non wetting fluids were used, the interface in the tube may be pushed downwards.
Under equilibrium conditions, i.e. after the tube has pierced the original interface,
the adhesion tension around the periphery (2πr) of the tube can be summed to give
the total force upwards. Since the interface is static, this force must be balanced by
the forces in the column of water drawn up the tube and the equivalent column of oil
outside the tube, i.e. at point C, the force (or pressure) must be the same in the tube
as outside, therefore the excess force produced by the column of water is balanced
by the adhesion tension.
net force upwards = 2πr σwoCosθ
(16)
net force downwards = (ρwgh - ρogh)πr2 = gh(ρw - ρo)πr2
(17)
the interface is at equilibrium, therefore
2πr σwoCosθ=gh(ρw - ρo)πr2
(18)
The capillary pressure is the difference in pressure across an interface, therefore in
terms of pressure (the Pc, force acting on area pr2)
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41
gh( ρw − ρo )πr 2 2πrσ wo Cosθ
=
= Pc
πr 2
πr 2
gh( ρw − ρo ) =
2σ wo Cosθ
r
It can be seen from the equations, capillary pressure can be defined both in terms of
curvature and in terms of interfacial tension, as expressed by the hydrostatic head.
Pc =
where
Pc
σ
θ
rc
h
ρw
ρo
2σCosθ
= gh( pw po )
rc
=
=
=
=
=
=
=
(19)
capillary pressure
surface tension
contact angle
radius of the tube
height of interface
the density of water
the density of oil.
For a distribution of capillaries, therefore, the capillary pressure will give rise to a
distribution of ingress of wetting fluid into the capillaries. The relative position of the
capillary rise is given with respect to the free water level, FWL, i.e. the point of zero
capillary pressure. Figure 32 illustrates the effect of three different capillary radii on
the rise of water. Figure 33 shows the behaviour for a full assembly of capillaries
and alongside the associated capillary pressure curve. In this figure it is important
to note five aspects.
•The free water level-the position of zero capillary pressure
•The oil -water contact
• The 100% water saturation at a distance above the free-water level due to
the capillary action of the largest tube.
•The irreducible level representing the limit if mobile water saturation
•The different radii segregate the capillary pressure and therefore the height to
which the water is drawn into the oil zone.
The zone of varying water saturation with height above the 100% free water oil
contact is called the transition zone.
The formation containing irreducible water will produce only hydrocarbons whereas the
transition zone of varying water saturation will produce water and hydrocarbons.
The shape of the capillary pressure curves in the transition zone will depend on the
nature of the rock.
42
fundamental Properties of Reservoir Rocks
θ
oil
oil
oil
oil
θ
θ
h
FREE WATER
LEVEL
WATER
WATER
Irreducible Water
Oil
Pc
Oil water contact
Transition Zone
Figure 32 Capillary Rise in Distribution of Capillaries
Water
OWC
Water
0
FWL
0%
100%
Sw
So
100%
0%
Free water level
Figure 33 Capillary Pressure Curve
It must be remembered that although concepts of capillary pressure were formulated
in terms of fine bore tubes, application in practice deals with a complex network of
interconnected pores in a matrix carrying surface chemical properties as illustrated
in figure 1 of the pore cast of the pore space.
The height at which a wetting liquid will stand above a free level is directly proportional
to capillary pressure which is related to the size and size distribution of the pores.
It is also proportional to interfacial tension and the cosine of the contact angle and
Institute of Petroleum Engineering, Heriot-Watt University
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inversely proportional to the tube radius and difference in fluid density. The smaller
the pores ie. the lower the permeability, the higher the capillary pressure.
7.2 Fluid distribution in reservoir rocks
Water is retained by capillary forces as hydrocarbons accumulate in productive
reservoirs. The water is referred to as connate or interstitial water and in water wet
rocks it coats the rock surfaces and occupies the smallest pores, whereas hydrocarbons
occupy the centre of the larger pores. The magnitude of the water saturation retained is
proportional to the capillary pressure which is controlled by the rock fluid system.
Rock Fluid
Property
Wettability
Rock / Fluid Property
2σCosθ
Sw _
~ Pc =
re
Rock Property
(Permeability and Porosity)
Water wet, coarse grained sand and oolitic and vuggy carbonates with large pores
have low capillary pressure and low interstitial water contents. Silty, fine grained
sands have high capillary pressures and high water contents.
Reservoir saturation reduces with increased height above the hydrocarbon-water
contact. At the base of the reservoir there will usually be a zone of 100% water
saturated rock. The upper limit of this is referred to as the water table or water
oil contact (WOC). However, there is a non identifiable level, the free water level
representing the position of zero capillary pressure.
Figure 34 shows the capillary pressure curve for a reservoir where the water saturation
reduces above the aquifer. The 100% water saturation continues some distance above
the free water level corresponding to the largest pores of the rock, hD. Above this level
both the oil and water are present and the reservoir water saturation decreases with
increased height above the hydrocarbon water contact, since the larger pores can no
longer support the water by capillary action and the water saturation falls. Between
the 100% WOC and the irreducible saturation level is termed the transition zone.
44
fundamental Properties of Reservoir Rocks
Oil
Sand
Grain
Pc
h
Transition Zone
WOC
FWL
hp
0%
Water Saturation
100%
Water
Figure 34 Capillary Pressure Curve for Porous media
Consider the capillary pressure curves for the two rocks in figure 35. The first sample
(case 1) has a small range of connecting pore sizes. The second sample (case 2) has a
much larger range of connecting pore sizes, although the largest pores are of similar
size in both cases. Also, in case 2, the irreducible water saturation is reached at low
capillary pressure, but with the graded system, a much larger capillary pressure is
needed.
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Pc = (Pw - Po) gh
High Pc needed to reach limiting
water saturation.
Case 1
Irreducible (or non - communicating)
water approach at low Pc
Case 2
hI
Largest connecting pores
about the same size.
Therefore simular hD
h
X
h
Irreducible
water saturation
Water saturation
D
100%
Figure 35 Capillary Pressure Curves for Different Rocks
In addition to water transition zones, there can also be an oil/gas transition zone, but
this is usually less well defined.
Height Above Water Level
Rock wettability influences the capillary pressure and hence the retentive properties
of the formation. Oil wet rocks have a reduced or negligible transition zone, and may
contain lower irreducible saturations. Low fluid interfacial tension reduces the transition
zone, while high interfacial tension extends it. Figure 36 illustrates this effect.
Interfacial Tension Effect
High Interfacial Tension
A
Low Interfacial Tension
0
100
Water Saturation: Percent Pore Space
Figure 36 Interfacial Tension Effect
Saturation history influences the capillary pressure water saturation relationship
and therefore the size of the transition zone. Drainage saturation results from the
drainage of the wetting phase (water) from the rock as the hydrocarbons accumulate.
It represents the saturation distribution which exists before fluid production. The
level of saturation is dictated by the capillary pressure associated with the narrow
pore and is able to maintain water saturation in the large pore below. Imbibition
46
fundamental Properties of Reservoir Rocks
Height Above Water Level
saturation results from the increase in the wetting phase (water) and the expulsion of
the hydrocarbons. In this case the saturation is determined by the large pore reducing
the capillary pressure effect and preventing water entering the larger pore. This is
the situation which occurs both when natural water drive imbibes into the formation
raising the water table level and in water injection processes. Clearly the two saturation
histories generate different saturation height profiles. Figure 37 shows the drainage
and imbibition effects on capillary rise.
Drainage
Drainage
A
Imbibition
0
100
Water Saturation: Percent Pore Space
Imbibition
Figure 37 Saturation History Effect
Height Above Water Level
A large density difference between water and hydrocarbons (water-gas) suppresses
the transition zone. Conversely, a small density difference (water-heavy oil) increases
the transition zone. Figure 38 shows the differences in density for water/heavy oil and
water/gas on capillary rise. Transition zones between oil and gas are not significant
because of the large density difference between oil and gas.
Fluid Density Difference Effect
Small Density Difference
(Water-Heavy Oil)
A
Large Density Difference
(Water Gas)
0
100
Water Saturation: Percent Pore Space
Figure 38 Fluid Density Effect
7.3. Impact of Layered Reservoirs
A characteristic of reservoirs is the various rock types making up the reservoir section.
Each rock type has its own capillary pressure characteristics. Wells penetrating such
formations will show a water saturation distribution reflecting the specific capillary
effects of each formation type. In some cases a 100% water saturation will be above
a lower water saturation associated with a lower elevation material with a higher
permeability, Figure 39.
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For example well A would only indicate 100% water. Well B would penetrate the
transition zone of the top layer then a region of 100% water saturation. The saturation
profiles for well B and C are illustrated in figure 39. The transition zone of the next
layer 2, followed by an interfacial of 100% saturation associated with layers 2, 3 and
4 then into 100% for the next two layers. Well D penetrates through the top and next
layer at the irreducible saturation level, into the transition zone for layer three, then
into irreducible saturation for the 4th layer.
B
C
D
SHALE
K=
K=
25
0=
10
0=
0=
FWL
0
30
m
40
K=
K=
d
d
5m
md
200
Free Water Level
100%
d
m
19 0
STO
0
5
=1
Water saturation
profile well C only
D
SAN
Transition
zone
NE R
ES.
Water saturation
profile
Well B only
Height
A
SHALE
FWL
0%
Sw
100%
1
2
3
4
100% Water Level
Figure 39 Capillary Effects in Stratified Formations
8 EFFECTIVE PERMEABILITY
8.1Definition
The idea of relative permeability provides an extension to Darcy’s Law to the
presence and flow of more than a single fluid within the pore space. When two or
more immiscible fluids are present in the pore space their flows interfere. Specific
or absolute permeability refers to permeability when one fluid is present at 100%
saturation. Effective permeability reflects the ability of a porous medium to permit
the passage of a fluid under a potential gradient when two or three fluids are present
in the pore space. The effective permeability for each fluid is less than the absolute
permeability. For a given rock the effective permeability is the conductivity of each
phase at a specific saturation. As well as the individual effective permeabilities being
less than the specific permeability, their sum is also lower.
48
fundamental Properties of Reservoir Rocks
If measurements are made on two cores having different absolute permeabilities
k1 and k2, there is no direct way of comparing the effective permeability kw and ko
curves since for the two cores they start at different points k1 and k2. This difficulty
is resolved by plotting the relative permeability krw and kro where
Relative Permeability =
permeability to one phase when one or more phases are present
permeability to one phase alone
kr =
ke
k
Relative permeability is dimensionless and is reported as a fraction or percentage.
On relative permeability plots the curves start from unity in each case, so direct
comparisons can be made.
A typical set of effective permeability curves for an oil water system is shown in
figure 40 and for a gas oil system in figure 41.
1.0
0.9
Relative Permeability
0.8
0.7
k ro
0.6
k rw
0.5
0.4
0.3
0.2
0.1
0
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
S , Water Saturation, Fraction
W
Figure 40 Relative permeability curves for water-oil sysrem
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Connate Water plus Residual Oil Saturation
Relative Permeability, Fraction of Absolute
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
k rg
k ro
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Liquid Saturation = SO + SWO, %
Figure 41 Relative permeability curves for gas-oil sysrem
The following points are to be noted:
The introduction of a second phase decreases the relative permeability of the first
phase: for example, kor drops as Sw increases from zero. Secondly, at the point where
the relative permeability of a phase becomes zero there is still a considerable saturation
of the phase remaining in the rock. The value of So at kro = 0 is called the residual oil
saturation and the value of Sw at krw = 0 is called the irreducible water saturation.
The shapes of the relative permeability curves are also characteristic of the wetting
qualities of the two fluids (figure 42). When a water and oil are considered together,
water is almost always the wetting phase. This means that the water, or wetting phase,
would occupy the smallest pores while the non-wetting phase, or oil phase, would
occupy the largest pores. This causes the shape of the relative permeability curves
for the wetting and non-wetting phase to be different.
50
fundamental Properties of Reservoir Rocks
Relative Permeability, %
100
90
Water-Wet Drainage
80
Water-Wet Imbibition
70
Oil-Wet Drainage
(Decreasing S w )
(Increasing S w )
(Increasing S w )
K ro
60
50
40
Krw
30
20
10
0
0
10
20
30
40
50
60
70
80
90 100
Water Saturation, S W
Figure 42 Oil and Water Relative Permeability Curves for Water-Wet and Oil-Wet
Systems (Core Laboratories Inc)
This is illustrated by looking at the relative permeability to one phase at the irreducible
saturation of the other phase. The relative permeability to water at an irreducible oil
saturation of 10% (90% water) is about 0.6, figure 40, whereas the relative permeability
to the non-wetting phase, oil, at the irreducible water saturation of 0.3 approaches 1.0. In this case it is 0.95. One practical effect of this observation is that it is normally
assumed that the effective permeability of the non-wetting phase in the presence of
an irreducible saturation of the wetting phase is equal to the absolute permeability.
Consequently, oil flowing in the presence of connate water or an irreducible water
saturation is assumed to have a permeability equal to the absolute permeability.
Similarly, gas flowing in a reservoir in the presence of irreducible water saturation
is assumed to have a permeability equal to the absolute permeability.
Relative permeability characteristics are important in the displacement of hydrocarbons
by water, and in the displacement of oil and water by gas. Such displacements occur
during primary and secondary recovery operations, as well as during coring and core
recovery.
Relative permeability data when presented in graphical form are often referred to as
drainage or imbibition curves. (figure 42)
Imbibition relative permeability is displacement where the wetting phase saturation
is increasing. For example, in a water flood of a water wet rock, or coring with a
water base mud.
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Drainage relative permeability is where the non-wetting phase saturation is increasing.
For example, gas expulsion of oil during primary depletion or gas expansion of fluids
during core recovery, and the condition existing in the transition zone at discovery.
Water displacement of oil differs from gas displacement of oil since water normally
wets the rock and gas does not. The wetting difference results in different relative
permeability curves for the two displacements.
8.2 Water displacement of oil
Prior to water displacement from an oil productive sand interstitial water exists as
a thin film around each sand grain with oil filling the remaining pore space. The
presence of water as previously stated has little effect on the flow of oil, and oil
relative permeability approaches 100%. Water relative permeability is zero.
Water invasion results in water flow through both large and small pores as the water
saturation increases. Imbibition relative permeability characteristics influence the
displacement. Oil saturation decreases with a corresponding decrease in oil relative
permeability. Water relative permeability increases as water saturation increases.
Oil remaining after flood-out exists as trapped globules and is referred to as residual
oil. This residual oil is immobile and the relative permeability to oil is zero. Relative
permeability to water reaches a maximum value, but is less than the specific permeability
because the residual oil is in the centre of the pores and impedes water flow.
8.2.1 Water-oil relative permeability
Accumulation of hydrocarbons is represented by drainage relative permeability curves
as the water saturation decreases from 100% to irreducible. Water relative permeability
reduces likewise from 100% to zero while oil relative permeability increases.
Subsequent introduction of water during coring or water flooding results in a different
set of relative permeability curves - these are the imbibition curves. The water curve
is essentially the same in strongly water wet rock for both drainage and imbibition.
The oil phase relative permeability is less during imbibition than during drainage.
The oil remaining immobile after a waterflood is influenced significantly by the
capillary pressure and interfacial tension effects of the system. It is of note that a
high residual oil saturation is a result of the oil ganglia being retained in the large
pores as a result of capillary forces. Figure 43 illustrates the pore doublet model
illustrating how oil can be trapped in a large pore. The forces to displace this droplet
have to overcome capillary forces and are too great to use pressure through pumping.
The force required can be reduced by reducing the interfacial tension which is the
basis for many enhanced oil recovery methods; for example, surfactant and miscible
flooding.
52
fundamental Properties of Reservoir Rocks
Water In
Oil
Advancing water
Water In
Water penetrating
smaller pores due to
capillary forces
Oil
Trapped oil
Water
Water In
Figure 43 Pore Doublet Model
An important perspective in a displacement process is the concept of mobility ratio.
This relates the mobility of the displacing fluid relative to that of the displaced fluid. It is therefore a ratio of Darcy’s Law for each respective fluid at the residual saturation
of the other fluid. In the context of water displacing oil.
M = mobility ratio =
where krw
kro
krw ©/µ w
kro ©/µ o
(20)
is the relative permeability at residual oil saturation
is the relative permeability at the irreducible water saturation.
These relative permeabilities are sometimes referred to as end point relative
permeabilities. When M is less than 1 this gives a stable displacement whereas when
M is greater then 1 unstable displacement occurs.
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53
This topic is covered extensively in the chapter on immiscible displacement
8.3 Gas displacement of oil and gas-oil relative permeability
Gas is a non-wetting phase and it initially follows the path of least resistance through
the largest pores. Gas permeability is zero until a ‘critical’ or ‘equilibrium’ saturation
is reached (figure 41).
Gas saturation less than the critical value is not mobile but it impedes the flow of
oil and reduces oil relative permeability. Successively smaller pore channels are
invaded by gas and joined to form other continuous channels. The preference of gas
for larger pores causes a more rapid decrease of oil relative permeability than when
water displaces oil from a water wet system. Figure 44 shows the alteration of relative
permeability as gas comes out of solution and flows at increasing saturation through
the oil reservoir. These gas/oil relative permeability curves are very significant in
relation to the drive mechanism of solution gas drive, which we will discuss in a
subsequent chapter.
54
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Relative Permeability: Percent
20
40
60
80
100
Gas
Characteristic Sand During Oil Displacement
by Gas @ 5% Gas saturation
250 md.
183 md.
0.0md.
183/250 = 0.73
0.0/250 = 0.0
Gas Saturation: Percent Pore Space
0
Krg
Kro
Gas Saturation: 5% of Pore Space
Specific Permeability
(Ks):
Effective Permeability to Oil (Ko):
Effective Permeability to Gas (Kg):
Relative Permeability to Oil (Kro) =
Relative Permeability to Gas (Krg) =
0
20
40
60
80
Water
0
20
40
60
80
20
40
Krg
60
80
100
Characteristic Sand During Oil Displacement
by Gas @ 20% Gas saturation
250 md.
52 md.
10md.
52/250 = 0.21
10/250 = 0.04
Gas Saturation: Percent Pore Space
0
Kro
Gas Saturation: 20% of Pore Space
Specific Permeability
(Ks):
Effective Permeability to Oil (Ko):
Effective Permeability to Gas (Kg):
Relative Permeability to Oil (Kro) =
Relative Permeability to Gas (Krg) =
Relative Permeability: Percent
100
20
40
60
Krg
80
100
Characteristic Sand During Oil Displacement
by Gas @ 45% Gas saturation
250 md.
6.2 md.
70md.
6.2/250 = 0.025
70/250 = 0.28
Gas Saturation: Percent Pore Space
0
Kro
Gas Saturation: 45% of Pore Space
Specific Permeability
(Ks):
Effective Permeability to Oil (Ko):
Effective Permeability to Gas (Kg):
Relative Permeability to Oil (Kro) =
Relative Permeability to Gas (Krg) =
0
20
40
60
80
100
Relative Permeability: Percent
100
Oil
fundamental Properties of Reservoir Rocks
Figure 44 Gas Oil Relative Permeabilities ( Core Lab)
55
Rock Properties Measurement
CONTENTS
1. INTRODUCTION
1.1 Core Analysis
1.2 Core Definitions
2. SAMPLE PREPARATION
2.1 Whole Core Scanning
2.2 Core Cleaning
3. POROSITY MEASUREMENTS
3.1 Methods
3.2 Whole core versus conventional versus
sidewall samples
4. PERMEABILITY
4.1 Introduction
4.2 Impact of Stress
4.3 Steady State Permeability Methods
4.4 Unsteady State Permeability Measurements
5. FLUID SATURATION
5.1 Gas saturation
5.2 Oil saturation by retort
5.3 Water saturation
6. CAPILLARY PRESSURE
6.1 Introduction
6.2 Capillary Pressure Measurement Techniques
6.2.1 Porous Diaphragm (figure 22)
6.2.2 Centrifuge method ( Figure 23)
6.2.3 Dynamic method ( Figure 24)
6.2.4 Mercury Injection ( Figure 25)
6.3 Use of Laboratory Capillary Pressure Data
for Reservoir
6.4 Averaging capillary pressure data
7. EFFECTIVE PERMEABILITY
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•
List the various types of recovered core.
•
Describe briefly the various methods of measuring porosity and permeability.
•
Briefly describe the various stress conditions that can be imposed on a rock
sample.
•
Understand how to convert laboratory based capillary pressure measurement
data to field related values of capillary pressure.
•
Be able to determine the saturation distribution in a well made up of different
rock types given capillary pressure data.
Derive the Leverett J function and be aware of the major tortuosity related assumption
in its derivation.
Rock Properties Measurement
1. INTRODUCTION
1.1 Core Analysis
In this chapter we will focus on the laboratory based methods used to determine
some of the parameters outlined in the previous chapter. The topic is also covered
in other modules of the overall Petroleum Engineering programme in the context of
the specific module. Core recovery is covered in drilling and rock properties are also
covered in the Petrophysics module.
Cores obtained from the reservoir formation contain a considerable amount of
information about the nature of the rocks themselves and various properties. They
are also a source of material for investigating rock behaviour with respect to fluid
displacement and its reaction to various fluid types. Cores are recovered from the formation of interest using an annular shaped coring
bit. The integrity of the recovered core depends on the nature of the rock and can
vary from rock which is well formed to that which is friable in character or even is
so unconsolidated that it would form a pile of sand on the rig floor when recovered
from the core barrel. The core from the core barrel provides a record, over the well
section recovered, of the properties of the formation. Figure 1 illustrates the wide
range of measurements and procedures carried out on core samples 1.
A comprehensive document on the procedures for generating some of the rock
properties through laboratory measurement is the API Recommended Practices for
Core Analysis 2. APR RP40 which was revised in 1998. This API document goes
into detail beyond that covered in this overview chapter
Routine Core Plug
Analysis
Slabbed Core
•
•
•
•
Photograph
Sedimentology
Lithology
Samples
Government or
Regulatory Board
Sampling
•
•
•
•
Curation
Porosity
Permeability
Grain Density
As-Received
Saturations
Special Core Analysis
Thin Sections
•
•
•
•
•
•
•
•
•
•
•
•
Detail Pore Structure
Diagenesis
Porosity Type
Environmental
Evidence
Preserved /Restored State
Capillary Pressure
Relative Permeabilty
Electrical Properties
Acoustic Properties
Compressive Properties
Clay Chemistry Effects
Specific Tests
Small Samples
• Grain Size Distribution
• Mineral Analysis
• X-Ray and SEM
Analysis
• Bio-Dating and
Association
Calbration of Wireline Logs
Figure 1 Data Obtained From Cored Wells 1.
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As covered in the previous chapter there are a number of properties in relation to
measurements possible on the cores as shown in the figure 1. In core analysis the
measurements can be divided into two parts; routine measurements which cover;
fluid saturations, porosity and permeability; special core analysis which covers a
wide range of measurements and special tests of special interest to the organisation
commissioning the testing. In this chapter we will focus on routine core analysis
and also cover briefly capillary pressure measurements.
1.2 Core definitions
Before examining some of the methods it is important to define the various core types
used in examining rock properties and their reaction to the transmission of fluids.
These definitions come from the API recommended RP 402.
Fresh Core
Any newly recovered core material preserved as quickly as possible at the wellsite
to prevent evaporative losses and exposure to oxygen. The fluid type used for coring
should be noted, e.g., fresh state ( oil-based drilling fluid), fresh state ( water -based
drilling fluid).
Preserved Core.
Similar to fresh core but some period of storage is implied. Preserved core is protected
from alteration by a number of techniques, from simple mechanical stabilisation using
bubble wrap or similar, freezing the core to lock in fluids which would otherwise
evaporate ( in this case the freezing may alter some of the rock properties), enclosure
in heat -sealable plastic laminates, and dips and coatings.
Cleaned Core.
Core from which the fluids have been removed by solvents. The cleaning process
(the specification and sequence of solvents, temperatures, etc) should be specified.
Some solvents could damage the fabric of the rock and special cleaning procedures like critical point drying might be required for example with rocks containing friable
clays (figure 2).
Figure 2 Sandstone contains illite.
Rock Properties Measurement
Restored - State Core
This is core that has been cleaned and then reexposed to reservoir fluids with the
intention of reestablishing the reservoir wettability condition. The conditions of
exposure to the crude oil, especially initial water saturation, temperature and time ,
can all affect the ultimate wettablity.
Pressure - Retained Core
This is material that has been kept, so far as possible, at the pressure of the reservoir
in order to avoid change in the fluid saturations during the recovery process.
2. SAMPLE PREPARATION
2.1 Whole Core Scanning
Prior to subdivision of the whole core for the various types of analysis a number of
procedures can take place to record the characteristics of the whole core and to relate
it to indirect down hole measurements. The purpose of this core examination and
description is to recognise lithological, depositional, structural and diagenetic features
of the whole core or slabbed core. Qualitative and quantitative core descriptions
provide the basis for routine core analysis sampling, facies analysis, and further
reservoir studies such as reservoir quality and supplementary core analysis. Besides
visual examination and generating a photographic record, these techniques provide
a means of relating to downhole measurements and to identify features of the core
which might otherwise if undetected generate unrepresentative data in subsequent
analysis.
The following analysis might be carried out on whole core. A core gamma log, an xray analysis, a computer tomography CT scan and or an Nuclear Magnetic Resonance
NMR Scan.
Within a rock are naturally occurring gamma-ray emitters which can give a measurable
gamma-ray response that can be recorded with depth. If such a measurement can be
made on the whole core in the laboratory this whole core laboratory based measurement
can be used as depth check to relate to open hole measurements. Figure 3.
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Scintillometer
Recorder
Lead Shield
Conveyor Belt
Core
Figure 3 Natural gamma scan on whole core. (Corelab).
A number of X-Ray techniques can be used which include, fluoroscopy, x-radiography
and computerised tomography (CT) scanning. In one method a continuous analysis is
where an attenuated x-ray beam directed through the core impinges on a fluorescent
screen and the captured image is recorded by video camera. In x-radiography the
attenuation of the beam is captured and recorded on sensitive film. In this procedure
the core is stationary. The advances in CT scanning in medical applications have
been used in CT scanning where the attenuated beam directed in multiple directions
by a rotating beam enables a reconstruction of density variations within the core. The
resolution of the image depends on the thickness of the beam and the size of pixel
used to construct the image. A sketch of CT scanning and the principal on which it
is based is shown in figure 4
Sample for
measurement
h
Io
I
Narrow incident
beam
Attenuated
beam
Particle
or energy
detector.
Shield
I = Ioe -µh
µ is a function of bulk density and atomic number
Figure 4(a) Computer aid tomography on whole core. Principal of attenuation.
Rock Properties Measurement
Rotating energy
source and detector
Intensity profiles
Reconstruction algorithm
in computer.
Figure 4(b) Reconstructed cross section.
The main benefit of Nuclear Magnetic Resonance, (NMR) imaging is that it is used
to provide a reconstruction of the fluids within a core, based on the frequency of the
excitation energy associated with a nudei. This excitation energy is supplied by an
oscillating magnetic field. The high energy attenuation associated with CT scanning
does not enable the distinctive density variations as possible with those from NMR
scanning.
These scans are able to identify localised variations in a core which if captured in
subsequent core analysis measurements could give rise to anomalous results.
2.2 Core Cleaning
Sample preparation is an important consideration in core analysis. Prior to samples
or plugs being used for the determination of porosity or permeability they must be
thoroughly cleaned to extract all of the oil and brine and then be properly dried, with
the exception of saturation measurements for the determination of porosity. This is
generally carried through flushing, flowing or contacting with various solvents to
extract hydrocarbons, water and brine.
Solvent extraction using centrifuge, Soxlet and Dean Stark refluxing solvent extractors
are commonly used to remove both oil and brine. No standard solvents are used and
organisations use their own preferences (figure 5).
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Measurement of
collected water
Core plug
Figure 5 Porous diaphragm capillary-pressure system.
Care needs to be taken to dry the samples particularly when hydrateable minerals are
present in the sample that break down at high temperatures. The drying procedure
is critical in that the interstitial water must be removed with no mineral alteration.
Humidity -controlled ovens are used when drying clay bearing samples to maintain
the proper state of hydration. Critical point can drying be used to clear core continuing
delicate clays like illite (see Phase Behaviour chapter - section 8.1).
3. POROSITY MEASUREMENTS
3.1 Methods
Figure 6 illustrates the methods used for routine determination of porosity. Rock Properties Measurement
Vacuum Gauge
Valve
Displacement Pump
Boyles Law Porosimeter
Pore Volume Determination
Water
Oil
Gas
Porosity
Pressure Gauge
Outlet Valve
Gas Inlet Valve
Sight Glass
Sample in Place,
Stopcock Open
Core Sample
Micrometer Scale
Mercury
Plunger
Washburn Porosimeter
Sample
Chamber
Reference
Volume
Valve
Kobe Porosimeter
Pressure
Gauge
Valve
Grain Volume Determination
Resaturation
Figure 6 Porosity measurement methods (Corelab)
(a) Bulk Volume
In all porosity methods a bulk core sample volume has to be determined and this may
be carried out either by displacement of liquid or by callipering a shaped sample and
computation by the appropriate formula. Figure 7 shows the displacement method,
and figure 8 shows a mercury displacement pump.
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Adjustable fork
Thermometer
Reference mark
Core plug
Mercury vessel
Single pan
balance
_ 0.01 gm
+
Weighted
base
Figure 7 Archimedes mercury immersion apparatus (API)2
Pressure
read-out
Volume
read-out
Sample
chamber
Displacement
plunger
Figure 8 Volumetric mercury displacement pump (API)2
(b) Summation of fluids
This method involves the independent determination of oil, gas and pure water
volumes of a fresh core sample. The oil and water can be obtained by retort ( Figure
9) and the gas by mercury injection. The pore volume is determined by summing the
three independent volumes.
10
Rock Properties Measurement
Thermocouple
Insulated Oven
Heating Elements
Sample Cup
Screen
Condensing Tube
Water Bath
Water Inlet
Temperature
Controller
Receiving Tube
(c) Gas transfer
Figure 9 Oven retort (API)2
(i) The Boyles Law based porosity determination method involves the compression
of a gas into the pore space or the expansion of gas from the pores of a prepared
sample. Depending on the instrumentation and the procedure, either pore volume
or grain volume can be determined. Figure 10 shows a typical set up for this and
is the most common method for measuring the grain volume. It involves setting up
a pressure in a known reference volume and then expanding the pressure into the
space containing the sample. With suitable calibration the grain volume is determined
using the ideal gas relation that PV=constant.
P
Sample
chamber
Reference
volume
P1
Gas in
Pressure
regulator
Figure 10 Boyle's law porosimeter.
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11
(ii) The Washburn-Bunting method involves the vacuum extraction and collection
of the gas contained in the pores of a prepared sample. The method measures pore
volume.
(d) Liquid resaturation
The pores of a prepared sample are filled with a liquid of a known density. The
increase in weight of the sample divided by the fluid density is a measure of the
pore volume.
(e) Grain density
Total porosity is determined by this method as compared with effective porosity. The
sample is reduced to grain size after the dry weight and bulk volume are determined.
Grain volume is determined and subtracted from the bulk volume to yield the total
pore volume.
3.2 Whole core versus conventional versus sidewall samples
As well as coring using a coring bit and core barrel, it is also possible to recover samples
of the formation using wireline tools, these are termed sidewall coring. There are two
types of sidewall coring devices. One is based on exploding a core plug shaped piece
into the formation. Clearly samples recovered by this technique may be suitable for
mineral description but are not so suited to porosity and permeability analysis as a
result of the damage generated by the explosive force of the sampling device. Sidewall
corers which cut into the formation do not suffer from such mechanical damage.
Whole core porosities tend to be slightly lower than small plug samples in certain
rock types. The whole core is likely to include tighter material than would be included
in a more carefully sampled plug.
For samples with medium to high porosity, sidewall and conventional samples agree
within one or two percent. During sidewall sampling low porosity highly cemented
materials tend to shatter and yield values greater than the true porosity.
4. PERMEABILITY
4.1 Introduction
The API recommended practice for the determination of permeability is also detailed
in API RP 40 which is a considerable improvement on API RP27.
There are essentially two approaches to measuring the permeability, the steady state
method where the pressure drop for a fixed flow rate is measured, generally a gas, or
the unsteady state method where the flow in the transient regeme is measured.
In the latter there are two types of test , the ‘pulse-decay’ method where two pressures
are set up and downstream of the contained sample. A slight increase in the upstream
pressure is imposed and the decay of this pressure through the sample is monitored.
The advent of very high speed data acquisition systems and accurate pressure
12
Rock Properties Measurement
transducers has made it possible to monitor these transient flow conditions. The other
approach is the pressure fall off method where a relatively low upstream pressure is
set and the decay of this pressure is monitored as it is released through the core to
the downstream open to atmosphere.
4.2 Impact of Stress
Over recent years the impact of reservoir stresses on rock properties and therefore the
interest in measuring rock properties under realistic stresses has grown in particular
in relation to permeability. Stress effects also have an impact on other properties
included porosity . In describing the various approaches to permeability measurement
we will also look at various procedures for imposing stress on the samples.
In figure 19 of the previous chapter we identified the various stress directions in
the context of permeability measurement. Figure 11 illustrates the core recovered
from a vertical well and the natural stresses imposed. It is important to distinguish
the different possible stress loadings that can be applied to core plugs and also the
configuration of the stresses in the natural state. In the natural state the stresses can
be considered to be resolved in three principal directions. The vertical direction being
the major principal stress and the two horizontal directions the two minor principal
stresses. Figure 11a
Core plug
for horizontal
k measurement
Major
principal stress
Core plug
for vertical
k measurement
Whole core
Minor principal stresses
Inch
Formation
Figure 11 (a) Core recovered from vertical well and stress orientation in the reservoir.
If a core plug is recovered from a whole core recovered from a vertical well then the
stress orientations in a permeability test would be as shown in the sketch below. Figure
11b and 11c. These figures demonstrate that for a cylindrical horizontal core plug
it is difficult to impose a distinctive major principal stress on the core plug different
from one of the minor principal stresses whereas for a vertical orientated core plug
such distinctive stresses can be applied.
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13
Major principal stress
Major principal stress
Minor
Minor principal
stress
principal
stress
Minor principal stress
Minor principal stress
Major principal stress
Major principal stress
Figure 11 (b) Stress orientation for horizontal core plug.
Major principal stress
Major principal stress
Minor principal stress
Minor principal stress
Figure 11 (c) Stress orientation from vertical core plug
In requesting reservoir stresses to be applied to core plug measurements it is important
to examine that the stresses applied actually represent those which the rock would
be subjected to in the formation. The various modes of stressing a rock are shown
in figure 12 a-d
Isostatic Stress. Figure 12a. Under isostatic stress loading, equal stress is applied
to the sample in all directions, and sample strain can occur on all axes. Excessive
porosity reduction typically occurs when the imposed isostatic stress is equal to the
vertical reservoir stress ( i.e., the overburden stress).
14
Rock Properties Measurement
A
σ1
∆D
∆L
Sample
Isostatic stress
σ1
σ1
L
σ1
D
Figure 12 (a) Isostatic Stress
Triaxial Stress. Figure 12b. Under the true triaxial stress conditions, unequal stress
is applied to the three major axes of the sample. In the general case, strains will be
different on each axis. Typically a cube or rectangular prism -shaped sample will
be used.
σ1
∆L1
Triaxial stress
σ2
σ3
∆L
∆L
Figure 12 (b) Triaxial Stress
Biaxial Stress. Figure 12c. Biaxial stress loading conditions are a special case of triaxial
stress loading. In the biaxial stress loading of a cylinder , the stress parallel to the
cylinder’s axis is different from the stress applied around the cylinder’s circumference. Strains can occur parallel to both the axis and diameter of the cylinder.
C
σ1
∆D
∆L
Sample
Biaxial stress
σ
σ
L
σ1
D
Figure 12 (c) Biaxial Stress
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15
Uniaxial Strain. Figure 12d. Uniaxial strain compression is a special case of biaxial
stress loading; the stress applied to the circumference is just sufficient to maintain
the diameter constant as the stress parallel to the cylinder axis is increased. Strain
occurs only parallel to the axis of the cylinder.
σ1
∆L
Sample
Uniaxial stress
σ
σ L
σ1
D
Figure 12 (d) Uniaxial Stress
4.3 Steady State Permeability Methods
The most conventional permeability measurement approach has been to use the
measurement of the pressure drop associated with a fixed flow rate. To determine
specific permeability nitrogen or air is usually caused to flow through a prepared sample
of measured dimensions. The pressure differential and flow rates are measured and
the permeability calculated from the Darcy equation. A schematic set up is shown
in the sketch below . Figure 13
End view showing
radial stress
Pressure
transducer
P1
Differential
Pressure.
∆p
_
+
L
Pressure
regulator
P
D
qr @ Pr, Tr
Pa
Flow meter
Sample holder
Figure 13 Schematic of steady state permeability measurement 2
The confining of the core in this case shows a Hassler type core holder where the radial
stress is low and is applied to ensure that flow of gas does not by-pass the core.
16
Rock Properties Measurement
Figure 14 shows a high pressure core holder designed to impose reservoir stresses.
The slideable inlet tube enables the strain of the stress core to be taken up. The stress
loading for this arrangement is isostatic.
Cavity for
Hydraulic Oil
Slidable Inlet to Produce Confining Rubber
Tube
Sleeve
Stresses
End Plug
Inlet Port for
Confining Oil
Cylindrical
Core Plug
Outlet
Flow Tube
Retaining
Ring
End Plug
Figure 14 High pressure core holder for stress condition, isostatic 2
Figure 15 shows a sophisticated core holder where a different axial stress can be
applied compared to the radial stress. In this arrangement the end faces of the core
plug need to be machined accurately to ensure that the loading of the axial stress
is distributed over the whole face. If not the core is liable to fragment. The stress
loading for this core plug is biaxial.
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17
Inlet Flow Port
Rubber Sleeve
Port for Oil to Produce
Radial Confining Stress,
or Vacuum to Dilate Sleeve
Core Plug
Reach Rod, X
Outlet
Ports
Port for Oil to Produce
Radial Confining Stress,
or Vacuum to Dilate Sleeve
Large Piston of Axial Stress
Intensifier
Cavity for
High Pressure
Nitrogen for
Axial Stress
N
Figure 15 High pressure core holder or biaxial loading 1.
Using a core plug removed from a horizontal well core it is possible using biaxial
stress loading to somewhat simulate the stress conditions, by considering the two minor
principal stresses as equal. However using biaxial stress conditions for a conventional
plug from a vertical well recovered core, then the stress conditions imposed do not reflect those in the formation. The radial stress is a combination of the major principal
stress and one of the minor principal stresses and in the equipment these are equal. If however, one is interested in measuring the vertical permeability from a sample
extracted from the whole core then biaxial stress conditions will reflect more readily
the reservoir stress condition.
A recent innovation has been the true triaxial cell 2 (Figure 16). In this arrangement a series of axial tubes are hydraulically pressured between the confining rubber sleeve
of the core and the core holder body. This enables a stress pattern to be established
to represent a more realistic stress condition reservoir stress conditions.
18
Rock Properties Measurement
Platen
Threaded
end cap
Trapped tube
A
A
Core
Rubber
sleeve
Aluminium
cell body
Hydraulically
pressured tubes
Maximum principal stress
σ
σ
1
1
1
σ
1
Face of
core plus
σ
Section AA
Figure 16 True trixial cell.
Although liquids could be used in permeability measurements it is common to use a
gas. Gas permeabilities need to be corrected for the Klinkenberg effect and reported
as equivalent liquid permeabilities.
The samples for analysis may be either the consolidated piece used for the porosity
determination or another sample but clearly it must be extracted and cleaned to
ensure that no water or oil are present. If interstitial water is very saline then it may
be necessary to remove salt.
Another recent innovation has been the probe permeameter. These devices were
initially invented to meet the need for a device to give indications of permeability of
an outcrop. The application of rock outcrops as analogues of subsurface formations
has been very valuable in developing geological/ reservoir modelling procedures. The
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19
examination of the various levels of permeability measurement , (upscaling) , have
demonstrated the value of being able to measure the permeability over a small area
which the probe permeameter affords. Figure 17 shows an arrangement of a typical
probe permeameter . As well as back pack mounted version for use in outcrop studies
they can also be laboratory mounted and can automatically scan the permeability
variations in a slab of rock.
Flow
meter
Pressure
regulators
Pressure
transducer
Rock being
examined
ri
ro
Figure 17 Schematic of steady state probe permeameter.
The API RP40 document also describes a radial steady-state apparatus, figure 18,
where flow is from the outer to the inner radius. In this set up the preparation is not
easy and axial stresses are not balanced by radial stresses.
20
Rock Properties Measurement
Rubber Gaskets
Calibrated
Gas Burette
rw
P1
Mercury Manometer
Springs
re
P
Pivot Ball
Piston
L
Regulators
Air Supply
Figure 18 Radial flow steady state permeameter 2.
4.4 Unsteady State Permeability Measurements
The advent of high speed computers and data acquisition systems has enabled the
application of unsteady state permeability measurements. The principles are similar
to the behaviour of a well during a well test and the analysis of the pressures during the
unsteady state draw down or build-up period. Figure 19 gives a schematic of a pressurefall off system. An upstream gas reservoir of different volumes, to accommodate a
wide range of permeabilities, is pressured and then released to atmosphere via flow
through the core. The pressure just upstream of the core is accurately monitored. Full
details of the calculation procedure presented by Jones are given in the API RP40 practise document 2.
Institute of Petroleum Engineering, Heriot-Watt University
21
VT
Fill
Vent
P1
VP
Pc
Hydrostatic
confining
pressure
Figure 19 Schematic of pressure - fall of gas permeameter 2.
In the pulse decay method for permeability measurement a configuration of equipment
is as shown in figure 20. It consists of an upstream and downstream reservoir. The
two gas reservoirs are filled to a pressure. When equilibrium is reached with all valves
open, the joining valves are closed and the pressure in the upstream gas reservoir is
increased by 2-3% of the pressure set in the vessels. The valve 1 is then opened and the
pressure time behaviour of the transient flow behaviour is monitored. This procedure
lends itself to very low permeability values, 0.1-millidarcies to 0.01 microdarcies.
Calculation procedures are also given in the API practise document.
Valve Fill/vac.
+_
V1
Valve 1
∆p
P
V
VP
Pc
Figure 20 Pulse decay apporatus axial flow of gas.
5. FLUID SATURATION
Core analysis is sometimes used to measure the fluid saturations associated with
the core. Because of the large pressure variations between the reservoir and the
surface these saturations are not too representative of the values that would exist
in the formation, unless precautions have been taken to prevent evaporation during
pressure decline. Such precautions could be the application of pressure coring where
the down hole pressure is held in the core barrel as it is recovered to surface. At the
22
Rock Properties Measurement
surface prior to releasing the pressure the core in its container is frozen. It is then
slipped and stored in a frozen state. During controlled thawing of the core the fluids
produced and retained enable downhole saturation to be obtained.
5.1 Gas saturation
Conventional and sidewall core samples have gas saturation measured by injecting
mercury into the gas filled portions of the pores. The gas is compressed into a small
volume or forced into solution in the liquids in the pores using a mercury pump.
Measurement of the volume of mercury penetrated is a measure of the gas content
of the sample.
5.2 Oil saturation by retort
Oil distilled at atmospheric pressure gives a measure of the oil content of the plug.
The distillate is collected in a calibrated receiver. Temperatures up to 6500C are used
(Figure 9).
5.3 Water saturation
Samples can have their water content determined by atmospheric distillation
concurrently with the oil content determination. A distinction should be made between
the pore water and the water of hydration or crystallisation.
Water saturation can also be measured by a solvent refluxing method (Dean-Stark)
(figure 20). Toluene is the most commonly used solvent. The oil content of the sample
is obtained by difference of the weight of the sample before and after extraction and
drying less the weight of the water removed during solvent extraction.
Measurement of
collected water
Core plug
Figure 21 Dean Stark Apparatus
Institute of Petroleum Engineering, Heriot-Watt University
23
6. CAPILLARY PRESSURE
6.1 Introduction
The general laboratory procedure for capillary pressures to saturate a core sample
with a wetting phase and measure how much wetting measurement phase is displaced
from the sample when it is subjected to some given pressure of non-wetting phase.
Displacement takes place when the oil or non-wetting phase just exceeds the capillary
pressure corresponding to the largest pore. In other words the capillary force will
hold the water in the largest pore until the oil pressure is larger than the capillary
pressure of the largest pore.
The volume of the fluid displaced at a particular pressure also represents the pore
volume of all pores of that particular size. Once this pore volume has been displaced at
a particular pressure the pressure is increased and the new pore volume measured.
A plot of water volume displaced versus the displacement pressure will represent
a plot of the capillary pressure versus the percentage of the pores with a capillary
pressure greater than the subject capillary pressure.
Clearly a rock which contains a variety of pore sizes will have a capillary pressure
curve which is not discontinuous but is a smooth curve.
Since capillary pressure, Pc =
2σCosθ
r
the curve can be calibrated to represent pore size versus percentage of pores less than
the subject pore size.
6.2 Capillary Pressure Measurement Techniques
There are four main methods for capillary pressure measurement
(i)
(ii)
(iii)
(iv)
Desaturation or displacement through a porous diaphragm.
Centrifuge or centrifugal method.
Dynamic capillary pressure method.
Mercury injection method.
6.2.1 Porous Diaphragm (figure 22)
In the porous diaphragm method there is a permeable membrane of uniform pore
size distribution containing pores of such a size that the selected displacing fluid will
not penetrate the diaphragm when the pressures applied to the displacing phase are
below some selected maximum pressure of investigation. Pressure applied to the
assembly is increased by small increments. The core is allowed to approach a state
of static equilibrium at each pressure level. The saturation of the core is calculated
at each point defining the capillary pressure curve. Any combination of fluids can be
used: gas, oil and/or water.
This procedure is closest to the actual saturation in the reservoir but the method is
time consuming varying from 10 to 40 days for a single sample.
24
Rock Properties Measurement
Nitrogen Pressure
Saran Tube
Crude Oil
Neoprene Stopper
Nickel-Plated
Spring
Core
Kleenex Paper
Scale of
Squared Paper
Seal of
Red Oil
Ultra-Fine
Fritted Glass
Disk
Brine
Figure 22 Porous diaphragm capillary-pressure system.
6.2.2 Centrifuge method ( Figure 23)
The high accelerations in a centrifuge increase the field of force on a sample subjecting
it to an increased gravitational force. The core plug is mounted in a modified centrifuge
tube as shown and the desaturation of the sample is monitored with a strobe light.
When the sample is rotated at various constant speeds a complete capillary pressure
curve can be obtained. The advantage of the method is the increased speed of obtaining
the data in that the complete curve can be established in a few hours.
Seal Cap
O-Ring
Core Holder Body
Core
Support Disk
Window
Tube Body
Figure 23 Centrifuge for determination of capillary pressure curves 5.
6.2.3 Dynamic method ( Figure 24)
A dynamic method has been used where a simultaneous steady-state flow of two
fluids is established in the core. The saturation is varied by regulating the quantity
of each fluid entering the core and the pressure difference between the two fluids
gives the capillary pressure.
Institute of Petroleum Engineering, Heriot-Watt University
25
Gas
inlet
Gas
outlet
∆po
∆pg
pc
Core
To atmosphere
Oil burette
Oil inlet
Porcelain
plate
Figure 24 Dynamic capillary pressure equipment 5.
6.2.4 Mercury Injection ( Figure 25)
The most common procedure for determination of capillary pressure is using mercury
injection. The procedure was developed to accelerate the determination of the capillary
pressure-saturation relationship. Mercury is the non-wetting fluid. The core sample
is inserted into the mercury chamber of a mercury pump or a mercury porosimeter
and evacuated. Mercury is then injected into the core under pressure. The volume of
mercury injected at each pressure determines the non-wetting phase saturation. This
procedure is continued until the core sample is filled with mercury or the injection
pressure reaches some predetermined value. The procedure is used in a number of
industries to determine the pore size characteristics of the porous media.
The main advantages are that the test takes considerably less than the diaphragm
method, a matter of one or two hours. The disadvantages are the difference in wetting
properties and permanent loss of the core sample. Also there is concern on the
pore size to pressure relationship since the desaturation of some large pores may be
determined by access via smaller pores.
26
Rock Properties Measurement
0-00 psi Pressure Guage
0-,000 psi Pressure Guage
Regulating Valve
Lucite Window
Lucite Window
To
Atmosphere
Cylinder
U-Tube
Manometer
Figure 25 Mercury injection porosimeter 5.
6.3 Use of Laboratory Capillary Pressure Data for Reservoir
Saturation Distribution.
As we have noted above, laboratory capillary pressure tests can be made with a
variety of fluids that differ from reservoir fluids. It is necessary therefore to convert
laboratory based results to be applicable to the field where the fluids might be different. We will examine the procedure for converting air-mercury data to water-oil data for
application in field determinations of saturation profiles. As shown previously, capillary pressure saturation data can be converted to height
saturation data:
h=
Pc
( ρw − ρo )g
(1)
Air/mercury capillary pressure curves are comparable in shape to air/brine or oil/brine
capillary pressure curves.
When converting capillary pressure curves to an equivalent height, the difference in
interfacial tension and contact angle between the laboratory and reservoir systems
must be accounted for. For example
surface tension (σ) of water = 70 dynes/cm
surface tension (σ) of mercury= 480 dynes/cm
contact angle (θ) water/solid = 0 degrees
contact angle (θ) mercury/solid = 140 degrees
Institute of Petroleum Engineering, Heriot-Watt University
27
Pc =
2σCosθ
r
(2)
At corresponding saturations therefore
Pcair / mercury 480Cos140
=
≅5
Pcair / water
70Cos0
Pc air/mercury = 5 Pc air/water (3)
The interfacial tension and contact angle values will depend on the characteristics of
the fluids. The relationship between Pc mercury/air and Pc oil/water is often taken
as 10:1 but these interfacial tension and contact angle values should be checked
before converting data.
Pc air / mercury = 10 Pc water / oil (4)
The equations below give the procedure for generating a height saturation profile for
the reservoir from a laboratory based Pc vs saturation capillary pressure data.
Pc L (σCosθ ) R
Pc R
(σCosθ ) L
h=
=
( ρw − ρh )g ( ρw − ρh )g (5)
where:
h = height in feet above the free water level corresponding to zero capillary
pressure
PcR = capillary pressure at initial reservoir conditions (psi)
PcL = capillary pressure in the laboratory (psi)
(σCosθ)R = interfacial tension cosine of the contact angle (initial reservoir
conditions)
(σCosθ)L = interfacial tension cosine of the contact angle (laboratory conditions)
ρw = density of water at initial reservoir conditions
ρh = density of hydrocarbon at initial reservoir conditions
It should be noted that the interfacial tension of an oil/water system is approximately
10 times greater than that for an oil/gas system and that consequently capillary forces
are more important for the former system.
28
Rock Properties Measurement
EXERCISE 1 – Calculation of water saturation distribution in a layered reservoir.
The purpose of this exercise is to show that in a well, the water saturation not only
varies with the height above the free water level, but also due to variations in rock
properties.
A well penetrates a reservoir which from cuttings is known to consist of rock types
A and B from which a set of air-mercury measured capillary pressure curves are
available, taken in a nearby well. Figure E1. During logging the lowest 100% Sw
was found at the bottom of the well in rock type B as indicated in the figure E2.
The porosity at this level is 15%.
Specific gravities of the water and oil are 1.03 and 0.80 respectively at reservoir
conditions. The density of water is 62.4 lbm/ft3.
Questions
1. Determine the Free Water level and locate it on figure E2.
2. Construct the water saturation profile.
3. Estimate permeabilities
4. Which intervals would you recommend for completion based on the criteria
Sw<50% and k<0.1mD.
What is the net pay (cumulative thickness having Sw<50%).
Institute of Petroleum Engineering, Heriot-Watt University
29
h
(lt)
(mD) Pc. (%) 1
(psi)
1
10
.
10
.0
0
type A rock
type B rock
00
10
100
0
0
0
0
Pore space unoccupied by mercury
Figure E1 Capillary pressure curves from nearby well.
30
100%
Rock Properties Measurement
Rock
type
Porosity
10
h
(ft) 1%
(1 cm for
10 ft)
0
Saturations
Oil
100
Water
Unit No.
100%
0
k
(mD)
A
%
B
1%
10%
A
1%
B
%
%
A
%
%
1%
B
10%
A
1%
B
10%
100 Sw
in B
type rock
found at
this level
1%
Figure E2 Opposite
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31
6.4 Averaging capillary pressure data
Capillary pressure measurements are not part of routine core analysis and a
comprehensive set of capillary pressure data is not always available. Leverett4 in 1941
generated a function which related capillary pressure to porosity and permeability,
which is commonly termed the Leverett J Function. The application of this function
was to be able to generate capillary pressure information when laboratory data was
not available. Capillary pressure data are obtained from core samples which represent
an extremely small part of the reservoir. The ‘J’ function is used to combine all the
capillary data to classify a particular reservoir.
The theory behind the J Function is outlined below and is based on figure 26 considering
flow through a core, which is assumed to be a bundle of capillary tubes.
Lcap
Lcore
Figure 26 Model of flow for Leverett J Function.
The laminar flow of fluid through a pipe is given by Poiseuille’s equation:
q=
πr 4 ∆P
8µL cap
(6)
For n tubes
qn =
nπr 4 ∆P
8µL cap
(7)
The porosity of the bundle of tubes is
φ=
nπr 2
A
(8)
and the permeability is
k=
32
qµL core
A∆P
(9)
Rock Properties Measurement
If φA is substituted for nπr and then
r2 =
8K L cap
φ L core
(10)
L cap
L core is the tortuosity of the bundle of tubes.
On the assumption that the reservoir rock has the same tortuosity at all points,
then
1
 K 2
r = constant  
φ
(11)
and substituting for r in the definition of capillary pressure gives:,
Pc =
2σCosθ
1
 K 2
constant  
φ
or
(12)
1
 K 2
Pc  
φ
1
=
=J
constant σCosθ
(13)
Sometimes the J function is written without the Cosθ term.
The capillary pressure measurements can therefore be normalised for differences in
permeabilities, porosities and fluids and used to measure the capillary pressure, i.e.
the J function is obtained independent of k, φ, σ and θ.
A set of capillary pressure data from a set of 9 core plugs taken from different depths
in a well is shown in figure 27 and shows the wide variation in shape of these curves
reflecting the different pore characteristics as given in the table below.
CAPILLIARY PRESSURE vs WATER SATURATION (Sw)
Sample
No.
1
Permeability
mD
0.
.10
.
0.
.
1,100.00
.00
.00
.10
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Porosity
%
1.
.
0.
0.
.0
.
.
.
1.
33
1
1
1
1
1
1
11
1
Pc (PSIG)
10
1
0
10
0
0
0
0
Sw %
0
0
0
0
100
Figure 27 Set of capillary pressure curves.
A plot of the J function for a set of capillary pressure curves is given in figure 28 and
shows the impact of bringing together different rocks under one curve
34
Rock Properties Measurement
100
1100
1000
00
00
_
1
k )
_
Pc( ϕ
00
00
00
00
00
00
100
10
10
0
0
0
0
Sw %
0
0
0
0
100
Figure 28 Leverett J Function
The data for figure 27 however would not generate such a good function. The big
assumption in Leverett's model is that of constant tortuosity. Clearly different
rock types will have different tortuosities as a result of the pore characteristics and
composition of the rock. However within a rock type the J function could be a useful
route to obtain capillary pressure data if porosity, permeability and saturation data
is available.
Examination of field data has shown that by plotting J versus a better correlation
(Sw − Swc )
(1 − Swc ) is obtained suggesting that the S
the various rocks. Figure 29
wc
reflects the tortuosity variations within
Institute of Petroleum Engineering, Heriot-Watt University
35
LEGEND
0
Different reservoir sand sequence
in a formation
Dimensionless Capillary Pressure Pc
σ
K
φ
0
1
1
1
1
10
0
0
0.1
0.
0.
0.
0.
0.
0.
0.
(
0.
Normalised Wetting Phase Saturation Sw* = Sw-Swc
1-Swc
1.0
(
Figure 29 Modified Leverett J Function Curves.
7. EFFECTIVE PERMEABILITY
It is not the intention of these notes to review in detail the various approaches
to measuring effective permeabilities to multiphase systems. There has been
considerable activity in this area for gas - oil, oil - water, and three phase gas - oil
- water systems.
There are two approaches to measuring relative permeability, using an unsteady state
method or a steady state method.
36
Rock Properties Measurement
In the unsteady state method, a displacement process is set up where one fluid displaces
another and the flow rates and pressure drops are monitored as a function of time
for a fixed rate process. The saturations are obtained by calculation the remaining
volumes of the respective fluids. It is more difficult to generate relative permeabilities
as a function of saturation in this way and some would consider the method is more
suited to generate end-point effective permeability values.
In the steady state method a range of constant rate tests are set up and the pressure
drop noted when equilibrium has been achieved. Figure 30 gives a sketch of a typical
steady state set up.
Oil recycle system
Differential
pressure
transducer
∆P
Differential
pressure
transducer
∆P
Oil
Brine
Composite core
Oil - water
separator and
production monitor
Brine recycle
system
Pressure
control
system
Figure 30 Steady state relative permeability.
The focus is again on three phase relative permeability which has been the subject
of many papers and correlations. It is however of great interest now that large WAG,
water - alternating gas injection processes are being used to improve recovery.
Institute of Petroleum Engineering, Heriot-Watt University
37
Solution to Exercise
EXERCISE 1 – Calculation of water saturation distribution in a layered reservoir.
The purpose of this exercise is to show that in a well, the water saturation not only
varies with the height above the free water level, but also due to variations in rod
properties.
A well penetrates a reservoir which from cuttings is known to consist of rock types
A and B from which a set of air-mercury measured capillary pressure curves are
available, taken in a nearby well. Figure E1. During logging the lowest 100% Sw
was found at the bottom of the well in rock type B as indicated in the figure E2. The
porosity at this level is 15%.
Specific gravities of the water and oil are 1.03 and 0.80 respectively at reservoir
conditions. The density of water is 62.4 lbm/ft3.
QUESTIONS
1. Determine the Free Water level and locate it on figure E2.
2. Construct the water saturation profile.
3. Estimate permeabilities
4. Which intervals would you recommend for completion based on the criteria Sw<50%
and k<0.1mD.
What is the net pay (cumulative thickness having Sw<50%).
SOLUTION
1. The first step is to convert the air-mercury capillary pressure data to oil-water.
Pc air/mercury = 10Pc water/oil (equation 4, page 26)
PcR = h (ρw - ρo) g (equation 5, page 27)
Conversion values:
Pcair/hg = 10 Pc water oil -
Pc
lb f
in 2
lb f 144in 2
lbm
= h ftx (1.03 − 0.8) x 62.4 3 xg
2
2
in
ft
ft
 lb f 144in 2 
lbm
Pc oil / water  2
 = h( ft ) x (1.03 − 0.8) 62.4 3 xg
2
n
ft 
ft
38
Rock Properties Measurement
1 lbf = 1 lbm xg
Pc oil/water psi = 0.1 ft oil/water
∴Pc air/mercury = 1 ft oil/water
The capillary pressure curves can now be rescaled. Figure E3.
Plotting hft = Pc air/mercury (psi) versus 0 - 100 % water saturation.
2. Free water level
This occurs in rock type B. φ = 15%. From capillary pressure curve 100% water
saturation at 15 psi i.e. 15 ft.
Free water level is 15 ft below this position, as indicated on Figure E4. The free water level now provides the basis for the water saturation profile
determination.
3. Water Saturation Profile
The water saturation value is determined at each level where the rock properties
change but noting where the 100% water saturation value occurs for each rock type. At the first change, the height is 20ft from rock type B, 15% φ to type B 10% φ
From the capillary pressure curves the respective saturations are 75% and 100%
Figure E4. For rock type B 10%, the 100% water saturation level is at 27ft when the
saturation decreases. The next rock change is at 41ft above the Free Water Level,
from rock type B 10% to type B 14% with a water saturation value of 73% and
44%. The 44% is based on an estimate of the capillary pressure curve for a value
of porosity of 14% between the 15% and 10% curves. This process is continued
through all the depths of the rock property changes and the total saturation profile
generated.
4. The estimates of permeability are based on porosity permeability trends from the
limited data given for the various rock types of the capillary pressure curves. In
unit 1 rock type B 15% the permeability is 35mD Unit 2, B 10% the permeability
is 15mD Unit 3 B 14%, interpolation suggests a value around 32mD and so on
through the units.
5. Completion intervals according to the criteria Sw<50% and k>0.1mD are
shaded on the figure E4.
6.
Net pay adds up to around 125ft.
Institute of Petroleum Engineering, Heriot-Watt University
39
0'
h
(lt)
(mD) Pc. (%) 1
(psi)
1
10
.
10
.0
0
type A rock
type B rock
00'
00
10'
10
100'
100
0'
0
0'
0'
0'
0
0'
1 psi
0
0
Water saturation
Figure E3 Capillary pressure curves from nearby well
40
100%
Rock Properties Measurement
Rock
type
h
(ft)
1%
Porosity
10
100
Water
Unit No.
0
00
A
10
B
Saturations
Oil
A
B
0
A
100
B
0
B
1'
100% WL
k
(mD)
0.0
1
0.0
1
1
1
1
1
0.
1
11
1
10
0.1
0.0
0.0
10
1
10 mm
0
1
A
100%
0.
1
1
FWL
Figure E4
Institute of Petroleum Engineering, Heriot-Watt University
41
REFERENCES
1. Archer. S., Wall. C., Petroleum Engineering Principles and Practice, Graham
and Trotman 1986
2. Recommended Practices for Core Analysis. American Petroleum Institute.
Recommended Practise 40. Second Edition , Feb 1998.
3. Smart. B,
4. Leverett. M,C., Capillary Behaviour in Porous Solids. Trans AIME 1941
5. Amyx et al Petroleum Reservoir Engineering McCranhill 1960
42
Permeability - Its Variations
CONTENTS
1 INTRODUCTION
2 AVERAGE PERMEABILITIES FOR SEVERAL
LAYERS
2.1 Beds in Parallel
2.2 Layers in Series - Linear Flow
3 MODELLING HETEROGENEOUS SYSTEMS
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•
Appreciate and understand that permeability is an anisotropic property.
•
Derive equations to enable the calculation of average permeability for; layers
in parallel-linear and radial flow, layers in series – linear and radial flow.
•Describe briefly the impact of layered reservoirs in the context of modeling
water oil displacement in heterogeneous reservoirs.
Permeability - Its Variations
1 INTRODUCTION
Although we have defined permeability as a rock property, seldom, if ever, is a homogeneous reservoir encountered in actual practice. In many cases the reservoir will
be found to contain several distinct units or layers of varying rock properties.
Even on a local scale the value of permeability is not necessarily the same in all directions. Permeability is an anisotropic property, (Figure 1) i.e. its value is dependent
on direction. Porosity is an isotropic property however.
Ky
Kx
Kz
Figure 1 Permeability An Isotropic Property.
The sedimentary nature of rocks is such that vertical permeability is less than horizontal permeability and horizontal permeabilities in the principal directions will also
be different.
On a reservoir scale, thin streaks of very low permeability material can reduce the
effective vertical permeability to a value lower than the actual rock values would
indicate. Whereas core analysis represents microscale observations, data obtained
from well tests represent microscale behaviour.
Figure 2 represents the variation in permeability observed using a mini-permeameter
in samples from the Leman gas field.
Institute of Petroleum Engineering, Heriot-Watt University
2
1.0
15.5
Scale 1
(cm)
0
1.0
1.0
12.0
17.5
2.5
>0.5
36.5
"Sample plug"
Poros 17.5%
Perm. 19 mD
a
15.0
b
c
d
1.0
e
<0.5
38.5
21.0
Location of minipermeability measurement
Permeability in mD
f
>0.5
20.0
g
h
Figure 2 Effect of scale of observation and measurement on permeability data from a
Rottliegende aeolian sand cross bed set in the Leman gas field 1.
Although the conditions for use of Darcy's Law state that the rock should be homogeneous and isotropic, in reality reservoirs do not conform to this restraint. If one
examined the variation say in a core as illustrated in figure 3, very large variations
in permeability occur and vary according to the scale of measurement. Conventional
core analysis takes a sample at around 1 per foot, probe permeability is able to sample
at much closer intervals. A well test result can reflect the permeability over tens of
feet. There is considerable effort taking place now in developing up-scaling methods
for representation of permeability for different applications. Permeability - Its Variations
Statistical Analysis of Rock Property Evaluation
K
Log Permeability
K
300
180
25
160
200
Figure 3 Statistical Analysis of Rock Property Evaluation
Waren and Price2 demonstrated that the most probable behaviour of a heterogeneous
systems tends towards that of the geometric mean.
kG = (k1 x k2 x k3 ...kn )
1/ n
(1)
Values for average permeability can be generated by considering the formation being
made up as a composite with different layers. There are two main types of layering
to be considered: linear and radial.
2 AVERAGE PERMEABILITIES FOR SEVERAL LAYERS
Simple geometry systems can be dealt with as follows:
2.1 Beds in Parallel
The horizontal system reflects the sedimentary nature of the rock: the rock material
may have been segregated as it was deposited giving different sizes, shapes etc. to
different layers in the formation. The reason for determining an average permeability
is in rationalising the permeability measured on small samples in the laboratory with
the measurements made for example by well test analysis. Well test analysis cannot test small sections of the reservoir (which would be uneconomical) on the same
scale as the laboratory tests. The results are therefore representative of flow through
several layers rather than only one.
Linear Flow
Consider the simple linear beds in parallel. Figure 4.
Institute of Petroleum Engineering, Heriot-Watt University
P2
P1
Q1
Q2
Q3
A1
K1
A2
A3
h1
K2
h2
K3
h3
L
Figure 4 Radial Flow in Parallel
The average permeability can be developed using the Darcy flow equation:
QT = Q1 + Q2 + Q3 Q1 =
k1 A1 ( P1 − P2 ) k2 A2 ( P1 − P2 )
,
...etc.
µL
µL
QT = k ∑ Ai ( P1 − P2 ) / µL
=
k1 A1 ( P1 − P2 ) k2 A2 ( P1 − P2 ) k3 A3 ( P1 − P2 )
+
+
µL
µL
µL
k=
∑k A
∑A
i
(2)
(3)
(4)
(5)
i
i
(6)
If all the beds have the same width the A ∝ h so k is the arithmetic average:
kA =
∑k h
∑ h i i
i
(7)
This equation is commonly used to determine the average permeability of a reservoir
from core analysis.
Radial Circular Flow
This is the case of several superimposed layers flowing simultaneously in the well.
Each layer supplies a rate of Qi. The total rate of flow is QT = Σ Qi. In Figure 5:
Permeability - Its Variations
re
rw
pe
pw
h1
Q1 K1
h2
Q2 K2
h3
Q3 K3
hT
Figure 5 Radial Flow in Parallel
Qi =
2πhi ki ( Pe − Pw )
r
µ ln e
rw
Q = ∑ Qi =
k=
(8)
2πhT k ( Pe − Pw ) 2π ( Pe − Pw )
=
( k1 + k2 + k3 + .......) (9)
r
r
µ ln e
µ ln e
rw
rw
∑h k
i i
hT
(10)
This value can be compared with that obtained through well flow tests or pressure
build-up tests.
2.2 Layers in Series - Linear Flow
In this case the reservoir may have been severely folded or faulted and the originally
horizontal layers are now vertical. It is assumed that the flow is now through each of
the layers towards the well. In this case, assuming a three layered system, the total
flow rate is constant through all of the layers and the total pressure drop is now the
sum of the pressure drops across each layer. Similarly, the total pressure drops across
each layer. Similarly, the total length is the sum of the lengths of the individual layers,
and the area open to flow is constant. Figure 6 illustrates.
Institute of Petroleum Engineering, Heriot-Watt University
P1
P2
P4
P3
Q
A
K1
K2
K3
L1
L2
L3
Figure 6 Linear Flow in Series
The average permeability of linear beds in series is obtained by adding the pressure
drop across each bed.
(P1 - P4) = (P1 - P2) + (P2 - P3) + (P3 - P4)
P1 − P4 =
QµL1 QµL2 QµL3 QµL
+
+
=
k1 A1
k2 A2
k3 A3
kA (11)
(12)
For beds of equal cross-sectional area then: k=
L
L
∑ ki
i
(13)
This mean is the harmonic average permeability kH.
Radial Circular Flow
In addition to natural lateral variations in permeability, wellbore damage can reduce
the permeability in the vicinity of the wellbore; also cleaning techniques, such as
acidising can increase the permeability in the vicinity of the wellbore.
For example in Figure 7:
Permeability - Its Variations
re
r1
rw
pw
k1
P1
k2
Pe
Figure 7 Radial Flow in Series
Q1 =
Q2 =
2πk 1 h( P1 − Pw
r
µ ln 1
rw
2πk2 h( Pe − P1 )
r
µ ln e
rw
(14)
(15)
2πkavg .h( Pe − Pw )
r
µ ln e
rw
(16)
Total flow:
QT =
Total pressure drop:
i.e.
( Pe − Pw ) = ( Pe − P1 ) + ( P1 − Pw ) re
r
r
Q2 µ ln e Qµ ln 1
rw
rw
r1
=
+
2πkavg.h
2πk2 h
2πk1h
(17)
QT µ ln
(18)
At steady-state flow:
QT = Q1 = Q2 Institute of Petroleum Engineering, Heriot-Watt University
(19)
ln re / rw ln r1 / rw ln re / r1
=
+
kavg.
k1
k2 kavg. =
ln re / rw
ln r1 / rw ln re / r1
+
k1
k2
(20)
(21)
3 MODELLING HETEROGENEOUS SYSTEMS
Permeability variations can often be traced from well to well throughout the reservoir,
there by enabling a layered reservoir system to be developed. Many fields demonstrate
this layering phenomenon, leading to very large variations in permeability.
For example Figure 8 shows the considerable variation in permeability for the various
sand units making up the Brent sands in the North Viking Graben area of the North sea.
Using an average value for the permeability can lead to large errors and misleading
results in reservoir modelling.
Lower
ness
11
10
Tarbert
Upper ness
Etive
Rannoch
9
8
7
Permeability (Darcies)
6
5
4
3
2
1
9200
9300
9400
Measured Depth (Feet)
9500
Figure 8 Permeability Variation in North Sea Field
In some cases it is not possible to correlate permeabilities from well to well, and it is
more difficult to put together a reservoir model to be used to examine flow behaviour
with such a reservoir.
Modelling reservoirs with average reservoir properties can only be valid if:
• reservoir sands are homogeneous
• random variations in reservoir properties occur across the sand
10
Permeability - Its Variations
• ordered distributions of the properties observed in one well do not correlate with
other wells.
Representation of permeabilities as a function of depth used to be presented on a log
permeability -versus- depth scale. Such a representation can lead to an observation
that an average permeability might be appropriate when in reality when presented in
a true linear form the differences in permeability are more distinct. Fluid behaviour
in reservoirs obeys a linear law, as against log permeability!
Figure 9 shows the permeability variation in a reservoir when plotted on a linear scale
as against the log scale, for a section of a reservoir. It would be easy to wrongly estimate
an average permeability looking at the log permeability presentation, which from the
linear permeability presentation such as average clearly cannot be interpreted.
Permeability Distributions
10
20
30
40
Log (Permeability mD)
20
40
Normal log scale
60
80
Thickness (ft)
1000 2000 3000 4000 5000
Permeability (mD)
20
40
60
Linear scale
80
Thickness (ft)
Figure 9 Permeability Distribution on a Log and Linear Scale.
The difference in, for example, the behaviour of a waterflood when modelled as a
homogeneous system as against a layered system is illustrated this process is shown
schematically by figure 10.
The example demonstrates that for these heterogeneous systems with large permeabilities contrasts, that in water flooding water flows preferentially down the high
permeability section.
The significance in these simplifications is demonstrated in the context of modelling
the behaviour of a waterflood, when modelling as a homogeneous system as against
Institute of Petroleum Engineering, Heriot-Watt University
11
a layered system. The illustration presented in the subsequent example is that of the
waterflooding of a reservoir with the permeability contrasts as illustrated in figure 9.
Figure 10 illustrates the process where injected water sweeps through the reservoir.
In the example given the mobility ratio of the process is less than 1. That is the injected water is less mobile than the displaced oil. If the reservoir was homogeneous
then the permeability can be considered as an average value and the displacement
simulated on a one dimensional basis. The key issues for the reservoir engineer are;
when does the water injected arrive at the producing wells and secondly what is the
oil recovery.
Gas
Limited flaring
Re-inject
qo sales
Separation
Injection
Pump
qwi
qwp
Purify /
Dump /
inject
qo + qwp
Sea Level
Seawater
for injection
Sea Bed
Reservoir
Figure 10 Water Injection Process
Figure 11 gives the one dimensional reservoir simulation process and figure 12 the
predicted outcome of oil production, water-cut and oil recovery. As is shown the
piston like displacement arrives at the production well around 2500 days after the
start of injection at which the oil recovery is around 48%. A very good project if this
is an accurate simulation of the process.
12
Permeability - Its Variations
_
k
average
Oil
Water
Figure 11 1-D Displacement Process
WATER CUT
1.0
0.8
0.6
1D
0.4
0.2
0
2000
4000
6000
TIME (days)
µw = 0.5 cp
OIL RATE
1.0
2D =
kv = kh
kv = 0.1kh
{
1D =
1D
0.5
0
2000
RECOVERY FACTOR
0.5
0.4
0.3
4000
6000
TIME (days)
49.7%
1D
0.2
0.1
0
2000
4000
6000
TIME (days)
Figure 12 Oil Rate, Water Cut and Recovery
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13
Figure 13 a,b, c and d shows the process for a layered case where the permeability
variations have been included in the simulation and the process is now two dimensional. The permeability distribution is as in figure 9 with the highest permeability
in the centre. Figure 13a shows the simulation after 600 days and the values on the
contour lines are values of water saturation. The figure clearly identifies the water
moving through the high permeability zone. In figure 13b after 1200 days water has
broken through into the producing well and this well is 'cutting' water over a short
interval. After 2400 days, figure 13c, the high permeability layer is taking the majority
of the water and a larger interval of the producing well is 'cutting' water.
0
50
20
40
60
70
600 Days
50
1000
2000
3000
0
µw = 0.5 cp
kv = kh
50
20
70
1200 Days
40
50
60
0
50%
20
70%
2400 Days
40
60
Impact of gravity 50%
1000
2000
3000
0
50%
µw = 0.5 cp
kv = kh
Attic oil
20
70%
40
4800 Days
50%
60
Figure 13 (a,b,c and d) Displacement simulation in layered reservoir
It is also interesting to note an increase in the water saturation along the base of the
reservoir figure 13c and 13d. This is due to the impact of gravity, the density difference
of the denser water causing the water to move towards the base of the reservoir. After
14
Permeability - Its Variations
4800 days this gravity segregation perspective is clearly seen as the low permeability
zone is being swept by this gravity impact. At the top of the reservoir no such benefit
is generated and oil is unswept. This unswept oil is sometimes termed attic oil.
Figure 14 gives the oil rate, water cut and recovery for the simulation where the
various permeability layers have been identified in the 2-D simulation. They show
that the project is not as attractive as that forecast by the previous 1-D simulation.
The two lines in the 2 D case show the impact of vertical permeability, a significant
perspective in relation to gravity flow.
WATER CUT
1.0
0.8
0.6
0.4
1D
2D
0.2
0
2000
4000
6000
µw = 0.5 cp
OIL RATE
1.0
2D =
2D
0
2000
RECOVERY FACTOR
0.5
0.4
0.3
kv = kh
kv = 0.1kh
{
1D =
1D
0.5
TIME (days)
4000
6000
49.7%
1D
TIME (days)
48.8%
2D
39.0%
0.2
0.1
0
2000
4000
6000
TIME (days)
Figure 14 Predictions for Homogeneous 1D and Heterogeneous Flooding 2D
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15
Clearly the main impact is that the field starts producing water after around 1200
days around half the time of the 1D prediction and the recovery at breakthrough is
only around 25%.
The example clearly demonstrates that for these heterogeneous systems with large
permeability contrasts, that in water flooding water flows preferentially down the
high permeability section.
Premature water breakthrough occurs resulting in deferment of oil production anticipated from an average value of permeability.
With such a result the reaction is what can be done to improve the process? The immediate reaction is not to complete the high permeability layers, forcing the fluids
through the more restricted lower permeabilities. This suggestion is illustrated in figure
15, where the simulation clearly shows that there is little impact. Near the injection
point there is evidence of some displacement into the lower permeability zone, but
once into the formation the water finds the easiest route the fluids move through the
central high permeability zone. The only way to impact the displacement would be
to reduce the permeability of the high value layer some distance into the formation.
The use of time setting polymers could provide such a fluid diversion. Clearly there
are technical risks associated with such a process.
0
20
1000
2000
3000
With high permeability layers not complete
50%
70%
40
50%
60
Figure 15 Water Injection Profile When High Permeability Layer Not Completed.
The recovery of the attic oil is a challenge. Gravity segregation based methods have
been suggested where the injection of a light fluid, for example nitrogen, would have
a similar impact on the unswept oil in the upper layers as the water has on the base
layers. In the example presented it has been assumed that there is strong pressure and
flow communication across the layers if this were not the case then the flow profiles
would be significantly different.
Although the example of water flooding has been used, the phenomena will also
occur in gas injection schemes, being therefore very relevant to the development of
gas condensate reservoirs by gas cycling.
The example has illustrated the importance of permeability contrast in a formation. The
topic is further covered in some depth in the chapter on immiscible displacement.
16
Permeability - Its Variations
Figure 16 a and b illustrate the geological process which result in the permeability
decreasing with depth (16a) and the permeability increasing with depth (16b) 3.
Considerable activity in the nineties was focused on the impact of a range of geological scenarios on permeability and the development of realistic reservoir simulation
models. The subject of upscaling, unheard of in the seventies is now an integral part
of building realistic reservoir flow models.
Coarser Sediment in
Shallow Turbulent Water
Sea Level
Increasing
Increasing
Wave Energy
Wave Energy
Sea Level
Coarser Sediment in
Shallow Turbulent Water
ain
e
Siz
n
itio
os
ep ize
D
us in S Advance
Inc
ion
eo ra
sit
an G
po of Bar
ult sing
e
m
Si crea
sD
Advance
ou
In
ne
of Bar
lta
u
Sim
Fine Sediment in
Deeper Quiet Water
Fine Sediment in
Deeper Quiet Water
r
gG
in
s
rea
GR
Profile (A)
GR
Profile (B)
GR
Profile (A)
Permeability
Injection
Injection
Permeability
Oil
Oil
Water
Depth
Depth
GR
Profile (B)
Water
Production
Production
(a) Favorable
(a) Favorable
Figure 16(a) Effect of Favourable Permeability in Waterflooding.
Levee
a
nokf B
siBoan
E
nrof
Er o si o
Channel
Migration
Channel
Migration
nk
Strong
Current
Strongarse
Co
Current
Permeability
Permeability
Levee
e
Weak
Fin
Current
on Growthinofe
Weak
siti
F
tion )
po
e
Point Bar
e
cre
Current
D
Ac e Lin
n
nd
o
a
m
i
l of (Ti
it
Growth
S
a
s
r
o
tion )
ate aBar
ce
ep
LPoint
e
cre
Ac e Lin
urf
dD
S
n
m
Sa
al
(Ti
ter
La face
r
Su
rse
a
Co
GR
GR
Injection
Injection
Depth
Depth
Oil
Oil
(a) Unfavorable
(a) Unfavorable
Production
Water
Production
Water
Figure 16(b) Effect of Unfavourable Permeability in Waterflooding 3.
Institute of Petroleum Engineering, Heriot-Watt University
17
REFERENCES
1. Van Veen, F.R., "Geology of the Leman Gas Field Petroleum and the Continental
Shelf of N.W Europe." (Woodland, A.W., ed.) Supplied Science Pub. Barkins
1975. 223
2. Warren, J.E., Price, K. S., "Flow in Heterogeneous Porous Media" Society of
Petroleum Engineering Journal 1961. 153 - 169.
3. Archer, J. S., Wall, C. C., Petroleum Engineering. Graham and Trotman 1988
London
18
10
Fluid Flow In Porous Media
CONTENTS
1 Introduction
2 Characterisation and Modelling of
Flow Patterns
2.1
Idealised Flow Patterns
2.2 General Case
2.2.1 Linear Horizontal Model of a Single Phase
Fluid
2.2.1.1 Linearisation Of Partial Differential Flow
Equation For Linear Flow
2.2.1.2 Conditions of Solution
2.2.2 The Radial Model
2.2.2.1 Range Of Application And Conditions Of
Solution
2.3 Characterisation of the Flow Regimes by
their Dependence on Time
3 Basic Solutions of the Constant
Terminal Rate Case for Radial
Models
3.1 The Steady State Solution
3.2 Non-Steady State Flow Regimes and
Dimensionless Variables
3.3 Unsteady State Solution
3.3.1 General Considerations
3.3.2 Hurst and Van Everdingen Solution
3.3.3 The Line Source Solution
3.3.3.1 Range of Application and Limitations to
Use
3.3.4 The Skin Factor
3.4 Semi-Steady-State Solution
3.4.1 Using The Initial Reservoir Pressure, Pi
3.4.2 Generalised Reservoir Geometry: Flowing Equation under Semi-Steady State
Conditions
3.5 The Application of the CTR Solution in
Well Testing
4. The Constant Terminal Pressure
Solution
5. Superposition
5.1 Effects of Multiple Wells
5.2
Principle of Superposition and Approximation of Variable - Rate Pressure
Histories
5.3 Effects of Rate Changes
5.4 Simulating Boundary Effects (Image
Wells)
6. Summary
Solutions to Exercises
LEARNING OBJECTIVES:
Having worked through this chapter the student will be able to:
• Understand the nature of fluid flow in a porous medium and the relation between
time, position and saturation
•Understand the assumptions used in the derivation of the diffusivity equation
• Understand the characterisation of the reservoir flow regime on the basis of
time
•Understand the application of the solutions of the diffusivity equation to steady
state flow, semi-steady state flow and transient flow
•Understand the use of the line source solution in radial systems to determine the
pressure at any point in a reservoir under transient flow conditions
•Understand the application of line source solution to multiple well/ multiple rate
histories in a transient flow reservoir
•Understand the basis of well test analysis, and use of the line source solution to
determine the reservoir permeability and skin factor
•Understand the application of semi-steady state solutions to determine reservoir
boundaries and their influence on flow rates.
10
Fluid Flow In Porous Media
1 Introduction
The ability to determine the productivity of a reservoir and the optimum strategy to
maximise the recovery relies on an understanding of the flow characteristics of the
reservoir and the fluid it contains. The physical means by which fluid diffuses through
a rock (or any other porous medium) depends on the interaction between the fluid (and
its properties) and the rock (and its properties). In terms of energy, the process may
at first sight appear to be similar in concept to the application of the general energy
equation to flow through pipes, although in this case the container through which the
fluid flows is made of very small tubes. It is precisely because of the geometry and
dimensions of the tubes that the application of the general energy equation would be
impossible: the description of a real pore network in a whole reservoir would be too
complex. Coupled with this is the interaction between the material of the tubes (or
pores) and the fluids. Surface chemistry effects start to dominate the flow when very
small tubes are considered and when multiphase flow occurs in them. Thus, complex
force fields are produced from not only the viscous pressure drop but also the effects
of surface tension and capillary pressure.
The combination of these factors dictates the nature of the fluid flow and one of the
initially unusual aspects is the time taken for pressure to change in the reservoir or for
fluid to migrate from one location to another. For instance, if a large body of water,
such as a swimming pool were drained, for all intents and purposes, the level of water
in the swimming pool would be the same as the water drained out. It would take an
appreciable amount of time for the water to drain (i.e. it would not be instantaneous),
but the pressure or level of the water in the pool would be the same at all locations of
the pool. The pressure in the pool would equilibrate almost immediately. Contrast this
with, for example, a water saturated reservoir rock in which the water could flow, but
where the permeability of the reservoir and the compressibility and viscosity of the
water dictated that the transfer of the water through the reservoir was not instantaneous
(as in a swimming pool), but took an appreciable time. In this case pressure changes
in one part of the reservoir may take days, even years to manifest themselves in other
parts of the reservoir. In this case, the flow regime would not be steady state while
the pressure was finding its equilibrium and a major problem, therefore, would be
that Darcy’s Law could not be applied until the flow regime became steady state. In
some way, the diffusion through the reservoir needs to be examined: Darcy’s Law
is one expression of that diffusion process, but time dependent scenarios must also
be examined.
To illustrate this, consider the following model of a linear reservoir with a well at
the left side (Figure 1).
Institute of Petroleum Engineering, Heriot-Watt University
10 vertical tubes, 100mm diameter, arranged linearly
Tube number 1
2
3
4
5
6
7
8
9
10
Initial
water
profile
Outlet
(constant flowrate)
Profile
after
time t
Interconnecting, small diameter pipes
height of water in tubes
top of tubes
0
-50
time, t after start of flow
-100
t=0
t=1
t=2
t=3
t=4
t=5
t=6
t=7
-150
-200
bottom of tubes -250
1
2
3
4
7
6
5
tube number
8
9
10
Figure 1 Model of a linear reservoir and the pressure response measured after different
times
Each tube contains water, the height of which represents the pressure at that part of
the reservoir. The tubes are connected to each other at the base by a small diameter
tube which restricts the flow. Under initial conditions, the height of the fluid is
identical in each of the tubes (assuming the model is level). The outlet at one end is
at a lower level than the model and when it is opened the fluid immediately drains
from the model and the level of the water in the tubes decreases. The energy to
drive this system is the potential energy stored in the height of the water columns:
there is no high pressure inlet to the model. As is shown in figure 1, to reduce the
pressure in the model, the fluid needs to be expelled, but because of the permeability
of the rock (the restrictions in the bottoms of the tubes) it takes time for the fluid in
the tubes nearest the outlet to move (or expand in the case of pressurised fluid in a
reservoir) and therefore it takes time for the pressure to change. When the flow is
started from the outlet, there is an immediate reduction in the pressure in tube 1 and
this pressure perturbation moves through the rest of the fluid at a rate dictated by
the rock permeability and fluid properties. This produces a variation in the pressure
along the model. The pressure profile takes time to develop from the outlet (at tube
1) to the tube farthest from the outlet (tube 10) and at time, t=1, the pressure in tube
10 is still equal to the pressure at the initial time, t=0. This is termed a transient flow
condition as the fluid is trying to reach pressure equilibrium. When the fluid in tube
10 starts to expand and flow, all of the fluid in the whole model is now expanding
and flowing to the outlet. Tube 10 represents the limit of the fluid volume: there are
no more tubes behind to supply fluid at the initial pressure. Therefore, as the pressure
10
Fluid Flow In Porous Media
perturbation moves through the model from tube 1 to tube 10, the rate of pressure
change in the fluid is not limited by the volume of the fluid: it is as if the volume of
fluid was infinite in extent. During the transient period, the reservoir is often referred
to as infinite acting.
On inspection, a profile has been developing across the tubes during the transient
period. At the end of the transient period, the fluid in all of tubes is expanding producing
a decline in the pressure in all of the tubes. The shape of the pressure profile across
all of the tubes remains essentially constant and as time continues, the profile sinks
through the model until the water in the tube nearest the outlet empties. During this
time, the water in the model has not been replaced so steady state conditions have not
been achieved, however, since the gradient between the pressures in each adjacent tube
is not changing, the system can be considered to be in pseudo-steady state or semisteady state: the pressure gradient is constant but the absolute pressure is declining.
This mimics the situation in a real reservoir where the pressure is perturbed around
a well and the pressure disturbance moves out into the rest of the reservoir until it
reaches the outer boundary. If this is sealing and no flow occurs across the boundary,
then the reservoir pressure will decline (neglecting any injection into the reservoir)
in a pseudo-steady state manner. If the boundary is nonsealing (i.e. it is the water
oil contact and the aquifer water is mobile) then the aquifer water will flow into the
reservoir and a steady state will be achieved if the flowrates match.
The flow described in this model is trivial, but it illustrates the problem of applying
Darcy’s Law to real reservoirs: the effect of time on flow may be considerable and
if only steady state flow relationships were available then either permeability of the
reservoir would remain unknown or unrealistic flow periods would be required to
measure an essentially simple rock property.
2 Characterisation and Modelling of Flow Patterns
The actual flow patterns in producing reservoirs are usually complex due mainly to
the following factors:
(i) The shapes of oil bearing formations and aquifers are quite irregular
(ii) Most oil-bearing and water bearing formations are highly hetereogenous with
respect to permeability, porosity and connate water saturation. The saturations
of the hydrocarbon phases can vary throughout the reservoir leading to different
relative permeabilities and therefore flow patterns
(iii)The wellbore usually deviates resulting in an irregular well pattern through the
pay zone
(iv)The production rates usually differ from well to well. In general, a high rate well
drains a larger radius than a lower rate well
(v) Many wells do not fully penetrate the pay zone or are not fully perforated
Institute of Petroleum Engineering, Heriot-Watt University
There are essentially two possibilities available to cope with complexities of actual
flow properties.
(i) The drainage area of the well, reservoir or aquifer is modelled fairly closely
by subdividing the formation into small blocks. This results in a complex series
of equations describing the fluid flow which are solved by numerical or seminumerical methods.
(ii) The drained area is modelled by a single block to preserve the global features and
inhomogeneities in the rock and fluid properties are averaged out or substituted
by a simple relationship or pattern of features (such as a fracture set, for example).
The simplifications allow the equations of flow to be solved analytically.
The analytical solutions will be examined in this chapter.
2.1 Idealised Flow Patterns
There are a number of idealised flow patterns representing fluid flow in a reservoir: linear, radial, hemispherical, spherical. The most important cases are the linear and
radial models since both of them can be used to describe the water encroachment
from an aquifer into a reservoir, and the radial model can be used to describe the flow
of fluid around the wellbore.
In the following sections, dealing mainly with oil, the compressibility of the flowing
fluid may depend on the pressure. It will always be assumed that the product of
compressibility and pressure, cP, is smaller than one, i.e. cP<<1. If it is not (as in the
case of a gas) then the pressure dependence of compressibility must be taken into
account.
2.2 General Case
Consider the co-ordinate system shown in figure 2. The X and Y coordinates form a
horizontal plane with the Z coordinate perpendicular to this plane. The flow velocity,
U, is a vector with components Ux, Uy, Uz.
Z
Y
U
Ux
Uz
Uy
X
Figure 2 The specification of the flow velocity in a Cartesian co-ordinate system
10
Fluid Flow In Porous Media
The components of the flow velocity vector, U are:
Ux = -(kx/µ)(δP/δx)
Uy = -(ky/µ)(δP/δy)
Uz = -(kz/µ)(δP/δz+ρg)
(2.1)
where
k = permeability (m2) in the direction of X, Y, Z. The Z direction has an elevation
term, ρg, included to account for the change in head.
P = pressure (Pa)
µ = viscosity (Pas)
ρ = density (kg/m3)
g = acceleration due to gravity (m/s2)
U = flow velocity (m/s) = (m3/s/m2)
These components are similar to Darcy’s law in each of the three directions.
2.2.1 Linear Horizontal Model of a Single Phase Fluid
In this geometry, the flow is considered to be along the axis (in the x direction) of
a cuboid of porous rock. The total length of the cuboid is L and fluid flows into the
rock at the left end (x=0) and exits at the right end (x=L). There is no flow in the
other directions at any time i.e. Uy = Uz =0 for all values of x, y, z and time, t (in a real
reservoir, there may be flows in different directions in different parts of the reservoir
and there may be cross flows from different layers within the reservoir). The rock is
100% saturated with the fluid.
The flow equations are:
 k   ∂P 
U x = −   
 µ   ∂x  δ (Uρ )
 δρ 
= −φ
;0 ≤ x ≤ L
 δt 
δx
(2.2a)
(2.‑2b)
where
k = permeability (in the X direction), (mD)
ρ = density, (kg/m3)
U = flow velocity (m/s)
t = time (s)
φ = porosity
µ = viscosity, Pas
P = pressure, Pa
x = distance, (m)
The latter equation is obtained from a mass balance as follows (figure 3):
Institute of Petroleum Engineering, Heriot-Watt University
flowrate, qout
dx
area, A
x=L
x+dx
porosity, φ
X axis
x
flowrate, qin
x=0
isometric view
dx
flowrate, q in
flowrate, qout
x=0
x
x+dx
x=L
X axis
plan view
Figure 3 Flow into and out of a cuboid of porous rock
In Figure 3, fluid flows into the end of the cuboid at position x=0, through the rock
only in the X direction and out of the cuboid at x=L. In the middle of the cuboid,
an element from position x to position x+dx is examined. The bulk volume of the
element is the product of the area, A and the length, dx, i.e. the bulk volume = A*dx.
The pore volume of the element is therefore the product of the bulk volume and the
porosity, φ, i.e. the pore volume = A*dx*φ. If the flow was steady state then the
flowrates into and out of the volume (qin and qout) would be identical and Darcy’s Law
would apply. If the flow rates vary from the inlet of the volume to the outlet, i.e. qin
≠ qout then either the fluid is accumulating in the element and qin > qout or the fluid is
depleting from the element qout > qin (which is possible in a pressurised system since
the pressure of the fluid in the element may reduce causing it to expand and produce
a higher flow rate out of the element). Therefore, there is a relationship between the
change in mass, m, along the cuboid and the change in density, ρ, over time as the
mass accumulates or depletes from any element. In terms of mass flowrate,
10
Fluid Flow In Porous Media
Mass flow rate through the area, A
= qρ ((m3/s)*(kg/m3) = kg/s)
Mass flow rate through the area, A at position x
= (qρ)x
Mass flow rate through the area, A at position x+dx = (qρ)x+dx
Mass flowrate into a volume element at x minus mass flowrate out of element at x +
dx
=(qρ)x - (qρ)x+ dx
The mass flow rate out of the element is also equal to the rate of change of mass
flow in the element,
i.e.
(qρ ) x + dx = (qρ ) x +
δ ( qρ )
* dx
δx
Therefore the change in mass flow rate = −
δ ( qρ )
* dx
δx
i.e. if the change in mass flowrate is positive it means the element is accumulating
mass; if the change is negative it is depleting mass.
This must equal the rate of change of mass in the element with a volume = A*dx*φ
The rate of change of mass is equal to
hence −
δ ( qρ ) 1
δρ
=φ
∂x A
δt
δρ
Aφdx
δt
since the flow velocity, U = q/A, this becomes
−
δ (Uρ )
δρ
=φ
∂x
δt
or
δ (Uρ )
δρ
= −φ
∂x
δt (2.2b)
Substituting the parameters of equation 2.2a in 2.2b gives
δ kρ δP
δρ
= −φ
δx µ ∂x
δt (2.3)
Equation 2.3 shows the areal change of pressure is linked to the change in density
over time. Realistically, it is pressure and time that can be measured successfully in
a laboratory or a reservoir, therefore a more useful relationship would be between
the change in pressure areally with the change in pressure through time. The density
can be related to the pressure by the isothermal compressibility, c, defined as:
Institute of Petroleum Engineering, Heriot-Watt University
c=−
1  δV 
V  δP  T
where V is the volume (m3) and P is the pressure (Pa).


