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Reservoir engineering handbook
Book · November 2017
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Contents
Chapter -1Reservoir rock properties

Porosity















Measurement of porosity
Direct measurement in the laboratory
Bulk Volume
Determination the bulk volume volumetrically
The Russell volumeter
Sand grain volume
Pore volume
Gas Expansion porosimeter
The Washburn-Bunting porosimeter
Boyle’s Law porosimeter
Saturation method
Mercury injection Method
Indirect Method
The sonic (acoustic) log
The formation density log

Rock compressibility
 Definition
 Significance
 Applications

Fluid saturation








Definition
Factor affecting fluid saturation of the cores
Determination of fluid saturation from a rock sample
Retort method
The other method of determining fluid
Saturation is by extraction with a solvent
Another method of determining water saturation is to use a centrifuge
There is another procedure for saturation determination that is used in conjunction with either
of extraction method.


Interstitial water saturation
Oil based Mud

Rock resistivity
 Definition
 Significance
[1]
Dr.Jawad R.R. Alassal Kirkuk U.
 Mathematical formulations

Permeability











Introductory theory
Flow system of simple geometry horizontal flow
Vertical flow
Radial Flow
Flow system of combinations of beds
Measurment of permeability
Perm plug method
Permeameter
Whole-core measurments
Factora affecting permeability measurments
Effect of Gas Slippage on Permeability measurement

Capillary pressure







Laboratory measurement of capillary pressure
Porous Diaphragm
Mercury injection
Centrifuge method
Dynamic method
Converting laboratory data
Averaging Capillary pressure data

Electrical Conductivity of Fluid Saturation Rock
Chapter -2Capillary Properties
 Saturation
 Wettability
 Surface and Interfacial Tension
 Capillary Pressure
 Capillary Pressure of Reservoir Rock
 Capillary Hysteresis
 Leverett J-Function
 Converting laboratory Capillary Pressure data
Chapter -3 Gas Properties
 Classification of gas
 Early gas laws
 Ideal gas mixture
[2]
Dr.Jawad R.R. Alassal Kirkuk U.







Apparent molecular weight
Real Gases
Real gas mixture
Correction of non hydrocarbon impurities
Gas formation volume factor
Gas compressibility
Gas viscosity
 Gas-Water System
 Solubility of natural gas in water
 Solubility of water in natural gas
 Gas hydrate
 Pressure-Temperature Diagram
 Gas Reservoirs
 Direct Calculatio of compressibility factor
Chapter -4 Reservoir fluid Properties




Single component system
Two component system
Three component system
Multi component system
 Classification of Reservoirs And Reservoir Fluids
 Properties of Crude oil systems








Crude oil Gravity
Specific gravity of the solution gas
Gas Solubility
Bubble-Point Pressure
Oil formation volume factor
Isothermal compressibility coefficient of cude oil
Crude oil density
Crude oil Viscosity
 Properties Of Reservoir Water
 Water Viscosity
 Pressure Volume Temperature (PVT)
Chapter-5 The Material Balance Equation (MBE)
[3]
Dr.Jawad R.R. Alassal Kirkuk U.
Preface
This book explains the fundamentals of reservoir engineering and their practical
application in conducting a comprehensive field study. Chapter 1 reviews Reservoir
rock properties. Chapter 2 Capillary Properties , while Chapter 3 presents Gas properties
and PVT. The Reservoir fluid properties discussed in Chapter 4 and Material Balance
Equation discussed in chapter 5.
Acknowledge
I express my thanks to the my best student in petroleum engineering
Shaho Abdulqader MohamedAli
who helped collect and organize this book ,and for some of my student
Ali Yahya jarjies
Sara Sajad
Mohammed Namiq
[4]
Dr.Jawad R.R. Alassal Kirkuk U.
Reservoir Engineer Should be familiar with these Abbreviations which are used in the
field
Definitions:
Work Over :
M.I = move in
RU = Rig up
W/O = work over
ND Xmass = nipple down (open) Christmas Tree
NU BOPS = nipple Up(Fix) blow out preventer
BPV
= Back pressure valve
RD
= Rig down.
M.O
= move out.
LD
= lay down
EOT
= End of tubing.
CSg
= casing.
Tbg.
= Tubing
WOU
= work over unit.
P/U
= Pick up.
POOH
= pull out of hole.
RIH
= run in hole.
GLMs
= gas lift mandrels.
RDMO
= Rig down and move out.
P.T
= Pressure test.
8Rd
= Type of teeth of the bit.
3STDS .DCs = 3 stands of Drill collar.
PBTD
= Plug Back Total Depth.
Pkr
= Packer .
D.O.C
= Drill out cement.
W.O.C
= Wait on cement.
sks
= sack
Injection Wells:
IAI = Air injection well.
IGI = Gas injection well.
IOI = Other fluid injection well.
ISI = Steam injection well.
[5]
Dr.Jawad R.R. Alassal Kirkuk U.
IWI = Water injection well.
Producing wells:
PAC = Producing through annulus .
PAG = Producing through annulus .( Gas)
PAL = Producing through annulus (by gas lift).
PAN = Producing through annulus ( natural flow)
PAP = Producing through annulus by pumping.
PBC = Producing condensate through tubing and casing (Both)
PBG = Producing Gas through tubing and casing (Both)
PBN = Producing through tubing and casing ( Natural flow).
PTC = Producing condensate through tubing.
PTG = Producing Gas through tubing.
PTL = Producing through tubing ( by gas lift ).
PTN = Producing through tubing ( Natural flow ).
Abandoned Wells :
AD = Abandoned Dry.
Injection Wells Shut – In :
AE = Abandoned Economic reasons.
SAI = Shut – in Air injection well.
AG = Abandoned high gas oil ratio.
SGI = Shut – in Gas injection well.
AM = Abandoned Mechanical reasons.
SOI = Shut – in Other Fluid injection well.
AO = Abandoned Other reasons.
SO = Shut – In other reasons.
AW = Abandoned high Water oil ratio.
SM = Shut – In order of Ministry .
Shut – In Wells :
SE
= Shut – In Economic reasons.
SW = Shut – In excessive water.
ST
= Shut – In Temporarily .
SG = Shut – In excessive gas.
SS
= Shut – In reservoir study.
SZ = Shut – In rezoning.
SSI = Shut – in Steam injection well.
SR = Shut – In work over.
SWI = Shut – in Water injection well.
SR = Shut – In well head repair.
SD = Shut – In dead.
[6]
Dr.Jawad R.R. Alassal Kirkuk U.
Duties of Reservoir Engineers :
1. Review previous geological, engineering and simulation studies.
2. Provide periodic reservoir performance analysis (including preparing of various of
isobaric maps and various production performance maps.
3. Prepare forecasting and update reserves.
4. recommended development – drilling locations in coordination with the area
geologist in order to increase the recovery factor and accelerate production of
proven reserves.
5. Follow up of drilling activities and prepare testing, coring and logging programs as
needed.
6. Perform log evaluation and recommend completion intervals for new wells.
7. Recommend workovers for shut-in well or producing wells to improve productivity.
8. Review injection rates and revise these rates to achieve the most practical effective
exploration for each reservoir.
9. Recommend various production and tests and perform the necessary analysis as
needed.
10. Recommend and follow – up of laboratory, engineering and simulation studies
required for each reservoir.
Familiar with various enhanced recovery techniques and potential application in
various reservoirs.
[7]
Dr.Jawad R.R. Alassal Kirkuk U.
Chapter -1Reservoir Rock Properties
The reservoir rocks are the porous media which containing the fluids in gas or liquid
form.
The properties of these rocks should be determined according to a certain order :
1. Storage Capacity (porosity)
2. Circulation Capacity (permeability)
3. Capillary pressure and Saturation
4. Electrical Properties.
Analysis of the core sample:
There are two types of analysis:
 Conventional Core analysis
A . For fresh or preserved samples (complete results)

Porosity

Permeability

Gas , water ,oil saturation
B. For weathered or extracted samples:

Porosity

Permeability
Conventional core analysis can be carried out on samples of very different sizes
1. On small samples collected at regular interval
2. On whole cores, especially in the cases of fissured deposit.
3. On samples from side wall cores.
 Special core analysis
1. The study of capillary pressure.
2. Measurement of formation factor and resistivity ratio.
3. Two phase flows and the study of relative permeabilities.
4. Study of wettability tests.
[8]
Dr.Jawad R.R. Alassal Kirkuk U.
Porosity
Porosity is defined as the ratio of the void space in a rock to the bulk volume (BV) of
that rock, multiplied by 100 to express in percent. Porosity may be classified according
to the mode of origin as primary and secondary. An original porosity is developed
during the deposition of the material, and later compaction and cementation reduce it to
the primary porosity. Secondary porosity is that developed by some geologic process
subsequent to deposition of the rock. Primary porosity is typified by the intergranular
porosity of sandstones and the intercrystalline and oolitic porosity of some limestones.
Secondary porosity is typified by fracture development as found in some shales and
limestones and the vugs or solution cavities commonly found in limestones (see Fig.2).
Rocks having primary porosity are more uniform in their characteristics than rocks in
which a large part of the porosity is induced. For direct quantitative measurement of
porosity, reliance must be placed on formation samples obtained by coring.
The total volume of a sample – termed the bulk volume – is the sum of both the
grain volume, Vg, and the pore volume, Vp.
Vb
=
Vp + Vg
Eq.1
Where
Vb :
bulk volume, cm3
This leads to a basic rock property called the porosity, which is defined as the fraction
of the bulk volume of a sample that is pore space. In mathematical terms,
= Vp / Vb
Eq. 2
Where
:
absolute porosity, fraction
Knowledge of the porosity of a sample of rock enables us to estimate its pore volume
since its bulk volume can be easily determined. If this sample is representative of the
reservoir, i.e., the reservoir rock is the same everywhere, then the pore volume of the
entire reservoir can be estimated by Eq. 2 , using the sample’s porosity. The bulk
volume of the reservoir is computed by simple arithmetic from its area and average
thickness. Since the maximum volume of fluids a reservoir can contain is equal to
its pore volume, porosity becomes a primary property that must be determined most
accurately.
[9]
Dr.Jawad R.R. Alassal Kirkuk U.
Unit cells of two systematic packings of uniform spheres are shown in Fig.1 the porosity
for cubical packing (the least compact arrangement) is 47.6% and for rhombohedral
packing (the most compact arrangement) is 25.96%. Considering cubical packing, the
porosity may be calculated as follows. The unit cell is a cube with sides equal to 2r
where r is the radius of the sphere. Therefore, Vb =(2r)3 = 8r3 where Vb is the bulk
volume. Since there are 81/8spheres in the unit cell, the sand-grain volume, Vs is given by
The porosity, Ø, is given by
where Vp is PV. Therefore,
(
)
= 47.6%
Fig.1—Unit cells and groups of uniform spheres for cubic and rhombohedral packing
Porosity is seldom constant within a reservoir; it varies with location and depth.
To compute the average porosity, a simple weighted-mean formula is applied:
̅
∑ ̅̅̅
Eq.3
Where
̅
̅
()
[10]
Dr.Jawad R.R. Alassal Kirkuk U.
()
The summation should include all sections of the reservoir.
Example
The contour map below depicts porosity variation within a reservoir. The lines
connect points of equal porosity. If the reservoir is 10 miles long , 3 miles wide, and
100 feet thick, its average porosity is estimated as follows:
There are 4 sections within the reservoir, the area and average porosity of each are
computed to be:
Section
Area (ft2)
Avg. Porosity (%)
1
155,399,147
12.5
2
277,261,208
17.5
3
122,322,723
22.5
4
14,303,684
25.0
Total
569,286,762
Since the reservoir thickness is uniform, section areas only are used in the computation.
̅
∑̅
= ( 9,904,413,847 / 569,286,762 )
= 17.4%
[11]
Dr.Jawad R.R. Alassal Kirkuk U.
(a) Secondary porosity by solution
(b) Secondary porosity by dolomitization of calcite
(c) Secondary porosity by fractures
Fig. 2: Examples of secondary porosity
[12]
Dr.Jawad R.R. Alassal Kirkuk U.
Measurment of porosity
Porosity can be measured by two methods
1. Direct methods in the laboratory.
2. In direct methods by using logs.
1.Direct meaturment in the laboratory
A.Bulk volume