The density equals mass per unit volume ρ =
c=−
m
, hence:
V
ρ δ ( m / ρ ) 1 δρ
=
m δP
ρ δP (Quotient Rule, constant mass system)
(2.4)
Since
δρ δρ δP
δP
=
= cρ
δt δP δt
δt (from above)
then
(2.5)
This is the partial differential equation for the linear flow of any single phase fluid
in a porous medium which relates the spatial variation in pressure to the temporal
variation in pressure. If it were applied to a laboratory core flood, it could describe
the pressure variation throughout the core from the initial start of the flood when the
flowrate was increased from zero to a steady rate (the transient period) as well as the
steady state condition when the flow into the core was balanced by the flow out of
the core. Inspection of the equation shows that it is non-linear because of the pressure
dependence of the density, compressibility and viscosity appearing in the coefficients
kρ
µ and φcρ. The pressure dependence of the coefficients must be removed before
simple solutions can be found, i.e. the equation must be linearised. A simple form of
linearisation applicable to the flow of liquids such as undersaturated oil is to assume
their compressibility is small and constant. More complex solutions are required for
more compressible fluids and gasses.
2.2.1.1 Linearisation Of Partial Differential Flow Equation For Linear
Flow
Assuming that the permeability and viscosity terms do not depend on location (i.e.
distance along the cuboid), then
δ  δP   φµcρ  δP
ρ
=
δx  δx   k  δt The left hand side can be expanded to:
10
(2.6)
10
Fluid Flow In Porous Media
 δ 2P 
δρ δP
+ ρ 2 
 δx 
δx δx
Using equation 2.4 and since
δρ δρ δP
=
δx δP δx
the above becomes
cρ(δP/δx)2 + ρ(δ2P/δx2).
Usually c(δP/δx)2 is neglected compared to δ2P/δx2 since the pressure gradient is
small, and substituting gives
δ 2 P  φµc  δP
=
δx 2  k  δt (2.7)
This is termed the linear diffusivity equation
The assumption is made that the compressibility is small and constant, therefore
φµc
the coefficients k are constant and the equation is linearised. In equation (2.7)
k
φµc is termed the diffusivity constant. For liquid flow, the above assumptions are
reasonable and have been applied frequently, but can be applied only when the
product of the compressibility and pressure is much less than 1, i.e. cP <<1.0. Thus
the requirement for small and constant compressibility. The compressibility in this
case is the saturation weighted compressibility, i.e. the effect of the oil, water and
formation compressibilities:
c = coSo + cwSwc + cf
(2.8)
where
c is the saturation weighted compressibility
co is the compressibility of oil
cw is the compressibility of the connate water
cf is the compressibility of the formation (pore volume)
So is the oil saturation
Swc is the connate water saturation
2.2.1.2 Conditions of Solution
The solution of the equation requires initial conditions and the boundary
conditions.
(i) Initial Solution Condition.
At time t = 0, the initial pressure, Pi, must be specified for every value of x.
Institute of Petroleum Engineering, Heriot-Watt University
11
(ii) Boundary Conditions.
At the end faces x = 0 and x = L, the flow rate or pressure must be specified for
every value of time, t.
Solutions of the linear diffusivity equation are needed when dealing with linear flow
from aquifers. For solutions dealing with well problems a radial model is required.
2.2.2 The Radial Model
Figure 4 illustrates the geometry of this model in which the flow occurs in horizontal
planes perpendicular to the Z axis (i.e. in planes parallel to the XY plane) within a
layer of constant height, h. The flow is radial and is either towards the Z axis or away
from it.
Z
Y
h
rw
re
X
Z
radial element
h
wellbore
qρr+dr
qρr
dr
r
section in the XZ plane
Figure 4 Radial horizontal flow geometry geometry
At a distance r from x-axis, the flow velocity, U is now radius dependent:
U = q/2πrh
12
(2.9)
10
Fluid Flow In Porous Media
From Darcy’s Law (taking account of the flow direction and the co-ordinate
direction):
U=
k δP
µ δr (2.10)
The mass balance gives:
δ ( qρ )
δρ
= 2πrhφ
δr
δt (2.11)
Eliminating U and q through equations 2.9 to 2.11 gives the non-linear equation:
1δ
k δP
δP
ρr
= φcρ
r δr µ δr
δt (2.12)
Making assumptions as for linear flow, linearises the equation to:
(2.13)
This is termed the radial diffusivity equation
2.2.2.1 Range Of Application And Conditions Of Solution
The two main systems to which the radial diffusivity equation can be applied are
water influx and wellbore production although there are others.
(a) In the case of water encroachment from an aquifer into a reservoir, the inner
boundary corresponds to the mean radius of the reservoir, the outer boundary to the
mean radius of the aquifer.
(b) In the case of the pressure regime around a wellbore, the inner boundary corresponds to the wellbore radius, rw, the outer boundary to the boundary of the drainage area. In general the wellbore radius, rw is a mathematical concept, however, the
following are widely treated as valid:
Open hole, drilled close to gauge
Well cased, cemented and perforated
Slotted liner with gravel pack
Out-of-gauge hole
:
:
:
:
rw = 1/2 drill bit diameter
rw = 1/2 drill bit diameter
rw = 1/2 outer diameter (OD) of the liner
rw = average radius from caliper log
The solution of the equation requires the initial conditions and the boundary
conditions.
(i) Initial Solution Condition.
At time t=0, the initial pressure, Pi, must be specified for every point of the range
of equation 2.13, i.e. in the reservoir or in the aquifer.
Institute of Petroleum Engineering, Heriot-Watt University
13
(ii) Boundary Conditions
The boundaries consist of the outer and inner boundaries. The number of solutions
depend on the number of boundary conditions, but in the main there are a few sensible
conditions representing the majority of reservoir performance.
Outer Boundary
(a) If there is no flow across the outer boundary it is a closed system and the flow
velocity, U will equal zero. The pressure gradient, δP/δr will also be zero
(b) If there is flow across the outer boundary, the reservoir pressure will
be maintained at a constant value equal to the initial reserv oir pressure, Pi.
Inner Boundary
There are two main cases for the inner boundary which represent either maintaining
a constant pressure or a constant flow rate. These are representative of possible flow
regimes in the reservoir during either water flooding or production from a well.
(a) Constant Terminal Rate Case (C.T.R.)
This can be applied to a wellbore in which the production rate of the well is held
constant and the pressure varies through time. It can also be applied to water
encroachment in which the influx rate of water from the aquifer into the
reservoir across the initial oil-water contact is constant.
(b) Constant Terminal Pressure Case (C.T.P.)
Applied to a wellbore, the flowrate is varied to maintain a constant bottom hole
pressure in the producing well. In the case of water influx, the pressure at the initial
oil water contact of the reservoir remains constant and the flow rate varies.
2.3 Characterisation of the Flow Regimes by their Dependence on Time
To apply the diffusivity equation to real reservoirs requires careful consideration
of the boundary conditions. It will be shown that for most practical purposes, the
solutions to the diffusivity equation can be grouped according to the flow regime
that they represent: steady-state, semi-steady-state (pseudo steady state) or unsteady
state (transient).
Steady-state refers to the situation in which the pressure and the rate distribution
in the reservoir remain constant with time. Unsteady state is the situation in which
the pressure and/or the flow rate vary with time. Semi-steady is a special case of
unsteady state that resembles steady-state flow. These differences in the flow regimes
have ramifications in practical reservoir engineering since working solutions to the
diffusivity equation are usually limited to a particular flow regime. For instance, in
a pressure build up test in a well, the determination of an accurate average reservoir
pressure will depend strongly on the flow regime the well is in and therefore which
working solution is used.
14
10
Fluid Flow In Porous Media
3 Basic Solutions of the Constant Terminal Rate
Case for Radial Models
In this flow regime, one of the conditions for solution of the diffusivity equation is that
the flow rate is constant. This can be applied to the flow of oil towards a full length
perforated well, and to the flow of water to a producing reservoir from an aquifer.
The flow can be described approximately as the radial flow of a single phase from the
outer radius ‘b’ of a right hollow cylinder towards its inner radius, ‘a’. It is assumed
that the cylinder consists of a homogeneous porous medium.
In the case of drainage by a well, ‘a’ is the radius of the well, rw and ‘b’ is the radius
of the external boundary, re. The flow rate, q at radius, r = rw is the production rate of
the well. In the case of natural water influx into a reservoir, ‘a’ is the mean reservoir
radius, ‘b’ is the mean aquifer radius, and q is the volume flow rate of water across
the initial oil-water contact.
The radial constant terminal rate case is determined by the following system of
equations:
1 δ δP
φµc δP
(r ) =
;a ≤ r ≤ b
r δr δr
k δt
(3.1)
 2πrkh   δP 
q=
;r = a

 µ   δr 
(3.2)
with the initial condition that the pressure at all points is constant
a ≤ r ≤ b, t = 0; P = Pi = constant
(3.3)
and the boundary conditions that at the wellbore the flowrate is constant after the
production starts
r = a, t ≥ 0 : q = constant
(3.4)
and at the outer boundary, the pressure is either a constant (and equal to the initial
pressure) in the case of pressure maintenance
r = b, t ≥ 0 : P = Pi = constant
(3.5a)
or there is a sealing boundary with no flow across it in which case the pressure
gradient at the boundary is zero
r = b, t ≥ 0:
δP
= 0
δr
(3.5b)
The solution of the equations 3.1 to 3.4 and 3.5a & equations 3.1 to 3.4 and 3.5b are
well known and can be referenced in “Pressure buildup and flow tests in wells” by
Institute of Petroleum Engineering, Heriot-Watt University
15
CS Matthews and DG Russell, SPE Monograph Volume 1. These are too complex for
most practical applications and asymptotic solutions which are fair approximations
of the general solution are used, i.e. simple solutions which approximate certain flow
regimes can be used. The problem is to identify accurately which flow regime and
therefore which asymptotic solution should be used. The steady state solution is the
simplest and is the same as Darcy’s Law. The non-steady state solutions involve a
time element and are conveniently expressed in dimensionless form.
3.1 The Steady State Solution
If a well is produced at a constant flow rate, q, and if the pressure at the external radius,
re is maintained constant, flow will finally stabilise to steady state conditions.
δP
= 0 for all values of radius,
i.e. flowrate, q = constant and the pressure gradient,
δ
t
r and time, t
therefore,
dr  2πkh 
δP dP
and the flow equation becomes q
=
=
 dP
r  µ 
δr dr
integrating between the limits rw and r gives:
qµ   r 
P − Pw = 
ln
 2πkh   rw 
(3.6)
Integrating between the limits rw and re gives:
qµ   re 
Pe − Pw = 
ln
 2πkh   rw 
(3.7)
which is identical to the relationship described for a radial system by Darcy’s Law.
In this case, the pressure at the external radius of the reservoir is required and the
only way to measure it in the reservoir would be to drill a well at the external radius.
This is uneconomic, therefore a mean reservoir pressure,P , is used. It is found from
routine bottom hole pressure measurements and well tests conducted on the wells
in a reservoir, it includes the effect of the area of influence of each well. In simple
terms, the volume drained by each well is used to weight the bottom hole pressure
measurements made in the well; all of the weighted pressures of all of the wells in
the reservoir are then averaged. Figure 5 shows a well in a reservoir and its area of
influence. Volumetrically, this volume is drained by the well and the mean reservoir
pressure,P , is related to the pressure, P of elements of volume, dV being drained.
The total volume is V.
16
10
Fluid Flow In Porous Media
initial pressure profile
wellbore
Pi
P
h
pressure profile due
to production rate, q
Pwf
rw
re
element of volume, dV, at radius, r and at pressure, P
Figure 5 Pressure distribution around a well
re
1
∫ PdV
V rw
(3.8a)
where dV = 2πrhφdr (3.8b)
P=
The volume of the well’s drainage zone, V = π(re2-rw2)hφ
and considering rw <<re, V ~ π re 2hφ 2
P= 2
r e
re
∫ Prdr
rw
(3.9)
from equation 3.6,
qµ   r 
P = Pw + 
ln
 2πkh   rw 
P=
2 re
qµ   r 
P +
ln
rdr
2 ∫ w
 2πkh   rw 
r e rw
P - Pw =
2  qµ  re  r 
∫ ln  rdr
r 2e  2πkh  rw  rw 
re
re 1  r 2 
2 qµ   1 2 r 
P - Pw = 2 
r
ln
−
∫   dr

rw  rw rw r  2 
r e  2πkh   2
P - Pw =
assuming
2  qµ    r 2e re r 2w rw   r 2e r 2w  
−
  ln − ln  −

r 2e  2πkh    2 rw 2 rw   4 4  
r 2w
is negligible
4
Institute of Petroleum Engineering, Heriot-Watt University
17
P - Pw =
2  qµ   r 2e re r 2e 
 ln − 
r 2e  2πkh   2 rw 4 
qµ   re 1 
P - Pw = 
ln −
 2πkh   rw 2 
(3.10)
EXERCISE 1
A well produces oil at a constant flowrate of 15 stock tank cubic metres per day (stm3/d).
Use the following data to calculate the permeability in milliDarcys (mD).
Data
porosity, φ
19%
formation volume factor for oil, Bo
1.3rm 3/stm 3 (reservoir cubic metres per
stock tank cubic metre)
net thickness of formation, h, 40m
viscosity of reservoir oil, µ 22x10-3 Pas
wellbore radius, rw 0.15m
external radius, re 350m
initial reservoir pressure, Pi 98.0bar
bottomhole flowing pressure, Pwf93.5bar
qreservoir = qstock tank x Bo
1bar = 105 Pa
EXERCISE 2
A well produces oil from a reservoir with an average reservoir pressure of 132.6bar.
The flowrate is 13stm3/day. Use the following data to calculate the permeability.
Data
porosity, φ,23%
formation volume factor for oil, Bo
1.36rm3/stm3
net thickness of formation, h 23m
viscosity of reservoir oil, µ
14x10-3 Pas
wellbore radius, rw 0.15m
external radius, re 210m
average reservoir pressure,
132.6bar
bottomhole flowing pressure, Pwf
125.0bar
18
10
Fluid Flow In Porous Media
EXERCISE 3
A reservoir is expected to produce at a stabilised bottomhole flowing pressure of 75.0
bar. Use the following reservoir data to calculate the flowrate in stock tank m3/day.
Data
porosity, φ
formation volume factor for oil, Bo
net thickness of formation, h
viscosity of reservoir oil, µ
wellbore radius, rw external radius, re average reservoir pressure,
bottomhole flowing pressure, Pwf
permeability, k
28%
1.41rm3/stm3
15m
21x10-3 Pas
0.15m
250m
83.0bar
75.0bar
125mD
3.2 Non-Steady State Flow Regimes and Dimensionless Variables
As mentioned previously, dimensionless forms of the diffusivity equation have found
wide application in the description of flow through porous media. They “normalise”
the equation for use with many different reservoirs and allow general solutions to
be found which can be applied to specific data to determine the specific solution for
a particular reservoir. In such a way, general plots of, for example, the difference in
pressure from the reservoir to the wellbore through time can be constructed which
can then be used to determine the actual pressure difference for a specific reservoir.
It should be noted that solutions for a radial flow reservoir can only be sensible if the
dimensionless variables and diffusivity equation have been developed for a radial
flow reservoir.
If the dimensionless variables are defined as:
dimensionless time, rD
:
dimensionless time, tD
:
dimensionless pressure, PD
(at a dimensionless radius
and at a dimensionless time)
:
rD =
r
rw
tD =
kt
φµcrw2
 2π kh 
PD ( rD ,t D ) = 
 ( Pi − Pr,t )
 qµ 
where
r = radius in question
rw = wellbore radius
k = permeability
t = time in question
φ = porosity
µ = viscosity
Institute of Petroleum Engineering, Heriot-Watt University
19
c = compressibility
h = thickness of the reservoir
Pi = initial reservoir pressure
Pr,t = pressure at the specified radius and time
then the radial diffusivity equation becomes
1 δ  δPD  δPD
 rD
=
rD δrD  δrD  δt D (3.11)
There are other definitions of dimensionless variables, such as a dimensionless external
radius, which may be used in particular instances.
3.3 Unsteady State Solution
The constant terminal rate (CTR) solution can be obtained in several forms, using
different assumptions and methods of mathematical analysis. The various solutions
overlap, and all of them have particular uses and limitations.
3.3.1 General Considerations
flowrate, q
zero flowrate
bottomhole flowing pressure, Pwf
time
(a)
Pi
transient
late transient
semi - steady state
(b)
time
Figure 6 Wellbore pressure response to a change in flowrate
Figures 6a and 6b show the response of a reservoir at a wellbore when a flow rate, q, is
suddenly applied. The pressure of the flowing fluid in the wellbore, Pwf falls from the
initially constant value, Pi (static equilibrium) through time and the constant terminal
rate (CTR) solution of the diffusivity equation describes this change as a function
of time. The CTR solution is therefore the equation of Pwf versus t for a constant
production rate for any value of the flowing time. The pressure decline, Figure 6(b),
20
10
Fluid Flow In Porous Media
can normally be divided into three sections depending on the value of the flowing
time and the geometry of the reservoir or part of the reservoir being drained by the
well. This figure represents the pressure change at the wellbore through time which
is equivalent to the pressure change (or change in the height of water) in the cylinder
nearest the outlet in the model represented in Figure 1.
Initially, the pressure response can be described using a transient solution which
assumes that the pressure response at the wellbore during this period is not affected
by the drainage boundary of the well and vice versa. This is referred to as the infinite
reservoir case, since during the transient flow period, the reservoir appears to be infinite
in extent with no limits to the fluid available to expand and drive the system.
The transient period is followed by the late-transient when the boundaries start to
affect the pressure response. This is analogous to the pressure disturbance having
moved along the line of tubes in the model in figure 1. The nature of the boundaries
affects the type of solution used to describe the pressure change since a well may drain
an irregularly shaped area where the boundaries are not symmetrical or equidistant
from the well.
The next phase in the pressure decline is termed semi-steady state or pseudo steady
state where the shape of the pressure profile in the reservoir is not changing through
time and the wellbore pressure is declining at a constant rate. It is analogous to the
model depicted in figure 1 where the level of water in all of the tubes is falling and
no additional water is being added to tube 10 to maintain absolute pressure profile. If
the pressure profile developed in the reservoir around the well had remained constant,
true steady state conditions would have occurred and the steady state solutions as
mentioned in the previous section would have applied.
3.3.2 Hurst and Van Everdingen Solution
The constant terminal rate solution for all values of the flowing time was presented
by Hurst and van Everdingen in 19492. They solved the radial diffusivity equation
using the Laplace transform for both the constant terminal rate and constant terminal
pressure cases. The full equation contains, as one of its components, an infinite
summation of Bessel functions which are required to describe the complex wellbore
pressure response during the late transient period. Simple solutions can be obtained
for the transient and semi-steady state flow.
The solution describes pressure drop as a function of time and radius for fixed values
of external radius, re, and wellbore radius, rw, rock and fluid properties. It is expressed
in terms of dimensionless variables and parameters as:
PD = f(tD,rD,reD) (3.12)
where
tD = dimensionless time
rD = dimensionless radius
reD = re/rw = dimensionless external radius.
Institute of Petroleum Engineering, Heriot-Watt University
21
If the reservoir is fixed in size, i.e. reD is a particular value, then the dimensionless
pressure drop, PD, is a function of the dimensionless time, tD and dimensionless radius,
rD. The pressure in a particular reservoir case can then be calculated at any time and/or
radius. One of the most significant cases is at the wellbore since the pressure can be
measured routinely during production operations and compared to the theoretical
solutions. The determination of a reservoir pressure at a location remote from a well
may be required for reasons of technical interest, but unless a well is drilled at that
location, the actual value cannot be measured.
At the wellbore radius, r=rw (or rD=1.0)
PD = f(tD, reD)
(3.13)
−
t
2
∞
e α m D J1 (α m reD )
2t D
3
i.e. PD ( t D ) = 2 + lnreD − + 2 ∑ 2 2
2
reD
4
m =1 α m J1 (α m reD ) − J1 (α m )
2
(
)
(3.14)
where
αm are the roots of J1 (α m reD )Y1 (α m ) − Ji (α m )Y1 (α m reD ) = 0
J1 and Y1 are Bessel functions of the first and second kind.
This series has been evaluated for several values of dimensionless external radius,
reD, over a wide range of values of dimensionless time, tD. The results are presented
in the form of tables (from Chatas, AT3, “A Practical Treatment of non-steady state
Flow Problems in Reservoir Systems,” Pet. Eng. August 1953) in “Well Testing” by
J Lee, SPE Textbook series, Vol 1. A summary of the use of the tables for constant
terminal rate problems is as follows in Table 1. It reports the dimensionless pressure
at some dimensionless time for various configurations of reservoir. It is the solution
to equation 3.14.
22
10
Fluid Flow In Porous Media
Table
2
3
Presents
Valid for
i
PD as a function of tD <1000 (from table)
infinite acting reservoirs
ii
PD ≅ 2
iii
PD ≅ 0.5 (intD + 0.80907) for 100< tD <0.25 reD2
(an extension of the table)
infinite acting reservoirs
iv
PD as a function of tD <0.25 reD2 (from table)
finite reservoirs
i
PD as a function of tD for 1.5< reD2 <10 (from table)
finite reservoirs, but if the
value of tD is smaller
than that listed for a given
value of reD then the
reservoir is infinite acting
and therefore table 2 is used.
ii
PD =
tD
for tD <0.01 (an extension of the table)
π
4
4
2
2( t D + 0.25) (3reD − 4reD lnreD − 2reD − 1)
−
2
2
2
reD − 1
4( reD
− 1)
infinite acting reservoirs
finite reservoirs
2
25 < t D and 0.25reD
< tD
iii
PD ≅
2t D
3
2
+ lnreD − for reD
>> 1
2
reD
4
finite reservoirs
The solutions summarised in table 1 are applicable to a well flowing at a constant
rate or to a reservoir and aquifer with a constant flowrate across the oil water contact.
Most problems involving flow at a well involve relationship 2(iii) and 3(iii). It can
be seen that in using these solutions, the pressure can be calculated anywhere in
the reservoir as long as the flow rate is known. If the pressure in the reservoir at a
location where the flow rate is unknown is required then an alternative solution is
needed (the Line Source solution).
EXERCISE 4
A reservoir at an initial pressure, Pi of 83.0bar produces to a well 15cm in diameter.
The reservoir external radius is 150m. Use the following data to calculate the pressure
at the wellbore after 0.01 hour, 0.1 hour, 1 hour, 10 hours and 100hours of production
at 23stm3/d
Data
porosity, φ21%
formation volume factor for oil, Bo
1.13rm3/stm3
net thickness of formation, h 53m
viscosity of reservoir oil, µ
10x10-3 Pas
wellbore radius, rw 0.15m
external radius, re 150m
initial reservoir pressure, Pi83.0bar
permeability, k
140mD
compressibility, c
0.2x10-7Pa-1
Institute of Petroleum Engineering, Heriot-Watt University
23
EXERCISE 5
An experiment on a cylindrical sand pack is conducted to examine the wellbore pressure
decline. The sand pack is filled with pressurised fluid which is withdrawn from the
wellbore at a constant flowrate of 0.1m3/d. There is no flow at the external boundary.
Calculate the wellbore pressure at times 0.001 hour, 0.005 hour and 0.1 hour after
the start of production. The figure below indicates the sand pack.
fluid production
flow to the
wellbore
closed top, bottom
and side
sand pack with
fluid filled pore space
Data
porosity, φ25%
net thickness of formation, h 0.2m
viscosity of fluid, µ 2x10-3 Pas
wellbore radius, rw 0.2m
external radius, re 2m
initial reservoir pressure, Pi2bar
permeability, k
1200mD
compressibility, c
0.15x10-7Pa-1
EXERCISE 6
A discovery well is put on test and flows at 2.9stm3/d. Using the following data. calculate
the bottomhole flowing pressure after 5 minutes production.
Data
porosity, φ
17%
net thickness of formation, h 40m
viscosity of reservoir oil, µ 14x10-3 Pas
formation volume factor of oil, Bo
1.27rm3/stm3
wellbore radius, rw 0.15m
external radius, re 900m
initial reservoir pressure, Pi200bar
permeability, k
150mD
compressibility, c
0.9x10-9Pa-1
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10
Fluid Flow In Porous Media
3.3.3 The Line Source Solution
This solution assumes that the radius of the wellbore is vanishingly small relative to
the mean radius of the reservoir. It allows the calculation of the pressure at any point
in an unbounded reservoir using the flowrate at the well. The benefits are clear in
that no flow rates other than those measured in the producing well are required and
from which the pressure at any location can be calculated. The disadvantage is that
the solution works for infinite acting reservoirs only and if barriers are met, then the
solution fails to represent the true flow regime. The technique of superposition can
be used to combine the effect of more than one well in an infinite acting reservoir
and this technique can represent the effect of a barrier. The barrier is equivalent to the
pressure disturbance produced by a second, imaginary well producing at the same rate
and having the same production history as the real well with both these wells in an
infinite acting reservoir. This solution has found a lot of use in well test analysis.
In constant terminal rate problems, the flowrate at the well was given by
 2πrhk   δP 
q=

 µ   δr  r = r w
(3.15)
and for a line source, the following boundary condition must hold:
lim
δp
qµ
2y =
y → 0 δy 2πkh for time, t > 0.
Using the Boltzman Transformation
y=
φµcr 2
4kt and substituting into the diffusivity equation
φµc δP 
 1 δ δP
(r ) =
 r δr δr
k δt 
gives
y
d 2 p dp
+ (1 + y) = 0
dy 2 dy
with the boundary conditions
p → pi as y → ∝
lim
δp
qµ
2y =
y → 0 δy 2πkh
If p′ =
dp
dy then
Institute of Petroleum Engineering, Heriot-Watt University
25
y
dp′
+ (1 + y)p′ = 0
dy
Separating the variables and integrating gives
lnp’ = -lny - y +C
i.e. p′ =
dp C1 − y
=
e dy y
(3.16)
where C and C1 are constants of integration. Since
lim
δp
qµ
lim
2y =
=
2C1e − y
y → 0 δy 2πkh y → 0
then C1 =
qµ
and equation 3.16 becomes
4πkh
dp
qµ e − y
=
dy 4πkh y which is integrated to give
y
qµ e − y
p=
dy + C 2
4πkh ∫∞ y
or
∞
qµ e − y
p=−
dy + C 2
4πkh ∫y y
which can be rewritten as
p=
qµ
Ei(-y) + C 2
4πkh
Applying the boundary condition that p → pi as y → ∝ then C2 = pi and the line
source solution is obtained:
p i − p(r,t) = −
qµ 
φµcr 2 
Ei()
4πkh 
4kt  (3.17)
The term Ei(-y) is the exponential integral of y (the Ei function) which is expressed
as
∞
e−y
dy
y
y
Ei( − y) = − ∫
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10
Fluid Flow In Porous Media
It can be calculated from the series
 yn 
Ei( − y) = γ + lny − 

 n!n 
where γ = 0.5772157 (Euler’s Constant). On inspection of the similarities in the Ei
function and the ln function, it can be seen that when y <0.01, Ei(-y) = g + lny and
the power terms can be neglected. Therefore,
Ei( − y) = ln(1.781y) = ln(γy)
(γ = 1.781 = eγ = e 0.5772157 )
Solutions to the exponential integral can be coded into a spreadsheet and used with
the line source solution. Practically, the exponential integral can be replaced by a
simpler logarithm function as long as it is representative of the pressure decline. The
25φµcr 2
t>
k
limitation that y<0.01 corresponds to time, t, from the start of production
. The equation can be applied anywhere in the reservoir, but is of significance at the
wellbore (i.e. for well test analysis) where typical values of wellbore radius, rw, and
reservoir fluid and rock parameters usually means that y<0.01 very shortly after
production starts. Therefore the line source solution can be approximated by
qµ  γφµcr 2 
P = Pi +
 ln

4πkh 
4kt 
or, since -ln(y) = ln(y-1)
P = Pi −
qµ 
4kt 
 ln

4πkh  γφµcr 2  (3.18)
qµ 
4kt 
 ln

4πkh  γφµcrw2  (3.19)
and if the pressure in the wellbore is of interest,
Pwf = Pi −
The values of exponential integral have been calculated and presented in Matthews
and Russel’s Monograph and are produced in Table 4. The table presents negative
values, i.e. -Ei(-y). For values of y<0.01, the ln approximation can be used. For values
>10.9, the decline in pressure calculated is negligible.
3.3.3.1 Range of Application and Limitations to Use
The Ei function has limitations on its application: it cannot represent the initial flow
into a wellbore since the assumption that the wellbore is a line is obviously not
the case and some time has to elapse for the relative size of the wellbore to have a
Institute of Petroleum Engineering, Heriot-Watt University
27
negligible effect on the flow and expansion of the fluid in the majority of the reservoir. The reservoir must also be infinite acting. Analysis of real reservoir performance has
shown that the Ei function is valid for:
(i)
flowing time, t >100 φµcrw2/k
(3.20)
where rw is the wellbore radius. The value of 100 has been derived form the analysis
of the responses of real reservoirs; it can be varied according to the nature of a specific
well and reservoir. The time involved here is not the same as the dimensionless time,
tD calculated for other models of fluid flow in a reservoir (e.g. the input parameters for
the Hurst and van Everdingen solutions require the dimensionless time at the radius
where the dimensionless pressure drop is required - this may be the wellbore and rw
would be used or it may be some other radius).
(ii) t < φµcre2/4k
(3.21)
where re is the external radius. The reservoir boundaries begin to effect the pressure
distribution in the reservoir after this time, the infinite acting period ends and the line
source solution does not represent the fluid flow.
EXERCISE 7
A well and reservoir are described by the following data:
Data
porosity, φ
19%
formation volume factor for oil, Bo
1.4rm3/stm3
net thickness of formation, h
100m
viscosity of reservoir oil, µ
1.4x10-3 Pas
compressibility, c2.2 x10-9Pa-1
permeability, k
100mD
wellbore radius, rw 0.15m
external radius, re 900m
initial reservoir pressure, Pi 400bar
well flowrate (constant)
159stm3/day =
skin factor 0
159
24x3600 stm3/second
Determine the following:
(1) the wellbore flowing pressure after 4 hours production
(2) the pressure in the reservoir at a radius of 9m after 4 hours production
(3) the pressure in the reservoir at a radius of 50m after 4 hours production
(4) the pressure in the reservoir at a radius of 50m after 50 hours production
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Fluid Flow In Porous Media
EXERCISE 8
A well flows at a constant rate of 20stm3/day. Calculate the bottomhole flowing pressure
at 8 hours after the start of production.
Data
porosity, φ25%
formation volume factor for oil, Bo
1.32rm3/stm3
net thickness of formation, h 33m
viscosity of reservoir oil, µ 22.0x10-3 Pas
compressibility, c
0.6x10-9Pa-1
permeability, k340mD
wellbore radius, rw 0.15m
external radius, re 650m
initial reservoir pressure, Pi 270bar
well flowrate (constant)20stm3/day
skin factor 0
EXERCISE 9
Two wells are drilled into a reservoir. Well 1 is put on production at 20stm3 /day. Well
2 is kept shut in. Using the data given, calculate how long it will take for the pressure
in well 2 to drop by 0.5bar caused by the production in well 1. Well 2 is 50m from well 1.
Data
porosity, φ
18%
formation volume factor for oil, Bo
1.21rm3/stm3
net thickness of formation, h 20m
viscosity of reservoir oil, µ
0.8x10-3 Pas
compressibility, c43x10-9Pa-1
permeability, k85mD
wellbore radius, rw 0.15m
external radius, re 1950m
initial reservoir pressure, Pi 210bar
well flowrate (constant)20stm3/day
skin factor 0
Distance well 1 to well 250m
Institute of Petroleum Engineering, Heriot-Watt University
29
EXERCISE 10
A well in a reservoir has a very low production rate of 2stm3/day. Calculate the flowing
bottomhole pressure after 2 years production.
Data
porosity, φ
16%
formation volume factor for oil, Bo
1.13rm3/stm3
net thickness of formation, h
10m
viscosity of reservoir oil, µ 5x10-3 Pas
compressibility, c
14x10-9Pa-1
permeability, k
10mD
wellbore radius, rw 0.15m
external radius, re 780m
initial reservoir pressure, Pi 86bar
well flowrate (constant)2stm3/day
skin factor 0
EXERCISE 11
A well is put on production at 15stm3/day. The following well and reservoir data are
relevant.
Data
porosity, φ21%
formation volume factor for oil, Bo
1.2rm3/stm3
net thickness of formation, h 23m
viscosity of reservoir oil, µ 5x10-3 Pas
compressibility, c22 x10-9Pa-1
permeability, k
130mD
wellbore radius, rw 0.15m
external radius, re 800m
initial reservoir pressure, Pi
120bar
well flowrate (constant)
15stm3/day
skin factor 0
Determine the following:
(1) the wellbore flowing pressure after 2 hours production
(2) the pressure in the reservoir at a radius of 10m after 2 hours production
(3) the pressure in the reservoir at a radius of 20m after 2 hours production
(4) the pressure in the reservoir at a radius of 50m after 2 hours production
3.3.4 The Skin Factor
The analysis of fluid flow encountered thus far has assumed that a constant permeability
exists within the reservoir from the wellbore to the external boundary. In reality,
the rock around the wellbore can have higher or lower permeability than the rest of
the reservoir. This results from formation damage which may occur during drilling
30
10
Fluid Flow In Porous Media
and completion (where the wellbore fluids alter the wettability of the near wellbore
formation as fluid leaks off into it, or solids suspended in the drilling fluids are
deposited in the pore spaces and become trapped thereby physically hindering the
flow of fluid and reducing the permeability) or during production (where sand or
precipitates from the hydrocarbon fluids or from formation brines can alter wettability
and plug pore spaces). Alternatively, wellbores intersecting fractures may exhibit
enhanced permeabilities as the fractures offer much greater conductive paths to the
fluids around the wellbore, thus enhancing the permeability. This situation may also
be required as part of the reservoir management: hydraulic fractures or acidising
workovers are performed on wells to bypass zones of reduced permeability which
have developed during production.
bottomhole following pressure, Pwf
In these cases, the Ei equation fails to model the pressure drop in these wells properly
since it uses the assumption of uniform permeability throughout the drainage area
of the well up to the wellbore. Figure 7 shows the effect of a reduction in permeability
around a wellbore. The skin zone does not affect the pressures in the rest of the
formation remote from the wellbore, i.e. it is a local effect on the pressure drop at
the wellbore.
pressure profile if no skin zone was present
Pwf(no skin)
∆P skin
Pwf(skin)
actual pressure profile through skin zone
skin zone
permeability, Ks
rw
permeability, K
rs
radius, r
∆P skin = Pwf(no skin) - Pwf(skin)
Figure 7 Variation of the permeability around the wellbore changes the local pressure
profile
It can be shown that if the skin zone is considered equivalent to an altered zone of
uniform permeability, ks, with an outer radius, rs, the additional drop across this zone
(∆Ps) can be modelled by the steady-state radial flow equation. It is assumed that
after the pressure perturbation caused by the start of production has moved off into
the rest of the formation, the skin zone can be thought of as being in a steady state
flow regime. The pressure drop associated with the presence of a skin is therefore
the difference in the bottomhole flowing pressures at the well when skin is present
and when skin is not present, i.e.
Institute of Petroleum Engineering, Heriot-Watt University
31
∆Ps =
r 
r 
r 
qµ
qµ
qµ k
ln s  −
ln s  =
( − 1)ln s 
2πk s h  rw  2πkh  rw  2πkh k s
 rw 
(3.22)
Equation 22 simply states that the pressure drop in the altered zone is inversely
proportional to the permeability, ks rather than to the permeability, k of the rest of the
reservoir and that a correction to the pressure drop in this region must be made.
When this is included in the line source solution it gives the total pressure drop at
the wellbore:
Pi − Pwf = −
k
 rs 
qµ
qµ 
Ei( − y) + ∆Ps = −
Ei( − y) − 2 − 1 ln 
4πkh
4πkh 
 k s  rw 
(3.23)
If at the wellbore the logarithm approximation can be substituted for the Ei function,
then:
Pi − Pwf = −
k
 r 
qµ  γφµcrw2
) − 2 − 1 ln s 
ln(
4πkh 
4kt
 k s  rw 
(3.24)
k
 r
s =  − 1 ln s
 k s  rw (3.25)
A skin factor, s, can then be defined as:
and the drawdown defined as:
Pi − Pwf = −

qµ  γφµcrw2
ln(
) − 2s

4πkh 
4kt
 (3.26)
Equation 3.26 shows that a positive value of skin factor will indicate that the permeability
around the well has been reduced (by some form of formation damage). The absolute
value reflects the contrast between the skin zone permeability and the unaltered zone
permeability and the depth to which the damage extends into the formation. Part of the
essential information from a well test is the degree of formation damage (skin factor)
around a well caused by the drilling and completion activities. Alternatively, a well
may have a negative skin factor, i.e. the permeability of the skin zone may be higher
than that of the unaltered zone, caused by the creation of highly conductive fractures
or channels in the rock. The extent of the damage zone cannot be predicted accurately
and there may be variations vertically in the extent of the damage zone therefore this
simple model may not characterise the near wellbore permeability exactly.
An altered zone near a particular well affects only the pressure near that well, i.e. the
pressure in the unaltered formation away from the well is not affected by the existence
of the altered zone around the well.
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10
Fluid Flow In Porous Media
EXERCISE 12.
A discovery well is put on well test and flows at 286stm3/day. After 6 minutes production,
the well pressure has declined from an initial value of 227bar to 192bar. Given the
following data, calculate the pressure drop due to the skin, ∆Pskin , and the mechanical
skin factor.
Data
porosity, φ,28%
formation volume factor for oil, Bo
1.39rm3/stm3
net thickness of formation, h, 8.5m
viscosity of reservoir oil, µ
0.8x10-3 Pas
compressibility, c2.3 x10-9Pa-1
permeability, k
100mD
wellbore radius, rw 0.15m
external radius, re 6100m
initial reservoir pressure, Pi 227bar
bottomhole flowing pressure
after 6 minutes
192bar
well flowrate (constant)286stm3/day
EXERCISE 13
A reservoir and well are detailed in the following data. Use this data to calculate the
skin factor around the well after producing for 1.5 hours.
Data
porosity, φ23%
formation volume factor for oil, Bo
1.36rm3/stm3
net thickness of formation, h 63m
viscosity of reservoir oil, µ
1.6x10-3 Pas
compressibility, c
17 x10-9Pa-1
permeability, k243mD
wellbore radius, rw 0.15m
external radius, re 4000m
initial reservoir pressure, Pi 263.0bar
bottomhole flowing pressure
after 6 minutes260.5bar
well flowrate (constant)
120stm3/day
3.4 Semi-Steady-State Solution
Once the initial pressure perturbation produced by bringing a well onto production
has moved through the reservoir and met the boundaries, the infinite-acting nature of
Institute of Petroleum Engineering, Heriot-Watt University
33
the fluid changes to become finite acting. As stated previously, this is termed pseudo
steady state or semi steady state because the pressure drop with time is the same at
all points around the flowing well, i.e.
δP dP
=
= constant
δt dt
and where there is no flow across the outer boundary at r = re of the drainage zone,
i.e.
δP
= 0 at r = re
δr
In a similar manner to the steady state flow regime, the pressure difference between
the wellbore and, say, the external radius, or the pressure difference between the
wellbore pressure and the initial pressure, or the pressure difference between the
wellbore pressure and the average reservoir pressure can be calculated depending on
the physical measurements which have been taken. Usually, an average pressure is
known in a reservoir and this is used to determine the pressure drop. Figure 8 shows
the pressure profile in the reservoir and the values which may be relevant.
well with constant flow rate, q
calculated average pressure
pressure profile in reservoir
Pi
Pe
Pwf
rw
flowing pressure, P
height of formation
initial pressure
re
radius, r
Figure 8 Pressure profile in a reservoir under semi steady state flow conditions
Under semi steady state conditions, the pressure profile can be averaged over the
volume of the reservoir. This gives the average reservoir pressure at a particular time
in the stage of depletion of the reservoir. If there are several wells in a reservoir, each
well drains a portion of the total volume. For stabilised conditions, the volume drained
by each well is stable and in effect the whole reservoir can be subdivided into several
portions or cells. The average pressure in each cell can also be calculated from the
stabilised pressure profile. The calculation of the average pressure is determined from
the material balance of the initial pressure and volume of fluid and its isothermal
compressibility. The expansion of the fluid in each cell manifests itself as a volume,
or flow rate, at the well, i.e.
34
10
Fluid Flow In Porous Media
cV ( Pi − P ) = qt (3.27)
where V = pore volume of the radial cell; q = constant production rate; t = total
flowing time, c = isothermal compressibility.
q=
dV
dt
dV qdt
dt
=
=q
dP dP
dP
since c = −
q = − cV
1  dV 
V  dP  T
dP
dt
dP
q
=−
dt
cV
(3.28)
which, for the drainage of a radial cell, can be expressed as
dP
q
=−
dt
cπre2 hφ (3.29)
Substitution of equation 3.29 in the radial diffusivity equation
1 δ δP
φµc δP
(r ) =
r δr δr
k δt
gives
1 δ δP
φµc q
(r ) = −
r δr δr
k cπre2 hφ
which is
1 δ δP
qµ
(r ) = − 2
r δr δr
πre hk
Integration gives
Institute of Petroleum Engineering, Heriot-Watt University
35
dP
qµr 2
r
=−
+ C1
dr
2πre2 kh
(3.30)
dP
= 0 therefore
at the outer boundary the pressure gradient is zero, i.e. r
dr
qµ
Ci =
2πkh and substitution into equation 3.30 gives
dP
qµ  1 r 
=
 − 
dr 2πkh  r re2  (3.31)
When integrated, this gives
r
[P]
Pr
Pwf
qµ 
r2 
=
lnr
−
2πkh 
2re2  r
w
or
Pr − Pwf =
qµ  
r2  
rw2 
lnr
−
−
lnr
−
w
2πkh  
2re2  
2re2 
Pr − Pwf =
qµ  r
r2 
ln
−


2πkh  rw 2re2 
(3.32)
rw2
2
The term 2 re is considered negligible, and in the case where the pressure at the
external radius, re is considered (including the skin factor, s, around the well),
Pe − Pwf =
qµ  re 1 
 ln − + s
2πkh  rw 2  (3.33)
If the average pressure is used, then the volume weighted average pressure of the
drainage cell is calculated as previously in the steady state flow regime, i.e.
P=
2
r2e
re
∫ Prdr
rw
(3.9)
where rw and re are the wellbore and external radii as before, and P is the pressure in
each radial element, dr at a distance r from the centre of the wellbore. In this case,
r
2 qµ e  r
r2 
P − Pwf = 2
r
ln
−

 dr
re 2πkh r∫w  rw 2re2 
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10
Fluid Flow In Porous Media
and integrating gives
(i)
re
re
r
e
 r
 r2
r
1 r2
r
ln
dr
=
ln
−
2 r 
∫r  rw 
∫r r 2 dr
w


rw
w
w
r
re
re
2 qµ e  r
r2 
 r2
rer 2 
r
P
−
P
=
r
ln
−
dr
2

=  ln 2  qµ−   r
r wf re2 2πkh ∫  rw 2re2 
r
P − Pwf2 = r2w  r
−
 dr
w
∫4rln
rw
w
re 2πkh
rw 2re2 
rw 
r2 r
r2
r
≈ e ln e − e
re
 r
 r2
 e re 1 r2
r
2
r
4
w
re
re
r re ln
dr =  ln  − ∫
dr
 r
 r2
r  ∫  1 rr2 
2
r
r
2


w
w
rw
rw
∫r r  ln rw  dr =  2 ln rw  r rw− r∫ r 2 dr
(ii)
re
w
w
r
w
r 2

 r2  e
r
re
=  ln  −  
r 2
re
 r 2  re
re
r
2
 2 rw  rw  4  rw
 r4 
r3
re=
 ln  −  
dr
=
≈
 8r 2 
 2 rw  rw  4  rw
∫ 2re2
re2 re re2
 e  rw 8
rw
≈
ln −
2
2
r
r
r
2 rw 4
≈ e ln e − e
2 with
rw inclusion
4r e 3 of the skin
and substitution into equation 3.32
factor
gives
r
 r 4  e re2
r
r
re
 r 4  e re2 ∫ 2 dr =  2  ≈
r3 
2re
 8re  r w 8
q
µ
r
rw
dr
=
 32   ≈
P − Pwf = ∫ 2re2 ln e −
+
s
8r
 8
w kh 
2rπ
rw  4e  r w (3.34)
qµ 
re
3

− wf
Pwf
+ swhereas
 ln with−time,

The pressure differences (Prqµ
- Pwf), (Per- Pwf),3( P -P
) do=not change
e
2πkh  rw 4 
P
−
P
=
ln
−
+
s


Pr, Pe, Pw and do change.
wf
2πkh 
 q 
rw
4

 q 
P = Po +   (t o − t )
 cV 
EXERCISE 14P = Po + 
 (t o − t )
 cV in
 a reservoir which is in asemi-steady
A well has been on production
state flow regime.
qt 

P =flowing
Pi −  pressure,
For the following data, calculate
P
 qt  the bottomhole
wf
 
P = Pi −  
 cV 
cV
Data
formation volume factor for oil, Bo
1.62rm3/stm3


net thickness of formation, h 72m P - P = qµ  ln re − 3 + 2kt 


i
wf
qµ
re 3 -3 Pas2kt 2πkh  r
viscosity of reservoir oil, µ
4 φµcre 2 
 ln 1.2x10

w
Pi - Pwf =
− +
permeability, k 2πkh
4 φµcre 2 
 r123mD
w
wellbore radius, rw 0.15m
external radius, re 560m
qµ  re 3 
average reservoir pressure, 263.0bar
P −3 Pwf =
 ln − 


qµ
re 3
well flowrate (constant)216stm
/day
2πkh
 rw 4 
P − Pwf =
 ln − 
skin factor
2πkh  rw 0 4 
 re 3
 ln − 
 re 3
 rw Pi 4
− Reservoir Pressure,
3.4.1 Using The
 ln Initial
r
4


2
If the pressure dropwfrom initial pressure conditions
is required then equation
3.27


12 
re 3  1 2  re 
3  1   re 
may be written as:



3


2ln
−

=
ln


−
=
1
re 3  1   re 2  3  r 1 2   re2  r  2  2  2  ln r 
w

 2ln −  =  ln   −  =w
ln  − lnw e  
2
rw 2  2   rw 
2  2   rw 
  
 r 2
Institute of Petroleum Engineering, Heriot-Watt University
37   e 
 r 2 
e
1   rw 
  
=
ln  3
1   rw  
 
2
q
P = Po +   ( t o − t )
 cV 
(3.35)
qt
P = Pi −  
 cV  (3.36)
where q is the volume flow rate, c is the isothermal compressibility, V is the original
volume to is a reference time after which flow starts, t is the flowing time, Po is the
pressure at the reference time and P is the pressure at time t after the flow starts. P
is the average reservoir pressure after time, t. Subtracting equation 36 from equation
34 gives
Pi - Pwf =
qµ  re 3
2 kt 
ln
−
+
2πkh  rw 4 φµcre 2 
(3.37)
3.4.2 Generalised Reservoir Geometry: Flowing Equation under SemiSteady State Conditions
The key aspect of the radial flow equation under semi-steady state conditions is that
the boundary of the reservoir has an effect on the flow regime. The pressure decline
is influenced by the fact that there is a finite limit to the amount of fluid present in
the reservoir. The equations developed have been for radial geometries. However,
the semi-steady state flow regime in non-radial reservoirs can be examined by
the radial equation if the shape of the reservoir can be attributed to a factor which
encapsulates the relative position of a producing well in a volume of reservoir fluid.
This non-symmetrical geometry can be described by the Dietz shape factor (given
the symbol CA ) as follows.
Using the average reservoir pressure and assuming no skin factor, the pressure drop
is described by equation 34 as
P − Pwf =
qµ  re 3 
 ln − 
2πkh  rw 4   re 3 
 ln − 
rw 4  as
Expressing the terms 
38
(3.34)
10
Fluid Flow In Porous Media
2
2
 23  
1
re 3  1   re 
3  1   re 
2ln
−
=
ln
−
=
ln
−
ln



 
  

e 
2
rw 2  2   rw 
2  2   rw 
  
 r  2 
 e  
r 
1
= ln  w3 
2
 
 e2  
   




2
1  (πre ) 
= ln
2   2 23  
  πrw e  
 
 
The area drained (for a radial geometry) is πre2 therefore the logarithm term
becomes



 
2

(4A)
 ( 4πre )  = 

3

  (1.781 x 31.6 x rw2 ) 
2 2


  4πrw e  
 
 
where A is the area drained,
radial drainage area) =31.6.
and Dietz shape factor, CA (for a well in a
The final form of the generalised semi steady state inflow equation for an average
reservoir pressure is
P − Pwf =

qµ  1
4A
+ s
 ln
2
2πkh  2 γC A rw
 (3.38)
For the pressure drop between initial reservoir pressure conditions and some bottom
hole flowing pressure during semi steady state flow, equation 3.37 can be expressed
as
Pi − Pwf =
qµ 1
4A
2πkt
( ln
+
)
2
2πkh 2 γC A rw φµcA
or
Pwf = Pi −
qµ 1
4A
2πkt
( ln
+
)
2
2πkh 2 γC A rw φµcA
(3.39)
(3.40)
Institute of Petroleum Engineering, Heriot-Watt University
39
In a convenient dimensionless form, this can be expressed as
 kt  rw2
2πkh
1
4A
(P - Pwf ) = ln
+ 2π 

qµ
2 γC A rw2
 φµcrw2  A
or
PD t D =
1
4A
rw2
ln
+
2
π
t
D
2 γC A rw2
A (3.41)
The term involving the wellbore radius can be accommodated by using the following
modified dimensionless time
t DA = t D
rw2
A
in which case
PD t D =
1
4A
ln
+ 2πt DA
2 γC A rw2
The calculation of the Dietz shape factors and their limitations in use is presented
in Lee and reproduced in Table 5. There are a series of common simple shapes with
wells located close to certain barriers and the shape factors associated with them.
There are also values of tDA which indicate the use of the shape factors.
(i) The infinite system solution with less than 1% error for tDA < X in this case, X
is the value of the maximum elapsed time during which a reservoir is infinite acting
and the Ei function can be used. The time, t is calculated by
t < t DA
φµcA
k
This time is different to that quoted earlier in the section on the line source solution
and reflects the subjective decision as to the acceptable accuracy of the solution using
the Ei function.
(ii) The solution with less than 1% error for tDA > X in this case, the semi steady
state solution can be used with the results having an error less than 1% for an elapsed
time, t
t > t DA
φµcA
k
(iii) The solution which is exact for tDA > X in this case, the semi steady state
solution can be used with the results being exact for an elapsed time, t
t > t DA
40
φµcA
k
10
Fluid Flow In Porous Media
For a real reservoir under semi steady state conditions, the volume of reservoir drained
by a well can be determined from its flow rate, and this volume correlated to the
structural map of the reservoir to determine the shape. The values of shape factor can
then be used to locate the position of the well relative to the boundaries of the area
being drained. This is not an exact procedure and variations in the heterogeneity of
the reservoir can alter the pressure responses, however, it is an analytical step in the
characterisation of the reservoir.
EXERCISE 15
For each of the following geometries, calculate the time in hours for which the reservoir
is infinite acting
Geometry
1. Circle
2. Square
3. Quadrant of a square
Data
Area of reservoir, A
1618370m2
viscosity of reservoir oil, µ
1.0x10-3 Pas
permeability, k
100mD
porosity, φ,20%
compressibility, c
1.45 x10-9Pa-1
The times are calculated by the dimensionless time, diffusivity of the reservoir and the
area of the reservoir. The dimensionless time accounting for the reservoir drainage
area is found for the conditions in Table 5.
3.5 The Application of the CTR Solution in Well Testing
The study of fluid flow so far has related the pressure drop expected as a result of a
flow rate from a well in a reservoir. If the appropriate parameters, such as porosity,
permeability and fluid viscosity are known, then for a particular flow regime, such
as unsteady state, the pressure drop at a certain distance from the well at a certain
time after production starts can be calculated.
In reality, only flow rates and pressures at wells can be measured directly, and the most
important unknown factor in the diffusivity equation is the permeability. Therefore,
rather than calculate a pressure drop for a given set of conditions, the pressure drop
can be continuously measured and the permeability calculated.
This is part of the objectives of well testing and for illustration, the following example
calculates the permeability and skin factor for a well in a reservoir. It is important to
note that these examples all assume that an initially undisturbed reservoir is brought
on production, i.e. that there has been no previous production in the reservoir therefore
the pressure is at its initial value. In well test analysis, the previous history of a well
must be accounted for. The section on superposition will introduce the concepts of
a multi-rate history for a well.
Institute of Petroleum Engineering, Heriot-Watt University
41
EXERCISE 16
A well is tested by producing it at a constant flow rate of 238stm3/day (stock tank) for
a period of 100 hours. The reservoir data and flowing bottomhole pressures recorded
during the test are as follows:
Data
porosity, φ
18%
formation volume factor for oil, Bo
1.2rm3/stm3
net thickness of formation, h 6.1m
viscosity of reservoir oil, µ
1x10-3 Pas
compressibility, c2.18 x10-9Pa-1
wellbore radius, rw 0.1m
initial reservoir pressure, Pi 241.3bar
well flowrate (constant)238stm3/day
Time (hours)
0.0
1.0
2.0
3.0
4.0
5.0
7.5
10.0
15.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
100.0
Bottomhole
flowing pressure
(bar)
241.3
201.1
199.8
199.1
198.5
197.8
196.5
195.3
192.8
185.2
180.2
176.7
173.2
169.7
166.2
162.7
159.2
1. Calculate the effective permeability and skin factor of the well.
2. Make an estimate of the area being drained by the well and the Dietz shape
factor.
(Refer to solution to exercise 16 on page 93)
42
10
Fluid Flow In Porous Media
EXERCISE 17
An appraisal well is tested by producing at a constant rate of 200stm3/day for 107
hours. The following table of flowing bottomhole pressures and time were recorded
during the test. Using the data,
1. calculate the permeability and skin factor of the well
2. estimate the shape of the drainage area
Data
porosity, φ22%
formation volume factor for oil, Bo
1.3rm3/stm3
net thickness of formation, h 21m
viscosity of reservoir oil, µ
1.9x10-3 Pas
compressibility, c4.3 x10-9Pa-1
wellbore radius, rw 0.15m
initial reservoir pressure, Pi 378.7bar
well flowrate (constant)200stm3/day
Time (hours)
Bottomhole
flowing pressure
(bar)
0.0
1.1
2.1
3.2
4.3
5.4
8.0
10.7
16.1
21.4
32.1
42.8
53.5
64.2
74.9
85.6
96.3
107.0
378.7
326.41
324.7
323.8
323.1
322.1
320.5
318.8
315.5
312.2
305.6
300.8
296.0
291.2
286.3
281.5
276.7
271.9
Institute of Petroleum Engineering, Heriot-Watt University
43
4. The Constant Terminal Pressure Solution
In the constant terminal rate solution of the diffusivity equation, the rate is known to
be constant at some part of the reservoir and the pressures are calculated throughout
the reservoir. Conversely, in the constant terminal pressure solution, the pressure is
known to be constant at some point in the reservoir, and the cumulative flow at any
particular radius can be calculated. The constant terminal pressure solution is not as
confusing as the constant terminal rate solution simply because less is known about
it. Only one constant terminal pressure solution is available, so there is no decision
to be made over which to use as in the case of the constant terminal rate solutions.
Hurst and Van Everdingen produced the solutions for cases of an infinite radial system
with a constant pressure at the inner boundary and for constant pressure at the inner
boundary and no flow across the outer boundary. These can model, for example, a
wellbore whose bottomhole flowing pressure is held constant whilst flow occurs
in the reservoir, or they can model a reservoir surrounded by an aquifer. The same
geometrical and property conditions apply as for the constant terminal rate solutions:
a radial geometry of constant thickness with a well in the centre, and with fixed rock
and fluid properties throughout, however, in this case there is a pressure drop from
an initial pressure to some constant value. In the case of aquifer encroachment, the
radius of the “well” is the radius of the initial oil water contact. The constant terminal
pressure solution is most widely used for calculating the water-encroachment (natural
water influx) into the original oil and gas zone due to water drive in a reservoir. This
topic is covered in the chapter on water influx.
5. Superposition
In the analyses so far, the well flow rate has been instantly altered from zero to
some constant value. In reality, the well flowrates may vary widely during normal
production operations and of course the wells may be shut in for testing or some other
operational reason. The reservoir may also have more than a single well draining it
and consideration must be taken of this fact. In short, there may be some combination
of several wells in a reservoir and/or several flowrates at which each produce. The
calculation of reservoir pressures can still be done using the previous simple analytical
techniques if the solutions for each rate change, for example, are superposed on each
other. In other words, the total pressure drop at a wellbore can be calculated as the
sum of the effects of several flowrate changes within the well, or it may be the sum
of the effects caused by production from nearby wells.
There is also the possibility of using infinite acting solutions to mimic the effects
of barriers in the reservoir by using imaginary or image wells to produce a pressure
response similar to that caused by the barrier.
Mathematically, all linear differential equations fulfill the following conditions:
(i) if P is a solution, then C x P is also a solution, where C is a constant.
(ii) if both P1 and P2 are solutions, then P1 + P2 is also a solution.
44
10
Fluid Flow In Porous Media
These two properties form the basis for generating the constant terminal rate and
constant terminal pressure cases. The solutions may be added together to determine the
total effect on pressure, for example, from several applications of the equation. This is
illustrated if a typical problem is considered: that of multiple wells in a reservoir.
5.1 Effects of Multiple Wells
In a reservoir where more than one well is producing, the effect of each well’s pressure
perturbation on the reservoir is evaluated independently (i.e. as though the other wells
and their flow rate/ pressure history did not exist), then the pressure drop calculated
at a particular well at a particular time is the simple addition of all of the individual
effects superimposed one effect upon the other. Consider 3 wells, X, Y and Z, which
start to produce at the same time from an infinite acting reservoir (figure 9).
Well X
Well Y
Well Z
Flowrate, qx
Flowrate, qy
Flowrate, qz
rxy
rzy
∆P caused by well X
independent of well Y
or well Z
∆P caused by well Z
independent of well Y
or well Z
Initial Pressure, Pi
No Barrier Detected
∆P caused by well Y
independent of well X
or well Z
No Barrier Detected
Actual well pressure,
Pwf
Pressure in well Y after flowing time, t
Figure 9 The superposition of pressure changes from several wells
Superposition shows that:
(Pi-Pwf)Total at Well Y
= (Pi -P)Due to well X + (Pi-P)Due to well Y
+ (Pi-P)Due to well Z
Assuming unsteady state flow conditions, the line source solution can be used to
determine the pressure in well Y. It is assumed here that the logarithm function can
be used for well Y itself and that there will be a skin around the well. The effects of
wells X and Z can be described by the Ei function. There is no skin factor associated
with the calculation of pressure drop caused by these wells, since the pressure drop
of interest is at well Y (i.e. even if wells X and Z have non-zero skin factors, their
skin factors affect the pressure drop only around wells X and Z). The total pressure
drop is then:
Institute of Petroleum Engineering, Heriot-Watt University
45
(Pi − Pwf )total at well Y =
2


− q Y µ   γφµcrwY
ln

 − 2SY 
 
4πkh 
4kt 

+
2

− q X µ   φµcrXY
Ei


 
4πkh 
4kt  
+
2

− q Z µ   φµcrZY
Ei

 
4πkh 
4kt  
(5.1)
where
qY is the flowrate from well Y
qX is the flowrate from well X
qZ is the flowrate from well Z
rwY is the radius of well Y
rXY is the distance of well Y from the X well
rZY is the distance of well Z from the X well
the rest of the symbols have their usual meaning
This technique can be used to examine the effects of any number of wells in an infinite
acting reservoir. This could be to predict possible flowing well pressures amongst a
group of wells, or to deliberately use the interaction between wells to check reservoir
continuity. These interference tests and other extended well tests are designed to
characterise the reservoir areally rather than to determine only the permeability and
skin factor around individual wells.
EXERCISE 18
Two wells, well 1 and well 2, are drilled in an undeveloped reservoir. Well 1 is completed
and brought on production at 500stm3/day and produces for 40 days at which time Well
2 is completed and brought on production at 150stm3/day. Using the data provided,
calculate the pressure in Well 2 after it has produced for 10 days (and assuming Well
1 continues to produce at its flowrate). Therefore, Well 1 produces for 50days when
its pressure influence is calculated; Well 2 produces for 10 days when its pressure
influence is calculated.
The wells are 400m apart and the nearest boundary is 4000m from each well.
Data
porosity, φ,21%
formation volume factor for oil, Bo
1.4rm3/stm3
net thickness of formation, h, 36m
viscosity of reservoir oil, µ
0.7x10-3 Pas
compressibility, c8.7 x10-9Pa-1
permeability, k80mD
wellbore radius, rw (both wells)
0.15m
initial reservoir pressure, Pi
180.0bar
Well 1 flowrate (constant)500stm3/day
Well 2 flowrate (constant)
150stm3/day
skin factor around both wells
0
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Fluid Flow In Porous Media
5.2 Principle of Superposition and Approximation of Variable - Rate
Pressure Histories
The previous section illustrated the effect of the production from several wells in a
reservoir on the bottomhole flowing pressure of a particular well. Of equal interest
is the effect of several rate changes on the bottomhole pressure within a particular
well. This is a more realistic situation compared to those illustrated previously where
a well is simply brought on production at a constant flowrate for a specific period
of time. For instance, a newly completed well may have several rate changes during
initial cleanup after completion, then during production testing then finally during
production as rates are altered to match reservoir management requirements (for
example limiting the producing gas oil ratio during production). A simple pressure
and flowrate plot versus time would resemble Figure 10.
flowrate, q
q2
q1
(q2 - q1)
bottomhole flowing pressure, Pwf
t1
time, t
continuation of the effect of q1 in the reservoir
initial reservoir pressure
∆P associated with (q1 - 0)
∆P associated with (q2 - q1)
t1
time, t
Figure 10 Effect of flowrate changes on the bottomhole flowing pressure
The well has been brought onto production at an initial flowrate, q1. The bottomhole
flowing pressure has dropped through time (as described by the appropriate boundary
conditions and the flow regime) until at time t1, the flowrate has been increased to q2
and this change from q1 to q2 has altered the bottomhole flowing pressure (again as
described by the boundary conditions and the flow regime). The total (i.e. the real
bottomhole flowing pressure) is calculated by summing the pressure drops caused by
the flowrate q1 bringing the well on production, plus the pressure drop created by the
flowrate change q2 - q1 for any time after t1. During the first period (q1) the pressure
drop at a time, t, is described by
∆P( t ) = Pi - Pwf = ∆PD ( t )
qiµ
2πkh (5.2)
where ∆PD(t) is the dimensionless pressure drop at the well for the applicable boundary
condition.
Institute of Petroleum Engineering, Heriot-Watt University
47
For times greater than t1, the pressure drop is described by
∆P( t ) =
qiµ
(q − q)
∆PD ( t ) + 2
µ ∆PD ( t - t1 )
2πkh
2πkh
(5.3)
In this case, the pressure drop is that caused by the rate q1 over the duration t, plus
the pressure drop caused by the flowrate change q2 - q1 over the duration t - t1. In fact,
the pressure perturbation caused by q1 still exists in the reservoir and is still causing
an effect at the wellbore. On top of that, the next perturbation caused by flowrate
change q2 - q1 is added or superposed to give the total pressure drop ( at the wellbore
in this case).
In mathematical terms:
qiµ
2πkh (5.4)
qiµ
q − q1
∆PD (t) + 2
µ ∆PD (t - t1 )
2πkh
2πkh
(5.5)
0 ≤ t ≤ t1: ∆P(t) = ∆PD (t)
t > t1 : ∆P(t) =
In this 2nd equation, the first term is ∆P from flow at q1 : 2nd term is the incremental
term ∆P caused by increasing rate by an increment (q2-q1). These expressions are
valid regardless of whether q2 is larger or smaller than q1 so that even if the well is
shut in, the effects of the previous flowrate history are still valid.
The dimensionless pressure drop function depends as mentioned on the flow regime
and boundaries. If unsteady state is assumed and the line source solution applied,
then
∆PD =
Pi − Pwf
1
−φµcr 2 w
= − Ei (
)
qµ / 2πkh
2
4 kt
(5.6)
and the equation for time, t less than or equal to t1 would be as expected
∆P(t) = -
q1µ
−φµcr 2 w
Ei (
)
4πkh
4 kt
(5.7)
For times greater than t1 the additional pressure drop is added to give
∆P(t) = -
q1µ
−φµcr 2 w
(q2 − q1 ) µ
−φµcr 2 w
Ei (
) Ei (
)
4πkh
4 kt
4πkh
4 k (t − t1 )
(5.8)
This approach can be extended to many flowrate changes as illustrated in figure 11.
48
10
Fluid Flow In Porous Media
flowrate, q
different flow rates
q3
q4
q2
q1
Bottomhole flowing pressure, Pwf
time, t
pressure responses cused by rate changes
time, t
Figure 11 Multi rate pressure response in a wellbore
This leads to a general equation
∆P(t) =
q1µ
(q − q1 ) µ
(q − q 2 ) µ
∆PD (t) + 2
∆PD (t − t1 ) + 3
∆PD (t − t 2 ) + ...
2πkh
2πkh
2πkh
(q − q n −1 ) µ
+ n
∆PD (t − t n −1 )
2πkh
or
∆P(t) =
n

q1µ 
q i − q i −1
∆
P
(t)
+
∆PD (t − t i −1 )
∑
D

2πkh 
q1
i=2
 (5.9)
(5.10)
This is the general form of the principle of superposition for multi rate history wells.
For the specific case where the well is shut in and the pressure builds up, an additional
term is added to reflect this. Assuming that the well was shut in during the nth flowrate
period, the pressure builds during the shut in time, ∆t (i.e. ∆t starts from the instant
the well is shut in) back up towards the initial reservoir pressure according to
Pi − Pws =
n
 q µ
q1µ 
q i − q i −1
∆
P
(t)
+
∆PD (t n-1 − t i −1 + ∆t) − n-1 ∆PD ( ∆t)
∑
D

2πkh 
q1
1= 2
 2πkh
Institute of Petroleum Engineering, Heriot-Watt University
(5.11)
49
where
Pws is the shut in bottomhole pressure
tn-1 is the total producing time before shut in
∆t is the closed in time from the instant of shut in.
5.3 Effects of Rate Changes
The application of superposition to a well with several rate changes is illustrated as
follows. A well is known to have the flowrate history as presented in figure 12. It is
seen that the well is brought onto production at a flowrate, q1 and this is maintained
constant until time, t1 at which the flowrate is decreased to q2. This second flowrate
continues until time t2 when the flowrate is increased to q3. In terms of the reservoir, it
is assumed that the reservoir is in unsteady state flow regime and the line source can
be used to describe the pressure drop caused by the flowrate changes. In this case, the
first flow rate change is when the well is brought on production, so the change from
zero to q1 causes the first pressure perturbation to move into the reservoir.
It is the bottomhole flowing pressure, Pwf, that is of interest, and it can be calculated
using the line source solution. There is the possibility of a skin zone around the well,
so this must be accounted for. If no other flowrate change occurred, then eventually
unsteady state would give way to either semi steady state or steady state conditions
and the bottomhole flowing pressure would either decline at a steady rate or (if steady
state) would remain constant at some level. Assuming that this did not occur and that
unsteady state conditions still existed when the flowrate was changed to q2 then the
change q2 - q1 would cause a second pressure perturbation that would move out into
the reservoir, following the first one created when the well was put on production. The
reservoir is still in unsteady state conditions i.e. the first pressure perturbation has not
met any barriers so the reservoir fluid still reacts as if it were an infinite volume and
this behaviour is still causing a decline in the pressure at the wellbore even though a
second pressure perturbation has been created and is moving out into the reservoir.
The pressure drop due to this flowrate change can be calculated by the line source
solution and added to that produced by bringing the well onto production.
Eventually at time t2, the flowrate is changed again. This time, the pressure perturbation
caused by q3 -q2 follows the first and second perturbations into the reservoir, and
again, as long as the reservoir fluid still behaves as if it were infinite in volume, the
pressure drop created by this flowrate change can be added to the changes produced
by the others to give the total pressure drop.
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10
Fluid Flow In Porous Media
real well flowrate history
flowrate, q
q3
q1
q2
t1
time, t
t2
equivalent flowrate effects in the reservoir
q1
flowrate, q
time, t
t1
q2 - q1
time, t
q3 - q2
t2
time, t
Figure 12 The equivalence of flowrate changes in a reservoir
The pressure drop produced by bringing the well onto production is calculated by the
logarithmic approximation of the Ei function (it is assumed that the checks have been
made to the applicability of the Ei function and its logarithmic approximation).
∆P1 = ( Pi − Pwf )1

− q1µ   γφµcrw2 
=
 − 2s
ln
4πkh 
4kt 

The next pressure drop is that produced by the flowrate change q2 - q1 at time, t1. It
is still the bottomhole flowing pressure that is to be determined, therefore any skin
zone will still exist and still need to be accounted for. The second pressure drop is:
Institute of Petroleum Engineering, Heriot-Watt University
51
∆P2 = ( Pi − Pwf )2 =

−(q 2 - q1 ) µ   γφµcrw2 
ln
 − 2s
4πkh   4k(t - t1 ) 

And finally the third pressure drop is:
∆P3 = ( Pi − Pwf )3 =

−(q 3 - q 2 ) µ   γφµcrw2 
ln
 − 2s
4πkh   4k(t - t 2 ) 

The total pressure drop at the wellbore caused by all of the flowrate changes is
(Pi - Pwf )= ∆P1 + ∆P2 + ∆P3
EXERCISE 19
Two wells are brought on production in an undeveloped reservoir. Using the data below,
calculate the bottomhole flowing pressure in each well. Well 1 produces at 110stm3/day
for 27 days at which time Well 2 starts production at 180stm3/day and both produce
at their respective rates for a further 13 days when the bottomhole flowing pressures
are calculated. Therefore Well 1 produces for 40 days when its pressure influence
is calculated; Well 2 produces for 13 days when its pressure influence is calculated.
Data
porosity, φ,
19%
formation volume factor for oil, Bo
1.2rm3/stm3
net thickness of formation, h, 36m
viscosity of reservoir oil, µ
1x10-3 Pas
compressibility, c
10 x10-9Pa-1
permeability, k
110mD
wellbore radius, rw (both wells)
0.15m
external radius, re7000m
initial reservoir pressure, Pi 250.0bar
Well 1 flowrate (constant)
110stm3/day
Well 2 flowrate (constant)
180stm3/day
skin factor around both wells
0
The wells are 350m apart.
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Fluid Flow In Porous Media
EXERCISE 20
A well is completed in an undeveloped reservoir described by the data below. The well
flows for 6 days at 60 stm3/day and is then shut in for a day. Calculate the pressure
in an observation well 100m from the flowing well.
Data
porosity, φ,
19%
formation volume factor for oil, Bo
1.3rm3/stm3
net thickness of formation, h, 23m
viscosity of reservoir oil, µ 0.4x10-3 Pas
compressibility, c3 x10-9Pa-1
permeability, k50mD
wellbore radius, rw (both wells)
0.15m
external radius, re6000m
initial reservoir pressure, Pi
180.0bar
flowrate (constant)60stm3/day
skin factor around well
0
The observation well is 100m from the flowing well.
EXERCISE 21
A well in a reservoir is brought on production at a flowrate of 25stm3/day for 6 days.
The production rate is then increased to 75stm3/day for a further 4 days. Calculate,
using the data given, the bottomhole flowing pressure at the end of this period, i.e.
10 days.
Data
porosity, φ,21%
formation volume factor for oil, Bo
1.31rm3/stm3
net thickness of formation, h, 20m
viscosity of reservoir oil, µ 0.6x10-3 Pas
compressibility, c8 x10-9Pa-1
permeability, k75mD
wellbore radius, rw (both wells)
0.15m
external radius, re5000m
initial reservoir pressure, Pi 200.0bar
1st flowrate (constant)25stm3/day
1st flowrate period6days
2nd flowrate (constant)75stm3/day
2nd flow period4days
skin factor around well
0
5.4 Simulating Boundary Effects (Image Wells)
One of the intriguing possibilities of the application of the principle of superposition
to reservoir flow is in simulating reservoir boundaries. It is clear that when a well in
a reservoir starts production, there will be a period where the flow regime is unsteady
while the reservoir fluid reacts to the pressure perturbation as if the volume of the
reservoir was infinite (i.e. an infinite acting reservoir).
Institute of Petroleum Engineering, Heriot-Watt University
53
Once the boundaries are detected, there is a definite limit to the volume of fluid
available and the pressure response changes to match that of, for example, semi
steady state or steady state flow. This assumes that the pressure perturbation reaches
the areal boundary at the same time, i.e. if the well was in the centre of a circular
reservoir, the pressure perturbation would reach the external radius at all points
around the circumference at the same time (assuming homogeneous conditions). If
the well was not at the centre then some parts of the boundary would be detected
before all of the boundary was detected. This means that some of the reservoir fluid
is still in unsteady flow whilst other parts are changing to a different flow regime.
This would appear to render the use of the line source solution invalid, however, the
effect of the nearest boundary in an otherwise infinite acting reservoir has the same
effect as the interaction of the pressure perturbations of two wells next to each other
in an infinite acting reservoir.
Therefore if an imaginary well is placed at a distance from the real well equal to
twice the distance to the boundary, and the flowrate histories are identical, then the
principle of superposition can be used to couple the effect of the imaginary well
to the real well in order to calculate the real well’s bottomhole flowing pressure.
Figure 13 illustrates the problem and the effect of superposition. Figure 14 shows a
simplification of the model.
Infinite Acting Reservoir
Well with pressure perturbations
moving out into the reservoir
Pressure perturbations hit a fault
at the edge of the reservoir
Real Reservoir
Infinite Acting Reservoir
Well with pressure perturbations
moving out into the reservoir
Imaginary well mimics the
effect of the fault
Imposition of an Imaginary Well
Figure 13 The pressure effect of the barrier in the real reservoir can be represented by an
imaginary well
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10
Fluid Flow In Porous Media
Can be modelled as
drainage boundary
between wells
L
L
L
actual well
image well
reservoir
boundary
Figure 14 Representation of the boundary by a real well and an image well
This shows a plane-fault boundary in an otherwise infinite acting reservoir, as in
the top of figure 13. To determine the pressure response in the well, the line source
solution can be used until the pressure perturbation hits the fault. Thereafter there
are no solutions for this complex geometry. However, the reservoir can be modelled
with an infinite acting solution if a combination of wells in an infinite-acting system
that limit the drainage or flow around the boundary is found. The bottom of figure 13
indicates 1 image well with the same production rate as the actual well is positioned
such that the distance between it and the actual well is twice the distance to the fault
of the actual well. No flow occurs across the plane midway between the two wells
in the infinite-acting system, and the flow configuration in the drainage area of each
well is the same as the flow configuration for the actual well. Pressure communication
crosses the drainage boundary, but there is no fluid movement across it and the problem
of the flow regime has been resolved: the real well can be thought of as reacting to
the flowrate in it and to the pressure drop produced by the imaginary well on the
opposite side of the fault. The pressure drop is therefore:
Pi − Pwf = −

 −φµc(2L)2 
qµ  γφµcrw2
qµ
ln(
)
−
2s
−
Ei




 4πkh 

4πkh 
4kt
4kt
where the symbols have their usual meaning, and L is the distance from the real well
to the fault. The skin factor is used in the actual well, but not in the other (image)
well since it is the influence of this image well at a distance 2L from it that is of
interest.
Institute of Petroleum Engineering, Heriot-Watt University
55
EXERCISE 22
A well in a reservoir is produced at 120 stm3 /day for 50 days. It is 300m from a
fault. Using the data given, calculate the bottomhole flowing pressure in the well and
determine the effect of the fault on the bottomhole flowing pressure.
Data
porosity, φ,
19%
formation volume factor for oil, Bo
1.4rm3/stm3
net thickness of formation, h, 20m
viscosity of reservoir oil, µ
1x10-3 Pas
compressibility, c9 x10-9Pa-1
permeability, k
120mD
wellbore radius, rw
0.15m
external radius, re4000m
initial reservoir pressure, Pi 300.0bar
flowrate (constant)
120stm3/day
flowrate period, t50days
distance to fault, L300m
skin factor around well
0
There are other examples of the use of image wells to mimic the effect of boundaries
on flow. The larger networks require computer solution to relieve the tedium. To
complicate the simple fault boundary described earlier, consider the effect of a well
near the corner of a rectangular boundary. In this case, there are more image wells
required to balance the flow from the real well. Figure 15 shows the boundary and
the image wells.
image well 1
L2
L2
image well 3
L1
L1
R3
L1
L1
L2
Actual Well
L2
Boundary
image well 2
Figure 15 Representation of a well at the intersection of two boundaries
Four pressure drop terms are required to determine the pressure at the actual well.
The total pressure drop then is the sum of the pressure drops caused by all of the
wells at the actual well.
56
Figure 17
Representation of a well
surrounded by boundaries
10
Fluid Flow In Porous Media
Pi - Pwf = (∆P)rw + (∆P)2L1 + (∆P)2L2 + (∆P)r3
(Pi-Pwf)Total at the actual well
= (Pi -P)at the actual wellbore radius, rw
+ (Pi-P)Due to image well 1 at distance 2L1
+ (Pi-P)Due to image well 2 at distance 2L2
+ (Pi-P)Due to image well 3 at distance R3
The number and position of image wells can become complex.
actual well
image wells
i7
i6
i3
i2
i1
i4
i5
parallel equidistant
boundaries
Figure 16 Representation of an actual well between two barriers
In the apparently simple geometry of an actual well surrounded by two equidistant
barriers, such as illustrated in figure 16, the flow can be balanced as before by defining
image well, i1 on the right. On the left side, the barrier is balanced by image wells
i2 and i3 (because seen from i2, there is a barrier with 2 wells on the other side - a
real well and an image well). Now there is an imbalance in production across the
right barrier, so image wells i4 and i5 are added. This unbalances the left barrier and
image wells i6 and i7 are added. This should continue to infinity, however, since the
line source solution is known to have little influence above a certain distance from
the actual well, the number of image wells used can be fixed with no error in the
approximation.
Even more complex patterns can be devised. Mathews, Brons and Hazebroek
(Matthews, CS, Brons, F and Hazebroek, P, A Method for the Determination of Average
Pressure in a Bounded reservoir. Trans. AIME.201) studied the pressure behaviour of
wells completely surrounded by boundaries in rectangular shaped reservoirs. Figure
17 shows the network of wells set up to mimic the effect of the boundaries.
Institute of Petroleum Engineering, Heriot-Watt University
57
Boundary
EXERCISE 23
A well in a reservoir is producing close to two intersecting faults as shown below.
Using the data given, calculate the bottomhole flowing pressure after 32 days and
indicate the effect of the faults on the bottomhole flowing pressure. The production
rate is constant at 100stm3 /day
fault
L1
70m
fault
L2
well
120m
Data
porosity, φ,22%
formation volume factor for oil, Bo
1.5rm3/stm3
net thickness of formation, h, 36m
viscosity of reservoir oil, µ
1x10-3 Pas
compressibility, c9 x10-9Pa-1
permeability, k89mD
wellbore radius, rw
0.15m
external radius, re6000m
initial reservoir pressure, Pi 240.0bar
flowrate (constant)
100stm3/day
flowrate period, t32days
distance to fault, L170m
distance to fault, L2
120m
skin factor around well
0
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Fluid Flow In Porous Media
EXERCISE 24
A well is 80m due west of a north-south fault. From well tests, the skin factor is 5.0.
Calculate the pressure in the well after flowing at 80stm3/day for 10 days.
Data
porosity, φ,25%
formation volume factor for oil, Bo
1.13rm3/stm3
net thickness of formation, h, 23m
viscosity of reservoir oil, µ
1.1x10-3 Pas
compressibility, c
10.1 x10-9Pa-1
permeability, k
125mD
wellbore radius, rw
0.15m
external radius, re6000m
initial reservoir pressure, Pi 210.0bar
flowrate (constant)80stm3/day
flowrate period, t32days
distance to fault, L80m
skin factor around well5.0
6. Summary
The basic partial differential equation expressing the nature of fluid flow in a porous
rock has been illustrated in the context of petroleum reservoirs. Only oil and water
have been used as the simplifications for solving the diffusivity equation have
required the compressibility of the fluid to be small and constant. This is the reason
that the compressibility of the fluid in the examples has not changed with pressure
as would be expected. So, for instance, the same value of compressibility is used for
the fluid at the wellbore which may be under a lower pressure than the same fluid at,
for example, the external radius of the reservoir.
In gasses, the same diffusion process occurs, but the pressure dependence of the
gas is accommodated by various mathematical devices which again lead to simple
working solutions.
The assumptions made concerning the geological structure and the petrophysical
properties of the rock may appear radical: to assume a reservoir is circular, horizontal
and has identical permeability in all directions is a great simplification of the problem.
Yet these simple analytical solutions allow an appreciation of the role of the fluids
and the rock in a producing reservoir. For more realistic treatments of real reservoirs,
approximations to the diffusivity equation are made from which simple algebraic
relationships can be formed. This process is encapsulated in reservoir simulation
where the reservoir (with its properties) is subdivided into small blocks within which
the flow equations have been approximated by simple relationships. These can then
be solved by a process of iteration to achieve an acceptable result. The great potential
of this process is the ability to represent the shape of the reservoir and the changing
properties, vertically and horizontally, throughout the reservoir.
Institute of Petroleum Engineering, Heriot-Watt University
59
Figure 18 summarises the route taken through the analytical solutions for radial flow
regimes examined in this chapter. The number of solutions is mathematically infinite;
only a few are suitable for real reservoir problems.
The subject of Well Testing is considerable and is covered in the separate module
with that title.
Summary of the application of analytical solutions of the Diffusivity equation in this chapter
Flow regimes based on reservoir geometry
radial
traditional assumption
for most reservoirs
linear
spherical
hemispherical
specific reservoir
aquifer influxes
short time application
formation testing devices
thin layers
short time application
formation testing devices
thin layers
constant terminal rate
constant terminal pressure
Hurst and Van Everdingen solutions
pressure at a specified radius
pressure at a specified time for a known flowrate
flowrate at a specified radius and time
outer boundary sealing
outer boundary non-sealing
asymptotic solutions
(radial geometry)
based on flow regime
unsteady state
semi-steady state
line source solution
pressure based on wellbore flowrates
steady state
initial reservoir pressure solution
average reservoir pressure solution
Ei function
ln approximation to Ei function
sealed outer boundary
average reservoir pressure solution
initial reservoir pressure solution
Figure 18 Summary of basic solutions to diffusivity equation
60
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Fluid Flow In Porous Media
Solutions to Exercises
EXERCISE 1
A well produces oil at a constant flowrate of 15 stock tank cubic metres per day
(stm3/d). Use the following data to calculate the permeability in milliDarcys (mD).
Data
porosity, φ
formation volume factor for oil, Bo
net thickness of formation, h, viscosity of reservoir oil, µ wellbore radius, rw external radius, re initial reservoir pressure, Pi bottomhole flowing pressure, Pwf
qreservoir = qstock tank x Bo
1bar = 105 Pa
19%
1.3rm3/stm3 (reservoir cubic metres per stock tank cubic metre)
40m
22x10-3 Pas
0.15m
350m
98.0bar
93.5bar
Solution EXERCISE 1
the steady state inflow equation (accounting for fluid flowrate at reservoir conditions
in m3/s and pressure in Pa) is
Pe − Pwf =
qµBo  re 
ln 
2πkh  rw 
k=
r 
qµBo
ln e 
2π (Pe − Pwf )h  rw 
k=
15x22x10 −3 x1.3
350.00 
ln
5
24x3600x2πx(98.0 − 93.5)x10 x40  0.15 
= 341x10 −15 m 2
= 341mD
Institute of Petroleum Engineering, Heriot-Watt University
(3.7)
61
EXERCISE 2
A well produces oil from a reservoir with an average reservoir pressure of 132.6bar.
The flowrate is 13stm3/day. Use the following data to calculate the permeability.
Data
porosity, φ,
23%
3
formation volume factor for oil, Bo
1.36rm /stm3
net thickness of formation, h 23m
viscosity of reservoir oil, µ 14x10-3 Pas
wellbore radius, rw 0.15m
external radius, re 210m
average reservoir pressure, 132.6bar
bottomhole flowing pressure, Pwf
125.0bar
Solution EXERCISE 2
the steady state inflow equation (accounting for fluid flowrate at reservoir conditions
in m3/s and pressure in Pa) is
P − Pwf =
qµBo  re 1 
 ln − 
2πkh  rw 2 
k=
 re 1 
qµBo
 ln − 
2π P − Pwf h  rw 2 
k=
13 x14 x10 −3 x1.36
 ln 210.00 − 1 
5
24 x 3600 x 2π (132.6 − 125.0) x10 x 23 
0.15
2
(
)
k = 176 x10 −15 m 2
k = 176 mD
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Fluid Flow In Porous Media
EXERCISE 3
A reservoir is expected to produce at a stabilised bottomhole flowing pressure of 75.0
bar. Use the following reservoir data to calculate the flowrate in stock tank m3/day.
Data
porosity, φ
formation volume factor for oil, Bo
net thickness of formation, h viscosity of reservoir oil, µ wellbore radius, rw external radius, re average reservoir pressure, bottomhole flowing pressure, Pwf
permeability, k
28%
1.41rm3/stm3
15m
21x10-3 Pas
0.15m
250m
83.0bar
75.0bar
125mD
Solution EXERCISE 3
the steady state inflow equation (accounting for fluid flowrate at reservoir conditions
in m3/s and pressure in Pa) is
P − Pwf =
q=
q=
qµBo  re 1 
 ln − 
2πkh  rw 2 
( P − P )2πkh
wf
 r 1
µBo  ln e − 
 rw 2 
(83.0 − 75.0) x10 5 x 2π 125 x10 −15 x15
250.00 1 
21x10 −3 x1.41x  ln
−

0.15
2
q = 46 x10 −6 stm 3 / s
q = 4.0 stm 3 / day
Institute of Petroleum Engineering, Heriot-Watt University
63
EXERCISE 4
A reservoir at an initial pressure, Pi of 83.0bar produces to a well 15cm in diameter.
The reservoir external radius is 150m. Use the following data to calculate the pressure
at the wellbore after 0.01 hour, 0.1 hour, 1 hour, 10 hours and 100hours of production
at 23stm3/d
Data
porosity, φ
formation volume factor for oil, Bo
net thickness of formation, h viscosity of reservoir oil, µ wellbore radius, rw external radius, re initial reservoir pressure, Pi
permeability, k
compressibility, c
21%
1.13rm3/stm3
53m
10x10-3 Pas
0.15m
150m
83.0bar
140mD
0.2x10-7Pa-1
Solution EXERCISE 4
Using Hurst and Van Everdingen’s solution for Constant Terminal Rate, the
dimensionless external radius and the dimensionless time are calculated and used
with the appropriate solution to determine the dimensionless pressure drop. The
dimensionless pressure drop is then turned into the real pressure drop from which
the bottomhole flowing pressure is calculated.
reD =
re 150.00
=
= 1000
rw
0.15
tD =
kt
140x10 -15 xt
=
= 0.148t
φµcrw2 0.21x10x10 -3 x0.2x10 −7 x0.152
time
(hour)
time
(second)
0.01
0.10
1.00
10.00
100.00
36
360
3600
36000
360000
tD
(0.148t)
5.3
53.3
532.8
5328.0
53280.0
PD
1.3846
2.4146
3.5473
4.6949
5.8462
expression
table 2
table 2
table 2
0.5(lntD +0.80907)
0.5(lntD +0.80907)
the bottomhole flowing pressure, Pwf is found from re-arrangement of the dimensionless
 2πkh 
PD = 
 ( pi − pwf )
q
µ


pressure
. Accounting for the oil formation volume factor,
Bo, the bottomhole flowing pressure, pwf, is:
64
10
Fluid Flow In Porous Media
Pwf = Pi −
qµBo
PD
2πkh
Pwf at 0.01hour = 83.0x10 5 −
23x10x10 −3 x1.13
x1.3846 = 82.1x10 5 Pa
−15
24x3600x2π 140x10 x53
i.e. Pwf at 0.01 hour =82.1bar
similarly for the rest of the times
time
(hour)
PD
Pwf
(bar)
0.00
0.01
0.10
1.00
10.00
100.00
0
1.3846
2.4146
3.5473
4.6949
5.8462
83.0
82.1
81.4
80.7
80.0
79.2
Institute of Petroleum Engineering, Heriot-Watt University
65
EXERCISE 5
An experiment on a cylindrical sand pack is conducted to examine the wellbore
pressure decline. The sand pack is filled with pressurised fluid which is withdrawn
from the wellbore at a constant flowrate of 0.1m3/d. There is no flow at the external
boundary. Calculate the wellbore pressure at times 0.001 hour, 0.005 hour and 0.1
hour after the start of production. The figure below indicates the sand pack.
fluid production
flow to the
wellbore
closed top, bottom
and side
sand pack with
fluid filled pore space
Data
porosity, φ
net thickness of formation, h viscosity of fluid, µ wellbore radius, rw external radius, re initial reservoir pressure, Pi
permeability, k
compressibility, c
25%
0.2m
2x10-3 Pas
0.2m
2m
2bar
1200mD
0.15x10-7Pa-1
Solution EXERCISE 5
Using Hurst and Van Everdingen’s solution for CTR, the dimensionless external
radius and the dimensionless time are calculated and used with the appropriate
solution to determine the dimensionless pressure drop. The dimensionless pressure
drop is then turned into the real pressure drop from which the bottomhole flowing
pressure is calculated.
66
reD =
re 2.0
=
= 10
rw 0.2
tD =
kt
1200 x10 −15 xt
=
= 4t
φµcrw2 0.25 x 2 x10 −3 x 0.15 x10 −7 x 0.2 2
10
Fluid Flow In Porous Media
time
time
tD
(hour)
(second)
(4t)
PD
expression
0.001
3.6
14.4
1.808
table 2
0.005
18.0
72.0
3.048
table 3 since tD
is not less than
2
0.25reD i.e. not
infinite acting
0.100
360.0
1440.0
30.35
PD =
2t D
3
+ lnreD −
2
reD
4
the bottomhole flowing pressure, Pwf is
Pwf = Pi −
Pwf
qµ
PD
2πkh
at 0.001 hour
i.e. Pwf
= 2 x10 5 −
at 0.001 hour
0.1x 2 x10 −3
x1.814 = 1.97 x10 5 Pa
−15
24 x 3600 x 2π 1200 x10 x 0.2
= 1.97bar
similarly for the rest of the times
time
(hour)
0
0.001
0.005
0.100
PD
0
1.808
3.048
30.35
Pwf
(bar)
2.00
1.97
1.95
1.53
Institute of Petroleum Engineering, Heriot-Watt University
67
EXERCISE 6
A discovery well is put on test and flows at 2.9stm3/d. Using the following data.
calculate the bottomhole flowing pressure after 5 minutes production.
Data
porosity, φ
net thickness of formation, h viscosity of reservoir oil, µ formation volume factor of oil, Bo
wellbore radius, rw external radius, re initial reservoir pressure, Pi
permeability, k
compressibility, c
17%
40m
14x10-3 Pas
1.27rm3/stm3
0.15m
900m
200bar
150mD
0.9x10-9Pa-1
Solution EXERCISE 6
Using Hurst and Van Everdingen’s solution for CTR, the dimensionless external
radius and the dimensionless time are calculated and used with the appropriate
solution to determine the dimensionless pressure drop. The dimensionless pressure
drop is then turned into the real pressure drop from which the bottomhole flowing
pressure is calculated.
reD =
re 900
=
= 6000
rw 0.15
tD =
kt
150 x10 −15 x 5 x 60
=
= 934
φµcrw2 0.17 x14 x10 −3 x 0.9 x10 −9 x 0.152
time
time
(minutes)
(second)
5
300
tD
PD
expression
934
3.826
table 2
the bottomhole flowing pressure, Pwf is
Pwf = Pi −
Pwf
qµ
PD
2πkh
at 5 min utes
i.e. Pwf
= 200 x10 5 −
at 5 min utes
2.9 x14 x10 −3 x1.27
x 3.826 = 199.39 x10 5 Pa
−15
24 x 3600 x 2π 150 x10 x 40
= 199.4bar
similarly for the rest of the times
Line Source Solution to CTR
68
10
Fluid Flow In Porous Media
EXERCISE 7
A well and reservoir are described by the following data:
Data
porosity, φ
formation volume factor for oil, Bo
net thickness of formation, h viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw external radius, re initial reservoir pressure, Pi 19%
1.4rm3/stm3
100m
1.4x10-3 Pas
2.2 x10-9Pa-1
100mD
0.15m
900m
400bar
well flowrate (constant)
skin factor 159stm3/day =
stm3/second
24x3600
0
159
Determine the following:
(1) the wellbore flowing pressure after 4 hours production
(2) the pressure in the reservoir at a radius of 9m after 4 hours production
(3) the pressure in the reservoir at a radius of 50m after 4 hours production
(4) the pressure in the reservoir at a radius of 50m after 50 hours production
Solution EXERCISE 7
The line source solution is used to determine the pressures required at the specified
radii and at the specified times (i.e. using the flowrate measured at the wellbore, the
pressures at the other radii and times are calculated by the line source solution). SI
units will be used so time will be converted to seconds. Checks are made to ensure
that:
(i)
there has been adequate time since the start of production to allow the line
source solution to be accurate
(ii) the reservoir is infinite acting.
Thereafter, the choice of Ei function or ln approximation to the Ei function has to
be made.
A. Check Ei applicability
line source not accurate until
Institute of Petroleum Engineering, Heriot-Watt University
69
t>
100φµcrw2
k
100x0.19x1.4x10 -3 x2.2x10 −9 x0.152
t>
100x10 -15
(3.20)
t >13.2s
time is 4 hours, therefore line source is applicable.
B. Check reservoir is infinite acting
the reservoir is infinite acting if the time,
t<
φµcre2
4k (3.21)
i.e.
0.19x1.4x10 −3 x2.2x10 −9 x900
t<
4x100x10 -15
2
t < 1185030s
t < 329 hours
therefore line source solution is applicable.
(1) the bottomhole flowing pressure after 4 hours production, Pwf at 4 hours
(i)
check ln approximation to Ei function
25φµcrw2
t>
k
the ln approximation is valid if the time,
t>
25x0.19x1.4x10 −3 x2.2x10 −9 x0.15
100x10 -15
2
t > 3.3s
therefore ln approximation is valid.
Pwf = Pi +
qµBo  γφµcrw2 
ln

4πkh  4kt  (taking account of the conversion from stock
(ii)
tank to reservoir conditions via the formation volume factor for oil, Bo, flow rates
in reservoir m3/s and pressures in Pascal).
70
10
Fluid Flow In Porous Media
qµBo
159x1.4x10 −3 x1.4
=
= 28703
4πkh 24x3600x4πx100x10 −15 x100
φµcr 2 0.19x1.4x10 −3 x2.2x10 −9 r 2
=
= 101597 x10 −9 r 2
-15
4kt
4x100x10 x4x3600
Pwf = 400x105 + 28703xln(1.781x 101597x10-9x0.152)
= 400x105 - 356249
= 39643751Pa
=396.4bar
(2) the pressure after 4 hours production at a radius of 9m from the wellbore
(i) check ln approximation to Ei function
25φµcr 2
t>
k
the ln approximation is valid if the time,
t>
25x0.19x1.4x10 −3 x2.2x10 −9 x92
100x10 -15
t > 11850s
t > 3.3hours
therefore ln approximation is valid.
2
qµBo  γφµcr 
P = Pi +
ln
4πkh  4kt  (taking account of the conversion from stock tank
(ii)
to reservoir conditions via the formation volume factor for oil, Bo and also the fact
that the radius, r, is now at 9m from the wellbore).
qµBo
159x1.4x10 −3 x1.4
=
= 28703
4πkh 24x3600x4πx100x10 −15 x100
φµcr 2 0.19x1.4x10 −3 x2.2x10 −9 r 2
=
= 101597 x10 −9 r 2
4kt
4x100x10 -15 x4x3600
P = 400x105 + 28703xln(1.781x 101597x10-9x92)
= 400x105 - 121209
= 39878791Pa
= 398.8bar
Institute of Petroleum Engineering, Heriot-Watt University
71
(3) the pressure after 4 hours production at a radius of 50m from the wellbore
(i) check ln approximation to Ei function
25φµcr 2
t>
k
the ln approximation is valid if the time,
t>
25x0.19x1.4x10 −3 x2.2x10 −9 x50 2
100x10 -15
t > 365750s
t > 101.6 hours
therefore ln approximation is not valid and the Ei function is used.
2
qµBo  φµcr 
P = Pi +
Ei −
4πkh  4kt  (taking account of the conversion from stock tank
(ii)
to reservoir conditions via the formation volume factor for oil, Bo and also the fact
that the radius, r, is now at 50m from the wellbore).
qµBo
159x1.4x10 −3 x1.4
=
= 28703
4πkh 24x3600x4πx100x10 −15 x100
φµcr 2 0.19x1.4x10 −3 x2.2x10 −950 2
=
= 0.254
4kt
4x100x10 -15 x4x3600
P = 400x105 + 28703xEi(-0.254)
Ei(-0.254) = -1.032 (by linear interpolation of the values in Table 4)
P = 400x105 +28703x-1.032
= 400x105 -29622
= 39970378Pa
= 399.7bar
(4) the pressure after 50 hours production at a radius of 50m from the wellbore
(i) check ln approximation to Ei function
25φµcr 2
t>
k
the ln approximation is valid if the time,
t>
25x0.19x1.4x10 −3 x2.2x10 −9 x50 2
100x10 -15
t > 365750s
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10
Fluid Flow In Porous Media
t > 101.6 hours
therefore ln approximation is not valid and the Ei function is used.
2
qµBo  φµcr 
P = Pi +
Ei −
4πkh  4kt  (taking account of the conversion from stock
(ii)
tank to reservoir conditions via the formation volume factor for oil, Bo and also the
fact that the radius, r, is now at 50m from the wellbore and the time is now 50hours
after start of production).
qµBo
159x1.4x10 −3 x1.4
=
= 28703
4πkh 24x3600x4πx100x10 −15 x100
φµcr 2 0.19x1.4x10 −3 x2.2x10 −950 2
=
= 0.020
4kt
4x100x10 -15 x4x3600
P = 400x105 + 28703xEi(-0.020)
Ei(-0.020) = -3.355
P = 400x105 +28703x-3.355
= 400x105 -96300
= 39903700Pa
= 399.0bar
Summary
time
radius
(hours) (m)
pressure
(bar)
0
4
4
4
50
400.0
396.4
398.8
399.7
399.0
all
0.15
9.00
50.00
50.00
Institute of Petroleum Engineering, Heriot-Watt University
73
EXERCISE 8
A well flows at a constant rate of 20stm3/day. Calculate the bottomhole flowing
pressure at 8 hours after the start of production.
Data
porosity, φ
formation volume factor for oil, Bo
net thickness of formation, h viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw external radius, re initial reservoir pressure, Pi well flowrate (constant)
skin factor 25%
1.32rm3/stm3
33m
22.0x10-3 Pas
0.6x10-9Pa-1
340mD
0.15m
650m
270bar
20stm3/day
0
Solution EXERCISE 8
The line source solution is used to determine the pressures required at the specified
radius and at the specified time. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
Thereafter, the choice of Ei function or ln approximation to the Ei function has to
be made.
A Check Ei applicability
line source not accurate until
t>
100φµcrw2
k
t>
100 x 0.25 x 22 x10 −3 x 0.6 x10 −9 x 0.152
340 x10 −15
t > 21.8s
time is 8 hours, therefore line source is applicable.
B Check reservoir is infinite acting
the reservoir is infinite acting if the time, t <
i.e.
74
φµcre2
4k
10
Fluid Flow In Porous Media
t<
0.25 x 22 x10 −3 x 0.6 x10 −9 x 650 2
4 x 340 x10 −15
t<
1025184s
t<
285 hours
therefore line source solution is applicable.
(i) check ln approximation to Ei function
25φµcr 2
the ln approximation is valid if the time, t >
k
t>
25 x 0.25 x 22 x10 −3 x 0.6 x10 −9 x 0.152
340 x10 −15
t > 5.5s
therefore ln approximation is valid.
Pwf = Pi +
qµBo  γφµcrw2 
ln

4πkh  4 kt  (taking account of the conversion from stock
(ii)
tank to reservoir conditions via the formation volume factor for oil).
qµBo
20 x 22 x10 −3 x1.32
=
= 47677
4πkh 24 x 3600 x 4πx 340 x10 −15 x 33
φµcrw2 0.25 x 2210 −3 x 0.6 x10 −9 0.152
=
= 1896 x10 −9
−15
4 kt
4 x 340 x10 x8 x 3600
Pwf = 270x105 + 47677xln(1.781x 1896x10-9)
= 270x105 - 600663
= 26399337Pa
= 264.0bar
Institute of Petroleum Engineering, Heriot-Watt University
75
EXERCISE 9
Two wells are drilled into a reservoir. Well 1 is put on production at 20stm3 /day.
Well 2 is kept shut in. Using the data given, calculate how long it will take for the
pressure in well 2 to drop by 0.5bar caused by the production in well 1. Well 2 is 50m
from well 1.
Data
porosity, φ
formation volume factor for oil, Bo
net thickness of formation, h viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw external radius, re initial reservoir pressure, Pi well flowrate (constant)
skin factor Distance well 1 to well 2
18%
1.21rm3/stm3
20m
0.8x10-3 Pas
43x10-9Pa-1
85mD
0.15m
1950m
210bar
20stm3/day
0
50m
Solution EXERCISE 9
The line source solution is used to determine the time equivalent to the specified
pressure drop at well 2. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
Thereafter, the choice of Ei function or ln approximation to the Ei function has to be
made.
A Check Ei applicability
line source not accurate until
100φµcrw2
t>
k
100 x 0.18 x 0.8 x10 −3 x 43 x10 −9 x 0.152
t>
85 x10 −15
t > 164 s
it is expected that the time will be in excess of 164 seconds therefore the line source
solution is acceptable
76
10
Fluid Flow In Porous Media
B Check reservoir is infinite acting
the reservoir is infinite acting if the time,
t<
φµcre2
4k
i.e.
t<
0.18 x 0.8 x10 −3 x 43 x10 −9 x1950 2
4 x85 x10 −15
t < 69250235s
t < 19236 hours
t < 802 days
therefore line source solution is applicable.
C check ln approximation to Ei function
the ln approximation is valid if the time,
t>
25φµcr 2
k
25 x 0.18 x 0.8 x10 −3 x 43 x10 −9 x 0.152
t<
85 x10 −15
t > 41s
therefore it is assumed that the ln approximation is valid.
Now,
Pi − Pat 50 m
from well 1
Pi − Pat 50 m from
qµBo
−
4πkh
well 1

P − Pat 50 m from
e i
qµBo

−


4πkh
=−
qµBo  γφµcr502 m 
ln

4πkh  4 kt 
 γφµcr502 m 
= ln

 4 kt 
well

2
 = γφµcr50 m

4 kt


Institute of Petroleum Engineering, Heriot-Watt University
77
t=
t=
t=
γφµcr502 m

P − Pat 50 m from
4 ke i
qµBo

−


4πkh
well





1.781x 0.18 x 0.8 x10 −3 x 43 x10 −9 x 50 2


5
0
.
5
x
10


4 x85 x10 −15 xe
−3

20 x 0.8 x10 x1.21
−

 24 x 3600 x 4π 85 x10 −15 x 20 
27.57 x10 −9
3.4 x10 −13 xe −4.77
t = 9561863s
t = 2656hours
t = 111 days
This time is within the limits for the use of the ln approximation to the Ei function and
within the limits to the reservoir being infinite acting therefore the result is correct.
78
10
Fluid Flow In Porous Media
EXERCISE 10
A well in a reservoir has a very low production rate of 2stm3/day. Calculate the flowing
bottomhole pressure after 2 years production.
Data
porosity, φ
formation volume factor for oil, Bo
net thickness of formation, h viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw external radius, re initial reservoir pressure, Pi well flowrate (constant)
skin factor 16%
1.13rm3/stm3
10m
5x10-3 Pas
14x10-9Pa-1
10mD
0.15m
780m
86bar
2stm3/day
0
Solution EXERCISE 10
The line source solution is used to determine the pressures required at the wellbore
after 2 years production. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
Thereafter, the choice of Ei function or ln approximation to the Ei function has to
be made.
A Check Ei applicability
line source not accurate until
t>
100φµcrw2
k
t>
100x0.16x5x10 -3 x14x10 −9 x0.152
10x10 -15
t > 2520s
t > 0.7 hours
time is 2 years, therefore line source is applicable.
B Check reservoir is infinite acting
the reservoir is infinite acting if the time,
t<
φµcre2
4k
Institute of Petroleum Engineering, Heriot-Watt University
79
i.e.
0.16x5x10 −3 x14x10 −9 x780 2
t<
4x10x10 -15
t < 170352000s
t < 5.4 years
therefore line source solution is applicable.
(i) check ln approximation to Ei function
the ln approximation is valid if the time,
t>
25φµcr 2
k
25x0.16x5x10 −3 x14x10 −9 x0.152
t>
10x10 -15
t > 630s
therefore ln approximation is valid.
Pwf = Pi +
qµBo  γφµcrw2 
ln

4πkh  4kt  (taking account of the conversion from stock
(ii)
tank to reservoir conditions via the formation volume factor for oil).
qµBo
2x5x10 −3 x1.13
=
= 104077
4πkh 24x3600x4πx10x10 −15 x10
0.16x5x10 −3 x14x10 −9 0.152
φµcrw2
=
= 99.89 x10 −9
-15
4kt
4x10x10 x2x365x24x3600
Pwf = 86x105 + 104077xln(1.781x 99.89x10-9)
= 86x105 - 1617567
= 6982433Pa
= 69.8bar
80
10
Fluid Flow In Porous Media
EXERCISE 11
A well is put on production at 15stm3/day. The following well and reservoir data are
relevant.
Data
porosity, φ
formation volume factor for oil, Bo
net thickness of formation, h viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw external radius, re initial reservoir pressure, Pi well flowrate (constant)
skin factor 21%
1.2rm3/stm3
23m
5x10-3 Pas
22 x10-9Pa-1
130mD
0.15m
800m
120bar
15stm3/day
0
Determine the following:
(1) the wellbore flowing pressure after 2 hours production
(2) the pressure in the reservoir at a radius of 10m after 2 hours production
(3) the pressure in the reservoir at a radius of 20m after 2 hours production
(4) the pressure in the reservoir at a radius of 50m after 2 hours production
SOLUTION EXERCISE 11
The line source solution is used to determine the pressures required at the specified
radii and at the specified time. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
Thereafter, the choice of Ei function or ln approximation to the Ei function has to
be made.
A Check Ei applicability
line source not accurate until
100φµcrw2
t>
k
t>
100x0.21x5x10 -3 x22x10 −9 x0.152
130x10 -15
t >400s
time is 2 hours, therefore line source is applicable.
Institute of Petroleum Engineering, Heriot-Watt University
81
B Check reservoir is infinite acting
the reservoir is infinite acting if the time,
t<
φµcre2
4k
i.e.
t<
0.21x5x10 −3 x22x10 −9 x800 2
4x130x10 -15
t < 28430769s
t < 7897 hours
therefore line source solution is applicable.
(1) the bottomhole flowing pressure after 2 hours production, Pwf at 2 hours
(i) check ln approximation to Ei function
the ln approximation is valid if the time,
t>
25φµcrw2
k
25x0.21x5x10 −3 x22x10 −9 x0.152
t>
130x10 -15
t > 100s
therefore ln approximation is valid.
Pwf = Pi +
qµBo  γφµcrw2 
ln

4πkh  4kt  (taking account of the conversion from stock tank
ii)
to reservoir conditions via the formation volume factor for oil, Bo).
qµBo
15x5x10 −3 x1.2
=
= 27724
4πkh 24x3600x4πx130x10 −15 x23
φµcr 2 0.21x5x10 −3 x22x10 −9 r 2
=
= 0.0062r 2
4kt
4x130x10 -15 x2x3600
Pwf = 120x105 + 27724xln(1.781x 0.0062x0.152)
= 120x105 - 230117
= 11769883Pa
= 117.70bar
(2) the presure after 2 hours production at a radius of 10m from the wellbore
(i) check ln approximation to Ei function
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10
Fluid Flow In Porous Media
25φµcr 2
t>
k
the ln approximation is valid if the time,
t>
25x0.21x5x10 −3 x22x10 −9 x10 2
130x10 -15
t > 444231s
t > 123hours
therefore ln approximation is not valid and the Ei function is used.
2
qµBo  φµcr 
P = Pi +
Ei −

4
π
kh
 4kt  (taking account of the conversion from stock
(ii)
tank to reservoir conditions via the formation volume factor for oil, Bo and also the
fact that the radius, r, is now at 10m from the wellbore).
qµBo
15x5x10 −3 x1.2
=
= 27724
4πkh 24x3600x4πx130x10 −15 x23
φµcr 2 0.21x5x10 −3 x22x10 −9 r 2
=
= 0.0062r 2 = 0.0062 x10 2 = 0.62
-15
4kt
4x130x10 x2x3600
P = 120x105 + 27724xEi(-0.62)
Ei(-0.62) = -0.437
P = 120x105 +27724x-0.437
= 120x105 -12115
= 11987885Pa
= 119.88bar
(3) the pressure after 2 hours production at a radius of 20m from the wellbore
(i) check ln approximation to Ei function
25φµcr 2
t>
k
the ln approximation is valid if the time,
t>
25x0.21x5x10 −3 x22x10 −9 x20 2
130x10 -15
t > 1776923s
t > 493hours
therefore ln approximation is not valid and the Ei function is used.
Institute of Petroleum Engineering, Heriot-Watt University
83
2
qµBo  φµcr 
P = Pi +
Ei −
4πkh  4kt  (taking account of the conversion from stock tank
(ii)
to reservoir conditions via the formation volume factor for oil, Bo and also the fact
that the radius, r, is now at 20m from the wellbore).
qµBo
15x5x10 −3 x1.2
=
= 27724
4πkh 24x3600x4πx130x10 −15 x23
φµcr 2 0.21x5x10 −3 x22x10 −9 r 2
=
= 0.0062r 2 = 0.0062 x 20 2 = 2.48
4kt
4x130x10 -15 x2x3600
P = 120x105 + 27724xEi(-2.48)
Ei(-2.48) = -0.026 (by linear interpolation between adjacent values in the tables)
P = 120x105 +27724 x -0.026
= 120x105 -721
= 11999279Pa
= 119.99bar
(4) the pressure after 2 hours production at a radius of 50m from the wellbore
(i) check ln approximation to Ei function
25φµcr 2
t>
k
the ln approximation is valid if the time,
t>
25x0.21x5x10 −3 x22x10 −9 x50 2
130x10 -15
t > 11105769s
t > 3085hours
therefore ln approximation is not valid and the Ei function is used.
P = Pi +
2
qµBo  φµcr 
Ei −
4πkh  4kt  (taking account of the conversion from stock tank
(ii)
to reservoir conditions via the formation volume factor for oil, Bo and also the fact
that the radius, r, is now at 50m from the wellbore).
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Fluid Flow In Porous Media
qµBo
15x5x10 −3 x1.2
=
= 27724
4πkh 24x3600x4πx130x10 −15 x23
φµcr 2 0.21x5x10 −3 x22x10 −9 r 2
=
= 0.0062r 2 = 0.0062 x 50 2 = 15.5
-15
4kt
4x130x10 x2x3600
P = 120x105 + 27724xEi(-15.5)
Ei(-15.5) is less than 1.56x10-6 therefore is assumed zero
P = 120x105 +27724x0
= 120x105 -0
= 12000000Pa
= 120.00bar
The following figure illustrates the nature of the infinite acting reservoir in that the
pressure at 50m after 2 hours production is still the initial pressure of 120bar.
Pressure v Distance
120.0
Pressure (bar)
119.5
119.0
118.5
118.0
117.5
117.0
0
10
20
30
40
50
60
Distance from centre of well (m)
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85
EXERCISE 12
A discovery well is put on well test and flows at 286stm3/day. After 6 minutes
production, the well pressure has declined from an initial value of 227bar to 192bar.
Given the following data, calculate the pressure drop due to the skin, ∆Pskin , and the
mechanical skin factor.
Data
porosity, φ,
formation volume factor for oil, Bo
net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw external radius, re initial reservoir pressure, Pi bottomhole flowing pressure
after 6 minutes
well flowrate (constant)
28%
1.39rm3/stm3
8.5m
0.8x10-3 Pas
2.3 x10-9Pa-1
100mD
0.15m
6100m
227bar
192bar
286stm3/day
SOLUTION EXERCISE 12
The line source solution is used to determine the skin factor at the wellbore after 6
minutes production. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line
source solution to be accurate
(ii) the reservoir is infinite acting.
Thereafter, the choice of Ei function or ln approximation to the Ei function has to be
made.
A Check Ei applicability
line source not accurate until
100φµcrw2
t>
k
100x0.28x0.8x10 -3 x2.3x10 −9 x0.152
t>
100x10 -15
t > 11.6s
time is 6 minutes, therefore line source is applicable.
B Check reservoir is infinite acting
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10
Fluid Flow In Porous Media
t>
the reservoir is infinite acting if the time,
φµcre2
4k
i.e.
t<
0.28x0.8x10 −3 x2.3x10 −9 x6100 2
4x100x10 -15
t < 47926480s
t < 555 days
therefore line source solution is applicable.
(i) check ln approximation to Ei function
the ln approximation is valid if the time,
t>
t>
25φµcrw2
k
25x0.28x0.8x10 −3 x2.3x10 −9 x0.152
100x10 -15
t > 2.9s
therefore ln approximation is valid.
Pi - Pwf = −

qµBo   γφµcrw2 
 − 2s
ln
4πkh 
4kt 
 (taking account of the conversion from
(ii)
stock tank to reservoir conditions via the formation volume factor for oil).
−
qµBo
286x0.8x10 −3 x1.39
=−
= −344610
4πkh
24x3600x4πx100x10 −15 x8.5
−3
−9
2
γφµcrw2 1.781x0.28x0.8x10 x2.3x10 x0.15
=
= 143371x10 −9
-15
4kt
4x100x10 x6x60
Pi - Pwf = (227-192)x105Pa = 35 x105Pa
 γφµcrw2 
Pi - Pwf
2s =
+ ln

qµBo
 4kt 
4πkh
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87
(227 -192)x10 5
+ ln(143371x10 -9 )
344610
2s = 10.2 − 8.9
2s =
s = 0.65
qµB
4πkh
∆Ps = 2x0.65x344610 = 447993Pa = 4.5bar
∆Ps = 2s
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Fluid Flow In Porous Media
EXERCISE 13
A reservoir and well are detailed in the following data. Use this data to calculate the
skin factor around the well after producing for 1.5 hours.
Data
porosity, φ
formation volume factor for oil, Bo
net thickness of formation, h viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw external radius, re initial reservoir pressure, Pi bottomhole flowing pressure
after 6 minutes
well flowrate (constant)
23%
1.36rm3/stm3
63m
1.6x10-3 Pas
17 x10-9Pa-1
243mD
0.15m
4000m
263.0bar
260.5bar
120stm3/day
SOLUTION EXERCISE 13
The line source solution is used to determine the skin factor at the wellbore after 1.5
hours production. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
Thereafter, the choice of Ei function or ln approximation to the Ei function has to
be made.
A Check Ei applicability
line source not accurate until
100φµcrw2
k
100x0.23x1.6x10 -3 x17x10 −9 x0.152
t>
243x10 -15
t>
t > 58s
time is 6 minutes, therefore line source is applicable.
B Check reservoir is infinite acting
the reservoir is infinite acting if the time,
t<
φµcre2
4k
Institute of Petroleum Engineering, Heriot-Watt University
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i.e.
t<
0.23x1.6x10 −3 x17x10 −9 x4000 2
4x243x10 -15
t < 102979424s
t < 1192 days
therefore line source solution is applicable.
(i) check ln approximation to Ei function
the ln approximation is valid if the time,
t>
t>
25φµcr 2
k
25x0.23x1.6x10 −3 x17x10 −9 x0.152
243x10 -15
t > 14.5s
therefore ln approximation is valid.
Pi - Pwf = −

qµBo   γφµcrw2 
 − 2s
ln
4πkh 
4kt 
 (taking account of the conversion from
(ii)
stock tank to reservoir conditions via the formation volume factor for oil).
−
qµBo
120x1.6x10 −3 x1.36
=−
= −15710
4πkh
24x3600x4πx243x10 −15 x63
−3
−9
2
γφµcrw2 1.781x0.23x1.6x10 x17x10 x0.15
=
= 47762 x10 −9
-15
4kt
4x243x10 x1.5x3600
Pi - Pwf = (263.0-260.5)x105Pa = 2.5 x105Pa
 γφµcrw2 
Pi - Pwf
2s =
+ ln

qµBo
 4kt 
4πkh
(263.0 - 260.5)x10 5
+ ln( 47762x10 -9 )
15710
2s = 15.9 − 10.0
2s =
s = 2.95
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10
Fluid Flow In Porous Media
EXERCISE 14
A well has been on production in a reservoir which is in a semi-steady state flow
regime. For the following data, calculate the bottomhole flowing pressure, Pwf
Data
formation volume factor for oil, Bo
net thickness of formation, h viscosity of reservoir oil, µ permeability, k
wellbore radius, rw external radius, re average reservoir pressure, well flowrate (constant)
skin factor
1.62rm3/stm3
72m
1.2x10-3 Pas
123mD
0.15m
560m
263.0bar
216stm3/day
0
SOLUTION EXERCISE 14
Substitute the values into the semi-steady state flow equation
P − Pwf =
qµBo   re  3 
ln  − + s
2πkh   rw  4 
Pwf = P -
qµBo   re  3 
ln  − + s
2πkh   rw  4 
Pwf = 263x10 5 −
216x1.2x10 −3 x1.62
  560.00  3

ln
− + 0
−15

24x3600x2xπ 123x10 x72   0.15  4

Pwf = 25647120Pa
Pwf = 256.5bar
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91
EXERCISE 15
For each of the following geometries, calculate the time in hours for which the
reservoir is infinite acting
Geometry
1.Circle
2.Square
3. Quadrant of a square
Data
Area of reservoir, A
viscosity of reservoir oil, µ permeability, k
porosity, φ,
compressibility, c
1618370m2
1.0x10-3 Pas
100mD
20%
1.45 x10-9Pa-1
The times are calculated by the dimensionless time, diffusivity of the reservoir and
the area of the reservoir. The dimensionless time accounting for the reservoir drainage
area is found for the conditions in Table 5.
SOLUTION EXERCISE 15
1. Circle
For infinite acting reservoirs, time,
t < t DA
φµcA
k
t < 0.1x
0.2x1x10 −3 x1.45x10 −9 x1618370
100x10 -15
t < 469327s
t < 130hours
2. Square
For infinite acting reservoirs, time,
t < t DA
φµcA
k
t < 0.09x
0.2x1x10 −3 x1.45x10 −9 x1618370
100x10 -15
t < 422395s
t < 117hours
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Fluid Flow In Porous Media
4. Quadrant of a square
For infinite acting reservoirs, time,
t < t DA
φµcA
k
t < 0.025x
0.2x1x10 −3 x1.45x10 −9 x1618370
100x10 -15
t < 117332s
t < 33hours
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93
EXERCISE 16
A well is tested by producing it at a constant flow rate of 238stm3/day (stock tank) for
a period of 100 hours. The reservoir data and flowing bottomhole pressures recorded
during the test are as follows:
Data
porosity, φ
formation volume factor for oil, Bo
net thickness of formation, h viscosity of reservoir oil, µ compressibility, c
wellbore radius, rw initial reservoir pressure, Pi well flowrate (constant)
Time (hours)
0.0
1.0
2.0
3.0
4.0
5.0
7.5
10.0
15.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
100.0
18%
1.2rm3/stm3
6.1m
1x10-3 Pas
2.18 x10-9Pa-1
0.1m
241.3bar
238stm3/day
Bottomhole
flowing pressure
(bar)
241.3
201.1
199.8
199.1
198.5
197.8
196.5
195.3
192.8
185.2
180.2
176.7
173.2
169.7
166.2
162.7
159.2
1. Calculate the effective permeability and skin factor of the well.
2. Make an estimate of the area being drained by the well and the Dietz shape factor.
SOLUTION EXERCISE 16
The description of the test is such that this is the first time the well has been put on
production and the reservoir pressure will decline at a rate dictated by the solutions
of the diffusivity equation. The pressure decline has been recorded at the wellbore (as
in the table of data) and it is expected that there will be an unsteady state (transient)
period initially followed by a semi steady state or steady state flow period. It is
thought to be an isolated block therefore there would be a depletion of the reservoir
pressure under semi steady state conditions expected. The initial unsteady state or
transient flow period can be used to determine the permeability and skin factor of
the well, and the subsequent semi steady state flow period can be used to detect the
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10
Fluid Flow In Porous Media
reservoir limits. SI units will be used at reservoir conditions, therefore flowrates are
in m3/s and the formation volume factor for oil is used to convert from stock tank to
reservoir volumes. The pressure related items are in Pascal.
1. The permeability and skin factor can be determined from the initial transient
period using the line source solution:
Pwf = Pi −

qµ 
4kt 
+ 2s
 ln
2
4πkh  γφµcrw 

or
Pwf = m lnt + c
(3.19)
Examining the data, the following are constant:
initial pressure, Pi, permeability, k, , porosity, φ, viscosity, µ, compressibility, c,
wellbore radius, rw, and skin factor, s. Both permeability and skin factor are unknown
(but they are known to be constant). Therefore in equation 3.26, there is a linear
relationship between the bottom hole flowing pressure, Pwf and the logarithm of
time, lnt, the slope of the relationship, m, equal to
m=
qµ
4πkh
From this, the unknown value, i.e. the permeability, k, can be calculated. Once the
permeability is known, the equation 3.26 can be rearranged to determine the other
unknown, the skin factor, as:
2s =
 4kt 
Pi − Pwf
− ln

m
 γφµcrw2 
Any coherent set of data points can be used to determine the permeability and skin,
however, it is not clear when the data represent the line source solution. Therefore all
of the pressure data are plotted and a linear fit attached to those data which show the
linear relationship between the bottom hole flowing pressure, Pwf and the logarithm
of time, lnt. Table 7 and figure 9 illustrates this.
Institute of Petroleum Engineering, Heriot-Watt University
95
Time (hours)
0.0
1.0
2.0
3.0
4.0
5.0
7.5
10.0
15.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
100.0
Bottomhole
flowing pressure
(bar)
241.3
201.1
199.8
199.1
198.5
197.8
196.5
195.3
192.8
185.2
180.2
176.7
173.2
169.7
166.2
162.7
159.2
In time
0.0
0.7
1.1
1.4
1.6
2.0
2.3
2.7
3.4
3.7
3.9
4.1
4.2
4.4
4.5
4.6
Pressure - time data (log to base e)
Bottom hole flowing pressure, Pwf (bar)
210
slope = 1.98 bar/unit
200
190
P
180
170
160
150
0
1
2
3
4
In flowing time, t (hours)
5
The plots of bottomhole flowing pressure show that the transient period (for which
the logarithm approximation is valid) lasts for approximately 4 hours and from the
plot, the slope, m, can be determined to be 1.98bar/log cycle. Substituting this into
the equation gives:
k=
96
qµBo
238 x 1.2 x 1x10 −3
=
= 218x10 −15 m 2 = 218mD
4πmh 24x3600x4πx1.98x10 5 x 6.1
10
Fluid Flow In Porous Media
(converting from stock tank cubic metres/day to reservoir cubic metres/second and
from bar to Pascal producing a permeability in terms of m2 which is then converted
to mD).
To determine the skin factor, the slope, m, of the line is theoretically extrapolated
to a convenient time. This is usually a time of 1 hour. The bottomhole pressure
associated with this time is calculated and this is used to determine a pressure drop
(P­i - Pwf ) during the time (t1 hour - t 0). This is then equal to the pressure drop calculated
from the ln function plus an excess caused by the skin. In this case, a real pressure
measurement was recorded at time 1 hour. This is not necessarily the same number
as calculated from the extrapolation of the linear section of the relationship since the
real pressure recorded at time 1 hour may not be valid for use with the Ei function.
Although it was recorded. It may have been too early for the Ei function to accurately
approximate the reservoir flow regime.
In this case P1 hour =201.2bar and therefore (by rearranging equation 3.26)
2s =
 4kt  241.3 − 201.2


Pi − P1 hour
4x218x10 -15 x3600
− ln
=
−
ln



 1.781x0.18x1x10 −3 x2.18x10 −9 x0.12 
m
1.98
 γφµcrw2 
2s=20.25-13.02 = 7.23
s=3.6
2. To determine the area drained and the shape factor, the data from the semi steady
state flow regime are required. From equation 3.29, there will be a linear relationship between bottomhole flowing pressure and time. This is related to the area of
the drained volume and the shape factor.
To determine the gradient of the pressure decline, the bottomhole flowing pressure
and time are plotted using Cartesian co-ordinates as in figure 10:
Pressure- time data
Bottom hole flowing pressure, Pwf (bar)
210
200
190
slope = 0.35 bar/hour
180
170
160
150
0
20
40
60
80
100
120
Flowing time, t (hours)
Institute of Petroleum Engineering, Heriot-Watt University
97
From the plot, the gradient is determined to be -0.35bar/hour or -9.72Pa/s. This is
related to the volumetric compressibility of the reservoir, i.e.
dP
q
=−
dt
cAhφ
where q is the flowrate, c is the compressibility, A is the area of the reservoir, h is the
thickness and φ is the porosity. Taking account of the formation volume factor, Bo,
A=−
A=−
qBo
dP
chφ
dt
238 x 1.2
24 x 3600 x 2.18x10 -9 x 6.1 x 0.18 x - 9.72
A = 142076m2
The semi steady state inflow equation is
Pwf = Pi −
qµ 1
4A
2πkt
( ln
+
+ s)
2
2πkh 2 γC A rw φµcA
The linear extrapolation of this line to small values of t gives the specific value of
Pwf of 194.2 bar at t=0. In reality, at t=0, the flowrate has not started, so this will be
named P0. Inserting this value in equation 3.39 at t=0, converting bar to Pascal and
including the skin factor gives:
Pi − P0 =

qµ  4A
 ln 2 − lnC A + 2s
4πkh  γrw

i.e.
 4 x 142076

(241.3 − 194.2) x10 5 = 1.98 x10 5  ln
−
lnC
+
2
x
3.62
A
2

 1.781x0.1

17.28 + 7.24 - 23.79 = 0.73 = lnCA
CA = 2.08
From Table 5, this is close to the configuration in the figure below.
1
2
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Fluid Flow In Porous Media
EXERCISE 17
An appraisal well is tested by producing at a constant rate of 200stm3/day for 107
hours. The following table of flowing bottomhole pressures and time were recorded
during the test. Using the data,
1. calculate the permeability and skin factor of the well
2. estimate the shape of the drainage area
Data
porosity, φ
formation volume factor for oil, Bo
net thickness of formation, h viscosity of reservoir oil, µ compressibility, c
wellbore radius, rw initial reservoir pressure, Pi well flowrate (constant)
22%
1.3rm3/stm3
21m
1.9x10-3 Pas
4.3 x10-9Pa-1
0.15m
378.7bar
200stm3/day
Time (hours)
Bottomhole
flowing pressure
(bar)
0.0
1.1
2.1
3.2
4.3
5.4
8.0
10.7
16.1
21.4
32.1
42.8
53.5
64.2
74.9
85.6
96.3
107.0
378.7
326.41
324.7
323.8
323.1
322.1
320.5
318.8
315.5
312.2
305.6
300.8
296.0
291.2
286.3
281.5
276.7
271.9
SOLUTION EXERCISE 17
(1) The permeability and skin factor can be calculated from the transient flow period
using the line source solution (if the reservoir is in transient flow) since
Pwf = Pi −

qµBo   4kt 
+ 2s
ln
2
4πkh   γφµcrw 

y = c1 + m
y = mx + c
X
+ c2
Institute of Petroleum Engineering, Heriot-Watt University
99
therefore, m is the gradient of the line Pwf versus lnt. Calculate the values as in the
table below and plot Pwf versus lnt to obtain the straight line section when the well
is in transient flow.
Time
Bottomhole
flowing
pressure
(bar)
In time
(hours)
0.0
1.1
2.1
3.2
4.3
5.4
8.0
10.7
16.1
21.4
32.1
42.8
53.5
64.2
74.9
85.6
96.3
107.0
378.7
326.4
324.7
323.8
323.1
322.1
320.5
318.8
315.5
312.2
305.6
300.8
296.0
291.2
286.3
281.5
276.7
271.9
0.1
0.8
1.2
1.5
1.7
2.1
2.4
2.8
3.1
3.5
3.8
4.0
4.2
4.3
4.4
4.6
4.7
It can be seen that the slope changes after about 5 hours, therefore the data until 5
hours is used to determine a straight line fit giving the figure below.
Bottom hole flowing pressure versus ln time
pressure (bar)
bottomhole flowing
330.0
320.0
310.0
y = -2.4161x + 326.6
300.0
290.0
slope
intercept
280.0
270.0
0.0
1.0
2.0
3.0
ln time
(i) Permeability
From this the slope is 2.42 bar/log cycle therefore
100
4.0
5.0
10
Fluid Flow In Porous Media
m=
qµBo
4πkh
k=
qµBo
200x1.9x10 −3 x1.3
=
= 89.5x10 −15 m 2
5
4πmh 24x3600x4π 2.42x10 x21
k = 90mD
ii) Skin factor
Extrapolation of the line to a time of 1 hour gives the pressure, P1 hour as 326.6bar.
2s =
Pi - Pwf(1hour)
 4kt 
− ln

m
 γφµcrw2 
2s =


378.7 − 326.6
4x90x10 -15 x1x3600
− ln
−3
−9
2
2.42
 1.781x0.22x1.9x10 x4.3x10 x0.15 
2s = 21.5 - ln(17993.4)
2s = 21.5 - 9.8 = 11.7
s = 5.9
(2) Area drained
This is obtained from the semi-steady state part of the flow. A plot of linear pressure
decline with time indicates this flow regime (i.e. an expansion of a fixed volume of
fluid) and this is shown in the figure below.
Bottom hole flowing pressure versus time
bottomhole flowing
pressure (bar)
390.0
370.0
slope
350.0
y = -0.45x + 320.05
intercept
330.0
310.0
290.0
270.0
0.0
50.0
100.0
150.0
time (hours)
Institute of Petroleum Engineering, Heriot-Watt University
101
The linear section of the data appears to be present after about 50 hours, therefore
this section is used to determine the slope and the extrapolated initial pressure.
Since the pressure decline rate is related to the volume, the area, A, of the drainage
cell can be calculated assuming a constant thickness, h, and a constant porosity.
dP
qBo
=−
dt
cAhφ
A=−
qBo
dP
chφ
dt
dP
-0.45x1x10 5
= −0.45bar / hour =
= −12.5Pa / s
dt
1x3600
200x1.3
A=−
= 12118m 2
-9
24x3600x4.3x10 x21x 0.22x -12.5
The semi-steady state pressure decline is
Pwf = Pi −

qµ  1
4A
2πkt
+
+ s
 ln
2
2πkh  2 γC A rw φµcA 
and extrapolation of the line to small values of time gives a pressure, Po of 320.8bar.
Insertion of these values at time = 0 gives
Pi − P0 =

qµ  4A
 ln 2 − lnC A + 2s
4πkh  γrw

i.e.


4 x 12118
(378.7 − 320.05) x10 5 = 2.42 x10 5  ln
− lnC A + 2 x 5.9
2
 1.781x0.15

58.7x105 = 2.42x105 (14.01- lnCA +11.8)
lnCA =14.01+11.8-24.27 = 1.54
CA = 4.7 which from Table 5 is close to
1
2
102
10
Fluid Flow In Porous Media
EXERCISE 18
Two wells, well 1 and well 2, are drilled in an undeveloped reservoir. Well 1 is
completed and brought on production at 500stm3/day and produces for 40 days at
which time Well 2 is completed and brought on production at 150stm3/day. Using the
data provided, calculate the pressure in Well 2 after it has produced for 10 days (and
assuming Well 1 continues to produce at its flowrate). Therefore, Well 1 produces for
50days when its pressure influence is calculated; Well 2 produces for 10 days when
its pressure influence is calculated.
The wells are 400m apart and the nearest boundary is 4000m from each well.
Data
porosity, φ,
formation volume factor for oil, Bo
net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw (both wells)
initial reservoir pressure, Pi Well 1 flowrate (constant)
Well 2 flowrate (constant)
skin factor around both wells
21%
1.4rm3/stm3
36m
0.7x10-3 Pas
8.7 x10-9Pa-1
80mD
0.15m
180.0bar
500stm3/day
150stm3/day
0
SOLUTION EXERCISE 18
The line source solution is used to determine the bottomhole flowing pressure at Well
2 after 10 days production, accounting for the effect of 50days production from Well
1. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
line source not accurate until
100φµcrw2
k
(3.10)
100x0.21x0.7x10 -3 x8.7x10 −9 x0.152
t>
80x10 -15
(3.21)
t>
t >36s
time is 50 days, therefore line source is applicable.
Institute of Petroleum Engineering, Heriot-Watt University
103
B Check reservoir is infinite acting
the reservoir is infinite acting if the time,
t<
φµcre2
4k
i.e.
t<
0.21x0.7x10 −3 x8.7x10 −9 x4000 2
4x80x10 -15
t < 63945000
t < 740 days
therefore line source solution is applicable.
The bottomhole flowing pressure at Well 2 is the sum of the pressure drops caused
by its production and by the pressure drop generated by the production of Well 1.
Pwf at Well 2 = Pi -∆Pwell2 flowing for 10 days - ∆Pwell1 flowing for 40+10 days 400m away
(A) At 10 days, contribution to pressure drop from production from Well 2
check ln approximation to Ei function
25φµcr 2
t>
k
the ln approximation is valid if the time,
t>
25x0.21x0.7x10 −3 x8.7x10 −9 x0.152
80x10 -15
t > 9s
therefore ln approximation is valid.
Pwf = Pi +
qµBo  γφµcrw2 
ln

4πkh  4kt 
qµBo  γφµcrw2 
Pi - Pwf = −
ln

4πkh  4kt 
−
qµBo
150x0.7x10 −3 x1.4
=−
= −47011
4πkh
24x3600x4πx80x10 −15 x36
−3
−9
2
γφµcrw2 1.781x0.21x0.7x10 x8.7x10 x0.15
=
= 185 x10 −9
-15
4kt
4x80x10 x10x24x3600
104
10
Fluid Flow In Porous Media
Pi - Pwf = -47011x ln(185x10-9)
Pi - Pwf = -47011x -15.5
Pi - Pwf =728671Pa
(B) At 10 days production from well 2, well 1 has been producing for 50 days and
its contribution to pressure drop at Well 2 is calculated as follows.
check ln approximation to Ei function
t>
t>
25φµcr 2
k
25x0.21x0.7x10 −3 x8.7x10 −9 x400 2
80x10 -15
t > 63945000s
t > 740 days
therefore ln approximation is not valid and the Ei function is used.
2

qµBo  φµcr1-2
Pi - Pwf at Well2 caused by Well 1 = −
Ei −

4πkh 
4kt 
qµBo
500x0.7x10 −3 x1.4
−
=−
= 156704
4πkh
24x3600x4πx80x10 −15 x36
2
0.21x0.7x10 −3 x8.7x10 −9 x 400 2
φµcr1-2
=
= 0.148˚˚˚
4kt
4x80x10 -15 x50x24x3600
Ei(-0.148) = -1.476
Pi - Pwf at Well 2 caused by Well 1 = -156704x-1.476
Pi - Pwf at Well 2 caused by Well 1 = 231295Pa
Pwf Well2 = 180.0 - 7.3 - 2.3
Pwf Well2 = 170.4bar
Institute of Petroleum Engineering, Heriot-Watt University
105
EXERCISE 19
Two wells are brought on production in an undeveloped reservoir. Using the data
below, calculate the bottomhole flowing pressure in each well. Well 1 produces at
110stm3/day for 27 days at which time Well 2 starts production at 180stm3/day and
both produce at their respective rates for a further 13 days when the bottomhole
flowing pressures are calculated. Therefore Well 1 produces for 40 days when its
pressure influence is calculated; Well 2 produces for 13 days when its pressure
influence is calculated.
Data
porosity, φ,
formation volume factor for oil, Bo
net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw (both wells)
external radius, re
initial reservoir pressure, Pi Well 1 flowrate (constant)
Well 2 flowrate (constant)
skin factor around both wells
19%
1.2rm3/stm3
36m
1x10-3 Pas
10 x10-9Pa-1
110mD
0.15m
7000m
250.0bar
110stm3/day
180stm3/day
0
The wells are 350m apart.
SOLUTION EXERCISE 19
The line source solution is used to determine:
the bottomhole flowing pressure at well 2 flowing for 13 days plus the pressure
influence on it of well 1 flowing for 40 days
the bottomhole flowing pressure at well 1 flowing for 40 days plus the pressure
influence on it of well 2 flowing for 13 days
Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
line source not accurate until
100φµcrw2
t>
k
100x0.19x1x10 -3 x10x10 −9 x0.152
t>
˚
110x10 -15
t >39s
106
10
Fluid Flow In Porous Media
time is 13 days, therefore line source is applicable.
B Check reservoir is infinite acting
φµcre2
t<
4k
the reservoir is infinite acting if the time, i.e.
0.19x1x10 −3 x10x10 −9 x7000 2
t<
4x110x10 -15
t < 211590909s
t < 2449 days
therefore line source solution is applicable.
C PRESSURE DROP AT WELL 2
The bottomhole flowing pressure at Well 2 is the sum of the pressure drops caused
by its production and by the pressure drop generated by the production of Well 1.
Pwf at Well 2 = Pi -∆Pwell2 flowing for 13 days - ∆Pwell1 flowing for 27+13 days 350m away
A) At 13 days, contribution to pressure drop from production from Well 2
check ln approximation to Ei function
25φµcr 2
t>
k
the ln approximation is valid if the time,
t>
25x0.19x1x10 −3 x10x10 −9 x0.152
110x10 -15
t > 10s
therefore ln approximation is valid.
Institute of Petroleum Engineering, Heriot-Watt University
107
Pwf = Pi +
qµBo  γφµcrw2 
ln

4πkh  4kt 
Pi - Pwf = −
−
qµBo  γφµcrw2 
ln

4πkh  4kt 
qµBo
180x1x10 −3 x1.2
=−
= −50238
4πkh
24x3600x4πx110x10 −15 x36
−3
−9
2
γφµcrw2 1.781x0.19x1x10 x10x10 x0.15
=
= 154 x10 −9
-15
4kt
4x110x10 x13x24x3600
Pi - Pwf = -50238x ln(154x10-9)
Pi - Pwf = -50238x -15.7
Pi - Pwf =788737Pa = 7.9bar
(B) At 13 days, contribution to pressure drop at Well 2 from production from Well
1
check ln approximation to Ei function
25φµcr 2
t>
k
the ln approximation is valid if the time,
t>
25x0.19x1x10 −3 x10x10 −9 x350 2
110x10 -15
t > 52897727s
t > 612 days
therefore ln approximation is not valid and the Ei function is used.
Pi - Pwf at Well2 caused by Well 1 = −
2

qµBo  φµcr1-2
Ei −

4πkh 
4kt 
qµBo
110x1x10 −3 x1.2
−
=−
4πkh
24x3600x4πx110x10 −15 x36
2
0.19x1x10 −3 x10x10 −9 x350 2
φµcr1-2
=
4kt
4x110x10 -15 x40x24x3600
Ei(-0.153) = -1.447
Pi - Pwf at Well 2 caused by Well 1 = -30701x-1.447
Pi - Pwf at Well 2 caused by Well 1 = 44424Pa = 0.4bar
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10
Fluid Flow In Porous Media
Pwf Well2 = 250.0 - 7.9 - 0.4bar
Pwf Well2 = 241.7bar
D PRESSURE DROP AT WELL 1
The bottomhole flowing pressure at Well 1 is the sum of the pressure drops caused
by its production and by the pressure drop generated by the production of Well 2.
Pwf at Well 1 = Pi -∆Pwell1 flowing for 40 days - ∆Pwell2 flowing for 13 days 350m away
(A) At 40 days, contribution to pressure drop from production from Well 1
check ln approximation to Ei function
25φµcr 2
t>
k
the ln approximation is valid if the time,
t>
25x0.19x1x10 −3 x10x10 −9 x0.152
110x10 -15
t > 10s
therefore ln approximation is valid.
Pwf = Pi +
qµBo  γφµcrw2 
ln

4πkh  4kt 
Pi - Pwf = −
−
qµBo  γφµcrw2 
ln

4πkh  4kt 
qµBo
110x1x10 −3 x1.2
=−
= −30701
4πkh
24x3600x4πx110x10 −15 x36
−3
−9
2
γφµcrw2 1.781x0.19x1x10 x10x10 x0.15
=
= 50.1x10 −9
-15
4kt
4x110x10 x40x24x3600
Pi - Pwf = -30701x ln(50.1x10-9)
Pi - Pwf = -30701x -16.8
Pi - Pwf =515777Pa = 5.2bar
(B) At 40 days, contribution to pressure drop at Well 1 from production from Well
2
check ln approximation to Ei function
Institute of Petroleum Engineering, Heriot-Watt University
109
the ln approximation is valid if the time,
t>
t>
25φµcr 2
k
25x0.19x1x10 −3 x10x10 −9 x350 2
110x10 -15
t > 52897727s
t > 612 days
therefore ln approximation is not valid and the Ei function is used.
Pi - Pwf at Well1 caused by Well 2
2

qµBo  φµcr1-2
=−
Ei −

4πkh 
4kt 
qµBo
180x1x10 −3 x1.2
−
=−
= −50238
4πkh
24x3600x4πx110x10 −15 x36
2
0.19x1x10 −3 x10x10 −9 x350 2
φµcr1-2
=
= 0.471
4kt
4x110x10 -15 x13x24x3600
Ei(-0.471) = -0.597
Pi - Pwf at Well 1 caused by Well 2 = -30701x-0.597
Pi - Pwf at Well 1 caused by Well 2 = 18329Pa = 0.2bar
Pwf Well1 = 250.0 - 5.2 - 0.2bar
Pwf Well1 = 244.6bar
110
10
Fluid Flow In Porous Media
EXERCISE 20
A well is completed in an undeveloped reservoir described by the data below. The well
flows for 6 days at 60 stm3/day and is then shut in for a day. Calculate the pressure
in an observation well 100m from the flowing well.
Data
porosity, φ,
formation volume factor for oil, Bo
net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw (both wells)
external radius, re
initial reservoir pressure, Pi flowrate (constant)
skin factor around well
19%
1.3rm3/stm3
23m
0.4x10-3 Pas
3 x10-9Pa-1
50mD
0.15m
6000m
180.0bar
60stm3/day
0
The observation well is 100m from the flowing well.
SOLUTION EXERCISE 20
The line source solution is used to determine the pressure in the observation well
after 6 days production from the flowing well then 1 day shut in at the flowing well.
Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
line source not accurate until
100φµcrw2
t>
k
100x0.19x0.4x10 -3 x3x10 −9 x6000 2
t<
50x10 -15
t>10.3s
time is 6 days, therefore line source is applicable.
B Check reservoir is infinite acting
Institute of Petroleum Engineering, Heriot-Watt University
111
the reservoir is infinite acting if the time,
t<
φµcre2
4k
i.e.
t<
0.19 x 0.4 x10 −3 x10 x10 −9 x 7000 2
4 x110 x10 −15
t < 41040000s
t <475 days
therefore line source solution is applicable.
The pressure drop at the observation well is described by
Pi − Pobs well = −
 φµcr 2  
 φµcr 2 
µBo 
q
Ei
−
+
(q
−
q
)Ei


 1
−

2
1
 4kt 
4πkh 
 4k(t − t1 )  
Checking for the validity of the ln approximation,
the ln approximation is valid if the time,
t>
t>
25φµcr 2
k
25x0.19x0.4x10 −3 x3x10 −9 x100 2
50x10 -15
t > 1140000s
t > 13 days
therefore ln approximation is not valid.
µBo
0.4x10 −3 x1.3
−
=−
= - 35982857
4πkh
4πx50x10 −15 x23
2
0.19x0.4x10 −3 x3x10 −9 x100 2
φµcr1-2
=
= 0.019
4kt
4x50x10 -15 x7x24x3600
2
0.19x0.4x10 −3 x3x10 −9 x100 2
φµcr1-2
=
= 0.132
4k(t - t1 ) 4x50x10 -15 x(7 - 6)x24x3600
Ei(-0.019) = -3.405
Ei(-0.132) = -1.576
112
10
Fluid Flow In Porous Media
60
0 - 60
Pi − Pobs well = −35982857
x − 3.405 +
x − 1.576
24x3600
 24x3600

[
Pi − Pobs well = −35982857 −2.36x10 −3 + 1.09x10 −3
Pi - Pobs well
Pobs well
= 45698Pa
= 180.0 - 0.5
]
= 0.5bar
= 179.5bar
Institute of Petroleum Engineering, Heriot-Watt University
113
EXERCISE 21
A well in a reservoir is brought on production at a flowrate of 25stm3/day for 6 days.
The production rate is then increased to 75stm3/day for a further 4 days. Calculate,
using the data given, the bottomhole flowing pressure at the end of this period, i.e.
10 days.
Data
porosity, φ,
formation volume factor for oil, Bo
net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw (both wells)
external radius, re
initial reservoir pressure, Pi 1st flowrate (constant)
1st flowrate period
2nd flowrate (constant)
2nd flow period
skin factor around well
21%
1.31rm3/stm3
20m
0.6x10-3 Pas
8 x10-9Pa-1
75mD
0.15m
5000m
200.0bar
25stm3/day
6days
75stm3/day
4days
0
SOLUTION EXERCISE 21
The line source solution will be used to assess the effects of variables rates on the
bottomhole flowing pressure. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
line source not accurate until
t>
100φµcrw2
k
100x0.21x0.6x10 -3 x8x10 −9 x0.152
t>
75x10 -15
t > 30.3s
time is 10 days, therefore line source is applicable.
B Check reservoir is infinite acting
the reservoir is infinite acting if the time,
i.e.
114
t<
φµcre2
4k
10
Fluid Flow In Porous Media
t<
0.21x0.6x10 −3 x8x10 −9 x5000 2
4x75x10 -15
t < 84000000s
t < 972 days
therefore line source solution is applicable.
Checking for the validity of the ln approximation,
25φµcr 2
t>
k
the ln approximation is valid if the time,
t>
25 x 0.21x 0.6 x10 −3 x8 x10 −9 x 0.152
75 x10 −15
t > 7.6s
therefore ln approximation is valid.
Pi − Pwf = −
−
 γφµcrw2  
 γφµcrw2 
µBo 
q
ln
+
(q
−
q
)ln


 1


2
1
 4kt 
4πkh 
 4k(t − t1 )  
µBo
0.6x10 −3 x1.31
=−
= −41698595
4πkh
4πx75x10 −15 x20
−3
−9
2
γφµcrw2 1.781x0.21x0.6x10 x8x10 x 0.15
=
= 155.8 x10 −9
-15
4kt
4x75x10 x10x24x3600
1.781x0.21x0.6x10 −3 x8x10 −9 x 0.152
γφµcrw2
=
= 389.6 x10 −9
4k(t - t1 )
4x75x10 -15 x(10 - 6)x24x3600
25
(75 - 25)
Pi − Pwf = −41698595
xln(155.8x10 -9 ) +
ln(389.6x10 -9 )
24x3600
 24x3600

Pi - Pwf = -41698595x(-0.00454 -0.00854)
Pi - Pwf = 545418Pa
Pwf = 200.0 - 5.5 = 194.5bar
Institute of Petroleum Engineering, Heriot-Watt University
115
EXERCISE 22
A well in a reservoir is produced at 120 stm3 /day for 50 days. It is 300m from a
fault. Using the data given, calculate the bottomhole flowing pressure in the well and
determine the effect of the fault on the bottomhole flowing pressure.
Data
porosity, φ,
formation volume factor for oil, Bo
net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw
external radius, re
initial reservoir pressure, Pi flowrate (constant)
flowrate period, t
distance to fault, L
skin factor around well
19%
1.4rm3/stm3
20m
1x10-3 Pas
9 x10-9Pa-1
120mD
0.15m
4000m
300.0bar
120stm3/day
50days
300m
0
SOLUTION EXERCISE 22
The line source solution will be used to assess the effects of the rate and the boundary
on the bottomhole flowing pressure. Using an image well 600m from the real well (i.e.
2x distance to the fault) with identical pressure and rate history as the real well, the
effect of the boundary on the infinite acting reservoir can be overcome. The bottomhole
flowing pressure in the real well will be the pressure drop caused by the production
from the real well plus a pressure drop from the image well 600m away.
The line source solution will be used. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
line source not accurate until
100φµcrw2
k
100x0.19x1x10 -3 x9x10 −9 x0.152
t>
120x10 -15
t>
t >32s
time is 50 days, therefore line source is applicable.
116
10
Fluid Flow In Porous Media
B Check reservoir is infinite acting
the reservoir is infinite acting if the time,
t<
t<
φµcre2
4k
0.19x1x10 −3 x9x10 −9 x4000 2
4x120x10 -15
t < 57000000s
t <660 days
therefore line source solution is applicable.
Checking for the validity of the ln approximation, for the real well
the ln approximation is valid if the time,
t>
t>
25φµcr 2
k
25x0.19x1x10 −3 x9x10 −9 x0.152
120x10 -15
t > 8s
therefore ln approximation is valid.
Checking for the validity of the ln approximation, for the image well
25φµc(2L)2
t>
k
the ln approximation is valid if the time,
t>
25x0.19x1x10 −3 x9x10 −9 x600 2
120x10 -15
t > 128250000s
t> 1484 days
therefore ln approximation is not valid.
For this case, then, the ln approximation will predict the bottomhole flowing pressure
around the real well, but the effect of the image well 600m away will need to be
predicted by the Ei function.
Institute of Petroleum Engineering, Heriot-Watt University
117
2
qµBo  γφµcrw2  qµBo  φµc(2L) 
Pi − Pwf = −
ln
Ei −
−
4πkh  4kt  4πkh 
4kt 
qµBo
120x1x10 −3 x1.4
−
=−
= −64473
4πkh
24x3600x4πx120x10 −15 x20
−3
−9
2
γφµcrw2 1.781x0.19x1x10 x9x10 x 0.15
=
= 33.1x10 -9
-15
4kt
4x120x10 x50x24x3600
2
φµc(2L)
0.19x1x10 −3 x9x10 −9 x600
00 2
=
= 0.297
4kt
4x120x10 -15 x50x24x3600
Ei(-0.297) = -0.914
Pi − Pwf = −64473x ln(33.1x10 -9 ) − 64473x − 0.914
Pi - Pwf = 1110466 + 58928 =1169394Pa = 11.7bar
Pwf = 300.0 - 11.7 = 288.3bar
The fault 300m away pulled the bottomhole flowing pressure down by an extra
58928Pa or 0.6bar.
118
10
Fluid Flow In Porous Media
EXERCISE 23
A well in a reservoir is producing close to two intersecting faults as shown below.
Using the data given, calculate the bottomhole flowing pressure after 32 days and
indicate the effect of the faults on the bottomhole flowing pressure. The production
rate is constant at 100stm3 /day
fault
L1
70m
fault
L2
well
120m
Data
porosity, φ,
formation volume factor for oil, Bo
net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw
external radius, re
initial reservoir pressure, Pi flowrate (constant)
flowrate period, t
distance to fault, L1
distance to fault, L2
skin factor around well
22%
1.5rm3/stm3
36m
1x10-3 Pas
9 x10-9Pa-1
89mD
0.15m
6000m
240.0bar
100stm3/day
32days
70m
120m
0
SOLUTION EXERCISE 23
The line source solution will be used to assess the effects of the rate and the boundary
on the bottomhole flowing pressure. Three image wells with identical pressure and
rate histories as the real well will be used as shown below.
image well 1
image well 3
L1
fault
L1
70m
fault
L2
well
r3
L2
120m
Institute of Petroleum Engineering, Heriot-Watt University
image well 2
119
The three image wells balance the effect of the flow (and therefore the pressure
disturbance) from the real well. The pressure disturbances are superposed onto the
real well, i.e. the bottomhole flowing pressure in the real well will be the pressure
drop caused by the production from the real well plus a pressure drop from the image
wells.
The line source solution will be used. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
line source not accurate until
t>
100φµcrw2
k
100x0.22x1x10 -3 x9x10 −9 x0.152
t>
89x10 -15
t > 50s
time is 32 days, therefore line source is applicable.
B Check reservoir is infinite acting
φµcre2
t<
4k
the reservoir is infinite acting if the time,
t<
0.22x1x10 −3 x9x10 −9 x6000 2
4x89x10 -15
t < 200224719s
t < 2317 days
therefore line source solution is applicable.
Checking for the validity of the ln approximation, for the real well
25φµcr 2
t>
k
the ln approximation is valid if the time,
t>
25x0.22x1x10 −3 x9x10 −9 x0.152
89x10 -15
t > 13s
120
10
Fluid Flow In Porous Media
therefore ln approximation is valid.
Checking for the validity of the ln approximation, for the image well 1
25φµc(2L1)2
t>
k
the ln approximation is valid if the time,
t>
25x0.22x1x10 −3 x9x10 −9 x140 2
89x10 -15
t > 10901124s
t > 126 days
therefore the ln approximation is not valid and the Ei function is used. The distances
to image wells 2 and 3 are greater, therefore they must also need to use the Ei
function.
The distance r3 is
r3 = (2L1)2 + (2L2)2
r3 = (140)2 + (240)2
r3 = 277.8m
Pi -Pwf = ∆Pwell + ∆Pimage well 1 + ∆Pimage well 2 + ∆Pimage well 3
Pi − Pwf = −
qµBo  γφµcrw2 
ln