linear measurements

displacement method
fluid displaced by sample measured volumetrically or gravimetrically
Prevent fluid penetration into pore space by:
i. coating the rock with paraffin
ii. saturating the rock with the fluid into which it is immersed
iii. using mercury
Determination the bulk volume volumetrically :
A varity of specially constructed pycnometer were used . An electric pycnometer
from which VB can be read directly (see Fig.3).
The sample is immersed in the core chamber , Which causes arises in the level of the
connecting U tube .The change in level is sensed by the micrometer scale .Either dry or
saturated samples may be used in the device.
[13]
Dr.Jawad R.R. Alassal Kirkuk U.
Fig.3—Electric pycnometer
The Russell volumeter (see Fig.4 ) . This provides for direct reading of VB A
saturated sample is placed in the sample bottle after a zero reading is established with
fluid in the volumeter . the resulting increase in volume is the bulk volume . Only
saturated or coated samples may be used in the device.
Fig. 4 Russell volumeter for determining grain and bulk volumes of rock sample
[14]
Dr.Jawad R.R. Alassal Kirkuk U.
B.Sand-Grain Volume (VG). The various porosity methods usually are distinguished
by the means used to determine the VG or VP. Several of the oldest methods of porosity
measurement are based on the determination of VG.
The VG may be determined from the dry weight of the sample and the sand-grain
density. For many purposes, results of sufficient accuracy may be obtained by using the
density of quartz (2.65 g/cm3) as the sand-grain density.
C. Pore volume
pore volume can be measured directly :
1.By measuring the air volume contained in the pores.
2. By weighting a liquid filling the pores
3. By mercury injection
All methods of measuring PV yield "effective" porosity. The methods are based on
either the extraction of a fluid from the rock or the introduction of a fluid into the pore
space of the rock.
 Gas expansion porosimeter
Another method utilizes gas expansion to fill the pore space of the sample, and it
requires a special instrument called a porosimeter (Fig. 5). In this method, the
sample is first placed in a chamber and placed under vacuum to remove all air within
it. Then gas, usually helium, is allowed to expand from a container of known volume
and initial pressure into the chamber. By application of Boyle’s law, the final
volume of the gas is computed from its final pressure. The increase in the gas
volume is equal to the connected pore volume of the sample plus the dead volume in
the chamber and tubing.
Gas Container
Pressure gauge
Valve
Core and core chamber
Fig. 5: Gas expansion porosimeter
[15]
Dr.Jawad R.R. Alassal Kirkuk U.
To vacuum pump
and gas sourc
 The Washburn-Bunting porosimeter
measured the volume of air extracted from the pore space by creating a partial vacuum
in the porosimeter by manipulation of the attached mercury reservoir. In the process,
the core is exposed to contamination by mercury and, therefore, is not suitable for
additional tests. The previously described Stevens method is a modification of the
Washburn-Bunting procedure especially designed to prevent mercury contamination of
the samples.
Fig .6 Washburn bunting porosimeter
A number of other devices have been designed for measuring PV; these include the
Kobe porosimeter, the Oil well Research porosimeter, and the mercury-pump
porosimeter. Kobe and Oilwell Research porosimeters are Boyle's-Iaw-type
porosimeters designed for use with nitrogen or helium with negligible adsorption on
rock surfaces at room temperature. The mercury-pump porosimeter is designed so that
the BV may be obtained as well as the PV.
The saturation method of determining porosity consists of saturating a clean dry
sample with a fluid of known density and determining the PV from the gain in weight
of the sample. The sample usually is evacuated in a vacuum flask to which the
saturation fluid may be admitted by means of a separatory funnel. If care is exercised
to achieve complete saturation, this procedure is believed to be one of the best
available techniques for intergranular materials. Example Problem 5 illustrates the
saturation technique of measuring PV.
[16]
Dr.Jawad R.R. Alassal Kirkuk U.
 Boyle's Law Porosimeter
A cleaned and dried sample is placed in a sample chamber that has an air pressure P1
(Figure 7, Boyle's Law Porosimeter, Grain Volume Determination).
P2
P1
2
V1
Figure. 7
When the valve that connects the sample chamber with the reference volume chamber
is opened, the air will expand isothermally and equilibrate on pressure P2. We further
assume that the volume of the sample chamber is V, and that the core sample has a
grain volume Vg. The sample chamber will therefore contain a volume V-Vg of gas. If
this gas is brought isothermally from pressure P1 to P2 by adding reference volume dV,
we can then apply the Law of Boyle and deduce:
[
]
Eq.4
[
]
( )
From this formula the grain volume, Vg and the resulting pore volume can be
calculated. Application of this technique results in a measurement of the porosity of
interconnected pores only. This means that isolated pores, which are common in
carbonate rocks, will not be accounted for. Additionally, by drying the samples, any
clay present will have lost its water content also leading to an inaccurate porosity
measurement.
[17]
Dr.Jawad R.R. Alassal Kirkuk U.
 Saturation method
Estimating Vg requires simple procedures and, usually, available equipment. On
the other hand, direct measurement of the pore volume provides a more accurate
porosity value. However, this requires some additional instruments. A simple
method starts with weighing the sample in air followed by placing the sample in a
vacuum flask (Fig. 8-a) for a few hours. Water is then introduced into the flask
gradually until the sample is completely submerged (Fig. 8-b). The sample is then
removed from the flask, shaken to remove excess water and then weighed quickly.
The increase in the mass of the sample is equal to the mass of water introduced into
the sample, and the volume of the water is equal to the connected pore volume. Care
must be taken to minimize water evaporation; and if the rock contains clay
minerals that absorb water, another liquid – oil, mercury, or KCl brine – must be
used instead.
liquid
reservoir
Vacuum gauge
Vacuum valve
(a) under vacuum
(b) saturated
Fig. 8: Pore volume measurement by the liquid saturation method
[18]
Dr.Jawad R.R. Alassal Kirkuk U.
Example
A sandstone core plug is 1 inch in diameter, 2 inches long, and has a mass of 56.6
grams. When completely saturated with water, its mass increases to 60.9 grams.
Compute the rock porosity.
Mass of water in sample = 60.9 – 56.6 = 4.3 grams
Volume of water = 4.3 / 1 = 4.3 cm3
Vb = π (0.5 x 2.54) 2 2 2.54
ϕ = 4.3 / 25.74 = 16.7%
= 25.74 cm3
note: The absolute porosity of the sample computed from its estimated grain volume is
17%.
 Mercury injection Method :
The principle consists of forcing mercury under relatively high pressure in to the rock
pores , An apparatus of the Ruska type is shown in Fig.9 .The volume pump gives
firstly VB and then the volume of mercury injected in term of pressure Fig.9 .The
pressure limit depends upon the tensile strength of the core sample.
Fig.9 Mercury injection method
2.Indirect methods
Indirect methods of estimating porosity rely on measurement of other rock and fluid
properties. These measurements are carried out in the well employing special
instruments as part of well logging operations. Therefore, no core samples are needed
and the porosity is estimated for the rock as it exists in the reservoir. Two of the most
common well logs are discussed below.
[19]
Dr.Jawad R.R. Alassal Kirkuk U.
 The sonic (acoustic) log
In this log, the instrument - the sonde - generates sound waves, which travel through
the reservoir - in the vicinity of the well bore - and are detected by the sonde
upon their return. The time lapse between generation and detection – travel time - is
recorded continuously versus the depth of the instrument. Since travel time is related
to porosity by
Eq.5
where
Δtlog : sound travel time in the reservoir as measured by the log, μs
Δtma : sound travel time in the grain material of the reservoir, μs
Δtf : sound travel time in the fluids of the reservoir, μs
and since Δtma and Δtf are usually known for the reservoir, the porosity can be
estimated at all depths.
 The formation density log
Another logging sonde emits gamma rays, which mostly pass through the reservoir
rock and fluids, but some are scattered back into the well bore and are detected by the
sonde . The fraction of scattered gamma rays is used to compute the bulk density rock and fluids - of the reservoir, which is related to porosity by
Eq.6
ρlog : bulk density of the reservoir as measured by the log, g/cm3
ρma : density of the grain material of the reservoir, g/cm
ρf
: density of the fluids of the reservoir, g/cm3
Since laboratory values of the porosity are more reliable, these are used to correct logestimated values at the same depth(s) of the core sample (s) , and then the same
correction is applied to the entire thickness of the reservoir. It should be noted that
both logs provide estimates of the absolute porosity.
A sonic log measured travel time of 58 μs for a formation. If the formation is primarily
limestone (46 μs) and contains oil only (190 μs), compute the rock porosity.
=
58 − 46
Note: This value is the average absolute porosity of the formation
190 − 46
= 8.3%
[20]
Dr.Jawad R.R. Alassal Kirkuk U.
Examples
1. A core sample coated with paraffin was immersed in a Russell tube .The dry sample
weight 20 gm . the dry sample coated with paraffin weighted 20.9 gm ,the coated
sample displaced 10 cc of liquid . Assume the density of solid paraffin is 0.9gm/cc.
What is the bulk vol. of the sample.
Solution
Weight of paraffin = 20.9 - 20 =0.9 gm
Volume of paraffin = 0.9/0.9 = 1cc
Bulk vol. of sample = 10-1 = 9cc
2.Water-saturated sample immersed in water
A is the weight of dry sample in air =20gm
B is the weight of saturated sample in air =22.5 gm
C is the weight of saturated sample in water at 40o F =12.6 gm
Calculate the bulk volume
Solution
Weight of water displaced = B-C = 22.5-12.6 =9.9 gm
Volume of water displaced = 9.9 /1.0 = 9.9 cc
The bulk volume = 9.9 cc
2. Dry sample immersed in Mercury pycnometer :
A is the weight of dry sample in air =20.0gm
B is the weight of pycnometer filled with mercury at 20oC =350 gm
C is the weight of pycnometer filled with mercury at 20oC and sample =235.9 gm
(density of mercury = 13.546 gm/cc)
Calculate the bulk volume .
Solution
Weight of sample + weight of pycnometer filled with mercury =20+350=370 gm
Weight of mercury displaced = 370 – 235.9 = 134.1 gm
Volume of mercury displaced = 134.1/13.546 = 9.9 cc
The bulk volume = 9.9 cc
3. A is the weight of dry crushed sample in air = 16.gm
B is the weight of sample absorbed water = 16.1 gm
C is the weight of pycnometer filled with water at 40oF = 65 gm
D is the weight of pycnometer filled with water and sample = 75 gm
Calculate the grain volume .
[21]
Dr.Jawad R.R. Alassal Kirkuk U.
Solution:
Wieght of pycnometer filled with water plus weight of crushed sample is =65+16 = 81
gm
Weight of water displaced = 81 -75 = 6 gm
Volume of water displaced = 6/1 = 6 cc
Therefore the grain volume = 6 cc
Sand grain density = 16/6 = 2.67 gm /cc
If there is a core sample its dry weight = 20 gm
Therefore its grain volume = 20/ 2.67 = 7.5 cc
If the bulk volume of the sample = 9.9 cc , then its porosity = (9.9-7.5) / 9.9 = 24.2 %
4. Effective porosity by the saturation method :
A is the weight of dry sample in air = 20 gm
B is the weight of saturated sample in air = 22.5 gm
The bulk volume = 9.9 cc
Density of saturated fluid (water) = 1 gm /cc
Calculate the effective porosity .
Solution :
Weight of water in pore spaces A-B = 22.5 -20 = 2.5 gm
Volume of water in pore space = 2.5/1 = 2.5 cc
Therefore the pore volume = volume of water = 2.5 cc
Porosity = Vp / VB = 2.5 / 9.9 = 25.3 %
5. A full diameter core ( D = 3in. , L = 6in.) is placed in a core chamber of Boyle’s
law device .Each cell has a volume 1000cc .After evacuated cell one is pressured
to 50 psig , the value is opened the two cell are connected and the resulting pressure
is 28.1 psig .Calculate the porosity of the core .
Solution :
( )
(
)
[22]
Dr.Jawad R.R. Alassal Kirkuk U.
6.A core sample is saturated with oil , gas and water , the initial weight of the saturated
sample is 224.14 gm .After the gas is displaced by water ,the weight is increased to
225.9 gm.The sample is then placed in soxhlet distillation apparatus only a 4.4 cc of
water is extracted after drying the core sample , the weight is now 209.75gm .The
sample bulk volume is 95 cc , Find the porosity Ø , So , Sg , Sw
Solution :
Weight of water displaced gas = 225.9-224.14 = 1.76 gm
Vol. of water = 1.76 cc
Vol. of water initialy in the core is = 4.4-1.76 =2.64 cc
Weight of water =2.64 gm
Weight of oil and water=Weight of saturated core with oil ,gas, water –weight of dry
core sample
=225.9-209.75 =16.15gm
Weight of oil = 16.15-4.4
=11.75 gm
Vol. of oil = 11.75/0.85 = 13.82 cc
Pore volume = gas volume + vol. of oil + water volume
=18.22
[23]
Dr.Jawad R.R. Alassal Kirkuk U.
ROCK COMPRESSIBILITY
Definition
The compressibility of a substance is defined as the shrinkage of a unit volume of the
substance per unit increase in pressure. In equation form:
(
)
Eq.7
Where
C : Compressibility psi -1
V : volume ft 3
T : temperature Fo
P : pressure , psi
Since volume always decreases with increase in pressure, the partial derivative in Eq.
7 is negative. Therefore, the minus sign is added to give a positive compressibility
value. Note that compressibility varies slightly with temperature, which explains
why the derivative is fixed at a given T. All minerals found in sedimentary rocks are
elastic, i.e., they show compressibilities of various magnitudes. Quartz, for example,
has a compressibility of about 2 x 10-8 psi-1 at 68 °F.
Significance
Buried deep under the surface, reservoir rock is exposed to a large overburden
pressure, Pob, created by the weight of all rock strata above it. This pressure is
transmitted from grain to grain at the points of contact. Another pressure that is
exerted on the grains of the rock is the pressure of the fluids within the pore space.
This pore pressure, Pp, is usually equal to the hydrostatic pressure at the depth of
the reservoir and is mostly independent of Pob. One can imagine a wall built with
hollow bricks. The Pob felt by the bottom bricks corresponds to the weight of the
wall while Pp within the cavities of the bricks is the atmospheric pressure of air that
can move freely through them.
Reservoir engineers are interested in rock compressibility because of its effect on
porosity, both in laboratory measurement and during the life of the reservoir. Let us
first look into laboratory measurement.Suppose a core sample is cut from a
sandstone reservoir 6000 feet deep, where a reasonable Pob would be 6000 psia. If a
quartz grain within the sample has a reservoir volume of 1 mm3, its volume in the
laboratory where Pob is atmospheric, would increase by
[24]
Dr.Jawad R.R. Alassal Kirkuk U.
ΔV =
- c V ΔP
= - 2 x 10-8 x 1 x (14.7 - 6000) = 0.00012
mm3
Remember, this is the expansion of one grain only; millions of other grains would
expand as well. Since an increase in the grain volume causes an equal decrease in
the pore volume of the core sample, an error, though very insignificant, would be
inherent in the laboratory- determined porosity. However, a more significant error is
introduced because of another effect. As Pob increases, the grains are also brought
closer together because of compaction. Taking the core sample to the surface
reverses this process and causes significant increases in both bulk and pore volumes.
Both increases, though, are difficult to relate to the matrix (grain material)
compressibility.
Similarly, Pp in the laboratory is very low compared to its value in the reservoir,
which causes further expansion of the grains and some collapse in the bulk volume.
This phenomenon is similar to deflating a piece of sponge. Thus, an added error is
introduced in porosity.
The resultant effect of reduction in Pob and Pp on porosity is difficult to quantify
analytically. Many workers in the field have proposed empirical correlations that
predict variation of porosity with pressure. An example is the one presented by Fatt1
and shown in Fig.10. In his correlation, Fatt defined the net overburden pressure, Pob,
net , as
Eq.8
and the pore volume compressibility as
(
)
Eq.9
Assuming that changes in Vb are small compared with changes in Vp, Eq.9 can
be rewritten as
(
)
[25]
Dr.Jawad R.R. Alassal Kirkuk U.
Eq.10
Therefore, once cp is estimated from Fig.10 Eq. 10 can be used to estimate
reduction in porosity from lab to reservoir as follows
Eq.11
Fig.10 Variation of pore-volume compressibility with net overburden pressure
Example
A core plug was obtained from a reservoir where the overburden pressure is 5000 psi
and the pore pressure is 2200 psi. If the plug shows a porosity of 18% in the
laboratory, what is the porosity under reservoir conditions? Assume the pore
compressibility follows curve E in Fig.10
Pob,net = Pob – 0.85 Pp
= 5000 – 0.85 x 2200 = 3130 psi
-6
-1
From Fig.10 , cp = 10 x 10 psi
Δ
= - cp
lab ΔPob,net
-6
= - 10 x 10 x 0.18 x 3130=0.0056
Therefore , porosity under reservoir conditions is
[26]
Dr.Jawad R.R. Alassal Kirkuk U.
0.18 – 0.0056
=
=
0.1744 or 17.44%
Note: Under laboratory conditions, net overburden pressure is usually negligible
.
Pore-volume compressibility can be measured in the laboratory by measuring the
variation in the pore volume of a core sample at different conditions of Pob and Pp.
The pore volume of the sample is first measured at atmospheric pressure and the
reservoir temperature. The saturated sample is then loaded into a core holder (Fig. 01),
which is a device that allows application of different combinations of overburden and
pore pressures independently. At each set of Pob and Pp, the liquid that is squeezed out
of the sample is collected, and its volume is used to compute the current porosity. A
plot of φ versus Pob,net would yield the pore volume compressibility as depicted in Fig.
00.
Fig.01: Schematic of a core holder
[27]
Dr.Jawad R.R. Alassal Kirkuk U.
Fig. 00: Computation of pore-volume compressibility
Applications
Rock compressibility – matrix, bulk or pore – is useful in correcting laboratorymeasured porosity as discussed above. However, other reservoir calculations also
require compressibility.
In some well testing techniques, the production rate of an oil well is changed
suddenly while its pressure is monitored over a period of time.
The transient
pressure response of the well is influenced by fluid as well as rock compressibilities,
and these values are needed for an accurate interpretation of the test results.
Some reservoir engineering calculations require reservoir pore pressure and total
production data for the estimation of reserves. But since the pore volume of the
reservoir changes as Pp decreases with production, cp is needed to correct the pore
volume from its initial value.
[28]
Dr.Jawad R.R. Alassal Kirkuk U.
FLUID SATURATION
Definition
The pore space of a petroleum reservoir is never filled completely with hydrocarbons;
water is always present in the liquid state, and the hydrocarbons can exist in one or
more states – gas, solid or liquid. The saturation of a given fluid is defined as the
fraction of the pore space occupied by that fluid. In equation form:
Sf = Vf / Vp
Eq.12
Where
Sf : saturation of the fluid, fraction or percent
Vf : total volume of fluid in reservoir or core sample, cm3
Vp : total pore volume of reservoir or core sample, cm3
The sum of all fluid saturations in a reservoir is obviously equal to
unity
1 = Sw + So + Sg
Eq.13
Knowledge of the average saturation of a fluid, say oil, in a reservoir allows the
reservoir engineer to estimate the total volume of the fluid in the reservoir simply
through application of Eq.12. This is the prime utility of saturation. Another, equally
important utility, is how some fluid-dependent rock properties vary with saturation
as will be seen in the following chapters.
The saturation of a fluid in a reservoir seldom remains constant. Water can enter the
reservoir either naturally - from an adjacent aquifer - or artificially by water injection.
Oil saturation decreases with oil production and the replacement of oil by another fluid
such as water. Gas saturation could increase with gas injection into the reservoir or as
gas evolves naturally from the oil when the pressure drops. The saturations of the
different fluids in a reservoir are, therefore, measured periodically employing direct
and indirect methods.
[29]
Dr.Jawad R.R. Alassal Kirkuk U.
Factors Affecting Fluid Saturations of Cores
The core samples delivered to the laboratory for fluid- saturation determinations are
obtained from the ground by either rotary, sidewall, or cable-tool coring. In all cases,
the fluid content of these samples has been altered by two processes. First, in the case
of rotary drilling, the mud column exerts a greater pressure at the formation wellbore
surface than the fluid in the formation. The differential pressure between the mud
column and the formation fluids causes mud and mud filtrate to invade the formation,
thus flushing the formation with mud and its filtrate. The filtrate displaces some of the
original fluids. This displacement process changes the original fluid contents of the inplace rock. Second, as the sample is brought to the surface, the confining pressure of
the fluid column is constantly decreasing. The reduction of pressure permits the
expansion of the entrapped water, oil, and gas. Gas, having the greater coefficient of
expansion, expels oil and water from the core. Thus, the content of the core at the
surface has been changed from that which existed in the formation. Because the
invasion of the filtrate precedes the core bit, it is not possible to use pressurized core
barrels to obtain undisturbed samples.
Drill cuttings, chips, or cores from cable-tool drilling also have undergone definite
physical changes. If little or no fluid is maintained in the wellbore, the formation adjacent to the well surface is depleted because of pressure reduction. As chips are
formed in the well, they may or may not be invaded, depending on the fluids in the
wellbore and the physical properties of the rock. In all probability, fluid will permeate
this depleted sample, resulting in flushing. Thus, even cable-tool cores undergo the
same two processes as was noted in the case of rotary coring, although in reverse
order.
Sidewall cores from either rotary- or cable-tool-drilled holes arc subjected to these
same processes.
[30]
Dr.Jawad R.R. Alassal Kirkuk U.
Determination of Fluid Saturations from Rock Samples
Methods for measuring fluid saturations of cores :-
1.Retort method:SATURATION AFTER MUD FLUSHING BEFORE PRESSURE REDUCTION
Oil
Oil
67.6
53.4
Water
46.6
Water
32.4
Gas
34.8
Oil
26.7
Water
38.5
Water base mud
(a)
Oil
]
50.9
Water
49.1
Oil
32.9
Gas
25.6
Filtrate
18.0
Oil
26.7
Water
49.1
Water
47.7
Saturation after mud flushing before
pressure reduction
Oil base mud
(b)
Fig. 12—Typical changes in saturations of cores flushed with water-based and oil-based muds.
This method takes a small rock sample and heats the sample to vaporize the water and
the oil, which is condensed and collected in a small receiving vessel. The retort
method has several disadvantages as far as commercial work is concerned. The water
of crystallization within the rock is driven off, causing the water-recovery values to be