4πkh  4kt 
−
2
qµBo  φµc(2L1) 
Ei −

4πkh 
4kt

−
2
qµBo  φµc(2L2) 
Ei −

4πkh 
4kt

−
qµBo  φµcr32 
Ei −

4πkh  4kt 
evaluating the groups
−
qµBo
100x1x10 −3 x1.5
=−
= −43120
4πkh
24x3600x4πx89x10 −15 x36
−3
−9
2
γφµcrw2 1.781x0.22x1x10 x9x10 x0.15
=
= 80.6 x10 −9
-15
4kt
4x89x10 x32x24x3600
ln(80.6x10-9) = -16.3
Institute of Petroleum Engineering, Heriot-Watt University
121
φµc(2L1)2 0.22x1x10 −3 x9x10 −9 x140 2
=
= 0.039
4kt
4x89x10 -15 x32x24x3600
Ei(-0.039) = -2.706
φµc(2L2)2 0.22x1x10 −3 x9x10 −9 x240 2
=
= 0.116
4kt
4x89x10 -15 x32x24x3600
Ei(-0.116) = -1.689
−3
−9
2
φµcr32 0.22x1x10 x9x10 x277.8
=
= 0.155
4kt
4x89x10 -15 x32x24x3600
Ei(-0.155)
=
-1.436
Pi - Pwf
Pi - Pwf
Pi - Pwf
Pwf = -43120 x -16.3
-43120 x -2.706
-43120 x -1.689
-43120 x -1.436
= 702856 + 116683 + 72830 + 61920
= 954289Pa = 9.5bar
= 240.0 - 9.5 = 230.5bar
The effect of the boundary is to pull the bottomhole flowing pressure down by an
extra 2.5bar.
122
10
Fluid Flow In Porous Media
EXERCISE 24
A well is 80m due west of a north-south fault. From well tests, the skin factor is 5.0.
Calculate the pressure in the well after flowing at 80stm3/day for 10 days.
Data
porosity, φ,
formation volume factor for oil, Bo
net thickness of formation, h, viscosity of reservoir oil, µ compressibility, c
permeability, k
wellbore radius, rw
external radius, re
initial reservoir pressure, Pi flowrate (constant)
flowrate period, t
distance to fault, L
skin factor around well
25%
1.13rm3/stm3
23m
1.1x10-3 Pas
10.1 x10-9Pa-1
125mD
0.15m
6000m
210.0bar
80stm3/day
32days
80m
5.0
SOLUTION EXERCISE 24
The fault can be represented by an image well twice the distance from the real well
as the fault is. The pressure effect this image well has on the real well augments the
pressure drop in the well caused by the production, however, there is an additional
pressure drop over the skin zone around the real well which must be taken into
account.
The line source solution will be used. Checks are made to ensure that:
(i) there has been adequate time since the start of production to allow the line source
solution to be accurate
(ii) the reservoir is infinite acting.
A Check Ei applicability
line source not accurate until
100φµcrw2
k
100x0.25x1.1x10 -3 x10.1x10 −9 x0.152
t>
125x10 -15
t>
t > 50s
time is 10 days, therefore line source is applicable.
B Check reservoir is infinite acting
φµcre2
t<
4k
the reservoir is infinite acting if the time,
Institute of Petroleum Engineering, Heriot-Watt University
123
t<
0.25x1.1x10 −3 x10.1x10 −9 x6000 2
4x125x10 -15
t < 199980000s
t < 2315 days
therefore line source solution is applicable.
Checking for the validity of the ln approximation, for the real well
the ln approximation is valid if the time,
t>
t>
25φµcr 2
k
25x0.25x1.1x10 −3 x10.1x10 −9 x0.152
125x10 -15
t > 13s
therefore ln approximation is valid.
Checking for the validity of the ln approximation, for the image well
25φµc(2L)2
t>
k
the ln approximation is valid if the time,
t>
25x0.25x1.1x10 −3 x10.1x10 −9 x160 2
125x10 -15
t > 14220800s
t> 165 days
therefore ln approximation is not valid.
 qµBo  φµc(2L)2 
qµBo   γφµcrw2 
Pi − Pwf = −
Ei −
 − 2s −
ln
4πkh   4kt 
4kt 
 4πkh 
qµBo
80x1.1x10 −3 x1.13
−
=−
= −31857
4πkh
24x3600x4πx125x10 −15 x23
−3
−9
2
γφµcrw2 1.781x0.25x1.1x10 x10.1x10 x 0.15
=
= 257.6 x10 −9
4kt
4x125x10 -15 x10x24x3600
φµc(2L)2 0.25x1.1x10 −3 x10.1x10 −9 x160 2
=
= 0.165
4kt
4x125x10 -15 x10x24x3600
[
]
Pi − Pwf = −31857 ln(257.6x10 −9 ) − 2x5.0 − 31857xEi( −0.165)
124
10
Fluid Flow In Porous Media
Ei(-0.165) = -1.383
Pi - Pwf = (-31857x[-15.2 - 10]) - (31857x -1.383)
Pi - Pwf =802796 +44058
Pi - Pwf = 846854Pa = 8.5bar
Pwf = 210.0 - 8.5 = 201.5bar
tD
pD
tD
0
0.0005
0.001
0.002
0.003
0.004
0.005
0.006
0.007
0.008
0.009
0.01
0.015
0.02
0.025
0
0.0250
0.0352
0.0495
0.0603
0.0694
0.0774
0.0845
0.0911
0.0971
0.1028
0.1081
0.1312
0.1503
0.1669
0.15
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.2
1.4
2.0
3.0
4.0
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0.1818 5.0
0.2077 6.0
0.2301 7.0
0.2500 8.0
0.2680 9.0
0.2845 10.0
0.2999 15.0
0.3144 20.0
30.0
40.0
50.0
pD
tD
pD
60.0
70.0
80.0
90.0
100.0
150.0
200.0
250.0
300.0
350.0
400.0
450.0
500.0
550.0
600.0
2.4758
2.5501
2.6147
2.6718
2.7233
2.9212
3.0636
3.1726
3.263
3.3394
3.4057
3.4641
3.5164
3.5643
3.6076
1.3625
650.0
1.4362
700.0
1.4997
750.0
1.5557
800.0
1.6057
850.0
1.6509
900.0
1.8294
950.0
1.9601 1000.0
2.1470
2.2824
2.3884
3.6476
3.6842
3.7184
3.7505
3.7805
3.8088
3.8355
3.8584
0.3750
0.4241
0.5024
0.5645
0.6167
0.6622
0.7024
0.7387
0.7716
0.8019
0.8672
0.9160
1.0195
1.1665
1.2750
Table 2 pD vs. tD - Infinite radial system, constant rate at inner boundary
Institute of Petroleum Engineering, Heriot-Watt University
125
reD = 1.5
tD
pD
0.06
0.08
0.10
0.12
0.14
0.16
0.18
0.20
0.22
0.24
0.26
0.28
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.251
0.288
0.322
0.355
0.387
0.420
0.452
0.484
0.516
0.548
0.580
0.612
0.644
0.724
0.804
0.884
0.964
1.044
1.124
1.204
1.284
1.364
1.444
reD = 2.0
tD
pD
0.22
0.24
0.26
0.28
0.30
0.32
0.34
0.36
0.38
0.40
0.42
0.44
0.46
0.48
0.50
0.60
0.70
0.80
0.90
1.0
2.0
3.0
5.0
0.443
0.459
0.476
0.492
0.507
0.522
0.536
0.551
0.565
0.579
0.593
0.607
0.621
0.634
0.648
0.715
0.782
0.849
0.915
0.982
1.649
2.316
3.649
reD = 2.5
tD
pD
0.40
0.42
0.44
0.46
0.48
0.50
0.52
0.54
0.56
0.58
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
2.0
3.0
4.0
5.0
0.565
0.576
0.587
0.598
0.608
0.618
0.628
0.638
0.647
0.657
0.666
0.688
0.710
0.731
0.752
0.772
0.792
0.812
0.832
1.215
1.506
1.977
2.398
reD = 3.0
tD
pD
0.52
0.54
0.56
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.0
1.2
1.4
1.6
2.0
3.0
4.0
5.0
0.627
0.636
0.645
0.662
0.683
0.703
0.721
0.740
0.758
0.776
0.791
0.806
0.865
0.920
0.973
1.076
1.328
1.578
1.828
reD = 3.5
tD
pD
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.25
2.50
2.75
3.0
4.0
5.0
6.0
0.802
0.830
0.857
0.882
0.906
0.929
0.951
0.973
0.994
1.014
1.034
1.083
1.130
1.176
1.221
1.401
1.579
1.757
reD = 4.0
tD
pD
1.5
1.6
1.7
1.8
1.9
2.0
2.2
2.4
2.6
2.8
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
8.0
9.0
10.0
0.927
0.948
0.968
0.988
1.007
1.025
1.059
1.092
1.123
1.154
1.184
1.255
1.324
1.392
1.460
1.527
1.594
1.660
1.727
1.861
1.994
2.127
Table 3 pD vs. tD - Finite radial system with closed exterior boundary, constant rate at inner
boundary
126
10
Fluid Flow In Porous Media
reD = 4.5
tD
pD
reD = 5.0
tD
pD
tD
reD = 6.0
pD
reD = 7.0
tD
pD
reD = 8.0
tD
pD
reD = 9.0
tD
pD
reD = 10.0
tD
pD
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
1.023
1.040
1.056
1.702
1.087
1.102
1.116
1.130
1.144
1.158
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
1.167
1.180
1.192
1.204
1.215
1.227
1.238
1.249
1.259
1.270
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
1.275
1.322
1.364
1.404
1.441
1.477
1.511
1.544
1.576
1.607
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
11.0
1.436
1.470
1.501
1.531
1.559
1.586
1.613
1.638
1.663
1.711
8.0
8.5
9.0
9.5
10.0
10.5
11.0
11.5
12.0
12.5
1.556
1.582
1.607
1.631
1.653
1.675
1.697
1.717
1.737
1.757
10.0
10.5
11.0
11.5
12.0
12.5
13.0
13.5
14.0
14.5
1.651
1.673
1.693
1.713
1.732
1.750
1.768
1.786
1.803
1.819
12.0
12.5
13.0
13.5
14.0
14.5
15.0
15.5
16.0
17.0
1.732
1.750
1.768
1.784
1.801
1.817
1.832
1.847
1.862
1.890
3.0
3.2
3.4
3.6
3.8
4.0
4.5
5.0
5.5
6.0
7.0
8.0
9.0
10.0
11.0
12.0
13.0
14.0
15.0
1.171
1.197
1.222
1.246
1.269
1.292
1.349
1.403
1.457
1.510
1.615
1.719
1.823
1.927
2.031
2.135
2.239
2.343
2.447
4.0
4.2
4.4
4.6
4.8
5.0
5.5
6.0
6.5
7.0
7.5
8.0
9.0
10.0
11.0
12.0
13.0
14.0
15.0
1.281
1.301
1.321
1.340
1.360
1.378
1.424
1.469
1.513
1.556
1.598
1.641
1.725
1.808
1.892
1.975
2.059
2.142
2.225
9.0
9.5
10.0
11.0
12.0
13.0
14.0
15.0
16.0
17.0
18.0
19.0
20.0
25.0
30.0
1.638
1.668
1.698
1.757
1.815
1.873
1.931
1.988
2.045
2.103
2.160
2.217
2.274
2.560
2.846
12.0
13.0
14.0
15.0
16.0
17.0
18.0
19.0
20.0
22.0
24.0
26.0
28.0
30.0
1.757
1.810
1.845
1.888
1.931
1.974
2.016
2.058
2.100
2.184
2.267
2.351
2.434
2.517
13.0
13.5
14.0
14.5
15.0
17.0
19.0
21.0
23.0
25.0
30.0
35.0
40.0
45.0
1.776
1.795
1.813
1.831
1.849
1.919
1.986
2.051
2.116
2.180
2.340
2.449
2.658
2.817
15.0
15.5
16.0
17.0
18.0
19.0
20.0
22.0
24.0
26.0
28.0
30.0
34.0
38.0
40.0
45.0
50.0
60.0
70.0
1.835
1.851
1.867
1.897
1.926
1.955
1.983
2.037
2.096
2.142
2.193
2.244
2.345
2.446
2.496
2.621
2.746
2.996
3.246
18.0
19.0
20.0
22.0
24.0
26.0
28.0
30.0
32.0
34.0
36.0
38.0
40.0
50.0
60.0
70.0
80.0
90.0
100.0
1.917
1.943
1.968
2.017
2.063
2.108
2.151
2.194
2.236
2.278
2.319
2.360
2.401
2.604
2.806
3.008
3.210
3.412
3.614
Table 3 (continued)
Institute of Petroleum Engineering, Heriot-Watt University
127
-Ei(-y),0.000<0.209,interval=0.001
y
0
1
2
3
0.00
+ ∞
6.332 5.639 5.235
0.01
4.038 3.944 3.858 3.779
0.02
3.355 3.307 3.261 3.218
0.03
2.959 2.927 2.897 2.867
0.04
2.681 2.658 2.634 2.612
0.05
2.468 2.449 2.431 2.413
0.06
2.295 2.279 2.264 2.249
0.07
2.151 2.138 2.125 2.112
0.08
2.027 2.015 2.004 1.993
0.09
1.919 1.909 1.899 1.889
0.10
1.823 1.814 1.805 1.796
0.11
1.737 1.729 1.721 1.713
0.12
1.660 1.652 1.645 1.638
0.13
1.589 1.582 1.576 1.569
0.14
1.524 1.518 1.512 1.506
0.15
1.464 1.459 1.453 1.447
0.16
1.409 1.404 1.399 1.393
0.17
1.358 1.353 1.348 1.343
0.18
1.310 1.305 1.301 1.296
0.19
1.265 1.261 1.256 1.252
0.20
1.223 1.219 1.215 1.210
4
4.948
3.705
3.176
2.838
2.590
2.395
2.235
2.099
1.982
1.879
1.788
1.705
1.631
1.562
1.500
1.442
1.388
1.338
1.291
1.248
1.206
5
4.726
3.637
3.137
2.810
2.568
2.377
2.220
2.087
1.971
1.869
1.779
1.697
1.623
1.556
1.494
1.436
1.383
1.333
1.287
1.243
1.202
6
4.545
3.574
3.098
2.783
2.547
2.360
2.206
2.074
1.960
1.860
1.770
1.689
1.616
1.549
1.488
1.431
1.378
1.329
1.282
1.239
1.198
7
4.392
3.514
3.062
2.756
2.527
2.344
2.192
2.062
1.950
1.850
1.762
1.682
1.609
1.543
1.482
1.425
1.373
1.324
1.278
1.235
1.195
8
4.259
3.458
3.026
2.731
2.507
2.327
2.178
2.050
1.939
1.841
1.754
1.674
1.603
1.537
1.476
1.420
1.368
1.319
1.274
1.231
1.191
9
4.142
3.405
2.992
2.706
2.487
2.311
2.164
2.039
1.929
1.832
1.745
1.667
1.596
1.530
1.470
1.415
1.363
1.314
1.269
1.227
1.187
-Ei(-y),0.000<2.09,interval=0.01
0.0
+ ∞
4.038 3.335
0.1
1.823 1.737 1.660
0.2
1.223 1.183 1.145
0.3
0.906 0.882 0.858
0.4
0.702 0.686 0.670
0.5
0.560 0.548 0.536
0.6
0.454 0.445 0.437
0.7
0.374 0.367 0.360
0.8
0.311 0.305 0.300
0.9
0.260 0.256 0.251
1.0
0.219 0.216 0.212
1.1
0.186 0.183 0.180
1.2
0.158 0.156 0.153
1.3
0.135 0.133 0.131
1.4
0.116 0.114 0.113
1.5
0.100 0.099 0.097
1.6
0.086 0.085 0.084
1.7
0.075 0.074 0.073
1.8
0.065 0.064 0.063
1.9
0.056 0.055 0.055
2.0
0.049 0.048 0.048
2.681
1.524
1.076
0.815
0.640
0.514
0.420
0.347
0.289
0.243
0.205
0.174
0.149
0.127
0.109
0.094
0.081
0.071
0.061
0.053
0.046
2.468
1.464
1.044
0.794
0.625
0.503
0.412
0.340
0.284
0.239
0.202
0.172
0.146
0.125
0.108
0.093
0.080
0.070
0.060
0.052
0.046
2.295
1.409
1.014
0.774
0.611
0.493
0.404
0.334
0.279
0.235
0.198
0.169
0.144
0.124
0.106
0.092
0.079
0.069
0.060
0.052
0.045
2.151
1.358
0.985
0.755
0.598
0.483
0.396
0.328
0.274
0.231
0.195
0.166
0.142
0.122
0.105
0.090
0.078
0.068
0.059
0.051
0.044
2.027
1.309
0.957
0.737
0.585
0.473
0.388
0.322
0.269
0.227
0.192
0.164
0.140
0.120
0.103
0.089
0.077
0.067
0.058
0.050
0.044
1.919
1.265
0.931
0.719
0.572
0.464
0.381
0.316
0.265
0.223
0.189
0.161
0.138
0.118
0.102
0.088
0.076
0.066
0.057
0.050
0.043
2.0<y<10.9,interval=0.1
y
0
1
2
4.89x10-2 4.26x10-2
3
1.3x10-2 1.15x10-2
4
3.78x10-3 3.35x10-3
5
1.15x10-3 1.02x10-3
6
3.60x10-4 3.21x10-4
7
1.15x10-4 1.03x10-4
8
3.77x10-5 3.37x10-5
9
1.24x10-5 1.11x10-5
10
4.15x10-6 3.73x10-6
128
2
3.72x10-2
1.01x10-2
2.97x10-3
9.08x10-4
2.86x10-4
9.22x10-5
3.02x10-5
9.99x10-6
3.34x10-6
2.959
1.589
1.110
0.836
0.655
0.525
0.428
0.353
0.295
0.247
0.209
0.177
0.151
0.129
0.111
0.096
0.083
0.072
0.062
0.054
0.047
3
3.25x10-2
8.94x10-3
2.64x10-3
8.09x10-4
2.55x10-4
8.24x10-5
2.70x10-5
8.95x10-6
3.00x10-6
4
2.84x10-2
7.89x10-3
2.34x10-3
7.19x10-4
2.28x10-4
7.36x10-5
2.42x10-5
8.02x10-6
2.68x10-6
5
2.49x10-2
6.87x10-3
2.07x10-3
6.41x10-4
2.03x10-4
6.58x10-5
2.16x10-5
7.18x10-6
2.41x10-6
6
2.19x10-2
6.16x10-3
1.84x10-3
5.71x10-4
1.82x10-4
5.89x10-5
1.94x10-5
6.44x10-6
2.16x10-6
7
1.92x10-2
5.45x10-3
1.64x10-3
5.09x10-4
1.62x10-4
5.26x10-5
1.73x10-5
5.77x10-6
1.94x10-6
8
1.69x10-2
4.82x10-3
1.45x10-3
4.53x10-4
1.45x10-4
4.71x10-5
1.55x10-5
5.17x10-6
1.74x10-6
9
1.48x10-2
4.27x10-2
1.29x10-3
4.04x10-4
1.29x10-4
4.21x10-5
1.39x10-5
4.64x10-6
1.56x10-6
10
Fluid Flow In Porous Media
Table 4 (opposite) Values of the exponential integral, -Ei(y)
Use Infinite System
Solution With
Less Than
1% Error
for tDA<
Exact
for tDA>
Less Than
1% Error
for tDA>
31.62
0.1
0.06
0.10
31.6
0.1
0.06
0.10
27.6
0.2
0.07
0.09
27.1
0.2
0.07
0.09
21.9
0.4
0.12
0.08
0.098
0.9
0.60
0.015
30.8828
0.1
0.05
0.09
12.9851
0.7
0.25
0.03
4.5132
0.6
0.30
0.025
3.3351
0.7
0.25
0.01
1
21.8369
0.3
0.15
0.025
1
10.8374
0.4
0.15
0.025
1
4.5141
1.5
0.50
0.06
1
2.0769
1.7
0.50
0.02
1
3.1573
0.4
0.15
0.005
In Bounded
Reservoirs
60º
1/3
CA
1
4
3
2
2
2
2
2
Table 5 Shape factors for various single-well drainage areas10
Institute of Petroleum Engineering, Heriot-Watt University
129
Use Infinite System
Solution With
Less Than
1% Error
for tDA<
CA
Exact
for tDA>
Less Than
1% Error
for tDA>
0.5813
2.0
0.60
0.02
0.1109
3.0
0.60
0.005
1
5.3790
0.8
0.30
0.01
1
2.6896
0.8
0.30
0.01
1
0.2318
4.0
2.00
0.03
1
0.1155
4.0
2.00
0.01
2.3606
1.0
0.40
0.025
In Bounded
Reservoirs
1
2
1
2
4
4
4
4
1
5
Table 5 (continued)
130
11
Drive Mechanisms
CONTENTS
1 DEFINITION
2 NATURAL DRIVE MECHANISM TYPE
2.1 Depletion Drive Reservoirs
2.2 Water Drive
2.3 Compaction Drive
2.4 Gravity Drainage
2.5 Depletion Type Reservoirs
2.5.1 Solution Gas Drive
2.5.2 Gas Cap Drive
2.6 Water Drive Reservoirs
2.7 Combination Drives
3 RESERVOIR PERFORMANCE OF DIFFERENT
DRIVE SYSTEMS
3.1 Solution Gas Drive
3.1.1 Solution Gas Drive, Oil Production
3.1.2 Solution Gas Drive, Gas / Oil Ratio
3.1.3 Pressure
3.1.4 Water Production, Well Behaviour,
Expected Oil Recovery and Well Location
3.2 Gas Cap Drive
3.2.1 Oil Production
3.2.2 Pressure
3.2.3 Gas / Oil Ratio
3.2.4 Water Production, Well Behaviour,
Expected Oil Recovery and Well Locations
3.3 Water Drive
3.3.1 Rate Sensitity
3.3.2 Water Production, Oil Recovery
3.3.3 History Matching Aquifer Characteristics
3.3.4 Well Locations
4 SUMMARY
4.1 Pressure and Recovery
4.2 Gas / Oil Ratio
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•
Define reservoir drive mechanism.
•
Describe briefly with the aid of sketches a depletion drive reservoir.
•
Describe briefly with the aid of sketches a water drive reservoir.
•
Describe briefly with the aid a sketches a gravity drainage.
•
Describe briefly with the aid of sketches solution gas drive distinguishing
behaviour both above and below the bubble point.
•
Describe briefly with the aid of sketches gas cap drive .
•
Describe briefly with the aid of sketches the reservoir performance
characteristics of a solution gas drive reservoir.
•
Describe briefly with the aid of sketches the reservoir performance
characteristics of a gas drive reservoir.
•
Describe briefly with the aid of sketches the reservoir performance characteristics
of water drive reservoir.
•
Describe briefly with the aid of sketches the rate sensitivity aspect of water
drive reservoir.
•
Summarise the characteristics of solution gas drive, gas cap drive and water
drive reservoirs.
11
Drive Mechanisms
RESERVOIR DRIVE MECHANISMS
In the previous chapters we have considered the physical properties of the porous
media, the rock, within which the reservoir fluids are contained and the properties and
behaviour of the fluids. In this chapter we shall examine the various methods used to
calculate the performance of different reservoir types, we will introduce the various
drive mechanisms responsible for production of fluids from a hydrocarbon reservoir.
In this qualitative description of the way in which reservoirs produce their fluids we will
see how the various basic concepts come together to give understanding to the various
driving forces responsible for fluid production. One of the main preoccupation’s of
reservoir engineers is to determine the predominant drive mechanism, for dependant
on the drive mechanism different recoveries of oil can be achieved.
As well as presenting natural drive mechanisms we will also review various artificial
drive mechanisms.
1 DEFINITION
A reservoir drive mechanism is a source of energy for driving the fluids out through
the wellbore. It is not necessarily the energy lifting the fluids to the surface, although
in many cases, the same energy is capable of lifting the fluids to the surface.
2 NATURAL DRIVE MECHANISM TYPES
There are a number of drive mechanisms, but the two main drive mechanisms are
depletion drive and water drive. Other drive mechanisms to be considered are
compaction drive and gravity drive. These drive mechanisms are natural drive
energies and are not to be confused with artificial drive energies such as gas injection
and water injection.
2.1 Depletion Drive Reservoirs
A depletion type reservoir is a reservoir in which the hydrocarbons contained are
NOT in contact with a large body of permeable water bearing sand. In a depletion
type reservoir the reservoir is virtually totally enclosed by porous media and the only
energy comes from the reservoir system itself. Figures 1 and 2 illustrate the types of
accumulations which can give rise to depletion drive characteristics.
In figure 1 the hydrocarbons are enclosed in isolated sand lenses which have been
generated by a particular depositional environment. Over geological time the
hydrocarbons have found their way into the porous media. The surrounding rocks may
have permeability but it is so low as to prevent energy transfer from other sources. In figure 2 is illustrated another depletion type reservoir where a mature reservoir has
been subjected to faulting, resulting in the isolation of a part of the reservoir from
the rest of the accumulation. In a total field system, such a situation can give rise to
parts of the reservoir having different drive mechanism characteristics.
Institute of Petroleum Engineering, Heriot-Watt University
Gas
Oil
Water
Figure 1 Depletion reservoir: No aquifer. Isolated sand lenses
Gas
Oil
Water
Figure 2 Depletion reservoir: Aquifer limited by faults
11
Drive Mechanisms
2.2 Water Drive
Gas
Oil
Water
Figure 3 Water drive: Active aquifer
A water drive reservoir is one in which the hydrocarbons are in contact with a large
volume of water bearing sand. There are two types of water drive reservoirs. There
are those where the driving energy comes primarily from the expansion of water as
the reservoir is produced, as shown in figure 3 The key issue here is the relative
size and mobility of the water of the supporting aquifer relative to the size of the
hydrocarbon accumulation.
Water drive may also be a result of artesian flow from an outcrop of the reservoir
formation, figure 4. In this situation either surface water or seawater feeds into the
outcrop and replenishes the water as it moves into the reservoir to replace the oil. The
key issues here are the mobility of the water in the aquifer and barriers to flow from
the outcrop to the reservoir. It is not often encountered, and the water drive arising
from the compressibility of an aquifer, figure 3, is the more common.
Outcrop
of sand
Oil well
Water flow
Figure 4 Reservoir having artesian water drive.
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2.3 Compaction Drive
Figure 5 illustrates another drive mechanism, compaction drive. Although not a common
drive energy, the characteristics of its occurrence can be dramatic. Compaction drive
occurs when the hydrocarbon formation is compacted as a result of the increase in the
net overburden stress as the reservoir pore pressure is reduced during production. The
nature of the rock or its degree of consolidation can give rise to the mechanism. For
example a shallow sand deposit which has not reached its minimum porosity level due
to consolidation can consolidate further as the net overburden stresses increase as fluids
are withdrawn. The impact of the further consolidation can give rise to subsidence at
the surface. This phenomena of compaction with increasing net overburden stress is
not restricted to unconsolidated sands, since chalk also demonstrates this phenomena. One of the spectacular occurrences of compaction drive is that associated with the
Ekofisk Field, in the Norwegian sector of the North Sea. This is a very undersaturated
chalk reservoir. The field was developed on the basis of using depletion drive down
to near the bubble point and then to inject sea water to maintain pressure above the
bubble point. During this period of considerable pressure decline, the net overburden
stress was increasing, causing the formation to compact to an extent that subsidence
occurred at the seabed. In an offshore environment such uniform subsidence can go
undetected, as was the case for Ekofisk. The magnitude of the subsidence has been
such that major jacking up of the structures has been required. Old land
surface
New land
surface
Oil
Figure 5 Compaction drive
2.4 Gravity Drainage
Gravitational segregation or gravity drainage can be considered as a drive mechanism. Figure 6 illustrates a situation where the natural density segregation of the phases
can be responsible for moving the fluids to the well bore. Gravity drainage is where
the relative density forces associated with the fluids cause the fluids, the oil, to drain
down towards the production well. The tendency for the gas to migrate up and the oil
to drain down clearly will be influenced by the rate of flow of the fluids as indicated
by their relative permeabilities. Gravity drainage is generally associated with the
later stages of drive for reservoirs where other drive mechanisms have been the more
11
Drive Mechanisms
dominant energy in earlier years. Gravity drainage can be significant and effective
in steeply dipping reservoirs which are fractured.
Of the drive mechanisms mentioned the major drive mechanisms are depletion drive,
which are further classified into solution gas drive and gas cap drive and water drive. Gravity Drive typically is active during the final stages of a depletion reservoir.
Closed in
Z
1000
Initial
GOC
Present
GOC
ΟWC
Gas
Oil
Water
Inactive aquifer
Gravity drive typically is active during the final stage
of a depletion reservoir.
Figure 6 Gravity drive
2.5. Depletion Type Reservoirs
In depletion drive reservoirs the energy comes from the expansion of the fluids in
the reservoir and its associated pore space. There are two types of depletion drive
reservoirs, solution gas drive reservoirs and gas cap drive reservoirs. In solution
gas drive reservoirs there are two stages of drive mechanism where different energies
are responsible for fluid production. 2.5.1. Solution Gas Drive
In solution gas drive reservoirs the initial condition is where the reservoir is
undersaturated, i.e. above the bubble point. Production of fluids down to the bubble
point is as a result of the effective compressibility of the system. When considering
pressure volume phase behaviour, in the chapter on phase behaviour, we observed
a small increase in volume of the oil for large reductions in pressure, for oil in the
undersaturated state. Associated connate water also has a compressibility as has the
pore space within which the fluids are contained. This combined compressibility
provides the drive mechanism for depletion drive above the bubble point. Perhaps
this part of the depletion drive should be called compressibility drive. The low
compressibility causes rapid pressure decline in this period and resulting low recovery. Of the three compressibilities, although it is the oil compressibility which is the
larger, the impact of the other compressibility components, the water and the pores,
should not be neglected.
Institute of Petroleum Engineering, Heriot-Watt University
As pressure is reduced, oil expands due to compressibility and eventually gas comes
out of solution from the oil as the bubble point pressure of the fluid is reached. The
expanding gas provides the force to drive the oil hence the term solution gas drive.
It is sometimes called dissolved gas drive (Figure 7). Gas has a high compressibility
compared to liquid and therefore the pressure decline is reduced. Solution gas drive
only occurs once the bubble point pressure has been reached.
Initially no gas cap
and Oil above Pb
Figure 7 Solution gas drive reservoir
2.5.2. Gas Cap Drive
Another kind of depletion type is where there is already free gas in the reservoir,
accumulated at the top of the reservoir in the form of a gas cap (Figure 8), as
compared to the undersaturated initial condition for the previous solution gas drive
reservoir. This gas cap drive reservoir, as it is termed, receives its energy from the
high compressibility of the gas cap. Since there is a gas cap then the bottom hole
pressure will not be too far away from the bubble point pressure and therefore solution
gas drive could also be occurring. The gas cap provides the major source of energy
but there is also the expansion of oil and its dissolved gas and the gas coming out of
solution. The oil expansion term is very low and is within the errors in calculating the
two main energy sources. The two significant sources of driving energy are ;
(1)
Gas cap expansion
(2)
Expansion of gas coming out of solution
11
Drive Mechanisms
Gas cap present initially
Oil at interface is at Pb
Gas cap
Oil
Oil may be above Pb
With production - Gas cap expansion
Solution gas liberation
Figure 8 Gas cap drive reservoir
2.6 Water Drive Reservoirs
Water drive reservoirs are also of two types. There is an edge water drive reservoir. The reservoir is thin enough so that the water is in contact with the hydrocarbons at
the edge of the reservoir (Figure 9). The other type of water drive reservoir is the
bottom-water-drive reservoir; where the reservoir is so thick or the accumulation so
thin that the hydrocarbons are completely underlain by water (Figure 10).
Edge water
Figure 9 Edge water drive reservoir
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Bottom water
Water coning
Figure 10 Bottom water drive reservoir
2.7 Combination Drives
‘Pure’ types of reservoirs are those reservoirs where only one drive system operates, for
example, depletion drive only - no water drive or water drive only - no gas drive.
It is rare for reservoirs to fit conveniently into this simple characterisation. In many
of them a combination of drive mechanisms can be activate during the production
of fluids. Such reservoirs are called combination drives (Figure 11). In the case in
figure 11, which is not unusual, we have a gas cap with the oil accumulation underlain
by water providing potential water drive. So both free gas and water are in contact
with the oil. In such a reservoir some of the energy will come from the expansion
of the gas and some from the energy within the massive supporting aquifer and its
associated compressibility.
Gas Cap
Oil zone
Water
Original condition
Water
Gas Cap
Oil zone
Water
50 % Depleted
Water
Figure 11 Combination water and gas - cap drive
Sometimes it may be only water drive in the above situations. If the hydrocarbons
are taken out at a rate such that for every volume of oil removed water readily moves
in to replace the oil, then the reservoir is driven completely by water. On the other
10
11
Drive Mechanisms
hand there may be only depletion drive. If the water does not move in to replace the
oil, then only the gas cap would expand to provide the drive.
3 RESERVOIR PERFORMANCE OF DIFFERENT DRIVE SYSTEMS
Having considered the basic aspects of the drive types we will now examine their
respective characteristics in relation to production, recovery and pressure decline issues.
The performance of different types of reservoirs in relation to the daily production,
gas/oil ratio and water production can give some indication of the type of drive
mechanism operative in the reservoir.
3.1 Solution Gas Drive
In the first part of solution gas drive, in what we termed compressibility drive, within the
reservoir no production of gas occurs and the fluid moves as a result of decompression
of the three components oil, water and pore space. The pressure reduction is rapid in
relation to volumes produced. The gas to oil ratio produced at the surface is constant
since the reservoir at this stage is above its bubble point pressure.
Once the bubble point is reached gas comes out of solution. Initially the gas bubbles
are small and isolated. The size and number of the bubbles increase until they reach
a critical saturation when they form a continuous phase and become mobile. At this
stage the gas has relative permeability. The impact of the first bubbles of gas on the
oil is very significant. The relative permeability to the oil is reduced by the presence
of the non wetting gas. (See gas-oil relative permeabilties in chapter 7. Figure 44) As
the increase in saturation of gas increases at the expense of oil saturation, the relative
permeabilties move in the same directions giving rise to reduced well productivity to
oil and increased productivity to gas, figure 12. That is the oil relative permeability
decreases and the gas relative permeability increases. The gas although providing
the displacing medium is effectively leaking out of the system. Not only does the gas
progress to the wellbore, depending on vertical permeability characteristics it will
move vertically and may form a secondary gas cap. If this occurs it can contribute
to the drive energy. Well location and rate of production can be used to encourage
gas to migrate to form such a gas cap as against being lost through production from
the wellbore. Vertical gas
migration
Rs< Rsi
Gas relative permeability
Rs< Rsi
Rs< Rsi
Oil relative permeability
Figure 12 Schematic of solution gas drive.
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11
We will now review the various production profiles, specific to the drive mechanisms
but before doing so we will review the various phases of production. Production
Plateau phase
Decline phase
Production
build up
Abandonment
0
Time
Figure 13 Phases in production.
Production Phases (figure 13)
The first phase, production build up, which may exist or not depending on the drilling
strategy is the increased production as wells are brought on stream. Clearly, as in some
cases, wells might be predrilled through a template and then all brought on stream
together when connected to production facilities, such a build up of production will,
therefore, not occur. The next stage represents the period when the productivity of the production facility
is at its design capacity and the wells are throttled back to limit their productivity. This period is called the plateau phase when production is maintained at the design
capacity of the facilities. Typical production rates for the plateau period cannot be
presented since it depends on the techno-economics of the field. Clearly for a field
with a very large front loaded capital investment there is an incentive to have a high
production rate during the plateau phase , say 20% of the STOIIP, whereas for a
lower cost onshore field 5% might be acceptable. Governments will also impose
their considerations on this aspect as well.
A time will come when the reservoir is no longer able to deliver fluids to match the
facilities capacity and the field goes into the decline phase. This phase can be delayed
by methods to increase production. Such methods could include artificial lift, where
the effort required to lift the fluids from the reservoir is carried out by a downhole
pump or by using gas lift to reduce the density of the fluid system in the well.
There comes a time when the productivity of the reservoir is no longer able to
generate revenues to cover the costs of running the field, This abandonment time
again is influenced by the size and nature of the operation. Clearly a single, stripper
well, carrying very little operational costs, can be allowed to produce down to very
low rates. A well, as part of a very high cost offshore environment however, could
be abandoned at a relatively high rate when perhaps the water proportion becomes
too high or the productivity in relation to all production is not sufficient to meet the
associated well and production costs.
We will now review the performance characteristics of the various mechanisms in
light of the forgoing production phases.
12
11
Drive Mechanisms
3.1.1 Solution Gas Drive, Oil Production ( Figure 14 )
After a well is drilled and production starts for a solution gas drive reservoir, the
pressure drops in the vicinity of the well. The initially pressure drop is rapid as flow
results from the low compressibility of the system above the bubble point. Pressure
continues to decline and solution gas drive becomes effective as gas comes out of
solution. Mobility of gas occurs and the reduced mobility to oil and resulting decreasing
oil relative permeability further causes the pressure to decline and productivity to oil
flow decrease. Initially when all wells are on stream the oil production is high but
the production rapidly declines and there is a short plateau and decline phase until
an economic limit is reached.
Reservoir
Pressure
Oil
G.O.R Prod
Oil
Prod
G.O.R
Reservoir
Pressure
Time-Year
Figure 14 Production for solution gas drive
A good analogy for this type of reservoir is the champagne bottle opened by a champion
to spray the contents over enthusiastic supporters - a short lived high production
scenario followed by rapid decline!
3.1.2 Solution Gas Drive, Gas/Oil Ratio
The distinctive characteristic of the solution gas drive mechanism is related to the
producing gas to oil ratio. When the reservoir is first produced the GOR being
produced may be low corresponding to the RSi value of the reservoir liquid. If the
reservoir is highly undersaturated there will be a period when a constant producing
GOR occurs 1-2 in figure 15.
When the bubble point is reached in the near well vicinity, the initial gas which
comes out of solution is immobile and therefore oil entering the wellbore is short of
the previous level of solution gas. Theoretically at the surface the producing GOR
level is less than the original GOR 2-3 in figure 15.
Institute of Petroleum Engineering, Heriot-Watt University
13
As the pressure further reduces the released gas becomes mobile and moves at a
velocity greater than its associated oil due to the relative permeability effects. Oil
enters the well bore, with its below bubble point solution GOR value, but also gas
enters the well bore from oil which has not yet arrived. The net effect is that at the
surface the producing GOR increases rapidly as free gas within the reservoir, which
has come out of solution, moves ahead of the oil 3-4 in figure 15. Producing GOR.
As the pressure continues to decline the productivity of the well continues to decline
from the combined impact of reducing relative permeability and drop in bottom hole
pressure. The production GOR goes though a maximum as oil eventually is produced
into the well bore with a low solution GOR and the associated gas which has come
out of solution has progressed much faster to the well and contributed to earlier gas
production 4-5 in figure 15. GOR constant
above bubble
point pressure
1
Rsi
2
4
5
3
Pb
Pressure
Figure 15 Producing GOR for solution gas drive reservoir
When the pressure drops below the bubble point throughout the reservoir a secondary
gas cap may be produced and some wells have the potential of becoming gas
producers.
3.1.3 Pressure
At first the pressure is high but as production continues the pressure makes a rapid
decline.
3.1.4 Water Production, Well Behaviour, Expected Oil Recovery and Well
Location
Since by definition there is little water present in the reservoir there should be no
water production to speak of. Because of the rapid pressure drop artificial lift will
be required at an early stage in the life of the reservoir. The expected oil recovery
from these types of reservoirs is low and could be between 5 and 30% of the original
oil-in-place. Abandonment of the reservoir will depend on the level of the GOR
and the lack of reservoir pressure to enable production. Well locations for this drive
mechanism are chosen to encourage vertical migration of the gas, therefore the wells
producing zones are located structurally low, but not too close to any water contact
which might generate water through water coning. Figure 16.
14
11
Drive Mechanisms
Secondary
gas cap
Oil water contact
Figure 16 Well location for solution gas drive reservoir.
3.2 Gas Cap Drive
Whereas for a solution gas drive reservoir where we have a reservoir initially in an
undersaturated state, for a gas cap drive reservoir, figure 7, the initial condition is a
reservoir with a gas cap. Since the gas oil contact will be at the bubble point pressure
the pressures within the oil accumulation will not be higher than this only so far as
relates to the density gradient of the fluid. It is the gas cap, with its considerable
compressibility, which provides the drive energy for such fields, hence the name. To get flow in the wells it is likely that gas will come out of solution in the near well
bore vicinity and therefore some degree of solution gas drive will also take place. A
good analogy for this type of reservoir is the plastic chemical dispenser fitted with a
pump to maintain gas pressure above the dispensed liquid.
Gas Cap
Water
Oil zone
Original condition
Water
Gas Cap
Oil zone
Water
50 % Depleted
Water
Figure 17 Gas-cap drive
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15
3.2.1 Oil Production
The producing characteristics for a gas cap drive reservoir are illustrated in Figure 18.
Although the production may be high as in the solution gas drive, the oil production
still has a significant decline but not as rapid as for solution gas drive. This decline in
oil production is due to the reducing pressure in the reservoir but also from the impact
of solution gas drive on the relative permeability around the well bore. If the well is
allowed to produce at too fast a rate, the very favourable mobility characteristics of the
gas, arising from its low viscosity compared to the oil, are such that preferential flow
can cause gas breakthrough into the wells and the well is then lost to oil production. Indeed it is this condition which will determine well abandonment.
3.2.2 Pressure
Pressure
G.O.R
500
Oil Prod (1000)
With an associated gas cap a loss of volume of fluids from the reservoir is associated
with a relatively low drop in pressure because of the high compressibility of the gas. In solution gas drive much of the driving gas is produced, but with a gas cap the fluid
remains till later in the life of the reservoir. The pressure drop for a gas cap system
therefore declines slowly over the years. The decline will depend on the relative
size of the gas cap to the oil accumulation. A small gas cap would be 10% of the oil volume whereas a large gas cap would be 50% of the volume.
5000
10
Gas Break
through
Pressure
Oil
Prod
Rate
5
250
2500
G.O.R
BSW %
20
10
0
0
0
1
2
3
4
5
6
7
Time-Year
Figure 18 Reservoir performance gas - cap drive.
3.2.3 Gas/Oil Ratio
During the early stages of replacement of oil by gas a 100% replacement takes place. Later on gas by-passes oil and a reduced displacement efficiency. In the early stages
the GOR remains relatively steady increasing slowly as the impact of solution gas
drive generates gas from oil still to reach the well bore. The increasing mobility of
the gas is such that there is an increasing GOR both from dissolved gas and by-pass
gas and eventually the well goes to gas as the gas cap breaks through.
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11
Drive Mechanisms
3.2.4 Water Production, Well Behaviour, Expected Oil Recovery and Well
Locations
Like solution gas drive there should be negligible water production. The life of the
reservoir is largely a function of the size of gas cap but it is likely to be a long flowing
life. The expected oil recovery for such a system is of the order of 20 to 40% of the
original oil-in-place. The well locations, similar to solution gas drive, are such that
the production interval for the wells should be situated away from the gas oil contact
but not too close to the water oil contact to risk water coning.
3.3 Water Drive
The majority of water drive reservoirs predominantly get their drive energy from the
compressibility of the aquifer system. The effectiveness of water drive depends on
the ability of the aquifer to replace the volume of the produced oil. The key issues
with a water drive reservoir are therefore the size of the aquifer and permeability.
This is because the only way for a low compressibility system to be effective is
for its relative size to the oil accumulation to be large, and the permeability of the
aquifer to water to enable flow though the aquifer and into the oil zone. These key
issues set a considerable challenge to the reservoir engineer since to predict water
drive behaviour, requires such information, which in pre production periods can
only be obtained from exploration activity to determine the extent and properties of
the aquifer. It is difficult to obtain justification to expend such exploration costs in
determining the size of a water accumulation!
3.3.1. Rate Sensitivity.
The characteristic features of natural water drive reservoirs are strongly influenced by
the rate sensitivity of these reservoirs. If oil production from the formation is greater
than the replacement flow of the aquifer then the reservoir pressure will drop and
another drive mechanism will contribute to flow, for example solution gas drive.
Three sketches below illustrate the various types of production profiles for different
aquifer types and the influence of rate sensitivity. In figure 19 we have the artesian
type aquifer where there is communication to surface water though an outcrop. In
this case if oil is produced at a rate less than the aquifer can move water into the oil
zone, then the reservoir pressure, as measured at the original oil water contact, remains
constant. The producing gas-oil ratio also remains constant since the reservoir is
undersaturated. These reservoirs will enable a plateau phase, however as in all water
drive reservoirs the decline of the reservoirs is not due to productivity loss through
pressure decline but the production of water. The encroaching aquifer with perhaps
its favourable mobility will preferentially move through the oil zone and if there
are high permeability layers will move through these. Eventually the water-cut, the
proportion of water to total production becomes too high and the well is abandoned
to oil production.
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17
Outcrop
of sand
Oil well
Water flow
Pi
Reservoir pressure
Oil production rate
Production
GOR
Rsi
Water production
Time
Figure 19 Producing characteristics for artesian water drive.
Figure 20 illustrates a more typical water drive reservoir where the drive energy comes
from the compressibility of the aquifer system. In this case if the oil withdrawal
rate is less then the rate of water encroachment from the aquifer then the reservoir
pressure will slowly decline, reflecting the decompression of the total system , the
oil reservoir and the aquifer. Clearly this pressure decline is related to the size of
the aquifer. The larger the aquifer the slower the pressure decline. As with all water
drive reservoirs productivity of the wells remains high resulting from the maintained
pressure, however the productivity of the well to oil reduces as water breakthrough
occurs. So a characteristic of water drive reservoirs is the increasing water production
alongside decreasing oil production.
18
11
Drive Mechanisms
Pi
Rsi
Oil production rate
Reservoir
pressure
Production
GOR
Water production
Time
Figure 20 Producing characteristics for water drive (confined aquifer).
Figure 21 illustrates the rate sensitive aspect of water drive reservoirs. If the oil
withdrawal rate is higher than the water influx rate from the aquifer then the oil
reservoir pressure will drop at a rate greater than would be the case with aquifer
support alone, as the compressibility of the oil reservoirs supports the flow. If this
pressure drops below the bubble point then solution gas drive will occur, as evidenced
by an increase in the gas-oil ratio. Cutting back oil production to a rate to less than
the water encroachment rate restores the system to water drive, with the gas-oil ratio
going back to its undersaturated level.
When two drive mechanisms function as above then we have what is termed
combination drive ( water drive and solution gas drive).
Water drive reservoirs have good pressure support. The decline in oil production is
related to increasing water production as against pressure decline.
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19
2000 10000
0
50
250
ratio
Oil production rate
Oil production
GOR
25
Bsw
69
Water
0
70
71
72
73
74
75
Producing gas / oil
PROD
Water production
B/d
psi
Reservoir pressure
0
GOR
500
Reservoir pressure
Ps
1000 5000
BSW
0
Figure 21 Reservoir performance - Water drive.
3.3.2 Water Production, Oil Recovery
Because there is a large aquifer associated with the oil reservoir unlike depletion
drive systems, water production starts early and increases to appreciable amounts. This water production is produced at the expense of oil and continues to increase
until the oil/water ratio is uneconomical. Total fluid production remains reasonably
steady. The expected oil recovery from a water drive reservoir is likely to be from 35
to 60% of the original oil-in-place. Clearly these recovery factors depend on a range
of related aspects , including reservoir characteristics for example the heterogeneity
as demonstrated by large permeability variations in the formation.
3.3.3. History Matching Aquifer Characteristics.
Predicting the behaviour of water drive reservoirs in particular the rate of water
encroachment is not straightforward. The topic is covered in a later chapter, but a
significant perspective as mentioned previously is that data is required of the aquifer
to carry out the calculations. In particular the size and geometry of the aquifer and its
permeability and compressibility characteristics. Since such information is generally
not available during the exploration and development phase, the characteristics of the
aquifer are only determined once production has been operational and the support from
the aquifer can be calculated from production and pressure data. (History Matching). Getting such information may require producing a significant proportion of the
formation say 5% of the STOIIP. RFT surveys have provided a very effective way of
determining the aquifer strength as well as the communicating layers of the formation.
Pressure depth surveys taken in an open hole development well after production has
started will give indications of pressure support in the formation
Because water drive, through pressure maintenance provides the most optimistic
recoveries, artificial water drive is often part of the development strategy because of
the uncertainties of the pressure support from the associated aquifer. In the North Sea
for example many reservoirs have associated aquifers. The risk of not knowing either
20
11
Drive Mechanisms
the extent or activity of the aquifers is such that many operators are using artificial
water drive systems to maintain pressure so that solution gas drive does not occur
with the consequent loss of oil production.
3.3.4. Well Locations
Well locations for water drive reservoirs are such that they should be located high in
the structure to delay water breakthrough. 4 SUMMARY
The following summaries and tables give the main features associated with the
various drive mechanisms.
4.1 Pressure and Recovery
Water-drive -pressure declines slowly and abandonment occurs when the water cut
is too-high at around 50% of recovery, but depends on local factors.
Gas-cap drive - the pressure shows a marked decline and economic pressures
are reached around 20% of the original pressure when about 30% of the oil is
recovered.
Solution- gas drive - the pressure drops more sharply and at 10% of the pressure
reaches, an uneconomical level of recovery at about 10% of the oil-in-place.
4.2 Gas/Oil Ratio
Water drive - the curve for a water drive system shows a gas/oil ratio that remains
constant. Variations from this indicate support from solution gas drive or other
drive mechanisms
Gas-cap drive - for this drive the gas/oil ratio increases slowly and continuously.
Solution- gas drive - the curve for a solution gas drive reservoir shows that the
gas/oil ratio increases sharply at first then later declines.
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21
SOLUTION GAS DRIVE RESERVOIRS
Characteristics
1. Reservoir Pressure
2. Gas/Oil Ratio
3. Production Rate
4. Water Production
5. Well Behaviour
6. Expected Oil Recovery
Trend
Declines rapidly and continuously
First low then rises to a maximum and
then drops
First high, then decreases rapidly and
continues to decline
None
Requires artificial lift at early stages
5-30% of original oil-in-place
GAS CAP DRIVE RESERVOIRS
Characteristics
1. Reservoir Pressure
2. Gas/oil ratio
3. Production Rate
4. Water Production
5. Well Behaviour Cap
6. Expected Oil Recovery
Trend
Falls slowly and continuously
Rises continuously
First high, then declines gradually
Absent or negligible
Long flowing life depending on size of gas cap
20 to 40% of original oil-in-place
WATER DRIVE RESERVOIRS
Characteristics
1. Reservoir Pressure
2. Gas/Oil Ratio
3. Water Production
4. Well Behaviour
5. Expected Oil Recovery
Trend
Remains high
Remains steady
Starts early and increases to appreciable
amounts
Flow until water production gets excessive
up to 60% original oil-in-place.
Figures 22 and 23 give the pressure and gas-oil ratio trends for various drive
mechanism types
22
11
Drive Mechanisms
Reservoir pressure trends for reservoirs under various drives.
Reservoir pressure - percent of original
100
80
Water drive
60
Gas cap drive
40
20
0
Dissolved
gas drive
0
20
40
60
80
Oil produced - percent of original oil in place
100
Figure 22
Reservoir gas - oil ratio trends for reservoirs under various drives.
5
GOR MCF /BBL
4 Dissolved
gas drive
Gas cap drive
3
2
1
Water drive
0
0
20
40
60
80
Oil produced - percent of original oil in place
100
Figure 23
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23
12
Vapour Liquid Equlibria
CONTENTS
1 INTRODUCTION - THE IMPORTANCE OF
VAPOUR-LIQUID EQUILIBRIUM
2 IDEAL SOLUTIONS
2.1 Raoult's Law
2.2 Dalton's Law
2.3 Ideal Equilibrium Ratio
3 NON IDEAL SYSTEMS
4 EQUATIONS FOR CALCULATING
EQUILIBRIUM RELATIONS
4.1 Vapour - Liquid Calculations
4.2 Dew - Point Calculation
4.3 Bubble Point Calculation
5 SEPARATOR PROBLEMS
5.1 Gas / Oil Ratio
5.2 Oil Formation Volume Factor
5.3 Optimum Pressure of Separator System
5.4 Example of Separator Problem
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•
Define equilibrium ratio.
•
Derive equations for vapour-liquid equilibrium calculations for real systems
and explain the application of the equations.
•
Derive and explain the use of equations to determine the dew point pressure
and bubble point pressure of a fluid mixture.
•
Describe in general terms the impact of separator conditions the gas-oil ratio
and oil formation volume factor.
12
Vapour Liquid Equlibria
1 INTRODUCTION - THE IMPORTANCE OF VAPOUR-LIQUID
EQUILIBRIUM
The multiphase perspectives of hydrocarbon mixtures in and produced from reservoir
accumulations are an important aspect in reservoir management, well productivity,
facility design and pipeline transport.
Predicting the relative amount of the phases and their respective physical properties
is an essential element in all of the above operations. The topic of vapour-liquid
equilibria is also at the heart of many subsequent and other process operations and
therefore there have been a range of approaches into the solution of the problem.
The behaviour in reservoirs of multicomponent mixtures and their production to
surface has provided one of the most rigorous challenges to design engineers because
of the complex and unique nature of the fluids and in many cases their behaviour
near the critical point.
Figure 1, and 2 illustrate the complex nature of oil and gas production where,
particularly in a major offshore province, as well as onshore, a number of reservoirs
produce into a common transport line and associated treatment facilities (Figure 1). Each of the fields with their unique composition clearly contribute to a compositional
blend entering a treatment facility (Figure 2), where further separation occurs. CUSTOMERS
ONSHORE FACILITIES
OFFSHORE FACILITIES
OFFSHORE FACILITIES
WELLS
WELLS
RESERVOIR
RESERVOIR
Figure 1 The total system
Institute of Petroleum Engineering, Heriot-Watt University
CHEMICAL
FEEDSTOCK
FUEL
OIL SALES
& TAXES
TERMINAL
Composition = (TA*CA)*(TB*CB)+(TC*CC)
TA + TB + TC
Throughput = TA + TB + TC
FIELD A
FIELD C
Composition CA
Throughput TA
Composition CC
Throughput TC
FIELD B
Composition CB
Throughput TB
Figure 2 Complexity of allocation of produced oil to supply fields
The allocation of revenue based on quality of product and oil injected into a common
pipeline provides a considerable challenge to metering and compositional analysis.
To the reservoir engineer, the main issues are the multiphase behaviour in the formation
and the relationship between the fluids in the reservoir and those produced at surface
conditions.
The critical element in reservoir simulation is the grid block where the saturations
and flow behaviour of the respective fluids; gas, oil and water are required. The
grid block therefore can be considered as a ‘separator’ and vapour-equilibrium
calculations are required out to determine the relative amounts of the phases which
lead to saturation values and relative permeability of the phases, and the composition
of these phases which lead to important physical property values of density, viscosity and
interfacial tension.
12
Vapour Liquid Equlibria
In the previous chapter the considerations of the relative amounts of gas and liquid
were considered in the simplistic two component black oil approach. In this chapter
we will consider approaches to vapour-liquid equilibrium from a compositional model
consideration both from an ideal behaviour perspective and then the consideration
of real systems.
On a pressure temperature phase diagram of a multi-component mixture, the area
bounded by the bubble point and dew-point curves defines the conditions for gas and
liquid to exist in equilibrium. It is an over-simplification to describe the system as
involatile oil with associated solution gas. The behaviour of the individual components
and their influence on the composition of the mixture need to be considered.
If a sample of bubble point fluid is brought to surface to separator conditions, the fluid
enters the two phase region at a temperature and pressure much lower than reservoir
conditions. In the separator the liquid and gas phases, in equilibrium, are withdrawn
separately. Large volumes of gas are formed at these separator conditions, and the
liquid volume shrinks substantially because of decreased temperature and conversion
of some of the fluid into the gas phase. The separator liquid is collected in the stock
tank, at which additional temperature and pressure drop may occur, more gas may
be released depending on the separator conditions to stock tank conditions.
If Vo is the volume of liquid at reservoir conditions and Vst is the volume of stock
tank oil. The oil formation volume factor Bo is :
Bo =
Vo
Vst
If Vg and Vst are the total volume of gas and oil collected from the separator and stock
tank. The solution gas to oil ratio is :
Rs =
Vg
Vst
The volume factors can be determined directly in the laboratory or from equilibrium
calculations.
In addition to separator calculations vapour liquid equilibrium data can be used
for:
• Reservoir calculations
• Two phase pipeline flow calculations
• Process calculations
Although phase behaviour considerations are required throughout the production
process from reservoir to refinery the context of this particular chapter is in relation
to reservoir predictions. When reservoir fluids undergo phase alteration as a result of
changes in pressure, temperature or composition it is considered that these changes are
Institute of Petroleum Engineering, Heriot-Watt University
slow and therefore the resulting separate phases are in equilibrium, i.e the properties
of the phases are not changing with respect to time.
For multicomponent phase behaviour predictions thermodynamic principles have
been applied to provide the predictive tools. Many works have been written which
provide the foundation of the topic. Danesh1 in his text provides a good review of
the topic both with respect to the foundation principles and the equations used.
Vapour-liquid equilibrium calculations have been somewhat restricted to analysis of
behaviour with the separate areas, i.e. reservoir and wells, surface separation, pipelines,
onshore treatment and refinery operations. Increasingly in the modern multidisciplinary
approach to technical management there is an interest in the integrated perspective of
vapour liquid equilibrium. For example, what is the impact on the quality of product
exiting from a multifield oil transport line of a pressure change in field X. Such
integrated perspectives provide a considerable technical and commercial challenge
to the various technical disciplines which have been separately involved.
2 IDEAL SOLUTIONS
Before we consider the behaviour of real systems we will first examine the behaviour
of an ideal solution, where no chemical interaction occurs and where no inter-molecular
forces occur when mixing components.
These ideal solutions result in no heating effects when ideal solutions are mixed and
the volume of the mixture equals the sum of the volumes the pure components would
occupy at the same pressure and temperature.
2.1 Raoult’s Law
Raoult’s equation states that the partial pressure of a component in the gas is equal
to the product of the mole fraction, xj in the liquid, multiplied by the vapour pressure
of the pure component pvj.
pj = xjpvj
(1)
where pj is the partial pressure of component j in the liquid with a composition xj and pvj is the vapour pressure of the pure component j.
2.2 Dalton’s Law
Dalton’s law states that the partial pressure of a component pj is equal to the mole
fraction of that component in the gas, yj times the total pressure of the system p,
i.e.
pj = yjp
(2)
where yj is the composition of the vapour and p is the pressure of the system
12
Vapour Liquid Equlibria
2.3 Ideal Equilibrium Ratio
By combining the above two laws,
yjp = xjpvj
(3)
y j pvj
=
xj
p
(4)
i.e. the ratio of the component in the vapour and liquid phases is given by the ratio of
the vapour pressure of the pure component to the total pressure of the system. This
ratio is termed the Equilibrium ratio, Kj .
If n is the total number of moles of mixture and zj is the mole fraction of component
j in the mixture.
zjn = xjnL + yjng
(5)
where nL and ng are the moles of liquid and gas such that nL + ng = n
From equation 4.
z j n = x j nL + x j
∴ xj =
pvj
ng
p
zjn
p
nL + vj ng
p
(6)
Σxj by definition = 1.0
zjn
= 1.0
p
j =1 n + vj .n
L
g
p
c
c
∴∑ xj = ∑
j =1
(7)
Similarly:
c
zjn
= 1.0
p
j =1 n +
.
n
g
L
pvj
c
∑y = ∑
j =1
j
(8)
If a basis of one mole of mixture is used i.e. n g + n L = 1.0
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zj
= 1.0
 pvj

1 + ng 
− 1
 p

zj
= 1.0
∑ yj = ∑
 p

1 + nL 
− 1
 pvj

∑x = ∑
j
(9)
(10)
Using these equations in a trial and error method the compositions of vapour and
liquid streams in a flash separation can be determined.
The equilibrium ratio Kj is defined as the ratio of the composition of j in the vapour
to liquid phase, i.e.
Kj =
yj
xj
(11)
Clearly Kj is defined at a particular pressure and temperature.
Other names for Equilibrium ratio, include K-factors, K-values, equilibrium vapour
-liquid distribution ratios.
Fugacity
Lewis2 introduced the concept of fugacity, for use in equilibrium calculations, to
extrapolate or correct vapour pressures. This is required since a pure component only
has a vapour pressure up to its critical point. The fugacity is a thermodynamic quantity
defined in terms of the change in free energy in passing from one state to another.
For an ideal gas , the fugacity is equal to its pressure, and the fugacity of each
component is equal to its partial pressure. The ratio of fugacity to pressure is termed
the fugacity coefficient, φ. For a multicomponet system,
φ=
fi
Pzi
(12) All systems behave as ideal gases at very low pressures, therefore φ > 1 when P >
0
When fugacities are not 1 , then this gives an indication of non-ideality.
Fugacity has been imagined (Danesh)1 as a measure of the escaping tendency of
molecules from one phase to an adjacent phase. In multicomponent systems, if
the fugacity of a component in adjacent phases is the same, the two phases will
be in equilibrium with no net transfer of molecules from one phase to another. At
equilibrium therefore
fg = fL.
(13)
12
Vapour Liquid Equlibria
The fugacity coefficient, φ of a pure component can be calculated from the following
general equation (Danesh).
p
v
 z − 1
1  RT
ln φ = ∫ 
− P dv
 dp = ( z − 1) − ln z +
∫
 v



p
RT
o
∞
(14)
The ratio of the fugacity at the state of interest to that at a reference state is called
the activity εi = fi/fio
The activity is a measure therefore of the fugacity contribution or activeness of the
component in a mixture. fi = εifio .
The ratio of activity to concentration is called the activity coefficient Θi, where
Θi = εi/xi
Therefore fi=Θixifio
(15)
3 NON IDEAL SYSTEMS
Ideal solution assumptions cannot be applied to the systems relevant to multicomponent
hydrocarbon fluids in reservoir flow, transport and processing conditions. The ideal
assumptions only apply to low pressures and moderate temperatures, chemically and
physically similar components and behaviour below the critical point.
Different methods have been developed for treating vapour-liquid equilibrium for
non ideal systems.
The previous K value is based on both ideal and ideal solutions laws. To extend
the principle of equilibrium ratio to multicomponent hydrocarbon mixtures to the
pressures and temperatures relevant to petroleum engineers, methods of treating non
ideal systems need to be established.
The subject of non ideal equilibrium ratios are treated later in the text. We assume
in this section that K values are available either from whatever source, experimental,
NGPSA data charts, or from equations of state and other predictive methods.
4 EQUATIONS FOR CALCULATING EQUILIBRIUM RELATIONS
4.1 Vapour-Liquid Calculations
The calculations for determining the amount of liquid and vapour present in a mixture
when the pressure and temperature is known are obviously important, for example,
in optimising the performance of a separator process.
The equilibrium equations which are used for a process separator are the same as
those within a grid block or element of a reservoir simulator.
Figure 3 represents such a separation element.
Institute of Petroleum Engineering, Heriot-Watt University
V
yj
F
T&P
zj
L
xj
Figure 3 Vapour-liquid separation in an element
F
L
V
zj
xj
yj
=
=
=
=
=
=
total moles of system both liquid and gas
total moles of material within liquid phase
total moles of material within vapour phase
mole fraction of jth component in the mixture
mole fraction of jth component in the liquid
mole fraction of jth component in the vapour
It is common to express the feed F as 1.0 or 100 moles and express L and V as
fractions or percentages of F.
i.e. F = 1 = L + V
(16)
For component j
zjF = xjL + yjV
For F = 1.0 mole
zj = xjL + yjV
(17)
The equilibrium ratio:
Kj =
yj
xj
(18)
By definition:
m
m
m
∑x = ∑y = ∑z
j =1
10
j
j =1
j
j =1
j
=1
(19)
12
Vapour Liquid Equlibria
where m is the number of components.
Replacing yj by Kjxj in (17)
zj = xjL + xjKjV
zj = xj (L + KjV)
dividing both sides by L + KjV
zj
L + K jV
xj =
(20)
and:
m
m
zj
∑ x = ∑ L + K V = 1.0
j =1
j
j =1
j
(21)
similarly:
zj
= 1.0
j =1 V + L K j
m
m
∑ yj = ∑
j =1
(21a)
by multiplying (21) by V we get:
m
∑L
j =1
V
zj
+ Kj
=V
(22)
and (21a) in the same way:
m
∑
j =1
zj
=V
L
+1
K jV
(22a)
These equations are the key equations in vapour-liquid equilibrium calculations and
their use is the same whether in those calculations to determine phase behaviour in
a separator or those which take place within the numerous grid blocks of a reservoir
simulator. Clearly in the latter the amount of calculations is considerable since each
grid block can be considered a separator. In a large compositional based simulation
a study thousands of grid blocks will be used.
The method of calculation is therefore as follows for each separation element:
(1) Select Kj for each component at the temperature and pressure of the system;
(For the determination of K see the later section.)
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11
(2)
Assume a vapour liquid split i.e. V&L such that V + L = 1.0;
(3)
Calculate either V, L, ∑xj or ∑yj from equation 21, 21a, 22 and 22a;
(4)
Either:
(i) check V&L calculated against assumed V or L;
(ii) determine if ∑xj or ∑yj = 1.0;
(5) Repeat the calculation until assumed value is calculated value or until ∑xj
and ∑yj = 1.0.
It can be understood therefore that in a compositional reservoir simulator a considerable
amount of computational time is taken up because of these iterative calculations at
each grid block. In a black oil simulator no such iteration takes place the specific
pressure and temperature provide the direct phase values either from a PVT report
or an empirical black oil correlation.
This phase equilibrium perspective can also be used to calculate the reservoir saturation
pressures for a particular temperature, ie. the dewpoint and bubble point pressures.
4.2 Dew-Point Calculation
The dew-point is when the first drop of liquid begins to condense. At this point
the composition of the liquid drop is higher in heavier hydrocarbons whereas the
composition of the vapour is essentially the composition of the system: Figure 4.
Vapour
yj = Zj
Liquid Xj
Vapour
Liquid
Figure 4 Conditions at the Dew Point
At the dew point therefore: zj = yj
(22)
or: zj = xjKj
The mixture at the dew-point is therefore in equilibrium with a quantity of liquid
having a composition defined by the above equation. Clearly:
m
m
zj
∑x = ∑ K
j =1
j
j =1
= 1.0
j
Similarly for the bubble point.
12
(23)
12
Vapour Liquid Equlibria
4.3 Bubble Point Calculation
The bubble point is when the first bubble of gas appears. At this point the composition
of this bubble of gas is higher in lighter hydrocarbons whereas the composition of
the liquid is essentially the composition of the system. Figure 5.
Vapour = Yj
Liquid
Xj = Zj
Vapour
Liquid
Figure 5 Conditions at the Bubble Point.
At the bubble point therefore: zj = xj
or:
(24)
yj
Kj
zj =
The mixture at the bubble point is therefore in equilibrium with a quantity of liquid
having a composition defined by the above equation.
Also:
m
m
∑y = ∑z K
j =1
j
j =1
j
j
= 1.0
(25)
The dew-point and bubble point when either temperature or pressure are known are
determined by trial and error techniques until the above relationships are satisfied.
The dew-point pressure or bubble point pressure are estimated, K values obtained
and equations 23 or 25 checked. If the summation ≠ 1, different pressure values are
tried until convergence is reached. When convergence is reached the respective dew
point or bubble point pressure has been obtained.
5 SEPARATOR PROBLEMS
In a separator a stream of fluid is brought to equilibrium at separator temperature and
pressure. Vapour and liquid are separated within the unit and continue as separate
streams. Several separators can be operated in series each receiving the liquid phase
from the separator operating at the next higher pressure.
Institute of Petroleum Engineering, Heriot-Watt University
13
Each condition of pressure and temperature at which vapour and liquid are separated
is called a stage-separation. Hence a process using one separator and a stock tank is a
two stage process a three stage process has two separators and one stock tank. (Figure 6).
Separator calculations are performed to determine the composition of products, the
oil formation-volume factor and the volume of gas released per barrel of oil and to
determine the optimum separator conditions for the particular conditions of fluid.
Using equilibrium calculations already derived we can calculate the separation
achieved at each stage, the composition of the phases separated, the gas/oil ratio, and
the oil formation volume factor.
Vapor
Vent
Feed
To pipeline / tanker
Liquid
Separator
at P sep and T sep
Stocktank
at P st and T st
Two-stage separation
1st stage vapor
2nd stage vapor
Vent
Feed
To pipeline / tanker
Liquid
1st stage separator
at (P sep ) 1 and (T sep ) 1
Liquid
2nd stage separator
at (P sep ) 2 and (T sep ) 2
Stocktank
at P st and T st
Three-stage separation
Figure 6 Schematic drawing of separation process.
5.1 Gas/Oil Ratio
Gas is removed from each stage so that the solution GOR can be calculated for each
stage or combination of stages.
Total gas / oil ratio =
14
sum of gas volumes (SCF)
= RT
volume of stock tank oil (bbl)
12
Vapour Liquid Equlibria
(a) Calculation for Stock Tank Oil, STO.
If n1 moles enter first stage, moles of liquid entering 2nd = n2 = n1L1.
where L1 = separation in stage one based on basis of one mole feed.
Number of moles entering third stage n3 = n2L2 = L2L1n1
If third stage is the stock tank then:
nST = L3n3 = L3L2L1n1
nST is the moles of liquid in stock tank for n1 moles into first separator:
m
∴ nST = n1 ∏ Li
(26)
i =1
m = number of stages
Li = mole fraction of liquid off ith stage
n1 = moles of feed to first stage
∴ If n1 = 1
then:
m
n ST = ∏ Li
(27)
i =1
m
n ST = ∏ Li
i =1
= mole fraction of STO in the feed.
(b) Calculation of Total Gas
ngi = number of moles off stage i
ng1 = V1n1
ng2 = V2n2 = V2L1n1
ng3 = V3n3 = V3L2L1n
or in general for total gas:
m
m
i −1
i =1
i =1
j =1
ngT = ∑ ngi = n1 ∑ Vi ∏ L j
∴ If nj = 1
m
i −1
i =1
j =1
n gT = ∑ Vi ∏ L j
(28)
ngT = mole fraction of total gas in the feed
Institute of Petroleum Engineering, Heriot-Watt University
15
Total gas volume per mole of feed =ngT Vmcu ft where Vm is the molar volume
5.2 Oil Formation Volume Factor
Volume of stock tank oil per mole of feed
(VST )m =
n ST MST
ρST
(29)
MST = molecular weight of stock tank oil
n sT= moles of STO per mole of feed
ρST = density of STO at standard conditions lb/bbl
Total gas to oil ratio RT =
RT =
n gT Vm n gT Vm ρST
=
(VST )m n ST MST
(30)
where RT is the total gas - oil ratio.
If the feed to the first stage is a single-phase liquid into its point of entry into the
production stream then Bo can be calculated.
ρ res. = density of feed (lb/bbl)
Volume of reservoir oil per mole = Vres = Mres/ρres
Oil formation volume factor Bo =
Bo =
(Vres ) =
VST
Mres ρST
ρres MST nST
where
Mres = molecular weight of reservoir fluid =
and
nST =
(31)
lb.res. fluid
lb.mole. fluid
lb.mol. stock tan k fluid
lb.mol.res. fluid
5.3 Optimum Pressure of Separator System
The operating conditions of pressure and temperature of a separator influence the
amount of gas and stock tank oil produced. Change in these valves will change the
GOR and the Bo. In quoting these values therefore it is important to keep note of
the associated separation conditions of pressure and temperature. A number of units
in series also influence these parameters. It is the role of the process designers to
optimise the operating conditions of such limits and the number of units required.
16
12
Vapour Liquid Equlibria
1.36
33.4
580
1.34
560
1.32
540
1.30
33.2
33.0
32.8
32.6
32.4
API g
ravity
600
Separator + stock tank gas-oil ratio, scf/bbl
33.6
520
1.28
Total
500
480
tio
gas-oil ra
olum
Formation v
0
20
40
100
60
80
First-stage separator pressure, psi
r
cto
e fa
120
Formation volume factor
stock tank gravity, ˚API at 60˚F
It is the equilibrium characteristics of the individual components as a function
of temperature, pressure and composition which influence these total separation
characteristics for the mixtures at each separation stage.
1.26
1.24
140
Figure 7 Effect of separator pressure in a two-stage separation process (Amyx, Bass &
Whiting)
Figure 7 illustrates the influence of a change of pressure for a two-stage separation
process on GOR, Bo and the density of stock tank oil.
Equilibrium ‘flash’ calculations, which the above are called, are used in many other
applications. In reservoir engineering, flash calculations are at the core of compositional
simulation.
5.4 Example of Separator Problem (McCain)4
The following example is taken from McCain’s text on Petroleum Fluids and the
values for K used in the calculations come from the NGPSA sources5.
Calculate the gas-to-oil ratio, stock-tank oil gravity and formation-volume factor
which will result from a two-stage separation of the hydrocarbon mixture below. Use separator conditions of 76˚F and 100 psig. Assume that the mixture is a liquid
at its bubble point at reservoir conditions of 2,695 psig and 220˚F.
Institute of Petroleum Engineering, Heriot-Watt University
17
Component
Mole Fraction
Methane
Ethane
Propane
i-butane
n-butane
i-pentane
n-pentane
Hexanes
Heavier
0.3378
0.0694
0.0982
0.0133
0.0299
0.0125
0.0193
0.0299
0.3897
1.0000
Gravity °API at 60°F
60°/60°
Molecular
Weight
0.8859
263
28.2
Step 1: Calculate the composition and quantities of separator gas and liquid using
equation 21.
zj
∑ x = ∑ L + K V = 1.0
j
j
Component
of feed to
separator
Component,
mole
fraction zj
C1
C2
C3
i-C4
n-C4
i-C5
n-C5
C6
C7+ *
0.3378
0.0694
0.0982
0.0133
0.0299
0.0125
0.0193
0.0299
0.3897
1.000
* Used K for C9
Equilibrium
ratio at
114.7 psia
and 76°F Kj
20.8
4.07
1.16
0.495
0.343
0.142
0.108
0.0334
0.00150*
x=
V=0.42
0.03626
0.03031
0.09202
0.01688
0.04129
0.01954
0.03086
0.05033
0.67117
0.98866
zj
L+KjV
V=0.43
0.03551
0.02991
0.09188
0.01699
0.04167
0.01985
0.03131
0.05117
0.68291
1.0015
V=0.4291
0.03557
0.02995
0.09189
0.01698
0.04164
0.01978
0.03127
0.05109
0.68184
1.00001
The summation equals 1.0 when V1 = 0.4291 and L1 = 0.5709 and the compositions
of the separator gas and liquid are:
Component
C1
C2
C3
i-C4
n-C4
i-C5
n-C5
C6
C7+
xj =
zj
L+KjV
0.0356
0.0299
0.0919
0.0170
0.0416
0.0198
0.0313
0.0511
0.6818
1.0000
yj=kjxj
0.7399
0.1219
0.1066
0.0084
0.0143
0.0028
0.0034
0.0017
0.0010
1.0000
Step 2: Calculate the compositions and quantities of stock tank and liquid using equation
21, noting that the composition of the feed to the stock tank is the composition of
the liquid from the separator.
18
12
Vapour Liquid Equlibria
Component
of feed to
separator
Component,
mole
fraction zj
C1
C2
C3
i-C4
n-C4
i-C5
n-C5
C6
C7+*
0.0356
0.0299
0.0919
0.0170
0.0416
0.0198
0.0313
0.0511
0.6818
1.0000
Equilibrium
ratio at
14.7 psia
and 76°F, Kj
161
30.7
8.15
3.27
2.30
0.90
0.675
0.20
0.0089*
V=0.13
0.00163
0.00615
0.04763
0.01313
0.03559
0.02006
0.03286
0.05703
0.78264
0.99654
zj
L+KjV
V=0.14
V=0.1351
0.00152
0.00580
0.04593
0.01290
0.03519
0.02008
0.03279
0.05755
0.79164
1.00340
0.00157
0.00597
0.04674
0.01301
0.03538
0.02007
0.03274
0.05729
0.78720
1.00000
xj=
* Used K for C9
The summation equals 1.0 so VST = 0.1351 and LST = 0.8649 and the compositions
of the stock tank gas and liquid are:
xj
0.016
0.0060
0.0467
0.0130
0.0354
0.0201
0.0327
0.0573
0.7872
1.0000
yj = Kjxj
0.2534
0.1831
0.3810
0.0425
0.0814
0.0181
0.0221
0.0115
0.0070
1.0001
Step 3: Calculate the density and molecular weight of the stock tank oil.
Component
of stock
tank oil
C1
C2
C3
i-C4
n-C4
i-C5
n-C5
C6
C7+
Total
Component,
mole
fraction
xj
Molecular
weight
0.0016
0.0060
0.0467
0.0130
0.0354
0.0201
0.0327
0.0573
0.7872
1.0000
16.0
30.1
44.1
58.1
58.1
72.2
72.2
86.2
263
Mj
Weight
xjMj
0.026
0.181
2.059
0.755
2.057
1.451
2.361
4.939
207.034
220.863
∑C3+ = 220.656
∑C2+ = 220.837
MSTO = 220.863 lb
lb mole
Institute of Petroleum Engineering, Heriot-Watt University
Liquid
density at
60°F and
14.7 psia
ρoj
31.66
35.12
36.45
38.96
39.35
41.30
55.25
Liquid
volume at
60°F and
14.7 psia
xjMj / ρoj
0.0650
0.0215
0.0564
0.0372
0.0600
0.1196
3.7472
4.1069
cu.ft. C3+
lb.mole STO
19
220.656
lb
= 53.73
cu ft
Density of propane plus = 4.1069
0.181
= 0.001
Weight fraction ethane in ethane plus = 220.837
0.026
= 0.0001
Weight fraction methane in STO = 220.863
ρSTO = 53.73
γ STO =
o
API =
lb
cu.ft. From chapter 6, figure 13.
53.73
= 0.861
62.4
141.5
− 131.5 = 32.8
0.861
Step 4: Calculate gas to oil ratio
R SP =
V1  lb moles sep gas  
SCF sep gas   5.615
lb mole STO 
379
×
ρSTO