too high. The second error that occurs from retorting samples is that the oil, when
heated to high temperatures, has a tendency to crack and coke. This change of
hydrocarbon molecules tends to decrease the liquid volume. The fluid wetting
characteristics of the
sample surface may be altered during the retorting process as a result of the two
previous factors. Before retorts can be used, calibration curves must be prepared for
[31]
Dr.Jawad R.R. Alassal Kirkuk U.
water and oils of various gravities to correct for losses and other errors. These curves
can be obtained by running 'blank' runs (retorting known volumes of fluids of known
properties). The retort is a rapid method for determining fluid saturations and, if the
corrections are used, yields satisfactory results. It gives both water and oil volumes
such that the oil and water saturations can be calculated from the following equations.
Picture
Of
A
Conventional
retort
Core lab. 1983
Sg =1-Sw-So
where
Sw
So
Sg
Vw
Vp
Vo
=
=
=
=
=
=
water saturation,
oil saturation,
gas saturation,
water volume, cm3,
pore volume, cm3, and
oil volume, cm3.
[32]
Dr.Jawad R.R. Alassal Kirkuk U.
Fig.13 Retort distillation apparatus
2.The other method of determining fluid saturation is by extraction with
a solvent :Extraction may be accomplished by a modified ASTM distillation method or a
centrifuge method. In the standard distillation test, the core is placed such that a vapor
of either toluene, pentane, octane, or naphtha rises through the core. This process
leaches out the oil and water in the core. The water and extracting fluid are condensed
and collected in a graduated receiving tube. The water settles to the bottom of the
receiving tube because of its greater density, and the extracting fluid refluxes over the
core and into the main heating vessel. The process is continued until no more water is
collected in the receiving tube. The water saturation may be determined directly by
The oil saturation is an indirect determination. The oil saturation as a fraction of PV is
given by
where
[33]
Dr.Jawad R.R. Alassal Kirkuk U.
Wcw= weight of wet core, g,
Wcd= weight of dry core, g,
Ww= weight of water, g,
Vp = PV, cm3, and
𝞺o= density of oil, g/cm3.
The gas saturation is obtained in the same manner as the retort.
3. Another method of determining water saturation is to use a
centrifuge:A solvent is injected into the centrifuge just off center. Because of centrifugal force, it
is thrown to the outer radii and forced to pass through the core sample. The outflow
fluid is trapped and the quantity of water in the core is determined. The use of the
centrifuge provides a very rapid method because of the high forces that can be applied.
In both extraction methods, at the same time that the water content is determined, the
core is cleaned in preparation for the other measurements such as porosity and
permeability
4.There is another procedure for saturation determination that is used in
conjunction with either of the extraction methods:The core as received from the well is placed in a modified mercury porosimeter in
which the BV and gas volume are measured. The volume of water is determined by
one of the extraction methods. The fluid saturations can be calculated from these data.
In connection with all procedures for determination of fluid content, a value of PV
must be established in order that fluid saturations may be expressed as percent of PV.
Any of the porosity procedures previously described may be used. Also, the BV and
gas volume determined from the mercury porosimeter may be combined with the oil
and water volumes obtained from the retort to calculate PV, porosity, and fluid
saturations.
Porosity, permeability, and fluid-saturation determinations are the measurements
commonly reported in routine core analysis.
Interstitial Water Saturations
Essentially, three methods are available to the reservoir engineer for the determination
of interstitial water saturations. These methods are
(1) Determination from cores cut with oil-based muds,
[34]
Dr.Jawad R.R. Alassal Kirkuk U.
(2) Determination from capillary-pressure data, and
(3) Calculation from electric- log analysis .
Oil-Based Mud. The obtaining of water saturations by using oil-based muds has
been discussed. A correlation between water saturation and air permeability for cores
obtained with oil-based muds is shown in Fig. below.
Fig. 14—Relation of air permeability to the water content of the South Coles Levee cores.
A general trend of increasing water saturation with decreasing permeability is
indicated. It is accepted from field and experimental evidence that the water content
determined from cores cut with oil-based mud reflects closely the water saturation as it
exists in a reservoir, except in transition zones where some of the interstitial water is
replaced by filtrate or displaced by gas expansion.
Fig.15 below shows permeability/interstitial-water relationships reported in the
literature for a number of fields and areas. There is no general correlation applicable to
all fields; however, an approximately linear correlation between interstitial water and
the logarithm of permeability exists for each individual field. The general trend of the
correlation is decreasing interstitial water with increasing permeability.
[35]
Dr.Jawad R.R. Alassal Kirkuk U.
Fig. 15—Interstitial water vs. permeability relationships.
Measurement of Saturation
Direct methods
Direct measurement methods rely simply on the removal of all liquids – by
evaporation or extraction - from a core sample and determining their individual
volumes. Dividing each fluid volume by the pore volume of the sample yields the
saturation of that fluid. One device used commonly for this purpose is the Modified
ASTM Extraction Unit (Fig.12). The procedure starts with placing the core sample in
a paper thimble and weighing them together. The thimble is then placed in the flask,
the heater is turned on, and water flow through the condenser is started. As the
toluene in the flask boils, its vapors rise and exit the flask , condense in the condenser
and accumulate in the condenser’s graduated tube. Once the tube is full, excess
toluene refluxes back into the flask flowing through the thimble. Oil present in the
core sample is extracted by the refluxing toluene and ends up dissolved in the
bulk toluene mass.
[36]
Dr.Jawad R.R. Alassal Kirkuk U.
Condenser
Graduated tube
Water
Thimble
Core sample
Flask
Toluene
Heater
Fig.16 ASTM extraction unit
The water present in the sample evaporates and condenses back into the graduated
tube. Since water is heavier than toluene, it sinks to the bottom of the tube rather than
returning to the flask. Extraction is continued until no more water accumulates in the
tube at which point heating is stopped and the volume of water is read. When the
unit cools down, the sample and thimble are removed and placed in a vacuum oven
to dry, after which they are both weighed again. The mass of the oil is computed
by a mass balance on the core sample before and after extraction as follows
mo = Δmc – Vw ρw
Eq.13
Vo = mo / ρo
Eq.14
Where
mo : mass of oil extracted, g
Δmc : reduction in mass of core sample, g
Vo : volume of oil extracted, cm3
Vw : volume of water extracted, cm
Example
A sandstone core sample 4 in . long, 1 in. diameter with an absolute porosity of
23% was cleaned in an extraction unit. The reduction in the sample’s mass was 7.4 g,
and 3.2 ml of water were collected. If the oil and water densities are 0.88 and 1.02
g/cm3, respectively, compute the fluid saturations.
mw =
mo =
3.2 x 1.02 = 3.26 g
7.4 – 3.26 = 4.14 g
[37]
Dr.Jawad R.R. Alassal Kirkuk U.
Vo =
4.14 / 0.88 = 4.7 cm3
Vp =
( )
Sw =3.2 / 11.84 = 0.27 = 27%
So = 4.7 / 11.84 = 0.397 = 39.7%
Sg=100 – 27 – 39.7 = 33.3%
Indirect methods
Fluid saturations can also be estimated through indirect methods
such as
measurement of rock resistivity and well logging.
ROCK RESISTIVITY
Definition
Resistivity is a measure of the resistance of a substance to the flow of
electrical current. It is equal to the resistance of a sample of the substance having a
volume of unity. For samples with regular geometric shapes, the resistivity can be
computed as follows
R = rA/L
Eq.15
where
R : resistivity of sample, Ωm
r : resistance of sample, Ω
A : cross-sectional area of sample, m2
L : length of sample, m
The resistance is measured by a simple electric circuit (Fig. 17) where
r
= V/I
[38]
Dr.Jawad R.R. Alassal Kirkuk U.
Fig.17 Measurement of electrical resistance
Significance
Most rock minerals have very high resistivities ; so do all crude oils and natural gas.
However, water, especially salt water, is an excellent conductor and, thus, shows low
resistivity. It is this resistivity contrast that is utilized by reservoir engineers to look
for oil and estimate its saturation within reservoir rock. To illustrate this principle, let
us follow a simple experiment. Suppose we have an empty rubber tube fitted with a
circuit to measure resistivity (Fig.18). The copper plates are only to ensure good
electrical contact with whatever substance that fills the tube’s cavity. With only air
filling the tube, the resistivity would be near infinite as air is an excellent insulator. If
sweet water fills the tube, the resistivity would drop to 30 Ωm, since water is a
better conductor. Replacing sweet water with brine - salt water - the resistivity
would drop to about 1 Ωm. Again brine is a good conductor. Now suppose we
insert in the tube a sandstone core sample saturated completely with the brine. The
resistivity would increase to 95 Ωm. The reason is that only part of the crosssectional area of the core sample, namely the brine-saturated pores, is available for
current flow since the sand grains are nonconductors. Another reason is that the
current has to travel a longer distance between the terminals, having to follow a
tortuous path between the grains. Finally, suppose we reduce the water (brine)
saturation in the sample by replacing some of it with crude oil. The resistivity would
jump to 300 Ωm, as oil is a poorer conductor than brine. If we repeat the last step
with different water saturations, we can construct a graph of resistivity versus water
saturation, which can be used to estimate other water saturations in the core sample by
simply measuring its resistivity. However, such a graph would be limited to that
particular sample. To generalize this technique to other core sampleor the reservoir as
a whole we need to develop a basic theory of rock resistivity.
[39]
Dr.Jawad R.R. Alassal Kirkuk U.
Fig. 18: Measurement of resistivity
Mathematical formulations
Let us begin with some definitions:
R w : resistivity of reservoir water, Ωm
R o : resistivity of reservoir rock saturated with reservoir water, Ωm
R t : resistivity of reservoir rock saturated with both oil and water, Ωm
Define the formation factor, F, as
F = Ro / Rw
Eq.16
Obviously, the formation factor is always greater than unity since the only
difference between the two resistivities is the presence of the rock matrix. The more
tortuous the path of current flow, the larger R o becomes. Also, the smaller the
fraction of pore area, relative to the total area of the rock, the larger R o becomes.
One can, therefore, conclude that the porosity of the rock has a considerable
influence on F, and indeed it does. The two parameters have been found by
Archie to relate to each other according to the following general correlation:
–m
Eq.17
F = C
where
C : tortuousity constant
m : cementation factor
The two parameters, C and m , vary with the type of rock. For clean sandstone, a
widely used form of Archie’s correlation is the Humble equation:
Eq.18
F = 0.62 –2.15
[40]
Dr.Jawad R.R. Alassal Kirkuk U.
To introduce the effect of other fluids within the rock, let us define the resistivity
index, I, as
I = Rt / R o
Eq.19
Again, the resistivity index is always greater than unity since the only difference
between the two resistivities is the presence of oil within the rock matrix. Therefore,
it must be related to Sw. Archie was able to correlate this effect by his second
equation:
I = 1/
Eq.20
where n is called the saturation exponent and varies with the type of rock; a
common value used is 2. Combining Eq. 16 and 19, the following relationship is
derived:
1/I =
F Rw / Rt
Eq.21
Substituting for I and F, Equs. 16 and 20, in Eq. 21 yields
Rw / Rt
Eq.22
Equation 21 is the basic tool of the theory of rock resistivity. All that is needed is
the values of the three constants, C , m and n, which are usually fixed for a given
reservoir and can be determined from simple laboratory experiments. Equation 22 is
the basis of the laboratory resistivity test and the resistivity log,.
Example
Consider the reservoir whose resistivity log is shown in Fig. 19. The reservoir
water resistivity is 1.2 Ωm. If the Humble equation applies to this reservoir, and
the saturation exponent is 2.2, estimate the oil saturation at 4226 ft where the porosity
is 24%.
At 4226 ft, the resistivity is approximately 400 Ωm. Applying Eq. 22 to this depth
Rw / Rt
(
)
Sw
=
0.232 = 23.2 %
So
=
76.8 %
= 0.04
[41]
Dr.Jawad R.R. Alassal Kirkuk U.
Fig.19 Resistivity log
Permeability
Introductory Theory
It is the purpose of this section to discuss the ability of the formation to conduct fluids.
In the introduction to API Code 27, it is stated that permeability is a property of the
porous medium and a measure of the medium's capacity to transmit fluids. The
measurement of permeability, then, is a measure of the fluid conductivity of the
particular material. By analogy with electrical conductors, the permeability represents
the reciprocal of the resistance that the porous medium offers to fluid flow.
The following equations for flow of fluids in circular conduits are well known.
Poiseuille's equation for viscous flow:
[42]
Dr.Jawad R.R. Alassal Kirkuk U.
Eq.23
Fanning's equation for viscous and turbulent flow:
Eq.24
where
v = fluid velocity, cm/s,
d = diameter of conductor, cm,
p = pressure loss over length L, dynes/cm 2,
L = length over which pressure loss is measured, cm,
µ = fluid viscosity, Pa s,
𝞺= fluid density, g/cm3, and
f = friction factor, dimensionless.
A more convenient form of Poiseuille's equation is
Eq.25
where r is the radius of the conduit, cm, q is the volume rate of flow, cm3/s, and the
other terms are as previously defined.
If a porous medium is conceived to be a bundle of capillary tubes, the flow rate qt
through the medium is the sum of the flow rates through the individual tubes. Thus
∑
Eq.26
where nj is the number of tubes of radius rj. If
( )∑
Eq.27
is treated as a flow coefficient for the particular grouping of tubes the equation reduces
to
Eq.28
Where
[43]
Dr.Jawad R.R. Alassal Kirkuk U.
Eq.28
( )∑
The pore structure of rocks does not permit simple classification of the flow channels.
Therefore, empirical data are required in most cases.
In 1856, Darcy investigated the flow of water through sand filters for water
purification. His experimental apparatus is shown schematically in Fig. 26.16.22 Darcy
interpreted his observations to yield results essentially as given in Eq. below
Eq.30
(
)
Eq.31
where
s = distance in direction of flow, always positive,
us = volume flux across a unit area of the porous medium in unit time along flow path
s, z = vertical coordinate, considered positive downward, cm,
p = density of the fluid,
g = acceleration of gravity,
= pressure gradient along s at the point to which u refers,
µ=viscosity of the fluid
k=permeability of the medium , and
[44]
Dr.Jawad R.R. Alassal Kirkuk U.
Flow Systems of Simple Geometry Horizontal Flow.
Horizontal rectilinear steady-state flow is, common to virtually all measurements of
permeability. If a rock is 100% saturated with an incompressible fluid and is horizontal
then dz/ds=0, dp/ds= dpldx
Which on integration becomes
Considering steady rectilinear flow of compressible fluids
or, for steady-state flow,
Vertical Flow: Fig.20 shows three sandpacks in which linear flow occurs in the
vertical direction. First consider Case 1 (Fig. 20)—when the pressure at the inlet and
outlet are equal (free flow), such that only the gravitational forces are driving the
fluids.
The flow is then defined by
Next consider Case 2—the case of downward flow when the driving head (difference
in hydraulic head of inlet and outlet) is h (Fig. 20). We know that
Therefore
[45]
Dr.Jawad R.R. Alassal Kirkuk U.
(
)
When the flow is upward and the driving head is h, Case 3 (Fig. 26.18), and z is
defined as positive downward,
(
)
And
(
)
Then
Fig. 21—Sand model for radial flow
of fluids to central wellbore.
Fig. 20—Sand model for vertical flow.
Radial Flow: A radial-flow system, analogous to flow into a wellbore, is idealized in
Fig. 21. If flow is considered to occur only in the horizontal plane under steady-state
conditions, an equation of flow may be derived from Darcy's law to be
(
[46]
Dr.Jawad R.R. Alassal Kirkuk U.
)
Eq.32
Average Permeability of combination layers
1.Linear Flow : Parallel combination
Fig.18 shows the flow
Qt = Q1+Q2+Q3
Ht = h1+h2+h3
(
)
(
)
(
)
(
)
From these equations we can get
∑
Eq.33
∑
Fig. 22—Linear flow—parallel combination of beds.
[47]
Dr.Jawad R.R. Alassal Kirkuk U.
Eq.33 is also used for Radial flow Fig.23
Fig. 23—Radial flow—parallel combination of beds.
2. Linear flow : Series
Combination (Fig.24)
Qt = Q1+Q2+Q3
P1 - P2 = ∆P1+∆P2+∆P3
L = L1 + L2 + L3
(
Fig.24 Linear flow , series combination of beds
)
( )
( )
( )
( )
Eq 1 = Eq 2 + Eq 3 + Eq 4
Therefore
[48]
Dr.Jawad R.R. Alassal Kirkuk U.
Eq.34
∑
The same reasoning can be used in the evaluation of the average permeability for the
radial system (Fig. 25) so as to yield
Eq.35
∑
Fig. 25 Radial flow—series combination of beds.
Example :
What is the equivalent permeability of four beds in series having equal formation
thickness under the following conditions (1) for a linear system and (2) for a radial
system if the radius of the penetrating well bore is 6 in and the radius of the effective
drainage is 2000 ft ?
Bed
1
2
3
4
Length of bed ,ft
250
250
500
1000
Horizontal , K , md
25
50
100
200
[49]
Dr.Jawad R.R. Alassal Kirkuk U.
For radial
250
500
1000
2000
Solution
Assume bed 1 adjacent the well bore
Linear system :
∑
∑
Radial system
Bed
1
2
3
4
Length of bed ,ft
250
250
500
1000
Horizontal , K , md
25
50
100
200
∑
[50]
Dr.Jawad R.R. Alassal Kirkuk U.
For radial
250
500
1000
2000
Example
Flow rate = 1000cc of air at 1 atmosphere and 70o F in 500 sec.
Pressure downstream side of core = 1 atmosphere , flowing temperature = 70o F
Viscosity of air at 70oF = 0.02 cp
Cross-sectional area of core = 2 cm2
Length of the core = 2 cm
Pressure up stream on core = 1.45 atmosphere
Calculate the permeability of the core sample
̅̅
̅
̅
̅
̅
̅
̅
Assuming that the data indicated above were obtained but water was used as the
flowing fluid , compute the permeability of the core . The viscosity of water at test
temperature was 1.0 cp
[51]
Dr.Jawad R.R. Alassal Kirkuk U.
Relative Permeability
• Absolute permeability: is the permeability of a porous medium saturated with a
single fluid (e.g. Sw=1)
• Absolute permeability can be calculated from the steady-state flow equation (1D,
Linear Flow; Darcy Units):
Commonly, reservoirs contain 2 or 3 fluids
• Water-oil systems
• Oil-gas systems
• Water-gas systems
• Three phase systems (water, oil, and gas)
To evaluate multiphase systems, must consider the effective and relative
Effective permeability: is a measure of the conductance of a porous medium for
one fluid phase when the medium is saturated with more than one fluid.
The porous medium can have a distinct and measurable conductance to each phase
present in the medium
Effective permeabilities: (ko, kg, kw)
ko A  o
o L
• Oil
qo 
• Water
qw 
• Gas
qg 
kw A  w
w L
k g A  g
g L
Relative Permeability is the ratio of the effective permeability of a fluid at a given
saturation to some base permeability
• Base permeability is typically defined as:
absolute permeability, k
air permeability, kair
[52]
Dr.Jawad R.R. Alassal Kirkuk U.
effective permeability to non-wetting phase at irreducible wetting phase saturation
[e.g. ko(Sw=Swi)]
because definition of base permeability varies, the definition used must always be:
confirmed before applying relative permeability data
noted along with tables and figures presenting relative permeability data
kro( 0.5,0.3) 
• Oil
ko ( 0.5,0.3)
k
So =0.5
• Water
k rw( 0.5,0.3) 
k w( 0.5,0.3)
• Gas
k rg ( 0.5,0.3) 
k g ( 0.5,0.3)
Sw =0.3
Sg = 0.2
k
k
Example :
Calculating permeability ( Data from Lab)
The following laboratory data are given from relative permeability tests.
Cross – sectional area of the core = 5 cm2 , Core length =3 cm
Pressure at out let face of the core = 1 atm , Pressure at inlet face of the core = 2 atm
Viscosity of the water = 1.0 cp , Viscosity of oil = 1.25 cp
Saturation and rate data are as follows :
Saturation
Water
100
90
80
60
40
30
Flow Rate cc/sec
Oil
0
10
20
40
60
70
Water
0.5
0.3
0.15
0.03
0.01
0.00
1. What is the absolute permeability ?
2. What is the permeability to oil at water saturation of 30% ?
3. Calculate the relative values ?
[53]
Dr.Jawad R.R. Alassal Kirkuk U.
Oil
0.00
0.00
0.01
0.15
0.25
0.38
Solution :
The absolute permeability can be calculated only when the core is 100% saturated with
one fluid , so use data when Sw = 100%
At Sw=30% Oil flow rate is 0.38 cc /sec
Calculation of Relative permeability
1
Sw
100
90
80
60
40
30
2
Kw
0.3
0.18
0.09
0.18
0.006
0.000
3
Ko
0.000
0.000
0.0075
0.0750
0.1875
0.2850
4
Krw
1.00
0.60
0.30
0.06
0.02
0.00
5
Kro
0.00
0.00
0.0255
0.250
0.625
0.950
Col. 1 given in the table , Col. 2 from Kw equation , Col. 3 from Ko equation
Col. 4 Kw/0.3 , Col. 5 Ko/0.3
[54]
Dr.Jawad R.R. Alassal Kirkuk U.
Measurement of Permeability
The permeability of a porous medium may be determined from samples extracted from
the formation or by in-place testing. The procedures discussed in this section pertain to
the permeability determinations on small samples of extracted media.
Two methods are used to evaluate the permeability of cores. The method most used on
clean, fairly uniform formations uses small cylindrical samples called perm plugs that
are approximately 3/4 in. in diameter and 1 in. in length. The second method uses fulldiameter core samples in lengths of 1 to 1 1/2ft. The fluids used with either method may
be gas or any nonreactive liquid.
Perm-Plug Method. As core samples ordinarily contain residual oil, water, and gas,
it is necessary that the sample be subjected to preparation before the determination of
the permeability. The residual fluids normally are extracted by retorting or solvent
extraction. The core is dried before permeability measurements are taken. Air
commonly is used as the fluid in permeability measurements. The requirement that the
permeability be determined for conditions of viscous flow is best satisfied by obtaining
data at several flow rates and plotting results. For conditions of viscous flow, the data
should plot a straight line, passing through the origin. Turbulence is indicated by
curvature of the plotted points. The slope of the straight-line portion of the curve is
equal to from which the permeability may be computed. To obtain k in darcies, q must
be in cm3/s,A in cm2, p1 and p2 in atm , L in cm, and µ in cp.
Permeameter designed for the determination of the permeability of rocks with
either gas or liquid is illustrated in Fig. 25. Data ordinarily are taken from this device
at only one flow rate. To assure conditions of viscous flow, the lowest possible rate
that can be measured accurately is used.
Fig. 25—Ruska universal permearneter.
[55]
Dr.Jawad R.R. Alassal Kirkuk U.
Whole-Core Measurement. The core must be prepared in the same manner as