L1L ST  lb mole STO  
lb mole sep gas   M STO
STB

RSP = 2130
V1ρSTO
L1 LST MSTO
Similarly:
RST =
2130V2 ρSTO
L2 MSTO
RT = RST + RSP
RSP =
SCF
(2130)(0.4291)(53.73)
= 450
STB
(0.5709)(0.8649)(220.9)
RST =
SCF
(2130)(0.1351)(53.73)
= 81
STB
(0.8649)(220.9)
RT = 531
SCF
STB
Step 5: Calculate the density and molecular weight of the reservoir liquid at reservoir
conditions.
20
12
Vapour Liquid Equlibria
Component Component,
of reservoir mole
liquid
fraction xj
Molecular
weight Mj
C1
C2
C3
i-C4
n-C4
i-C5
n-C5
C6
C7+
16.0
30.1
44.1
58.1
58.1
72.2
72.2
86.2
263
0.3378
0.0694
0.0982
0.0133
0.0299
0.0125
0.0193
0.0299
0.3897
1.0000
Weight xjMj
5.405
2.089
4.331
0.773
1.737
0.902
1.393
2.577
102.491
Mor =121.699
lb / lb mole res oil
∑C2+ = 116.294
∑C3+ = 114.205
Liquid
density at
60°F and
14.7 psia
ροj
31.66
35.12
36.45
38.96
39.35
41.30
55.25
Liquid
volume at
60°F and
14.7 psia
xjMj / ροj
0.1368
0.0220
0.0477
0.0232
0.0354
0.0624
1.8550
2.1825 cu.ft.C3+
lb.mole res oil
114.205
lb
= 52.33
cu ft
Density of propane plus = 2.1825
2.089
= 0.018
Weight fraction ethane in ethane plus = 116.294
5.405
= 0.044
Weight fraction methane in reservoir oil = 121.699
ρpo = 49.1 lb.cu ft.
From figure 13, Chapter 6.
From figure 14 chapter 6
compressibility correction 49.1 + 0.8 = 49.9 at 60˚F and 2710 psia.
From figure 15 chapter 6
thermal expansion correction 49.19 - 3.86 = 46.04 at 220˚F and 2710 psia.
ρor = 46.04 lb/cu ft.
Step 6: Calculate formation volume factor using equation:
Bo =
M res ρSTO
ρ res M STO L1L ST
Bo =
(121.7)(53.73)
(46.04)(220.9)(0.5709)(0.8649)
Bo = 1.302
res bbl
STB
Institute of Petroleum Engineering, Heriot-Watt University
21
Integration of the Black-Oil and Compositional Approach
The example above illustrates the combination of the compositional based prediction
of phase volumes and associated properties and that based around the black-oil
model, centred around parameters of oil formation volume factor and gas-oil ratio.
By such a combination, the weaknesses of the simple two component black-oil
model which is at the heart of describing oil field parameters, can be overcome by
using compositional derived values rather than using perhaps inappropriate empirical
correlations and charts.
Determination of K valves
In the procedures developed, it has been assumed that values for K are available. In
the next chapter we will examine the procedures for determining K.
22
12
Vapour Liquid Equlibria
REFERENCES
(1) Danesh, A, "PVT and Phase Behaviour of Petroleum Reservoir Fluids." 1998
Elsevier. pp 105-206
(2)Prausnitz,J.M., Lichtenthaler,R.N., and d Azevedo,E.G. “ Molecular
Thermodynamic of Fluid -Phase Equilibria’ 2nd Edition, Prentice Hall Inc, NY.,
(1986)
(3) Smth,J.M. and Van Ness,H.C. “ Introduction to Chemical Engineering
Thermodynamics”, Third Edition, McGraw-Hill ( 1975)
(4) McCain,W.D. “The Properties of Petroleum Fluids” Petroleum Publishing Co.
Tulsa 1973
(5) GPSA: Engineering Dat Book, Gas Processors Association. Tulsa Oklahoma,
Gn Education 1972
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23
Equilibrium Ratio Prediction and Calculation
CONTENTS
1 INTRODUCTION
1.1 Importance of the Critical Point
1.2 Equilibrium Ratio
2 EMPIRICAL PREDICTION METHODS FOR
EQUILIBRIUM BEHAVIOUR
2.1 Black-Oil System
2.2 Correlation of Experimental Data Through K
Values
2.2.1 Method of Calculating Convergence Pressure
3 METHODS BASED ON THERMODYNAMIC
PRINCIPLES
3.1 Methods Based on Empirical Equations of
State of Fluid Phase Theory
3.2 Prediction of Vapour-Liquid Equilibrium
4. THE COMPOSITION DATA REQUIREMENTS
FOR THE APPLICATION THERMODYNAMIC
EQUILIBRIUM PREDICTION METHODS TO
MULTI-COMPONENT HYDROCARBON MIXTURES
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•
Comment briefly on the approach to vapour equilibrium ( VLE ) calculations
in terms of ; the black oil approach, empirical K values and EOS..
•
Describe briefly the application of convergence pressure based K values in VLE
calculations.
•Be able to determine the correct convergence pressure for VLE application .
•
Comment on the evolution of EOS in VLE application . (The student would not
be expected to reproduce the equations.)
•
Describe briefly the application of EOS in VLE calculations.
Equilibrium Ratio Prediction and Calculation
VAPOUR-LIQUID EQUILIBRIUM PREDICTIONS OF
PHYSICAL PROPERTIES
1 INTRODUCTION
In the previous chapter we introduced the concept of vapour-liquid equilibrium
and its role in attempting to predict the behaviour of multi-component hydrocarbon
mixtures in the two phase region, in particular the ability to predict phase densities
and composition as a function of pressure and temperature. In the chapter we also
reviewed some of the basic equations and procedures which are at the foundation of
these multicomponent calculations
In petroleum engineering applications the situation is complicated in that often very
little detailed information is available on the hydrocarbon mixture. That is, either
individual components are not identified, or, because it is a multi-component mixture
certain component concentrations are difficult to measure accurately even though
they can greatly influence the properties of the mixture.
There are two distinctive features of vapour equilibrium calculations in the context
of petroleum engineering compared to other sectors like process engineering. In
process engineering the calculations are mainly related to either a single process unit
or a few in series, whereas in reservoir engineering the calculations are often part of
multi grid block reservoir simulation where the numbers of “process units” can be
considered in hundreds or even thousands. In process engineering a full compositional
description is available for the prime objective is to determine compositional split
to obtain specific compositional objectives. In petroleum engineering however, the
objective is not to determine full compositional description of produced phases, but
to determine phase volumes and phase properties. The challenge is to carry this out
with a minimal composition description without compromising the quality of physical
property predictions.
In addition for petroleum engineering applications the calculation method used
should be:
(1) applicable to multi-component mixtures;
(2) be accurate in predicting thermodynamic equilibrium and volumetric properties
over a wide range of conditions of temperature and pressure and specifically be
accurate around and beyond the critical point;
(3) preferably require only pure component data or binary data which is either
readily available from the literature or derivable from available data.
1.1 Importance of the Critical Point
The prediction of the true critical properties of a multi-component system is an
important aspect of the general problem of predicting the overall phase behaviour
of the system. The critical state is the unique condition about which the liquid and
vapour phases are defined, and hence it has theoretical as well as practical significance. In hydrocarbon processing and producing operations, a knowledge of the critical
condition is of particular significance because many of these operations take place
under conditions which are at or near the bubble point or upper dew point regions
and are frequently accompanied by isobaric or isothermal retrograde phenomena.
Institute of Petroleum Engineering, Heriot-Watt University
Fluid property predictions and design calculations in this region are often the most
difficult to make, and a knowledge of a precise location of the critical point for the
system under study is of the utmost assistance.
From a theoretical point of view, the derivatives of many of the thermodynamic and
transport properties take on a special significance as the critical state is approached. In an empirical way the critical state has formed an integral part of many useful
generalised correlations such as those based on the theorem of corresponding states
or the convergence pressure concept in vapour-liquid equilibrium calculations.
In many ways the characteristics of the critical state that make it theoretically and
practically significant are also the characteristics that make it one of the more difficult
conditions to measure experimentally. The very fact that density differences between
phases vanish, that the rate of volume change with respect to pressure approaches
infinity, or that infinitesimal temperature gradients can be responsible for a transition
from 100% liquid to 100% vapour all make the critical condition one of the more
difficult to measure or observe accurately. For obvious economic reasons, it is a
condition that cannot be obtained by experiment in any practical way for the many
systems for which it is required. Consequently, many attempts have been made to
develop methods for predicting the critical properties based on generalised empirical
or semi-empirical procedures.
1.2 Equilibrium Ratio
In the previous chapter we presented the concept of equilibrium ratio, the measure of
how a component is distributed between different phases however we didn’t explain
how these distribution values are obtained.
The literature abounds with methods and equations associated with phase equilibrium
prediction. The methods can be broadly classified as:
(1) methods which involve empirical curve fitting to experimental data;
(2) methods which are based on thermodynamic principles.
2 EMPIRICAL PREDICTION METHODS FOR EQUILIBRIUM
BEHAVIOUR
There are a range of approaches to expressing the vapour liquid distribution behaviour
of reservoir fluid hydrocarbon systems, as outlined below.
2.1 Black-Oil System
The black-oil system as described in the chapter on liquid properties treats the fluids
as two components, stock tank oil and solution gas. This concept and associated
calculations were covered in that chapter.
2.2 Correlation of Experimental Data Through K Values
A complicating factor in the application of equilibrium ratios in the context of reservoir
fluids is that the distribution of a component between phases is not only influenced by
the temperature and pressure but is also influenced by the composition. Calculating
therefore how a unique fluid like that from a new exploration well performs in these
Equilibrium Ratio Prediction and Calculation
separation calculations is not straightforward, since no one before would have been
able to carry out tests on the fluid!!
An approach is required that the distribution of each component of the mixture between
the liquid and vapour phases be experimentally determined for a range of temperatures
and pressures. The resultant measurements are expressed as an equilibrium ratio Kj,
for component ‘j’ defined as:
where:
Kj
yj
xj
Kj =
yj
xj
(1)
= equilibrium ratio
= mole fraction of component ‘j’ in the vapour phase
= mole fraction of component ‘j’ in the liquid phase
This approach is really only useful for light hydrocarbons. The data is usually expressed
graphically as a plot of Kj versus pressure for a constant temperature.
The oil industry has relied on experimentally determined equilibrium ratios although
increasingly over recent years the move has been towards Equation of State derived K
values. Clearly the empirical K values are obtained from known mixtures, the challenge
is to determine the applicability of these empirical values to new mixtures.
At high pressures the equilibrium ratio is a function of temperature, pressure and also
the composition of the mixture. This compositional influence is of great significance.
Hence, pure component K-values measured for one mixture cannot be accurately
applied to another mixture. Figure 1 and 2 present the K values for a heavy oil and at
the other extreme a condensate at a temperature of 200˚F. The K values converge to
unity at 5,000 psia and 4,000 psia respectively. This point is called the convergence
pressure. Close examination of these two sets of K values demonstrate that these
distribution coefficients do not have the same values at particular pressures and
temperatures, confirming this compositional influence on K values.
The effect of composition on the K values is achieved by applying the concept of
convergence pressure. The concept of convergence pressure arose from the observation
that for light hydrocarbon mixtures the isothermal component K-values converge to
unity at a specific pressure known as the convergence pressure Pc, see Figures 3 for
a binary mixture and 4 for a light oil . This convergence pressure is a measure of the
composition of the mixture and can be calculated from a knowledge of the effective
boiling point of the lightest and heaviest components in the mixture.
If the temperature at which the equilibrium ratios are presented is the critical temperature
then the convergence pressure would be the critical pressure.
At a given temperature, as the system pressure increases, the K-values of all components
of the system converge to unity when the system pressure reaches the convergence
pressure. In other words, it is the pressure for a system at a given temperature when
vapour-liquid separation is no longer possible. Naturally, it is equally impossible to
have a vapour-liquid separation at a given temperature in which the system pressure
is greater than the convergence pressure.
Institute of Petroleum Engineering, Heriot-Watt University
100
M
et
ha
10
Et
K=1
He
Pr
op
an
es
ne
an
e
ta
ne
s
xa
ne
s
e
H
pt
an
es
Pe
n
Bu
t
ha
ne
0.1
0.01
10
an
d
He
av
ie
r
100
1,000
Pressure, psia
10,000
Figure 1 Equilibrium Ratios at 200˚F for Low-Shrinkage Oil(Amyx Bass & Whiting)1
40
20
10
8
6
Et
ha
Equilibrium Ratio, K
4
2
1
0.8
0.6
0.4
0.2
0.1
0.08
0.06
Pr
o
Bu
ne
pa
ne
tan
es
Pe
nta
ne
s
He
xa
nes
0.04 He
pta
ne
sa
0.02
nd
0.01
100
M
et
ha
ne
Heavier
1,000
Pressure, psia
10,000
Figure 2 Equilibrium Ratios at 200˚F for a Condensate Fluid (Amyx Bass & Whiting)1
Equilibrium Ratio Prediction and Calculation
CONSTANT
TEMPERATURE
Component 1
In Kj
Kj - 1.0
Component 2
In Pc
In P
Figure 3 Variation of K Values with pressure for a typical Binary Hydrocarbon System
M
10.0
et
ha
ne
Equilibrium Constant, K
h
Et
an
Pr
e
op
Bu
an
ta
e
ne
Pe
s
n
ta
H
ex
ne
an
s
es
Variations of Equilibrium
Constant With Pressure
at 200º F
1.0
H
0.1
0.01
10
ep
ta
ne
s
an
d
He
av
ie
r
100
1,000
10,000
Pressure, psia
Figure 4 Variation of K Values with pressure for a crude oil containing Light
Hydrocarbons
Institute of Petroleum Engineering, Heriot-Watt University
The apparent convergence pressure is related to the composition of the fluid. This
empirical basis has found favour in the industry for many years as a source of K
value data despite the ability to calculate the information based on equations of
state. However with the increasing availability of computational power and greater
confidence in equations of state the use of convergence pressure based K values is
diminishing.
A comprehensive set of equilibrium ratios is published in the GPSA Engineering Data
Book 2. The K values for single components are presented in the NGAA book for a
range of convergence pressures. K values for a convergence pressure of 5,000 psia
are at the back of this chapter for 2 pure components. Figure 9 and 10.
2.2.1 Method of Calculating Convergence Pressure
As was indicated in figures 2 and 3 different mixtures present different K values as
represented by different convergence pressures. The challenge is therefore to select
the convergence pressure value appropriate to the fluid for which we are seeking to
determine its phase separation characteristics.
The procedure is to convert our mixture into a two component mixture based on
methane then seek to identify a compound which has the same physical properties
as our heavier C2+ component
The chart, Figure 5 from the NGPSA Manual, can be used for this2. The method is
to take as the lightest component present in any significant quantity (1.0% or greater
in a raw mixture; it is usually methane) as the light component of a two component
system. The heavy component is estimated from the composition of the remaining
components. A visual estimate is usually good enough. Join the heavy and light
component together as shown on the Figure 5 and read off PK against the operating
temperature.
It has been established that the convergence pressure of systems as encountered
generally in natural gas processes is a function of the temperature and the composition
of the liquid phase. This presumes that the liquid composition had already been known
from a flash calculation using a first approximate guess for convergence pressure. Therefore, the method of calculating convergence pressure is an iterative procedure.
This calculation is suggested:
Step 1: Assume the liquid phase composition or make an approximation. (If there
is no guide, use the total feed composition).
Step 2: Identify the lightest hydrocarbon component which is present at least 0.1
mole % in the liquid phase.
Step 3: Calculate the weight average critical temperature and critical pressure for
the remaining heavier components to form a pseudo binary system. (A shortcut
approach good for most hydrocarbon systems is to calculate the weight average Tc
only).
Step 4: Trace the critical locus of the binary consisting of the light component and
pseudo-heavy component. When the averaged pseudo heavy component is between
two real hydrocarbons, interpolation of the two critical loci must be made.
Triple Phase Locus
-300
100
200
300
400
500
1,000
900
800
700
600
en
Nitrog
2,000
-200
-100
Triple Phase Locus
Met
h an
e
3,000
yle
ne
Eth
4,000
ne
0
Eth
a
5,000
CO2
C1-nC9
P
r
op
yle
ne
10,000
9,000
8,000
7,000
6,000
an
e
100
Pr
op
-C 2
H2
I-B
-C 2
H2
e
uta
n
C -Kens
ol
1
H2S
7
SO2
7
C3-nC
C -nC
2
C63-nC6
NH3
C -n
1 C
10
C -nC
1
C
nC 4-n 10
200
300
Temperature, ºF
10
C2 -nC
NBu
ta
ne
-I P
en
ta
ne
NP
en
ta
ne
N
He
xa
ne
20,000
400
NH
ep
ta
ne
N
-O
ct
an
e
Institute of Petroleum Engineering, Heriot-Watt University
Benzene
500
600
nC9
nC7-nC19
Methyl- Toluene
Cyclohexane
ne
-C 3
H2
ca
nC 6
H 2NDe
Convergence Pressures for Hydrocarbons
(Critical Locus)
nC13
800
e
nC15
nC14
an
ec
ad
x
He
n-
nC12
700
nC11
nC7-nC23
nC5-nC24
nC5-nC22
nC5-nC18
nC5-nC16
H2O
nC17
900
Kensol
nC18
Equilibrium Ratio Prediction and Calculation
Figure 5 Convergence pressure for Hydrocarbons.2
Convergence Pressure, PSIA
20,000
C thru' Mid-Continent
1
Crude
10,000
9,000
8,000
7,000
6,000
Absorber O
5,000
4,000
C thru' Gaso
lin
1
3,000
e
,C
ru
de
C
tin
en
t
on
Li
gh
t
Le
an
O
il
M
ol
.
wt
.1
Cr
200
61
ud
e
300
M
id
-C
Absorbers, Crude Flashing Towers
400
Di
st
illa
te
500
G
as
ol
in
e
ne
tha
1,000
900
800
700
600
1
2,000
Me
Convergence Pressure, PSIA
verhead
C 1 thru' Distillate Crude
100
-100
0
100
200
300
400
500
600
700
800
900
Temperature, ºF
Figure 6 Pseudocritical properties of Heptane Plus. (NGPSA 5th. edition 1957)
Step 5: Read the convergence pressure (ordinate) at the temperature (abscissa)
corresponding to that of the desired flash conditions, from Figure 5.
Step 6: Using the convergence pressure determined at Step 5, together with the
system, obtain K-values for the components from the appropriate convergencepressure K-charts.
Step 7: Make a flash calculation with the feed composition and the K-values from
Step 6.
Step 8: Repeat Steps 2 through 7 until the assumed and calculated convergence
pressures check within an acceptable tolerance, or until the two successive calculations give the same light and pseudo heavy components check within an acceptable
tolerance.
When the convergence pressure so determined is between the values for which charts
are provided, interpolation between charts may be necessary depending on how close
the operating pressure is to the convergence pressure.
If K-values do not change rapidly with PK(PK>>P) then the set of charts nearest to
calculated Pic may be used. Otherwise, a crossplot of K values versus PK all at constant
temperature and pressure, must be constructed for interpolation.
For those components characterised as a C7+ fraction Katz has suggests using a K
value of 15% of that of C7+. Danesh3 makes reference to other correlations to estimate
the critical properties of C7+ fractions. McCain6 has also presented pseudo critical
properties of C7+ as a function of molecular weight and specific gravity, his correlation
figures are shown below in figure 7.
10
Equilibrium Ratio Prediction and Calculation
1700
Pseudocritical temperature, °R
1600
1500
Specific gravity of heptanes plus = 1.0
1400
.95
1300
.85
1200
.75
.90
.80
.70
1100
1000
900
100
150
200
250
Molecular weight of heptanes plus
300
500
Pseudocritical pressure, °R
450
400
1.0 = Specific gravity of heptanes plus
350
.95
.90
300
.85
.80
250
.75
.70
200
150
100
100
150
200
250
Molecular weight of heptanes plus
300
Figure 7 Pseudocritical properties of Heptane Plus6.
5
The following example has been presented by McCain in his text . and is helpful as
a worked example in determining the appropriate convergence pressure charts
Example: McCain5. The gas-liquid equilibrium of a high-shrinkage crude oil has
been calculated. The composition of the liquid phase formed at 75˚F and 100 psia
is given below. A convergence pressure of 2000 psia was used to determine the
equilibrium ratios for the calculations. What value of convergence pressure should
have been used for this mixture?
Institute of Petroleum Engineering, Heriot-Watt University
11
Component
Composition of liquid
mole fraction
Methane
Ethane
Propane
i-butane
n-butane
i-pentane
n-pentane
Hexane
Heavier*
0.0356
0.0299
0.0919
0.0170
0.0416
0.0198
0.0313
0.0511
0.6818
1.0000
* Molecular weight and gravity of C7+
assumed to be 263 and 0.886
Solution
First, the composition of the liquid must be expressed as weight fraction.
Component
Composition of
liquid mole
C1
C2
C3
i-C4
n-C4
i-C5
n-C5
*C6
*C7+
0.0356
0.0299
0.0919
0.0170
0.0416
0.0198
0.0313
0.0511
0.6818
1.0000
Molecular
weight Mj
16.0
30.1
44.1
58.1
58.1
72.2
72.2
86.2
263
Weight xMj
0.5696
0.9000
4.0528
0.9877
2.4170
1.4296
2.2599
4.4048
179.3134
196.3348
Composition of
liquid, weight
fraction
xjMj/ΣxjMj
0.0029
0.0046
0.0206
0.0050
0.0123
0.0073
0.0115
0.0224
0.9133
0.9999
* Molecular weight and gravity of C7+ assumed to be 263 and 0.886 respectively
Second, adjust weight fraction to exclude methane and calculate weighted-average
critical properties.
Component Composition
excluding
Critical
°R
temperature wjtcj
°R Tcj
psia
pressure, wjpcj
psia pc
psia pcj
C2
C3
i-C4
n-C4
i-C5
n-C5
C6
*C7+
549.8
665.7
734.7
765.3
828.8
845.4
899.3
1360
707.8
616.3
529.1
550.7
490.4
488.6
445.4
240
methane, wj
0.0046
0.0207
0.0050
0.0123
0.0073
0.0115
0.0225
0.9160
0.9999
* Critical properties of C7+ from figure 7.
12
2.53
13.78
3.67
9.41
6.05
9.72
20.23
1245.76
wt avg Tc=
1311°R, 851°F
Critical
3.26
12.76
2.65
6.77
3.58
5.623
10.02
219.84
wt avg pc=
265 psia
Equilibrium Ratio Prediction and Calculation
The calculated values of weight averaged Tc and Pc are close to Kensol and the mid
continent crude point of figures 5 and 6. The location of the temperature 75°F with
the methane - component (Kensol or mid - continental crude) is at a convergence
pressure of around 10,000 psia, a value much greater than the assumed value of 2,000.
The calculator needs to be repeated using the higher convergence pressure related K
value data. The calculation procedure will converge when the estimated convergence
pressure is the same as the calculated convergence pressure.
3 EQUATION OF STATE BASED EQUILIBRIUM CALCULATIONS
3.1 Methods Based on Empirical Equations of State of Fluid Phase
Theory
The thermodynamic properties of a pure fluid may be represented by an equation of
state of the generalised form:
f (P,V,T) = 0
where the pressure P, temperature T and molar volume V are related by a mathematical
function. Most equations of state have been developed by fitting an analytical expression
to pure component PVT data. To extend the application of the developed equation
to mixtures, the parameters of the equation must take into account the composition
of the mixture and for simplicity require only the insertion of pure component data.
Since most equations are effectively only mathematical models, the equations tend
in general to be more complex, that is contain a large number of parameters, as the
required level of accuracy increases. The basic assumption in the development of an
equation of state is that at a critical point:
 ∂2 P 
 ∂P 
  =  2 = 0
 ∂V  T  ∂V  T
Equations of state can be used for the following purposes:
1representation of PVT data to assist data smoothing and improve
interpolation;
2 prediction of vapour-liquid equilibria of mixtures especially at high
pressures;
3
prediction of gas phase properties of pure fluids and their mixtures using a
minimum amount of experimental data.
In the gas property chapter we reviewed the topic of equations of state. Currently
although a number of different equations could be used, the industry favours two, the
Peng Robinson equation of state and the Soave modification of the Redlich Kwong
equation of state.
Institute of Petroleum Engineering, Heriot-Watt University
13
3.2 Prediction of Vapour-Liquid Equilibrium
An equation of state capable of predicting behaviour in the liquid phase and vapour
phase is sufficient in itself for vapour-liquid equilibrium predictions. Unfortunately
as we have indicated although modifications to the Van der Waals type equation have
predicted vapour properties the equations in general have not been so accurate in
predicting the liquid behaviour.
Many vapour-liquid prediction methods have therefore gone to an equation of state,
for example the Redlich-Kwong equation, for the prediction of the vapour phase
fugacity and have employed a liquid theory to evaluate the liquid phase properties.
We will now go through the steps in the application of equation of states in vapour
liquid equilibrium calculations. It is worth noting that the steps described are those
which would be taking place within each grid block of a compositional reservoir
simulator. The implication therefore is a very large computational load for large
numbers of grid blocks.
The application of equation of states is based on the fact that at
equilibrium, the fugacity of the gas and liquid phases are identical i.e:
fiv = fiL
(2)
where fiv and fiL are the fugacities of component in the vapour and liquid phases
respectively.
The ratio of the fugacity to pressure is called the fugacity coefficient where φi = fi/(pzi)
(3)
where zi is the composition of the component in the system.
The fugacities can be expressed for a vapour and liquid therefore by:
fiL = φi xip
and:
fiv = φi yip
p
yi & xi
- is the system pressure;
- are the mole fractions of i in the vapour and liquid phase.
Therefore:
fiL
xp
φ
y
Ki = iL = i = i
fig
φig
ni ( )
yi p
The fugacity coefficients for the liquid and gas can be calculated using the following
equation
14
Equilibrium Ratio Prediction and Calculation
ln φi =
V

1  RT  ∂p 

 dV − ln z
−
 
RT ∫∞  V  ∂ni  T , V , n 
i 

(4)
McCain5 and also Ahmed6 give good descriptions of the application of EOS in
equilibrium calculations.
At the present time the preferred EOS are the 3 parameter Peng-Robinson (PR) 8,9
and the Soave-Redlich Kwong (SRK)10. Other equations exist and are more accurate
in predicting some properties. The considerable investment in binary interaction
parameters for the preferred equations is such that there is a reluctance to use some
recently developed EOS’s. Danesh's3 text gives a good review of the EOS.
Since the applications of the equations are applied to mixtures, mixing rules are required
to determine the values for the parameters in the particular EOS being used.
We will use the Peng Robinson equation as our basis but others could be used.
The Peng Robinson equation is
p=
RT
aT
−
Vm − b Vm (Vm + b) = b(Vm − b)
(5)
The equation is set up for both liquid and gas using the following mixing rules to
calculate b, and aT . The rules are presented in the context of gas ie. y, clearly for
liquids, x values are used.
and
b = ∑ yi bi
i
aT = ∑ ∑ yi y j (aTi aTj )0.5 (1 − kiy )
i
j
(6)
(7)
where kij are the binary interaction coefficients, and kii = kjj = o and kij = kji.
The value of bi and aTi for the individual components are calculated as follows
bi = 0.07780
and
RTci
Pci
aTi = aciαi and
ac i = 0.45724
R2 T 2 ci
Pci
Institute of Petroleum Engineering, Heriot-Watt University
(8)
(9)
(10)
15
αi is a temperature dependant factor where
α i 0.5 = 1 + m(1.0 − Tri0.5 )
(11)
where
m = 0.37476 + 1.54226ω i − 0.26992ω i2
(12)
The Peng Robinson equation can be written as a cubic equation in terms of z, the
compressibility factor,
z3 - (1-B)z2 + (A-2B -3B2)z - (AB - B2 -B3) =0
(13)
where
A=
aP
R2 T 2
B=
bP
RT
(14)
The solution to the above equation gives three roots for z. The highest value is the
liquid phase z factor and when the equation is solved using vapour compositions then
the lowest root is the z factor for the vapour.
If the Peng Robinson equation is combined with the fugacity coefficient equation the
following equation results for the fugacity coefficient of each component.
 z + (21.5 + 1) B 
A
ln φi = − ln( z − B) + ( z − 1) Bi′ − 1.5 ( Ai′ − Bi′) ln 
1.5

2
 z − (2 + 1) B 
Where B'i = bi/b
(15)
(16)
and
Ai′ =
1
aT


0.5
0.5
2 aTi ∑ yi aTi 1 − kij 
1


(
)
(17)
Following all these steps independently for the liquid and gas phases the fugacities
of the gas and liquid phases can be calculated.
fLi = xipφLi and fvi = yipφvi
When fLi=fvi then equilibrium is achieved and calculations are complete.
Having presented these equations we will now describe the process to calculate K
values , vapour liquid equilibrium ratios and compositions given a system composition,
temperature and pressure.
16
Equilibrium Ratio Prediction and Calculation
Step 1: Estimate the K values of the system. The Wilson11 equation is good for
this purpose.
p 
T 

Ki =  ci  exp 5.37(1 + ω i )1 − ci  

 p
T 

(18)
wi = acentric factor for component i
Step 2: Carry out vapour equilibrium calculations using estimated K values using
the iterative procedure outlined previously. That is estimate the V/L ratio and iterate until convergence is obtained, that is when compositions sum to unity. We now
have liquid and vapour compositions to use in equation of state calculations.
Step 3: Using liquid compositions, calculate the A & B values for the EOS and then
solve the z-value form of the EOS, to determine z. The lowest root (value) is the
z value for the liquid.
Step 4: Calculate the compositional coefficients A'i and B'i for the liquid components
and calculate the fugacity coefficients of the components of the liquid
Step 5: Repeat steps 3 & 4 using the vapour phase compositions.
Step 6: Calculate fgi and fLi fgi=yipφi and fLi= xipφI . Check if fgi= fLi.
If this value is greater than 10-12 then the whole process has to be repeated from step
1, except that the K values used are the calculated K values arising from step 5 i.e.
Rather than set up the tolerance check on fugacity
equivalence the tolerance can be
εi = fLi - fgi
based on K values.
A value of ε of 0.001 can be used for the sum of the errors.
Ki =
εi =
φ Li
φ gi
(K
E
i
−K
E
i
K K
)
C 2
i
C
i
(19)
(20)
The iteration is complete when these tolerance limits are met and the compositions
of the respective phases are those which have been been determined at the last
iteration. Calculations can then proceed to provide volumetric and density data for
the respective phases.
Danesh3 has given a flow diagram for the above flash calculation.
An example follows to illustrate the calculation process.
Institute of Petroleum Engineering, Heriot-Watt University
17
Start
Input zi, P, T,
Component Properties
Estimate Ki, Using Eq.(18)
Calculate xi, yi, Using Figs. 20-22 (Ch 12)
Set-Up EOS
For Liquid
Set-Up EOS
For Vapour
Calculate ZL
Calculate ZV
Calculate fiL
Calculate fiV
Adjust Ki =
Ki old (fiL/fiV)
NO
NO
Is
Σ(1- fiL/fiV)2<10-12
?
YES
YES
Print xi,yi,vL,vV,nL,nV
End
Figure 8 Flow chart of flash calculations using equation of state.2
EXERCISE 1.
Calculate the liquid and vapour phase composition when the mixture with the composition given in Table 1 is flashed to 2000 psia and 100°C. Use the Peng- Robinson
equation of state with binary interaction parameters given in Table 2.
Table 1: Multicomponent system.
Component
Methane
n-Pentane
n-Decane
n-Hexadecane
Composition
mole fraction
0.55100
0.11400
0.14600
0.18900
Table 2: Binary interaction parameters of Peng- Robinson Eq.
18
Equilibrium Ratio Prediction and Calculation
No.
1
2
3
4
Component
Methane
n-Pentane
n-Decane
n-Hexadecane
1
0.0000
0.0236
0.0489
0.0600
2
3
4
0.0000
0.0000
0.0070
0.0000
0.0000
0.0000
4. THE COMPOSITION DATA REQUIREMENTS FOR THE APPLICATION OF THERMODYNAMIC EQUILIBRIUM PREDICTION METHODS
TO MULTI-COMPONENT HYDROCARBON MIXTURES
In designing a versatile phase equilibrium prediction computer program, the available
experimental data for the mixture whose performance is to be simulated, must be
considered. Most laboratory analyses of crude oil are predominantly concerned with
the composition. This is normally reported as a percentage weight. This presents
few problems for lighter hydrocarbons such as methane which can have no isomers
nor for other light hydrocarbons which do have isomers, for example butane, since
the number of hydrocarbons having the same number of carbon atoms is relatively
few. However, for heavier hydrocarbons the number of hydrocarbons which possess
the same number of carbon atoms can be very large due to the presence of not only
aliphatic or straight chain compounds but also of cyclic compounds such as aromatics
or naphthenes, or combinations of these structural types. Obviously the structure of
molecules greatly influences the forces between them and hence their deviations from
ideality. This is most significant when binary interaction parameters are employed in
the mixing. The analysis of the heavier components in crudes is further complicated
by the fact that most of the individual components are present in very small quantities
which makes them difficult to identify and quantify.
Ideally what is required is a full analysis of a crude oil in which all components
are quantified and identified so that the effects on the mixture behaviour due to
their structural and chemical properties can be taken into consideration. The phase
equilibrium prediction method which is employed, must match the quality and extent
of the experimental data available, for example, the use of a highly accurate method
might be invalidated if the composition analysis has been carried out on a basis of
generalised crude fractions such as boiling range. However, this is where flexibility
must be built into the program.
At present there is a tendency to increase the extent to which a crude oil composition
is analysed and this can only serve to make phase equilibrium prediction more
accurate. Condensate well fluids, however, are still generally reported on a basis of
a heptanes plus weight fraction, that is the collective weight fraction of heptane and
heavier components, because these heavier components are generally only present
in minute quantities.
It is thus desirable that any proposed phase equilibrium method should be able to
utilise detailed experimental data based on identifying the individual components
of a crude oil as well as data derived from a more generalised approach as discussed
above.
Institute of Petroleum Engineering, Heriot-Watt University
19
Most of the work carried out on fluid phase equilibria has concerned itself solely with
binary or tertiary systems and occasionally mixtures containing pure components whose
behaviour is quite well understood. Much of the published work on the adaptation
of experimental data for prediction techniques has been related to petroleum refinery
operations.
20
Equilibrium Ratio Prediction and Calculation
Pressure, PSIA
10
1,000
2
30
4
50
6 7 8
9 100
2
300 4 500 6
7 8 91,000
9
8
7
6
2
3,000 4
6 7 8 9
Plotted from 1947 tabulation
of G. G. Brown, University of
Michigan. Extrapolated and
drawn by The Fluor Corp. Ltd.
in 1957.
5
4
9
8
7
6
5
4
3
3
2
2
100
100
9
8
7
6
9
8
7
6
5
5
4
4
3
3
Temperature °F
2
2
30
50
40 0
0
25
10
0
10
0
9
8
7
6
9
8
7
6
0
20
K= y/x
10,000
1,000
10
0
80
5
4
5
4
60
40
20
3
2
3
2
0
-2
0
-4
1.0
1.0
0
9
8
7
6
9
8
7
6
-6
0
5
5
4
4
3
3
2
2
0.1
0.1
9
8
7
6
9
8
7
6
5
5
4
4
3
3
2
2
10
K= y/x
2
30
4
50
6 7 8 9 100
2
300 4 500 6
Pressure, PSIA
7 8 91,000
2
3,000 4
6 7 8 9
ETHANE
10,000
CONV. PRESS. 5000 PSIA
Figure 9 K value chart for Ethane.
Institute of Petroleum Engineering, Heriot-Watt University
21
Pressure, PSIA
10
1,000
2
30
4
50
6 7 8 9 100
300 4 500 6
2
7 8 9 1,000
9
8
7
6
3,000 4
6 7 8 9
Plotted from 1947 tabulation
of G. G. Brown, University of
Michigan. Extrapolated and
drawn by The Fluor Corp. Ltd.
in 1957.
5
4
10,000
1,000
9
8
7
6
5
4
3
3
2
2
Temperature oF
100
100
50
9
8
7
6
0
45
9
8
7
6
0
40
0
38
0
36
0
34
0
32
0
30
0
28
0
26
0
24
0
22
0
200
5
4
3
2
10
K= y/x
2
9
8
7
6
5
4
3
5
4
3
2
10
9
8
7
6
5
4
3
18
0
2
K= y/x
2
16
0
14
0
1.0
9
8
7
6
1.0
12
9
8
7
6
100
5
80
3
0
5
4
3
2
4
2
60
40
0.1
9
8
7
6
0.1
9
8
7
6
20
5
5
4
4
0
3
3
2
2
-20
10
2
30
4
50
6 7 8 9 100
2
300 4 500 6
Pressure, PSIA
7 8 9 1,000
Figure 10 K value chart for Octane.
22
2
3,000 4
6 7 8 9
OCTANE
10,000
CONV. PRESS. 5000 PSIA
Equilibrium Ratio Prediction and Calculation
Solution to Exercise
EXERCISE
Calculate the liquid and vapour phase composition when the mixture with the composition given in Table 1 is flashed to 2000 psia and 100°C. Use the Peng- Robinson
equation of state with binary interaction parameters given in Table 2.
Table 1: Multicomponent system.
Component
Composition
mole fraction
0.55100
0.11400
0.14600
0.18900
Methane
n-Pentane
n-Decane
n-Hexadecane
Table 2: Binary interaction parameters of Peng- Robinson Eq.
No.
1
2
3
4
Component
Methane
n-Pentane
n-Decane
n-Hexadecane
1
0.0000
0.0236
0.0489
0.0600
2
3
4
0.0000
0.0000
0.0070
0.0000
0.0000
0.0000
SOLUTION
Step 0: Calculate the coefficients of components in the mixture, using the properties
(critical temperature Tc and critical pressure Pc) of pure compounds:
R2 T 2 ci
aci = 0.45724
Pci RT
bi = 0.07780 ci
Pci If ω <0.5
:
m= 0.37464 + 1.54226 ω - 0.26992 ω2
If ω >0.5
:
m= 0.3796 + 1.485 ω - 0.1644 ω2 - 0.01667ω3
(10)
(8)
(12)
w - acentric factor from thermodynamic property tables for pure components
(
)
(
α i = 1.0 + m 1.0 − T r )
2
(11)
ai = ci αi
Component Tci
K
Pci
atm
C1
n-C5
n-C10
n-C16
45.39
33.26
20.82
13.82
190.6
469.7
617.7
723
ωi
acentric
factor
0.0115
0.2515
0.4923
0.7174
(9)
aci
αi
ai
bi
2.4632
20.4230
56.4253
116.4577
0.7112
1.1686
1.5327
1.9176
1.7518
23.8663
86.4825
223.3236
.0268
.0902
0189
.3340
Institute of Petroleum Engineering, Heriot-Watt University
23
Step 1: Select initial values for equilibrium ratios (K-values) and calculate the trial
composition of liquid and vapour phases at equilibrium. The procedure is iterative as
follows.
Iteration One
Initial K-values can be calculated from Wilson Equation:
Pci
T
Exp (5.37)(1.0 + ω i )1.0 − ci ))
P
T
Ki =
Component
Methane
n-Pentane
n-Decane
n-Hexadecane
Feed (Zi)
0.55100
0.11400
0.14600
0.18900
(15)
K-Values (Ki)-Wilson Eq.
4.7644
4.290 E-02
7.9917 E-04
1.7768 E-05
Step 2: Determine the trial value of phase fraction (vapour fraction here) by solving
the vapour equilibrium equation:
N
∑L
i=l
V
Zi
+ KL
=V
(5)
Vapour Fraction (Vf) = 0.4378
Use the determined vapour fraction (Vf) to calculate the liquid and the vapour phase
composition from material balance:
Component
Methane
n-Pentane
n-Decane
n-Hexadecane
Feed (Zi)
0.55100
0.11400
0.14600
0.18900
Liquid (xi)
0.2081
0.1962
0.2595
0.3362
Vapour (yi)
0.9914
0.0084
0.0002
0.000006
Step 3, 4 and 5: Calculate composition dependent coefficients for compressibility factor
(Z - factor) calculations for both liquid and vapour:
Liquid:
N
N
( ) (1. − k )
a = ∑ ∑ xi x j ai a j
i=l J =l
0.5
ij
(7)
N
b = ∑ xi bi
i=l
(8)
N
Vapour:
24
N
( ) (1. − k )
a = ∑ ∑ yi y j ai a j
i=l J =l
0.5
ij
(7)
Equilibrium Ratio Prediction and Calculation
N
b = ∑ yi bi
i=l
aP
A= 2 2
RT
Phase
Liquid
Vapour
(8)
B=
bP
RT
a
74.88118
1.83396
(14)
b
0.18470
0.02737
A
10.86872
0.26619
B
0.82089
0.12165
Calculate Z - factors of the liquid and vapour phases (Peng - Robinson EOS):
Z3 - (1 - B)Z2 + (A - 2B - 3B2)Z - (AB - B2 - B3) = 0.
(13)
Vapour Phase, take the highest root: Zv = 0.92034
Liquid Phase, take the smallest root: Zl = 0.96602
Step 6: Calculate the fugacity of each component in the liquid and vapour phase:
ln φi = − ln( Z − B) + ( Z − l )
 bi 
bi
A  1  0.5 N
0.5
−
2 ai ∑ y j a j l − kij  −  x

b 2 2B  a 
j =l
 b
(
)
 Z + ( 2 + 1) B 
ln 

 Z − ( 2 − 1) B 
(15)
fi v = yi Pφi v
Fugacity of components in the vapour phase:
fi l = xi Pφi l
Fugacity of components in the Liquid phase:
Liquid
φil
fil
.23811E+01 .67429E+02
.66312E-01 .17708E+01
.19023E-02 .67189E-01
.43075E-04 .19707E-02
Component
Methane
n-Pentane
n-Decane
n-Hexadecane
Vapour
φiv
fiv
.90058E+00 .12150E+03
.36516E+00 .41838E+00
.14686E+00 .41455E-02
.51625E-01 .41965E-04
End of Iteration One
Ki =
φil
φiv Next Iteration:
(16)
Iterate till the fugacity of every component in the liquid phase is equal to that of vapour
phase (iterate from Step 2.):
Final Iteration
Component
Methane
n-Pentane
n-Decane
n-Hexadecane
Feed (Zi)
0.55100
0.11400
0.14600
0.18900
K-Values (Ki)
2.44097
0.19991
1.6774 E-02
1.25099 E-03
Institute of Petroleum Engineering, Heriot-Watt University
25
Then: Vapour Fraction (Vf) = 0.2710
Component
Methane
n-Pentane
n-Decane
n-Hexadecane
Feed (Zi)
0.5510
0.1140
0.1460
0.1890
Liquid (xi)
0.3962
0.145665
0.199039
0.259147
Vapour (yi)
0.967236
0.029100
0.003339
0.000324
Calculate composition dependent coefficient for z - factor calculations for both liquid
and vapour:
N
N
( ) (1. − k )
a = ∑ ∑ xi x j ai a j
Liquid:
i=l J =l
(7)
0.5
ij
N
b = ∑ xi bi
(8)
i=l
N
N
( ) (1. − k )
a = ∑ ∑ yi y j ai a j
Vapour:
i=l J =l
0.5
ij
N
(4)
b = ∑ yi bi
(8)
i=l
A=
aP
R2 T 2
Phase
Liquid
Vapour
B=
bP
RT
a
48.03656
2.11326
(14)
b
0.14799
0.02929
A
6.97233
0.30673
B
0.6577
0.1301
Calculate z - factors of Liquid and Vapour and Vapour Phase (Peng - Robinson
EOS):
Z3 - (1 - B)Z2 + (A - 2B - 3B2)Z - (AB - B2 - B3) = 0.
(3)
Vapour Phase, take the highest root: Zv = 0.90039
Liquid Phase, take the smallest root: Zl = 0.81539
Calculate the fugacity of each component in the liquid and vapour phase:
 bi 
bi
A  1  0.5 N
0.5
ln φi = − ln( Z − B) + ( Z − l ) −
2 ai ∑ y j a j l − kij  −  x

b 2 2B  a 
j =l
 b
(
 Z + ( 2 + 1) B 
ln 

 Z − ( 2 − 1) B 
Fugacity of components in the vapour phase:
Fugacity of components in the liquid phase:
26
fi v = yi Pφi v
fi l = xi Pφi l
)
Equilibrium Ratio Prediction and Calculation
Component
Methane
n-Pentane
n-Decane
n-Hexadecane
Molecular volume
P2
Phase
Liquid
Vapour
Liquid
φil
fil
.22032E+01 .11881E+03
.66155E-01 .13105E+01
.19873E-02 .53832E-01
.45319E-04 .15983E-02
Vm 2
Vapour
φiv
fiv
.90269E+00 .11882E+03
.33093E+00 .13106E+01
.11847E+00 .53831E-01
.36226E-01 .1598E-02
2 RT
P
PM
2 RT
2
099039
.81539
Vm gr mole/cc
0.05451
0.04936
Ma g/mole/cc
103.86
18.165
Ps/cc
0.56613
0.08967
Compositions are for final iteration
Component
Methane
n-Pentane
n-Decane
n-Hexadecane
Feed (Zi)
0.5510
0.1140
0.1460
0.1890
Liquid (xi)
0.3962
0.145665
0.199039
0.259147
Institute of Petroleum Engineering, Heriot-Watt University
Vapour (yi)
0.967236
0.029100
0.003339
0.000324
27
REFERENCES
1. Amyx, J.W, Bass, D.M, Whiting, RL, Petroleum Reservoir Engineering, McGraw
Hill. New York 1960
2. Gas Processors Suppliers Association. “Engineering Data Book 9th Edition.
Tulsa
3. Danesh, A, PVT and Phase Behaviour of Petroleum Reservoir Fluids. 1998
Elsevier. pp 66-77
4. Wilson,G: “A Modified Redlich-Kwong EOS, Application to General Physical
Data Calculations”, Paper Nu 15C, presented at the AIChE 65th National Meeting
(May 1968)
5. McCain, W.D. The Properties of Petroleum Fluids Penwell. 1st. Edition 1973
Starling, K.E., A.I.Ch.E.Jnl, 1972, 18, 6, 1184-1189.
6. McCain, W.D. The Properties of Petroleum Fluids. Pennwell, 2nd Ed., 1990,
p 414-436.
7. Ahmed, T. Hydrocarbon Phase Behaviour, Gulf Pub., 1989.
8. Peng, D.Y., Robinson, D.B., I.E.C. Fundamentals 1976, 15, 1, 59-64.
9. Jhaveri B.S. and Youngren G.K. Three-Parameter Modification of the Peng-Robinson Equalising State to Improve Volumetric Predictions SPE 13118 (1984)
10. Soave, G., Chem. Eng. Sci., 1972, 27, 1197-1203.
11. Wilson, G.M., "A Modified Redlich - Kwong EOS Application to General
Physical Data Calculations." Paper No 15c presented at the AIChE 65th National
Meeting (May 1968) AI ChE.
28
PVT Analysis
CONTENTS
1
SCOPE
2
SAMPLING
2.1 Subsurface Sampling
2.2 Surface Sampling
3
SAMPLING WET GAS AND GAS
CONDENSATE SYSTEMS
3.1 Introduction
3.2 Phase Behaviour
3.3 Sampling Gas And Gas Condensate
Reservoirs
3.4 Separator Sampling Points
3.5 Sample Details
4
APPARATUS
5
RECOMBINATION OF THE SURFACE OIL AND GAS SAMPLES
6
PVT TESTS
6.1 Flash Vaporisation
6.2 Differential Vaporisation
6.3 Separator Tests
6.4 Viscosity
6.5 Hydrocarbon Analysis
7
WAX AND ASPHALTENES
7.1 Wax Crystallization Temperature
8
SUMMARY OF RESULTS PRIOVIDED BY AN
OIL SAMPLE PVT TEST
8.1 Interfacial Tension IFT
8.2 IFT Measurements Methods
9
RETROGRADE CONDENSATION
10 UNDERSTANDING PVT REPORTS
11 PURPOSE OF THE PVT REPORT
12 OIL SAMPLE PVT STUDY
12.1 Separator Tests of Reservoir Fluids
12.2 Fluid Properties at Pressures Above
The Bubble Point Pressure
12.3 Total Formation Volume of Original
Oil Below The Bubble Point Pressure
12.4 Differential Liberation Tests
12.5 Calculation of Gas - Oil Ratios Below
The Bubble Point.
12.6 Calculation of Formation Volume
Below The Bubble Point
12.7 Viscosity Data
13 GAS CONDENSATE PVT REPORT
14 HIGH PRESSURE / HIGH TEMPERATURE,
HP/HT, FLUIDS
15 MERCURY
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•
Describe briefly the scope of PVT analysis
•
Describe the sampling options for oil systems
•
Describe the impact of well flow interruption on the sampling of ‘wet’ and gas
condensate systems
•
List the main items of equipment in PVT analysis.
•
List and describe the five main PVT tests for oils systems and their
application.
•
Explain the two basic PVT tests for gas condensates, constant mass and constant
volume depletion.
•
Be able to use an oil PVT analysis report to calculate;
•
The gas-oil ratio and oil formation volume factor at the bubble point
pressure.
•
Determine the bubble point pressure from a set of P vs. V relative volume test
data
•
Calculate oil formation volume factors above the bubble point.
•
Determine the total formation volume factors above and below the bubble
point.
•
Determine the oil formation volume factors and gas-oil ratios for pressures
below the bubble point pressure.
PVT Analysis
1 SCOPE
Reservoir fluid analysis provides some of the key data for the petroleum engineer. The quality of the testing, therefore, is important to ensure realistic physical property
values are used in the various design procedures. As important is the quality of
the samples collected to ensure that the fluids tested are representative of the field. Clearly, any subsequent high quality testing is of little value if the sample is not
representative.
PVT-analysis of a reservoir fluid comprises the determination of:
(a) the correlation between pressure and volume at reservoir temperature;
(b) various physical constants that enter into reservoir engineering calculations,
such as viscosity, density, compressibility etc;
(c) the effect of separator pressure and temperature on oil formation volume factor,
gas/oil ratio etc;
(d) the chemical composition of the most volatile components.
The physical properties measured depend on the nature of the fluid under evaluation.
For a dry gas, the key parameters are, the composition, the specific gravity, the gas
formation volume factor, compressibility factor, z , and viscosity as a function of
pressure. The isothermal gas compressibility can be determined from the z value
with pressure variation.
For a wet gas, then all of the above parameters for a gas are required. However there
are some variations because of the production of liquids with gas. The formation
volume factor used is the gas condensate formation volume factor, Bgc, which is
the reservoir volumes of gas required to produce one stock tank volume (barrel or
M3) of condensate. The composition, specific gravity and molecular weight of the
produced condensates and produced gas are required. The composition and apparent
specific gravity of the reservoir gas are obtained by recombining the values for the
gas and condensates.
For an oil system, the following information is required; the bubble point pressure
at reservoir temperature, the composition of the reservoir and produced fluids, the
formation volume factor, the solution gas to oil ratio, the total formation volume factor,
and viscosity, all as a function of pressure. The coefficient of isothermal compressibility
of oil. The impact of separation on the above properties. The impact of operating
below the bubble point on the formation volume factor and solution GOR.
For a gas condensate system, properties measured reflect those for both wet gas and
oil analysis. The related property to the saturation pressure for the oil, the bubble
point pressure, for a gas condensate is the dew point pressure. Above the dew point
the compressibility characteristics of the gas are required, and the impact of allowing
the reservoir to drop below the dew point is another important evaluation.
Institute of Petroleum Engineering, Heriot-Watt University
2 SAMPLING
The value to be attached to the laboratory determinations depends on whether the
sample investigated is representative of the reservoir contents or at least of the drainage
area of the well sampled. Therefore, the composition of the fluid entering the well
and of the samples taken should be identical with that of the fluid at all other points
in the drainage area. It is therefore desirable to take samples early in the life of the
reservoir in order to minimise errors caused by differences in relative movement of
oil and gas after solution gas has been liberated.
The taking of samples can be accomplished either by sub-surface sampling or by
surface sampling. There have been considerable advances in recent years in this
area aimed at taking samples under down hole pressure conditions. A brief survey
of these methods is given below, for detailed descriptions of the techniques, which
is outside the scope of this section, service companies providing sample collection
service should be consulted.
2.1 Subsurface Sampling
In this case a subsurface sampler is lowered into the well and kept opposite the
producing layer for a sufficiently long time, Figure 1. Many types of bottom hole
samplers have been devised and described in the literature.
Subsurface samples can only be representative of the reservoir contents when the
pressure at the point of sampling is above or equal to the saturation pressure. If this
condition is not fulfilled, one should take a surface sample. Even at pressures close
to the saturation pressure there is a serious possibility of sample integrity being lost
as a result of the system going two phase during transfer to the sample chamber.
A simple test can be carried out in the field at the well site to find out whether a reliable
sample has been obtained. In this test the sample cylinder is pressurised either by using
a piston cell or using mercury as the displacing fluid. Whereas mercury was the most
common fluid to be used as a pressure transfer and volume change fluid, because of
toxic and other concerns its use is diminishing. From the relation between injection
pressure and volume of mercury injected, the following properties are derived:
(1) the pressure existing within the sampler when it is received at the surface;
(2) the compressibility of the material within the sampler;
(3) the bubble-point pressure of the contents of the sampler.
If two or three samples taken at short time intervals show the same measured properties,
it is highly probable that good samples have been obtained.
In recent years there have been considerable advances on downhole fluid
sampling.
The contents of the sampler are transferred by means of a mercury pump into a high
pressure shipping container. Two fillings of a sampler of 600 ml capacity are usually
sufficient for a complete PVT analysis. In recent years mercury has been replaced
using the application of piston cells by some companies. PVT Analysis
Subsurface sampling is generally not recommended for gas-condensate reservoirs
nor for oil reservoirs containing substantial quantities of water.
Lubricator
To wire line truck
Ground level
Tubing
Packer
Producing Formation
Bottom hole sampler
run in tubing
Subsurface sampling
Figure 1 Subsurface Sampling
2.2 Surface Sampling
A sample of oil and gas is taken from the separator connected with the well (Figures
2 - 5 give sketches of vertical and horizontal separators and the arrangement for
collecting different fluid samples). The surface oil and gas samples are recombined
in the laboratory on the basis of the producing GOR. Particular care, therefore, must
be exercised in the field to obtain reliable samples and accurate measurement of the
GOR and separator conditions.
In the case of two or three stage separation the samples are taken from the high
pressure separator.
Institute of Petroleum Engineering, Heriot-Watt University
Gas
outlet
Vertical
Separator
Gauge
Inlet
Sight
glass
Liquid
outlet
Momentum
absorber
Guage
Gas
outlet
Inlet
Sight
glass
Horizontal
Separator
Figure 2 Vertical and Horizontal Separators
Sample Vessel
Figure 3 Separator Gas Sampling
Liquid
outlet
PVT Analysis
Sample Vessel
Figure 4 Separator Liquid Sampling by Gas Displacement
Figure 5 Separator Liquid Sampling by Water Displacement
3. SAMPLING WET GAS AND GAS CONDENSATE SYSTEMS
3.1 Introduction
The use and value of any PVT study or other analysis of a reservoir fluid is dependant
on the quality of the sample collected from the reservoir. Many PVT reports show
a variation of results from fluids from the same well. Sampling wet gas and gas
condensate fluids can give rise to errors. During the sampling procedure it is often
possible to alter the conditions such that the fluids sample are not representative of
those within the reservoir, the characteristics of which are being assessed during the
PVT report. It is important, therefore, in sampling reservoir fluids to ensure that the
conditions during which the samples are being taken are not altered to give rise to
false samples.
3.2 Phase Behaviour
The behaviour characteristics of fluids are uniquely described by a phase diagram,
Figure 6. The hydrocarbon mixture with it’s own unique composition will have it’s
own phase diagram and phase envelope. Within the phase envelope the system is in
two phases, whereas outside the phase envelope a single phase exists. Institute of Petroleum Engineering, Heriot-Watt University
Liquid
Vapour
c
e
rv
Dewpoint curve
uid
ur
uid
Liq
Liq
%
ui
d
20
5%
Li
q
10
0%
uid
Vap
o
Liq
40
%
80
%
10
0%
Liq
80%
uid
Liq
70
uid
%
Liq
uid
Pressure
Bu
bb
le
po
in
t
cu
Temperature
Figure 6 Phase diagram for a reservoir fluid
At a particular point within the envelope the composition of each component in each
phase is constant. The separation of oil and gas as predicted by the phase diagram
results in each phase itself having a phase diagram. These phase diagrams intersect at
the separation temperature and pressure, the oil will exist at it’s bubble point and the
gas at it’s dew-point. Therefore, for example, in the separator, gas will be separated
and produce gas at it’s dew-point and the oil separated will be at it’s bubble point. Gas at dewpoint
Separator
Oil at bubble point
Figure 7 Conditions in a separator
For a given system, therefore, a change in the temperature or pressure within the
phase envelope will result in alteration of the system and therefore alteration in the
characteristics of the two phases produced. The behaviour just described, therefore,
will have implications on the way samples are taken and on the techniques used to
collect the sample, for example, from the separator.
PVT Analysis
3.3 Sampling Gas And Gas Condensate Reservoirs
The potential locations for sampling these reservoirs are shown in Figure 8, samples
could possibly be taken in the reservoir, at the bottom of the well, at the wellhead or in
the separator. The respective advantages and disadvantages are given in table 1. The
reservoir would be the most ideal sampling point but clearly this is impossible. 4. Separator
3. Wellhead
2. Bottom hole
1. Reservoir
Figure 8 Locations for sampling
Merits / Disadvantages
Location
For
Against
1. Reservoir
Ideal
Impossible
2. Bottom - hole 1- phase Representative?
Technology
Cost
Handling
3. Wellhead
Cost
2- phase?
Representative
4. Separator
Cost
Gas / liquid volumes
1- phase Separator conditions
Buffer
Representative?
Sampling
Volume
Table 1 Merits and disadvantages of sampling locations.
The bottom hole sample where the fluid is in single phase, probably, would be an ideal
situation. Present technology, however, is such that it is often difficult to produce a
single phase sample representative of the fluids bottom hole. The necessary pressure
drop to get the fluid from the bottom hole into the sample container can often give
Institute of Petroleum Engineering, Heriot-Watt University
rise to a two phase situation and an unrepresentative collection of these two phases. Bottom hole samples are also more costly to collect. The wellhead from a cost point
of view could be the most suitable location point, however, again the question of the
representative nature of the sample is a concern. As a result of the lower pressure
and temperature it is likely that the single phase fluid at the bottom of the well has
gone into the two phase region at the wellhead and therefore the relative proportions
of liquid to gas would be unknown and their sampling would be difficult to produce a
representative sample. The most common sample location is the separator. Considerable
care, however, has to be taken to ensure that the samples taken from the separator are
those representative of the reservoir from which the fluids derive.
The well behaviour can significantly influence the nature and characteristics of the
fluids which eventually arrive from the separator, for example, figure 9 in a flowing
well gas condensate entering the wellbore as it travels to the surface will experience
a drop in pressure likely to give rise to retrograde liquid behaviour in the wellbore. The flow must be sufficient to lift this uniform liquid and gas fluid to the surface. If
the flow is slow it is possible that some liquid may fall back therefore altering the
overall composition moving up the wellbore. Gas
Condensate
Flow must be sufficient
to lift liquid to surface
Condensate
Gas
Retrograde liquid due to fall
in pressure and temperature
Gas
Gas
Condensate
Figure 9 Flowing well.
10
PVT Analysis
If the reservoir is shut in after flow then considerable changes can rise in the composition
of the fluid in the wellbore. The reservoir gas flowing into the wellbore sets up a
new equilibrium with condensed retrograde liquid which has rained down within the
wellbore. This separation in the wellbore gives rise to a lean gas near the top of the
well with a more than rich mixture at the bottom of the well. Figure 10.
Gas
Condensate
Lean gas
Gas
Retrograde liquid rains
down in the well
Reservoir gas flowing into vicinity
of well sets up new equilibrium
with condensed retrograde liquid.
Gas flow
Pressure build up
Condensate
Figure 10 Shut well in after flow
When the well is flowing after a shut-in period, figure 11 and sampling takes place
there will be a variation in the compositions produced at the surface and therefore
unrepresentative samples collected from the separator. For example, in the early
period after shut-in the lean gas at the top of the well enters the separator producing
a fluid with a GOR higher than that representative of the reservoir. As the fluids at
the bottom of the well move to the surface much richer as a result of the liquid having
collected at the bottom of the well fluids they are produced with a GOR lower than
that of the representative reservoir fluid. It is important, therefore, for the well to be
flowed for a sufficient period for all the fluid within the well to have been displaced
and also that in the near wellbore region which also could have been influenced by
the pressure and compositional changes experienced during the shut-in period.
Institute of Petroleum Engineering, Heriot-Watt University
11
Gas
Condensate
Lean gas moves
to surface
Higher GOR
Excess liquid in well
moves to surface
Lower GOR
Equilibrium gas flows to
surface richer in heavier
components.
Lower GOR
Figure 11 Well flowed after shut in period
In assessing whether good samples have been taken it is important to know how
long it will take for unrepresentative samples to be displaced from the separator, the
wellbore and the near wellbore reservoir zone. For example, for a 12ft x 5ft diameter
separator with a liquid flowrate of the order of 200 barrels per day, it could take 1
hour to displace an 8.4bbls. If a contaminant enters a separator then the time to reach
1% of the original concentration could be over 4.5 hours. For example, if the tubing
is 4 inches in diameter with a length of 9000ft and an average tubing pressure of
5000psi and a temperature of 170°F and a gas flowrate of 5MM/scf/day, the volume
in the tubing would be 0.23MM standard cubic feet and the time to displace this gas
would be just over 1 hour. 3.4 Separator Sampling Points
A very practical aspect is often ignored in the design of separators and in particular
in relation to the sampling points associated with them. These sampling points are
often located primarily in relation to accessibility rather than the representative
nature of the fluids which can be extracted from them. For example, in the gas line,
12
PVT Analysis
figure 12a the sampling valve might be located on the lower portion of the valve. It
is likely that entrained liquids in the gas stream could collect at this point giving rise
to a very rich gas composition if these entrained liquids were collected in the sample
containers. Similarly sampling valves in the liquid line figure 12b could be located
on the upper portion of the line any gas which is carried through the line again will
collect in the dead volume of the sampling valve such that when samples are taken
the gas will enter the gas bottle of the container giving rise to an unrepresentative
sample. An alternative would be to locate both of these sampling valves on the side
of the pipe rather than at the top or bottom of the line. Figure 12c.
(a)
Separator
Liquid
(b)
(c)
Figure 12 Location of sampling points
3.5 Sample Details
In taking samples there are some obvious aspects which are often ignored. Clearly
details in relation to the sample should be taken in particular; the date and the time of
a sample, the identification of the cylinder into which the sample is to be collected,
the location of where the sample was taken, the temperature and pressure at which
the sample was taken, details of any other aspects which will be important for those
subsequently handling the sample, for example, the presence of any H2S etc. The
details of the sample including, for example, the gas to oil ratio during the separation
will be transmitted with a sample to the laboratory which will carry out the analysis. For example, if the sample identification cylinder has not been taken then if sample
details become separated from the sample cylinder then the sample would be wasted. Prior to any liquid samples being transported to the laboratory it is important to
reduce the pressure within the container to a value below the bubble point to ensure
that a two phase mixture is transported. Very high pressures can occur as a result of
a temperature rise on a single phase liquid sample. Such pressures could go over
safe working pressures of the vessel!
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13
4 APPARATUS
The apparatus required for PVT analysis consists of:
(a) apparatus for the transfer and the recombination of separator oil and gas
samples;
(b) apparatus for measuring gas-volumes and for performing separator tests;
(c) the PV cell and displacement pumps and dispensing cell;
(d) high pressure viscometer;
(e) gas chromatograph.
Diagrams of the layouts for different systems are given in Figures 13, 14 and 15.
5 RECOMBINATION OF THE SURFACE OIL AND GAS SAMPLES
Analysis
Separator
Viscometer
Trap
Bleed
Oil sample
bottle
To liquid
analysis
Vacuum gage
Gas meter
Displacement
fluid
To gas
analysis
Vacuum pump
PVT cell in
thermal bath
Bleed
Precision pressure
gage
Displacement
pump
Figure 13 Equipment for the Study of Subsurface Samples
14
Dead weight
tester
PVT Analysis
Analysis
Separator
Viscometer
Trap
Recombination
cell
Gas sample
bottle
Vacuum gage
Bleed
To liquid
analysis
Liquid
sample
bottle
Gas meter
To gas
analysis
Vacuum pump
Displacement
fluid
PVT cell in
thermal bath
Bleed
Precision pressure
gage
Displacement
pump
Dead weight
tester
Figure 14 Equipment for the Study of Subsurface Samples
Analysis
7
Trap
9
To liquid
analysis
Vacuum gage
11
10
Gas meter
Vacuum pump
Gas condensate
cell in thermal bath
To gas
analysis
3
Bleed
Precision pressure
gage
4
Figure 15 Equipment for the Study of Gas Condensate Samples
The GOR given by the field refers to the separator tank gas-oil ratio. In order to
recombine the separator gas and the separator oil in the correct ratio, the volumetric
ratio between tank oil and separator oil is determined in the laboratory
85% of the cylinder containing separator oil is occupied by oil. There is a gas cap
on top. This precaution is taken in view of the long transport time and the risk of
great fluctuations in temperature involved.
After the gas cap has been dissolved by pressing water into the cylinder at a pressure
higher than the separator pressure a given quantity of oil is then flashed at pressure
and room temperature through a separator operating under the same conditions as
Institute of Petroleum Engineering, Heriot-Watt University
15
the tank in the field i.e. at the same pressure and temperature. The collected tank oil
is weighed and the density determined, after which the volume of oil is known. The
volume and the density of the liberated gas (tank gas) are also determined. From the
above measurements, the tank gas/tank oil ratio is also known. If the volume of tank
oil is lower than the corresponding quantity of separator oil we speak of shrinkage,
in the opposite case of expansion; shrinkage occurs when a large quantity of gas
is produced and expansion occurs when a small quantity of gas is dissolved in the
separator oil. When the shrinkage or the expansion is known, recombination can
take place on a tank-oil basis.
6 PVT TESTS
At this stage we should remind ourselves of the main applications of the PVT data.
The three main application areas are;
• to provide data for reservoir calculations,
• to provide physical property data for well flow calculations
• for surface facility design.
Although all are cited as users of the data, the reservoir calculation requirement has
provided the main driving force for the tests to be carried out. In surface facility
design for example the more simplistic black oil approach around which the PVT
analysis is structured is considered too limiting, and the main data for this application
is the compositional analysis of the fluids.
Over recent years, as the data is subsequently applied to computer based simulation
tools, the ability to handle more complex descriptions of the fluids has led to more
extended compositional analysis, beyond the C7+ limit which was the basis for many
years. It is common practice in some PVT laboratories to measure co-position to C28
and then define a C29+ component
In reservoir calculations the PVT tests and subsequent report provides the source
of the reservoir engineering properties necessary to describe the behaviour of the
reservoir over its development and production. The tests conducted therefore have to
take into consideration the processes going on both above and below the saturation
pressure.
There are four main PVT tests for oil systems plus associated compositional
analysis:
(i)
(ii)
(iii)
(iv)
(v)
the flash vaporisation or relative volume tests.
the differential test.
the separator tests.
viscosity measurements
compositional measurements.
A simple layout of a PVT facility is given in figure 16.
16
PVT Analysis
PVT Facility
Valve 1
P
V1
Ps1
L1
Ts1
T
Valve 2
Displacement
fluid or piston
V2
Ps2
L2
Ts2
PVT Cell in
temperature
controlled bath
Figure 16 Simple Schematic of PVT Facility for Oil Tests.
6.1 Flash Vaporisation (Relative Volume Test)
By flash vaporisation is meant the determination of the correlation between pressure
and volume of a reservoir liquid at constant temperature (reservoir temperature) from
high pressure to the lowest possible pressure. The gas liberated below the point of
saturation remaining in equilibrium with the oil throughout the experiment. That
is, the system remains constant (Figure 17). The vaporisation process occurring in
the reservoir cannot be duplicated in the laboratory, since in the reservoir below the
bubble point the system does not remain constant as the increased mobility of the
gas causes it to move away from its associated oil.
The flash vaporisation test gives the relationship between P & V of a reservoir liquid
at constant (reservoir) temperature. Liberated gas remains in equilibrium with the
oil.
Gas
Oil
P
Hg
Thermal expansion
V2 − V1
β=
V2 (T2 − T1 )
V1 = volume of oil at T1
V2 = volume of oil at T2
Figure 17 Flash Vapourisation. Determination of the relationship between P & V of a
reservoir liquid at constant (reservoir) temperature. Liberated gas remains in equilibrium
with the oil.
Institute of Petroleum Engineering, Heriot-Watt University
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By plotting the volume of the system versus pressure a break is obtained in the slope. This occurs at the Bubble Point pressure.
To carry out a relative volume test run, the PV cell is set up as in Figure 18.
Valve 1
closed
P
V
Tres
P
Pb
Vb
V
PVT Cell
Pump
Figure 18 PVT Set Up for Flash Vapourisation /Relative Volume Test
The PVT cell is filled with a certain quantity of reservoir liquid at a pressure above the
estimated bubble point and room temperature. After the PV cell has been filled (about
90 ml), it is immersed in a temperature bath and heated to reservoir temperature.
During heating, it is necessary to maintain pressure by increasing the content volume
of the PV cell. When the pressure remains constant, the temperature in the cell is
equal to that of the bath. The thermal expansion factor (β) can then be calculated
from the volume withdrawn. It is equal to:
β=
V2 − V1
V2 (T2 − T1 )
where:
V2 = volume of the oil at reservoir temperature T2
V1 = volume of the oil at room temperature T1
The thermal expansion factor is expressed, for example, in ˚C-1. The pump reading
taken at the moment when the pressure became constant is the first reading for the
PV curve. The pressure is now reduced by gradually withdrawing small quantities of
transfer fluid from the PVT cell and after each withdrawal equilibrium is established
by shaking the cell. After each equilibration the pressure and the volume are read. By plotting the pressures against the volumes a curve is obtained showing a break
at the bubble point (Pb). Figure 18.
The saturation pressure or bubble point pressure is that pressure below which gas
is liberated. Hence, a two-phase system is formed, whereas above the bubble point
pressure a one phase system is present (undersaturated liquid).
18
PVT Analysis
The compressibility of the oil phase above the bubble point can now be calculated
from the graph.
c=
V2 − V1
V2 ( P2 − P1 )
where:
V2 = volume at pressure P2
V1 = volume at pressure P1
The compressibility is expressed in reciprocal atmospheres, psi, etc. After the
reservoir liquid has expanded to its maximum volume (dependent on the capacity
of the PV cell), the gas cap is removed at constant pressure (lowest possible flash
expansion pressure). Volume, density, gas expansion and gas compressibility factor
of the liberated gas are then determined successively.
The main objectives of the flash vaporisation test are to provide the reservoir bubble
point pressure and together with the information from the separator test, the formation
volume factors above the bubble point.
6.2 Differential Vaporisation
When the reservoir pressure falls below the bubble point the process of gas liquid
separation in the reservoir is one of a constant changing system. A PVT process has
been designed in an attempt to provide a means of in part simulating the changing
systems as separation occurs within the reservoir below the bubble point.
The differential vaporisation differs from the flash in that the liberated gas is removed
from the cell stepwise. At each step below the bubble point the quantity of gas, oil
volume, density, gas expansion and gas compressibility are determined. The objectives
of the differential test therefore are to generate PVT data for conditions below the
bubble point. Figure 19 below indicates the differential process.
Institute of Petroleum Engineering, Heriot-Watt University
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Gas
Oil
Gas
Oil
P = Pb
P1 < Pb
At each stage - volume of oil and gas, density and
composition of gas measured.
Remaining oil is called RESIDUAL OIL
P = P1
P < P1
Gas
Oil
Mercury /
displacing fluid
Figure 19 Differential Vaporisation. In differential vaporisation. At each successive
pressure drop liberated gas is removed.
The bubble point Pb is the starting pressure for the differential test. The next step
is to reduce the pressure in the PV cell by expansion of the PV cell volume. The
reduction of pressure causes the system to go two phase. All the gas phase is removed
at constant pressure by reducing the cell volume as gas is withdrawn. The volume
of the remaining oil is then determined. The cell pressure is then again dropped by
expansion of the PV cell and the above process repeated until the cell pressure has
been dropped to atmospheric pressure. The pressure steps for the tests cover a range
of around 8-10 steps. All the above steps have taken place at reservoir temperature.
The final stage is to reduce the cell to 60°F keeping the pressure at atmospheric
pressure. The final oil volume is measured. This remaining all is termed residual
oil to distinguish it from stock tank oil which although at the same pressure and
temperature conditions has got there by a different process.
The cumulative weight of the amounts of gas withdrawn are used in the calculation of
the densities of the oil phase in the differential vaporisation process. These densities
can also be determined directly if a pressure pycnometer is available.
Differential liberation is considered to be representative of the gas-liquid separation
process in the reservoir below the bubble point pressure.
Flash liberation is considered to take place between the reservoir and through the
separator.
Differential liberation tests are carried out therefore to obtain oil formation volume
factors and GOR’s that can be used to predict the behaviour of a reservoir when the
pressure has dropped below the bubble point pressure.
20
PVT Analysis
6.3 Separator Tests
The object of these tests are to examine the influence of separator pressure and
temperature on formation volume factor, gas/oil ratio, gas density and tank-oil density. These tests are not driven by those who will be responsible for the optimised separation
process if the field ultimately is developed. They are carried out to give an indication
of the oil shrinkage and GOR which occurs when the fluids are produced to surface
conditions. It should be emphasised at this stage that there is not a unique value for
the formation volume factor and solution gas-oil ratio. It depends on the stages and
conditions of separation through which the fluids pass. With the equipment available, a single test or a multiple separation test can be carried
out.
The systems is set up as shown in the schematic below, Figure 20.
Valve 1
Pb
V1
Ps1
L1
Ts1
Valve 2
Vres
Tres
V2
Ps2
L2
Ts2
60 F & 14.7 psia
PVT Cell
Pump
GOR = (V1+V2) / L2
Bo = Vres / L2
Figure 20 Schematic of a Two Stage Separator Test
The procedure for the separation test is as follows. The starting point is oil in the PVT
cell at its reservoir bubble point, ie. the same starting condition as the differential test. Fluid is displaced from the PVT cell ensuring that the PVT cell contents remain at
bubble point pressure. The gas and liquids are collected from the separation stage(s)
and their respective properties measured. The final stage is at stock tank conditions.
The resulting fluid is termed stock tank oil.
A single separator test is carried out by flashing reservoir liquid at bubble point pressure
and reservoir temperature through the separator operating at the average annual
temperature and at pressures which may be expected in the field. The difference in
results when using a single or a double separator is that in the former case the total
gas/oil ratio is higher, the shrinkage is greater and the density of the tank oil is higher
than in the latter case.
The main objectives of the separator test are in combination with the flash vaporisation
and differential tests to provide formation volume factor and solution gas-oil ratios
over a full pressure range above and below the bubble point. In quoting these values
it is important to recognise that the values are separator condition specific.
Institute of Petroleum Engineering, Heriot-Watt University
21
In the sketch figure 20 oil formation volume factor Bob is equal to Vres/L2 reservoir
volumes/stock tank volumes.
The solution to gas-oil ratio, Rsb, is equal to (V1 + V2)/L2 Standard cubic volume
(SCF or SM3)/Stock Tank volumes (STB or STM3).
6.4 Viscosity
The viscosity is measured at reservoir temperature and at different pressures both
above and below saturation pressure. It is important for viscosity measurements below
the bubble point to generate the fluid for study by a differential mode to simulate the
nature of the fluid that would exist at these conditions.
Viscosity measurements were largely carried out with a “rolling ball” high pressure
viscometer consisting of a highly polished-steel capillary of 1/4" dia. which can be
closed at the top by means of a plunger and is provided at the bottom with a contact
which is connected with an amplifier. A steel or platinum ball rolls in the capillary, its
diameter hence slightly smaller than that of the capillary. When the ball reaches the
bottom, it makes contact between the wall of the capillary and the point of contact,
as a result of which a circuit is closed and a whistling sound is heard. The time of
rolling is a measure of the viscosity. In recent years the pressure drop along a capillary tube of known length and internal
dimensions has been used. The viscosity being calculated using the Poiselle equation,
the laminar flow pressure drop equation for a pipe of a particular diameter and length. Although being used it is clearly restricted by operating under a fixed flow regime,
laminar and velocities to ensure that there is a sufficient pressure drop to measure
and not too large to influence physical properties.
6.5 Hydrocarbon Analysis
A hydrocarbon analysis is made of the methane to an upper paraffin fraction of the
recombined surface (or subsurface) sample. Historically this upper limit was C6
and the remainder lumped as a C7+ fraction. C10 . Higher C numbers are now used as
analytical methods enable even higher levels of characterisation. The plus component
having been separated in a distillation column needs to be characterised by its specific
gravity and its apparent molecular weight. The latter is achieved using a depression
of freezing point method. The different components are determined by means of
gas/liquid chromatography. If the description is based solely on paraffins then non
paraffinic components are added to the next higher paraffin.
The value of a higher characterisation is particularly helpful to process engineering
considerations, say up to C30 because at lower temperatures long chain hydrocarbons
will form a solid phase (such as wax) and adhere to surfaces. At reservoir conditions
they are in the liquid phase and therefore do not effect the reservoir flow process. In some cases the fluids produced in the final separation stage are identified to a
higher C number together with aromatic and napha components.
Compositions are also made of the produced gases from the various tests.
22
PVT Analysis
7 WAX AND ASPHALTENES
The formation of solid deposits during oil production is a concern. Some heavy
hydrocarbons as mentioned above at low temperatures can form solid phases waxes and
in transfer lines and process facilities. The wax formation temperature is therefore an
important measurement. Ashphaltene, is another solid phase of concern. Asphaltenes
are large molecules largely of hydrogen and carbon with sulphur, oxygen or nitrogen
atoms. Asphaltenes do not dissolve in oil but are dispersed as colloids in the fluid.
7.1 Wax Crystallization Temperature
Different techniques can be used for this. In the following procedure a sample of
separator oil is transferred to a vessel and then pre filtered by passing through a 0.5
micron filter. The crystallisation temperature (WTC) is measured by carrying out a
series of flow experiments where the oil is flowed through a fine filter (15 microns)
across which the pressure differential is measured. Prior to reaching the filter the oil
is flowed through a temperature equilibrium coil at the same temperature as the filter
at a temperature of 140°F. The temperature of the bath is gradually lowered and the
pressure difference measured. A sudden increased in pressure indicates the onset of
wax crystals building up on the filter giving an indication of the WCT.
From this rough indication constant temperature flows are taken at temperatures
just above and below the indicated WTC to give a more precise value. A plot of
differential pressure vs. flow indicates a more precise value of the WTC. Figures 21
& 22 below from a PVT report provided by Core Laboratories (UK) Ltd gives the
plots used to determine the WTC for a separator oil sample.
The appearance temperature is considered to be affected by super coiling, whereas
the disappearing temperature is considered to be the equilibrium valve. There is a
concern that the appearance temperature may not be so accurate as would be the case
for the disappearance temperature.
Institute of Petroleum Engineering, Heriot-Watt University
23
60
Differential Pressure, psig
50
40
30
Approximate Cloud Point
20
10
0
75
85
95
105
115
125
135
145
Temperature, °F
Figure 21 Graph of differential pressure v temperature during constant cooling.
(Core laboratories)
24
PVT Analysis
45
40
Differential Pressure, psig
35
30
90°F
88°F
86°F
25
20
15
10
5
0
0
5
10
15
20
Cumulative Volume Flow, cc
25
30
Figure 22 Graph of Cumulative Volume Flow Across Filter v Differential Pressure.
(Core laboratories)
8 SUMMARY OF RESULTS PROVIDED BY AN OIL SAMPLE PVT
TEST
The PVT analysis as described above furnishes the following data:
(1) Saturation pressure. ie. bubble point.
(2) Compressibility coefficient
c=
1 dV
V dP (specify unit of pressure change)
(3) Coefficient of thermal expansion
β=−
1 dV
V dT (specify unit of temperature change)
(4) Relative total volume of oil and gas: vt
Institute of Petroleum Engineering, Heriot-Watt University
25
(5) Cumulative relative gas volume: vg
(6) Relative oil volume: vo
(7) Gas formation or gas expansion factor: Bg or E, i.e. volume occupied under sc
(standard conditions) by that amount of gas which occupies unit volume at reservoir
temperature and pressure.
(8) Gas compressibility factor: z defined by pv = zn RT.
(9) Specific gravity of gases (air = 1)
(10) Liquid density: ρ
(11) Viscosity of the liquid phase at different pressures: µ
(12) Oil Formation Volume Factor or Shrinkage factor C, where C = ratio of the
volume of tank oil produced under specified separator conditions and then measured
under sc to the volume occupied by that quantity of oil and its dissolved gas under
reservoir conditions. depends on separator conditions.
(13) Solution gas to Oil Ratio or Gas-solubility factor D = that volume of gas liberated under separator conditions and measured under sc which is held in solution at
reservoir temperature and any particular pressure by that quantity of oil which will
occupy unit volume when produced; depends on separator conditions.
(14) Shrinkage of separator oil to tank oil.
(15) Tank-gas/tank-oil ratio.
(16) Hydrocarbon analysis of the reservoir and produced fluids.
Figure 23 below illustrates the volume relationship of fluids in an oil PVT tests to the
black oil description of volumes in a reservoir. In PVT analysis the basis of reference
is the bubble point, figure 23a whereas for the black oil system the reference state
is surface or stock tank conditions, figure 23c. The relationships between the two
are given in figure 23b.
V
V
V
Vt
Vt
B1
Cb
Vo
Cb
P
(a)
I
Vt
Pb P
Vi
Pi
I
P
Vo
Cb
(b)
l Vt Vi
Cb Cb Cb
Pb P Pi
Figure 23 Nomenclature PVT SYMBOLS
26
Bo
l
P
(c)
Bb
Pb
Bt
P
Bt
Pi
PVT Analysis
(a) All volumes relative to oil volume at bubble point
(b) Same as using black oil notations
(c) All volumes relative to oil volume at standard surface conditions
8.1 Interfacial Tension, IFT
Over recent years, particularly with respect to gas condensates, the impact and
importance of interfacial tension has developed.
Research has demonstrated that the IFT between hydrocarbon liquid and vapour phases
has a significant impact on residual condensate saturation and relative permeability.
Although not yet part of conventional PVT analysis the author considers that such
information will in future be required for PVT reports for more fluids where the
conditions are close to critical conditions and when dealing with gas condensates
fluids.
The determination of the interfacial tension of vapour-liquid systems is very important
therefore for these systems where such phase separation occurs in the reservoir. For
these forces will impact on the forces effective recovery, in particular the balance
between gravity forces. Recent research has shown that liquid drop out in retrograde
condensation is not necessarily immobile as previously considered. Recovery therefore
by gravity drainage, when IFT's are low can therefore be significant. IFT's are very
low for those fluids near critical conditions as the value of zero is approached.
8.2 IFT Measurement Methods
The most common approach of measuring IFT has been using the Pendant Drop
Method. In this method the liquid phase is suspended from a capillary tube in a vessel
containing its equilibrium vapour. Figure 24.
Tube
Tip
ds
Gas
de
Oil
Figure 24 Pendant Drop Method For IFT Measurement.
The dimensions of the suspended droplet are controlled by the balance between the
surface and gravity forces. The equation relating the various parameters from this
method is:
Institute of Petroleum Engineering, Heriot-Watt University
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s=
gde2 L
(ρ − ρ V ) )
l
Where s = interfacial tension (dyne/cm)
g = acceleration due to gravity
de = equitoral diameter drop
ds = diameter of drop measurement at height
de above the bottom of the droplet
rL, rV liquid and vapour phase densities
l = shape factor, which is r = ds/de
When the IFT is very low, the droplet size is very small which requires an extremely
fine diameter tube to suspend it. A thin wire may be required for such situations. When
conditions are close to the critical point the pendant drop method may not work.
For very low IFT values other methods have been used. A light scattering technique
where the propagation of thermally stimulated waves in a vapour-liquid system has
been used3. A very successful and low cost method was developed at Heriot-Watt,
based on the curvature of the vapour-liquid interface on the viewing window of a
PVT cell4. Figure 25.
Enlarged veiw
Viewing window
of PVT device
Vapour
Vapour
h
Liquid
q
Liquid film
Liquid
Cross section
of window
Figure 25 Rising film method for IFT measurement
The curvature is seen as a dark band of a specific thickness, due to the scattering
transmitted light across the viewing window of the PVT cell. The thickness, of the band
is directly proportional to the IFT. The equation relating the various measurements
and properties is as follows:
s=
gde2 L
(ρ − ρ V ) )
l
where q = contact angle (assumed zero at low IFT's)
Then s = (rL - rV)gh2/2
It has been applied successfully to IFT values as low as 10-3 mN/m, where it is difficult
to distinguish the vapour-liquid phases.
28
PVT Analysis
9 RETROGRADE CONDENSATION
In the phase behaviour chapter we described the characteristics of gas condensates,
ie, those reservoir fluids which show the phenomenon of retrograde condensation,
i.e. liquid is formed from a vapour when the pressure is reduced isothermally. In this
case the system saturation pressure is the dew-point pressure (the pressure at which
condensation starts) instead of the bubble point pressure for oil systems.
When studying the PVT behaviour of gas condensates, use is often made of a PVT
cell with a glass window to observe the dew point condition. Over recent years there
has been considerable developments in the types of PVT cells used in gas condensate
PVT measurements. It is not the intention of these notes to describe the different
cells. The key difference between an oil PVT cell compared to a condensate PV cell
is that the latter requires a window to observe the dew point condition.
This is carried out for two reasons:
(a) the dew-point pressure of most condensates cannot be detected by a sharp
change in the pressure-volume relation as the bubble point pressure can;
(b) the liquid phase constitutes only a small part of the total volume in the PV-cell A windowed cell makes accurate measurement of small liquid volumes possible.
In gas condensates there are large volumes of gas relative to liquid. The liquids
produced are generally small in volume. The relative small amounts of liquid can
yield errors in measurement.
There are three main aspects of a gas condensate study.
• Compositions of gas, liquid and reservoir fluids.
• Constant mass (composition) expansion
• Constant volume depletion
• Specialised tests. e.g. Interfacial Tension Measurements.
• Compositions of gas, liquid and reservoir fluids.
There are different approaches to compositional measurements in gas condensates.
It is only recently that it has been possible to measure the compositions of liquid and
gas phases at reservoir conditions and only now these direct sampling methods have
limited pressure applications. Compositions of condensate phases are often obtained
by blow down techniques and then recombining the compositions of the produced
gas and liquid in the proportion of these two phases. For liquid compositions, the
fluid is flashed at 120°F to obtain gas and liquid components. The resulting gas and
liquids are then analysed by gas chromatography techniques and the compositions
recombined in proportion to the ratio of gas to liquid produced. For gas compositions
gas chromatography is used. The composition of the recombined reservoir fluid
is obtained by recombining the liquid and gas compositions in proportion to the
condensate to gas ratio of the fluids.
Figure 26 below gives a schematic of the two main gas condensate tests.
Institute of Petroleum Engineering, Heriot-Watt University
29
(a) Constant Mass Study
Gas
Oil
Hg
Vg
Measurement of volume of gas
and oil (condensate) as a function
of preessure.
Vo
Dew point when first condensate
appears.
No fluids removed from the cell.
P = Pd
(b) Constant volume depletion
Gas removed at each step
Gas
Gas
V>Vi
Vi
Oil
Hg
P = Pd
Gas
Hg
P1 < Pd
Oil
Hg
P = P1
Vi
Pressure is reduced
below dew point by
expanding system.
Volume is put back to
original volume by
injection of mercury
at constant pressure
and removing some gas.
Volume of condensate
measured.
Figure 26 Schematic of constant mass and constant volume depletion gas condersale tests.
There are two basic tests for gas condensates, Figure 26. The constant mass study,
the purpose for which is the z value determination above the dew point and the
determination visually of the dew point pressure. The constant volume depletion test
is carried out in at attempt to examine the potential liquid drop out in the reservoir
by retrograde condensation.
The constant mass study and the constant volume depletion study can be compared
to the flash vaporisation test (relative volume test) and the differential test for oils
respectively.
In the constant mass study, Figure 26, the system remains constant. No fluids are
removed. A portion of the reservoir fluid is charged to the gas condensate PVT cell. The pressure is raised to that significantly above the anticipated dew point and then
the volume of the gas measured at reducing pressure steps. The dew point is observed
visually as liquid condenses on the PVT cell window. The retrograde liquid build
up is also measured at several pressures below the dew point, although the numbers
have little reservoir significance. Values of the compressibility factors z are obtained
above the dew point pressure.
30
PVT Analysis
The Constant Volume Depletion, CVD, test is carried out in an attempt to determine
the potential loss of liquids if the reservoir is depleted below the dew point. In the
test the same fluid at reservoir temperature as contained in the cell for the constant
mass test is used. The test consists of a series of pressure expansions and constant
pressure displacements to return the sample to a constant volume. This volume is
equal to the volume of the sample at the dew point pressure. This process is repeated
down to a low pressure.
Starting at the dew point the pressure is reduced by expansion of the cell. Liquids
condense and then gas is displaced at constant pressure until the volume returns to the
original volume. The quantity and compositions of the displaced fluids are determined
using gas chromatography methods. In one approach the produced gas from the cell
are pumped into a pre-weighed flask submerged in liquid nitrogen and condensed.
The condensed gas is then gradually allowed to return to ambient temperature. The
evolved gas and residual condensate are collected separately, weighed and analysed. These compositions are recombined mathematically to the gas - oil ratio to determine
the produced gas composition. The remaining condensate phase is measured and
expressed as a percentage of the sample volume at the dew point pressure. . The
pressure is then reduced again and the process repeated. A series of relative liquid
amounts are obtained and a liquid dropout curve generated. Figure 27.
Retrograde
condensate
RLV %
Ps
Figure 27 Constant Volume Depletion Liquid Drop Out Curve.
Special tests include the measurement of interfacial tension, described earlier which
has come to be realised an important factor in understanding the behaviour of gas
condensates. Different methods have been proposed including pendant drop and
rising film techniques. The latter method can be carried out as a matter of routine
during the CVD test by observing the thickness of the rising film of the condensate
on the window of the cell. The thickness of this film is directly proportionate to the
IFT and the method can determine IFT values lower than conventional pendant
drop methods.
10. UNDERSTANDING PVT REPORTS
Having considered the various aspects of PVT analysis we will now consider the PVT
report and examine how we can generate the various reservoir engineering parameters
of interest. We will remind ourselves of the reason for the report and then using a
PVT report go through the main tests and interpret the detail.
Institute of Petroleum Engineering, Heriot-Watt University
31
11 PURPOSE OF THE PVT REPORT
Although the PVT report can be a source of information for a variety of applications
from reservoir, through well to surface facility calculations, the reservoir engineering
application has provided the main basis and structure of the report. The report is
structured to provide the much of the black oil model information together with
limited compositional data. The material balance equation which is covered in a latter
chapter also provides a basis for the PVT report. It is the PVT report which is the
source of much of the data embodied in the material balance equation. Thus, some
of the tabular information is set up to satisfy that need.
The PVT report can be used for a range of purposes; from its use in determining the
potential prospects of a hydrocarbon accumulation to history matching a reservoir
which has been on production for some time. The report should therefore cover all
past, present and future situations which may require calculations. To do this with a
minimum of tables and curves, the data are normalised to a reference state and only
data for the reference state given. In PVT data reporting as indicated in earlier the
reference state is the bubble point. The petroleum engineer must then “work back”
from the reference state to the particular situation.
As described in previously the laboratory tests are carried out in an attempt to simulate
the processes which take place in a reservoir and through the production system. These will include the flash equilibrium separation of gas and oil in the surface traps
during production and for an oil below the bubble point the differential equilibrium
separation of gas and oil in the reservoir during pressure decline. In interpreting
the data the engineer needs to use both sets of data to provide the information for
reservoir calculations.
The PVT report is clearly specific to a particular fluid, collected from a specific well
under specific conditions. This sample may not be representative of the total field
system and therefore using subsequent reports it may be necessary to to adjust the
data for field application. In a PVT report therefore detail is given as to the manner
of obtaining the sample and the conditions that existed at the sampling time. Also,
the compositional analysis of the sample is given so that equilibrium calculations
can be made for conditions other than studied in the laboratory. In this area there has
been considerable progress in compositional analysis and although the report used in
this section only goes up to C7+ it is now common practise to characterise to higher
C numbers, even as light as e29.
We will now examine two PVT reports. The first is for an oil sample and is commonly
used in textbooks to illustrate the interpretation of a PVT report. The second is a PVT
study on a gas condensate fluid. It should be emphasised that reporting styles vary
from the different service providers. Both reports are provided by Core Laboratories.
Ltd.
32
PVT Analysis
12. OIL SAMPLE PVT STUDY
The first report is for an undersaturated oil from Texas field attributed to the Good
Oil Company. The report is given at the end of this chapter. Although not covered
in this report there may be a number of data sheets reporting the validation of the
samples used and selected in the PVT report. These sheets would include the various
gas to oil ratios when taking the samples and other information as part of the sample
validation. A text description is usually given to the report describing the various tests
conducted and the principle observations. The source of these principle observations
and calculations will be covered as in the following sections. The pages of the report
often include processed data .
To determine black oil parameters of oil formation volume factor and gas to oil ratios
as a function of pressure, a combination of tables of the report are required.
We will first look at the separator tests the results of which provide the basis for the
Bo and GOR values.
12.1 Separator Tests of Reservoir Fluids
In the separator test, page 7 of 15 of the report, oil at reservoir temperature and the
bubble point pressure in a PVT cell has been carefully displaced from the cell through
a series of pressure and temperature steps. The tests show what quantity of surface
gases and stock tank oil results when one barrel of bubble point oil is flashed through
a certain surface trap sequence. The tabulation also gives the ˚API gravity of the
stock tank oil and, in some instances, the gravity of gas coming from the primary
trap. There are four separate tests reported. The first where the first separation is at
50 psig and 75°F and the tank separation at 0 psig and 75°F, the other tests where
the first stage conditions are at 100 psig, 200 psig and 300 psig. The temperatures
and conditions for the final stage are the same for each test.
Column 1 and 2 give the pressure-temperature conditions of the surface trap tests
that were investigated. These should be specified by the reservoir engineer at the
time the test is planned . A difficulty here is that the engineer specifying to the PVT
service company these separation conditions is unlikely to be involved in the ultimate
optimised surface separation conditions if the field ultimately is developed.
Exercise PVT 1.
What is the solution gas-oil ratio and formation volume factor resulting from the separator test, at first stage of 300 psig and 2nd stage 0 psig and both at 75˚F ?
Solution.
This exercise illustrate the results using one of these tests, a two-stage separation;
a primary trap operating at 300 psig and 75˚F followed by a stock tank operating at
14.7 psia (0 psig) and 75˚F.
Institute of Petroleum Engineering, Heriot-Watt University
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When one barrel of bubble point oil; defined as oil saturated at 2620 psig and 220˚F in
footnote 3, on page 7 of 15 is flashed (processed) through this separation arrangement,
the stock tank has a quality of 40.1˚API (column 5). The formation volume factor
of the bubble point oil, Bob = 1.495 B/BSTO (column 6). This would have been the
volume of oil displaced from the PVT cell divided by the volume collected at the final
stage and then corrected for the thermal reduction from 75˚F to 60˚F. The source of
this bubble point pressure value will be indicated later.
Columns 3 and 4 show the surface gas-oil ratio from the first stage and the tank. The first stage ratio of 549 ft3/BSTO ( column 4) and the tank stage gas amount to
246 ft3/BSTO. It is important to read the footnotes of the report. Column 3 gives
the results in relation to the volumes at indicated P & T whereas column 4 gives the
volumes with respect to stock tank conditions of 14.65 psia and 60˚F. The solution
gas-oil ratio at bubble point conditions (2620 psig and 220˚F), is therefore Rsb is 549
+ 246 = 795 ft3/ BSTO when flashed through this surface trap arrangement. If we compare these results for the 50psig, 0psig arrangement we obtain a Bob of
1.481 B/BSTO and a solution GOR of 778 ft3/ BSTO.
Clearly therefore Rsp, Bob, ˚API all vary with the separation pressure-temperature
situation. There is not one unique result. When reporting Bo and GOR data for a
reservoir therefore it is important to report that these are for a specific separation
route or averaged for a series of tests. The latter is not so useful since it is not so
straightforward to calculate the result for a different separation route using VLE
methods.
The results from the separation test are based on the bubble point condition and to
obtain volumetric information at other pressures we require the results from other
tests.
12.2 Fluid Properties at Pressures Above The Bubble Point Pressure
The source of information for calculations for conditions above the bubble point
is a combination of two tests, the flash vaporisation test (relative volume test ) and
the separation tests. The results of the flash test are presented in table 4 of 15, titled
Pressure -Volume Relations at 220˚F.
Remember in this test, the contents of the cell at reservoir temperature have been
expanded and the volumes measured. None of the contents has been removed, the
system has remained constant. In the style presented here the expansion of the fluid
as measured has been plotted and then the intersection of the two slopes of the liquid
phase expansion and the two phase, gas/liquid phases, has been interpreted as the bubble
point pressure and bubble point volume. All the volumes have then been normalised
to this bubble point condition and presented as a relative volume. (column 2).
The first and second columns of the Reservoir Pressure Volume Relations Data on
page 4 give the pressure volume relations of the original fluid at 220˚F. Note that the
data are presented in terms of a unit volume at the bubble point condition.
This flash vaporisation test gives us therefore the reservoir temperature bubble point
pressure, which in this case is 2620psig. i,e the point where the relative volume is
1.0.
34
PVT Analysis
Column 2 gives the volume of the system at pressure per unit system volume at 2620
psig and 220˚F. These are listed as relative volumes, relative to the bubble point.
Column 3 presents what is called the Y function, this function should provide a straight
line or a slight curve and can be used to pick out anomalous data.
We will now see how we can use the relative volume data to provide us with some
formation volume factors above the bubble point.
Exercise PVT 2.
What is the formation volume factor and the density of the oil at the last reservoir
pressure measured.
Solution.
The well characteristics give the last reservoir pressure as 3954 psig. @ 8500 ft: We
obtain the oil formation volume at 3954 psig by multiplying the formation volume
factor at the bubble point by the relative volume (to the bubble point). Why multiply? Because:
Bo =
vol reservoir oil
vol bubble point oil
vol reservoir oil
=
×
vol stock tank oil
vol stock tank oil
vol bubble point oil
and the reference bubble point oil volume cancels out. Therefore Boi, the initial
formation volume factor is 1.495 x 0.9778 = 1.4618 when the 300 psig primary trap
is involved. It is a different value if another separation pressure is used. The 0.9778
was obtained by interpolation between 3500 and 4000 psig in column 2.
Reservoir oil density at pressures greater than 2620 psig also make use of the relative
volume data of column 2, page 4. The added information we have is the density of
the bubble point oil. This is given in the summary data on page 3 of the report. We
see here that the specific volume at the bubble point, vb = 0.02441 ft3/lb. This comes
from direct weight-volume measurements on the sample in the PVT cell. We can
now calculate the density, roi, of the initial reservoir oil as:
ρoi =
1
1
=
voi voi ⋅ vrel
ρoi =
1
= 41.89lbs / ft 3
0.02441x 0.9778
The compressibility of the oil above the bubble point can also be obtained from the
relative volume test. The definition of compressibility is:
Institute of Petroleum Engineering, Heriot-Watt University
35
1  ∂v 
Co = −  
v  ∂p  T
It makes no difference whether the volume units in the equation are relative volumes
to the bubble point, formation volumes, or specific volume values. To evaluate CO at
pressure p it is only necessary to graphically differentiate the p-vrel data in columns
∂v
1 and 2 to get ∂p at the pressure and divide by vrel. A less accurate value can be
obtained by the assumption:
Co = −
1  ∆v 
 