perm plugs. The core is then mounted in special holding devices as shown in Fig.
26. The measurements required are the same as the perm plugs but the
calculations are slightly different.
Fig. 26—Clamp - type permeameter for
large cores.
Fig. 27—Permeability constant of core sample L to hydrogen,
nitrogen, and C02 at different pressures (permeability constant
to isooctane, 2.55 md).
In the case of the clamp-type permeameter, the geometry of the flow paths is complex,
and an appropriate shape factor must be applied to the data to compute the
permeability of the sample. The shape factor is a function of the core diameter and the
size of the gasket opening. The shape factor affects the quantity LI A in the previous
equations.
Factors Affecting Permeability Measurements
In the techniques of permeability measurement previously discussed, certain
precautions must be exercised to obtain accurate results. When gas is being used as the
measuring fluid, corrections must be made for gas slippage. When liquid is the testing
fluid, care must be taken that it does not react with the solids in the core sample. Also,
corrections may be applied for the change in permeability because of the reduction in
confining pressure on the sample.
Effect of Gas Slippage on Permeability Measurements
Klinkenberg24 has reported variations in permeability determined by using gases
as the flowing fluid from that determined by using nonreactive liquids. These
variations were ascribed to slippage, a phenomenon well known with respect to
[56]
Dr.Jawad R.R. Alassal Kirkuk U.
gas flow in capillary tubes. The phenomenon of gas slippage occurs when the
diameter of the capillary openings approaches the mean free path of the gas.
The linear relationship between the observed permeability and the reciprocal of mean
pressure may be expressed as follows.
Eq.36
where
kL = permeability of the medium to a single liquid phase completely filling the pores of
the medium
kg = permeability of the medium to a gas completely filling the pores of the medium,
= mean flowing pressure of the gas at which kg was observed,
b = constant for a given gas in a given medium, and
m = slope of the curve.
(
)
Eq.37
where
k = permeability of the porous medium, Fs = shape factor, L = length of the sample,
and
, La = actual length of the flow path.
[57]
Dr.Jawad R.R. Alassal Kirkuk U.
Capillary Pressure
Capillary pressure may be thought of as a force per unit area resulting from the
interaction of surface forces and the geometry of the medium in which they exist.
Capillary pressure for a capillary tube is defined in terms of the interfacial tension
between the fluids, a, the angle of contact of the interface of these two fluids and the
tube, Ɵc, and the radius of the tube ,rt
This relationship is expressed in Eq.39
Eq.39
where the angle Ɵc is measured through the more dense fluid.
In a packing of spheres, the capillary pressure is expressed in terms of any two
perpendicular radii of curvature (these radii touch at only one point), r1 and r2, and the
interfacial tension of the fluids. This relationship is given in Eq. below
(
)
Eq.40
Comparing this Eq. with the equation for capillary pressure as determined by the
capillary-tube method, it is found that the mean radius r is defined by
Eq.41
It is practically impossible to measure the values of r j and r2; hence, they generally are
referred to by the mean radius of curvature and empirically determined from other
measurements on a porous medium.
The distribution of the liquid in a porous system depends on the wetting
characteristics. It is necessary to determine which is the wetting fluid so as to ascertain
which fluid occupies the small pore spaces . From packings of spheres, the wettingphase distribution within a porous system has been described as either funicular or
pendular in nature. In funicular distribution, the wetting phase is continuous,
completely covering the surface of the solid. The pendular ring is a state of saturation
in which the wetting phase is not continuous and the nonwetting phase is in contact
with some of the solid surface. The wetting phase occupies the smaller interstices. As
the wetting-phase saturation progresses from the funicular to the pendular-ring
distribution, the volume of the wetting phase decreases .
[58]
Dr.Jawad R.R. Alassal Kirkuk U.
We see that if rj and r2 decrease in size, the magnitude of the capillary pressure would
have to increase in value. Since r j and rj can be related to the wetting-phase saturation,
it is possible to express the capillary pressure as a function of fluid saturation when
two immiscible phases are within the porous matrix.
Laboratory Measurements of Capillary Pressure
Essentially, five methods of measuring capillary pressure on small core samples are
used. These methods are
(1) The desaturation or displacement process, through a porous diaphragm or
membrane (restored-state method of Welge).
(2) The mercury-injection method.
(3) The centrifuge or centrifugal method.
(4) The dynamic- capillary-pressure method, and
(5) The evaporation method.
Porous Diaphragm. The first of these, illustrated in Fig .28, is the displacement or
diaphragm method. The essential requirement of the diaphragm method is a permeable
membrane of uniform pore-size distribution containing pores of such 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.
Various materials including fritted glass, porcelain, and cellophane have been used
successfully as diaphragms. 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 may be used: gas, oil, and/or
water. Although most determinations of capillary pressure by the diaphragm method
are drainage tests, by suitable modifications imbibition curves similar to Leverett's
may be obtained.
[59]
Dr.Jawad R.R. Alassal Kirkuk U.
Fig. 28 Schematic of porous-diaphragm method of capillary
prerssure.
Mercury Injection. The mercury-capillary-pressure apparatus was developed to
accelerate the determination of the capillary-pressure/saturation relationship. Mercury
is normally a nonwetting fluid. The core sample is inserted in the mercury chamber
and evacuated. Mercury is forced into the core under pressure. The volume of mercury
injected at each pressure determines the nonwetting-phase saturation. This procedure is
continued until the core sample is filled with mercury or the injection pressure reaches
some predetermined value. Two important advantages are gained by this method: (1)
the time for determination is reduced to a few minutes, and (2) the range of pressure
investigation is increased because the limitation of the diaphragm's properties is
removed. Disadvantages are the difference in wetting properties and permanent loss of
the core sample.
[60]
Dr.Jawad R.R. Alassal Kirkuk U.
Fig.29 Capillary pressure cell for mercury injection
Centrifuge Method. A third method for determining capillary properties of
reservoir rocks is the centrifuge method. The high accelerations in the centrifuge increase the field of force on the fluids, in effect subjecting the core to an increased
gravitational force. By rotating the sample at various constant speeds, a complete
capillary-pressure curve may be obtained. The speed of rotation is converted into force
units in the center of the core sample , and the fluid saturation is read visually by the
operator. The advantage of the method is the increased speed of obtaining the data. A
complete curve may be established in a few hours, while the diaphragm method
requires days.
Fig.30 Centrifuge for determination of capillary properties of rock
[61]
Dr.Jawad R.R. Alassal Kirkuk U.
Dynamic Method. Brown32 reported the results of determining capillarypressure/saturation curves by a dynamic method. Simultaneous steady-state flow of
two fluids is established in the core. By the use of special wetted disks that permitted
hydraulic pressure transmission of only the selected fluid phase, the difference in the
resulting measured pressures of the two fluids in the core is the capillary pressure. The
saturation is varied by regulating the quantity of each fluid entering the core. Thus, it is
possible to obtain a complete capillary- pressure curve.
Fig.31 Dynamic Capillary pressure apparatus
Converting Laboratory Data. To use laboratory capillary-pressure data, it is
necessary to convert them to reservoir conditions. Laboratory data are obtained with a
gas/water or an oil/water system that normally does not have the same physical
properties as the reservoir water, oil, and gas.
Essentially two techniques, differing only in the initial assumptions, are available for
correcting laboratory capillary-pressure data to reservoir conditions.
[62]
Dr.Jawad R.R. Alassal Kirkuk U.
Fig 32 -Comparison of water saturation from capillary pressure and
electric log.
Eq.42
Or
Eq.43
where
= interfacial tension water/oil,
= interfacial tension water/gas,
= water/oil contact angle
= water/gas contact angle, subscript
R = reservoir conditions, and subscript
L = laboratory conditions.
Since the interfacial tensions enter as a ratio, pressure in any consistent units may be
used together with the interfacial tension in dynes/cm.
Averaging Capillary-Pressure Data. Two methods have been proposed for
correlating capillary-pressure data of similar geologic formations. The first correlating
procedure is a dimensionless grouping of the physical properties of the rock and the
saturating fluids. This function is called a J function34 and is expressed as
(
)
( )
[63]
Dr.Jawad R.R. Alassal Kirkuk U.
Eq.44
Where
Sw = water saturation, fraction of PV,
Pc = capillary pressure, dyne/cm2,
= interfacial tension, dyne/cm,
k =permeability, cm2, and
Ø=fractional porosity
Fig. 33—Series of capillary-pressure curves as a function of permeability.
Some authors alter the above expression by including the cosƟc. (where Ɵc is the
contact angle) as follows.
(
)
( )
Eq.44
The J function originally was proposed as a means of converting all capillary-pressure
data to a universal curve. There exist significant differences in correlation of the J
function with water saturation from formation to formation such that no universal
curve may be obtained, but the J function may be used to correlate the data from one
formation.
The second method of evaluating capillary-pressure data is to analyze a number of
representative samples and treat the data statistically to derive correlations that,
together with the porosity and permeability distribution data, may be used to compute
the interstitial-water saturations for a field. A first approximation for the correlation of
capillary-pressure data is to plot water saturation against the logarithm of permeability
[64]
Dr.Jawad R.R. Alassal Kirkuk U.
for constant values of capillary pressure. A straight line may be fitted to the data for
each value of capillary pressure, and average-capillary-pressure curves may be
computed from permeability-distribution data for the field. The resulting straight-line
equation takes the general form of
Eq.45
Sw=m log k + b
Fluid-distribution curves are reported for several values of permeability, ranging from
10 to 900 md in Fig. 33. These data also may be considered to be capillary-pressure
curves. The ordinate on the right reflects values of capillary pressure determined by
displacing water with air in the laboratory. The ordinates on the left include the
corresponding oil/water capillary pressure that would exist at reservoir conditions and
the fluid distribution with height above the free-water surface.
The results of the statistical correlation previously discussed applied to the capillarypressure data. The reader should note the linearity of the curves for each value of
capillary pressure and the tendency of all capillary-pressure curves to converge at high
permeability values. This behavior is what normally would be expected because of the
larger capillaries associated with high permeabilities.
To convert capillary-pressure saturation data to height saturation, it is necessary only
to rearrange the terms in Eq. 46 so as to solve for the height instead of the capillary
pressure—i.e.,
Eq.46
hfw, = height above the free-water surface, ft,
pw - density of water at reservoir conditions, Ibm/cu ft,
po = density of oil at reservoir conditions,Ibm/cu ft, and
Pc = capillary pressure at some particular saturation for reservoir conditions (it must be
converted from laboratory data first, psi).
[65]
Dr.Jawad R.R. Alassal Kirkuk U.
Electrical Conductivity of Fluid-Saturated Rocks
Porous rocks comprise an aggregate of minerals, rock fragments, and void space. The
solids, with the exception of certain clay minerals, are nonconductors of electricity.
The electrical properties of a rock depend on the geometry of the voids and the fluids
that fill the voids. The fluids of interest in petroleum reservoirs are oil, gas, and water.
Oil and gas are nonconductors. Water is a conductor when it contains dissolved salts.
Current is conducted in water by movement of ions and therefore may be termed
electrolytic conduction. The resistivity of a material is the reciprocal of conductivity
and commonly is used to define the ability of a material to conduct current. The
resistivity of a material is defined by the following equation
Fig. 34 Core-sample resistivity cell
Eq.47
where
p = resistivity. , r = resistance, A = cross-sectional area of the conductor.
L = length of the conductor.
And For electrolytes, p is commonly reported in fi-cm, and r is expressed in ohms, A in
cm2, and L in cm. In the study of the resistivity of soils and rocks, it has been found
that the resistivity may be expressed more conveniently in Q-m. To convert to Q-m
from Q-cm, divide the resistivity in Q-cm by 100. In oil field practice, the resistivity in
Q-m commonly is represented by the symbol R with an appropriate subscript to define
the condition to which R applies.
[66]
Dr.Jawad R.R. Alassal Kirkuk U.
Problems
1.Calculate the porosity of the sample described below :
Mass of dry sample = 104.2 gm
Mass of water saturated sample =120.2 gm
Mass of saturated sample immersed in water = 64.7 gm
Is this effective or the total porosity of the sample ? What is the most probable lithology
of the matrix material ? explain
2. A limestone sample weight 241 gm. The limestone sample coated with paraffin was
found to weight 249.5 gm , the coated sample when immersed in a partially filled
graduated cylinder displaced 125 cc of water the density of paraffin is 0.9 gm/cc , what
is the porosity of the sample ? Does the process measure total or effective porosity?
Why?(note : limestone grain density = 2.71 gm/cc)
3. Prove that the porosity of hexagonal packing of uniform sphere = 39.5% ?
4. A core 2.54 cm long and 2.54 cm diameter has a porosity of 22% .It’s saturated with
oil and water , Where the oil content is 1.5 cc , find the pore volume of the core? Also ,
find the saturation of water & oil?
5. The bulk volume of a sample was found to be equal 25 cc , in Laboratory using
mercury if the weight of dry sample in air = 33 gm ; weight of dry sample immersed in
mercury ; weight of mercury = 260 gm ; mercury density = 13.6 gm/cc ; Find the
weight of mercury used during experiment ?
6. Determine the porosity and lithology of core sample given the following data :
A-weight of dry core sample = 259.2 gm
B- weight of 100% water saturated core sample = 297 gm
C- weight of core sample in water = 161.4 gm
Define the terms absolute and effective porosity and decided which term is use when
characterizing the core sample ?
7. Calculate the porosity of a sandstone core sample given the data from the core analysis
Bulk volume of dried sample = 8.1 cc
Weight of dried sample = 17.3 gm
Sand grain density = 2.67
[67]
Dr.Jawad R.R. Alassal Kirkuk U.
8. Dolomite sample weight 344 gm , The dolomite sample coated with paraffin was
found to weight 349.5 gm the coated sample when immersed in a partially graduated
cylinder displaced 160 cc of water , The density of a paraffin is 0.97 gm/cc ; what is
the porosity of the sample ? Does the process measure total or effective porosity ?
Why ?
(note : dolomite grain density = 2.87 gm/cc)
9. The following information is available from a core sample :
Dry weight of sample = 427.3 gm
Weight of sample when saturated with water = 448.6 gm
Water density = 1 gm/cc
Weight of water saturated sample immersed in water = 269.6 gm
Find :
a) Porosity of the sample ?
b) Is the porosity effective or total ?Why?
c) Grain density?
10. The weight of a dry core sample is 25 gm , the core sample saturated with water and
it’s weight was 28gm . the core sample was dried again and coated with paraffin , the
weight of the coated sample was 26.7 gm .The coated sample was immersed in a
Russell tube and displaced a liquid volume of 11 cc .the density of solid paraffin is
0.85 gm /cc , density of water is 11 gm /cc .Calculate the porosity of this core sample.
11. A core sample is saturated with Oil, Gas and Water , the initial weight of the saturated
sample is 220 gm . After the gas is displaced by water , the weight is increased to
222gm . thw sample is then placed in soxhlet distillation apparatus , only 5cc of water
was extracted .After cleaning and drying the core , the weight was 210 gm .the sample
bulk volume is 100 cc ,
gm /cc . =0.85 gm/cc .Calculate the effective
porosity , Absolute porosity , the saturation of Oil , water and gas .
12.Four beds have an equal width and length ; find the average permeability ( Kavg.) of
these beds from the following data :
Bed No.
1
2
3
4
Pay-Zone thickness , ft
20
15
10
5
[68]
Dr.Jawad R.R. Alassal Kirkuk U.
Horizontal permeability ,md
25
50
100
200
B) If these Four beds have an equal width and thickness , find the average permeability
(Kavg.) of these beds from the following data
Bed No.
1
2
3
4
Length of bed ,ft
100
100
100
100
Horizontal permeability , md
25
50
100
200
C) If the system is radial with an effective drainage radius (1500 ft) and well bore radius of
( 0.5 ft). Find the effective permeability ( Kavg.) using the data below :
Bed No.
1
2
3
4
Length of bed ,ft
100
100
500
1000
Horizontal permeability , md
25
50
100
200
13.A rectangular grid block
meter . A one meter wide fracture runs along
the entire cell in y-direction at x=49.5 m , the fracture cover all thickness ,
Kf=10000md , Kma=2md .What is the average permeability of the entire system if flow
:
a- Across the fracture.
b- Parallel to fracture.
14.What is the average permeability of three beds connected in parallel:
Bed No.
1
2
3
Permeability , md
10
30
20
[69]
Dr.Jawad R.R. Alassal Kirkuk U.
Thickness , ft
10
20
30
15.Having the following data :
Penetration well bore radius = 3 in.
Effective Drainage radius = 1000 ft (Which is the boundary of the bed no.4)
Also :
Bed No.
1
2
3
4
Bed Radius , ft
…..
200
200
…..
Horizontal k ,md
200
100
25
50
What is the effective permeability of the above four beds in series ?
16.Cubic packing of sphere consists of two diameter of ratio 1-3 , the packing was such
that the pores formed by each eight large sphere contains two small sphere ; calculate the
porosity of the sample ?
17 . A bed containing two layers as showing below also given :
A = 1800 ft2 , h1=5 ft , h2 = 2 ft , K1=2000 md , K2 = 1000 md
Calculate the height of water that is needed to keep Q constant at 5000 gal/hr.
h
Water
1
2
18. Gas flow rate was 1250 cc in 100 sec at out let pressure of 1 atmosphere , the inlet
pressure was 1.5 atmosphere ,Gas viscosity = 0.025 cp , length of the core sample = 3 cm ,
the cross – sectional area = 2.5 cm2 .Assume constant temperature throughout the
experiment.
Calculate the permeability of the core sample.
[70]
Dr.Jawad R.R. Alassal Kirkuk U.
19. a) Given the following data from the core analysis report , calculate the average
permeability of the reservoir in case of parallel combination of layers , and linear flow.
Depth , ft
3998-4002
4002-4004
4004-4006
4006-4008
4008-4010
Permeability , md
200
130
170
180
140
b) if these beds are connected in series with lengths of (150 , 200 , 300 , 500 , and 200 ft)
respectively but with the same thickness , calculate the average permeability of the
reservoir by assuming the two system [ Linear , and Radial].
20. A core plug has : L =4 cm , A = 3cm2 , Q =0.5 cm3/sec and ∆P= 2 atm
a) Calculate the absolute permeability (ka) by using brine.
b) Recalculate (Ka) by using crude oil with
cm3/sec
=2cp , the crude flows at a rate of 0.25
c) Compare and discuss your results.
21. Drive an equation of radial flow in series have the same thickness . Also calculate
average permeability of bed connected in series if the flow is radial, all beds have the
same thickness , well bore radius = 6 in.
Bed
Radius , ft
Permeability,md
1
250
25
2
500
50
22.Having the following data :
Q air = 1000 cc in 500 seconds
Pinlet = 1.45 atm abs.
P outlet = 1.0 atm abs.
Cross sectional area(A) = 2 cm
Length (L) = 2cm
[71]
Dr.Jawad R.R. Alassal Kirkuk U.
3
1000
100
5
2000
200
= 0.02 cp.
T=70 oF = flowing temperature
1Darcy = 9.869
cm2
a) Calculate the permeability of the porous medium.
b) When water is used instead of air , calculate the permeability of the porous medium.
c) If a permeability of a circular opening 0.005 inches in radius is exists, calculate the
water flow rate in bbl/day.
d) If a permeability of a fracture 0.01 inches in thickness is exists, calculate the water
flow rate in bbl/day.
[72]
Dr.Jawad R.R. Alassal Kirkuk U.
Chapter 2
Capillary Properties
SATURATION
Saturation is defined as that fraction, or percent, of the pore volume occupied by a
particular fluid (oil, gas, or water). This property is expressed mathematically by the
following relationship:
Applying the above mathematical concept of saturation to each
Eq (1)
Eq (2)
Eq (3)
Where :
So = oil saturation
Sg = gas saturation
Sw = water saturation
Thus, all saturation values are based on pore volume and not on the
gross reservoir volume.
The saturation of each individual phase ranges between zero to 100
percent. By definition, the sum of the saturations is 100%, therefore
Sg + So + Sw = 1.0
Eq (4)
The fluids in most reservoirs are believed to have reached a state of
equilibrium and, therefore, will have become separated according to their density, i.e., oil
overlain by gas and underlain by water. In addition to the bottom (or edge) water, there
will be connate water distributed throughout the oil and gas zones. The water in these
zones will have been reduced to some irreducible minimum. The forces retaining the
water in the oil and gas zones are referred to as capillary forces because they are
important only in pore spaces of capillary size.
Connate (interstitial) water saturation Swc is important primarily
because it reduces the amount of space available between oil and gas. It is generally not
uniformly distributed throughout the reservoir but varies with permeability, lithology, and
[73]
Dr.Jawad R.R. Alassal Kirkuk U.
height above the free water table. Another particular phase saturation of interest is called
the critical saturation and it is associated with each reservoir fluid. The definition and the
significance of the critical saturation for each phase are described below.
Critical oil saturation, Soc
For the oil phase to flow, the saturation of the oil must exceed a certain value which is
termed critical oil saturation. At this particular saturation, the oil remains in the pores and,
will not flow.
Residual oil saturation, Sor
During the displacing process of the crude oil system from the porous
media by water or gas injection (or encroachment) there will be some
remaining oil left that is quantitatively characterized by a saturation
value that is larger than the critical oil saturation. This saturation value is called the
residual oil saturation, Sor. The term residual saturation is
usually associated with the nonwetting phase when it is being displaced by a wetting
phase.
Movable oil saturation, Som
Movable oil saturation Som is another saturation of interest and is
defined as the fraction of pore volume occupied by movable oil as
expressed by the following equation:
Som = 1 - Swc -Soc
where
Swc = connate water saturation
Soc = critical oil saturation
Critical gas saturation, Sgc
As the reservoir pressure declines below the bubble-point pressure, gas evolves from the
oil phase and consequently the saturation of the gas increases as the reservoir pressure
declines. The gas phase remains immobile until its saturation exceeds a certain saturation,
called critical gas saturation, above which gas begins to move.
Critical water saturation, Swc
The critical water saturation, connate water saturation, and irreducible
water saturation are extensively used interchangeably to define the maximum water
saturation at which the water phase will remain immobile.
[74]
Dr.Jawad R.R. Alassal Kirkuk U.
Average Saturation
Proper averaging of saturation data requires that the saturation values
be weighted by both the interval thickness hi and interval porosity . The average
saturation of each reservoir fluid is calculated from the
following equations:
∑
Eq(5)
∑
∑
Eq (6)
∑
∑
Eq(7)
∑
where the subscript i refers to any individual measurement and hi represents the depth
interval to which i , Soi , Sgi , and Swi apply.
[75]
Dr.Jawad R.R. Alassal Kirkuk U.
Example
Calculate average oil and connate water saturation from the following measurements:
Sample
hi,ft
Ø
So %
Swc %
1
1.0
10
75
25
2
1.5
12
77
23
3
1.0
11
79
21
4
2.0
13
74
26
5
2.1
14
78
22
6
1.1
10
75
25
Solution
Construct the following table and calculate the average saturation for
the oil and water phase:
Sample
hi , ft
Ø
Øh
So
SoØh
Swc
Swc Øh
1
1.0
.10
.100
.75
.0750
.25
.0250
2
1.5
.12
.180
.77
.1386
.23
.0414
3
1.0
.11
.110
.79
.0869
.21
.0231
4
2.0
.13
.260
.74
0.1924
.26
.0676
5
2.1
.14
.294
.78
.2293
.22
.0647
6
1.1
.10
.110
.75
.0825
.25
.0275
1.054
0.8047
Calculate average oil saturation by applying Equation 5:
Calculate average water saturation by applying Equation 6:
[76]
Dr.Jawad R.R. Alassal Kirkuk U.
0.2493
WETTABILITY
Wettability is defined as the tendency of one fluid to spread on or
adhere to a solid surface in the presence of other immiscible fluids. The concept of
wettability is illustrated in figure 1. Small drops of three liquids : mercury, oil, and
water—are placed on a clean glass plate
Figure 1
The three droplets are then observed from one side as illustrated in Figure .It is noted that
the mercury retains a spherical shape, the oil droplet develops an approximately
hemispherical shape, but the water tends to spread over the glass surface.
The tendency of a liquid to spread over the surface of a solid is an indication of the
wetting characteristics of the liquid for the solid. This spreading tendency can be
expressed more conveniently by measuring the angle of contact at the liquid-solid surface.
This angle, which is always measured through the liquid to the solid, is called the contact
angle θ. The contact angle θ is a measure of wettability, as the contact angle decreases,
the wetting characteristics of the liquid increase. Complete wettability would be evidenced
by a zero contact angle, and complete nonwetting would be evidenced by a contact angle
of 180°. There have been various definitions of intermediate wettability but, in much of
the published literature, contact angles of 60° to 90° will tend to repel the liquid.
Wettability is a major factor controlling the location, flow and distribution of fluids in the
oil bearing reservoir.
Changes in wettability have been shown to affect:
 Capillary pressure.
 Initial water saturation.
 Two- phase relative permeability.
 Three- phase relative permeability.
Water Wet:
Tendency of water to occupy the small pores and to contact the majority of the rock surface (θ =
0)
Oil Wet:
[77]
Dr.Jawad R.R. Alassal Kirkuk U.
Tendency of oil to occupy the small pores and to contact the majority of the rock surface (θ =
105 to 120)
Neutral –wet:
The rock has no strong preference for either oil or water (θ = 60 to 75)
Intermediate wet:
Different areas of the pore network have different wetting preference.
SURFACE AND INTERFACIAL TENSION
In dealing with multiphase systems, it is necessary to consider the
effects of the forces at the interface when two immiscible fluids are in contact. When
these two fluids are liquid and gas, the term surface tension is used to describe the
forces acting on the interface. When the interface is between two liquids, the acting
forces are called interfacial tension..
Consider the two immiscible fluids, air (or gas) and water (or oil) as
shown schematically in Figure 2. A liquid molecule, which is remote
from the interface, is surrounded by other liquid molecules, thus having a resulting net
attractive force on the molecule of zero. A molecule at the interface, however, has a force
acting on it from the air (gas) molecules lying immediately above the interface and from
liquid molecules lying below the interface .
Resulting forces are unbalanced and give rise to surface tension. The unbalanced
attraction force between the molecules creates a membrane- like surface with a
measurable tension, i.e., surface tension. As a matter of fact, if carefully placed, a needle
will float on the surface of the liquid, supported by the thin membrane even though it is
considerably more dense than the liquid
Figure 2. Illustration of surface tension.
[78]
Dr.Jawad R.R. Alassal Kirkuk U.
The surface or interfacial tension has the units of force per unit of length, e.g., dynes/cm,
and is usually denoted by the symbol σ. If a glass capillary tube is placed in a large open
vessel containing water, the combination of surface tension and wettability of tube to
water will cause water to rise in the tube above the water level in the container outside the
tube as shown in Figure 3.The water will rise in the tube until the total force acting to pull
the liquid upward is balanced by the weight of the column of liquid being
Figure 3. Pressure relations in capillary tubes.
Supported in the tube. After derivation:
Eq(8)
(
)
Eq(9)
Where :
σgw= the surface tension between air and water ,dynes/cm
[79]
Dr.Jawad R.R. Alassal Kirkuk U.
σow= the interfacial tension between oil and water ,dynes/cm
h =height to which the liquid is held, cm
g =acceleration due to gravity, cm/sec2
ρw=density of water, gm/cm3
ρo=density of water, gm/cm3
ρg=density of water, gm/cm3
r =radius , cm
CAPILLARY PRESSURE
The capillary forces in a petroleum reservoir are the result of the combined effect of the
surface and interfacial tensions of the rock and fluids, the pore size and geometry, and the
wetting characteristics of the system
When two immiscible fluids are in contact, a discontinuity in pressure exists between the
two fluids, which depends upon the curvature of the interface separating the fluids. This
pressure difference called the capillary pressure (pc)
The displacement of one fluid by another in the pores of a porous
medium is either aided or opposed by the surface forces of capillary pressure.
Denoting the pressure in the wetting fluid by pw and that in the nonwetting fluid by pnw, the capillary pressure can be expressed as:
Capillary pressure = (pressure of the nonwetting phase) -(pressure of the wetting
phase)
Eq(10)
That is, the pressure excess in the nonwetting fluid is the capillary
pressure, and this quantity is a function of saturation..
There are three types of capillary pressure:
• Water-oil capillary pressure (denoted as Pcwo)
• Gas-oil capillary pressure (denoted as Pcgo)
• Gas-water capillary pressure (denoted as Pcgw)
Applying the mathematical definition of the capillary pressure as
expressed by Equation 14, the three types of the capillary pressure can
be written as:
pcow = po -pw
Pcgo = pg - po
pcgw = pg - pw
where :
pg, po, and pw represent the pressure of gas, oil, and water, respectively.
Referring to Figure 3, the pressure difference across the interface
between Points 1 and 2 is essentially the capillary pressure, i.e.:
[80]
Dr.Jawad R.R. Alassal Kirkuk U.
pc = p1 -p2
Eq (11)
The pressure of the water phase at Point 2 is equal to the pressure at
point 4 minus the head of the water, or:
p2 = p4 -ghρw
Eq(12)
The pressure just above the interface at Point 1 represents the pressure of the air and is
given by:
p1 = p3 -ghρair
Eq (13)
It should be noted that the pressure at Point 4 within the capillary tube is the same
as that at Point 3 outside the tube. Subtracting Equation 12 from 13 gives
pc = gh (ρw - ρair) = gh∆ρ
Eq(14)
where
∆ρ is the density difference between the wetting and nonwetting
phase. The density of the air (gas) is negligible in comparison with the
water density. In practical units, Equation 14 can be expressed as:
)
=(
where
pc = capillary pressure, psi
h = capillary rise, ft
∆ρ = density difference, lb/ft3
In the case of an oil-water system, Equation 14 can be written as:
pc = g h (ρw - ρo) = g h ∆ρ
Eq(15)
and in practical units
=(
)(
)
• Gas-liquid system
(
=
)
Eq(16)
and
(
(
)
Eq(17)
)
Where
ρw = water density, gm/cm3
σgw = gas-water surface tension, dynes/cm
r = capillary radius, cm
θ= contact angle
h = capillary rise, cm
g = acceleration due to gravity, cm/sec2
[81]
Dr.Jawad R.R. Alassal Kirkuk U.
pc = capillary pressure, dynes/cm2
• Oil-water system
=
(
)
(
)
Eq(18)
and
(
Eq(19)
)
Example
Calculate the pressure difference, i.e., capillary pressure, and capillary
rise in an oil-water system from the following data:
θ=30° , ρw =1.0 gm/cm3 , ρo=0.75 gm/cm3
r =10-4 cm , σow = 25 dynes/cm
Solution
Step 1
Apply Equation 18 to give
Since 1 dyne/cm2 =1.45 ×10-5 psi, then
pc =6.28 psi
This result indicates that the oil-phase pressure is 6.28 psi higher than the water-phase
pressure
Step 2.
Calculate the capillary rise by applying Equation 19.
(
or:
=(
6.28 =0
)(
)
)
(
) , h = 58 ft
Capillary Pressure of Reservoir Rocks
The capillary pressure that exists within a porous medium between two immiscible
phases is a function of the interfacial tensions and the average size of the capillaries
[82]
Dr.Jawad R.R. Alassal Kirkuk U.
which, in turn, controls the curvature of the interface. In addition, the curvature is also a
function of the saturation distribution of the fluids involved.
Laboratory experiments have been developed to simulate the displacing forces in a
reservoir in order to determine the magnitude of the capillary forces in a reservoir and,
thereby, determine the fluid saturation distributions and connate water saturation. One
such experiment is called the restored capillary pressure technique which was developed
primarily to determine the magnitude of the connate water saturation. A diagrammatic
sketch of this equipment is shown in figure 4. Briefly, this procedure consists of saturating
a core 100% with the reservoir water and then placing the core on a porous membrane
which is saturated 100% with water and is permeable to the water only, under the pressure
drops imposed during the experiment, air is then admitted into the core chamber and the
pressure is increased until a small amount of water is displaced through the porous, semipermeable membrane into the graduated cylinder. Pressure is held constant until no more
water is displaced, which may require several days or even several weeks, after which the
core is removed from the apparatus and the water saturation determined by weighing. The
core is then replaced in the apparatus, the pressure is increased, and the procedure is
repeated until the water saturation is reduced to a minimum
The data from such an experiment are shown in Figure 5. Since the pressure required to
displace the wetting phase from the core is exactly equal to the capillary forces holding
the remaining water within the core after equilibrium has been reached, the pressure data
can be plotted as capillary pressure data. Two important phenomena can be observed in
Figure 5. First, there is a finite capillary pressure at 100% water saturation that is
necessary to force the nonwetting phase into a capillary filled with the wetting phase. This
minimum capillary pressure is known as the displacement pressure, pd. If the largest
capillary opening is considered as circular with a radius of r, the pressure needed for
forcing the wetting fluid out of the core is:
=
(
)
This is the minimum pressure that is required to displace the wetting phase from the
largest capillary pore because any capillary of smaller
[83]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 4. Capillary pressure equipment.
radius will require a higher pressure. As the wetting phase is displaced, the second
phenomenon of any immiscible displacement process is encountered, that is, the reaching
of some finite minimum irreducible saturation. This irreducible water saturation is
referred to as connate water
It is possible from the capillary pressure curve to calculate the average
size of the pores making up a stated fraction of the total pore space. Let pc be the average
capillary pressure for the 10% between saturation of 40 and 50%. The average capillary
radius is obtained from
[84]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 5. Capillary pressure curve.
=
(
)
The above equation may be solved for r providing that the interfacial
tension σ, and the angle of contact θ may be evaluated.
Figure 6 is an example of typical oil-water capillary pressure curves.
In this case, capillary pressure is plotted versus water saturation for four rock samples
with permeabilities increasing from k1 to k4. It can be seen that, for decreases in
permeability, there are corresponding increases in capillary pressure at a constant value of
water saturation. This is a reflection of the influence of pore size since the smaller
diameter pores will invariably have the lower permeabilities. Also, as would be expected
the capillary pressure for any sample increases with decreasing water saturation, another
indication of the effect of the radius of curvature of the water-oil interface
[85]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 6 Variation of capillary pressure with Permeability
Capillary Hysteresis
It is generally agreed that the pore spaces of reservoir rocks were originally filled with
water, after which oil moved into the reservoir, displacing some of the water and reducing
the water to some residual saturation. When discovered, the reservoir pore spaces are
filled with a connate- water saturation and an oil saturation. All laboratory experiments
are designed to duplicate the saturation history of the reservoir. The process of generating
the capillary pressure curve by displacing the wetting phase, i.e., water, with the
nonwetting phase (such as with gas or oil), is called the drainage process.
This drainage process establishes the fluid saturations which are found when the reservoir
is discovered. The other principal flow process of interest involves reversing the drainage
process by displacing the non-wetting phase (such as oil) with the wetting phase, (e.g.,
water). This displacing process is termed the imbibition process and the resulting curve is
termed the capillary pressure imbibition curve. The process of saturating and desaturating
a core with the nonwetting phase is called capillary hysteresis. Figure 7 shows typical
drainage and imbibitions capillary pressure curves.
Frequently, in natural crude oil-brine systems, the contact angle or wettability may change
with time. Thus, if a rock sample that has been thoroughly cleaned with volatile solvents
is exposed to crude oil for a period of time, it will behave as though it were oil wet. But if
it is exposed to brine after cleaning, it will appear water wet. At the present time, one of
[86]
Dr.Jawad R.R. Alassal Kirkuk U.
the greatest unsolved problems in the petroleum industry is that of wettability of reservoir
rock.
Figure 7. Capillary pressure hysteresis.
Initial Saturation Distribution In a Reservoir:
An important application of the concept of capillary pressures pertains
to the fluid distribution in a reservoir prior to its exploitation.
The capillary pressure-saturation data can be converted into height-saturation data by
arranging Equation 19 and solving for the height h above the free- water level
[87]
Dr.Jawad R.R. Alassal Kirkuk U.
Where
pc = capillary pressure, psia
∆ρ = density difference between the wetting and nonwetting phase, lb/ft3
h = height above the free-water level, ft
Figure 8 shows a plot of the water saturation distribution as a function of distance from
the free-water level in an oil-water system. It is essential at this point to introduce and
define four important concepts:
• Transition zone
• Water-oil contact (WOC)
• Gas-oil contact (GOC)
• Free water level (FWL)
Figure 8. Water saturation profile.
[88]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 9 Initial saturation profile in a combination-drive reservoir.
Figure 9 illustrates an idealized gas, oil, and water distribution in a reservoir. The figure
indicates that the saturations are gradually charging from 100% water in the water zone to
irreducible water saturation some vertical distance above the water zone. This vertical
area is referred to as the transition zone, which must exist in any reservoir where there is
a bottom water table. The transition zone is then defined as the vertical thickness over
which the water saturation ranges from 100% saturation to irreducible water saturation
Swc. The important concept to be gained from figure 9 is that there is no abrupt change
from 100% water to maximum oil saturation.
The creation of the oil-water transition zone is one of the major effects of capillary forces
in a petroleum reservoir.
Similarly, the total liquid saturation (i.e. oil and water) is smoothly
changing from 100% in the oil zone to the connate water saturation in the gas cap zone. A
similar transition exists between the oil and gas zone. figure 9 serves as a definition of
what is meant by gas-oil and water-oil contacts. The WOC is defined as the ―uppermost
depth in the reservoir where a 100% water saturation exists.‖ The GOC is defined as the
―minimum depth at which a 100% liquid, i.e., oil + water, saturation exists in the reservoir
Section A of figure 10 shows a schematic illustration of a core that is represented by five
different pore sizes and completely saturated with water, i.e., wetting phase. Assume that
we subject the core to oil (the nonwetting phase) with increasing pressure until some
water is displaced from the core, i.e., displacement pressure pd. This water displacement
will occur from the largest pore size
The oil pressure will have to increase to displace the water in the second largest pore. This
sequential process is shown in sections B and C of figure 10. It should be noted that there
[89]
Dr.Jawad R.R. Alassal Kirkuk U.
is a difference between the free water level (FWL) and the depth at which 100% water
saturation exists. From a reservoir engineering standpoint, the free water level is defined
by zero capillary pressure. Obviously, if the largest pore is so large that there is no
capillary rise in this size pore, then the free water level and 100% water saturation level,
i.e., WOC, will be the same. This concept can be expressed mathematically by the
following relationship:
Eq (20)
Where
pd = displacement pressure, psi
∆ρ = density difference, lb/ft3
FWL = free water level, ft
WOC = water-oil contact, ft
Figure 10. Relationship between saturation profile and pore-size distribution.
Example
The reservoir capillary pressure-saturation data of the X Oil reservoir is shown graphically
in Figure 11. Geophysical log interpretations and core analysis establish the WOC at 5023
ft. The following additional
data are available:
• Oil density = 43.5 lb/ft3
• Water density = 64.1 lb/ft3
• Interfacial tension = 50 dynes/cm
[90]
Dr.Jawad R.R. Alassal Kirkuk U.
Calculate:
• Connate water saturation (Swc)
• Depth to FWL
• Thickness of the transition zone
• Depth to reach 50% water saturation
Solution
a. From figure 11, connate-water saturation is 20%.
b. Applying Equation 20 with a displacement pressure of 1.5 psi gives
(
)
c. Thickness of transition zone =
(
(
)
)
d. Pc at 50% water saturation = 3.5 psia
Equivalent height above the FWL = (144) (3.5)/(64.1 - 43.5) = 24.5 ft
Depth to 50% water saturation = 5033.5 - 24.5 = 5009 ft
The above example indicates that only oil will flow in the interval
between the top of the pay zone and depth of 4991.5 ft. In the transition zone, i.e., the
interval from 4991.5 ft to the WOC, oil production would be accompanied by
simultaneous water production.
It should be pointed out that the thickness of the transition zone may
range from few feet to several hundred feet in some reservoirs. Recalling the capillary rise
equation, i.e., height above FWL,
h=
(
)
[91]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 11. Capillary pressure saturation data.
The above relationship suggests that the height above FWL increases with decreasing the
density difference ∆ρ.
From a practical standpoint, this means that in a gas reservoir having a gas-water contact,
the thickness of the transition zone will be a minimum since ∆ρ will be large. Also, if all
other factors remain unchanged, a low API gravity oil reservoir with an oil-water contact
will have a longer transition zone than a high API gravity oil reservoir.
Example
A four-layer oil reservoir is characterized by a set of reservoir capillary
pressure-saturation curves as shown in Figure 12. The following additional data are also
available
Layer
1
2
3
4
Depth, ft
4000 – 4010
4010 – 4020
4020 – 4035
4035 – 4060
Permeability, md
80
100
70
90
WOC =4060 ft
Water density =65.2 lb/ft3
Oil density =55.2 lb/ft3
Calculate and plot water saturation versus depth for this reservoir.
[92]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 12. Variation of pc with k.
Solution
1-Establish the FWL by determining the displacement pressure pd
for the bottom layer, i.e., Layer 4, and apply Equation 20:
• pd =0.75 psi
FWL = 4060 +
(
)
= 4070.8 ft
2- The top of the bottom layer is located at a depth of 4035 ft which
is 35.8 ft above the FWL. Using that height h of 35.8 ft, calculate
the capillary pressure at the top of the bottom layer.
(
)
)(
(
)
• From the capillary pressure-saturation curve designated for
Layer 4, read the water saturation that corresponds to a pc of
2.486 to give Sw = 0.23.
• Assume different values of water saturations and convert the
corresponding capillary pressures into height above the FWL by
applying Equation 19
(
)
[93]
Dr.Jawad R.R. Alassal Kirkuk U.
Sw
Pc , psi
h , ft
Depth = FWL - h
0.23
0.25
0.30
0.40
0.50
0.60
0.70
0.80
0.90
2.486
2.350
2.150
1.800
1.530
1.340
1.200
1.050
0.900
35.8
33.84
30.96
25.92
22.03
19.30
17.28
15.12
12.96
4035
4037
4040
4045
4049
4052
4054
4056
4058
3-The top of Layer 3 is located at a distance of 50.8 ft from the
FWL (i.e., h = 4070.8 -4020 = 50.8 ft). Calculate the capillary
pressure at the top of the third layer:
(
)
(
)(
)
• The corresponding water saturation as read from the curve designated for Layer 3 is
0.370
• Construct the following table for Layer 3.
Sw
0.37
0.40
0.50
0.60
Pc , psi
3.53
3.35
2.75
2.50
h , ft
50.8
48.2
39.6
36.0
Depth=FWL - h
4020
4023
4031
4035
4 -.
• Distance from the FWL to the top of Layer 2 is:
h = 4070.8 - 4010 = 60.8 ft
(
)
(
)(
)
• Sw at pc of 4.22 psi is 0.15.
• Distance from the FWL to the bottom of the layer is 50.8 ft that
corresponds to a pc of 3.53 psi and Sw of 0.15. This indicates
that the second layer has a uniform water saturation of 15%.
[94]
Dr.Jawad R.R. Alassal Kirkuk U.
5-. For Layer 1, distance from the FWL to the top of the layer:
• h = 4070.8 – 4000 = 70.8 ft