vavg  ∆p  T
For example, to get Co at 4500 psig using relative volume values of 500 psi on each
side of 4500 psig:
Co = −
Co =
1
(0.9639 − 0.9771)
0.9639 + 0.9771 (5000 − 4000)
2
1 0.0219
= 13.6(10 −6 )vol / vol / psi
0.9639 1000
The report also lists some compressibility numbers on page 3. These are not the
same as indicated above because they are changes in volume (in the pressure interval
indicated) per unit volume at the higher pressure. For example, the value of 13.48 (10-6)
for the 5000 and 4000 psi interval is obtained as:
−
1 (0.9771 − 0.9639)
0.9639 (5000 − 4000)
The compressibility data on page 2 are set up in this manner because of the way they
are used in one form of the material balance.
12.3 Total Formation Volume of Original Oil Below The Bubble Point
Pressure.
In the liquid properties chapter we introduced the total formation volume factor,
Bt. This factor is of little practical significance since it describes the volume of an
oil and its associated gas both above and below the bubble point, when the system
does not change. In reality below the saturation pressure the system changes as gas
and oil have different mobilities. In some forms of the material balance equation
Bt is used however to express oil volumes. We have just seen that to calculate the
formation volume factor of the oil above the bubble point we multiply the bubble
point formation volume by the relative volume given in column 2, page 4. If we
multiply Bob by vrel at pressures less than pb, we also get a formation volume factor, 36
PVT Analysis
the total formation volume Bt, of the original system. That is at p < pb we will have
two phases and the Bt is the volume in relation to both gas and liquid phases in
equilibrium at pressure p.
One form of the material balance equation makes use of the expansion of the original
oil between the initial system pressure and any subsequent pressure. This expansion
is given by the term:
Eo = N(Bt - Boi)
where N is the initial stock tank barrels in the reservoir and (Bt - Boi) is the expansion
per unit stock tank oil. Eo is, therefore, the expansion (bbl) of the original oil system. Sometimes we see the expansion equation written:
Eo = N(Bt - Bti)
Figure 28 below illustrates the change of Bt and Bo with pressure over the total
pressure range.
Formation Volume
Factor - B
Bt
Bti
(Rsi-Rs)Bg
5.61
Bo
Boi
p
pb
Pressure p
pi
Figure 28 Shape of Total Formation Volume Factor Bt and Oil Formation Volume Factor
Bo
Above pb it makes no difference whether we consider the formation volume to be a
total formation volume or an oil formation volume.
12.4 Differential Liberation Tests.
Previously we have considered what happens when reservoir fluid comes to the surface
and is separated into surface gas and oil products. In determining what happens in
the separator test we consider it as flash equilibrium conditions because we believe
that the action going on in the separator is essentially one in which the whole system
entering the trap immediately separates into two components - separator gas and
liquid. This constitutes the elements of a flash separation. However as we discussed
in section 6 in the reservoir the separation below the bubble point is different and the
differential test has been devised to to enable calculation of the appropriate volume
factors and GOR’s.
Institute of Petroleum Engineering, Heriot-Watt University
37
The standard PVT report includes data referred to as “the differential data”. These are
gas solubility and phase volume data taken in a manner to model what some people
believe happens to the oil phase in the reservoir during pressure decline. Basically,
the argument that differential liberation tests model the subsurface behaviour comes
primarily from two things:
(1) the reservoir pressure changes are not violent and large as are the pressure
changes in entering surface separators. The subsurface changes are more gradual
and might be considered to be a series of infinitesimal changes.
(2) because of the relative permeability characteristics of reservoir rock-fluid
systems, the gas phase moves toward the well at a faster rate than the liquid phase. Of consequence, the overall composition of the entire reservoir system is changing. These two ideas promote the idea that a test procedure modelled on a differential
process should be used to study subsurface behaviour. Because of experimental
limitations and time-cost considerations the laboratory cannot perform a true differential procedure. Instead, they perform a series, usually about ten, of stepwise
flashes at the reservoir temperature, commencing at the bubble point. Of course, the
greater the number of steps, the closer the true differential process is modelled.
The differential data are reported on page 5. Note that the table is headed by the title
“Differential Liberation at 220˚F”. Probably the best way to understand these data
is to explain again the manner of obtaining the values.
To begin with, the laboratory starts with a known volume of the original system in
the PVT cell. This may be of the order of 100-200cm3. The volume at the bubble
point pressure (2620 psig in this instance) is determined accurately as it is a reference
for all subsequent measurements.
b
c
Volume
Initial Bubble Point
a
2620 psig
b'
200° F
c'
2350 psig
60° F
Pressure
Resudual Oil
Figure 29 Volume Changes During Differential Liberation
38
PVT Analysis
Referring to page 5, we see that the first pressure step was to 2350 psig. At this
pressure the original system will be in two phases. Its volume would be at b on the
adjoining sketch. Figure 29.
The first step in altering the overall system composition is made at 2350 psig by
removing the gas phase from the PVT cell while maintaining constant pressure. The quantity of gas removed is determined by collecting it in a calibrated container. The volume that the gas phase occupied in the cell is determined by the amount of
mercury or non contacting fluid injected during the removal process. Also, the gas
gravity is measured on the sample bled off. The volume of liquid remaining in the
cell is shown at b' in the sketch.
The above procedure is repeated by taking the 2350 psig saturated liquid to 2100
psig (point c) and removing a second batch of gas at that pressure. Again the volume
of the displaced gas in the cell at 2100 psig is determined as is the gravity of the
removed gas. The volume of liquid phase remaining after the second gas removal
step is illustrated by point c' in the sketch.
This process of removing batches of equilibrium gas continues until the cell pressure
at the last displacement is 0 psig. As indicated by the differential data on page 5,
there were ten equilibrium removals, all at 220˚F. The final volume of liquid phase
remaining in the cell at 0 psig and 220˚F is corrected by thermal expansion tables
(or by cooling the cell) to 0 psig and 60˚F. This 0 psig/60˚F liquid is called residual
oil. Note that residual oil and stock tank oil are not the same fluids. They are both
products of the original oil in the system but are generated by different pressuretemperature routes.
Having now got to residual oil the data obtained are recalculated and presented on
the basis of a unit barrel of residual oil. By the time 0 psig and 220˚F had been
reached, the original system had liberated 854ft3/B residual oil. Column 2 expresses
the amount of gas in solution at the various pressures. This is the difference of the
854ft3 total liberated and the amount liberated between the original bubble point
pressure and that pressure.
It is important to understand why the solution gas-oil ratio determined from surface
flash by taking oil at its bubble point directly to separator and surface conditions
conditions compared to differential removal will be different, although the starting
and finishing conditions are the same. It is because the process of obtaining residual
oil and stock tank oil from bubble point oil are different. The first is a multiple series
of flashes at the elevated reservoir temperature ( the differential test); the second is
generally a one or two-stage flash at low pressure and low temperature (flash tests). The quantity of gas released will be different and the quantity of final liquid will
be different because the changing composition of remaining liquid at each stage
will influence the distributions of components between the phases. Also, the quality
(gravity) of the products will be different (compare ˚API of residual oil vs ˚API of
stock tank oil). The only thing that will be the same for the two processes is the total
weight of end products.
Column 3 are the relative volumes of the liquid phase measured during the differential
liberation of gas. Note that (per the footnote) these are volumes at pressure p are
Institute of Petroleum Engineering, Heriot-Watt University
39
expressed per unit volume of residual oil. Again, these relative volumes must not
be confused with formation factor volumes because formation factor volumes are
specified per barrel of stock tank oil. Note on page 5 that relative volumes start
at 1.000 at 0 psig/60˚F and that the value of 1.075 at 0 psig/220˚F is the thermal
expansion of 35.1˚API residual oil from 60˚F to 220˚F.
Above 2620 psig, the original bubble point, the system remained constant in
composition. Therefore, the relation of the relative oil volume at p to the bubble
point value, 1.600, must be the same as the relative volume in numbers in column
2, page 4 of the report.
The other data on page 5 are differential liberation that refer to the oil and gas phases
in the reservoir at 220˚F. Column 8 shows that the gravity of the gas liberated between
2620 psig and 2350 psig was 0.825. The next batch between 2350 psig and 2100 psig
was 0.818. The gas deviation (compressibility) factor of the first liberated gas was
0.846 at 2350 psig. The oil density at 2350 psig/220˚F was 0.6655 gm/cc.
Now we understand the basic difference between flash and differential data as given in
the standard PVT report, we can calculate flash solubilities and oil formation volume
factors below the bubble point from a combination of the differential and flash data. It is important to appreciate that there are two separation stages in separating the oil
from its original solution gas when the fluid in the reservoir has dropped below the
bubble point. The drop of the reservoir pressure from the bubble point pressure to
a lower pressure is considered to be by a differential process. The separation of gas
from the reservoir pressure to the surface is then by the flash process. 12.5 Calculation of Gas - Oil Ratios Below The Bubble Point.
If we examine the separator and differential tests there seems to be a confusing result
since the initial and final pressures are the same.
(a) Differential solubility data at the bubble point state (2620 psig/220˚F) and
eleven pressures below the bubble point pressure gives a bubble point value at
854ft3/B residual oil. All fluids at pressures above pb have this amount of gas.
(b) The flash solubility of the bubble point oil for four different surface trap situations, where these vary from 778ft3/B stock tank oil to 795ft3/B stock tank oil for
a 300 psig primary trap-tank situation. These are shown on the sketch. Figure 30.
The 59 ft3/B difference in values is not experimental error but is a result of the total
differential process of the test. In reality there is only a small differential element
in the early stages of depletion.
The pressure depletion starts at the bubble point and the solution GOR is that from
the flash separator tests. We now need to develop the GOR curve below this value.
We will use the 300 psig primary - 0 psig tank situation and will examine the GOR for the reservoir pressure of 1850 psig.
40
PVT Analysis
854 ft3/B residual oil
200°F
795 ft3/B stock tank oil
Rs ft3 /bbl
Differential
Two stage
surface flash
2620 psig
Pressure
Figure 30 Comparison of data from flash differential tests.
Exercise PVT 3.
Calculate the solution GOR at 1850 psig using the 300/0psig separator data.
Solution.
Looking at the differential liberation data in column 2, page 5, we see that 242ft3 of
gas has come out of solution, per barrel of residual oil, when the pressure declined
from 2620 psig to 1850 psig. 854 - 684. In other words, we can say that the 1850 psig
saturated oil contains less gas by this amount. If this liquid were taken to the surface
and processed through the traps, it would also show somewhat less gas solubilities
than the 795ft3/B stock tank oil that the bubble point oil shows; but it would not be
242ft3 less because we now have a different oil base.
If we let (∆Rs)diff be liberated gas-oil ratio by differential vaporisation, we can convert
this to a (∆Rs)flash as follows:
(∆Rs)diff = ft3/B Residual Oil
ft 3
B Re sidual Oil
ft 3
⋅
=
B Re sidual Oil B Bubble Po int Oil B Bubble Po int Oil
(1)
(2)
ft 3
B Bubble Po intoil
ft 3
⋅
=
B Bubble Po int Oil B Stock Tank Oil B Stock Tank Oil (3)
ft 3
= ( ∆Rs ) flash
B Stock Tank Oil
In equation (2)
B Re sidual Oil
1
=
B Bubble Po int Oil 1.600
In equation (3)
B Re sidual Oil
= 1.495
B Stock Tank Oil
Institute of Petroleum Engineering, Heriot-Watt University
(4)
41
Therefore:
1.495
(∆Rs ) flash = (∆Rs )diff ⋅ 1.600
and
1.495
= 795 − ( ∆Rs ) flash = 795 − 242 ⋅
( Rs )1850
flash
1.600
= 795 − 2261 = 569 ft 3 / BTSO
( Rs )1850
flash
Solution gas to oil ratio at 1850 psi = 569 scf/STB.
For those who prefer equations, this can be generalised as:
Bob
b / VR
( Rs ) flash = ( Rsb ) flash − ( Rs )diff ⋅ V
12.6 Calculation of Formation Volume Below The Bubble Point
Examination of the formation volume factors between the bubble point pressure
and surface pressures also show a distinct difference between the flash tests and the
differential data. This is illustrated by the figure (figure 31). The bubble point state
has a relative oil volume of 1.600 B/B residual oil. We also have the formation
volume factor at the bubble point state, Bob, with a value of 1.495 B/B stock tank oil
(300/0 separator combination).
Volume Factor
1.660
1.0
1.495
Differential
@ 200º F
Two Stage
Surface Flash
2620 psig
Pressure
Figure 31 Comparison of differential data with flash for volume factor volumes.
The result from the separator test is the correct value, since it is based on stock tank
volumes. The differential data is used to calculate the change in this separator value
below the bubble point.
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PVT Analysis
We can see that the relative oil volume and the formation volume factor at pressure
p can be related by transferring to the common point - the bubble point.
Let:
V/VR = relative oil volume at pressure p, B/B residual oil.
Then:
B Saturated Oil
B Re sidual Oil
⋅
B Re sidual Oil B Bubble Po int Oil
=
B Saturated Oil
B Bubble Po int Oil
B Bubble Po int
= Bo
B Stock Tank Oil
Therefore:
Bo = V VR .
Bob
Vb VR
Exercise PVT 4.
What is the oil formation volume factor at 1850 psig.
SOLUTION
At 1850 psig we would have: from page 5 relative volume of 1.479 B/B residual
oil
Bo1850 = 1.479
B1850
1 B residual
x
Bresidual 1.600 B at bubble
x 1.495
po int
B Bubble po int
B stock tan k oil
Bo1850 = 1.3819 B / B stock tan k oil
12.7 Viscosity Data
Page 6 of 15 presents the viscosity data for the fluid measured for the oil and calculated
for the gas. It should be noted that the pressure for the data below the bubble point are
the same as those for the differential tests, since the viscosity is also measured below
the bubble point having generated the fluid pressure through a differential process.
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13 GAS CONDENSATE PVT REPORT
Parts of a gas condensate report attributed to the Good Oil Exploration and
Production Company is given for a North Sea field. Not all the report is given,
information not included is the sample validation section. Table B1 is the summary
sheet giving information on the reservoir and the sampling details. Table C4 gives a
comprehensive compositional analysis of separator products and the calculated well
stream composition. The table C7 is the compositional analysis of the reservoir fluid
sample. This sample was generated by recombining the separator fluids and then
blowing down the recombined fluid to determine the fluid composition. This provides
a quality check against the calculated well stream composition obtained from the
composition of the separator fluids. Examination of the well stream and reservoir
fluid compositions show they are in close agreement.
The compositional analysis of these condensate fluids is far more comprehensive
than the oil report results. This is because the heavy fraction has a significant impact
on the properties of the condensate and therefore it is important to have the data for
modelling the behaviour of these fluids.
The two key tests for gas condensates is the constant mass test or the constant
composition expansion test and the constant volume depletion test. Table D1 is the
constant composition expansion test. In the oil test the saturation pressure is obtained
by a distinctive change in slope of the pressure volume curves for the single phase
and two phase regions. For gas condensates no such distinct change in slope occurs
at the dew point. The dew point is obtained by observation of the condensation on
the window of the PVT cell. The distinctive nature of the condensation depends on
how near to the critical temperature the fluid is. If the fluid is near to the critical
temperature than the dew point is clearly observed, whereas for those fluids away
from the critical point the dew point is less distinct. Colour changes of the fluid also
occur around the dew point, getting darker as the dew point is approached.
For the constant composition expansion at 275 °F on page D1 relative volume results
are given over a pressure range from 8,000 psi g to 1198 psig. In the same way as
for the oil PVT analysis volumes are related to those at the saturation pressure, the
dew point. At each pressure step not only is the relative volume calculated but also
the compressibility factor, z. The dew point was observed to be at 5454 psig. The
table also records pressure steps below the dew point and down to 4,900 psig the
retrograde liquid volume is also reported. Clearly these values are not of great value
since in the reservoir constant composition does not occur below the dew point
because the system is considered to be changing as immobile condensate separates
from its associated gas.
The constant volume depletion test data is given on D8. In this test a series of
depletion steps have been carried out. Starting at the dew point pressure of 5454
psig the pressure was dropped to 4900psig. Between 5454 psig and 4900 the liquid
proportion was measured. At 4900psig gas was removed at constant pressure until
the volume at the dew point was obtained. A series of further depletion steps were
carried out at 4000, 3100,2200,1400 and 726 psig. At each stage gas was removed
to bring the contents back to the dew point condition volume.
44
PVT Analysis
The important liquid drop-out curve is presented in the curve on page D9. The dotted
line is the relative volume data from the constant mass test. The curve shows a very
high maximum liquid drop out of 44% at a pressure of 3900psig.
14 HIGH PRESSURE / HIGH TEMPERATURE, HP/HT, FLUIDS
Over recent years as exploration activity has moved deeper into the sub-surface,
high pressure and high temperature hydrocarbon fluids have been found. A number
of these are now being produced providing a number of challenges in production
and design. Conventional PVT facilities do not cover the pressure and temperature
ranges covered by these fluids and special facilities are required. The temperature
and pressure ranges for such fluids are up to 250°C and 20,000 psi.
In order to handle these fluids and conditions and to enable visualisation of phenomena,
like for example IFT, very low volume apparatus are required. The move to Hp/HT
fluids has also put in question some of the physical property prediction methods which
have been based on pressure and temperature data not extending to HP/HT conditions.
An important feature of high pressure and high temperature is the role of water which
in conventional PVT practice is ignored. At these more extreme conditions recent
measurements are showing that the presence of water cannot be ignored because its
presence will contribute to the physical properties of the hydrocarbon fluids.
15 MERCURY
Historically the transfer fluid in PVT tests has been mercury. It has proved to be a
very effective fluid to generate variable volumes in PVT apparatus as well as being
non contamination with respect to the hydrocarbon fluids. Unfortunately health and
safety concerns with respect to personnel exposed to increased levels of mercury,
and it's incompatibility with certain materials e.g. Aluminium, are such that mercury
is being replaced by alternate systems. Such alternate systems are not as simple to
replace the "flexible metal" which mercury has proved to be. Although in these notes
we refer to mercury, the principals are the same, where for example the mercury is
replaced by a rigid piston driven by a safe fluid e.g. Water.
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Solutions to Exercises
Exercise PVT 1.
What is the solution gas-oil ratio and formation volume factor resulting from the
separator test, at first stage of 300 psig and 2nd stage 0 psig and both at 75˚F ?
Solution.
This exercise illustrate the results using one of these tests, a two-stage separation;
a primary trap operating at 300 psig and 75˚F followed by a stock tank operating at
14.7 psia (0 psig) and 75˚F.
When one barrel of bubble point oil; defined as oil saturated at 2620 psig and 220˚F in
footnote 3, on page 7 of 15 is flashed (processed) through this separation arrangement,
the stock tank has a quality of 40.1˚API (column 5). The formation volume factor
of the bubble point oil, Bob = 1.495 B/BSTO (column 6). This would have been the
volume of oil displaced from the PVT cell divided by the volume collected at the final
stage and then corrected for the thermal reduction from 75˚F to 60˚F. The source of
this bubble point pressure value will be indicated later.
Columns 3 and 4 show the surface gas-oil ratio from the first stage and the tank. The first stage ratio of 549 ft3/BSTO ( column 4) and the tank stage gas amount to
246 ft3/BSTO. It is important to read the footnotes of the report. Column 3 gives
the results in relation to the volumes at indicated P & T whereas column 4 gives the
volumes with respect to stock tank conditions of 14.65 psia and 60˚F. The solution
gas-oil ratio at bubble point conditions (2620 psig and 220˚F), is therefore Rsb is 549
+ 246 = 795 ft3/ BSTO when flashed through this surface trap arrangement. If we compare these results for the 50psig, 0psig arrangement we obtain a Bob of
1.481 B/BSTO and a solution GOR of 778 ft3/ BSTO.
Clearly therefore Rsp, Bob, ˚API all vary with the separation pressure-temperature
situation. There is not one unique result. When reporting Bo and GOR data for a
reservoir therefore it is important to report that these are for a specific separation
route or averaged for a series of tests. The latter is not so useful since it is not so
straightforward to calculate the result for a different separation route using VLE
methods.
The results from the separation test are based on the bubble point condition and to
obtain volumetic information at other pressures we require the results from other
tests.
Exercise PVT 2.
What is the formation volume factor and the density of the oil at the last reservoir
pressure measured.
Solution.
The well characteristics give the last reservoir pressure as 3954 psig. @ 8500 ft: We
obtain the oil formation volume at 3954 psig by multiplying the formation volume
factor at the bubble point by the relative volume (to the bubble point). Why multiply? Because:
70
PVT Analysis
Bo =
vol reservoir oil
vol bubble point oil
vol reservoir oil
=
×
vol stock tank oil
vol stock tank oil
vol bubble point oil
and the reference bubble point oil volume cancels out. Therefore Boi, the initial
formation volume factor is 1.495 x 0.9778 = 1.4618 when the 300 psig primary trap
is involved. It is a different value if another separation pressure is used. The 0.9778
was obtained by interpolation bewteen 3500 and 4000 psig in column 2.
Reservoir oil density at pressures greater than 2620 psig also make use of the relative
volume data of column 2, page 4. The added information we have is the density of
the bubble point oil. This is given in the summary data on page 3 of the report. We
see here that the specific volume at the bubble point, vb = 0.02441 ft3/lb. This comes
from direct weight-volume measurements on the sample in the PVT cell. We can
now calculate the density, roi, of the initial reservoir oil as:
ρoi =
1
1
=
voi vob ⋅ vrel
ρoi =
1
= 41.89lbs / ft3
0.02441× 0.9778
The compressibility of the oil above the bubble point can also be obtained from the
relative volume test. The definition of compressibility is:
1  ∂v 
Co = −  
v  ∂p T
It makes no difference whether the volume units in the equation are relative volumes
to the bubble point, formation volumes, or specific volume values. To evaluate COat
pressure p it is only necessary to graphically differentiate the p-vrel data in columns
∂v
1 and 2 to get ∂p at the pressure and divide by vrel. A less accurate value can be
obtained by the assumption:
Co = −
1  ∆v
 
vavg  ∆p T
It makes no difference whether the volume units in the equation are relative volumes
to the bubble point, formation volumes, or specific volume values. To evaluate COat
pressure p it is only necessary to graphically differentiate the p-vrel data in columns
∂v
1 and 2 to get ∂p at the pressure and divide by vrel. A less accurate value can be
obtained by the assumption:
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Co = −
1  ∆v
 
vavg  ∆p T
For example, to get Co at 4500 psig using relative volume values of 500 psi on each
side of 4500 psig:
1
(0.9639 − 0.9771)
C o = − 0.9639 + 0.9771
(5000 − 4000 )
2
Co =
1 0.0219
= 13.6(10−6 )vol / vol / psi
0.9705 1000
The report also lists some compressibility numbers on page 3. These are not the
same as indicated above because they are changes in volume (in the pressure interval
indicated) per unit volume at the higher pressure. For example, the value of 13.48 (10-6)
for the 5000 and 4000 psi interval is obtained as:
−
1 ( 0.9771− 0.9639)
0.9639 (5000 − 4000)
The compressibility data on page 2 are set up in this manner because of the way they
are used in one form of the material balance.
Exercise PVT 3.
Calculate the solution GOR at 1850 psig using the 300/0psig separator data.
Solution.
Looking at the differential liberation data in column 2, page 5, we see that 242ft3 of
gas has come out of solution, per barrel of residual oil, when the pressure declined
from 2620 psig to 1850 psig. 854 - 684. In other words, we can say that the 1850 psig
saturated oil contains less gas by this amount. If this liquid were taken to the surface
and processed through the traps, it would also show somewhat less gas solubilities
than the 795ft3/B stock tank oil that the bubble point oil shows; but it would not be
242ft3 less because we now have a different oil base.
If we let (∆Rs)diff be liberated gas-oil ratio by differential vaporisation, we can convert
this to a (∆Rs)flash as follows:
(∆Rs)diff = ft3/B Residual Oil
ft3
B Residual Oil
ft 3
⋅
=
B Re sidual Oil B Bubble Point Oil B Bubble Po int Oil
(1)
(2)
ft3
B Bubble Po int Oil
ft3
⋅
=
B Bubble Point Oil B Stock Tank Oil
B Stock Tank Oil (3)
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PVT Analysis
ft3
= (∆Rs ) flash
B Stock Tank Oil
(4)
In equation (2)
B Residual Oil
1
=
B Bubble Point Oil 1.600
In equation (3)
B Bubble Point Oil
= 1.495
B Stock Tank Oil
Therefore:
(∆Rs ) flash = (∆Rs )diff ⋅ 1.495
1.600
and
1.495
= 795 − ( ∆R s ) flash = 795 − 242 ⋅
( Rs )1850
flash
1.600
= 795 − 2261 = 569 ft 3 / BTSO
( Rs )1850
flash
Solution gas to oil ratio at 1850 psi = 569 scf/STB.
For those who prefer equations, this can be generalised as:
( Rs ) flash = ( R sb) flash − (R s ) diff ⋅
B ob
Vb / VR
Exercise PVT 4.
What is the oil formation volume factor at 1850 psig.
SOLUTION
At 1850 psig we would have: from page 5 relative volume of 1.479 B/B residual
oil
Bo1850 = 1.479
B 1850
1 Bresidual
B Bubble point
x
x1.495
Bresidual 1.600 B at bubble point
B stock tan k oil
Bo1850 = 1.3819 B / B stock tan k oil
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Material Balance Equation
CONTENTS
1. INTRODUCTION
2. LIST OF SYMBOLS
3. Material Balance for Gas Reservoirs
3.1 Dry gas, no water drive
3.2 Dry gas reservoir with water drive
3.3 Graphical Material Balance
3.4 Wet Gas Reservoirs
3.5 Gas Cap Expansion
4. MATERIAL BALANCE FOR OIL EXPANSION
4.1Above The Bubble Point
4.2 Gas Liberation Below the Bubble Point
4.3 Material Balance with Gas Cap and Water
Drive
4.4Effect of Pore - Volume Changes
4.4.1Compressibility Effecys
4.4.2Overburden Pressure
4.4.3Connate Water
5. THE GENERAL MATERIAL BALANCE EQUATION
6. MODIFICATIONS TO THE GENERAL EQUATION
7. DERIVATION OF THE MATERIAL BALANCE
EQUATION BY EQUATING SUBSURFACE VOLUME OF PRODUCED FLUIDS TO EXPANSION OF ORIGINAL FLUIDS PLUS PORE VOLUME REDUCTION
8ASSUMPTIONS IN MATERIAL BALANCE EQUATION
9. SIGNIFICANCE AND USAGE OF THE MATERIAL BALANCE EQUATION
10. SOURCES OF DATA TO BE USED IN THE MATERIAL BALANCE
11. LIMITATIONS OF THE MATERIAL BALANCE
12. CONCLUSION
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•Present a material balance (MB) equation for a dry gas reservoir with and without
water drive.
•Demonstrate the linear form of the MB equation for a gas reservoir with water
drive and comment on its application.
•Be able to derive the material balance equation including gas cap expansion,
water influx and core and water compressibility.
•Given the equation be able to identify the component parts of the MB equation,
eg. gas cap expansion etc.
•Comment briefly on the assumptions, significance , use, data and limitations of
the MB equation.
Material Balance Equation
1. INTRODUCTION
In the chapter on Drive Mechanisms we reviewed qualitatively the various drive
energies responsible for hydrocarbon production from reservoirs. In this and subsequent chapters we will introduce some reservoir engineering tools used in calculating reservoir behaviour. The petroleum engineer must be able to make dependable
estimates of the initial hydrocarbons in place in a reservoir and predict the future
reservoir performance and the ultimate hydrocarbon recovery from the reservoir. In
this chapter the material balance equation is presented.
The material balance equation is one of the basic tools in reservoir engineering.
Practically all reservoir engineering techniques involve some application of material balance. Although the principle of conservation of mass underlies the material
balance equation, custom has established that the material balance be written on a
volumetric basis, because oilfield measurements are volumetric and significant factors can only be expressed volumetrically.
The principle of conservation underpins the equation:
Mass of fluids originally in place = fluids produced + remaining reserves.
The equation was first presented by Schilthuis1 in 1936 and many reservoir engineering
methods involve the application of the material balance equation. Since the equation
is a volumetric balance, relating volumes to pressures, it is limited in its application
because of any time dependant terms. The equation provides a relationship with a
reservoir’s cumulative production and its average pressure. However when combined
with fluid flow terms, we have a basis to carry out predictive reservoir modelling, for
example to put a time scale to production figures.
Over recent years, as increasingly powerful computers have enabled the application of
large numerical reservoir simulators, some have looked down on the simple material
balance equation and the tank model of the reservoir which it represents.
Reservoir simulators however apply the material balance approach within each of
their multi-dimensional cells. The value of this classical tool is that it enables the
engineer to get a’feel’ of the reservoir and the contribution of the various processes
in fluid production. A danger of blind application of reservoir simulators is that the
awareness of the various components responsible for production might be lost to the
engineer using the simulation output in predictive forecasting.
The basic ‘material balance’ equation is presented as a volumetric reservoir balance
as follows:
The reservoir volume of original fluids in place = reservoir volume of fluids produced
+ volume of remaining reserves.
When fluids (oil, gas, water) are produced from an oil reservoir, which may or may
not have a primary gas cap, the pressure in this reservoir will drop below the original
value. As a consequence of this pressure drop, a number of things will happen:
• the pore volume of the reservoir will become smaller
• the connate water will expand
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•oil, if still undersaturated, will expand
•oil, if at or already below bubble point, will shrink while gas will come out of
solution
•free gas, if present, will expand
•water may start flowing into the reservoir, for instance, across the original oil/
water contact (OWC).
The question is now; if we start off with a “given” reservoir, and after some time we
have produced certain quantities of oil, gas and water, what can we say about the
average pressure in the reservoir, and what can we say about the average saturation
distribution? The answer to these questions can be obtained by considering our
reservoir at two stages:
(a)at the initial pressure pi,
(b)when we have produced certain amounts of oil, gas and water, by which time
the average pressure has declined to p (to be calculated).
Besides these natural phenomena the equation also has to be capable of handling
other factors affecting behaviour, for example injecting gas and or water.
There are a number of ways of developing the equation. We will look at two approaches, the first examining the equation as applied to specific reservoir types and
then a simple volumetric expansion approach.
The nomenclature to be used for the various terms is given below:
NOTE:
In the following derivations, volumes at standard conditions will be converted into
subsurface volumes and vice versa. Remember that to convert a volume from standard
conditions to reservoir conditions, one must multiply by a formation volume factor (B) and to convert from reservoir into standard conditions one must divide by a
formation volume factor.
2. LIST OF SYMBOLS
Symbols
Units
Units SI
Bg
Bo
Bt
Bw
cf
cw
G
Gp
Gps
Gpc
bbl/SCF
bbl/STB
bbl/STB
bbl/STB
vol/vol/psi
vol/vol/psi
SCF
SCF
SCF
SCF
M3/SCM
M3/SCM
M3/SCM
M3/SCM
vol/vol/Mpa
vol/vol/Mpa
SCM
SCM
SCM
SCM
Gas formation volume factor
Oil formation volume factor
Total formation volume factor
Water formation volume factor
Pore compressibility
Water compressibility
Initial gas-cap volume
Cumulative gas produced = Gps + Gpc
Cumulative solution gas produced
Cumulative gas cap produced
Material Balance Equation
m
Ratio initial reservoir free gas volume to
initial reservoir oil volume
N Stock tank oil initially in place
Np Cumulative tank oil produced
pAverage reservoir pressure
pi Initial reservoir pressure
Rp Cumulative gas/oil ratio
Rs Solution gas/oil ratio
SwAverage connate water saturation
We Cumulative water influx
Wp Cumulative water production
i
b
bbl/bbl
STB
STB
psi
psi
SCF/STB
SCF/STB
fraction
bbl or STB
bbl or STB
M3/M3
STM3
STM3
MPa
MPa
SCM/STM3
SCM/STM3
fraction
M3 or STM3
M3 or STM3
Other subscripts
at initial conditions
at bubble point
3. MATERIAL BALANCE FOR GAS RESERVOIRS
The simplest material balance equation is that applied to gas reservoirs. The compressibility of gas is a very significant drive mechanism in gas reservoirs. Its compressibility
compared to that of the reservoir pore volume is considerable. If there is no water
drive and change in pore volume with pressure is negligible (which is the case for a
gas reservoir), we can write an equation for the volume of gas in the reservoir which
remains constant as a function of the reservoir pressure p, the volume of gas produced
SCF, the original volume of gas, SCF, and the gas formation volume factor.
A representation of the equation for a gas drive reservoir with no water drive is
given below.
3.1 For a dry gas reservoir - no water drive:
Figure 1
G.Bgi = (G-Gp) Bg (1)
Bgi - based on zi, pi, Ti
Bg - based on z, p, T
Gp
G Bgi
=
Pi
(G - Gp)Bg
P
Figure 1 Material Balance For a Dry Gas Reservoirs no Water Drive
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N.B:
pV = znRT
If the gas reservoir is supported by water drive then as gas is produced water will
encroach into the gas pore space, and some of this water may be also be produced.
Figure 2 below illustrates the contact with a supporting aquifer. Because the mobility of gas is far greater than water, evidence in the form of produced water may be
delayed as the water keeps to the gas water contact. The support from the water would
be evidenced however by the pressure support given to the reservoir. In earlier years
this may not be so easy to detect.
3.2 For a dry gas reservoir with water drive
With water drive water will enter pore volume originally occupied by gas and some
water may be produced. Figure 2
Gp
GBgi
(G - Gp) Bg
=
Water
Wp
We - Wp
Water
Figure 2 Material Balance For a Dry Gas With Water Drive
GBgi = (G-Gp)Bg + We - Wp
(2)
EXERCISE 1
A gas reservoir without water drive contains 500 million standard cubic feet of gas
at an original pressure of 3,000psia. How much gas has been produced when the
reservoir pressure has declined to 2,900 psia. Use Bgi and Bg for the initial and
2,900psia pressure as 0.0010 and 0.0011 bbl/scf.
This simple example illustrates the significant amount of gas production associated
with a relatively small pressure decline.
3.3 Graphical Material Balance
One can use a graphical form of the material balance equation to analyse a gas reservoir and predict its behaviour especially if no water drive is present.
Material Balance Equation
G. Bgi = (G − Gp ) Bg
From equation 25 in Gas Pr operties chapter where Bg =
0.00504 zi T
p
 0.00504 zi T 
 0.00504 zT 
G

 = (G − Gp ) 
pi
p




G
zi
z
= (G − Gp )
pi
p
p  Gzi 
Gp = G −   
 z   pi 
(3)
hence plot of Gp vs p/z should give a straight line
G
X
Gp
Cumulative gas
production
O
X
X
p/z
Pi/Zi
Figure 3 Gp vs. p/z
If gas was ideal a plot of Gp vs p would be a straight line. It is often practice to do
this and get a relatively straight line, but caution has to be taken, since deviation from
a straight line could indicate additional energy support.
- when p/z = 0
- when Gp = 0
Gp = G the original gas in place
p/z = pi/zi
This procedure is often used in predicting gas reserves. Often the influence of water
drive is ignored resulting in a serious error in reserves.
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This simple analysis method for gas reservoirs has gained wide acceptance in the
industry as a history matching tool, to determine for example an estimate of initial
gas reserves based on production data. This figure, (figure 3 ), can then be compared
to estimates from exploration methods. It can also give indications of gas to be
produced at abandonment pressures. The following example exercise from Slider’s2
reservoir engineering text illustrates the application of the method
EXAMPLE
A dry gas reservoir has produced as follows:
Data
07-Jan-65
07-Jan-66
09-Jan-67
10-Jan-68
11-Jan-69
Data
Reservoir Temperature T=
Gas Gravity
SG=
Cumulative
production
MM SCF
0
1,800
3,900
5,850
9,450
Static Res.
Pressure
psia
3,461
3,370
3,209
3,029
100°F
0.68
1. Determine the original pressure and original gas in place.
2. What will be the average reservoir pressure at the completion of a contract calling
for delivery of 20 MM SCFD for 5 years (in addition to the 9,450 MM SCF
produced to 11-Jan-69?)
SOLUTION
To construct the graphical material balance plot we must first determine the P/Z
values.
Using figures 2 and 3 from the Gas Reservoir chapter for a gas gravity of:
SG=0.68
The pseudo-critical parameters are found to be:
Pseudo critical pressure (psia)
Ppc= 667.5 psia
Pseudo critical temperature (oR) Tpc=
385.0 °R
Graphical material balance
Cumulative
production
MM SCF
0
1,800
3,900
5,850
9,450
Reservoir
Pressure
psia
Pseudo reduced
Pressure
Temp.
Pr
Tr
Z
P/Z
(From Fig. 2)
3,461
3,370
3,209
3,029
5.19
5.05
4.81
4.54
0.796
0.790
0.778
0.765
1.45
1.45
1.45
1.45
4,348
4,266
4,125
3,959
Material Balance Equation
The P/Z vs. G, plot is shown in the following figure:
Graphical Material Balance
4500
y=-0.0522x + 4448.3
R2 = 0.9904
4400
P/Z (psia)
4300
4200
4100
4000
3900
3800
0
2,000
4,000
6,000
8,000
10,000
12,000
14,000
Cumulative gas production (MM scf)
Figure 4 (a)
Graphical Material Balance
4500
4000
y=-0.0522x + 4448.3
R2 = 0.9904
P/Z (psia)
3500
3000
2500
2000
1500
1000
500
0
0
2,000
4,000
6,000
8,000
10,000
12,000
14,000
Cumulative gas production (MM scf)
Figure 4 (b)
From the straight line of figure 1,
Slope
=
-0.0522
Intercept
=
4,448.3
Equation
=
P/Z= -0.0522 Gp + 4448.34
(1a)
Initial Pressure
From figure 4, at Gp = 0:
Pi/Zi= 4,448.3 psia
Institute of Petroleum Engineering, Heriot-Watt University
Now, dividing by Pc
i.e.
( Pi / Zi ) = Pri
Pc
Zi
Pri/Zi = 6.6642
From figure 5:
Zi= 0.81
Therefore:
Pi= 3,603
Original Gas in Place "G"
G can be calculated directly from Eq. (1a), when P/Z = 0
G= 85,217
MM SCF
Pressure at the Completion of the Contract
Gas production rate (contract)
=
Duration (contract) t
=
Cumulative volume (contract) =
At the end of the contract:
Total cumulative production
Gp
=
P/Z
=
Again, by dividing by Pc
Pr/Z =
20 MMSCFD
5 Years
36,500 MM SCF
45,950 MM SCF
2,050 psia from equation 1a
3.07
and from Figure 5
Z
=
0.775
Finally,
P
=
1,589
10
psia
1.05
1.10
1.
1.2015
1.25
1.30
1.35
1.40
1.45
1
1 .5 0
1 .6 0
1 .7
1.9.800
2
.0 0
2.2.20 0
40
Material Balance Equation
1.75
1.65
1.55
Used to obtain p and z when
analyses result in p/z answers.
1.45
(pr/z) = (p/z)/pc
1.35
p = (pr/z) z pc
z can be read from graph.
2.60
2.80
3.00
Gas deviation factor, z
1.25
1.15
Reduced temperature, Tr
1.05
3.00
2.80 2.60
2.40 20
2.
2.00
1.90
1.80
1.70
1.60
1.50
1.45
1.40
1.35
1.30
1.25
0.95
0.85
0.75
0.65
0.55
1.20
0.45
1.15
0.35
1.10
1.05
0.25
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0
(pr/z)
Figure 5 Gas Deviation Factor z vs. Pr/z (Slider2)
Great caution has to be taken when using this method. Water drive is considered to
be zero, that is the gas is being solely produced as a result of gas compressibility.
If water drive exists this will contribute to pressure support. If a plot of Gp vs p/z
deviates from linearity than that gives evidence of water drive support. Figure 6
from Dake illustrates this deviation. If a straight line is fitted to this data assuming
no pressure support from water then gas reserves are enhanced, beyond what they
are in actuality.
(b)
(a)
3500
3500
P/Z
P/Z
O
Gp
G
G'>G
2700
Gp
Figure 6 p/z Plots For A Water Drive Gas Reservoir3.
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11
We will consider this topic later If there is water drive then the equation;
GBgi=(G-Gp)Bg +We-WwBw
(4)
applies.
3.4 Wet Gas Reservoirs
Another aspect that needs to be considered with gas reservoirs is the treatment of wet
gas reservoirs. In these reservoirs production also includes liquids as well as gas,
although in the reservoir the liquids were in a gaseous state, figure 7. In the application of the material balance equation to these reservoirs it is important to convert oil
production to gas equivalent figures to add to the gas production figures.
Pi, Ti
Liquid
Bubble point line
Single phase
P
2 phase
Sep.
Dew point line
Mixture
Gas
T
Figure 7 Phase Diagram For a Wet Gas System
The equation already produced assumed that the formation of liquid condensate
causes insignificant error in the quality.
For condensate systems the Gp produced should include the produced condensate
and the produced water (originally dissolved in gas).
The volume of 1 STB of condensate of molecular weight Mo and specific gravity γo
follows from equation.
v=
znRT
P
z = 1.0 at p = 14.7 psia and T = 520˚R
12
(Density of water = 62.4 lb/ft3)
Material Balance Equation
∴ V / STB = 10.73
v = 133,000
psia SCF
520°R lb mole
lb
cu ft
x
x
x62.4γ o
x5.615
lb mole°R 14.7 psia M o lb
cu ft
STB
γ o SCF
M o STB
3.5 Gas Cap Expansion
If a gas reservoir is attached to an oil reservoir (figure 8 ), a similar equation to (2):
GBgi = (G-Gpc) Bg
can be written to describe the change in gas cap volume due to oil production and
production of gas. In this case it is suggested that some gas has been produced from
the gas cap, Gpc.
Gpc
GBgi
(G - Gpc) Bg
=
Oil
Gas cap exp.
Oil
Figure 8 Gas Cap Expansion
Change in gas cap volume is
(G - Gp)Bg - GBgi
(5)
4. MATERIAL BALANCE FOR OIL RESERVOIRS
4.1 Above the Bubble Point
Above the bubble point the production of the reservoir is due to the expansion of
the liquid (including water) in the reservoir as pressure declines, and the reduction
in pore volume due to the decrease in pressure. Assuming that the production is due
entirely to liquid oil expansion, a material balance for an oil reservoir is similar to
that for a gas reservoir.
NBoi = (N - Np)Bo
(6)
where N is the original oil volume in the reservoir and the Np is the volume of oil
produced both expressed in stock tank barrels.
Clearly this is a poor assumption but is useful in illustrating the equation development. Pore volume changes will be considered later in the context of pore space and
connate water.
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13
4.2 Gas Liberation Below the Bubble Point
When the oil in the reservoir reaches the bubble point pressure, gas will be liberated
and will continue to be liberated as the pressure declines. This is the mechanism of
solution gas drive described previously.
As we observed in this mechanism, the produced fluids are now oil with its contained
solution gas and gas which has come out of solution from the oil. Not all of this released free gas will be produced to the surface, some will remain in the reservoir.
The free gas in the reservoir can be written:
Free gas in reservoir = original gas in solution - remaining gas in solution - produced
gas(Gps).
= NRsi - (N - Np)Rs - Gps SCF
= (NRsi - (N - Np)Rs - Gps)Bg bbl
(7)
The volume of free gas and the remaining oil can now be added to the original oil
volume.
NBoi = (N - Np)Bo + (NRsi-(N - Np)Rs - Gps)Bg (8)
The equation can be written in terms of the original stock-tank volume in the reservoir.
N=
N p Bo + Bg (G ps − N p R s )
Bo − Boi + (R si − R s )Bg Np
NBoi
=
(9)
Gps
Free Gas
(NRsi-(N-Np)Rs-Gps)Bg
Oil
(N-Np)Bo
Figure 9 Material Balance For Solution Gas Drive
4.3 Material Balance with Gas Cap and Water Drive
The equation just developed assumes no change in reservoir volume. If fluid encroaches
into original oil bearing volume either from an expanding gas cap or an encroaching
water drive there will be loss to the reservoir volume.
Change in volume due to gas cap expansion.
= (G - Gpc)Bg - GBgi
14
(10)
Material Balance Equation
Change in volume due to water encroachment
= (We-Wp)
(11)
∴ Total change in volume = original oil volume - (oil volume + free solution gas)
Np
NBoi
=
Gpc
Gps
Wp
Gas cap expansion
Oil volume and free
solution gas
Net water encroachment
Material balance with gas cap and water drive
Figure 10 Material balance with gas cap and water drive
That is (We-Wp) + (G - Gpc)Bg - GBgi
(
= NBoi - (N - Np)Bo + (NRsi-(N - Np)Rs - Gps)Bg
N=
)
N p Bo + Bg (Gps − N p Rs ) − ((G − G pc )Bg − GBgi )) − (We − Wp )
Bo − Boi + (Rsi − Rs )Bg
(12)
(13)
N p Bo + Bg (Gps − N p Rs ) − ((G − G pc )Bg − GBgi )) − (We − Wp )
= production is separated into gas cap and solution gas, G and G . However,
TheNgas
ps
N p Bo + Bg (Gp − Np R
− Bgi−)R− (W
− Wp ) pc
Bos )−−GBG(B
(R
oi +g+G
si then:
s )Bge
= can
theNtwo
be combined so that
=G
p
pc
ps
Bo − Boi + (Rsi − Rs )Bg
N=
N p Bo + Bg (Gp − Np R s ) − G(Bg − Bgi ) − (We − Wp )
Bo − Boi + (Rsi − Rs )Bg
(14)
4.4 Effect of Pore-Volume Changes
4.4.1 Compressibility Effects
The compressibility of water is about 10-6 psia-1 compared to oil at around 10-5 psia-1
and although it is a low value it can contribute significantly to the hydrocarbon pore
volume change when pressure declines. Another contributing effect to reduction of
pore volume available to hydrocarbons is the compressibility of the pore volume
itself.
The reduction in pore volume with decline in pressure is due to two factors..
• reduction in the bulk volume of reservoir
• increase in volume of the reservoir grains
Figure 11 below illustrates the impact of overburden stress
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15
Overburden pressure
Pore pressure
- sand grains
- pore space
Figure 11 Cross Section of Sandstone Influence of Overburden and Pore Pressure2.
4.4.2 Overburden Pressure
A reservoir is subjected to an overburden pressure caused by the weight of the formation above the reservoir. It is equivalent to about 1 psi/ft of depth. These pressures
simply apply a compressive force to the reservoir rock. The pressure in the pore
space, the pore pressure, does not normally approach that of the overburden pressure
and is normally about 0.5 psi/ft. If the reservoir sands are highly unconsolidated
then this pressure could be higher as the overburden pressure is transmitted to the
fluids in the pore space.
When the pore pressure is reduced then the effective opposing pressure is increased
and the bulk volume is slightly reduced. At the same time since the rock grains are
compressible albeit only slightly the effect of a reduction in reservoir pressure will
be to expand the rock grains.
hence reduce Vb)
)
and increase Vs )
- decrease in porosity ø
1 ∆Vpr
.
V
∆p
p
Compressibility of rock cf =
∆Vpr = Cf∆pVp
Where Vp is the volume of the pores.
16
(15)
Material Balance Equation
4.4.3 Connate Water
Expansion of connate water can contribute to the reduction in pore volume for the
hydrocarbons.
The compressibility of the water can be expressed as
cw =
1 ∆Vpw
Vpw ∆p (16)
Where Vpw is the volume of water in the pores
Vpw = Vp x Swc
∴ ∆Vpw = cw∆pSwcVp
(17)
Total pore volume change:
∆Vp = ∆Vpw + ∆Vpr
∴ ∆Vp = (cf + cwSwc) ∆p Vp (18)
This term can be added to the material balance equation in the same way that water
encroachment can be considered.
Pore volume:
Vp =
NBoi
1 − Swc
Compressibility effect of water and pores
∆Vpw + pores =
NBoi
(cw Swc + c f )∆p
1 − Swc
(19)
If the original volume of pores also includes that associated with the gas cap, then
the pore volume is equal to
Vp = (1 + m)
NBoi
1 − Swc
where m is the ratio of the original reservoir gas cap volume to the original reservoir
oil volume. Some choose not to add this m factor term since if free gas is present as
a gas cap then the compressibility associated with the gas is far greater than the pore
and water compressibility values.
5. THE GENERAL MATERIAL BALANCE EQUATION
Combining all the effects of gas cap expansion, water encroachment, pore volume
changes and equating these to the volume changes associated with the oil gives the
general material balance equation below.
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17
net water inf lux
pore volume reduction
(We − Wp ) + (G − Gpc ) Bg − GBgi +
(C f + Cw Swc )∆pNBoi
(1 − Swc )
gas cap exp ansion
original oil
oil and free solution gas
(
= NBoi − ( N − N p ) Bo + ( NRsi − ( N − N p ) Rs − Gp ) Bg
N=
)
N p Bo + Bg (Gp − N p Rs ) − G( Bg − Bgi ) − (We − Wp )
Boi
Bo − Boi + ( Rsi − Rs ) Bg + (C f + Cw Swc )∆p
1 − Swc
The above equation is the general material balance equation.
In some texts the pores connate water compressibility term includes a product with
(1+m). This includes pore volumes associated with a gas cap as mentioned above.
The equation can be rearranged for different applications. The following useful
rearrangement by Archer4 , Figure 11, helps to identify the constituent parts of the
equation.
Present Oil Volume
(N-Np)Bo
=
Original Oil
Volume
NBoi
Free Solution Gas
[NRsi-(N-Np)Rs-Gps]Bg
Gas Cap Expansion
(G-Gpc)Bg-GBgi
Net Water Influx
We-WpBw
Rock(pore) and connate water expansion
NBoi (1+m)∆p
cwSwc+cf
1-swc
Injected volumes
WinjBw+GinjBg
Figure 12 Elements of the MB Equation
In the literature sometimes the equation is presented using the total formation volume
factor Bt and the ratio of the initial reservoir free gas volume to the initial reservoir
oil volume m, Gp, is also expressed as a function of produced gas - oil ratio Gp.
18
Material Balance Equation
• total formation volume factor Bt
where Bt= Bo + (Rsi-Rs)Bg
•
m=
GBgi
NBoi
• Gp = NpRp
Substituting these factors the general material balance equation because:
N=
N p ( Bt + ( Rp − Rs ) Bg ) − (We − Wp )
Bt − Bti + (c f + cw Swc )∆ p Bti / (1 − Swc ) + mBti
( Bg − Bgi )
Bgi
(21)
6 MODIFICATIONS TO THE GENERAL EQUATION
All the terms of the general equation as just presented may not be significant all the
time.
For example above the bubble point a number of the parameters will be zero.
Above the bubble point the solution gas - oil ratio is constant and therefore
Gp-NpRs = 0 since only solution gas will be produced above the bubble point.
If we are operating above the bubble point then there will for the majority of reservoirs
be no gas cap, and therefore the gas in place term, G or m, will also be zero. (Some
reservoirs with a compositional gradient can have a gas cap and also at the lower
part of the formation a different com position with undersaturated fluid.) The term
in the denominator, Rsi - Rs, will also be zero.
The equation therefore above the bubble point reduces to a simple equation, associated with compressibility terms of the oil ( the formation volume factors ) and those
of the connate water and pore space. We will consider this later when we examine
some applications of the equation.
When the reservoir is below the bubble point then, the terms described above being
zero in the undersaturated condition, have significance and are not zero. However the
term in relation to the compressibility of the connate water and pore space although
contributing to the overall balance is very small when compared to that from free gas
compressibility. It could be argued that the absolute changes in the water and pore
compressibility term is less than the errors associated with the free gas terms, when
the system is below the bubble point
The equation also includes a term with respect to water drive, We. Other terms can also
be added to include artificial drive, for example gas injection, Gi and water injection
Wi. Clearly when any of these three drive supports, natural or otherwise, are not active then clearly they are zero. Although there may not be any water drive, We or Wi
, there still could be water production as a result of mobilisation of connate water.
Institute of Petroleum Engineering, Heriot-Watt University
19
We have developed the equation by considering the impact of the various elements
involved in fluid production. An alternative derivation is based on the perspective that
the equation is an expression of the total compressibility of the reservoir system.
GBgi
m=
NBoi
7. DERIVATION
OF THE MATERIAL BALANCE EQUATION BY
EQUATING SUBSURFACE VOLUME OF PRODUCED FLUIDS TO
Rs )Bg ) − PLUS
(We − W
EXPANSION OFNORIGINAL
PORE
VOLUME REDUCp (Bt + (Rp −FLUIDS
p)
N=
(Bg − Bgi )
TION
Bt − Bti + (c f + c w Swc )∆ p B ti / (1− Swc ) + mBti
Bgi
The simple definition of compressibility, c , is
c =−
1 dv
v dP
The individual expressions of the compressibility of the oil, gas, water and rocks are,
co, cg, cw and cf. These compressibilities depend on the nature of the fluids and rocks
and between them have significant variations Gas is the most compressible down to
water and rock depending on its composition and nature.
Cg = 500x10-6
Co = 10x10-6 Cw = 3x10-6 Cf = 1x10-6 to
to
to
to
1500x10-6 psi-1
20x10-6 psi-1
5x10-6 psi-1
25x10-6 psi-1
It is better to consider rock compressibilities in the context of the reservoir behaviour
in terms of pore volume compressibility, since it is the pore volume which is available for containing fluids. If the compressibility is in terms of the change in pore
volume per unit bulk volume, dividing it by the porosity changes the numeric value
when expressed as pore volume per pore volume. For example, at a porosity of 20%,
a compressibility of 1x10-6 pore volume per bulk volume is 5x10-6 pore volume per
pore volume per psi.
Even these pore compressibilities are small but their significance should not be
neglected particularly above the bubble point.
This compressibility perspective as an alternative way to derive the material balance
equation is based on the following philosophy. Visualise again a reservoir at initial
pressure pi containing oil plus dissolved gas, a primary gas cap and connate water
(Figure 15)
20
Material Balance Equation
FLUID VOLUMES
Prim. Gas cap
Initial volumes
at pressure Pi
Oil + originally
dissolved gas
Fluid expansions
down to pressure P
Connate water
Water influx We
PORE VOLUMES
New pore volume
at pressure p
Reduction in total
pore volume down to
pressure P
Total pore volume
at pressure Pi
Figure 13 Material Balance By Equating Sub Surface Expansion To Fluid Production
Suppose that the pressure were reduced from pito p. Obviously this could not be
done without production, but let us see what effect such pressure reduction would
have. The volumes of the three phases will expand as shown in Figure 15. There
may also have been a water influx We. Also, the total available pore volume will
become smaller, through pore compressibility effects just described. Clearly, the new
fluid volumes, plus the water influx, do not fit any longer in the available pore space,
there is a shortage of space equivalent to the sum of the shaded areas in Figures 15.
Consequently, an equal volume of fluids can no longer be present in the formation,
and must therefore be the same as the reservoir volume at pressure p of the produced
fluids. With this in mind, we can state the material balance as follows:
Reservoir volume at pressure p of the produced fluids = expansion of primary gas
cap + expansion of oil plus originally dissolved gas + expansion of connate water +
water influx + reduction of total pore volume.
Institute of Petroleum Engineering, Heriot-Watt University
21
Or put in another way by Dake15
Underground withdrawal = expansion of the system + cumulative water influx.
The individual components in this equation can be quantified as follows:
Reservoir Volume at p of Produced Fluids
Production consists of:
Np STB of oil, Gp or Np RpSCF of gas and Wp STB of water.
The subsurface volume at pressure p of oil is Np Bo res. bbl. including an amount of
dissolved gas equivalent to Np Rs SCF.
We have produced Np Rp SCF, therefore the equivalent of Np Rp - Np Rs SCF must exist
as free gas in the reservoir at pressure p. Its subsurface volume is Np (Rp-Rs)Bg res.
bbl. Hence the reservoir volume at pressure of produced hydrocarbons ,HCPV is:
produced HCPV (p)
= Np Bo + Np (Rp-Rs)Bg (res.bbl.) = Np (Bo + (Rp-Rs) Bg) (res.bbl.)
(22)
Reservoir volume of produced water is about equal to WpBw barrels, hence:
Res. volume at p of produced fluids
=Np(Bo+(Rp-Rs)Bg) + Wp (res. bbl.)
(23)
Expansion oil + originally dissolved gas
The original volume of the oil is:NBoi, when the pressure is reduced the oil shrinks
and gas is liberated.
Volume change of oil is N(Bo-Boi).
During this process gas comes out of solution whose reservoir volume is:
N(Rsi-Rs)Bg.
Therefore the total change in volume is:
N[(Bo - Boi) + (Rsi- Rs)Bg]
Expansion of primary gas cap
The original hydrocarbon pore volume of the gas cap is:
mNBoi
The surface volume is;
mN
22
Boi
scf
Bgi
(24)
Material Balance Equation
At a lower pressure,p, the reservoir volume is:
mNBoi
Bg
Bgi
The expansion of the gas cap is therefore:
B

mNBoi  g − 1
 Bgi  (26)
Expansion of connate water
Connate water has a low compressibility but is significant in undersaturated conditions
The compressibility of the water is:
cw =
1 dVw
Vw dP
The expansion of the connate water with pressure decline is therefore:
dVw = cwVw∆p
Vw is the total volume of the water. This is a proportion Sw of the total pore volume.
The total pore volume is that associated with the oil and a gas cap. The pore volume
of the oil at a saturation of (1-Swc) or So is: NBoi, the pore volume including the water
is NBoi/(1-Swc).
For the gas cap, the hydrocarbon pore volume is mNBoi, and including water is
mNBoi/(1-Swc) .
The total pore volume associated with the gas cap and oil, including connate water,
is:
(1+m)NBoi.
The pore volume of connate water is:
(1 + m) NBoi Swc
(1 − Swc )
The expansion of the connate water is therefore:
(1 + m) NBti Swc cw ∆p
(1 − Swc )
Institute of Petroleum Engineering, Heriot-Watt University
(27)
23
Pore Volume Changes
The impact of pressure reduction on the pore volume is to reduce volume available
for hydrocarbons and therefore can be treated as an expansion term alongside the
expansion terms associated with the oil, gas and connate water.
The change in volume of the pores associated with the total pore volume is therefore:
(1 + m) NBti Swc cw ∆p
(1 − Swc )
(28)
Water influx
If there is an aquifer, then as pressure is reduced water influxes into the reservoir
volume. This water influx is We (res.bbl.)
We can now add all these expansion terms and make then equivalent to the reservoir
volume of produced fluids. This gives the full material balance equation.
reservoir volume of produced fluids
[
]
[
N p Bo + ( Rp − Rs ) Bg + Wp Bw = N ( Bo − Boi ) + ( Rsi − Rs ) Bg
(22)
]
Expansion of oil and dissolved gas
(24)
expansion of as cap
(26)
B
 (1 + m) NBoi (cw Sw + c f )∆p
+ mNBoi  g − 1 +
+ We
(1 − Swc )
 Bgi 
pore volume reduction
(27 + 28)
This simplifies to:
[
]
N p ( Bo + ( Rp − Rs ) Bg =
 ( B − Boi ) + ( Rsi − Rs ) Bg 
 Bg

 c S + cf  
NBoi  o
− 1 + (1 + m) w wc
 + m
 ∆p
Boi

 1 − Swc  
 Bgi 

+(We − Wp Bw )
Injection terms
(28)
If there is water injection Wi and or gas injection Gi these can also be added, to the
equation either as added to the expansion terms or subtracted from the production
terms.
24
Material Balance Equation
8 ASSUMPTIONS IN MATERIAL BALANCE EQUATION
The MB equation has some basic assumptions and limitations which can cause some
erros when applied to some reservoirs.
Pressure
The MB equation is a tank model treating the reservoir as a large tank at which the
pressure is constant throughout the reservoir at a particular time. It clearly ignores
pressure changes which may arise across the reservoir. In the radial flow section it
was clear that there are large pressure variations around the producing and injection
wells.
In order to apply the equation at a particular time an average pressure has to be selected being representative of the reservoir pressure at the particular time. All fluid
properties are evaluated at this pressure. In the next chapter we will discuss this
topic further.
Temperature
Changes in a reservoir generally take place at isothermal, constant temperature, conditions, unless major external temperatures are imposed thorough for example thermal
recovery processes and in some cases large cold water injection schemes.
Production Rate
When things happen is not part of the MB equation as there is no term present including time, for this permeability would be required. Rate sensitivity is therefore
not part of the equation and for those situations, for example in water drive, which
are dependant on rate of production the material balance equation requires the application of other equations.
Representative PVT data
The PVT measurements should be made in an attempt to reflect the behaviour in the
reservoir. Although this may not be totally possible conditions as near to the real
situation are used, for example in the differential test to reflect below bubble point
conditions.
Good production data
It is important in the application of the MB equation to have reliable production data
not only oil and gas but also water.
9. SIGNIFICANCE AND USAGE OF THE MATERIAL BALANCE
EQUATION
The material balance is roughly a relation between four quantities:
•oil and gas in place (N, m or G)
•production (Np, Rp, Wp)
•water influx (We)
•average reservoir pressure (pressure dependent PVT parameters and: p in pore/
water compressibility term).
Institute of Petroleum Engineering, Heriot-Watt University
25
This means that if three of these quantities are known, the fourth can be calculated.
Some examples illustrate this:
• If production and pressure date are known as a function of time, and oil and gas
in place is available from a volumetric estimate, the water influx We can be
determined as function of time. Its magnitude has a direct bearing on secondary
recovery plans.
• If there is no evidence of a natural water drive (We=0) the oil in place can
be calculated from production and pressure data. This may have an influence
on the geological interpretation (volumetric estimate) and thus on the further
development of the reservoir.
• For a known oil in place, the pressure at future dates can be calculated for
a postulated production plan (making some assumptions regarding the future
water influx). The result of this calculation may help in:
(a) deciding whether or when artificial lift facilities will be needed.
(b) estimating the reserve of the reservoir down to a certain abandonment pressure, also as function of the cumulative gas oil ratio Rp.
Dake3 has also examined the status of the various parameters of the equation with
respect to the application of the equation. He divides the parameters into should be
known and potential unknown.
Should be known
Potential unknown
Np
N
Rp
We
Wp
p
cw
Bo, Bg, Rs
Swc
m
Bw
cf
From this list there appears to be 6 knowns and 8 unknowns, demonstrating the
challenge facing reservoir engineering in needing sufficient independent equations
to solve to determine the number of unknowns. As Dake points out the situation in
reservoir simulation is even worse with more unknowns of reservoir geometry and
description in terms of porosity, and a variety of relative permeabilities.
In examining the ‘knowns’, he points out that although Np and Rp are generally the
best known, in old and remote fields good records may be such that oil, gas and water
production figures may not be so readily available. He points out that petrophysical
evaluation is always correct. So for example the connate water saturation Swc is obtained
by averaging its values over all intervals and wells associated with the analysis.
In relation to the unknowns, the material balance, once production and pressure
information is available, provides a useful route to upgrading the original estimate
of in place, STOIIP, N, which has previously been estimated from a combination of
26
Material Balance Equation
petrophysical related information. The material balance generated result provides a
more effective value since it would not include volumes in undrained or low permeability areas of the reservoir.
Waterdrive as was discussed in the drive mechanism chapter is a very effective drive
mechanism. In reservoir development it provides a major challenge in predicting its
role. To predict the influx of water from an aquifer requires a good characterisation
of the aquifer, its geometry and the important flow related properties. To determine
such for what is compared to the associated oil reservoir a very large system is very
costly and is difficult to justify, for something which only produces water! The nature
of waterdrive is best determined when its impact on actual reservoir performance
is observed. Clearly if water underlays the hydrocarbon formation as a bottom
water drive system, then the advancing water oil contact can be logged in the well.
However if edge water drive is occurring then actual well observations may not be
possible. Material balance provides an opportunity to determine the support from
water drive, but translating this information into specific aquifer characteristics is not
straightforward.
The size of gas cap although more accessible is not always easy to determine, since
it may be preferred for development reasons during drilling to drill through the gas
cap.
Up until recently, water and rock compressibility terms of cw and cf have largely been
assumed to be of little importance and their value if not readily available obtained
from text book type sources. Such assumptions can be very costly particularly for
those fields where compaction drive is very significant.
The material balance equation, a zero dimensional model, or tank model, requires
an average pressure and this average pressure is reflected implicitly in relation to
PVT parameters and explicitly in relation to compressibility of water and rock. This
average pressure determination may be obtained from a range of pressures from wells
within the drainage area. We will discuss this in the next chapter.
The material balance is also a backbone in all mathematical reservoir simulators,
where pressures in individual grid blocks are calculated (apart from production data)
on the basis of influxes from or effluxes to adjacent grid blocks. Over recent years
there has developed an perception by some that the ‘simple material balance’ approach
has been superseded by the more comprehensive reservoir numerical simulation, with
its potential of analysis at small dimension levels compared to the full field tank size
of the MB equation . Until his recent death, Dake and others have recognised the
value of the MB in 'feeling' the reservoir and also providing useful input to the many
uncertainties associated with implementing a full reservoir simulation study.
10. SOURCES OF DATA TO BE USED IN THE MATERIAL BALANCE
A range of sources provide the key data for the application of the MB equation. These
sources are also the source for other simulation tools.
Institute of Petroleum Engineering, Heriot-Watt University
27
PVT Data
From PVT reports of individual wells. Averaging and correcting PVT data prior to
use in and M.B. may also be required
Production Data
From well and reservoir records (data banks) or the subject of calculation.
Oil and Gas in Place
From volumetric estimates or subject of the calculation.
Connate Water Saturation
Sw : from petrophysics
Water Compressibility
cw : at oilfield temperatures and pressures: Should be determined.
4 to 5 x 10-5 atm-1 = about 3 x 10-6 psi-1
Pore Compressibility
In the past has often been assumed from texts. Should be measured.
Reservoir Pressures
From pressure surveys in the field, or subject of the calculation. In the next chapter
we will see how an average pressure can be obtained from a reservoir where there
are different drainage zones.
Water Influx
The subject of water influx, We is covered in a subsequent chapter.
11. LIMITATIONS OF THE MATERIAL BALANCE
A material balance is a zero dimensional mathematical model, in which fluid properties and pressures are averaged over the entire reservoir. Variations in initial fluid
properties, for instance, a change in bubble point either laterally or as function of
depth, as a result of compositional variations, cannot be handled adequately. The
degree to which the results of an M.B. calculation are invalidated depends on the
magnitude of such variations.
With the MB, the average saturation distribution (So, Sg, Sw with So + Sg + Sw = 1)
can be calculated. However, no conclusion may be inferred how a calculated gas
saturation is distributed, i.e. whether this free gas is spread more or less evenly over
the entire reservoir, or whether the gas is concentrated in some localised areas.
The most significant aspect of MB is that it does not contain time as a parameter. This
means that although an M.B. calculation may tell us what will happen, it cannot say
when it will happen. We can, for instance, calculate that the average pressure of a
given reservoir will drop by 1973 psi for an oil and gas production of 88 MM STB
and 59 MMM SCF respectively, but the material balance will not tell us whether this
28
Material Balance Equation
situation will be achieved in 1, 10 or 100 years. By combining the material balance
results with other methods for example well productivity equations time information
can be added to the production / pressure predictions from MB methods.
12. CONCLUSION
Summarising: the material balance is an important and indispensable reservoir engineering tool. As with other reservoir engineering tools it has its limitations of
which the user should be aware. Viewed against a somewhat wider background the
following quotation from Muskat (Reservoir Engineering News Letter September
1947). is still applicable:
“The materials balance method is by no means a universal tool for estimating
reserves. In some cases it is excellent. In others it may be grossly misleading.
It is always instructive to try it, if only to find out that it does not work, and
why. It should be a part of the stock in trade of all reservoir engineers. It will
boomerang if applied blindly as a mystic hocus-pocus to evade the admission
of ignorance. The algebraic symbolism may impress the old timer and help
convince a Corporation Commission, but it will not fool the reservoir. Reservoirs
pay little heed to either wishful thinking or libellous misinterpretation.
Reservoirs always do what they ought to do. They continually unfold a past
which inevitably defies all man-made laws. To predict this past while it is
still the future is the business of the reservoir engineer. But whether the
engineer is clever or stupid, honest or dishonest, right or wrong, the reservoir
is always right.”
Solutions to Exercises
EXERCISE 1
A gas reservoir without water drive contains 500 million standard cubic feet of gas
at an original pressure of 3,000psia. How much gas has been produced when the
reservoir pressure has declined to 2,900 psia. Use Bgi and Bg for the initial and
2,900psia pressure as 0.0010 and 0.0011 bbl/scf.
SOLUTION 1
Gas Material Balance with no water drive
GBgi = ( G-Gp)Bg
(5x108x0.001)=(5x108 -Gp)x0.0011
Gp = 4.55E+07 scf
Institute of Petroleum Engineering, Heriot-Watt University
equation 1
29
REFERENCES
1. Schilthuis, R.J. Active Oil and Reservoir Energy, Trans AIME, 118:33-52,
1936.
2. Slider, H.C., Petroleum Reservoir Engineering Methods, Petroleum Publishing
Co. Tulsa, 1976.
3. Dake,L.P. The Practise of Reservoir Engineering. Elsevier Ams. 1994
4. Archer,J.S and Wall,C.G. Petroleum Engineering , Principles and Practise,
Graham&Trotman .Ldn 1986
5. Drake L.P Principles of Reservoir Engineering. Elsevier 1978
30
Material Balance Equation Application
CONTENTS
1. INTRODUCTION
2. LINEAR FORM OF MB EQUATION
2.1
Short Hand Version of MB Equation
2.2
No Water Drive and No Original Gas Cap
2.3
Gas Drive Reservoirs, No Water Drive and
Known Gas Cap
2.4
Gas Drive Reservoirs with No Water Drive,
N and G Are Unknown
3. DEPLETION DRIVE OR OTHER
4. GAS FIELD APLICATIONS OF THE MATERIAL
BALANCE EQUATION
5. MATERIAL BALANCE EQUATION APPLIED
TO OIL RESERVOIRS
5.1
Depletion Drive Reservoirs
5.1.1 Above The Bubble Point
5.1.2 Solution Gas Drive
5.1.2.1 Instantaneous Gas - Oil Ratio
5.1.2.2 Oil Saturation Equation
5.2
History Matching
5.3
Performance Prediction Methods For Solution Gas Drive Reservoir Reservoirs
5.3.1 Solution Gas Drive Characteristics
5.3.2 Performance Prediction - Schilthuis Method
5.3.3 Tarner's Methods5
5.3.4 Tracy's Form of Tarner Method
5.3.5 Muskat Method
5.4
Data For Solution Gas Drive Predictions
5.5
Gas Drive Reservoirs
5.6
Average Reservoir Pressure
6. RESERVOIR PERFORMANCE AS A FUNCTION
OF TIME
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•Given the MB equation be able to present it in a short hand form as a basis for
use in linear forms.
•Using the various linear forms with sketches illustrate the MB equation for use
for:
• Reservoir with no water drive or gas cap.
• No water drive but with known gas cap.
•Comment with the aid of sketches the impact of water drive on the application
of MB equation in linear and other forms.
•
Derive and use a simplified MB equation for application to an oil reservoir above the
bubble point, in terms of recovery, and oil, rock and water compressibility.
•Derive the instantaneous gas-oil ratio equation and use to explain the producing
GOR of a solution gas drive reservoir.
•Comment on how the instantaneous GOR equation can be used to history match
the ratio of the gas to oil relative permeabilities.
•Describe the procedure for the application of the Tarner or Tracy Tarner methods
in predicting solution gas drive performance.
•Be able to use both Tracy and Tracy Tarner methods for predicting future
performance given prerequisite equations and data.
•Comment how with well performance information a time line can be included
in material balance predictions.
Material Balance Equation Application
1. INTRODUCTION
In the previous chapter we developed the material balance, MB, equation identifying
the various elements. In this chapter we will examine the application of the equation
to different reservoir types and examine some techniques developed to predict
reservoir performance. There is no one universal solution to the MB equation and in
this chapter we will examine the equation and its application with respect to different
reservoir types.
In recent times the computing power behind other reservoir engineering tools like
numerical simulation, has cast a shadow of a lack of confidence in the old material
balance approach. Laurie Dake1 up until his sudden death in 1999, was a strong
proponent of the material balance equation and was sometimes interpreted as having a
negative perspective with respect to reservoir simulation. As demonstrated in his text,
"The Practise of Reservoir Engineering", such interpretation is far from the truth. To
quote Professor Dake in his defence of the use of the MB equation “It seems no longer
fashionable to apply the concept of material balance to oilfields, the belief that it is
now superseded by the application of modern numerical simulation. Acceptance
of this idea has been a tragedy and has robbed engineers of their most powerful
tool for investigating reservoirs and understanding their performance rather than
imposing their wills upon them, as is often the case when applying numerical
simulation directly in history matching. ........There should be no competition
between material balance and simulation instead they must be supportive of one
another: the former defining the system which is then used as input to the model.
Material balance is excellent at history matching production performance but has
considerable disadvantages when it comes to prediction, which is the domain of
numerical simulation modelling.”2
One reason for perhaps a lack of appreciation of the equation might be the immediate
impression of complexity through its many terms. A significant step forward in the
equation which had been originally presented by Schilthuis2 in 1936 was by Odeh
and Havlena3, who in 1963 examined the equation in its various linear forms.
2. LINEAR FORM OF MB EQUATION
2.1 Short Hand Version of MB Equation
The material balance equation in itself is not a difficult concept to understand, the
difficulties lie in the application of the equation to real reservoir problems. The problem
which generally faces the engineer is the inadequate understanding of the reservoir
preventing knowing the extent of the driving mechanism or mechanisms.
In 1963 Havlena and Odeh3 presented a paper aimed at reducing the above problem.
Their method consists of re-arranging the material balance equation to result in an
equation of a straight line. The method requires the plotting of a variable group
versus another variable group with the variable group selection depending on the
drive mechanism.
Institute of Petroleum Engineering, Heriot-Watt University
Their technique is useful in that if a linear relationship does not exist for a particular
interpretation of the reservoir, then this deviation from linearity suggests that the reservoir
itself is not performing as anticipated and other mechanisms are involved.
Once linearity has been achieved, based on matching pressure and production data
then a mathematical model has been produced. This technique is referred to as history
matching, and the application of the model to the future enables predictions of the
reservoir’s future performance to be made.
The material balance equation can be written in the following form:
[
]
[
N p Bo + ( Rp − Rs ) Bg + Wp Bw = N ( Bo − Boi ) + ( Rsi − Rs ) Bg
]
B
 (1 + m) NBoi (cw Sw + c f )∆p
+ mNBoi  g − 1 +
+ We
(1 − Swc )
 Bgi 
(1)
In some instances water formation volume factors are not included, i.e. Wp, We and
WiBw
Havlena and Odeh simplified the equation into a short hand form:
F = NEo + NmEg + NEfw +We
(2)
The left hand side of equation 2 represents the production terms in reservoir volumes
and are denoted by F, i.e.
F = Np[Bo + Bg (Rp - Rs)] + Wp...bbl
(3)
The right hand side includes:
(i) the expansion of the oil and its originally dissolved gas, Eo, where:
Eo = Bo - Boi + (Rsi -Rs) Bg ....bbl/STB
(4)
(ii) the expansion of the pores and connate water Efw where:
E fw = (1 + m)
Boi
(c f + Sw cw )∆p.....bbl / STB
1 − Sw
(5)
(iii) expansion of the free gas Eg where:
B