(
)(
)
• Sw at the top of Layer 1 = 0.25
• The capillary pressure at the bottom of the layer is 3.53 psi with
a corresponding water saturation of 0.27.
6- Figure 13 documents the calculated results graphically. The figure
indicates that Layer 2 will produce 100% oil while all remaining
layers produce oil and water simultaneously.
Figure 13. Water saturation profile
Leverett J-Function
Capillary pressure data are obtained on small core samples that represent an extremely
small part of the reservoir and, therefore, it is necessary to combine all capillary data to
classify a particular reservoir. The fact that the capillary pressure-saturation curves of
nearly all naturally porous materials have many features in common has led to attempts to
devise some general equation describing all such curves. Leverett (1941) approached the
problem from the standpoint of dimensional analysis. Realizing that capillary pressure
should depend on the porosity, interfacial tension, and mean pore radius, Leverett defined
the dimensionless function of saturation, which he called the J-function, as
[95]
Dr.Jawad R.R. Alassal Kirkuk U.
(
)
√
Eq(21)
where
J(Sw) =Leverett J-function
pc =capillary pressure, psi
σ=interfacial tension, dynes/cm
k =permeability, md
ϕ=fractional porosity
In doing so, Leverett interpreted the ratio of permeability, k, to porosity, ϕ , as being
proportional to the square of a mean pore radius.
The J-function was originally proposed as a means of converting all
capillary-pressure data to a universal curve.
Example
A laboratory capillary pressure test was conducted on a core sample
taken from the XX Field. The core has a porosity and permeability
of 16% and 80 md, respectively. The capillary pressure-saturation data
are given as follows:
Sw
1.0
0.8
0.6
0.4
0.2
Pc , psi
0.50
0.60
0.75
1.05
1.75
The interfacial tension is measured at 50 dynes/cm. Further reservoir engineering analysis
indicated that the reservoir is better described at a porosity value of 19% and an absolute
permeability of 120 md. Generate the capillary pressure data for the reservoir.
Solution
1- Calculate the J-function using the measured capillary pressure data.
(
Sw
1.0
0.8
)
(
)√
Pc , psi
0.50
0.60
[96]
Dr.Jawad R.R. Alassal Kirkuk U.
J(Sw) = 0.096799 ( Pc)
0.048
0.058
0.6
0.4
0.2
0.75
1.05
1.75
0.073
0.102
0.169
2- Using the new porosity and permeability values, solve Equation
21 for the capillary pressure pc.
( )
√
(
)
√
(
)
3- Reconstruct the capillary pressure-saturation table.
Sw
J(Sw)
Pc = 9.192 J (Sw)
1.0
0.048
0.441
0.8
0.058
0.533
0.6
0.073
0.671
0.4
0.102
0.938
0.2
0.169
1.553
Some times we can relate Sw and J as:
Sw = a. Jb
Where:
a = 0.175381 x ϕ.286354
b = -0.461911 x ϕ.003849
and J relates Sw to height above contact as:
J = (0.21645 x (Water gradient – Oil gradient ) x height above FWL x ( √ )
Converting Laboratory Capillary Pressure Data
For experimental convenience, it is common in the laboratory determination of capillary
pressure to use air-mercury or air-brine systems, rather than the actual water-oil system
characteristic of the reservoir. Since the laboratory fluid system does not have the same
[97]
Dr.Jawad R.R. Alassal Kirkuk U.
surface tension as the reservoir system, it becomes necessary to convert laboratory
capillary pressure to reservoir capillary pressure. By assuming that the Leverett J-function
is a property of rock and does not change from the laboratory to the reservoir, we can
calculate reservoir capillary pressure as shown below:
(
)
(
)
(
)
(
)
Contact Angle
Interfacial Tension, σ dyne/ cm
Air- water
0
72
Oil- water
30
48
Air – Mercury
Air – oil
Reservoir
Water- oil
Water - gas
140
0
480
24
30
0
25 -30
50
System
Laboratory
Example:
Convert the laboratory – measured capillary pressure data to reservoir conditions and
determine the saturation distribution
Laboratory σ (air-water)
= 72 dyne/cm
θ (air- water)
= 00
Reservoir σ (oil-water)
= 24 dyne/cm
θ (oil- water)
= 300
ρw
= 65 lb/ft3
ρo
= 53 lb/ft3
Solution:
Using the above equation:
(Pc)res = (Pc)Lab x 24 cos 30/(72 cos 0) = 0.289 (Pc )Lab
Since : h = 144 Pc /Δρ
Then the height above the free water level can be estimated. The table below shows the
results of laboratory capillary pressure measurements using air and water:
[98]
Dr.Jawad R.R. Alassal Kirkuk U.
Sw , %
100
90
80
70
60
50
40
30
20
10
Lab measured Pc ,
air-water (psig)
2
3
4
5
6
7
8
10
27
75
Reservoir Pcow
Oil-water(psig)
0.578
0.867
1.16
1.45
1.73
2.02
2.31
5.8
7.8
21.7
Hight above free
water level , ft
6.9
10.4
13.9
17.4
20.8
24.2
27.7
70
94
260
The figure below is the plot of height and capillary pressure vs. saturation
Water – Wet
Oil – Wet
Connate Water Saturation Usually greater than 20 to Generally less than 15%
25 % of PV
of PV, frequently less
than 10%
Saturation at which oil
Greater than 50% water
Less than 50% water
and water permeabilities saturation
saturation
are equal ( crossover
saturation)
Relative permeabilities to Generally less than 30% Greater than 50% and
water at maximum water
approaching 100%
saturation i.e flood out
[99]
Dr.Jawad R.R. Alassal Kirkuk U.
Problems
1. Calculate the reservoir height of water ( 100% water saturated zone) above the FWL ,
using the following capillary pressure data :Pc laboratory = 18 psi , Swc = 35% , Swo = 24 dyne/cm
= 68 Ib/ft3 , Sgw = 72 dyne/cm ,
= 53 Ib/ft3 ,
If Water – Gas system was used during laboratory test While the true reservoir fluid are
Water – Oil ?
Also, Schematically draw the height Vs. Saturation
Showing the level and depth of each zone if the transition zone capillary pressure equal 3
psi , FWL depth = 1200 ft ?
2. An oil water capillary experiment on a core sample gives the following results :Pc
0
4.4
5.3
5.6
10.5
15.7
35
Sw %
100
100
90.1
82.4
43.7
32.2
29.8
Given that sample was taken from a point 100 ft above the oil-water contact ; What is the
expected water saturation at that elevation ? Also show the depth of each zone if the FWL
depth is 1200 ft , Swc = 25 ?
3. The capillary pressure data for a water – oil System are given :Sw
0.25
0.3
0.4
0.5
1
1
Pc
35
16
8.5
5
3
0
The core sample used in generalizing the caplillary pressure data was taken from a layer
that is characterized by an absolute permeability of 300 md and porosity of 17% .
Generate the capillary pressure data for a different layer that is characterized by a porosity
& permeability of 15% , 200md respectively ; the interfacial tension is measured at 35
dyne/cm
Also draw and determine the connate water saturation , FWL if you know that the WOC
depth = 3120 ft ?
[100]
Dr.Jawad R.R. Alassal Kirkuk U.
4. Given the following Hg injection data :PcL (psia)
0
5
10
20
30
40
100
200
S (Hg)
0
0
0.1
0.3
0.6
0.8
0.9
0.9
Also : -

Calculate the height of water sat. (0.5) above free water level ?

What is the D.P ?

Calculate thickness of T.Z. ?
5. A res. Rock showed the following Pc data :Pc (lab)
0
2
3
5
10
20
S (Hg)
0
0
0
0.1
0.4
0.7
Given
Pc corr. = PcL /12 ;
;
;
Find :
D.P

Largest pore radius

Sw at point (10 ft) above WOC ?
[101]
Dr.Jawad R.R. Alassal Kirkuk U.
6. For the following capillary pressure data by mercury injection :Pc(psi)
0
1.5
4.5
9
20
40
120
250
Hg vol
0
0.1
0.5
1.2
1.5
1.7
2
2
Pore vol. = 2.5 cm3 ;
;

Calculate the thickness of T.Z.

Calculate the size of largest pores in the core sample

Calculate the size of pores where mercury entered at press. Of (40 psi)
7. When oil was displacing water in pore the radius of curvature of the interphases was
0.01 cm and the angle of the inter phase
Calculate

Pressure associated with the pore

Average pore radius

For imbibition case Ɵ changes by
what’s Pc for this case .
8. Height of water in capillary tube (5cm) ; tube radius = 0.01 cm
Calculate Ɵ ?
9. A Laboratory capillary pressure test was conducted on a core sample taken from XX
field. The core has a porosity and permeability of 16% and 80 md, respectively.The
capillary-pressure saturation data are given as follows :
Sw
Pc , psi
1.0
0.50
0.8
0.60
0.6
0.75
0.4
1.05
0.2
1.75
[102]
Dr.Jawad R.R. Alassal Kirkuk U.
The interfacial tension is measured at 50 dyne / cm .Further – reservoir Engineering
analysis indicates that the reservoir is better described at a porosity value of 19% and an
absolute permeability of 120 md . Generate the capillary pressure data for the reservoir .
10. A reservoir has an oil zone thickness of 40 ft. Determine the thickness of the
following : Water zone , Transition zone, and Gas zone .Using the following information :
,
,
,
,
,
,
(
)
[103]
Dr.Jawad R.R. Alassal Kirkuk U.
,
Chapter - 3 GAS PROPERTIES
A gas is defined as a homogeneous fluid of low viscosity and density that has no definite
volume but expands to completely fill the vessel in which it is placed. Generally , the
natural gas is a mixture of hydrocarbon and nonhydrocarbon gases . The hydrocarbon
gases are normally found in a natural gas are methane , ethane , propanes , butanes ,
pentanes , and small amounts of hexanes and heavier . the nonhydrocarbon gases (i.e.,
impurities) include carbon dioxide , hydrogen sulfide , and nitrogen.
Knowledge of pressure-volume-temperature (PVT) relationships and other physical and
chemical properties of gases is essential for solving problems in natural gas reservoir
engineering .These properties include :
Apparent molecular weight , Ma
Specific gravity .
Compressibility factor , z
Density ,
Specific volume , v
Isothermal gas compressibility coefficient , Cg
Gas formation volume factor , Bg
Gas Expansion factor , Eg
Viscosity , µg
The above gas properties may be obtained from direct laboratory measurments or by
prediction from generalized mathematical expressions .
The Standard Condition (SC):
P= Pressure = 14.7 psia = 101.325 Kpa (SPE uses 100 Kpa)
T= Temperature = 600F =288.720K (SPE uses 2880 K)
[104]
Dr.Jawad R.R. Alassal Kirkuk U.
Gases are classified as:
 Ideal Gases
 Real gases
Ideal Gases : Is defined as one in which :
1. The Volume occupied by the molecules is small compared to the total volume.
2. All molecular collisions are elastic.
3. There are no attractive or repulsive forces among the molecules.
Early Gas Laws:
1. Boyle’s Law:
Robert Boyle (1627-1691), during the course of experiments with air, observed the
following relation between pressure and volume: if the temperature of a given
quantity of gas is held constant, the volume of gas varies inversely with the absolute
pressure. This relation, written as an equation, is
Vα
or PV = Constant , T =constant
In the application of Boyle's law, volume at a second set of pressure conditions is
generally desired.
2. Charle’s Law:
V α T or
= Constant , at low pressure.
3. Avogadro’s Law:
Under the same conditions of P and T, equal volumes of all ideal gases contain the
same number of molecules = 2.73x1026 molecules /lb-mole of ideal gas. Or at a given
P & T one molecular weight of any ideal gas occupies the same volume as one
molecular weight of another ideal gas.
One molecular weight in pounds of any ideal gas at 600 F & 14.7 psia occupies a
volume of 379.4 SCF.
The three gas laws, can be combined to express a relationship among P, V, & T,
called the ideal gas law.
[105]
Dr.Jawad R.R. Alassal Kirkuk U.
For a given quantity of gas
Or
a constant
=R
Where : Vm = the volume of one molecular weight of the gas at P & T.
For n moles of ideal gas:
PV = nRT
n=
or
since
Therefore
Specific Volume = 1/
Value of R is:
R=
=
= 10.73 psi scf/lb mole R0
Example:
Calculate the mass of methane gas contained at 1000 psia & 680F in a cylinder with
volume of 3.2 Cuf. Assume the methane is an ideal gas.
The molecular weight of methane is 16 lbm/lb-mole.
Solution:
PV = nRT ,
m=
HW.
Calculate the density of Methane at SC.
[106]
Dr.Jawad R.R. Alassal Kirkuk U.
Ideal Gas Mixtures:
For a mixture of ideal gases, we have to include two additional ideal gas laws:
Dalton’s Law: each gas in a mixture of gases exerts a pressure equal to that which it
would exert if it occupied the same volume as the total mixture. This pressure is called the
partial pressure. The total pressure is the sum of the partial pressures. This law is valid
only when the mixture and each component of the mixture obey the ideal gas law. It is
some times called the law of additive pressures.
Consider a mixture containing nA moles of component A, nB moles of component B and nC
moles of component C. The partial pressure exerted by each component of the gas mixture
may be determined with ideal gas equation:
PA = nA RT/V, PB = nB RT/V, PC = nC RT/V
According to Dalton’s law, the total pressure is the sum of the partial pressures :
P = PA + PB + PC
P = nA RT/V + nB RT/V + nC RT/V
∑
P=
=
Therefore:
∑
Where : is defined as the mole fraction of the jth component in the gas mixture; so the
partial pressure of the j component is :
Pj = P x yj
Amagat’s Law:
The total volume of gaseous mixture is the sum of the volumes that each component
would occupy at the given P & T. The volumes occupied by the individual components
are known as partial volumes. This law is correct only if the mixture and each component
obey the ideal gas law.
VA = nA RT/P, VB = nB RT/P, VC = nC RT/P
V = VA + VB + VC
[107]
Dr.Jawad R.R. Alassal Kirkuk U.
V = nA RT/P + nB RT/P + nC RT/P
∑
V=
=
∑
So the volume fraction is equal to the mole fraction.
Apparent Molecular Weight:
One of the main gas properties that is frequently of interest to engineers is the apparent
molecular weight . If yi represent the mole fraction of the ith component in a gas mixture ,
the apparent molecular weight is defined mathematically by the following equation :
Where Ma = apparent molecular weight of a gas mixture.
Mi = molecular weight of the ith component in the mixture.
yi = mole fraction of component i in the mixture.
Example:
Dry air is a gas mixture consisting essentially of Nitrogen, Oxygen, and small amount of
other gases. Compute the apparent molecular weight of air given its approximate
composition as:
Component
Nitrogen (N2)
Oxygen (O2)
Argon (A)
Mole Fraction(yi)
0.78
0.21
0.01
Σ = 1.00
Solution:
Ma = yN2 x MN2 + yO2 x MO2 + yA x MA
[108]
Dr.Jawad R.R. Alassal Kirkuk U.
Molecular Weight
28.01
32
39.94
= (0.78) x (28.01) +(0.21) x (32.00) +(0.01) x (39.94) = 28.97 ≈ 29.0
The specific gravity of a gas is defined as the ratio of the density of the gas to the density
of dry air taken at standard conditions of P & T
or =
Mg =
HW:
Calculate the gravity of a natural gas has the following composition:
Component
Methane
Ethane
Propane
n- butane
Ans. =.66
Mole fraction
0.85
0.09
0.04
0.02
Molecular weight
16
30.1
44.1
58.1
Example:
A gas well is producing gas with a specific gravity of 0.65 at a rate of 1.1 MMscf/ day.
The average reservoir pressure and temperature are 1500 psi and 1500 F. Calculate :
1. Apparent molecular weight.
2. Gas density at reservoir temperature
3. Flow rate in lb/day.
Solution:
1-The molecular weight is
Mg =
= 28.96 x 0.65 = 18.82
23- Because 1 lb – mole of any gas occupies 379.4 SCF at standard conditions, then the
daily number of moles that the gas well is producing can be calculated from:
n=
= 2899 lb- mol
[109]
Dr.Jawad R.R. Alassal Kirkuk U.
n = mass/ Ma
therefore:
m = 2899 x 18.82 = 54559 lb/day
HW.
A gas well is producing a natural gas with the following composition.
Component
Mole Fraction
Co2
0.05
C1
0.9
C2
0.03
C3
0.02
Assuming an ideal gas behavior, calculate:
1. Apparent molecular weight.
2. Specific gravity.
3. Gas density at 2000 psia and 1500 F
4. Specific volume at 2000 psia and 1500 F
REAL GASES:
At high pressure & temperature, the gases are no longer behave as ideal gases; so
correction must be made to account for the deviation from ideal behavior. The most
widely used correction method is the gas deviation factor, more commonly called the ZFactor, it is defined as the ratio of the actual volume occupied by a mass of gas at some P
& T to the volume of the gas would occupy if it behaved ideally :
Z=
P xVideal = nRT or P x Vactual / Z = nRT
Or P xV = ZnRT
The Z – factor varies with changes in gas composition, T, and P
[110]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure (1) shows Z as a function of P at constant T.
At very low pressure the molecules are relatively far apart and the conditions of ideal gas
behavior are more likely to be met at low P & T the factor approaches a value of 1.0
Example:
Calculate the mass of Methane gas contained at 1000 psia & 680 F in a cylinder with
volume of 3.2 cuft.
Solution:
From Figure 2
Z = 0.89
m =
m=(
(
)(
)(
)(
)(
)
)
[111]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 2(a)
[112]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 2(b)
[113]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 2(c)
[114]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 2(d)
[115]
Dr.Jawad R.R. Alassal Kirkuk U.
Real Gas Mixtures:
Z – factor charts are available for most of the single component light hydrocarbon gases,
but in practice a single component gas is rarely encountered. In order to get Z- factor for
natural gas mixtures, the law of corresponding state is used. This law states that the ratio
of the value of any intensive property ( properties are independent of the quantity of
material present: density, specific volume and Z- factor, while extensive properties such
as volume and mass are extensive; their values are determined by the total quantity of
matter present) to the value of that property at the critical state is related to the ratios of
the prevailing absolute T & P to the critical T & P by the same function for all similar
substances. This means that all pure gases have the same Z factor at the same values of
reduced P & T, where the reduced values are defined as:
,
Where:
TC & PC are the critical T & P for the gas respectively.
Example:
Calculate the density of ethane at 900 psia & 110 0 F , TC = 5500 R , PC = 708 psia, & M
= 30.1lb/ lb- mole.
Solution:
= ( 110 + 460)/ 550 = 1.04
= 900/708 = 1.27
From Figure 3 Z = 0.34
(
(
)(
)(
)
)(
)
[116]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 3: Z factor for pure hydrocarbon
The law of corresponding states have been extended to cover mixtures of gases that are
closely related chemically.
To obtain the critical point for multicomponent mixtures, the quantities of pseudo critical
T & pseudo critical P have been conceived:
Tpc = Σ yj Tcj , Ppc = Σ yj Pcj
,
[117]
Dr.Jawad R.R. Alassal Kirkuk U.
Example:
Calculate the pseudo critical T and pseudo critical P of the following natural gas mixture,
and calculate the volume occupied by 1 lb- mole at 1200F and 1500 psia.
Component
C1
C2
C3
n-C4
TC,0R
343.1
549.8
665.7
765.3
Mole Fraction
0.85
0.09
0.04
0.02
Solution:
Tpc = Σ yj Tcj = 383.0R , Ppc = Σ yj Pcj = 667.0 psia
= 580 / 383 = 1.51,
= 1500 / 667 = 2.25
Z= 0.813 from figure 4, V = ZnRT/ P = 3.37 cuft
[118]
Dr.Jawad R.R. Alassal Kirkuk U.
PC, psia
667.8
707.8
616.3
550.7
Figure 4
[119]
Dr.Jawad R.R. Alassal Kirkuk U.
HW:
A gas mixture consists of 50% C1 , 30% C2 and 20% C3 by weight.
Calculate the apparent molecular weight and the specific gravity of the mixture.
Component
C1
C2
C3
Mass, lb
50
30
20
The pseudo critical properties can be calculated using the following equations:
Tpc = 170.5 + 307.3
, Ppc = 709.6 – 58.7
For condensate fluid:
Tpc = 187 + 330 - 71.5
Ppc = 706 – 51.7 - 11.1
CORRECTION FOR NONHYDROCARBON IMPURTIES:
Natural gases frequently contain materials other than hydrocarbon, such as N 2, CO2, and
H2S. The presence of these impurities affects the value obtained from the Z-factor chart.
The procedure for obtaining the Z-factor for sour gas is:
1. Determine Ppc & Tpc for the gas using the gas composition.
2. Calculate the adjusted critical properties :
,
(
)
= 120 (A0.9 – A1.6) + 15 (B0.5 – B4)
B = mole fraction of H2S
A = mole fraction of CO2 +B
= correction factor,0F
3. Calculate the reduced properties using corrected critical properties :
[120]
Dr.Jawad R.R. Alassal Kirkuk U.
Tr = T /
, Pr = P/
4. Find Z factor from chart.
Or you can use the following correlation (Carr- Kobayashi – Burrows)
= TpC – 80 yCO2 + 130 yH2S -250 yN2
= PpC + 440 yCO2 + 600 yH2S -170 yN2
Tr = T /
, Pr = P/
Gas Formation Volume Factor ( βg):
βg is a conversion factor from surface conditions of P & T to insitu conditions.
βg = Volume at P & T/ Volume at SC
= ( Z T PSC)/( ZSC TSC P)
Using TSC = 5200F and PSC = 14.7 psia , ZSC = 1.0 , then
βg = 0.0283 ( Z T)/ P cuft/ Scf
Gas Expansion Factor (Eg)
Eg = 1/ βg = 35.37 P/(ZT) Scf/ cuft
Example:
A gas well is producing at a rate of 15000 ft3/day from a gas reservoir at an average
pressure of 2000 psia and temperature of 1400 F. The specific gravity is 0.72. Calculate
the gas flow rate in Scf/day.
Solution:
Calculate the pseudo properties; using the gas gravity
Tpc = 395.5 0R , Ppc = 668.4 psia
Calculate :Tpr = 600/395.5 = 1.52. , Ppr = 2000/ 668.4 = 2.99
Determine the Z- factor from Figure , Z = 0.78
Calculate the gas expansion factor ( Eg):
[121]
Dr.Jawad R.R. Alassal Kirkuk U.
Eg = 35.37 x 2000/(0.78 x 600) = 151.15 Scf/ cuft
The gas flow rate in Scf/ day is = 151.15 x 15000 = 2267000 = 2.267 MMScf/day.
Gas Compressibility:
The isothermal compressibility of agas is the measure of the change in volume per unit
volume with pressure change at constsnt temperature.
Cg = -
( )
Ideal gas compressibility:
V = nRT/ P , ( ) = - nRT/ P2
And therefore :
Cg = - 1/V x ( ) = - (P/nRT) x ( -nRT/P2) = 1/ P
Real gas compressibility:
V = ZnRT/ P , ( ) = nRT
Cg =
And the reduced compressibility has been defined as Cr
Cr = C x Pc
Figure 5 the Pseudo critical properties of a
natural gas.
[122]
Dr.Jawad R.R. Alassal Kirkuk U.
Gas Viscosity(µg):
The viscosity of a fluid is a measure of the fluid’s ability to flow, or the ratio of shearing
force to the shearing rate. The viscosity is usually expressed in Centipoises or Poises.
1 poise = 100 centipoises = 6.72 x10-2 ibm/ft-sec = 2.09 x10-3 lbf-sec/ft2 = 0.1 Kg/m-sec
Gas viscosity is difficult to measure experimentally and for engineering purposes µg can
be determined accurately enough from empirical correlations.
Carr – Kobayashi - Burrows Method: This method is graphical method and it is include
correction for non hydrocarbon impurities.
Procedure of calculation:
1. Calculate the pseudo critical pressure, pseudo critical temperature, and apparent
molecular weight from the specific gravity or the composition of the natural gas.
Correction to the nonhydrocarbon gases ( CO2, N2 and H2S) should be made if they
are present in concentrations greater than 5 mole percent.
2. Obtain the viscosity of the natural gas at one atmosphere and the temperature of
interest from figure 6. This viscosity as denoted by µ1must be corrected for the
presence of nonhydrocarbon components by using figure 7. The nonhydrocarbon
fractions tend to increase the viscosity of the gas phase
.
[123]
Dr.Jawad R.R. Alassal Kirkuk U.
Viscosity at 1 atm
Figure 7
[124]
Dr.Jawad R.R. Alassal Kirkuk U.
Lee – Gonzalez –Eakin Method: This method does not include correction for
nonhydrocarbon impurities; therefore, the value obtained will be for a pure hydrocarbon
gas. However, if the z- factor used in calculating the gas density has been corrected, the
Lee method is valid for sour gases.
µg = k x 10-4 Exp(X ρgY )
Where:
k=
(
)
X = 3.5 + 986/T + 0.01 M
Y = 2.4 – 0.2 X
T = 0R, µg = cp, M = Molecular weight, ρg = gm/cc
Example:
Using both the Carr and Lee methods, calculate the viscosity of 0.8 gravity gas at a
pressure of 2000 psia & temperature of 1500F. The gas contains 10% H2S and 10% CO2.
Solution:
Carr Method:
From figure 6 : µ1 = 0.0111 cp.
Correction for 10% H2S =+ 0.0003
Correction for 10% CO2 =+ 0.0006
µ1= 0.0111+ 0.0003+ 0.0006 = 0.0120 cp
Tc = 170.5 +307.3 (0.8) = 4160R
Pc = 709.6 – 58.7 (0.8) = 663 psia.
Tr = (460 +150)/416 = 1.47
Pr = 2000/ 663 = 3.02
Therefore Z = 0.791
Using figure 7 : µ/ µ1 = 1.61
µg = µ1 x (µ/ µ1) = 0.012 x 1.61 = 0.0193 cp
[125]
Dr.Jawad R.R. Alassal Kirkuk U.
Lee – Gonzalez –Eakin:
M=
k=
x Mair = 0.8 x 29 = 23.2
(
)
=
(
)
X = 3.5 + 986/T + 0.01 M
= 3.5 +986/610 +0.01 x 23.2 = 5.35
Y = 2.4 – 0.2 X
= 2.4 - .02 x 5.35 = 1.33
ρg SC =
ρair = 0.8 x 0.0764 = 0.06112 gm/cc
ρg = ρg SC x
= 0.1436gm/cc
µg = k x 10-4 Exp(X ρgY ) = 0.0177 cp
The kinematic viscosity (υ) = µg/ ρg
Gas – water System
In many cases the engineering design of gas production operations will involve natural
gas in contact with water. This water may be connate reservoir water or water produced
from other zone.
It is necessary to determine the water contained in the gas in the vapor state, the gas
dissolved in the water, and under what conditions of P & T gas hydrate will be formed.
Solubility of Natural Gas in Water:
The solubility of natural gas in water is very low at most pressures & temperature of
interest in gas production engineering.
The primary factors affecting the amount of gas that will be evolved from water saturated
with gas depends on P , T and solids content.
The correction factor for salinity may be calculated from:
[126]
Dr.Jawad R.R. Alassal Kirkuk U.
Where:
Y = salinity of water, ppm
X = 3.471/ T0.837
T = temperature, 0F
Rsw = gas solubility in brine, Scf/STB
Rswp = gas solubility in pure water, Scf/STB
Example:
Calculate the Scf of gas dissolved in brine containing 50000 ppm at pressure of 5000 psia
and temperature of 2000F
Solution:
From figure 8, Rswp = 20 Scf/ STB
X = 3.471 /(200)0.837 = 0.041
Rsw = Rswp ( 1 – XYx10-4) = 20 (1- 0.041 x 50000 x 10-4) = 15.9 Scf/STB
Figure 8
[127]
Dr.Jawad R.R. Alassal Kirkuk U.
Solubility of water in natural Gas:
The amount of water vapor contented in natural gas depends on P, T, and water salinity.
Figure 9 shows the mass or weight of fresh water contained in gas as a function of P & T.
The following equation can be used to correct for water salinity:
Ws/ Wsp = 1 – 2.87 x 10-8 Y1.266
Where:
Ws = brine content lbm/MMScf
Wsp = pure water content lbm/MMScf , from figure 8
Y = salinity of water, ppm
Example:
Determine the water content in a natural gas in contact with 100,000ppm brine at 3000
psia & 2000 F.
Solution:
From Figure 9: Wsp = 280 lbm/MMScf
Ws = Wsp (1- 2.87 x 10-8 x (100,000)1.266) = 263 lbm/ MMSCF
[128]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 9(a)
[129]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 9(b)
[130]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 9(c)
[131]
Dr.Jawad R.R. Alassal Kirkuk U.
Gas hydrates:
Gas hydrates are crystalline compounds, formed by the chemical combination of natural
gas and water under pressure at temperatures considerably above the freezing point of
water. In the presence of free water, hydrates will form when the temperature of the gas is
below a certain temperature, called the ― hydrate temperature‖
Hydrate formation would result in lower flow rates or some times may completely block
flow lines and surface equipment.
Conditions promoting hydrate formation are :
1. Gas at or below its water dewpoint with ―free‖ water present.
2. Low temperature.
3. High pressure.
Further discussion will be presented.
Pressure – Temperature Diagram