Eg = Boi  g − 1 ....bbl / STB
 Bgi 
(6)
Material Balance Equation Application
With the above terms the material balance equation can be written:
F = NEo + NmEg + NEfw + We
(7)
This equation as presented above neglects water or gas injection terms.
Using this equation as a basis, Havelena and Odeh manipulated the equation making
different assumptions to produce a linear function.
2.2 No Water Drive and No Original Gas Cap
In this condition We and Nm are zero and the equation becomes:
F = NEo
(8)
i.e. a plot of F vs Eo should produce a straight line through the origin Figure 1. This
is the simplest relation and is just a plot of observed production against determined
PVT parameters. The slope of the line gives the oil in place N.
F
N
O
Eo
Figure 1 F vs. Eo No Water Drive and No Gas Cap.
2.3 Gas Drive Reservoirs, No Water Drive and Known Gas Cap
Although We is zero, the gas cap has a volume as given by m, and the equation 7
becomes:
F = N(Eo + mEg)
(9)
A plot of F vs (Eo + mEg) should produce a straight line through the origin with a
slope N. Figure 2. If m is not known then by making assumptions for m a number
of plots can be generated with the linear slope being the correct value for m.
Institute of Petroleum Engineering, Heriot-Watt University
F
m
o
to
l
al
sm
N
m too
O
large
Eo +mEg
Figure 2 F vs (Eo +mEg) Gas Drive, With Known Gas Cap, But No Water Drive.
2.4 Gas Drive Reservoirs with No Water Drive, N and G Are Unknown
If there is uncertainty in both the size of the oil and gas accumulation then Havlena
and Odeh suggest the following form of the material balance equation, by dividing
both sides by Eo.
E
F
= N +G g
Eo
Eo (10)
where:
G = Nm
Boi
Bgi
A plot of F/Eo vs Eg/Eo should be linear with an intercept of N and a slope of mN.
Figure 3.
F/Eo
mN
N
F = N + G Eg
Eo
Eo
Eg /Eo
Figure 3 F/Eo vs. Eg/Eo
Material Balance Equation Application
2.5 Water Drive Systems
Water influx is discussed in greater detail in a later chapter and we will examine this
linearisation of the MB equation in that context then.
3 DEPLETION DRIVE OR OTHER
Determining the drive energies responsible for production is important in understanding
and predicting the future behaviour of a reservoir. The material balance equation can
be used in this short hand form to get an indication of whether a field is depleting
volumetrically or there is other energy supporting the system for example, water drive
or gas cap expansion. If we consider a reservoir which does not have an aquifer but
there is the possibility of a supporting aquifer. In this case the short hand version
of the MB equation is;
F = N(Eo + Efw) + We.....bbl
(11)
If both sides are divided by
Eo +Efw
Then the following equation results:
F
We
=N+
....STB
Eo + E fw
Eo + E fw
(12)
F
Eo + Efw
Dake1 points out that the right hand side has two unknowns, N and We, and suggest
that the MB in this form is a powerful tool in assessing whether there is a supporting
aquifer or not. He suggest plotting F/(Eo+Efw) vs Np, or time or pressure drop,
∆p. The plot will take different shapes dependant on the energy support . Figure 4
illustrates this.
+
x C
x x
x
x x
x x
x
x
x
+ + + + +
x + + +
+ + B
x+
A
N
O
Np, ∆p or time
Figure 4 Determining Drive Mechanism
The examples above in figure 4 give various scenarios as a result of plotting regular
production figures. Dake1 points out that for Curve A, the horizontal line, indicates
Institute of Petroleum Engineering, Heriot-Watt University
that the left and right hand side of the equation are constant, ie. We =0. and the pore
compressibility is constant. This is a solely depletion drive where the energy comes
from the compressibility of the oil and its originally dissolved gas. The intercept on
the y axis is also the constant term, N, the oil in place, the STOIIP. The other plots
indicate that pressure support is also coming from elsewhere, water drive, or abnormal
compaction or a combination of both. Perhaps in C there is a water drive from an
infinite aquifer, where the aquifer boundary has still to feel the pressure. Curve B
might be for a finite aquifer, where later in production, there is less support from the
aquifer. Another feature of this presentation is that back extrapolation of the B and
C curves also gives the in place oil volume, N.
4 GAS FIELD APPLICATIONS OF THE MATERIAL BALANCE
EQUATION
In the previous chapter we introduced the topic of the linear form of the MB equation for
a gas reservoir without water drive, the p/z plot. Craft and Hawkins4 in their text gave
a warning of the application of this approach both with respect to neglecting another
possible energy support such as water drive, and using Gp vs. p as against p/z.
In figure 5 below, the plots of Gp vs p or p/z illustrate the different values of gas in
place which can result from the linear extrapolation of the early production pressure
values. Using the Gp vs.p results in an under estimate of the gas in place whereas
neglecting water drive could result in a significant over estimate of gas in place.
5000
4000
Pressure or P/Z
Water drive
2000
Volumetric
Gp vs p
1000
0
Initial gas in place
3000
Volumetric
Gp vs p/z
Wrong
extrapolation
1
2
3
4
5
Cumulative Production, MMMSCF
6
7
Figure 5 Comparison of the Gp vs p/z and p for Volumetric and Water Drive Gas Reservoirs. (Craft and Hawkins4)
The material balance equation lends itself to gas field application in part because of
a more uniform pressure across the reservoir, because of the fluid diffusivity, k/φµc.
For gases because of low viscosity, this gives a diffusivity of the order of five times
that of oil. Because of such rapid equilibrium of pressure it is easy to neglect pressure
support from elsewhere. We will examine both the approach of Havlena and Odeh
and the long established p/z approach.
Material Balance Equation Application
Using the approach of Havlena and Odeh to gas reservoirs provides a history matching
approach to get a good estimate of gas initially in place to compare with previous
volumetric calculations and also determine if the reservoir is a volumetric depletion
reservoir or it is also supported with water drive.
Using the approach of fluid production being equal to the expansion of insitu gas +
water and pore expansion + water influx gives the following equation.
Fluid production
influx
= gas expansion + water expansion & pore compaction +water
Gp Bg + Wp Bw = G( Bg − Bgi ) + GBgi
(c S
w wc
+ cf
1 − Swc
) ∆p + W
e
(13)
Using the short hand version of Havlena and Odeh results in
F = Gp Bg + Wp Bw ...res.cu. ft
(
)
Eg = Bg − Bgi .....rcf / scf
E fw = Bgi
(c S
w wc
+ cf
1 − Swc
) ∆p.....rcf / scf
(14)
The short hand gas material balance equation becomes
F = G (Eg + Efw) + We
When dealing with gas reservoirs the water and pore compressibility terms can
generally be neglected because of the large gas compressibility.
The equation results in
F = GEg + We (15)
Dividing both sides by Eg gives
F
W
=G+ e
Eg
Eg (16)
This provides a useful form to examine the production figures of a gas reservoir.
The figure 6 below illustrates the possible resulting plots from plotting F/Eg vs Gp,
time, or ∆p.
Institute of Petroleum Engineering, Heriot-Watt University
Strong water drive
F/Eg
Moderate water drive
Volumetric depletion
G = initial gas in place
Gp
Figure 6 Material Balance Application to Determine G and Drive Mechanism
The three plots demonstrate how that if there is pressure support from an aquifer the
curves deviate from the horizontal line applicable to the volumetric depletion. The
intercept also provides a figure for gas initially in place.
This can be also be a useful tool in gas field development since if water drive by
this means is detected, then it could be some time before there is evidence of water
production. The advancing water will only reveal itself when the gas-water contact
reaches the lower limit of the gas production wells which are generally set high in
the structure. Another consideration is the mobility ratio for water displacing gas.
The mobility ratio for water displacing is
M=
krw ′ krg ′
/
µw µg
Where:
krw′ and krg′ are the end point relative permeabilities for water and gas.
M can be as low as 0.10 for this situation which interprets into the gas moving 100times
faster than the water. Clearly as the water table advances it will trap gas, in the form
of the residual gas saturation, Sgr. The higher the pressure of the reservoir the larger
the amount of gas which will be trapped. Dake1 developed an equation to determine
the volume of gas trapped behind an advancing water drive.
Gas initially in place =G = Vφ(1-Swc)/Bg (17)
The pore volume, PV, of the reservoir, Vφ can be expressed in terms of the gas in
place.
PV = GBgi
1
1 − Swc the movable gas volume, MGV, by water flooding is;
10
(18)
Material Balance Equation Application
MGV = PV (1 − Sgr − Swc ) = GBgi
(1 − Sgr − Swc )
1 − Swc
(19)
After an influx of WeBw into the gas reservoir, the proportion of this movable gas
volume swept by the water is:
α = We Bw / GBgi
(1 − S
gr
− Swc
1 − Swc
)
(20)
The volume of trapped gas within this swept proportion is therefore:
α ( PV )Sgr / Bg = α
GBgi Sgr
S
p/z
= αG gr
Bg 1 − Swc
1 − Swc pi / zi (21)
This equation can be used to determine the amount of gas that is being ‘lost’ to
production through an advancing water drive. This can be reduced by depleting the
gas reservoir at a higher rate taking advantage of the higher diffusivity of the gas
relative to that of the water.
The p/z approach is long established in gas reservoir engineering as a method to
determine gas in place and recovery at field abandonment.
Gas produced from Reservoir = Gas initially in reservoir - Gas remaining in reservoir
(scf)
c + cwc Swc


Gp = G −  GBgi − GBg f
∆p − We Bw  / Bg
1 − Swc


(22)
The above does not include the water production term, Wp. We is considered in the
above to be the net water influx.
As stated before the compressibility terms for water and pore are very small and can
be neglected which reduces the equation to:
Gp
B  WB 
= 1 − gi 1 − e w 
G
Bg 
GBgi 
(23)
From the equation of state and assuming constant temperature the gas formation
factors can be replaced with z/p, and the equation becomes:
 Gp 
1 − 

p pi
G
=
z zi  We Bw / Bgi 
1 −


 G
Institute of Petroleum Engineering, Heriot-Watt University
(24)
11
The term WeBw/GBgi is the proportion of the gas in place volume invaded by water,
and as can be seen in the equation the higher this proportion the higher the pressure
and vice versa to the extent that with depletion drive where there is no water drive
or compaction drive , the equation simplifies to:
p pi  Gp 
= 1 − 
z zi 
G  (25)
This is the well known linear equation of p/z vs. Gp which when plotted enables
the gas in place, G to be obtained when p/z =0. (Chapter 5.) If there is any other
pressure support the curve will deviate from linear as demonstrated in figure 7 below
by Dake1.
pi/zi
Material balance
at abandonment
Strong waterdrive
Depletion
p/z
Moderate waterdrive
Abandonment pressure
Surface compression
0
Gp
Gp=6
Figure 7 P/z vs Gp Plots For Depletion and Water Drive (Dake1)
The difficulty with the application of this approach is that in the early time periods,
the pressure support from the aquifer may not be felt and deciding what is the slope
of a linear line may take in points from pressures being supported by water drive,
leading to an over estimate of gas in place.
This danger has been known for some time, but the simplicity afforded by this p/z vs.
Gp plot can readily lead to erroneous interpretation. As pointed out by Dake1, warnings
by respected reservoir engineers of Craft & Hawkins 1959 ,Bruns & Fetkofitch 1965,
Agarwal, Al-Hussainy & Raimey1965, Dake 1978 were taken up by Cason in 1989 ,
when he said, “The theory showing that depletion drive gas reservoirs will exhibit
a straight p/z plot has been developed but the corollary; that a straight line p/z plot
proves the existence of depletion drive has not been proven”
As indicated above the high diffusivity of the gas in contact with an aquifer is such
that if gas is withdrawn at a rate greater than water can encroach into the gas reservoir
then the pressure will decline faster than if the gas production rate is slower enabling
the water drive to replace gas production . This rate effect also distorts the slope of
12
Material Balance Equation Application
the p/z plot for fields supported by water drive. Varying production rate is a common
characteristic in gas field management, where higher gas production in winter periods
will be followed by lower summer production. Clearly all these contributing factors
can offset other phenomena in p/z plots and lead to poor interpretation. The rate
effect is illustrated in figure 8 in the distortion generated with high winter and low
summer production.
low summer rate
p/z
high winter rate
0
Gp
Figure 8 Non Linear Impact of Gas Rate Production For Waterdrive Gas Fields.
Dake1 has also demonstrated how the p/z plot can be used to determine abandonment
conditions. When water drive reaches water break through, than gas will be remaining
in two forms. In the swept portion at a residual gas saturation and in portions bypassed by the water at the original saturation (1-Swc).
At abandonment gas production Gp = G -Trapped Gas - By-passed gas.
S


Gp = G − αGBgi gr + (1 − α )GBgi  / Bgab
1 − Swc


(26)
α is the volumetric sweep efficiency at the abandonment, pressure Bgab, gas formation
volume factor at abandonment.
When this equation is expressed in terms of p/z it becomes
 Gp 
1 − 

p
pi
G
=
zab zi  Sgr
1−α 
α
+
α  1 − Swc
(27)
This is plotted on the p/z vs Gp plot of figure 7. It indicates the maximum gas recovery
points for the strong water drive and the moderate water drive. Dake explains further
application of this in his text .
Institute of Petroleum Engineering, Heriot-Watt University
13
5 MATERIAL BALANCE EQUATION APPLIED TO OIL RESERVOIRS
5.1 Depletion Drive Reservoirs
We will now examine the application of the material balance to oil reservoirs and
present some methods using the material balance equation to predict reservoir
performance.
As was discussed in the section on drive mechanisms, a depletion drive reservoir has
two phases of depletion. The first stage above the saturation pressure the undersaturated
state, and that below the saturation pressure from where the drive mechanism gains
its name, solution gas drive.
5.1.1 Above the Bubble Point
Above the bubble point, in the undersaturated state, the system is all liquid and
without any other drive mechanism therefore the fluid production is purely a result
of the total compressibility of the system.
Although it appears complex, the material balance the equation is a sophisticated
version of the compressibility definition:
dV = C x V x ∆p
Production = Expansion of reservoir fluids.
If we take the full MB equation developed in the previous chapter then the equation
simplifies above the bubble point to:
 ( B − Boi ) (cw Swc + c+ 
N p Bo = NBoi  o
+
∆p
1 − Swc
 Boi
 (28)
• no gas cap
• aquifer small in volume We =Wp =0
• Rs=Rsi=Rp - all gas at surface dissolved in oil in reservoir.
co =
Oil compressibility is
( Bo − Boi )
Boi ∆p (29)
Replacing this for the oil term gives:
(

cw Swc + c f
N p Bo = NBoi co +
1 − Ssw

14
) ∆p

(30)
Material Balance Equation Application
Above the bubble point only oil and connate water exists , therefore:
So + Swc = 1
The MB equation for application above the bubble point becomes:
c S + c S + cf
N p Bo = NBoi  o o w wc
1 − Swc


 ∆p
 (31)
N p Bo = NBoi ce ∆p (32)
1
(co So + cw Swc + c f )
1 − Swc
(33)
or:
ce =
ce is the effective saturation weighted compressibility of the reservoir system.
The recovery at bubble point pressure is therefore:
N p Boi
=
(ce ∆p)
N
Bob
(34)
Clearly this fractional recovery expressed as Np/N can be determined from a
simple function of the change in oil formation volume factors, and water and pore
compressibilities for the conditions above the bubble point where
Np = cumulative oil produced
N = original oil in place
The equations above have neglected water production. This may not be the case as
a reduction in pressure could mobilise connate water. The equation in such a case
would be:
NpBo + WpBw = NBoiCe∆p
(35)
During this phase pressure declines very rapidly and the recovery is low. There are
some exceptions to this where the compressibilities of the rock are exceptionally large,
such that the rock compressibility provides the major energy source. Such is the case
in some Venezuelan fields and in the carbonate field of Ekofisk in the North Sea.
As production wells are brought on stream their productivity can put a time line to
the MB calculations of production vs .pressure. The time for the reservoir pressure
to drop to a safe position above the bubble point, then is a basis for determining the
time availability for water injection installations to arrest the pressure decline.
Institute of Petroleum Engineering, Heriot-Watt University
15
EXERCISE:
Solution Gas Drive - Undersaturated Oil Reservoir
Determine the fractional oil recovery, during depletion down to bubble point pressure, for the reservoir whose PVT parameters are listed in PVT Chapter 12. Assume a separator pressure of 300 psi. Other reservoir fluid properties are listed in
table below.
5.1.2 Solution Gas Drive
When the reservoir pressure drops below the bubble point, then solution gas drive is
effective, then the analysis is more complex as gas comes out of solution. Solution
gas drive is the most common drive mechanism in oil reservoirs but unfortunately
it is also very inefficient. It is often associated with other drive mechanisms for
example water drive and gas cap drive. In this next section however we will ignore
supplementary drive from these two sources. In order to use the material balance
equation to predict further production as a function of pressure decline, we need
to develop other independent equations, the instantaneous producing gas-oil ratio
equation and the saturation equation.
5.1.2.1 Instantaneous Gas - Oil Ratio
The instantaneous gas- oil ratio, R, is the ratio of gas production to oil production
at a particular point in the production time of the reservoir, that is at a particular
reservoir pressure.
The gas-oil ratio equation is based on the Darcy flow equation.
The instantaneous producing gas-oil ratio is:
R=
Gas producing rate, SCF / Day
Oil producing rate, STB / Day (36)
The gas production can come from gas in solution in the reservoir and from production
free gas in the reservoir which has come out of solution.
Free Gas =
qg
Bg
Solution gas = QoRs
where:
qg = free gas flow rate, res.bbls/day
Bg = gas formation-volume factor, bbls/SCF
Qo = oil flow rate, STB/day
Rs = gas solubility SCF/STB
Total gas production rate:
16
Material Balance Equation Application
Qg =
qg
+ Qo Rs
Bg
(37)
where:
Qg = total gas producing rate.
Oil producing rate is:
Qo =
qo
Bo (38)
where:
qo = reservoir oil flow rate, res.bbls/day
Combining equations 36,37 and 38 gives:
qg
+ Qo Rs
Bg
R=
qo / Bo (39)
since:
Qo =
qo
Bo qg
B
R = g + Rs
qo
Bo
(40)
(41)
qg and qo are reservoir flow rate where for a radial system in terms of Darcy’s law
gives:
qg =
2πkeg h∆p
µ g ln re / rw
and:
qo =
2πkeo h∆p
µo ln re / rw
Institute of Petroleum Engineering, Heriot-Watt University
17
Therefore in (41):
2πkeg h∆p
B µ lnr / k
R = g g e w + Rs
2πkeo h∆p
Bo µo lnre / kw
R=
Bo keg µ o
+ Rs
Bg keo µ g
(42)
This is the usual form of the instantaneous gas-oil ratio equation.
Instantaneous
GOR
When discussing solution gas drive (Chapter 10) we presented the typical shape for
the producing gas-oil ratio curve . The distinctive shape is again reproduced below
and shows the various phases.
Rsi
3
4
1
Pb
2
PNp
Figure 9 Instantaneous Gas - Oil Ratio R For Solution Gas Drive Reservoir.
Examination of this plot in the context of the instantaneous GOR equation is consistent
with its shape. Above the bubble point at 1, there is no free gas, therefore keg is zero,
and R=Rs=Rsi. Once gas comes out of solution from the bubble point, theoretically
there will be a short time when critical gas saturation has not been reached, keg is still
zero but R = Rs < Rsi (point 2). In reality it is unlikely that this reduction in GOR
is seen since the whole reservoir is not at uniform saturation. From 2 to 3 gas has
reached the critical gas saturation, and keg increases with increasing gas saturation,
while correspondingly keo decreases. Gas is very mobile compared to the oil and the
reservoir gives up its free gas while the oil moving at an increasingly slower rate
is depleted of its solution gas. The curve goes through a maximum. To understand
this negative slope is to appreciate that the gas formation factor is increasing with
decreasing pressure. At high pressures the change in Bg is small but at lower pressures
the change is larger and greater than the increasing values of keg and µg.
This instantaneous GOR should not be confused with the cumulative producing GOR,
Rp. or the solution GOR, Rs. The instantaneous GOR, R, is the ratio of the total oil and
gas production rates at a particular moment in time. The cumulative GOR, Rp is the
18
Material Balance Equation Application
ratio of the cumulative gas and cumulative oil produced up to a particular moment
in time. These two GOR's are related to by the following two equations.
Rp =
∫
RdN p
o
Rp =
(42)
Np
ΣRi ∆N pi
N p (43)
whereRi is the average instantaneous GOR over the period that ∆Npi of oil was
produced.
Another equation which is required in conjunction with the instantaneous gas-oil
ratio equation is the Oil Saturation Equation.
5.1.2.2. Oil Saturation Equation
This equation, the oil saturation equation, provides an average oil saturation for a
reservoir at any time . It is defined as:
So =
oil volume remaining
total pore volume
So =
( N − N p ) Bo
NBob / (1 − Swi )
(44)
where
So - oil saturation at any time
Swi = connate water saturation
N is the oil in place at the bubble point.
Np is the cumulative oil production below the bubble point.
This equation can be rearranged to
So = (1 −
Np Bo
)
(1− Swi )
N Bob
(45)
the oil saturation equation.
Equation 45 is important because the oil saturation as a function of pressure must
be known to determine the relative permeability associated with the instantaneous
GOR equation.
Equation 45 assumes that the oil saturation is uniform throughout the reservoir.
However due to vertical migration of free gas a secondary gas cap may be produced
and another equation may be required.
Institute of Petroleum Engineering, Heriot-Watt University
19
So ′ =
( N − N p ) Bo
( NBob − m©NBob ) / (1 − Swi )
So ′ =
( N − N p ) Bo (1 − Swi )
NBob (1 − m©)
N p  Bo

(1 − Swi )
1 −


N  Bob
So ′ =
1 − m©
(46)
where:
So = oil saturation with oil zone
m = ratio of secondary gas cap to original oil zone size
The big assumption in equation. 46 is that the oil saturation in the secondary gas
cap is zero.
Predicting the future performance of a reservoir is difficult not the least because there
are many uncertainties associated with the reservoir. For example we may not know
the drive mechanism responsible for fluid production which would have an impact
on the parameters included in the MB equation. We may have doubts in relation to
the quality of the laboratory data, such as relative permeability.
5.2 History Matching
A common tool in reservoir engineering is history matching, which operates on
the principle if your prediction model cannot be used to predict the past then its
application for the future will be in serious question. When past performance cannot
be calculated then either the input data is wrong or the model is incorrect or both
are not correct. It is likely that there are errors associated with the input data. The
instantaneous GOR equation can be used to provide a history matched set of relative
permeability data.
The instantaneous gas-oil ratio equation when rearranged in another form can be
used for generating relative permeability data from production data or validating
laboratory generated relative permeability data.
Keg
Bµ
= ( R − Rs ) o o
Keo
Bg µo (47)
One of the most difficulty aspects in predicting reservoir performance of solution
gas drive reservoirs is having relative permeabilities which represent the reservoir.
Laboratory values are specific to small samples of rock which clearly cannot be
representative of the whole field. There is even doubt on the laboratory procedures
used and the preparation procedures to obtain ‘reality’ of rocks and fluids. The above
equation provides where data is available a procedure to make use of past production
data to generate field relative permeabilities for use in performance predictions.
20
Material Balance Equation Application
Production data will provide data for R, and Np as a function of pressure. For each
pressure value, the PVT values would be available from the PVT report. The Np
value will enable the oil saturation to be calculated from the oil saturation equation,
and the corresponding gas/oil relative permeability ratio is calculated from the IGOR
equation (47).
5.3 Performance Prediction Methods For Solution Gas Drive Reservoir
Reservoirs
There are two phases to a solution gas drive reservoir, the period above the bubble
point where reservoir fluids remain single phase, with the gas remaining in solution. The other phase is that below the bubble point, where gas comes out of solution and
the productivity of the formation is influenced by the relative properties of the oil
and gas phases.
In predicting reservoir performance, these phases of production are treated separately.
The production and pressures from the initial pressure and pressure down to the
bubble point, and then the production and pressure profile from the oil in place at
the bubble point pressure.
5.3.1 Solution Gas Drive Characteristics.
Before looking at the various predictions it is worthwhile reviewing the characteristics of
solution gas drive reservoirs. When the subject was considered in the drive mechanism
chapter it was indicated that these reservoirs do not yield good characteristics from
an oil recovery perspective. They are characterised by
(1) Rapid pressure decline
(2) Water-free production
(3) Rapidly increasing gas-oil ratio
(4) Low ultimate recovery
There are a number of procedures which different authors have developed for predicting
performance, they are attributed to Schilthuis, the originator of the MB equation,
Tarner, Tracy& Tarner and Muskatt. The methods focus on the predictions below the
bubble because the predictions above the bubble point are straightforward.
In section 4.1.1. we derived the MB equation for application above the bubble point.
Such calculations need to done separately from those below the bubble point.
5.3.2 Performance Prediction - Schilthuis Method
Schilthuis rearranged the material balance equation to the following form:
Np =
[
]
N Bo − Bob + ( Rsi − Rs ) Bg + G( Bg − Bgi ) − Gp Bg
Bo − Bg Rs
(48)
All the PVT related properties are determined from the PVT report using the reservoir
pressure.
At the start of the prediction the values of the parameters at the bubble point are
known, N,G,Bob,Rsi and Bgi.
Institute of Petroleum Engineering, Heriot-Watt University
21
The gas production, Gp is obtained from the producing GOR and cumulative oil
production. This GOR however varies as the reservoir is depleted, so an incremental
procedure is used based on
j =n
Gp = ∑ Rj ∆N pj
j =1
(49)
Rj is the average producing GOR over the increment of production.
The GOR from the instantaneous gas-oil ratio equation is also a function of relative
permeability.
R=
Bo keg µo
+ Rs
Bg keo µ g
(42)
and the relative permeability is related to saturation which is related to cumulative
oil production through the saturation equation.
N  B

So = 1 − p  o (1 − Swi )

N  Bob
(45)
Schilthuis’s procedure is as follows:
1.
2.
3.
4.
Set a pressure below Pb- Bubble point pressure
Assume Gp and calculate Np from the MB equation above.
Calculate So from oil saturation equation above, using Np determined in 2.
Determine relative permeability ratio from laboratory or field generated
rel.perm vs So data.
5. Calculate Gp using instantaneous GOR eqn. and Gp equation above.
Ri and Ri+1 are the instantaneous GOR at the start end of the production interval.
Compare the calculation of Gp with predicted Gp and repeat steps, 2-5 if they do not
agree. When the values agree go to step 1 and set a lower pressure and continue with
the Np and Gp vs pressure decline prediction.
Tarner approach is not too different from the Schilthuis method .
5.3.3 Tarner’s method5
In Tarners5 method the material balance equation is arranged as follows in a form to
calculate Gp, where subscripts refer to the saturation condition.
N p Rp =
N ( Bo + ( Rsi − Rs ) Bg − Bob ) − N p ( Bo − Rs Bg )
Bg
(50)
NpRp = cumulative gas produced SCF.
.
The procedure is of a trial and error approach using the two independent equations,
instantaneous GOR and MB:
22
Material Balance Equation Application
1. Starting at the bubble point pressure.
2. Select a future pressure and assume a value of Np at that pressure. It may be convenient to express Np as a function of N. Solve the MB equation for NpRp
which is the cumulative gas production.
3. Using the assumed Np solve the oil saturation equation-equation 45 for So.
These enable keg/keo to be determined.
4. Calculate the instantaneous gas-oil ratio R from equation 42.
5. Calculate the gas produced during the pressure drop period, i.e.
( Ri + Ri +1 )
N p1
2
(51)
where Ri = instantaneous gas-oil ratio at start of period.
Ri+1 = instantaneous GOR at end of period
Np1 = cumulative oil produced at end of the period.
The assumption of the equation is that a plot of R vs Np is linear. Since it is unlikely
that such a relationship exists small pressure drops should be used. A figure of 50
psi. is suggested.
6. The total gas produced from the material balance equation and the instantaneous
GOR equation are compared, and the assumed value of Np adjusted and steps
2-6 repeated until the two values match to within an acceptable error. We then
go to the next step
(b) Step Two
1 A second pressure is selected and a new value of Np assumed Np2.
2 Solve MB for Np2. This is the cumulative gas production at the end of the second pressure.
G2 = N p 2 Rp 2 − N p1 Rp1 =
N ( Bo + ( Rsi − Rs ) Bg − Bob ) − N p 2 ( Bo − Rs Bg )
− N p1 Rp1
Bg
(52)
3 Calculate gas produced during 2nd step by removing from cumulative gas,
gas produced during step 1, G1
4 With the assumed Np2 a value of So is obtained from the saturation equation.
5 Instantaneous GOR calculated keg / keo above ratio determined.
Institute of Petroleum Engineering, Heriot-Watt University
23
6 Gas produced during second step:
(R i +1 + R i + 2 )
(N p2 − N p1 ) = G 2
2
(53)
where G2 is the total gas produced during step two.
7 If G2MB does not compare with G2GOR then a new value of Np2 is assumed, and
so on until convergence as before. By plotting Gmb and GGOR vs Np the point of
convergence can be determined.
The stepwise trial and error procedure is continued until the desired limit is
achieved.
Both of these procedures are not so well suited to computer application and Tracy
modified Tarners method to make it more suited to computer application.
5.3.4 Tracy's Form of Tarner Method
Tracy6 took the Schilhuis MB equation
N=
N p (Bo − R s Bg ) + G p Bg − (We − Wp )
Bo − Boi + (R si − R s )Bg + mBoi (Bg − Bgi ) / Bgi (54)
and suggested a shorthand version, where:
φn =
Bo − R s Bg
Bo − Boi + (R si − R s )Bg + mBoi (Bg − Bgi ) / Bgi φg =
Bg
Bo − Boi + (R si − R s )Bg + mBoi (Bg − Bgi ) / Bgi φw =
1
Bo − Boi + (R si − R s )Bg + mBoi (Bg − Bgi ) / Bgi (55)
(56)
(57)
For simplicity in presenting the equations if we assume no gas cap these become:
24
φn =
Bo − Rs Bg
Bo − Boi + ( Rsi − Rs ) Bg (58)
φg =
Bg
Bo − Boi + ( Rsi − Rs ) Bg (59)
φw =
1
Bo − Boi + ( Rsi − Rs ) Bg (60)
Material Balance Equation Application
These functions φn, φg and φw are only dependent on reservoir pressure and oil
properties, i.e. they can all be obtained from PVT data.
Using the above short-hand system the material balance equation can be written:
N = Npφn + Gpφg - (We-Wp) φw
(61)
If we assume no water encroachment or production:
N = Npφn + Gpφg
(62)
This is now a simplified form of the MB equation with the PVT related functions
conveniently grouped together.
Tracy considered two pressure conditions Pj and a lower pressure Pk and the oil
production ∆Np during this pressure interval.
Tracy differs from Tarner in estimating the producing GOR, Rk at the lower pressure
rather than the production ∆Np.
For the kth pressure:
N = Npkφnk + Gpkφgk
(63)
If N is set at 1.0 then the equation takes a fractional recovery form:
1.0 = Npkφnk + Gpkφgk
(64)
also:
1.0 = (Npj+∆Npk)φnk + (Gpj+∆Gpk) φgk
(65)
and:
1.0 = (Npj+∆Npk)φnk + (Gpj+R′avg x ∆Npk) φgk
(66)
where R′ avg is the estimated average gas-oil ratio between pressure Pj and Pk.
i.e.
Ravg
′ =
R j + R′ K
2
(67)
R′k can be estimated by extrapolating the plot of R versus pressure calculated at a
higher pressure.
Equation 66 can also be written in the form:
1.0 = Npjφnk+Gpjφgk+ ∆Npk(φnk+φgk R′avg)
(68)
solving for ∆Npk gives:
Institute of Petroleum Engineering, Heriot-Watt University
25
∆N pk =
1.0 − N pjφnk − Gpjφ gk
φnk + φ gk R′ avg
(69)
In Equation 69, the only unknown is R′avg. All the other parameters are uniquely
determined from the PVT data or have been calculated during previous pressure
steps.
Rk can also be estimated if the liquid saturation is known using the instantaneous
GOR equation, equation 42.
Rk =
Bo keg µo
+ Rs
Bg keo µ g
keg/keo is estimated from the liquid saturation since keg/keo=f(So)
So being obtained from the oil saturation equation:
N  B

So = 1 − p  o (1 − Sw )

N  Boi
Tracy observed that φn, φg and φw are smooth curved and above Pb they have infinite
slope. An example of these curves is given in figure 10.
0.10
.08
.06
100
80
60
.04
40
.02
20
.010
.008
.006
10
8
6
φg
.004
φn
4
2
.002
.001
0
5
10
15
20
25
Pressure, hundreds of psia
1
Figure 10 φg + φn Functions (Tracy6)
Using the above equations the step by step procedure of Tracy is as follows.
26
Material Balance Equation Application
Tracy’s Procedure
Set pressure step below Pb
(1) Estimate R´k
• above bubble point - Rs
• from extrapolation of previous trend
• from instantaneous GOR equation.
(2) Estimate R´avg (Equation 67)
(3) Determine functions φn and φg (Equation 55 and 56).
(4) Using equation 69 determine ∆Npk & Σ∆Np.
(5) Using Σ∆Np. Determine So from saturation equation (equation 45) and thereby
keg/keo from So vs keg/keo data.
(6) Calculate Rk from instantaneous GOR equation 42.
(7) Compare Rk with R´k. of step 1. If (Rk - R´k) is greater than tolerance, tolerance
limit set R´k = Rk and repeat steps 1 - 6.
(8) Estimate ∆Gp and Σ∆Gp. ∆Gp = ∆Np x Ravg.
(9) Estimate φg Gp + Np φn. This should be 1.00. If the error is greater than a preset
tolerance, calculations are repeated with an adjustment to Rk'.
(10) Set next pressure and repeat step 1-9.
5.3.5 Muskat Method
Morris Muskat, according to Dake1 did more in the 40’s and 50’s to evolve reservoir
engineering techniques in terms of well established physical principles and their
mathematical foundation. Muskat7 considered the oil remaining in the reservoir, Nr,
from which Np of oil had been produced. The remaining oil is:
Nr = N − N p =
Vp So
Bo (70)
Vp is the pore volume in rb (res bll). The change in this volume is;
dNr
1 dSo
S dBo
= Vp
− Vp 2o
dp
Bo dp
B o dp (71)
The total volume of dissolved and free gas is:
Gr = Vp
So Rs
V
+ (1 − So − Swc )
Bo
Bg Institute of Petroleum Engineering, Heriot-Watt University
(72)
27
The change in this volume with pressure is:
 S dR R dS
dGr
R S dB 1 − So − Swc dBg 
= V  o s + s o − s2 o o −

dp
B o dp
B2 g
dp   Bo dp Bo dp
(73)
The current producing GOR can be obtained by dividing eqn 73 by equation 71.
R=
∆G p ∆G r ∆G r / ∆p dG r / dp
=
=
=
∆N p ∆N r ∆N r / ∆p dN r / dp
So dR s R s dSo R sSo dBo
1 dSo 1 − So − Swc dBg
+
− 2
−
−
Bo dp Bo dp
B o dp Bg dp
B2g
dp
R=
1 So So dBo
−
Bo dp B2o dp
(75)
The producing GOR can also be obtained through the instantaneous GOR equation
R=
Bo keg µo
+ Rs
Bg keo µ g
(42)
This equation is equated to the eqn. 75 above to yield:
So Bg dRs So krg µo dBo (1 − So − Swc ) dBg
+
−
Bo dp Bo kro µ g dp
Bg
dp
dSo
=
k
µ
dp
1 + rg o
kro µ g
(76)
This equation can be expressed in an incremental form to yield:
So X ( p) + So Υ ( p)krg / kro − (1 − So − Swc ) Z ( p)
k µ
1 + rg o
kro µ g
(77)
Χ( p) =
Bg dRs
Bo dp (78)
Υ( p) =
1 dBo µo
Bo dp µ g (79)
Ζ( p) =
1 dBg
Bg dp (80)
∆So = ∆p
In which :
28
Material Balance Equation Application
These are all pressure related functions and can be obtained from PVT data.
So =
oil remaining ( N − N p ) Bo
=
(1 − Swc )
one PV
NBoi
(81)
Which gives
Np
So
Boi
=1−
N
1 − Swc Bo
(82)
Dake1 proposes structuring a table with the following steps:
Steps:
(i)
Pressure
(psia)
(ii)
X,Y,Z(p)
/psi
(iii)
So
PV
(iv)
krg/kro
PV
(v)
∆So
PV
(vi)
So
(vii) (viii)
Np/N GOR
scf/stb
(i) Pressure in steps below bubble point
(ii) Table of X,Y &Z(p) values, calculated at the average pressures between the
steps in I.
(iii) So prior to the pressure drop ∆p.
(iv) Relative permeability ratio at last value of So.
(v) ∆So determined using eqn. 77.
(vi) The lower value of So at reduced pressure.
(vii) The fractional recovery from bubble point. Equation 82
(viii) The GOR obtained from GOR eqn using krg/kro value obtained from new So
All the procedures are similar , and are very dependant on reservoir, fluid and rock
data. The quality of a material balance study of a reservoir is related to the quality
of the data. There clearly should be sufficient data both with respect to quantity
and quality as in any simulation study the quality of the output is directly related to
the quality of the input. Another challenge is the definition of the average reservoir
pressure. We will briefly look at these two data perspectives.
5.4 Data for Solution Gas Drive Predictions
Prior to carrying out the MB procedure it is important to test the data, if there is past
production data available. The methods can then be used to examine if they are
capable of predicting past performance. Clearly if the ‘cold data’ does not enable
past performance predictions to be matched, then there is an opportunity to adjust
some of the data to obtain a history match. The data can then be used with better
confidence to predict future performance.
Listed below are the data which are used in the various solution gas drive prediction
procedures. Some of them could be adjusted in history matching.
(a) Reservoir fluid information from PVT analysis:
(i) Oil formation-volume factor Bo
(ii) Gas solubility, Rs
Institute of Petroleum Engineering, Heriot-Watt University
29
(iii) Compressibility factor, z
z can be calculated from gas composition, or gas gravity.
(iv)Reservoir oil and gas viscosity, µo and µg
It is convenient to plot the data as a function of pressure.
(b) Past production:
(i) Oil production
(ii) Gas production
(iii) Water production
(iv) Net water influx
Again these are plotted versus pressure.
(c) Core analysis data:
(i) Laboratory relative permeability
keg/keo vs So or kew/keo vs So
(d) Geological, petrophyisical and simple core data:
(i) Original oil in place, N
(ii) Size of gas zone, m
(iii) Initial water saturation, Swi
(iv) Porosity.
5.5 Gas Drive Reservoirs
As discussed in chapter 10, Drive Mechanisms, Gas Drive is also a depletion type
reservoir. From the nature of the pressure gradients within the oil column it is also
likely that solution gas drive is also active when depleting a gas cap drive reservoir.
Cole8 has pointed out that gas drive is essentially a frontal drive displacing mechanism.
In this respect the high mobility of the displacing gas to that of the displaced oil is
such that in depleting the oil reservoir it is important to minimise the rate , to reduce
by-passing of oil by the advancing gas oil contact. The density differences of gas
and oil clearly help to offset the advancing mobility ratio effect.
Tarner's method can be used for Gas Cap drive reservoir predictions. The equation
however may need alteration to account for gas coming out of solution migrating
into the gas cap.
In Tarner's method the equation for Gas Production, Gp, becomes:
 B − Bgi 
N ( Bo + ( Rsi − Rs ) Bg − Bob ) + mBob  g
 − N p ( Bo − Rs Bg )
 Bgi 
N p Rp =
Bg
(83)
30
Material Balance Equation Application
5.6 Average Reservoir Pressure
The material balance approach is sometimes considered as the ‘tank model’. In
the application of the MB equation we are assuming that the pressure is uniformly
distributed across the reservoir. If there is uniform pressure decline in all the wells
in the reservoir then this pressure decline gives confidence for application of the MB
tool. Dake1 pointed out that if this equilibrium is not achieved, the MB approach can
still be used. He suggested that an average pressure can be determined to represent
a reservoir where there are large differential pressures across the reservoir. He
presents an averaging procedure for reservoirs where pressure equilibrium has not
been achieved.
Pressure
In the figure 11 from Dake1 are presented the pressures for equilibrium conditions
and the well positions and boundaries for a non equilibrium condition.
Time
Figure 11(a) Individual Well Pressures -Equilibruim In The Reservoir.
pj,qj,Vj
Figure 11(b) Well Positions and Drainage Boundaries
Institute of Petroleum Engineering, Heriot-Watt University
31
Pressure
Time
Figure 11c Well pressure for Non Equilibrium Wells1
In figure 11 c the wells have their own pressure declines. Dake1 presents a volume
weighting for the pressures within each drainage area.
pj, Vj and qj, represent the pressure , volume and reservoir rate for the drainage area
j.
The volume weighted average pressure is therefore:
p = ∑ p j Vj / ∑ Vj
j
j
(84)
Dake suggests that determination of the volumes of each drainage zone is subjective
and suggests a production rate alternative, based on the time derivative of the
compressibility equation.
dV= cV∆p
dVj
dp
= q j = cVj j = cVj p′ j
dt
dt
(85)
For a constant compressibility this becomes
Vj ∝ q j / p′ j (86)
Therefore equation 85 can be expressed as:
∑ p q / p′
p=
∑ q / p′
j
j
j
j
j
j
j
(87)
Since the material balance is applied at regular periods, say six months , then the
change in fluid withdrawal (UWj) can be used over a pressure drop ∆pj. Then equation
87 can be expressed as:
32
Material Balance Equation Application
∑ p ∆UW / ∆p
p=
∑ UW / ∆p
j
j
j
j
j
j
j
(88)
Dake comments that some might consider that this averaging approach somewhat
tedious and is better superseded by applying numerical simulation from the start. He
suggests that the two approaches are not exclusive and suggests that the MB approach is
a useful investigation tool prior to structuring a more complex simulation model.
6. RESERVOIR PREDICTIONS AS A FUNCTION OF TIME
None of the terms in the material balance equation have a time term and therefore,
the equation just provides a volume, pressure result. It does not indicate when, a
cumulative production is obtained at a particular pressure value. To do this we need
to incorporate another method where time is included. The productivity of the wells
could be used and has been suggested by Cole8.
On the basis of productivity index the reservoir performance as a function of time
can be predicted.
The productivity index of a well is:
J=
Qo
STB / psi
pe − pw
(89)
where:
Qo = Oil Producing rate, STB/day bbl/day
Pe = Reservoir pressure, psia
Pw = Well bore producing pressure, psia
Replacing Qo in equation 89 with the Darcy radial flow equation results in the
following equation:
J=
7.07keo h( pe − pw )
µo ln(re / rw ) Bo ( pe − pw ) (90)
For most reservoir producing conditions, the height and drainage area remains
constant and therefore:
7.07h
= Cons tan t = C
1nre / rw
(91)
Combining equations 90 and 91, and eliminating the identical (pe - pw) terms gives:
Institute of Petroleum Engineering, Heriot-Watt University
33
J=
Cko
µo Bo (92)
Jµo Bo
Ko (93)
or:
C=
Once determined, i.e. at initial conditions, this value, c, can then be used for future
performance predictions.
An important aspect in the material balance equation is the water influx term. In the
next chapter we will examine in more detail this topic
Solution to Exercise
EXERCISE:
Solution Gas Drive - Undersaturated Oil Reservoir
Determine the fractional oil recovery, during depletion down to bubble point pressure,
for the reservoir whose PVT parameters are listed in PVT Chapter 12. Assume a separator
pressure of 300 psi. Other reservoir fluid properties are listed in table below.
Data
Initial reservoir pressure
Reservoir temperature
Water compressibility
Rock compressibility
Water saturation
pi =
T =
cw =
cf =
Swc =
3,000
220
3.0E-06
8.6E-06
0.22
psi
ºF
psi-1
psi-1
Separator pressure
Psep =
300
psi
SOLUTION (A)
From PVT data
Bubble point pressure Oil formation factor at Pi:
Oil formation factor at Pb:
Pb
Boi
Bob
=
=
=
Separator
Pressure
FVF
psig
bblb/STB
34
300
1.495
2,620 psi
1.4844 bbl/STB
1.4950 bbl/STB
Material Balance Equation Application
Oil Compressibility:
Pressure
psig
co =
Bob − Boi
Boi ∆P 3,000
2,620
Relative
Volume*
bbl/bblb
0.9929
1.0000
FVF
Bo
bbl/STB
1.4844
1.4950
co = 1.88E - 05 psi-1
Effective, saturation-weighted compressibility of the system:
co =
(co So + cw Swc + cf )
(1 − Swc )
ce = 1.58E - 05 psi-1
Finally, the fractional recovery at bubble point pressure is:
Np
N
= ce ∆P
Boi
Bob
Np
N =0.0059 or 0.59% of the original oil in place
If rock and water compressibilities are ignored in undersaturated conditions then a
significant error results. Using the previous exercise calculate this error.
SOLUTION (B)
When water and rock compressibility is ignored:
Np Bo − Boi
=
N
Boi
Np 1.4950 − 1.4844
=
N
1.4844
Np
= 0.00714
N
Np
= 0.00714 or 0.71%
N
Institute of Petroleum Engineering, Heriot-Watt University
35
REFERENCES
1. Dake,L.P. “The Practise of Reservoir Engineering” Elsevier, Ams 1994
2. Schilthuis,R.J. Active Oil and Reservoir Energy. Trans., AIME,118:33-52,
1936.
3. Havelena,D and Odeh,A.S. The Material Balance Equation as an Equation of
a Straight Line. J.Pet.Tech. Aug. 896-900. Trans AIME. 1963
4. Craft and BC Hawkins M.F Applied Petroleum Reservoir Engineering.
Prenticetiall, New Jersey, 1959
5. Tarner,J. “How Different Size Gas Caps and Pressure Maintenance Programs
Affect Amount of Recoverable Oil”, Oil Weekly, June 2,1944 No2 32-34
6. Tracy, G.W. “ Simplified Form of the Material Balance Equation “ Trans AIME
294,243, 1955
7. Muskat,M. “The Production Histories of Oil Producing Gas-Drive Reservoirs’
J.Applied Physics,16,147, 1945.
8. Cole,F.W “Reservoir Engineering Manual” Gulf Publishing, Houston 1961.
9. Havlena,D. and Odeh,A.S. “ The Material Balance as the Equation of a Straight
Line- Oart II Field Cases”, JPT July1964
36
Water Influx
CONTENTS
1. Driving Force for Water Drive
2. Models for Water Encroachment
2.1 The Diffusivity Equation
2.2 States of Flow
2.3 Schilthuis Steady State
2.4 Hurst Modified Steady State
2.5 Van Everdingen and Hurst unsteady-state
3. Reservoir Performance Prediction
3.1 Unsteady-State Model - Van Everdingen & Hurst
3.2 Application to a Declining Pressure
3.3 What are the correct values ∆p?
4. History Matching Water Influx
4.1 Water Drive, No Gas Original Cap
4.2 Water Drive, Gas Cap of Known Size
5. RESERVOIR PERFORMANCE PREDICTION USING THE UNSTEADY STATE EQUATION AND THE MATERIAL BALANCE EQUATION
6. Fetkovitch Method for Water Influx
Determination
7. Carter-Tracy Water Influx Procedure
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•
Calculate the total water influx resulting from a known aquifer volume in terms
of total aquifer compressibility and pressure drop over the aquifer.
•
Sketch and describe the Schiltuis steady state model and the Van Everdingen
and Hurst Unsteady State Model for Water .
•
Sketch the progressive pressure profile for a constant boundary pressure.
•
Explain how a constant boundary pressure profile solution can be used for
declining pressure aquifer/ reservoir pressure.
•
Calculate given prerequisite equations the water influx as a function of time for
a declining pressure profile.
•
Describe and sketch the short hand linear forms of the MB equation for water
drive reservoirs for :
• Water drive and no gas cap
• Water drive and gas cap
• Describe the above for very small aquifers.
•
Describe briefly the combined application of the MB equation and unsteady
state water influx equation in predicting future production – pressure decline.
•
Describe briefly the application of the Fetkovitch method for water influx
determination.
Water Influx
RESERVOIR PERFORMANCE PREDICTION - WATER INFLUX
In the preceding chapters on drive mechanisms and material balance we identified the
positive characteristics of water drive. In this chapter we will examine the various
methods which can be used to predict the amount of natural water drive.
A large proportion of the world’s hydrocarbon reservoirs have an associated aquifer
which depending on production rates can provide a major part of the energy for producing the oil. The consideration is that the original reservoir system was occupied
with water and hydrocarbons have migrated in, displacing some of the water. The
hydrocarbon and aquifer are therefore part of the same reservoir system responding
to the various pressure changes resulting from the production of fluids. As pointed
out during the drive mechanism chapter if the aquifer bounds the edges of the oil zone
it is called edge water drive and if the water bounds the base of the oil reservoir it is
called bottom water drive. The characteristics of water drive are usually the most
efficient displacing agent naturally available in the reservoir. The most significant
characteristics of a water drive system are:
(1)
(2)
(3)
(4)
Pressure decline is very gradual.
Excess water production occurs in structurally low wells.
The gas-oil ratio normally remains steady during the life of the reservoir.
A good recovery of oil can be anticipated
1. DRIVING FORCE FOR WATER DRIVE
The driving force for water drive comes from the response to pressure being lowered as a result of oil production, and since the aquifer is part of this system it also
responds to this declining pressure. As pointed out in the material balance equation
chapter, fluid production is a response to the compressibility of the oil reservoir and
the same is true in most cases for aquifer water drives. The porous system representing the hydrocarbon reservoir and the aquifer are compressible. All its elements:
hydrocarbon, water, and rock expand as pressure declines. It is on the basis of this
compressibility that water encroachment is understood and calculated.
Water encroaches, (moves ) into a reservoir in response to pressure reduction resulting from well production. This pressure reduction causes:
(a) Expansion of the water due to pressure drop within the aquifer
(b) Expansion of hydrocarbons in the aquifer, if any
(c) Expansion of rock, which decreases porosity
(d) Artesian flow, if any, where the outcrop is located structurally higher than the
hydrocarbon accumulation, and the water is replenished at the surface
Institute of Petroleum Engineering, Heriot-Watt University
Clearly the amount of expansion or fluid encroachment is proportional to:
(a) The size of the aquifer
(b) The porosity and permeability of the rock
(c) The presence of any artesian water support
The amount of water flowing into the hydrocarbon reservoir is also influenced by
other factors:
(a) The cross sectional area between the water zone and the hydrocarbon
accumulation
(b) The permeability of the rock in the aquifer
(c) The viscosity of the water
It is clear therefore in predicting the performance of an aquifer a whole range of
aquifer reservoir characteristics are required.
The decline in pressure resulting from oil or gas production moves with a finite velocity
( related to fluid flow) into and through the aquifer. The reduction in pressure causes
the aquifer, water and rock to expand. As long as this moving pressure disturbance
has not reached the external limits of the aquifer, the aquifer will continue to provide
expansion water to the hydrocarbon reservoir. In describing the size of aquifers we
refer to infinite and finite aquifers. Clearly there is not an aquifer which extends to
an infinite extent! The terminology indicates, where in the time considerations of the
analysis, the pressure disturbance has not reached the external limits of the aquifer.
Although natural water drive provides very effective recovery characteristics “there
are still more uncertainties attached to this subject in reservoir engineering, than to
any other. This is simply because one seldom drills wells into an aquifer to generate reservoir characteristic. Instead these properties have frequently to be inferred
from what has been observed in the reservoir. Even more uncertain is the geometry
and the areal continuity of the aquifer itself. The reservoir engineer should therefore
consult both production and exploration geologists. Due to these inherent uncertainties the aquifer fit obtained from history matching is seldom unique and the aquifer
model may require frequent updating as more production and pressure data becomes
available.” Dake 1978.1
In relation to artesian aquifer support it is considered that their occurrence is probably
rare, although there are some reservoirs which purportedly have this type of drive. Due to faulting or pinchouts, most hydrocarbon reservoirs do not communicate to
a surface outcrop. In addition, for artesian flow to contribute substantially to water
influx into a reservoir there must be sufficient ground water moving into the outcrop
to replace the fluid withdrawals from the reservoir, and this water must move through
the entire distance from outcrop to reservoir at this same rate. Thus it is believed that
water influx is usually the result of expansion as a result of pressure drop.
The compression of the void spaces in the reservoir and aquifer rock as a result of
pressure decline in the pore spaces can affect reservoir performance and contribute to
water influx from an aquifer. The compression of the void spaces results in a reduction in the pore volume of the reservoir as withdrawals continue. Water Influx
From a practical standpoint it is usually difficult to separate the water expansion from
the rock compression. Therefore, these two effects, which are additive, are usually
combined into one term which, for convenience, is referred to as effective water
compressibility. The compressibility of water, as well as the compressibility of other
liquids, will vary slightly, according to the pressure and temperature imposed on the
water. Increasing the pressure will reduce the compressibility of water and increasing the temperature will increase the compressibility of water. The compressibility
of fresh water at one atmosphere pressure and 60ºF is 3.3 x 10-6 bbl/bbl/psi.
Effective water compressibilities which have been used in reservoir engineering calculations with good results vary from 1.0 x 10-6 bbl/bbl/psi to 1.0 x 10-4 bbl/bbl/psi.
The figure below illustrates an aquifer supported oil reservoir.
We
AQUIFIER
Wi
OIL
Oil/Water
Contact
Figure 1 Aquifer supported oil reservoir.
Assuming no restrictions due to permeability etc. the maximum water influx associated
with an aquifer system, We, the water influx from the aquifer, can therefore be related to
the volume of the aquifer, its effective compressibility and the pressure drop over it.
We = cWi(pi-p)
(1)
where:
Wi
pi
p
=
=
=
c
=
where cf =
cw
=
initial volume of water in aquifer (function of geometry)
initial aquifer/reservoir pressure
current reservoir pressure. Generally assumed to be pressure at original oil water contact.
total aquifer compressibility. cf +cw
pore compressibility
water compressibility
The main problem facing the reservoir engineer is determining the characteristics of
the aquifer; its geometry, size and flow characteristics.
Institute of Petroleum Engineering, Heriot-Watt University
The following example illustrates the water influx impact of a relatively low compressibility oil reservoir aquifer system.
Exercise 1
(a)Calculate the volume of water an aquifer of 35,000 ft. radius can deliver to a reservoir of 3,200 ft. radius rock and water compressibility under a 1,100 psi pressure drop throughout the aquifer. assume porosity = 0.22.
(b)Compare the available influx with the initial hydrocarbon volume of the reservoir.
Data:
Aquifer radius rw
Reservoir radius ro
Water compressibility cw
Rock compressibility cf
Reservoir thickness h
Pressure drop ∆P
Porosity φ
Water saturation swc
= 35,000 ft.
= 3,200 ft.
= 3.0E-06 psi-1
= 5.0E - 06 psi-1
= 45ft.
=
1,100 psi
=
0.22
=
0.25
2. MODELS FOR WATER ENCROACHMENT
As we have indicated water influx arises as a result of pressure changes decompressing
the aquifer. If the pressure in an aquifer can be calculated then the resulting volumetric
changes can be determined from the pressure / volume compressibility relationship The figure 2 below illustrates the pressure profiles in a reservoir aquifer system. It is
important to appreciate that these profiles are generated as a result of decompression
of the oil and aquifer as a result of well bore production.
(i) Aquifer boundary not affected by decreasing pressure
(ii) Aquifer boundary influenced by pressure decline.
(i)
(ii)
Pressure
OIl
RESERVOIR
Wellbore
AQUIFER
Reservoir outer boundary
Aquifer Outer Boundary
Figure 2 Pressure distribution in an oil reservoir aquifer system.
Water Influx
Before examining the different models we will review the development of equations
which enable the pressure, time and distance solution to be obtained
2.1. The Diffusivity equation
The diffusivity equation results from a combination of the continuity equation,
Darcy’s law.
Well
+
pp
q
+
dq
dp
q
h
r
dr
Figure 3 Radial flow segment
The flow rate at any radius r + dr is q, Figure 3. The rate of flow at radius r will be
larger by the amount dq caused by:
(i) Expansion of the fluid q due to pressure drop dp over element dr
(ii) Expansion of the fluid in the element due to pressure changing with time dp/
dt
The expansion of (i) is too small and can be neglected.
Volume of element
V = 2πrh φ dr
(2)
Change in volume dV due to pressure drop dp is:
dV = - cV dp = - c 2πrh φ dr dp
(3)
dv
∂p
= 2πrh φcdr
dt
∂t
(4)
dq =
∂q
∂p
= − 2πrh φc
∂r
∂t
(5)
Darcy's law for radial flow is
q=−
2πrkh ∂p
µ ∂r
Institute of Petroleum Engineering, Heriot-Watt University
(6)
Differentiating
∂q −2πkh
=
∂r
µ
 ∂ 2 p ∂p 
r ∂r 2 + ∂r 


(7)
Equating equation 5 and 7 gives
−2πrh φc
dp −2πkh
=
dt
µ
 d 2 p dp 
r dr 2 + dr 


Dividing by r gives
∂ 2 p 1 ∂p µcφ ∂p
+
=
∂r 2 r ∂r
k ∂t
(8)
This is the diffusivity equation and describes the flow of a slightly compressible
fluid in porous media. The pressure with respect to distance and time is related by
the parameters φ,µ,c and k.
This group of terms is referred to as the diffusivity constant,η where
1 φµc
=
η
k
(9)
The equation becomes:
∂ 2 p 1 ∂p 1 ∂p
+
=
∂r 2 r ∂r η ∂t
(10)
The name diffusivity equation comes from its application to the flow or diffusion of
heat. The equation is also applied in range of flow systems, heat, electricity as well
as flow in porous media as is the application in a reservoir situation.
In the radial diffusivity equation when applied to an aquifer hydrocarbon reservoir
system the inner boundary is the extent of the hydrocarbon reservoir and the outer
boundary is the limit of the aquifer . In the analysis of the flow and pressures in the
hydrocarbon reservoir the inner and outer boundaries are the radius of the well bore
and the radius of the reservoir.
2.2. States of flow
The diffusivity equation developed above shows that the pressure is a function of
time. As long as this exists, the pressure change with time δp/δt is not constant, and
the flow is termed unsteady state. All states of flow at the start are unsteady state.
During this period we need to analyse the pressure at elements across the radial
symetry and from that determine the resulting expansion After a time period,δp/δt
becomes constant; when this occurs the system is termed pseudo steady state and
fluid expansion can be obtained from a tank model concept.
Water Influx
All aquifers are finite in size, however there is a period of time when a pressure disturbance created by production from a well has not travelled far enough and reached
the boundary of the aquifer. During this time the aquifer behaves as being infinite
and unsteady state flow applies. After the boundary influences the behaviour of the
system pseudosteady state flow starts. The diffusivity equation demonstrates that the states of flow are influenced by the
initial conditions and the boundaries, the outer boundary having a significant influence.
In analysing behaviour, the two boundary conditions must be specified: the inner
boundary the oil-water interface, and the outer boundary the limit of the aquifer. Conditions may be constant pressure, constant rate, closed boundary etc. The initial
condition describes the condition of the system at time, t=0, where a uniform pressure
distribution exists. To solve the equation for water encroachment we need to specify
the boundary and initial conditions. In general for water influx calculations, the most common conditions are a closed
system, no flow at the outer boundary of the aquifer and constant rate or constant
pressure at the inner boundary. In general constant pressure is used in aquifer modelling, whereas in reservoir behaviour constant rate is assumed at the inner, well bore
boundary.
We will now consider the various aquifer models in light of the above discussion.
2.3. Schilthuis steady-state model
The simplest model is that due to Schilthuis. The assumptions associated with this
model are that the aquifer is very large such that the pressure remains constant , and
it has a high permeability such that there is no pressure gradient across the aquifer.
Craft and Hawkins (2) present a useful hydraulic analogue of this steady state model
as shown in Figure 4 below.
Pi
p
Sand-filled pipe
Aquifer
Production
Figure 4 Hydraulic analogue of the Schilthuis steady state model.2
In this model, the aquifer tank pressure remains constant, and could represent an artesian
type aquifer recharged with water or an aquifer large compared to the reservoir.
The reservoir is considered to be relatively small in size with high permeability such
that a flat pressure profile exists. The relative sizes should be at least 10-20:1 .
Institute of Petroleum Engineering, Heriot-Watt University
Initially both aquifer and reservoir tanks are at the original reservoir pressure and as
the reservoir is produced at constant rate the pressure in the reservoir drops. At any
instant, when the reservoir pressure has dropped to a value p, the rate of water influx
by Darcy’s law, will be proportional to the permeability of the sand in the pipe, the
cross-sectional area of the pipe, and the pressure drop (pi - p); and inversely proportional to the water viscosity and the length of the pipe, provided the pressure of the
aquifer tank remains constant. The maximum rate of influx occurs when p=0. If this
rate is greater than the reservoir production rate then at some intermediate pressure the
rate of influx will be equal to the rate of production and the pressure will stabilize, at
a steady-state value. This is an analog of steady-state water influx into a reservoir as
expressed analytically by the Schilthuis equation in which the constant, C, depends
upon the permeability and dimensions of the aquifer rock and the average viscosity
of the water in the aquifer.
Schilthuis equation for this is:
t
We = C ∫ ( pi − p)dt
Î
(11)
C is the aquifer constant and contains the unchanging components of Darcy’s Law,
(units vol/time/pressure).
Expressed in terms of rate of water influx:
dWe
= C( pi − p)
dt
(pi-p) is the boundary pressure drop .
(12)
Hurst(3) in 1943 proposed an equation which recognised that at least part of the aquifer
flow was transient.
2.4. Hurst Modified Steady State
The analogue of this is if the tank is neither very large nor replenished, then as production proceeds, the level in the aquifer tank falls and the potential of the aquifer
declines. If this decline is exponential then it may be represented by Hurst’s following equation in which the declining value of C is represented by C1/log at, and a is
a time conversion constant.
t
We = c1 ∫
o
( pi − p)dt
log e at
dWe c1 ( pi − p)dt
=
dt
log e at
2.5. Van Everdingen and Hurst unsteady-state
(13)
The following equation which we will develop later is the Van Everdingen and Hurst
unsteady-state equation , which is a model which has become generally accepted
10
Water Influx
in water influx modelling. Before developing the equations we will consider the
hydraulic analogue of Craft & Hawkins2.Figure 5.
pi
p4
p3
p2
p1
p
Aquifer
Reservoir
Figure 5 Hydraulic Analogue of Steady State Water Influx2
t
We = B∑ ∆pQ(t )
o
where B is the water influx constant in barrels per pounds per day per square inch,
∆p is the pressure decrement in pounds per square inch, and Q(t) is the dimensionless water influx, which is a function of the dimensionless time. This equation will
be discussed later.
The unsteady-state hydraulic analog is shown above where the reservoir tank on
the right is connected to a series of tanks of increasing size which are connected by
sand-filled pipes of constant diameter and sand permeability, but of decreasing length
between the larger tanks. Initially all tanks are at a common level or pressure Pi representing the original pressure across the system As production occurs, the pressure
in the reservoir tank drops causing water to flow from aquifer tank 1, and a resulting
lower pressure in tank 1. This pressure drop in tank 1 in turn generates flow from
tank 2, and so on. Clearly the pressure drops in the aquifer tanks are not uniform
but vary with time and production rate, and are progressive across the reservoir. An
illustration of these pressure profiles in a radial aquifer are shown in figures 6 and
7 for a constant rate of water influx and for a constant boundary pressure. Even if
there is an infinite number of aquifer tanks, it is evident that reservoir pressure can
never fully stabilise at constant production rate, because an ever-increasing portion
of the water influx must come from an ever-increasing distance.
Institute of Petroleum Engineering, Heriot-Watt University
11
T = 0 = Time
1
T
=
Reservoir Pressure
Pi
T
=
10
T
P1
00
=1
T=
00
10
P10
P100
P1000
RW
RE
Radius
Figure 6 Pressure Distributions in an aquifer at several time periods, for a constant rate of
water influx.2
1
T
=
10
T=
0
10
1
T=
P
RW
T = 0 = Time
T=
Reservoir Pressure
P1
000
T = ∞∞
Radius
RE
Figure 7 Pressure distributions in an aquifer at several time periods for a constant
boundary pressure.2
The water analog can also be illustrated by the series of concentric circles in the figure
8 below . The figure represents the cylindrical elements in an aquifer surrounding a
circular reservoir. An analysis of the pressure in each element will enable the amount
of expansion of water each element can produce as a result of effective compressibility in a pressure decline from pi to zero.
12
Water Influx
Reservoir
1
2
3
4
1
2
3
4
Aquifer
Reservoir
Figure 8 Cylindrical element in an aquifer surrounding a reservoir2
3. PERFORMANCE PREDICTION
Although a consideration of the nature of various aquifers lends support to the analytical expressions presented above, there is no certainty beforehand that any one of
the three will adequately represent the water influx into a particular reservoir, and
studies must be made to determine the most suitable expression.
Examination of the mechanics of water expansion into a hydrocarbon reservoir shows
that it must be an unsteady state process. However, for combination drive reservoirs,
where the water influx rate is small compared to the other driving forces, the use of
the Schilthuis steady state equation can usually be used with reliable results. The
water influx constant, C, can be determined from past production data, and then this
same value of C can be used to aid in predicting reservoir performance.
For active water drive reservoirs, the use of steady state water influx equation will
not usually result in reliable predictions of reservoir performance. As the pressure
drop due to water expansion moves out further into the aquifer, the expanding water
will not move into the hydrocarbon reservoir at the same rate, because for a given
pressure drop the water has to move a greater distance in order to enter the oil or gas
zone. The favoured approach of analysis is the unsteady state model of Van Everdingen
& Hurst (4).
3.1. Unsteady-State Model - Van Everdingen & Hurst
From the analog above it is clear the exact solution for water influx is the unsteady
state solution, where to access the water production we need to determine the pressure time distance profile across the aquifer.
The diffusivity equation in radial form expresses the relation between pressure and
radius and time for a radial system such as drainage from an aquifer, where the driving
potential of the system is the water expandability and the rock compressibility :
Institute of Petroleum Engineering, Heriot-Watt University
13
δ 2 p 1 δp φµc δp
+
=
δr 2 r δr
k δt
(8)
This diffusivity equation is the same basic equation as has been used to calculate
heat flow and electrical flow, as well as fluid flow through porous media. The term
( is usually defined as the diffusivity constant (η) and will be essentially constant
for any given reservoir.
where:
η=
k
µφc
An exact analytical solution of the diffusivity equation for specified boundary
and initial conditions define this pressure time profile and therefore will allow the
calculation of the rate of water influx into a reservoir, provided the proper data are
available. Van Everdingen & Hurst did this in 1949.4 Their analysis was for two cases:
(a) The Pressure case, where the pressure at the inside boundary is known and
the outside boundary is closed, or the reservoir is infinite; and we want to calculate
the water influx.
(b) The Rate case, where the rate is known at the inside boundary. At the outside
boundary there is no flow or the pressure is constant or the reservoir is infinite, and
we want to calculate the total pressure drop.
To enable their analysis to be applicable for different reservoirs they produced a more
general solution of the diffusivity equation by generating dimensionless functions.
Dimensionless time, tD, in place of real rime, t, and dimensional radius, rD, which is
re/ro where re is the radius of the aquifer and ro is the radius of the oil reservoir.
The dimenslionless form of the diffusivity equation is
1 δ  δpD  δpD
 rD
=
rD δrD  δrD  δrD
(14)
where:
tD =
kt
r
2πkh∆p
, rD = e , pD =
2
µφcro
ro
qµ
(15)
Since only basic equations have been utilised, the units for the above quantities
are:
tD
t
k
14
= time, dimensionless
= time, seconds
= permeability, darcy
Water Influx
µ
φ
c
ro
= viscosity, centipoise
= porosity, fraction
= effective aquifer compressibility, vol/vol/atmosphere
= reservoir radius, centimetres
Converting equation 15 to more commonly used units of t = days; k = millidarcies; µ= centipoises; φ = fraction; c = vol/vol/psi and r = feet; results in:
kt
µφcro2
t D = 6.323 x10 −3
(16)
The solution of equation 14 with constant terminal rate boundary conditions is used
in well testing. Hurst and Van Everdingen also derived the constant terminal pressure
solution which is used in water influx calculations.
q D (t D ) =
qµ
2πkh∆p
(17)
qD(tD) = dimensionless influx rate at rD=1.
It is the change in rate from zero to q due to a pressure drop ∆p applied at the outer
hydrocarbons reservoir boundary, ro , at time t=0.
Examining equation (17) from t = 0 to t
t
t
D
µ
dt
qdt
=
q D (t D )
dt D
∫
∫
2πkh∆p o
dt D
o
We µ
φ µcro2
= Q( t )
2πkh∆p
k
Where:
(18)
(19)
We = cummulative water influx
Q(t) = dimensionless water influx
(20)
This equation gives the cumulative water influx for a fixed pressure drop ∆p. The
equation applies in Darcy units. However, when oilfield units are used the following
equation applies.
We =
2πφcro2 h∆pQ( t )
= 1.119φcro2 h∆pQ( t )
5.615
(21)
where;
Institute of Petroleum Engineering, Heriot-Watt University
15
We = cumulative water influx, barrels
∆p = pressure drop, psi
Q(t) = dimensionless water influx
5.615 = conversion factor, cubic feet to barrels.
Equation (21) may also be expressed as:
We = B x ∆pQ(t)
(22)
where:
B = 1.119φcr2oh
(23)
This term B can be considered to be an aquifer characteristic, where the terms do
not change during the decline.
Van Everdingen and Hursts’ paper presented the solution of equation 8 in the form
of dimensionless time, tD, and dimensionless water influx Q(t). Their solution of the diffusivity equation can therefore be applied to any reservoir where the flow of
water into the reservoir is essentially radial in nature. They provided solutions for
external boundaries of an infinite extent and for those of limited extent. Tables 1 and
2 show the tabulated form of their solutions. In addition to being presented in tabular
form the dimensionless water influx as a function of dimensionless time, Dake1 also
reproduced Van Everdingen & Hurst data solutions in graphical form. The graphs
are presented in figures, 8a-e
16
Water Influx
Dimensionless water influx and dimensionless pressures for infinite radial aquifers (courtesy of SPE)3
tD
1.0 x 10-2
5.0 x 10-2
1.0 x 10-1
1.5 x 10-1
2.0 x 10-1
Qt
0.112
0.278
0.404
0.520
0.606
pD
0.112
0.229
0.315
0.376
0.424
tD
1.5 x 103
2.0 x 103
2.5 x 103
3.0 x 103
4.0 x 103
Qt
4.136 x 102
5.315 x 102
6.466 x 102
7.590 x 102
9.757 x 102
tD
1.5 x 107
2.0 x 107
2.5 x 107
3.0 x 107
4.0 x 107
Qt
1.828 x 106
2.398 x 106
2.961 x 106
3.517 x 106
4.610 x 106
tD
1.5 x 1011
2.0 x 1011
2.5 x 1011
3 . 0 x 1 0 11
4.0 x 1011
Qt
1.17 x 1010
1.55 x 1010
1.92 x 1010
2.29 x 1010
3.02 x 1010
2.5 x 10-1
3.0 x 10-1
4.0 x 10-1
5.0 x 10-1
6.0 x 10-1
0.689
0.758
0.898
1.020
1.140
0.469
0.503
0.564
0.616
0.659
5.0 x 103
6.0 x 103
7.0 x 103
8.0 x 103
9.0 x 103
11.88 x
13.95 x
15.99 x
18.00 x
19.99 x
102
103
103
103
103
5.0 x 107
6.0 x 107
7.0 x 107
8.0 x 107
9.0 x 107
5.689 x 106
6.758 x 106
7.816 x 106
8.866 x 106
9.911 x 106
5.0 x 1011
6.0 x 1011
7.0 x 1011
8.0 x 1011
9.0 x 1011
3.75 x 1010
4.47 x 1010
5.19 x 1010
5.89 x 1010
6.58 x 1010
7.0 x 10-1
8.0 x 10-1
9.0 x 10-1
1.0
1.5
1.251
1.359
1.469
1.570
2.032
0.702
0.735
0.772
0.802
0.927
1.0 x 104
1.5 x 104
2.0 x 104
2.5 x 104
3.0 x 104
21.96 x 102
3.146 x 103
4.679 x 103
4.991 x 103
5.891 x 103
1.0 x 108
1.5 x 108
2.0 x 108
2.5 x 108
3.0 x 108
10.95 x 106 1.0 x 1012
1.604 x 107 1.5 x 1012
2.108 x 107 2.0 x 1012
2.607 x 107
3.100 x 107
7.28 x 1010
1.08 x 1011
1.42 x 1011
2.0
2.5
3.0
4.0
5.0
6.0
7.0
8.0
9.0
1.0 x 101
2.442
2.838
3.209
3.897
4.541
5.148
5.749
6.314
6.861
7.417
1.020
1.101
1.169
1.275
1.362
1.436
1.500
1.556
1.604
1.651
4.0 x 104
5.0 x 104
6.0 x 104
7.0 x 104
8.0 x 104
9.0 x 104
1.0 x 105
1.5 x 105
2.0 x 105
2.5 x 105
7.634 x 103
9.342 x 103
11.03 x 104
12.69 x 104
14.33 x 104
15.95 x 104
17.56 x 104
2.538 x 104
3.308 x 104
4.066 x 104
4.0 x 108
5.0 x 108
6.0 x 108
7.0 x 108
8.0 x 108
9.0 x 108
1.0 x 109
1.5 x 109
2.0 x 109
2.5 x 109
4.071 x 107
5.032 x 107
5.984 x 107
6.928 x 107
7.865 x 107
8.797 x 107
9.725 x 107
1.429 x 108
1.880 x 108
2.328 x 108
1.5 x 101
2.0 x 101
2.5 x 101
3.0 x 101
4.0 x 101
9.965
1.229 x 101
1.455 x 101
1.681 x 101
2.088 x 101
1.829
1.960
2.067
2.147
2.282
3.0 x 105
4.0 x 105
5.0 x 105
6.0 x 105
7.0 x 105
4.817 x 104
6.267 x 104
7.699 x 104
9.113 x 104
10.51 x 105
3.0 x 109
4.0 x 109
5.0 x 104
6.0 x 109
7.0 x 109
2.771 x 108
3.645 x 108
4.510 x 108
5.368 x 108
6.220 x 108
5.0 x 101
6.0 x 101
7.0 x 101
8.0 x 101
9.0 x 101
2.482 x 101
2.860 x 101
3.228 x 101
3.599 x 101
3.942 x 101
2.388
2.476
2.550
2.615
2.672
8.0 x 105
9.0 x 105
1.0 x 106
1.5 x 106
2.0 x 106
11.89 x 105
13.26 x 105
14.62 x 105
2.126 x 105
2.781 x 105
8.0 x 109
9.0 x 109
1.0 x 1010
1.5 x 1010
2.0 x 1010
7.066 x 108
7.909 x 108
8.747 x 108
1.288 x 109
1.697 x 109
1.0 x 102
1.5 x 102
2.0 x 102
2.5 x 102
3.0 x 102
4.301 x 101
5.980 x 101
7.586 x 101
9.120 x 101
10.58 x 101
2.723
2.921
3.064
3.173
3.263
2.5 x 106
3.0 x 106
4.0 x 106
5.0 x 106
6.0 x 106
3.427 x 105
4.064 x 105
5.313 x 105
6.544 x 105
7.761 x 105
2.5 x 1010
3.0 x 1010
4.0 x 1010
5.0 x 1010
6.0 x 1010
2.103 x 109
2.505 x 109
3.299 x 109
4.087 x 109
4.868 x 109
4.0 x 102
5.0 x 102
6.0 x 102
7.0 x 1022
8 .0 x 1 0
9.0 x 102
1.0 x 103
13.48 x 101
16.24 x 101
18.97 x 101
21.60 x 101
24.23 x 101
26.77 x 101
29.31 x 101
3.406
3.516
3.608
3.684
3.750
3.809
3.860
7.0 x 106
8.0 x 106
9.0 x 106
1.0 x 107
8.965 x 105
10.16 x 106
11.34 x 106
12.52 x 106
7.0 x 1010
8.0 x 1010
9.0 x 1010
1.0 x 1011
5.643 x 109
6.414 x 109
7.183 x 109
7.948 x 109
Table 1 Dimensionless Qt and Pt vs. tD for infinite reservoir 4.
Institute of Petroleum Engineering, Heriot-Watt University
17
Table 4-7 Dimensionless water influx for finite outcroping radial aquifer (courtesy of SPE)3
tD
rD = 1.5
Qt
5.0 x 10-2 0.276
tD
rD = 2.0
Qt
tD
rD = 2.5
Qt
5.0 x 10-2 0.278
1.0 x 10-1 0.408
6.0 x 10-2 0.304
7.5 x 10-2 0.345
7.0 x 10-2 0.330
1.0 x 10-1 0.404
8.0 x 10-2 0.354
tD
rD = 3.0
Qt
3.0 x 10-1 0.755
1.00
1.5 x 10-1 0.509
4.0 x 10-1 0.895
2.0 x 10-1 0.599
5.0 x 10-1 1.023
1.25 x 10-1 0.458
2.5 x 10-1 0.681
9.0 x 10-2 0.375
1.50 x 10-1 0.507
tD
rD = 3.5
Qt
1.571
2.00
1.20
1.761
1.40
1.940
6.0 x 10-1 1.143
1.60
3.0 x 10-1 0.758
7.0 x 10-1 1.256
tD
rD = 4.0
Qt
tD
rD = 4.5
Qt
2.442
2.5
2.835
2.20
2.598
3.0
3.196
2.40
2.748
3.5
3.537
2.111
2.60
2.893
4.0
3.859
1.80
2.273
2.80
3.034
4.5
4.165
4.454
1.0 x 10-1 0.395
1.75 x 10-1 0.553
3.5 x 10-1 0.829
8.0 x 10-1 1.363
2.00
2.427
3.00
3.170
5.0
1.1 x 10-1 0.414
2.00 x 10-1 0.597
4.0 x 10-1 0.897
9.0 x 10-1 1.465
2.20
2.574
3.25
3.334
5.5
4.727
1.2 x 10-1 0.431
2.25 x 10-1 0.638
4.5 x 10-1 0.962
1.00
1.563
2.40
2.715
3.50
3.493
6.0
4.986
1.3 x 10-1 0.446
2.50 x 10-1 0.678
5.0 x 10-1 1.024
1.25
1.791
2.60
2.849
3.75
3.645
6.5
5.231
1.4 x 10-1 0.461
2.75 x 10-1 0.715
5.5 x 10-1 1.083
1.50
1.997
2.80
2.976
4.00
3.792
7.0
5.464
1.5 x 10-1 0.474
3.00 x 10-1 0.751
6.0 x 10-1 1.140
1.75
2.184
3.00
3.098
4.25
3.932
7.5
5.684
1.6 x 10-1 0.486
3.25 x 10-1 0.785
6.5 x 10-1 1.195
2.00
2.353
3.25
3.242
4.50
4.068
8.0
5.892
1.7 x 10-1 0.497
3.50 x 10-1 0.817
7.0 x 10-1 1.248
2.25
2.507
3.50
3.379
4.75
4.198
8.5
6.089
1.8 x 10-1 0.507
3.75 x 10-1 0.848
7.5 x 10-1 1.229
2.50
2.646
3.75
3.507
5.00
4.323
9.0
6.276
1.9 x 10-1 0.517
4.00 x 10-1 0.877
8.0 x 10-1 1.348
2.75
2.772
4.00
3.628
5.50
4.560
9.5
6.453
2.0 x 10-1 0.525
4.25 x 10-1 0.905
8.5 x 10-1 1.395
3.00
2.886
4.25
3.742
6.00
4.779
10
6.621
2.1 x 10-1 0.533
4.50 x 10-1 0.932
9.0 x 10-1 1.440
3.25
2.990
4.50
3.850
6.50
4.982
11
6.930
2.2 x 10-1 0.541
4.75 x 10-1 0.958
9.5 x 10-1 1.484
3.50
3.084
4.75
3.951
7.00
5.169
12
7.208
2.3 x 10-1 0.548
5.00 x 10-1 0.982
1.0
1.526
3.75
3.170
5.00
4.047
7.50
5.343
13
7.457
2.4 x 10-1 0.554
5.50 x 10-1 1.028
1.1
1.605
4.00
3.247
5.50
4.222
8.00
5.504
14
7.680
2.5 x 10-1 0.559
6.00 x 10-1 1.070
1.2
1.679
4.25
3.317
6.00
4.378
8.50
5.653
15
7.880
2.6 x 10-1 0.565
6.50 x 10-1 1.108
1.3
1.747
4.50
3.381
6.50
4.516
9.00
5.790
16
8.060
2.8 x 10-1 0.574
7.00 x 10-1 1.143
1.4
1.811
4.75
3.439
7.00
4.639
9.50
5.917
18
8.365
3.0 x 10-1 0.582
7.50 x 10-1 1.174
1.5
1.870
5.00
3.491
7.50
4.749
10
6.035
20
8.611
3.2 x 10-1 0.588
8.00 x 10-1 1.203
1.6
1.924
5.50
3.581
8.00
4.846
11
6.246
22
8.809
3..4 x 10-1 0.594
9.00 x 10-1 1.253
1.7
1.975
6.00
3.656
8.50
4.932
12
6.425
24
8.968
3.6 x 10-1 0.599
1.00
1.295
1.8
2.022
6.50
3.717
9.00
5.009
13
6.580
26
9.097
3.8 x 10-1 0.603
1.1
1.330
2.0
2.106
7.00
3.767
9.50
5.078
14
6.712
28
9.200
4.0 x 10-1 0.606
1.2
1.358
2.2
2.178
7.50
3.809
10.00
5.138
15
6.825
30
9.283
4.5 x 10-1 0.613
1.3
1.382
2.4
2.241
8.00
3.843
11
5.241
16
6.922
34
9.404
5.0 x 10-1 0.617
1.4
1.402
2.6
2.294
9.00
3.894
12
5.321
17
7.004
38
9.481
6.0 x 10-1 0.621
1.6
1.432
2.8
2.340
10.00
3.928
13
5.385
18
7.076
42
9.532
7.0 x 10-1 0.623
1.7
1.444
3.0
2.380
11.00
3.951
14
5.435
20
7.189
46
9.565
8.0 x 10-1 0.624
1.8
1.453
3.4
2.444
12.00
3.967
15
5.476
22
7.272
50
9.586
2. 0
1.468
3.8
2.491
14.00
3.985
16
5.506
24
7.332
60
9.612
2.5
1.487
4.2
2.525
16.00
3.993
17
5.531
26
7.377
70
9.621
3.0
1.495
4.6
2.551
18.00
3.997
18
5.551
30
7.434
80
9.623
4.0
1.499
5.0
2.570
20.00
3.999
20
5.579
34
7.464
90
9.624
5.0
1.500
6.0
2.599
22.00
3.999
25
5.611
38
7.481
100
9.625
7.0
2.613
24.00
4.000
30
5.621
42
7.490
8.0
2.619
35
5.624
46
7.494
9.0
2.622
40
5.625
50
7.497
10.0
2.624
Table 2 Dimentionless Qt Vs. tD for finite reservoir
18
Water Influx
4.0
reD = 3.5
=
r eD
reD = 4.0
reD =
3.5
∞
3.0
3.0
reD = 2.5
Qt 2.5
2.0
reD = 2.0
1.5
1.0
reD = 1.5
0.5
0
0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
tD
Figure 9a Graphical form of Qt vs. tD for infinite and finite reservoirs. Dake1.
Institute of Petroleum Engineering, Heriot-Watt University
19
20
reD = 7.0
reD = 8.0
18
re D =
∞
6
r eD =
16
.0
14
Qt
12
reD = 5.0
10
reD = 4.5
8
reD = 4.0
6
reD = 3.5
reD = 3.0
4
2
0
0
10
20
30
40
50
60
70
tD
Figure 9b Graphical form of Qt Vs. to for infinite and finite reservoirs. Dake1.
20
Water Influx
110
re
D
=
∞
100
90
80
Qt
70
60
50
reD = 10.0
40
reD = 9.0
reD = 8.0
30
reD = 7.0
20
reD = 6.0
reD = 5.0
10
0
0
20
40
60
80 100 120 140 160 180 200 220 240 260 280 300
tD
Figure 9c Graphical form of Qt Vs. to for infinite and finite reservoirs. Dake1
Institute of Petroleum Engineering, Heriot-Watt University
21
Figure 9d Graphical form of Qt Vs. to for infinite and finite reservoirs.
22
Qt
1
2
4
10
8
6
2
4
10
102
8
6
2
4
6 8 102
2
r eD
=
∞
4
tD
6 8 103
2
r eD
=
∞
on
(c
reD = 7.0
4
102
103
10
6 8 104
reD = 2.5
reD = 3.0
reD = 4.0
reD = 3.5
reD = 5.0
reD = 4.5
reD = 6.0
reD = 10.0
reD = 9.0
reD = 8.0
)
t'd
reD = 15.0
USE THIS SCALE FOR LINE reD = ∞ (CONT'D) ONLY
Water Influx
10
6 8 104
4
reD = 2.5
reD = 3.0
reD = 3.5
102
reD = 4.0
reD = 5.0
reD = 4.5
reD = 6.0
10
1
2
4
10
8
6
2
Qt
4
102
8
6
2
4
6 8 102
2
r eD
=
∞
4
tD
6 8 103
r eD
2
=
∞
on
(c
reD = 15.0
)
t'd
reD = 7.0
reD = 10.0
reD = 9.0
reD = 8.0
103
USE THIS SCALE FOR LINE reD = ∞ (CONT'D) ONLY
Figure 8e Graphical form of Qt Vs. to for infinite and finite reservoirs. Dake1.
Although the solutions are for a radial system the solution can be applied where the
influx is not full radial but can be considered a segment of such. One of the simplest
modifications which can be made is to determine the fraction of a circular area through
which water is encroaching, and the equation is modified to:
B = 1.119φcr2ohf
(24)
where:
f = fraction of the reservoir periphery into which water is encroaching. Figure 9
Institute of Petroleum Engineering, Heriot-Watt University
23
θ
Aquifer
OWC
Reservoir
Figure 10 Segment of radial water influx
The graphical solutions demonstrate clearly the finite time it takes for a pressure
disturbance to reach the limit of the aquifer, when in the figures Q(t) becomes constant. Dake' has indicated that this maximum value of Q(t) depends on the size of the
aquifer and is equal to:
Q(t) max = 0.5 (rwD2-1) for radial systems
And
Q(t) max = 1 for linear systems
(25)
(26)
It is significant to note that when these values for Q(t) are put in equation 20 for a full
radial system the following expression results.
 r2 − r2 
We = 2πφhcro2 ∆p0.5 e 2 o  = π (re2 − ro2 )hφc∆p
 ro 
(27)
Examination of this equation indicates that it is the total water influx resulting from
from the ∆p being instantly communicated throughout the aquifer.
Clearly for infinite acting radial aquifers there is no maximum Q(t) value since the
effect of the pressure drop is continually moving out into the aquifer. For an infinite
linear aquifer there is no plot of Q(t). The water influx can be directly calculated using the equation below: Dake1.
We = 2 hw
φkct
× ∆p
πµ
in field units this is:
24
(ccs)
(28)
Water Influx
We = 3.26 × 10 −3 hw
φkct
× ∆p
π
(bbls) (29)
t in the above equation is in hours.
Table 3 lists the summary of expressions for Hurst and Van Everdingen for both
radial and linear system1.
Radial Geometry
ro
re
f= θ/390°
θ
Oil reservoir
Aquifer
Linear Geometry
Aquifer
Oil reservoir
h
L
W
Figure 11 Parameters for radial and linear geometry.
Institute of Petroleum Engineering, Heriot-Watt University
25
RADIAL SYSTEM
r0 = radius of reservoir
DARCY UNITS
FIELD UNITS
kt
tD =
φµcr 20
t D = const ×
B = 2 π φ hcr 02 f
constant
+ - hrs.
t - days
t - yrs.
kt
φµcr 20
= 0.000264
= 0.00634
= 2.309
B = 1.119 fφhcr 20 (bbl/psi)
LINEAR SYSTEM
w = width of aquifer
L = length of aquifer
boundary to oil
reservior boundary
tD =
kt
φµcL2
t D = const ×
kt
φµcr 20 → L2
const. as for radial system
B = wLhφc
t in secs and B in cc/atm
B = 0.1781wLhφc
B-bbls/psi
Table 3 Summary of equations and constants for Van Everdingen and Hurst water influx
model
EXERCISE 2
Water Influx - Infinite Aquifer Extent
Calculate the water influx at the end of 1,2 and 5 years into a circular reservoir with
an aquifer of infinite extent. Effective water permeability is 120md, water viscosity
is 0.8 cp, effective water compressibility is 1.0 x 10-6 bbl/bbl/psi, the radius of the
reservoir is 2,400 ft. reservoir thickness is 35ft, porosity is 22%, initial reservoir
pressure is 4,500 psig and present reservoir pressure is 4,490 psig.
Data Table 1
Reservoir Radius
Effective water permeability
Effective water compressibility
Water viscosity
Reservoir thickness
Porosity
Initial reservoir pressure
Current reservoir pressure
Water encroaching factor
26
ro
kw
ce
µw
h
φ
Pi
P
f
= 2,400 ft
= 120 mD
= 1.0e-06 psi-1
= 0.8 cp
= 35 ft
= 0.22
= 4,500 psi
= 4,490 psi
= 1
Water Influx
This example shows that for a given pressure drop, doubling the time interval will not double the water influx. It also shows how to calculate the water influx as a result
of a single pressure drop.
3.2. Application to a Declining Pressure
The application of the Van Everdingen & Hurst model to water influx modelling is
by application of their constant terminal pressure solution; that is the boundary at the
reservoir aquifer contact is constant. For this constant pressure solution the rate and
the cumulative water influx is calculated. In the example above this was the condition and the effects of a fixed pressure drop were determined. In reality however
the pressure at the reservoir / aquifer boundary is declining continuously. How
can we apply this fixed terminal pressure drop solution to a situation where there is
a declining pressure? Figure 11 below gives such a declining pressure at the aquifer
/reservoir boundary
3800
3780
Pi
P1
3750
Boundary Pressure - Psia
3750
P2
3700
3700
P3
3650
3620
P4
3600
3550
3500
0
3
6
9
12
15
18
21
24
Time - Months
Figure 12 Declining pressure profile at the oil / aquifer boundary
Van Everdingen & Hurst proposed a method of calculating the results of a series
of successive pressure drops and adding the solutions together. By superimposing
the effects of a series of fixed pressure drops a steady declining pressure can be
simulated.
The method is illustrated in the Figure 12 where the progressive impact of a series
of fixed pressure drops is illustrated.
Institute of Petroleum Engineering, Heriot-Watt University
27
owc
Pi
owc
Pi
Pi
T=0
We0 = 0
owc
Pi
P1
owc
P1
P1
Pi
T=T
1
We1 = B∆P1QT1
owc
Pi
P1
P2
owc
P2
P2
P1
Pi
T=T
2
We1 = B∆P1QT2
We2 = B∆P2Q(T2-T1)
owc
Pi
P1
P2
P3
owc
P3
We = We1+We2
P3
P2
P1
Pi
We(1) = B∆P1Q(T3)
We(2) = B∆P2Q(T3-T1)
T=T
3
We(3) = B∆P3Q(T3-T2)
We = We1+We2+We3
Figure 13 Progression of fixed pressuredrops through the aquifer
In order to use the unsteady state method it is necessary to assume that the boundary
reservoir pressure declines in a series of steps. For example in figure 12 above it
is assumed that at the end of the first time period T1 the pressure at the reseservoir
aquifer boundary drops suddenly from pi to p1. It is further assumed that the pressure
stays constant for another time period, at the end of which it again drops suddenly
throughtout at the reservoir aquifer boundary to p2. These stepwise decreases in
reservoir pressure are continued for the length of time desired in the water influx
calculations.
If the boundary pressure in the reservoir is suddenly reduced from pi to p1, a pressure drop, will be imposed across the aquifer. Water will continue to expand and
the new reduced pressure will continue to move outward into the aquifer. Given a
sufficient length of time the pressure at the outer edge of the aquifer will finally be
reduced to p1.
If some time after the boundary pressure has been reduced to p1 a second pressure
p2 is suddenly imposed at the boundary, a new pressure wave will begin moving
28
Water Influx
outward into the aquifer as a result of the decompression resulting from the second
pressure drop, decompressing further that decompressed from the first pressure drop.
This new pressure wave will also cause water expansion and therefore encroachment
into the reservoir. However, this new pressure drop will not be pi - p2 but will be
p1 - p2. This second pressure wave will be moving behind the first pressure wave. Just ahead of the second pressure wave will be the pressure at the end of the first
pressure drop, p1. Since these pressure waves are assumed to occur at different times, they are entirely
independent of each other. Thus, water expansion will continue to take place as a
result of the first pressure drop, even though additional water influx is also taking
place as a result of one or more later pressure drops. In order to determine the total
water influx into a reservoir at any given time, it is necessary to determine the water
influx as a result of each successive pressure drop which has been imposed on the
reservoir and aquifer.
In calculating cumulative water influx into a reservoir at successive intervals, it is
necessary to calculate the total water influx from the beginning. This is required
because of the different times during which the various pressure drops have been
effective.
The aquifer term, B is usually a constant for a given reservoir. Thus where the water influx must be calculated for several different pressure drops, each of which has
been effective for varying lengths of time, instead of calculating the water influx for
each pressure step, the total water influx as a result of all the pressure steps can be
calculated as follows:
We1 = B x ∆p1 Q(t)1
We2 = B x ∆p2 Q(t)2
Wen = B x ∆p3 Q(t)n
Combining the above equations:
We = B(Σ∆p x Q(t))
(30)
This equation is the form usually used to calculate water influx.
3.3. What are the correct values ∆p?
By considering the gradual pressure drop to be a series of series of fixed pressure
drops we need a method which will go towards representing this. Clearly shorter
time periods are an advantage together with pressure drops which overlay the curve
as shown in the figure 13 below.
Institute of Petroleum Engineering, Heriot-Watt University
29
AVERAGE BOUNDARY PRESSURES
Pi
P1
∆P1
P2
∆P1 = 1/2 (Pi - P1)
∆P2 = 1/2 (Pi - P2)
∆P2
∆P3 = 1/2 (P1 - P3)
∆P3
P3
∆P4 = 1/2 (P2 - P4)
∆P4
∆P5
P4
0
1
2
3
4
5
TIME PERIODS
Figure 14 Calculation of time period pressure drops
Rather than use the entire pressure drop for the first period a better approximation is
to consider that one half of the pressure drop, eg. 1/2 (pi - p1), is effective during the
entire first period. For the second period the effective pressure drop then is one-half
of the pressure drop during the first period, 1/2 (pi - p1) plus one-half of the pressure
drop during the second period, 1/2 (p1 - p2), which simplifies to:
/2 (pi - p1) + 1/2 (p1 - p2) = 1/2(pi - p2)
1
Similarly, the effective pressure drop for use in the calculations for the third period
would be one-half of the pressure drop during the second period, 1/2(p1 - p2) plus one-half of the pressure drop during the third period, 1/2 (p2 - p3), which simplifies to
1
/2 (p1 - p3). The time intervals must all be equal in order to preserve the accuracy
of these modification.
EXERCISE 3
Water Influx -Infinite Aquifer - Declining Pressure Profile.
Data Table 1
Reservoir Radius
Effective water permeability
Effective water compressibility
Water viscosity
Reservoir thickness
Porosity
Initial reservoir pressure
Water encroaching factor
30
ro
kw
ce
µw
h
φ
Pi
f
= 2,500 ft
=
105 mD
=
1.0E06 psi-1
=
0.8 cp
= 30 ft
=
0.22
= 3,800 psi
=
1
Water Influx
The pressure time relationship is as follows:
Period
Time
Months
Pressure
psi
0
6
12
18
24
3,800
3,780
3,750
3,700
3,620
1
2
3
4
i.e. as figure 11.
4. HISTORY MATCHING WATER INFLUX
In carrying out water influx calculations it has become clear that a number of parameters have a big impact on the pressure support from an aquifer. Not least the
relative size and geometry of the aquifer when compared to the hydrocarbon reservoir.
In reservoir predictions many of these parameters are not available to the reservoir
engineer. Clearly it is not until production has commenced and the reservoir reacts
to fluid production can one determine the pressure support coming from various drive
energies. It is during this phase that the aquifer characteristics can be determined. In the previous chapter on the material balance equation applications we discussed
the Havelena and Odeh5 approach to history matching using a linearisation approach. We will now continue this in the context of reservoirs with a water drive, using their
basic equation:
F = NEo + NmEg + NEfw + We We have seen that water influx is due to expansion of the aquifer water and rock
as a result of a decline in pressure. Simply, a drop in reservoir pressure due to fluid
production is transmitted through the aquifer and the compressibility of the water
albeit small causes the water to expand and flow into the hydrocarbon reservoir.
Water Influx = Initial water volume x Pressure Drop x Aquifer Compressibility
We = (cw +cf ) Wi x ∆p
(31)
As we have seen this equation is generally not sufficient to describe water influx behaviour, in particular for reasonably sized aquifers, because of the finite time required
for the pressure effect to be felt throughout the aquifer. In water influx calculations
therefore it is necessary to include this time dependency as a result of fluid flow. In
Havlena and Odeh’s paper they recognise this time dependant water influx perspective, where they use the dimensionless water influx term to express We, ie:
We = B∑∆pQt
Institute of Petroleum Engineering, Heriot-Watt University
(30)
31
They then apply their short hand MB equation to examine different reservoir senarios
as follows:
4.1. Water Drive, No Gas Original Cap
Havlena and Odeh’s equation can be re-arranged as:
F
∑ ∆pQt
=N+B
Eo
Eo
(32)
∑ ∆pQt
A plot of F/Eo vs.
Eo
should give a straight line, as shown in figure 14.
∑∆pQt
correct
sm
all
B
W
e
to
o
incorrect
geometry
F/Eo
We
large
too
F
Eo
= N+B
∑∆pQt
Eo
N
∑∆pQt
Eo
Figure 15 Plot of F/Eo vs .
∑ ∆pQ
t
Eo
This line will be straight if the aquifer characteristics, B, and the radius of the aquifer
are correct. The intercept will be the oil in place, N, the slope B. Havalena and Odeh
suggest four other plots. Complete scatter, suggesting the calculations or basic data is in error. A systematically upward or downward curve suggesting that ∑∆pQtDis
too small or to large. This means that re/ro and/or td is too small or too large . An S
shaped curve indicates that a better fit might be obtained by assuming linear water
influx.
Once the assumed values give realistic behaviour then the model, which has been
obtained by history matching, can be applied in predicting future reference reser32
Water Influx
voir performance. The assumption is taken in such a situation that the reservoir and
associated aquifer continue to behave as before. Because of this large assumption
and that no aquifer model is likely to be unique the validity of the model should be
updated as more pressure and production data becomes available.
This linearisation approach has been used as a means of determining the extent of
a supporting aquifer. In figure 15 below the curves show the results for a range of
dimesionless radius’s for a field in the North Sea6. The results show a straight line
fit with an infinite aquifer.
16
o
x
x
14
NPBo + Wρ - Wi , MMMSTB
Eo
Re/Rw = 10
Re/Rw = 20
xx
x xx
x
x xx x
12
10
6
4
x
x x
2
x
x
x
x
x x
xx
x
x
x
x x
x
x x
xx
x
x
x x
x
x
x
o x
o x
o x
o x
Re/Rw = 00
x x
xx
xx
x
x
x
8
oo x
o xx x
oxxx
x
x
x
6 MONTHS
0
0
1.0
2.0
3.0
4.0
5.0
∑∆PQ(Qt) , MMPSI
Eo
Figure 16 Havalena and Odeh approach applied to Piper Field. 6
Very small Aquifers
For small aquifers, the assumption might be made that steady state flow exists as the
pressure drop is quickly transmitted through the aquifer. In this case;
We = B′ ∆p′
(33)
where: ∆p′ = pi - p
and: B′ = Wicw
(34)
Wi is the water volume in the aquifer.
The equation now becomes
F
∆p′
= N + B′
Eo
Eo
Institute of Petroleum Engineering, Heriot-Watt University
(35)
33
A plot of F/Eo vs ∆p'/Eo should give a straight line of slope B′ and intercept N. Havlena
and Odeh point out that the points on this graph will plot backwards as in figure 16
below This is because Eo increases faster than ∆p, therefore ∆p/Eo decreases as
the pressure decreases. In some situations a steady state water influx sets in after a
certain period. In this case the points plotted for the unsteady state period will plot
in forward sequence but when the steady state exists then the plotted points reverse
the plot backwards.
B'
F/Eo
F
Eo
= N + B'
∆p'
Eo
N
∆p'
Eo
Figure 17 Plot of F/Eo Vs. ∆p'/Eo for small aquifers
4.2. Water Drive, Gas Cap of Known Size
Using a similar approach to the treatment of We as before and applying it to a gas
cap drive reservoir Havlena and Odehs’ equation yields:
F
∑ ∆pQt
=N+B
Eo + mEg
Eo + mEg
(36)
where We = B∑∆pQ
Again this gives a straight line function ,figure 16 to 17, if the geometry of the aquifer
and time are assumed correct. If the line is not straight then assumptions regarding
the aquifer need to be modified as for water drive systems without a gas cap. 34
Water Influx
B
F
Eo + mEg
F
Eo + mEg
=N+B
∑∆pQt
Eo + mEg
N
∑∆pQt
Eo + mEg
Figure 18 Havalena and Odeh plot for water drive and known gas cap
Small Aquifers, Gas Cap of Known Size
For small aquifers, as for the system without a gas cap, a plot of the lefthand side
and the B term of the following equation should result in a straight line, the points
plotting in reverse sequence. Figure 18.
F
∆p′
= N + B′
Eo + mEg
Eo + mEg
Institute of Petroleum Engineering, Heriot-Watt University
(37)
35
F
Eo + mEg
F
Eo + mEg
= N + B′
Eo + mEg
N
Eo + mEg
Figure 19 Havlena and Odeh plot for small aquifers and known gas cap
5. RESERVOIR PERFORMANCE PREDICTION USING THE
UNSTEADY STATE EQUATION AND THE MATERIAL BALANCE
EQUATION
With both the water influx equation and the material balance equation there are
two unknowns, We and pressure. It is therefore not straightforward to predict future
performance. However by combining the material balance equation with the unsteady
state equation we are able to predict reservoir performance, but it is a tedious trial
and error approach.
The procedure is as follows:
(1) Collect all the available reservoir production and subsurface sample data.
(2) Using the production data we can calculate a value for the aquifer constant, B,
in the unsteady state equation.
In order to determine the validity of the constant B it is important to carry this calculation over a range of times. Values of B at these various times are calculated
using:
The Material Balance equation given in the Havlena and Odeh shorthand form:
F = NEo + NmEg + NEfw + We
36
Water Influx
We(MB) = F - N( Eo +mEg + Efw)
F are production terms.
Using the unsteady state equation and knowing the time and pressure drop,we can
calculate
∑∆pQ(t)
The value of the aquifer characteristic, B, is then obtained by division:
B=
We ( mb )
∑ ∆pQ(t )
(38)
The apparent value of B, is determined at several past production times and plotted
versus cumulative oil production. The best horizontal line is then drawn through the
various points. This is the value of B which can be used for all future calculations.
(3) Using the calculated aquifer characteristic, B, water influx over the past history
of the reservoir is calculated using both the unsteady state equation and the
material balance equation. Clearly these should reasonably agree since the past
production data has been used in calculating B.
(4) Using the trends of past production history of the reservoir the future production
trends of oil, gas and water production throughout the future prediction period can
be plotted and the following trial and error calculation steps carried out:
1. The first step is the estimation of the reservoir pressure at the end of the first
trial period (suggested 6 months). Gross water influx is calculated by
both equations, the MB and the Unsteady State Influx Equations. If the results agree then the first estimate of pressure is correct; if not, then another pressure
must be selected and the procedure repeated until convergence achieved.
2. This procedure is carried out for each time period until the desired range of
production has been covered
Different combinations of production rates of oil, gas and water should be used and
a complete prediction of pressure decline made for each set of values.
There are a number of factors to consider with respect to the limitation of the
performance prediction, in particular, the extent of the aquifer, the size of the constant
B, and the time intervals of the calculation. All of these are significant factors in
the calculations. For example in later time periods the pressure influence could
have reached the aquifer limit and therefore we have a finite aquifer. Using the MB
equation to calculate water influx may not be so accurate as small changes in pressure
can cause significant changes in PVT terms. Clearly shorter time periods generates
more accurate predictions. Three to six month periods are suggested.
Institute of Petroleum Engineering, Heriot-Watt University
37
6. FETKOVITCH METHOD FOR WATER INFLUX
DETERMINATION
Although the Hurst and van Everdingen unsteady state influx theory provides a well
established method for predicting water influx its application is somewhat tedious
particularly as a result of the superposing solutions for each time step and the trial
and error approach for history match and predicting future performance. In 1971
Fetkovitch7 produced an approach for finite aquifers.
The method of Fetkovitch7 models the aquifer flow in the same way as oil flow from
a reservoir into a well, and the results of the model follow closely those of van Everdingen and Hurst. Fetkovitch’s concept is that the productivity index approach can
be used to describe the water influx from an aquifer into a reservoir.
The inflow equation used is:
qw =
dWe
= J ( pa − p)
dt
(39)
where:
qw = water influx rate
J = aquifer productivity indix
p = reservoir pressure (at oil or gas water contact)
pa = average pressure in the aquifier
We = cumulative water influx volume
For a finite aquifer the total water influx arising from a total pressure drop is:
We = cWi ( pi − pa )
(40)
where:
pi = is the initial pressure in the aquifer
Wi = initial volume of water in the aquifer and reservoir
c
= total aquifer compressibility = cw + cf
therefore:

We 
pa = pi 1 −

 cWi pi 
(41)

W 
pa = pi 1 − e 
 Wei 
(42)
or:
38
Water Influx
where:
Wei = cWipi is the maximum possible expansion of the aquifer arising from a pressure drop of Pi.
Differentiating equation 42 with respect to time gives:
dWe Wei dpa
=
dt
pi dt
(43)
substituting in equation 39 gives:
dpa
Jp
= − i dt
pa − p
Wei
(44)
Integrating this equation with initial conditions when t = 0(We=0,pa=pi) and a pressure
drop ∆p=pi-p is imposed at the hydrocarbon reservoir boundary and assuming boundary
pressure remains constant over the period of interest:
ln( pa − p) = −
Jpi
dt
Wei
(45)
C= constant of integration which from initial conditions= 1n(pi-p)
therefore:
( pa − p) = ( pi − p)e
− Jpi t / Wei
(46)
substituting in equation 39 gives:
− Jpi t / Wei
dWe
= J ( pi − p)e
dt
(47)
Integrating this equation yields:
We =
(
− Jpi t / Wei
Wei
pi − p) 1 − e
(
pi
)
(48)
As t tends to infinity then:
We =
Wei
( pi − p) = cWi ( pi − p)
pi
(49)
which is the maximum amount of water influx that could occur once the pressure
drop has moved through the aquifer.
Institute of Petroleum Engineering, Heriot-Watt University
39
The equations developed so far assume a constant inner boundary pressure. Therefore to apply to a declining pressure a superposition principle should be applied as
for van Everdingen and Hurst. Fetkovitch however showed that a difference form
of the equation can be used.
The water influx for a time step ∆t1.
∆We1 =
(
Wei
− J ∆t / W
pi − p1 ) 1 − e pi 1 ei
(
pi
)
(50)
where p1 is the average reservoir boundary pressure during the first time step.
Similarly for ∆t2
∆We 2 =
(
Wei
− J ∆t / W
pi − p2 ) 1 − e pi 2 ei
(
pi
)
(51)
wherepa1 is the average aquifer pressure at the end of the first time interval.

∆We1 
pa1 = pi 1 −

Wei 

(52)
The general solution for the nth time step is:
∆Wen =
where:
pa( n−1)
)(
Wei
− J ∆t / W
pan−1 − pn 1 − e pi n ei
pi
(
)
n −1


∆Wej
∑


= pi 1 − 1

Wei 



(53)
(54)
The average reservoir boundary pressurePn is calculated as in the Van Everdingen
& Hurst method, i.e.:
pn =
pn −1 + pn
2
(55)
Equations 53 and 54 used in a stepwise technique by Fetkovitch have demonstrated
that the water influx calculated match closely results using the van Everdingen and
Hurst.
The aquifer productivity index J values are listed below1.
40
Water Influx
FLOWING CONDITIONS
RADIAL AQUIFERS
J (cc/sec/atm)
Semi-steady State
(drawndown expressed
asPa - P)
Steady State
(drawndown expressed
as Pi - P)
Field Unit Factor
2 πfkh
 re 3 
µ1n − 
 ro 4 
2 πfkh
re
ro
µ1n
7.08 × 10 −3
LINEAR AQUIFERS
J (cc/sec/atm)
3
khw
µl
...................(56)
khw
µl
...................(57)
1.127 × 10−3
Table 4 Values of the aquifer productivity index J. (Dake1)
The semi-steady state equations 56 are used in conjunction with Fetkovitch equations
53 and 54. The steady state equations, however, are applied differently. In the
steady state situation the assumption is made of the water influx from the aquifer
being replaced from another source thereby maintaining pressure Pi constant at the
external aquifer boundary. It is therefore unnecessary to use Fetkovitch’s method of
determining the average pressure. The aquifer productivity values are used directly
with the drawdown equation 47.
qw =
dWe
= J ( pi − p)e − Jpi t / Wei
dt
(47)
Since steady state assumes Wei as infinite:
qw =
i.e.
dWe
= J ( pi − p)
dt
(58)
t
We = J ∫ ( pi − p)dt
o
(59)
The expression of the above equation 57 is the same as the Schilthuis steady state
equation, equation 11.
Institute of Petroleum Engineering, Heriot-Watt University
41
Infinite Aquifers
It is not possible to apply the Fetkovitch method to very large aquifers, because of
transient flow effects. With large aquifers there is a finite time for the initial pressure
disturbance at the reservoir boundary to feel the effect of the aquifer boundary. It is necessary for large aquifers to use the Hurst and Van Everdingen for the first time
intervals. Dake1 has demonstrated that a combination of the two procedures, using van
Everdingen and Hurst for the first time periods covering the more transient time with
Fetkovitch for the later time periods provides an effective approach to water influx
calculations. The reader is recommended to read Dake's text where some examples
are worked through, using both the van Everdingen, Hurst and Fetkovitch method.
7. Carter & Tracy Water Influx Procedure
In 1960 Carter and Tracy8 published a procedure for calculating water influx which
eliminates the superposition calculations described previously to accommodate a
steady declining pressure for a constant terminal pressure solution. The results show
close approximation to the procedure of van Everdingen and Hurst4. They used the
constant terminal rate solution of the diffusivity equation and their equation is;
 B∆p − E
p′ 
e ( t Dj −1 ) ( t Dj )
( t Dj )

We t = We t
+
( Dj )
( Dj −1 )  p − t p′ t

Dj
 ( t Dj ) Dj −1

( )
B
We
(60)
= aquifer constant as previously = 1.119fφhcr2o (bbl/psi)
= cumulative water influx (bbls)
2.309
kt
φµcro2 where t is in years
tD
= dimensionless time = P(tD ) = dimensionless constant terminal rate solution of diffusivity equation
P'(tD) = time derivative dPtD /dtD.
The “j” refers to the present time step and the “j-1” refers to the previous time
step.
Rather than using the tables of PtD given by Van Everdingen and Hurst, Fanchi10 as
presented by Dake9 used a regression equation form of Van Everdingen and Hurst
functions as;
p(tD)=a0 + a1tD + a1tD + a2lntD + a3(lntD)2
The regression coefficients are given in the table 5 below
42
Water Influx
Carter-Tracy influence functions regression coefficients for the constant terminal rate case (17)
reD
Regression coefficients
a0
1.5
2.0
3.0
4.0
5.0
6.0
8.0
10.0
∞
0.10371
0.30210
0.51243
0.63656
0.65106
0.63367
0.40132
0.14386
0.82092
a1
1.66657
0.68178
0.29317
0.16101
0.10414
0.06940
0.04104
0.02649
-3.68 x 10-4
a2
a3
-0.04579
-0.01599
0.01534
0.15812
0.30953
0.41750
0.69592
0.89646
0.28908
-0.01023
-0.01356
-0.06732
-0.09104
-0.11258
-0.11137
-0.14350
-0.15502
0.02882
Table 5 Carter - Tracy influence functions regression coefficients for the constant terminal
rate case
The reader is referred to Dake's9 text to see worked examples on Dake's approach.
Solutions to Exercises
EXERCISE 1
(a) Calculate the volume of water an aquifer of 35,000 ft. radius can deliver to a reservoir of 3,200 ft. radius rock and water compressibility under a 1,100 psi pressure
drop throughout the aquifer. assume porosity = 0.22.
(b) Compare the available influx with the initial hydrocarbon volume of the reservoir.
Data:
Aquifer radius rw
Reservoir radius ro
Water compressibility cw
Rock compressibility cf
Reservoir thickness h
Pressure drop ∆P
Porosity φ
Water saturation swc
=
=
=
=
=
=
=
=
35,000 ft.
3,200 ft.
3.0E-06 psi-1
5.0E - 06 psi-1
45ft.
1,100 psi
0.22
0.25
SOLUTION (A)
Total aquifer compressibility ce 8.0E-06 psi-1
Aquifer Volume
Wi = π* (rw2 - ro2)*h*
Wi = 3.78E+10 ft.3
Institute of Petroleum Engineering, Heriot-Watt University
43
Water Influx
We =ce * Wi * ∆P
We = 3.325E+08 ft.3
SOLUTION (B)
Initial hydrocarbon volume Vhc = π* ro2 h* φ(1-Swc)
Vhc = 2.389E+08 ft.3
The water influx is capable of replacing the hydrocarbons pore volume.
EXERCISE 2
Water Influx - Infinite Aquifer Extent
Calculate the water influx at the end of 1,2 and 5 years into a circular reservoir with
an aquifer of infinite extent. Effective water permeability is 120md, water viscosity
is 0.8 cp, effective water compressibility is 1.0 x 10-6 bbl/bbl/psi, the radius of the
reservoir is 2,400 ft. reservoir thickness is 35ft, porosity is 22%, initial reservoir
pressure is 4,500 psig and present reservoir pressure is 4,490 psig.
Data Table 1
Reservoir Radius
Effective water permeability Effective water compressibility
Water viscosity
Reservoir thickness
Porosity
Initial reservoir pressure
Current reservoir pressure
Water encroaching factor
ro
kw
ce
µw
h
φ
Pi
P
f
=
=
=
=
=
=
=
=
=
2,400 ft
120 mD
1.0e-06 psi-1
0.8 cp
35 ft
0.22
4,500 psi
4,490 psi
1
B
=
1.119 φ ce ro2 h f
B
=
49.63 bbl/psi
Dimensionless time
td
=
6.323x10-3
After 1 year, i.e.
t
tD
=
=
365 days
273 days
SOLUTION
Water influx constant
44
k t
φ µ ce ro2
(24)
(16)
Water Influx
and the Dimensionless influx rate Q(t);
=> is extracted from the table for
infinite aquifers
Q(t)
=
98.0
Water influx We: We =
B ∆P Q(t)
We = 48,626 bbl
Results Summary
Time (t)
Years Days
tD
Q(t)
Current
(Interpolation) Pressure (psi)
Pressure
Drop (psi)
Water Influx
We (bbl)
After :
1 Year =>
2 Years =>
3 Years =>
1
2
5
365
730
1825
273
546
1,366
98.0
175.3
383.9
4,490
4,490
4,490
10
10
10
48,626
86,988
190,524
EXERCISE 3
Water Influx -Infinite Aquifer - Declining Pressure Profile.
Data Table 1
Reservoir Radius
Effective water permeability Effective water compressibility
Water viscosity
Reservoir thickness
Porosity
Initial reservoir pressure
Water encroaching factor
ro
kw
ce
µw
h
φ
Pi
f
=
=
=
=
=
=
=
=
2,500 ft
105 mD
1.0E06 psi-1
0.8 cp
30 ft
0.22
3,800 psi
1
The pressure time relationship is as follows:
Period
i.e. as figure 11.
1
2
3
4
Time
Months
Pressure
psi
0
6
12
18
24
3,800
3,780
3,750
3,700
3,620
Institute of Petroleum Engineering, Heriot-Watt University
45
SOLUTION
Water influx constant
B
=
1.119 φ ce ro2 h f
B
=
46.16 bbl/psi
=
k t
2
6.323x10-3 φ µ ce ro
td
Dimensionless time
Dimensionless influx rate Q(t);
nite
Water influx (24)
(16)
=> is extracted from the table for infiaquifers
We =
Σ
B*( ∆P*Q(t))
(30)
1. Water influx at the end of 6 months
B = 46.15875 bbl/psi
t = 182.5 days
tD = 110.15 => Q(t) =
Period
1
Time (t) (Duration)
Months
Days
6
182.5
tD
110.1
46.63
(By interpolation)
Pressure
Effective ∆p
Q(t)
(psi)
Pressure Drop
(From Table
(psi)
Infinite Aquifers)
3,780
10 *
∆P*Q(t) Water Influx
(psi)
We (bbl)
46.63
466
Σ(∆P*Q(t)) =
466
21,522
* Where:
∆P1 =
1
(P − P1 )
2 i
2. Water influx at the end of 12 months
B = 46.15875 bbl/psi
t = 365 days
tD = 220.3 => Q(t) =
82.06
(By interpolation)
The first pressure drop, Pi - P1, has been effectivefor one year, but the second pressure drop, Pi - P2, has been effective for only six months. Separate calculations must
be made for the two pressure drops because of this time difference and the results
added in order to determine the total water influx.
The equation:
Σ
We = B( ∆P Q(t))
46
Water Influx
will need to be used for each pressure drop.
Calculating Water Influx for a year: Period
1
2
Time (t) (Duration)
Months
Days
12
6
365.0
182.5
tD
220.3
110.1
Pressure
Effective ∆p
Q(t)
(psi)
Pressure Drop
(From Table
(psi)
Infinite Aquifers)
3,780
3,750
10
25*
82.06
46.63
Σ(∆P*Q(t)) =
∆P*Q(t) Water Influx
(psi)
We (bbl)
821
1,166
1,986
91,681
* Where:
∆P2 =
1
1
( Pi − P1 ) + ( P1 − P2 )
2
2
∆P2 =
1
( Pi − P2 )
2
3. Water influx at the end of 18 months
B = 46.15875 bbl/psi
t = 547.5 days
tD = 330.45 => Q(t) =
114.87 (By interpolation)
The first pressure drop, ∆P1, will have been effective the entire eighteen months, the
second pressure drop ∆P2, will have been effective for 12 months and the last pressure
drop, ∆P3, will have been effective for only 6 months. the table below summarises
the calculations:
Period
1
2
3
Time (t) (Duration)
Months
Days
18
12
6
547.5
365.0
182.5
tD
330.4
220.3
110.1
Pressure
Effective ∆p
Q(t)
(psi)
Pressure Drop
(From Table
(psi)
Infinite Aquifers)
3,780
3,750
3,700
10
25
40*
114.87
82.06
46.63
Σ(∆P*Q(t)) =
∆P*Q(t) Water Influx
(psi)
We (bbl)
1,149
2,051
1,865
5,065
233,800
* Where:
∆P3 =
1
1
( P1 − P2 ) + ( P2 − P3 )
2
2
∆P3 =
1
( P1 − P3 )
2
Institute of Petroleum Engineering, Heriot-Watt University
47
4. Water influx at the end of 24 months
B = 46.15875 bbl/psi
t = 730 days
tD = 440.60 => Q(t) =
146.23 (By interpolation)
The first pressure drop, ∆P1, has now been effective for the entire twenty four months,
the secondpressure drop, ∆P2, has been effectinve for eighteen months, the third pressure drop, ∆P3, has been effective for twelve months and the fourth pressure drop,
∆P4, has been effective for only six months. the table below summarises the water
influx calculations:
Period
Time (t) (Duration)
Months
Days
1
2
3
4
24
18
12
6
730.0
547.5
365.0
182.5
tD
440.6
330.4
220.3
110.1
Pressure
Effective ∆p
Q(t)
(psi)
Pressure Drop
(From Table
(psi)
Infinite Aquifers)
3,780
3,750
3,700
3,620
* Where:
48
∆P4 =
1
1
( P2 − P3 ) + ( P3 − P4 )
2
2
∆P4 =
1
( P2 − P4 )
2
10
25
40
65*
146.23
114.87
82.06
46.63
Σ(∆P*Q(t)) =
∆P*Q(t) Water Influx
(psi)
We (bbl)
1,462
2,872
3,282
3,031
10,647
491,452
Water Influx
REFERENCES
1. Dake ,L.P. “Fundamentals of Reservoir Engineering” Elsevier ISBN 0-44441667-6. 1978
2. Craft,B.C. & Hawkins,M.F. “Applied Petroleum Reservoir Engineering” PrenticeHall,NY. 1959
3. Pirson,S.J. “Elements of Oil Reservoir Engineering” 2nd Edition McGraw-Hill
Book Co. Inc (NY) 1958 pp608.
4. Van Everdingen,A.F. & Hurst,W, “ The Application of Leplace Transformation
to Flow Problems in Reservoirs”. Trans AIME (1949) Vol i86, 305.
5. Havlena,D. and Odeh,A.H. “The Material Balance as an Equation of a Straight
Line” J. Pet.Tech Aug.896-900. Tarns AIME 1963
6. Stewart,L. “Piper Field- Reservoir Engineering”. EUR 152. Paper presented at
the SPE European Offfshore Petroleum Conference. London .Oct 1980
7. Fetkovitch.M.J. “ A Simplified Approach to Water Influx Calculations-Finite
Aquifer Systems”, J.Pet.Tech.,July,814-828, 1971
8. Carter,R.D. and Tracy,G.W. “An improved Method for Calculating Water Influx”,
Trans AIME. Vol 219, p 415-417., 1960.
9. Dake.L.P. The Practise of Reservoir Engineering" Elsevier ISBN 0-444-820949, 1994
10.. Fanchi,J.R. "Analytical Representation of the van Everdingen-Hurst Aquifer
Influence Functions for Reservoir Simulation,SPE-Reservoir Engineering, June
1985.
Institute of Petroleum Engineering, Heriot-Watt University
49
Immiscible Displacement
CONTENTS
1.
INTRODUCTION
7. Application to Field Performance
2.
The Reason for Water Injection
2.1 Zone Isolation
2.2 Permeability
2.3 Oil Viscosity
2.4 Undersaturated Reservoirs
2.5 Overpressured Reservoirs
2.6 Reservoir Depth
2.7 Facility Design
2.8 Thermal Fracturing
2.9 Water Handling
8.
3.
BASIC WATERDRIVE THEORY
3.1 Introduction
3.2 Water-Oil displacement at Microscopic and
Macroscopic levels
3.3 Relative Permeability
3.4 Fractional Flow
4.
DISPLACEMENT THEORIES
4.1 Introduction
4.2 Buckley- Leverett Theory
4.3 Welge Analysis
4.4 Calculations for Oil Recovery
4.5 The Impact of Viscosity.
5. TWO DIMENSIONAL BEHAVIOUR- SEGREGATED FLOW
5.1 Introduction
6.
Coping with hetereogeneity
6.1 Introduction
6.2 Vertical Heterogeneity
6.3 Areal Hetereogeneity
6.4 Vertical Sweep Displacement Calculations
for Layered Reservoirs
6.5 Ordering of the Layers
6.6 Impact of Capillary Pressure in Homogeneous Systems
6.7 Impact of Permeability Distribution on Waterflooding
IMMISCIBLE DISPLACEMENT IN GAS DRIVE SYSTEMS
8.1 Mobility Ratio for Gas Oil Systems.
8.2 Gravity Segregation
8.4 Other parameters
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
•
•
Describe briefly the various benefits of water injection.
Present a simple equation for the fractional flow of water in terms of water and
oil flow rate.
• Comment briefly on the impact of ;angle of dip, capillary pressure, and velocity
on the fractional flow.
•Plot a set of relative permeabilties and identify end-point relative
permeabilities.
• Define mobility ratio and present an equation for it and calculate its value given
relative permeability data.
• Generate a fractional flow curve given relative permeability and viscosity data
for injected and displaced fluids.
•Derive the Buckley-Leverett Frontal Advance Equation.
• Show the shape of the fractional flow curve and its associated derivative curve
and the progressive saturation displacement profile for the following three types
of displacment;
• Water displacing viscous oil
• Water displacing avery light oil.
• Water displacing medium denisty oil
• Determine the breakthrough fractional flow, saturation and time for a diffuse
flow displacement process.
•Calculate the oil recovery at breakthrough for a water displacing oil process.
• Calculate the oil recovery after breakthrough, using an equation or the Welge
construction.
• Comment on the impact of oil viscosity on the fractional flow curve and the
displacement process.
• Show how the relative permeability curves for segregated flow are two straight
lines between end point relative permeabilities and saturation values.
• Generate a set of pseudo relative permeability values for segregated flow and
use them to generate a fractional flow curve for such flow conditions.
• Calculate breakthrough and subsequent fractional flow, saturation and recovery
for segregated flow conditions.
•Present an equation expressing the recovery in terms of vertical and areal
recovery efficiency.
• Generate pseudo relative permeabilties and subsequent fractional flows for
layered reservoir systems for both layers where cross flow exists and where no
cross flow occurs.
• Sketch the water flood profiles for different permeability distributions.
• Describe how a reservoir fractional flow curve can be generated and used to
justify increased water injection to improve oil recovery.
•Be aware of the application of immiscible displacement theory in gas oil and
dry gas wet gas displacement processes.
Immiscible Displacement
1. INTRODUCTION
In previous chapters we have examined the various fundamental properties associated with the behaviour of fluids when subjected to pressure and temperature changes
and the characteristics of reservoir porous media in relation to its pore volume and
transmission characteristics. At another extreme scale we have reviewed the various
drive mechanisms responsible for providing the energy to move hydrocarbons in a
reservoir. We have also examined the various volumetric methods used to relate the
volumes of fluids produced in relation to the overall pressure decline of the reservoir
and the original volumes in place and energy support provided by attached water
and gas.
It is the purpose of this next chapter to bring some of these topics together in the
context of those reservoirs where the principle drive mechanism is that associated
with the immiscible displacement of oil. The subject will be mainly presented in
the context of water displacing oil, and then later the application to gas displacing
oil will be covered.
The topic of water drive in the chapter on drive mechanisms showed that this drive
mechanism provided the highest recovery factor in relation to reservoir depletion.
For this reason therefore water drive provided by intervention, that is when water
is injected into the reservoir through injection wells, is common practise in oilfield
operation.
The modelling of water drive in reservoirs in relation to understanding the displacement
behaviour and associated recovery is generally carried out using computer based
numerical reservoir simulation at dimension scales considerable compared to the
scale at which the physics of immiscible displacement takes place. In this chapter we
will review and consider some of the important properties important to predicting the
displacement and then examine analytical techniques which can be used to provide
predictions of behaviour in immiscible displacement processes. The methods presented
are not intended to displace using reservoir simulation but as an encouragement to
use the methods to understand the contribution of the various parameters involved
rather than blindly use the numerical simulation results where the large sizes of the
‘grid blocks’ inevitably disperse behaviour which occurs at pore size, centimetre or
a few metres scale.
Most of the reservoir engineering texts cover this topic. The author considers that
Dake1,2 and the text of Chierici3 provide excellent detailed analysis of the topic
In the next sections; we will review some of the reasons for using water injection,
then review some of the basic properties used in prediction, derive the fractional
flow equation and then examine procedures used to determine the movement and
displacement of fluids within a reservoir.
2.The Reason for Water Injection
Water injection is the main intervention method used in reservoir development, primarily because of the associated recovery achieved and also the availability of the
injection fluid. Historically it is termed a secondary recovery process in recognition
Institute of Petroleum Engineering, Heriot-Watt University
of the application of using it after a reservoir has been depleted by its natural energy,
and the pressure has dropped below the bubble point. The perspective here is using
the injected water to displace some of the remaining oil and thereby recover more
oil. Water injection can have two benefits, and for this reason it being termed as a
secondary recovery process causes some confusion. Allowing a reservoir to fall
below the bubble point we have seen leads to solution gas drive and resulting low
recoveries. Keeping the reservoir above its saturation pressure by the injection of
another fluid maintains the energy of the process providing good well productivity
and more important keeps the reservoir fluid in single phase. This voidage replacement, pressure maintenance process using water injection has been common practise in major offshore oil sectors where there is plentiful supply of clean injection
material. Dake1, in his text, outlines the engineering benefits of water drive in the
context of the North Sea, a major area where water injection has been practised for
over twenty five years.
It should be recognised that with water injection come many technical challenges
not the least the fact that injected fluids will eventually arrive at the producing wells
where they will present a disposal challenge. Historically, returning the fluids from
where they came was the straightforward answer. With increasing concern of environmental contamination, disposal of produced oily water to the sea is being gradually
replaced by a recycle process where rather than a once through process, the water is
reinjected into the formation. This is termed produced water injection. Dake1 outlines
the following benefits of water injection:
2.1 Zone Isolation
Although a field might be supported by an active aquifer providing natural pressure
support, in some cases faulting within the reservoir structure, can result in zones
being isolated from pressure support, figure 1 . If no intervention was used the
zone would produce by its natural energy with a rapid loss of pressure and resulting
poor recovery resulting from solution gas drive. Also as we saw in the section on
water influx, in order to predict aquifer pressure support a considerable amount of
aquifer characteristics are required. The cost of collecting this information prior to
production are high and therefore it is not until oil production starts can the strength
of any aquifer be determined. In many offshore fields therefore water injection is
planned as part of the development since the associated facilities required and the
implication on platform design are such that a delayed decision to implement water
injection once production has started is very costly.
Immiscible Displacement
Zone Isolation
OIL
Sealing Fault
OIL
OIL
AQUIFER
AQUIFER
Figure 1 Zone isolation.
2.2 Permeability
A characteristic of a number of offshore oil producing regions, for example the North
Sea, is the moderate to high permeabilities, which enable production wells to be very
productive reducing the required number of wells. Since the major cost in offshore
production is the offshore structures then minimising well slots results in minimising
the number of platforms. Maintaining high productivity through pressure maintenance
can be obtained through water injection when good injectivities can be achieved.
2.3 Oil Viscosity
As discussed in the chapter on rock properties the displacing characteristics at pore
scale are influenced by the relative mobility of the two fluids, the fluid being displaced and the fluid displacing. The ratio of the mobility of the displacing fluid to
the displaced fluid is a ratio of Darcy’s law as applied to the system. The different
parameters being the permeability of the one fluid in the presence of the irreducible
saturation of the other, the end point relative permeability, and the viscosity of the
fluids. The mobility ratio for water displacing oil is expressed as, M where:
M=
krw
′ kro′
/
µ w µo (1)
In sectors like the North Sea, Dake1 points out that relatively low oil viscosities lead
to high flow rates and the favourable oil viscosity compared to water gives a mobility ratio for some North Sea reservoirs of less than 1. This means that at least at
microscopic level the water cannot move faster than the oil and therefore displaces
the oil in a piston like manner. If M is greater than one, the case where oil viscosities are higher, then the higher velocity of the water causes an increasing instability
and water fingers through the oil and breaksthrough early compared to piston like
behaviour. The behaviour is illustrated in the sketch below, Figure 2. As pointed
out this behaviour only relates to the microscopic scale, and at reservoir scale the
various heterogeneities and the influence of gravity will have a big impact on the
reservoir flooding behaviour.
Institute of Petroleum Engineering, Heriot-Watt University
Impact on Resistability Ratio on horizontal Displacement
Injection
Production
Water
Oil
M<1
Stable Displacement
Injection
Production
M>1
Unstable Displacement
Figure 2 Impact of mobility ratio on horizontal displacement.
2.4 Undersaturated Reservoirs
As pointed out in the material balance chapter, when a reservoir is above its bubble
point, i.e. it is undersaturated, then if there is no pressure support from an aquifer
the pressure declines rapidly. This pressure decline can be detected using pressure
surveys in an open hole well as the dynamic behaviour of the reservoir is reflected
in the various layers making up the formation. This is demonstrated in the context
of the Montrose field3. where following production the pressure depth profile was
determined for successive development wells through the oil zone and basal aquifer.
The pressure profile did not follow the original water pressure gradient established
during the evolution of the field but reflected the permeability variation and
communication between the various sand layers, figure 3 As more development wells
are drilled pressure surveys continue to confirm the layering of the formation ,figure
4 This powerful application of pressure surveys to determine the communication
characteristics of a reservoir enables waterfloods to be planned and simulated much
more effectively.
Immiscible Displacement
Gr%
0 100
Sw%
100 0
Reservoir pressure - psig
θ%
0 50
2500
3000
3500
4000
Top paleocene
Layer 1
Layer 2
True vertical subsea depth - metres
2500
8100
Original
pressure
gradient
8200
Layer 3
8300
2550
8400
Layer 4
8500
2600
8600
Layer 5
2650
8700
True vertical subsea depth - feet
Perforations
8800
14
24
18
22
16
20
Reservoir pressure - MPa
26
Figure 3 Pressure depth profile for montrose field well.3
2.5 Overpressured Reservoirs
In those areas where reservoirs are overpressured ( chapter 2), the overpressure provides extra energy support. This additional pressure enables high production rates
in the early time period and also information from the reservoir of the dynamics of
the various units making up the system particularly if pressure depth surveys are
carried out during this period. The overpressure provides an opportunity to ‘feel’ the
reservoir during the early production period without the reservoir dropping below
the bubble point and reducing oil recovery.
Institute of Petroleum Engineering, Heriot-Watt University
Reservoir pressure - psig
3000
3400
3200
A15 A11
A17
A18
2500
A6
A8
8000
Original
pressure
gradient
8100
8200
8300
2550
8400
8500
2600
8600
2650
symbol
2700
18
?Well number
22/17-A6
A8
A11
A15
A17
A18
20
Date
05/04/77
27/01/78
20/12/77
15/08/78
02/11/78
28/03/79
26
22
24
Reservoir pressure - MPa
8700
8800
True vertical subsea depth - feet
True vertical subsea depth - metres
2450
8900
28
9000
Figure 4 Pressure depth profiles for Montrose Field wells.3
2.6 Reservoir Depth
The cost of offshore production facilities are such that it is important to maximise
the functionality of each platform. If waterflooding is carried out then the water
flooding wells are generally at the extremities of the formation. The wells slots on the
platform therefore have to be capable of reaching these limits, figure 5. The deeper
accumulations provide the drillers with an easier task to reach the outer limits of the
reservoir using, deviated and vertical wells. The application of horizontal wells in
recent years also enables shallower accumulations to be reached.
Immiscible Displacement
Sea Level
Sea Bed
Crestal
Producer
OIL
WATER
Injector
Figure 5 Application of deviated wells from one structure to reach limits in the reservoir.
2.7 Facility Design
In planning water injection, at least two important considerations are required. The
injection perspective; where should injection take place in relation to the various zones
of the formation and the ability to inject in relation to formation characteristics, and
secondly and equally important the time and associated cost of handling the water
when it eventually arrives at the production wells.
2.8 Thermal Fracturing
It is beyond the scope of this text to go into detail, but in recent years, as experience
in large waterflood operations has been obtained, new insight is developing on how
reservoirs have reacted. Of great significance is the phenomena of thermal fracturing.
In waterfloods where large injection flow rates are required large pumps are utilised
which can handle the necessary capacity. In many offshore zones where the injection
water is cold, the reduction in temperature around the injection well reduces the
natural fracture gradient and the pumps capable of overcoming the resistance of flow
through the formation generate a pressure greater than the fracture gradient causing
the formation to successively fracture. This generates a high surface area for flow
and therefore injectivities have been maintained compared to those expected from
predictions using simple radial flow around a well. This thermal fracturing phenomena
enabling good injectivities to be maintained is causing some companies to consider
using forced fracturing associated with water injection where temperature gradients
in warmer regions or with warm injection fluids will not reduce the natural fracture
gradient. Such could be the case when reinjected produced water is being used. Institute of Petroleum Engineering, Heriot-Watt University
Although good injectivities can be achieved due to fracturing a greater understanding
of the stress sensitivity of the formation is required. The fracture will follow the
natural direction according to the natural stresses and strength characteristics of the
formation. A concern is that such fractures will cause the injected water to by-pass
the desired flood front and cause premature water breakthrough. Figure 6.
Thermal
Fracture
Producer
Produces
Premature
Water
Breakthrough
Injector
Desired Water
Injection Flood
Front
Figure 6 Impact of fracture on water injection flood profile.
2.9 Water Handling
The handling of water is a major technical challenge in the oil industry particularly in
offshore operations, where many operators as fields mature find themselves handling
more water than oil. This technical challenge is also increasing as water disposal
options in relation to reducing oil emissions become more limited. Those involved in
providing associated production and treating facilities require important information
from the reservoir engineer. A schematic layout of a typical offshore water injection
scheme is shown in figure 7. Some key information required is, when will water
breakthrough to the producing wells, and how much water will be increasingly be
produced? The water handling facilities required are not insignificant and therefore
good forecasts are important. A more demanding challenge to the reservoir engineer
and outside the scope of this text is how can we manage the reservoir to reduce water
production.
10
Immiscible Displacement
WATER INJECTION DESIGN
Injection
Water
Treatment
Gas/Oil/Water
Separation
Injection
Pump
Gas-sales or
re-inject
q o Sales
Treat/
qw Dump/
Inject
Sea Level
q o + qw
qi
Seawater
Sea Bed
Reservoir
Figure 7 Schematic of ffshore facilities for Water Injection.
Dake2 points out an equation which links the reservoir engineer to the production
engineer which is:
qwi = qoBo + qwpBw (rb/d)
(2)
where qwi
= injection rate (assume Bw = 1)
qo + qwp = produced fluids requiring separation
qwp
= produced water for disposal or reinjection
He points out that this simple equation is fundamental to the process.
In water injection, the injection rate, qwi is maintained constant, since this is the drive
in water drive, and is therefore under engineering control. The right hand side, the
fluids requiring facilities for treatment are under the control of the reservoir. Dake
points out that this equation is not just a statement of material balance but it can be
regarded as a ‘platform equation’ since it contains the key elements associated with
topside capacities.
If water breaksthrough prematurely then, since the water injection rate has to be
maintained to maintain the reservoir pressure, there is an inevitable reduction in oil
production.
Institute of Petroleum Engineering, Heriot-Watt University
11
Equations were presented by Dake2 which he used to illustrate the impact of these
reservoir considerations on production capacities. The injected water in the reservoir
provides two functions maintaining pressure and displacing oil. Until breakthrough,
only oil is produced, after water breakthrough an increasing watercut occurs.
This watercut or fractional flow is defined as:
fws =
qwp
qo + qwp where ‘s’ denotes surface conditions.
(3)
Expressing the equation in terms of water production and substituting in (2) gives:
fws
1 − fws (4)

B f 
qwi = qo  Bo + w ws 
1 − fws  
(5)
qwp = qo
and
In his text Dake2 gives an example of the use of these equations to highlight the
commercial impact underestimating water breakthrough and the serious impact of
not being able because of platform limitations to increase water handling facilities.
Behind these commercial calculations is the importance for the reservoir engineer to
predict the producing watercut as a function of oil recovery.
In this next section we will review the basic parameters which are used to predict the
displacement process and then present the basic theory of water drive presented by
Buckley and Leverett over 20 years ago. The theory is a combination of behaviour
at the microscopic scale and that at a macroscale and then is applied at a reservoir
scale.
3 BASIC WATERDRIVE THEORY
3.1 Introduction
Before examining the various methods used in predicting the behaviour of reservoirs
under a constant injection process, such as water drive or gas injection, we will review some of the important basic properties relevant to the application. The method
presented is applicable to both water injection and gas injection where an immiscible
displacement process occurs. An immiscible displacement process is where there is
no mixing of the respective injection and displaced phases at the pore level through
mass transfer of components. This is distinguished from a miscible displacement
process where the injected phase mixes with the displaced phase by mass transfer
of the components from the respective phases, for example in a CO2 enhanced oil
recovery process.
12
Immiscible Displacement
As in many reservoir engineering processes we are combining properties, measurements
and application over a huge range of physical scales. Such an example of this is in
immiscible displacement calculations in oilfield oil recovery predictions. It is important
to keep this relative scale perspective in mind so as not to make an unrealistic ‘jump”
in application of data beyond its significance.
In water-oil displacement considerations we are dealing with a process which takes
place at a range of scales. At pore level or the microscopic scale, where the isolation
and movement of the respective phases is dependant on fundamental properties such
as; interfacial tension, wettability, viscosity, pore size and shape to name the obvious. At a significant larger scale, the macroscopic scale, we measure behaviour and generate
properties at the laboratory level where fluid movement and displacement are examined
at core plug scale, such as permeability, relative permeability and capillary pressure. The field scale,or behavioural scale, where the impact of characteristics at another
quatum leap level of scale will impose behaviour on those measured at microscopic
and macroscopic scale. For example the heterogeneous characteristics of the various
layers of the formation giving rise to different mobilties within the layers and the
large thicknesses of the layers resulting in vertical segregation perspectives.
An illustration of these different perspectives is shown in figure 8, where the oil water displacement process is illustrated at two, microscopic, and reservoir behaviour
scales. This scale up perspective is considerable and should not be forgotten, if not
‘giant leaps of faith” might be made using data beyond its range of applicability. The
engineering of sub surface behaviour such as a water injection process can be compared to the engineering of an oil refining plant. In the later, the process takes place
in vessels and pipes of centimeters and metres size over an area of a some hectares.
In a reservoir, the pipes and vessels, ‘the pores” are of micron dimensions and are
considerable in number to cover depths of hundreds of metres with an area perhaps
of tens of square kilometres.
Over recent years, considerable effort has been put into scale-up considerations
in relation to reservoir simulation, where rock properties at microscopic level can
be combined with geological characteristics at various scales to provide greater
confidence in field scale predictions. This topic is covered in the Geology and
Reservoir Simulation modules .
3.2 Water-Oil displacement at Microscopic and Macroscopic levels
Figure 8(b) illustrates the remaining oil at microscopic level following displacement by water. This remaining or residual oil is held by the competition between
the interfacial tension forces and the viscous flow forces associated with fluid flow. This topic was covered briefly in the rock chapter, when the pore doublet model was
presented, explaining how the continuous phases of oil is broken leaving oil ganglia
held by capillary forces . The residual oil saturation, Sor, in the water swept rock
can be in a range of 10-40% of the pore space. Figure 7(a). At the field displacement level the nature of the reservoir formation and well locations causes some of
the rock to be unswept by the water. These leads to two residual oil saturations, in
the swept portions oil at residual oil saturation and in the unswept portions oil at
original oil saturation.
Institute of Petroleum Engineering, Heriot-Watt University
13
Swept
Zone
(a)
Water
Oil
Rock Grains
(b)
Water
Injection
Wells
Oil at Residual
Oil Saturation
By-Passed
Oil
Oil at Original
Oil Saturation
Oil
Producers
Figure 8 (a) microscopic displacement (b) Residual oil remaining after a water flood.
This microscopic behaviour illustrates its effects in macroscopic properties of relative
permeability, and capillary pressure curves.
3.3 Relative Permeability
Permeability is a macroscopic property of the rock describing its resistance to flow
in terms of fluid velocity, fluid viscosity, pore size and shape, and pressure gradient.
This flow resistance term comes from Darcy’s law:
u=
Q k Dp
=
A µ Dl
in relation to figure 9.
14
Immiscible Displacement
DP
Superficial
Fluid Velocity
A
Q
distance
Figure 9 Darcys' law for permeability
This equation is for single phase flow only and does not apply to flow resistance when
two phases ( for example oil and water) are present. For this purpose the concept of
relative permeability is used, which is a measure of the permeability of one of the
phases and is a function of the phase saturations.
For example the relative permeability of water, krw is expressed as follows;
krw =
kew
k (7)
where kew is the effective permeability to water calculated from Darcy’s law when
oil and water are present, and k is the absolute permeability (single phase).
Darcy’s Law in linear flow for the two fluids allowing for gravity effects in an inclined
configuration, figure 10, is
qo = −
kkro A  ∂po

+ ro gSinθ 


µo  ∂x
qw = −
kkrw A  ∂pw

+ rw gSinθ 


 µw
∂x
Institute of Petroleum Engineering, Heriot-Watt University
(8)
15
Crossectional Configuration of Water Injection in a Reservoir
qt
(Cross section view)
qi
h
qo
qw
y
x
z
ø
X
w
O
X
(Plan view)
Production
Injection
O
X
O
Figure 10 Configuration of water injection in a reservior
The relative permeabilities are a function of saturation and reflect the surface, and
wettability forces of the fluid-rock system. An example of relative permeabilitiy
curves for a water oil rock is given in figure 11.
k'ro
End point
relative
permability.
k'rw
kr
0
Swc
Irreducible
Water
Saturation
Sw
Water Saturation
1-Sor
1.0
Residual
Oil
Saturation
Figure 11 Relative permeability curves for an oil-water system
16
Immiscible Displacement
Identified on the curves are the two conditions at the limiting saturation of the respective phases, the end point relative permeabilties for oil and water k'ro and k'rw.
k'ro – the relative permeability to oil in the presence of irreducible water saturation
and
k'rw– the relative permeability to water in the presence of residual oil saturation.
Dake2 reminds his readers that rock relative permeabilities are obtained from one
dimensional core flooding experiments, where often a cleaned core is flooded with
water and then the oil displaced with water. Two types of experiments are then
used. A viscous displacement of oil with water or a steady state experiment where
co-injection of water and oil at increasing ratios of water to oil.
Dake2 also notes that the relative permeability data, used in subsequent reservoir engineering calculations are unlikely to be representative of field characteristics. They
have probably been carried out at flow rates orders of magnitude higher than in the
reservoir, often using a synthetic oil not necessarily representative of the reservoir
fluid, and with wetting characteristics probably different than in the reservoir. In the viscous displacement experiment the injected water, starting at the irreducible
connate water level, Swc, where the water is immobile, generates increasing saturations in the core as a result of displacing oil. This increases until the saturation in
the core, where there is no more oil mobile in the core and the saturation to water, is
1-Sor, where Sor is the residual oil saturation.
If we express the volume of the pores in the core plug as the pore volume, PV, then
the oil displaced from the core flood experiment, is the movable oil volume, MOV,
which is;
MOV = (1- Sor - Swc) PV
The importance of end point relative permeabilities was presented earlier in this
chapter in the context of mobility ratio, M, where ;
M=
krw
′ kro′
/
µ w µo (1)
At the end point conditions this represents the maximum velocity of the water flow
compared to the maximum velocity of the oil.
3.4 Fractional Flow
Considering flow in a core plug or a reservoir, the ratio of the flow of water at any
point is termed the fractional flow ,fw ,where:
fw =
qw
qw + qo Institute of Petroleum Engineering, Heriot-Watt University
(9)
17
The oil rate qo can also be expressed as;
qo = qt - qw
If the Darcy equations for water and oil are subtracted (using field units P in atmos.)
the equations become;
 µ
µ  qµ
Drg sin θ 
 ∂P
qw  w + o  = t o + A c −

 ∂x 1.0133 × 10 6 
 kkrw kkro  kkro
(11)
where;
∂Pc ∂po ∂pw
=
−
∂x
∂x
∂x which is the capillary pressure in the direction of flow and,
Dr = rw - ro is the density difference of water and oil.
If values for flow rates using Darcy’s Law are now substituted in fraction flow equation
(equation 9) it becomes;
fw =
1+
kkro A  ∂Pc
Drg sin θ 
−



qt µo ∂x 1.0133 × 10 6 
k µ
1 + ro w
krw µo
(12)
Dake2 has also simplified this equation as;
fw =
1− G
µ k
1 + w ro
µo krw where G is a positive gravity number; G = 4.886 × 10 −4
kkro ADrg sin θ
in field units.
qt µo
(13)
(14)
where ∆ρ is the specific gravity difference relative to water.
The above term not only considers gravity effects but also includes a velocity term
,v, which is qt/A
The impact of the various components of this equation is worthy of consideration.
The angle of dip. If water is being injected downdip than the gravity term, ∆ρgsinθ/
1,0133x106 will be positive, reducing the fractional flow of water and it would be
positive for gas being injected downdip in a gas displacing oil senario. The density
difference in gas displacing oil systems is larger and therefore the significance is
greater. If the dip angle is zero, ie. horizontal flow, then the gravity term is zero.
18
Immiscible Displacement
The impact of capillary pressure, is illustrated from the slope of the capillary pressure and saturation with distance curves, figure 12 since;
1-Sor
Pc
- dPc
Sw
+ dSw
- dSw
Swf
+ dX
Swc
Swc
Sw
1-Sor
X
Figure 12 Capillary pressure curve and saturation distribution as a function of distance
∂Pc dPc ∂Sw
=
⋅
∂x dSw ∂x
i.e. the capillary pressure term is also positive increasing the fractional flow, for a
water displacing oil system as the two function gradients are negative. The capillary
pressure term is often neglected because the saturation with distance profile is unknown
being the objective of the displacement calculation, which we will consider later.
Velocity. This velocity is the superficial velocity, the rate divided by the cross sectional
area,A. The actual velocity is larger because of the impact of porosity. The impact
of velocity is small. Dake2 notes that the value for G for an edge water drive ,typical
of the North Sea, is 0.22kro and a comparative bottom water drive is 10.29kro. This
demonstrates the stability of the bottom water drive, where piston like displacement
will inevitably occur.
If both the angle of dip, and capillary pressure effects are neglected the fractional
flow equation becomes;
fw =
1
µ k
1 + w ro
µo k rw
(15)
The fractional flow equation enables a fractional flow versus saturation curve to be
generated from relative permeability data. This curve is influenced by a number of
parameters not least the viscosity of the respective phases. Its shape varies but can
have a shape as given by figure 13 below.
Institute of Petroleum Engineering, Heriot-Watt University
19
1.0
ƒw
Swc
Sw
1-Sor
Figure 13 Fractional flow curve
EXERCISE 1
Plot the water-oil relative permeability from the following data set. Indicate the end
point relative permeabilities.
Sw
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
Krw
0.0000
0.0021
0.0095
0.0210
0.0347
0.0536
0.0788
0.1050
0.1386
0.1785
0.2184
0.2636
0.3150
Kro
0.8800
0.6710
0.5170
0.4070
0.3135
0.2420
0.1793
0.1320
0.0891
0.0550
0.0297
0.0110
0.0000
An interesting presentation was given by Mayer-Gurr4. as illustrated in figure 14
where the capillary pressure, relative permeability and fractional flow curves are
presented. The impact of various well locations are considered.
20
Capillary Pressure Height
Above Free Water Level
Immiscible Displacement
Oil & Connate Water
C
0
50
100
% Relative Permeability
A
Transition Zone
100
B
Free Water
Level
100% Water
k'rw
k'ro
50
kro
0
krw
50
100
50
100
% Water Cut
100
50
0
% Water Saturation
Figure 14 The relationship between capillary pressure, relative permeability and fractional flow in a reservoir4
The capillary pressure curves represents the transition zone saturation profile associated
with the advancing imbibition process as a result of water injection. If a well is located
at A, the well will only produce oil since although the water saturation is 10%, the
relative permeability to water is zero. At B, the 45 % saturation level the well will
produce both water and oil with a water cut of 50%. At location C, the advancing
water has isolated an irreducible oil saturation and the well produces only water.
The redistribution of the saturation profile giving rise to a height saturation function
is called vertical equilibrium, and depends on a number of factors, including; a large
vertical permeability, small reservoir thickness, a large density difference between
the injected and displaced fluids, high capillary forces, low fluid viscosities and
low injection rates. It is not the intention of this chapter to present the associated
procedures, which would be part of a full numerical simulation analysis.
Institute of Petroleum Engineering, Heriot-Watt University
21
Case
Oil Viscosity
"µo" (cp)
Water Viscosity
"µw" (cp)
1
2
35.0
4.5
0.5
0.5
3
0.4
1.0
EXERCISE 2
Water is to be injected into a horizontal core, with the relative permeability characteristics of table 1, to displace oil. Determine the mobility ratios, and the fractional
flow curves for the following three cases.
4.DISPLACEMENT THEORIES
4.1 Introduction
To model the displacement process a number of theories have been successfully applied. These theories are aimed at providing the important predictions of reservoir
performance including the proportion of hydrocarbons recovered. In the methods
presented , there are a number of assumptions.
The displacement is incompressible, which implies that steady state conditions exist,
that is the pressures within the reservoir at any point remain constant.
This will occur if, the following reservoir flows exists;
qt=qo+qw=qi
where
qt = the total flow rate in reservoir volumes/time.
qo = the oil flow rate in reservoir volumes/time.
qw = the water flow rate in reservoir volumes/time.
qi = the water injection flow rate in reservoir volumes/time.
Diffuse flow conditions exist. Diffuse flow means that the saturations at any point
in the direction of linear displacement are uniformly distributed over the thickness.
This diffuse flow assumption enables a one dimensional simple analysis to be used
for the displacement modelling. In a simple core flooding relative permeability test
such an assumption is not unreasonable. Diffuse flow can also be encountered in a
reservoir where the injection rates are high preventing the establishing of vertical
equilibrium and for low injection rates where the thickness of the reservoir is small
compared to the thickness of the transition zone.
22
Immiscible Displacement
4.2 Buckley- Leverett Theory
The theory that has established itself in reservoir engineering for displacement calculations is that by Buckley and Leverett in 19425. Their theory is for linear, immiscible,
one dimension displacement, in which the total flow rate is constant in every cross
section, (incompressible). The theory determines the velocity of a plane of constant
water saturation moving through a linear system, such as a core in a water flood test.
Figure 15. The theory is well founded on the conservation of mass principle.
dx
Water
qw rw x + dx
qw rw x
A
Water
+ Oil
ø
Porosity
L
Figure 15 Mass flow through a linear core.
Consider the linear system in which water is displacing oil. The systems has a porosity
of φ and we are considering the principle of conservation of mass around a volume
element of length, dx. Therefore;
Mass flow rate in –mass flow rate out =rate of increase of mass in the volume.
qw rw x − qw rw x + dx = Aφdx
∂x
( rw Sw )
∂t
(16)
or
∂
∂


qw rw x −  qw rw x + (qw rw )dx = Aφdx ( rw Sw )


∂x
∂t
(1)
This becomes
∂
∂
qw rw ) = − Aφ ( rw Sw )
(
∂x
∂t
(17)
Since we are assuming incompressible flow, ρw is a constant. Therefore;
∂qw
∂S
= − Aφ w
∂x t
∂t x (18)
The differential of water saturation is
Institute of Petroleum Engineering, Heriot-Watt University
23
dSw =
∂Sw
∂S
dx + w dt
∂x t
∂t x (19)
We are examining the advancement of a particular saturation value. Since Sw is
constant dSw=0.
Then
∂Sw
∂S dx
=− w
∂t x
∂x t dt sw (20)
Also
 ∂q ∂S 
∂qw
=  w ⋅ w
∂x t  ∂Sw ∂x  t (21)
Inserting equations 20 and 21 in equation 18 gives;
∂qw
dx
= Aφ
∂Sw t
dt Sw (22)
For incompressible flow, the total injection rate, qt is constant, and the water flow
rate is the total rate times the fractional flow, qw=qt x fw. Rearranging equation 22
therefore gives:
v sw =
dx
dt
q t ∂fw
s w aφ ∂sw
=
sw
(23)
where vSw is the velocity of the plane of saturation, Sw.
This is the Buckley-Leverett equation, and is also the equation of characteristics.
It indicates the velocity of a plane of saturation moving through the linear system.
It enables the calculation of Sw as a function of time and distance and indicates its
dependance on the derivative of the fractional flow curve.
Clierici6 has presented a very thorough analysis of the displacement process for three
fractional flow curves.
In understanding the use of the equation it is important to appreciate the initial
boundary conditions, for our injection process. These are;
Sw = Swi for 0 < x ≤ L,t = 0
Sw = 1 - Sor for x = 0, t ≥ 0
24
(24)
Immiscible Displacement
That is the system is at its initial connate water
If the initial conditions at t=0 are applied to the general equation;
vSw =
dx
q ∂fw
= t
dt Sw Aφ ∂Sw
Sw
and the equation is then integrated a general solution to the displacement process is
obtained which enables the calculation of Sw in terms of x and t.
[ x(S )]
w
t
= x0 ( Sw ) +
qt  dfw 

 t
φA  dSw  sw
(25)
This equation describes a series of straight lines, the characteristics, with an initial
qt  dfw 


φA  dSw  Sw
ordinate value of x0(Sw) and a slope of
Clierici considers three cases
Case 1 For viscous oils
In this case the viscosity of the displaced phase, the oil, is considerably greater
than the injected water phase. The fractional flow curve has a concave downward
shape, figure 16A and its gradient fw′ increases from Sw=1-Sor to a maximum value
at Sw=Swi+∆Swi. Figure 16B
x
1
A
C
Swi
fw
+
S iw
0
B
dfw
dSw
0 Swi S1 S2
1
w
dfw
dSw
Swi+ DSw
S1
xo(Siw)
S2
xo(S1)
Sor
�S
qt
Αθ
1 - S or
xo(S2)
0
0
t
Sw
Figure 16 Displacement of viscous oil by water6. A = Concave downwards fractional
flow curve B = Velocity of water saturation C = Characteristics of water saturations Sw
The velocity of saturation is therefore maximum where Sw is just greater than Swi
and decreases to a minimum at Sw=1-Sor . Figure 16C, The progression of water
profiles are shown in figure 17 and shows the fraction of water at breakthrough at the
Institute of Petroleum Engineering, Heriot-Watt University
25
producing end. As can be seen the breakthrough saturation is just greater than Swi and
explains why for a very viscous oil breakthough occurs with low water saturations
and then gradually increases until the saturation reaches an unacceptable level.
Water
Injector
1
Producer
Sor
Sw
t=0
0
t1
t2
tBT
Swi
0
L
X
Figure 17 Progressive saturation profile for a concave downwards fractional flow curve6.
Case 2, Very Light Oils
In this case when the oil is very light with a low relative viscosity and large gravitational effects for example with a highly dipping structure, and with very low velocity,
a concave upward fraction flow curve is generated, Figure 18A, resulting in a fw′ curve
decreasing from its value at Sw=1-Sor to a minimum value at Sw=Swi , figure 18B
x
1
C
A
fw
�Sw
S wi+
xo(Swi)
S1
0
B
dfw
dSw
S2
xo(S1)
0
Swi S1 S2
1
dfw
dSw
Swi+ DSw
1
-S
or
xo(S2)
Sor
qt
Αθ
0
0
t1
t2
t3 t
Sw
Figure 18 Displacement of oil by water for a concave upwards fractional flow curve (light
oil displacement).6 A = concave upwards fractional flow curve. B = velocity of water saturation. C = characteristics of water saturations Sw.
26
Immiscible Displacement
The implications of this are that the highest velocity is for the highest water saturation,
Sw=1-Sor and that saturations less than this cannot exist since they would be overtaken
by the Sw=1-Sor saturation. Figure 18C, There is therefore a quick build up of a shock
front with a saturation, Swf=1-Sor . The producing characteristics are shown in figure
19, where , until the shock front arrives water-free oil is produced and thereafter only
water is produced. The oil remaining in the reservoir with a saturation of Sor .
Water
Injector
1
Producer
Sor
Sw
0
t2
t1
t=0
t3
tBT
Siw
0
L
X
Figure 19 Progressive saturation profile for a concave upwards fractional flow curve.6
Case 3 Typical medium density oils.
Figure 20A presents the fractional flow curve for a medium density and viscosity oil,
where the displacement velocities are not unlike field values. The S shaped curve
generates the two curvatures we have considered in case 1 & 2. With the corresponding derivative values, fw' . The slope in the fw curve increases from its starting value,
Sw=1-Sor and then decreases. Figure 20B
x
1
C
A
fw
xo(Swi)
xo(S1)
0
B
dfw
dSw
S wi+ �Sw
S1
S wf
S2
xo(Sw,f)
xo(S2)
Sor
0
Swi S1 S2
1
xo(1-Sor)
0
0
Sw
1 - S or
t1
t2
t3
t
Figure 20 Displacement of oil by water for a rock with an S-shapeed fractional flow
curve (light oil displacement). A = S shaped fractional flow curve. B = velocity of water
saturation. C = characteristics of water saturations Sw.6
Institute of Petroleum Engineering, Heriot-Watt University
27
The development of the saturation, would be such that their would be a steady increase
in the velocity of the increasing saturation, but this would reach a maximum at a
saturation Swf, where Swi<Swf<(1-Sor). Behind this the velocities would decrease with
decreasing Sw, figure 20C.
.
The impact on the process is such that a shock front is developed, at the value Swf, the saturations greater than this moving at a lower velocity, behind this shock front
there is a steady increase in the saturations moving at decreasing velocity. This
process is illustrated in figure 21, which shows that water free oil is produced until
breakthrough at a saturation of Swf,and a breakthrough fractional flow of fwbt . The
saturation then climbs until it reaches the irreducible oil saturation level when only
water is produced.
Water
Injector
1
Sor
Sw
0
Producer
t1
t=0
t2
Siw
t3
tBT
Sw,f
0
L
X
Figure 21 Progressive Saturation for an S-shaped Saturation Curve6.
The velocity of the stabilised shock front can be calculated from a material balance
across the front. Chierici4 explains this using figure 22. He designates R to represent
conditions ahead of the front and L those behind. Firstly for case two, piston like
displacement
Front at time t
q w,L
S w,L
Front at time t + dt
q w,R
dxf
S w,R
x
Figure 22 Conditions ahead (R) and behind (L) Water Front6
28
Immiscible Displacement
If the velocity of the front vf=dxf/dt, then
qw,Ldt-qw,Rdt = Af(Sw,L - Sw,R)dxf
(26)
s
Since
qw= qtfw
vf =
dx f
q ( fw , L − fw , R )
= t
dt
Aφ ( Sw , L − Sw , R )
(27)
dx f ut ( fw , L − fw , R )
=
dt
φ ( Sw, L − Sw, R ) (28)
Since qt=Aut
Then
vf =
This is the Rankine-Hugoniot condition for the frontal velocity of shock fronts for
physical systems.
If we specify our limiting conditions then
Sw,L=1-Sor
fw,L=1
Sw,R=Swi
fw,R=0
Therefore:
(f
(S
w, L
w, L
− fw , R )
− Sw , R )
=
1
1 − Sor − Swi
= tan α
(29)
Where tan α is the angle on the Sw vs. fw curve joining (Swi,0) and (1-Sor,1)
Institute of Petroleum Engineering, Heriot-Watt University
29
1
fw
0
α
0
Swi
sw
1-Sor 1
Figure 23 Calculation of Front Velocity using Rankine-Hugonist conditions6
From the Buckley-Leverett equation, equation 23
vSw1− Sor =
dx
u ∂f
= t w
dt 1− Sor φ ∂Sw 1− S
or
Therefore from eqns. 23,28 & 29 we hav
vS1− Sor =
dx
u ∂f
u ( fw , L − fw , R ) ut
1
u
= t w
= t
=
= t tan α
dt 1− Sor φ ∂Sw 1− S
φ ( Sw, L − Sw, R ) φ 1 − Sor − Swi φ
or
(30)
If we apply in a similar fashion the Rankine-Hugoniot condition to case 3 for the
shock front we have for our limiting conditions ;
Sw,L=Swf
fw,L= fw(Swf)
Sw,R=Swi
fw,R=0
In equation 30, this gives us:
vSwf =
dx
u df
= t w
dt wf φ dSw
=
wf
fwf
ut ( fw , L − fw , R ) ut
u
=
= t tan β
φ ( Sw, L − Sw, R ) φ Swf − Swi φ
(31)
Examination of these equations generates a convenient graphical procedure to
determine the conditions at the shock front.
30
Immiscible Displacement
tan β =
dfw
dSw
=
wf
fwf
Swf − Swi
(32)
That is a tangent drawn to the fractional flow curve from the point (Swi,0) which meets
the curve at the conditions of the shock front. Figure 24 below.
1
fw(Swf)
dfw
tan β = dS S
w w,f
fw
0
β
Sw,t
Swi
1-Sor
Figure 24 Graphical Procedure for Determining the Conditions of the Shock Front
If we now consider the time, tbt it takes for this shock front to move though our linear system we generate a useful equation which we will use later in water injection
performance calculations.
tbt =
L Lφ Swf − Swi
=
vf
ut fw S
wf
(33)
where L is the distance from injector to producer.
4.3 Welge Analysis
In 1952 Welge(7) presented a method to obtain the average saturation behind the shock
front, which is useful in determining the oil recovery. Figure 25 gives the saturation
profile as the shock front breaks through at the producing end, a distance L from the
injection end.
Water is being injected at a rate qw
Institute of Petroleum Engineering, Heriot-Watt University
31
Time for breakthrough tbt
tbt =
L ALφ Swf − Swi
=
Vf
qw
fw
1-Sor
Sw
Sw
Swf
Swi
X1
X2
X
Figure 25 Water Saturation as a Function of Distance before Breakthrough
Before water arrives at the exit, the volume of oil produced is equivalent to the volume
water injected. Wi = qw x t. At breakthrough the volume of oil produced, Np is the
difference between the initial oil volume, (ALφ(1-Swi), less that remaining in terms
of an average saturation, Sw, at breakthrough, (ALφ(1-Sw))
Swf − Swi
= ALφ ( Sw − Swi )
fwf
(34)
Swf − Swi
fwf (35)
N pbt = qw tbt = ALφ
Therefore:
Sw − Swi =
from equation 32 this can be written as:
Sw − Swi =
1
 dfw 


 dSw  Swf
(36)
Combining equations 35 and 36 gives
1 − fwf = 1 −
Swf − Swi Sw − Swf
 df 
=
= Swbt − Swf  w 
Sw − Swi Sw − Swi
 dSw  Swf
rearranging this equation becomes:
32
[
]
(37)
Immiscible Displacement
Swbt − Swf +
1 − fwf
 dfw 


 dSw  Swf
(38)
This is Welge's equation for average saturation and and combining with equation
16 gives:
dfw
dSw
=
S wf
(1 − f ) =
w S wf
Sw − Swf
1
Sw − Swi (39)
There is also a graphical significance in the above equation. The line of the tangent
drawn previously at the breakthrough point cuts the line fw=1 at an x-axis of
.
The construction is illustrated in the figure 26
Sw
1.0
fwf
Swf, f
fw
Swi
Swf
wf
1-Sor
Figure 26 Determination of Average Saturation Behind Front
Institute of Petroleum Engineering, Heriot-Watt University
33
EXERCISE 3
Fractional Flow - Diffuse Flow Conditions
Oil is being displaced by water in a horizontal, direct line drive under the diffuse flow
condition. The rock relative permeability functions for water and oil are listed in table E.1. Pressure is being maintained at its initial value for which Bo = 1.36 bbl/STB
and Bw = 1.01 bbl/STB.
Data Summary
Oil formation volume factor
Water formation volume factor
Initial water saturation
Bo =
1.36 bbl/STB
Bw =
1.01 bbl/STB
Swc =
0.20
Draw the fractional flow curves and calculate the cumulative oil recovery at breakthrough for the following combinations.
Case
Oil Viscosity
"µo" (cp)
Water Viscosity
"µw" (cp)
1
2
35.0
4.5
0.5
0.5
3
0.4
1.0
4.4 Calculations for Oil Recovery
The objective of Welge’s work was to enable oil recoveries to be determined. We will
now develop the equations and methods to calculate oil recovery over the displacement
period both before water breakthrough and subsequent to it. We should not forget
at this stage we are discussing the displacement process and associated recovery in
a relatively small core plug, of oil by water.
In analysing recoveries it is convenient to express the volumes of fluids injected and
recovered in terms of pore volumes, (PV). For our core plug with a length L, cross
section area, A, and porosity, φ, the pore volume is.
PV = AfL.
(40)
Before water breakthrough, only oil exits from the core at a rate equivalent to the
rate of water being injected, since it is an incompressible system. At breakthrough
therefore the pore volumes of fluids involved are;
(
)
N pd bt = Wid bt = qid tbt = Sw bt − Swi (41)
Npdbt = pore volumes oil produced at water breakthrough. Widbt = pore volumes water
injected at water breakthrough
qid is the water injection rate expressed in pore volumes, ie.
34
Immiscible Displacement
qid =
qi
AφL (42)
tbt is the time taken for the water to breakthrough, which is;
tbt =
Wid bt
qid
(43)
Sw bt average water saturation at water breakthrough
At breakthrough , when x=L in the Buckley Leverett equation (equation 23), the
following convenient result exists;
xSwbt = L =
Wi dfw
Aφ dSw
S wbt (44)
(45)
Therefore ;
1
Wi
Wid =
= dfw
LAφ
dSw
S wbt
In equation 45 above therefore the oil recovery at breakthrough is also equal to the
inverse of the slope of the breakthrough characteristic, the slope of the line drawn from,
Swi, tangent to the fractional flow curve, figure 26. The oil recovery at breakthrough
is fully given by the equation below.
(
)
N pd dt = Wid bt = q id t bt = sw bt − swi =
1
dfw
dsw swbt
(46)
After breakthrough, the fractional flow saturation profile at the exit of the core increases as shown in figure until the irreducible oil saturation, Sor, is reached where
fw=1, as the flood front moves through the core. Figure 27
Institute of Petroleum Engineering, Heriot-Watt University
35
1 - Sor
Sw
Swe
Swbt= Swf
Swbt
X
0
L
Figure 27 Water Saturation profiles after breakthrough
From Welge’s equation 38 where the saturation value is now Swe, the average saturation remaining is now
(
Sw = Swe + 1 − fw
S we
) / dSdf
w
w S we
(47)
This can also be expressed, in terms of injected pore volumes of water, using the
Buckley Leverett relationship, so that
(
Sw = Swe + 1 − fw
S we
)W id
(48)
The oil recovered associated with this average saturation is given by;
(
N pd = s w −swi = (swe − swi ) + 1 − fw s
we
)W
id
(49)
Equation 47 also gives a useful construction to determine the average saturation.
Since;
dfw
dSw
=
S we
(1 − f )
(S
w
w S we
− Swe ) (50)
As for breakthrough the average saturation is the intersection at fw =1 of the tangent
drawn to the fractional flow curve at the exit fwe saturation value. Figure 28
36
Immiscible Displacement
Sw
fw= 1
Sw,bt
1-Sor
Sw
(1-fwe)
Swe,fwe
(Sw - Swe)
fw
dfw
dSw
Swbt
Swe
=
1-fwe
Sw - Swe
Figure 28 Welge construction for Determination of Average Water SaturationS after
breakthrough
The procedure for the oil recovery calculations is summarised below.
1. Generate a fractional flow vs. water saturation curve for the system to be studied,
using the appropriate relative permeability data.
2. Draw a tangent to the fractional flow curve from the initial Sw = Swi position at
fw=0.
At the point of tang