Figure 10 shows a typical Pressure – Temperature diagram for multicomponent system
with a specific overall composition. Although a different hydrocarbon system would have
a different phase diagram. The general configuration is similar.
Figure 10. Typical P- T diagram for a multicomponent system
[132]
Dr.Jawad R.R. Alassal Kirkuk U.
These multicomponent pressure-temperature diagram are essentially used to:
 Classify reservoir
 Classify the naturally occurring hydrocarbon systems\
 Describe the phase behavior of the reservoir fluid
To fully understand the significance of the pressure-temperature diagrams , it is
necessary to identify and define the following key points on these diagrams:
 Cricondentherm (Tct) : the Cricondentherm is defined as the maximum temperature
above which liquid cannot be formed regardless of pressure (point E) . The
corresponding pressure is termed the Cricondentherm pressure Pct.
 Cricondenbar (Pcb) : the Cricondenbar is the maximum pressure above which no
gas can be formed regardless of temperature (Point D) . the corresponding
temperature is called the Cricondenbar temperature Tcb .
 Critical point : the critical point for a multicomponent mixture is referred to as the
state of pressure and temperature at which all intensive properties of the gas and
liquid phases are equal (Point C) . At the critical point , the corresponding pressure
and temperature are called the critical pressure Pc and the critical temperature Tc of
the mixture.
 Phase envelope ( two-phase region) : the region enclosed by the bubble point
curve and the dew point curve (line BCA), wherein gas and liquid coexist in
equilibrium , is identified as the phase envelope of the hydrocarbon system .
 Quality lines : the dashed lines within the phase diagram are called quality lines .
They describe the pressure and temperature conditions for equal volumes of liquids.
Note that the quality lines converge at the critical point (Point C).
 Bubble –point curve : the bubble point curve (line BC) is defined as the line
separating the liquid-phase region from the two phase region.
 Dew-point curve : the dew point curve (line AC) is defined as the line separating
the vapor-phase region from the two phase region.
In general , reservoir are conveniently classified on the basis of the location of the
point representing the initial reservoir pressure pi and temperature T with respect to
the pressure -temperature diagram of the reservoir fluid .Accordingly , reservoir can
be classified into basically two types , these are :
 Oil reservoir : if the reservoir temperature T is less than the critical temperature Tc
of the reservoir fluid, the reservoir is classified as an oil reservoir.
 Gas reservoir : if the reservoir temperature is greater than the critical temperature of
the hydrocarbon fluid, the reservoir is considered a gas reservoir.
[133]
Dr.Jawad R.R. Alassal Kirkuk U.
Gas Reservoirs:
Reservoirs that yield natural gas can be classified into essentially four categories. These
are:
 Dry – Gas : The fluid exists as a gas both in the reservoir and the piping system.
The only liquid associated with the gas from a dry gas reservoir is water. The phase
diagram is shown in figure 11.
Figure 11 Phase diagram of Dry Gas
 Wet – Gas : The fluid initially exist as a gas in the reservoir and remaining in the
gaseous phase as pressure declines at reservoir temperature. However, in being
produced to the surface, the temperature also drops, causing condensation in the
piping system and separators. The phase diagram is shown in figure 12.
Wet – gas reservoir are characterized by the following properties‖
 Gas oil ratios between 60000 to 100000 Scf/STB
 Stock – tank oil gravity above 600 API
 Liquid is water – white in color.
[134]
Dr.Jawad R.R. Alassal Kirkuk U.
 Separator conditions , i.e.. separator pressure and temperature, lie within the two –
phase region.
Figure 12 phase diagram of wet Gas
 Retrograde Condensate gas: The fluid exists as a gas at initial reservoir
conditions. As reservoir pressure declines at reservoir temperature, the dew point
line is crossed and liquid forms in the reservoir, liquid also forms in the piping
system and separators. The phase diagram is shown in figure 13.
Retrograde Condensate – gas reservoir are characterized by the following properties:
 Gas – oil ratios between 8000 to 70000 Scf/ STB . generally, the gas- oil ratio for a
condensate system increase with time due to the liquid dropout and loss of heavy
components in the liquid.
 Condensate gravity above 500 API.
 Stock tank liquid is usually water – white or slightly colored.
[135]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 13. phase diagram of Retrograde Condensate Gas
 Associated Gas: many oil reservoirs exist at bubble point pressure of the fluid
system at initial conditions. Free gas can be produced from the gas cap of such a
system. Gas which is initially dissolved in the oil can also be produced as free gas
at the surface.
Adjustment of Properties For Condensate Mixture:
In many cases the properties of the total well or reservoir fluid will not be measured, but
the gas and condensate properties after separation will be known.
The specific gravity of the mixture can be calculated using the following equation:
Where:
= specific gravity of the mixture ( air =1).
= specific gravity of the separator gas.
= specific gravity of the condensate.
[136]
Dr.Jawad R.R. Alassal Kirkuk U.
= condensate vaporizing volume Scf/STB from figure 14
R = Producing gas – condensate ratio Scf/ STB
= 141.5/ ( 131.5 API)
Example:
Calculate the specific gravity of a reservoir gas given the following :
= 0.65 ,
= 0.75 , R = 3000 SCf/STB, separator pressure = 900 psia.
Solution:
From figure 14, VO = 1110.SCF/STB
= ( 0.65 + 4584 x( 0.78/ 30000)/ ( 1 + 1110/30000) = 0.74
Figure 14
[137]
Dr.Jawad R.R. Alassal Kirkuk U.
DIRECT CALCULATION OF COMPRESSIBILITY FACTORS
After four decades of existence, the Standing-Katz z-factor chart is still widely used as a
practical source of natural gas compressibility fac- tors. As a result, there has been an
apparent need for a simple mathemati- cal description of that chart. Several empirical
correlations for calculating z-factors have been developed over the years. The following
three empirical correlations are described below:
• Hall-Yarborough
• Dranchuk-Abu-Kassem
• Dranchuk-Purvis-Robinson
The Hall-Yarborough Method :
Hall and Yarborough (1973) presented an equation-of-state that accu- rately represents the
Standing and Katz z-factor chart. The proposed expression is based on the StarlingCarnahan equation-of-state. The coefficients of the correlation were determined by fitting
them to data taken from the Standing and Katz z-factor chart. Hall and Yarborough
proposed the following mathematical form:
[
]
[
(
Eq.1
) ]
where ppr = pseudo-reduced pressure
t = reciprocal of the pseudo-reduced temperature, i.e., Tpc /T
Y = the reduced density that can be obtained as the solution of
the followin g equation:
( )
(
(
)
(
[
(
(
)
(
)
(
)
[138]
Dr.Jawad R.R. Alassal Kirkuk U.
)
Eq.2
)
) ]
Equation 2is a nonlinear equation and can be conveniently solvedfor the reduced density
Y by using the Newton-Raphson iteration technique. The computational procedure of
solving Equation 2 at any specified pseudo-reduced pressure ppr and temperature Tpr is
summarized
in the following steps:
Step 1. Make an initial guess of the unknown parameter, Yk , where k is an iteration
counter. An appropriate initial guess of Y is given by the following relationship:
[
(
) ]
Step 2. Substitute this initial value in Equation 2 and evaluate thenonlinear function.
Unless the correct value of Y has been initial-ly selected, Equation 2will have a nonzero
value of F(Y):
Step 3. A new improved estimate of Y, i.e., Y k+1 , is calculated from the following
expression:
( )
( )
where
(
Eq.3
) is obtained by evaluating the derivative of Equation 2 at Y k , or:
( )
(
(
)
)
(
)(
) (
)
Eq.4
Step 4. Steps 2–3 are repeated n times, until the error, i.e., abs(Y k -Y k+1 ), becomes
smaller than a preset tolerance, e.g., 10-12
Step 5. The correct value of Y is then used to evaluate Equation 1 for the compressibility
factor.Hall and Yarborough pointed out that the method is not recommended for
application if the pseudo-reduced temperature is less than one.
The Dranchuk-Abu-Kassem Method
Dranchuk and Abu-Kassem (1975) derived an analytical expressionfor calculating the
reduced gas density that can be used to estimate the gas compressibility factor. The
[139]
Dr.Jawad R.R. Alassal Kirkuk U.
reduced gas density r r is defined as the ratio of the gas density at a specified pressure
and temperature to that of the gas at its critical pressure or temperature, or:
[
[
]
[
[
]
]
]
The critical gas compressibility factor z c is approximately 0.27 whichleads to the
following simplified expression for the reduced gas density:
Eq.4
The authors proposed the following eleven-constant equation-of-state for calculating the
reduced gas density:
( )
(
)
(
)
(
)
(
)(
)
[
]
Eq.5
With the coefficients R1 through R5 as defined by the following relations:
*
+
*
+
*
[
+
]
*
+
Eq.6
The constants A 1 through A 11 were determined by fitting the equation ,using nonlinear
regression models, to 1,500 data points from the Standing and Katz z-factor chart. The
coefficients have the following values :
A1 = 0.3265 , A2 = -1.0700 , A3 = - 0.5339 , A4 = 0.01569
[140]
Dr.Jawad R.R. Alassal Kirkuk U.
A5 = - 0.05165 ,
A6 = 0.5475
A9 = 0.1056 , A10 = 0.6134
, A7 = -0.7361 , A8 = 0.1844
, A11 = 0.7210
Equation 5 can be solved for the reduced gas density by applying the Newton-Raphson
iteration technique as summarized in the following steps:
Step 1 . Make an initial guess of the unknown parameter,
, where k is an iteration
counter. An appropriate initial guess of
, is given by the following relationship:
Step 2 . Substitute this initial value in Equation 5 and evaluate the nonlinear
function. Unless the correct value of
has been initially selected, Equation 5 will have
a nonzero value for the function f ( ).
Step 3 . A new improved estimate of
expression:
, i.e.,
, is calculated from the following
(
(
)
)
(
)
Where
( )
(
)
(
(
)
(
)
][(
[
)
)]
Step 4. Steps 2–3 are repeated n times, until the error, i.e., abs(
-
),becomes
smaller than a preset tolerance, e.g., 10-12 .
Step 5. The correct value of
is then used to evaluate Equation 4 for the compressibility
factor, i.e.,:
[141]
Dr.Jawad R.R. Alassal Kirkuk U.
The proposed correlation was reported to duplicate compressibility factors from the
Standing and Katz chart with an average absolute error of 0.585 percent and is applicable
over the ranges:
The Dranchuk-Purvis-Robinson Method
Dranchuk, Purvis, and Robinson (1974) developed a correlation based on the BenedictWebb-Rubin type of equation-of-state. Fitting the equation to 1,500 data points from the
Standing and Katz z-factor chart optimized the eight coefficients of the proposed
equations. The equation has the following form :
[
(
)
(
)]
Eq.7
With
*
+
*
+
*
+
[
]
[
where
values
]
is defined by Equation 5 and the coefficients A1 through A8 have the following
[142]
Dr.Jawad R.R. Alassal Kirkuk U.
A 1 = 0.31506237
, A 5 = -0.61232032
A 2 = -1.0467099
, A 6 = -0.10488813
A 3 = -0.57832720 , A 7 = 0.68157001
A 4 = 0.53530771 , A 8 = 0.68446549
The solution procedure of Equation 7 is similar to that of Dranchuk and Abu-Kassem.
The method is valid within the following ranges of pseudo-reduced temperature and
pressure:
[143]
Dr.Jawad R.R. Alassal Kirkuk U.
Problems
1. Calculate the volume 1 Ib-mole of ideal gas will occupy at :
a) 14.7 psia and 60oF
b) 14.7 psia and 32oF
c) 14.7 plus 10 oz and 80oF
d) 15.025 psia and 60oF
2. A 500 cuft tank contains 10 Ib of methane and 20 Ib of ethane at 90oF
a) How many moles in tank ?
b) What is the pressure of the tank in psia ?
c) What is the molecular weight of the mixture ?
d) What is the specific gravity of the mixture ?
3. A 50 cuft tank contains gas at 50 psia and 50o F .It is connected to another tank that
contains gas at 25 psia and 50oF.when the valve between two is opened ,the
pressure equalizes at 35psia at 50oF. what is the volume of the second tank?
4. A 2 cuft tank contains gas at 1000 pounds per square inch gauge (psig) and at
60°F. This tank is connected to another tank of unknown volume containing
the same gas at 14.7 psia and 60°F. The two tanks are allowed to come to
equilibrium, and when this is accomplished and pressure is 650 psi gauge and
the temperature is 60°F. What is the volume of the tank?
Atmospheric pressure is at 14.7 psia.
Assume gas behaves as a perfect gas.
5. (a) Two cylinders, one containing propane at 100 psig and the other
containing propane at 0 psig, are connected and the pressures allowed to equalize.
If the volume of each of the cylinders is 5 cuft and the temperature is held constant
at 90°F what would be the equalization pressure, the density of the propane under
[144]
Dr.Jawad R.R. Alassal Kirkuk U.
this pressure, and the weight of gas in the system. Assume gas behaves as a perfect
gas. (b) If the temperature of the system were reduced to 60°F, what would the
pressure be?
6. A 2 cuft tank contains propane at 1209 psig and 332°F; taking into account the
deviation from the gas laws, how many standard cubic feet of gas have been
withdrawn from the cylinder when the pressure has dropped to 602 psig and
332oF? Atmospheric pressure is at 14.7 psia.
7. For a gas of the following composition:
Composition
Mol %
Methane
92.67
Ethane
5.29
Propane
1.38
i-butane
0.18
n-butane
0.34
n-pentane
0.14
Calculate:
a. The specific gravity, the density, and the specific volume of the gas at 60°F
and 14.65 psia.
b. The composition on weight basis.
8. A sample of natural gas from the Bell Field has a specific gravity of 0.665
(air = 1.00). The carbon dioxide and nitrogen content are 0.10 and 2.07 mol %,
respectively. Calculate the gas deviation factor, z, at reservoir temperature of
213°F and reservoir pressure of 3250 psia.
[145]
Dr.Jawad R.R. Alassal Kirkuk U.
9. For a gas of the following composition:
Composition
Mol %
Methane
92.67
Ethane
5.29
Propane
1.38
i-butane
0.18
n-butane
0.34
n-pentane
0.14
Calculate:
a. The pseudocritical temperature and pressure
b. The gas deviation factor for (1) 400 psig and 80°F and (2) 2500 psig and 200°F.
c. The density and specific volume of the gas for each of the conditions in 5(b).
d. The number of standard cubic feet of gas per acre foot of sand of 25% porosity
and 10% connate water for each of the conditions given in 5(b). Atomspheric
pressure is at 14.7 psia.
10. A natural gas has the following composition:
Component
Mol %
C1
87.09
C2
4.42
C3
1.60
i-C4
0.40
n-C4
0.52
C5
0.46
C6
0.29
C7+
0.06
N2
4.76
CO2
0.40
100.00
[146]
Dr.Jawad R.R. Alassal Kirkuk U.
Assume a molecular weight of C7+, the same as for C7.) Calculate:
(a) Weight percent of each component in the gas.
(b)
Apparent molecular weight of the gas.
(c)
Specific gravity of the gas.
11. A natural gas has the following composition:
Component
Mol %
C1
86.02
C2
7.70
C3
4.26
i-C4
0.57
n-C4
0.87
i-C5
0.11
n-C5
0.14
C6
0.33
100.00
For pressure = 750 psia and temperature = 150°F, calculate:
(a) Pseudocritical temperature.
(b) Pseudocritical pressure.
(c) z-factor.
12. A natural gas has the following composition: 90% CH4; 5.0% C2H6; 5.0% N2.
(a) Calculate its pseudocritical temperature and pseudocritical pressure.
(b) Calculate the pseudoreduced temperature and pseudoreduced pres¬sure, and the gas
deviation factor for the natural gas at 100°F and 600 psia.
(c) 325,000 cuft of this natural gas is to be compressed from 14.7 psia and 60°F to 600
psia and 100°F. Calculate the volume the natural gas will occupy under the compressed
conditions.
[147]
Dr.Jawad R.R. Alassal Kirkuk U.
13. Calculate the gas reserve in a gas field of 2000 acres, with 40-ft sand thickness, 25%
porosity, 15% water saturation, a bottom-hole pressure of 3000 psi. gauge, a temperature
of 200°F, and barometric pressure at 14.7 psia. Composition of gas: methane, 94.63% ;
ethane, 2.54% ; propane, 1.46% ; isobutane, 0.46%, N-butane, 0.38%; pentanes, 0.36%;
and hex- anes plus, 0.17%.
14. In a recycling plant 250,000 cuft of gas at 2500 psi. gauge, temperature 100 T, is
compressed to 4000 psi. gauge and 150°F and put back into the sand at these conditions.
What is the volume of the gas as it is injected into the sand if the gas composition is 75%
methane, 15% ethane, and 10% propane?
15. A gas of the following composition leaves a high-pressure absorber at 500 psia and
80°F. The gas is then compressed and injected back into the reservoir at 3600 psia and
200°F. Calculate the volume 10 million scf of this gas would occupy in the reservoir.
Component
Mol%
Methane
93.3
Ethane
3.84
Propane
2.21
n-Butane
0.65
100.00
16. Show that :
(
∑(
)
)
[148]
Dr.Jawad R.R. Alassal Kirkuk U.
17. Determine the composition in weight fraction of the following gas :
Component
C1
C2
C3
C4
C5
Mole fraction
0.65
0.10
0.10
0.10
0.05
18. Determine the composition in mole fraction of the following gas
Component
C1
C2
C3
C4
C5
Weight fraction
0.40
0.10
0.20
0.20
0.10
19. A gas has the following composition :
Component
C1
C2
C3
C4
C5
C6
C7
Yi
0.75
0.07
0.05
0.04
0.04
0.03
0.02
Assuming an ideal gas behavior , calculate the following gas properties at 1000
psia and 100oF :
a. Apparent molecular weight
b. Specific gravity
c. Gas density
d. Specific volume
20. Calculate the compressibility factor of :
a. Methane
b. Ethane
c. Propane
[149]
Dr.Jawad R.R. Alassal Kirkuk U.
At a reduced pressure and temperature of 2 , and 1.6 , respectively .
21. Using the gas composition , pressure and temperature given in problem 20 , and
assuming a real gas behavior , calculate gas density .
22. Using the data given in problem 21 , recalculating the gas density by estimating
the pseudo-critical properties .
23. A sour natural gas has the following composition :
Component
CO2
H2S
N2
C1
C2
Yi
0.10
0.20
0.05
0.63
0.02
Determine the density of the gas mixture at 1000 psia and 110oF
1. Without making any any corrections to account for the presence of the non –
hydrocarbon component .
2. Using the Wichert-Aziz correction method .
24. Given the following gas composition ,
Component
C1
C2
C3
C4
C5
C6
C7
Yi
0.75
0.07
0.05
0.04
0.04
0.03
0.02
Calculate the isothermal gas compressibility at 1000 psia and 100oF by assuming :
a . An ideal gas behavior.
b. A real gas behavior.
[150]
Dr.Jawad R.R. Alassal Kirkuk U.
25. A gas well is producing at a rate of 15000 ft3 /day from a gas reservoir at a
bottom hole with flowing conditions of 1000 psia and 100o F . the specific gravity of
the gas is 0.903 . Calculate the gas flow rate in Scf/day.
26. Given the following gas composition
Component
C1
C2
C3
C4
Yi
0.850
0.055
0.035
0.010
Calculate the gas viscosity at 3000 psia and 150oF.
27. Given the gas composition
Component
N2
CO2
H2S
C1
C2
C3
Yi
0.05
0.05
0.02
0.80
0.05
0.03
Calculate the gas viscosity at 200oF and 3500 psia.
28. Rework problem 27 Calculate the gas viscosity by using the Lee-Gonzales-Eakin
method .
29. Assuming an ideal gas behavior , calculate the density of n-butane at 220oF
and 50 psia.
[151]
Dr.Jawad R.R. Alassal Kirkuk U.
30. Given the following gas
Component
C1
C2
C3
n-C4
n-C5
Weight Fraction
0.65
0.15
0.10
0.06
0.04
Calculate
a. Mole fraction of the gas
b. Apparent molecular weight
c. Specific gravity
d. Specific volume at 300 psia and 120oF by assuming an ideal gas behavior.
31. An ideal gas mixture has a density of 1.92 Ib/ft3 at 500 psia and 100oF . Calculate
the apparent molecular weight of the gas mixture.
32. Using the gas composition as given in problem 30 , and assuming real gas
behavior , calculate
a. Gas density at 2000 psia and 150oF
b. Specific volume at 2000 psia and 150oF
c. Gas formation volume factor in scf/ft3
33. A natural gas with a specific gravity of 0.75 has a gas formation volume factor of
0.00529 ft3/scf at the prevailing reservoir pressure and temperature . Calculate the
density of the gas.
34. A natural gas has the following composition
Component
C1
C2
C3
i-C4
n-C4
i-C5
n-C5
Yi
0.75
0.10
0.05
0.04
0.03
0.02
0.01
[152]
Dr.Jawad R.R. Alassal Kirkuk U.
Reservoir condition are 3500 psia and 200oF . Calculate
a. Isothermal gas compressibility coefficient
b. Gas viscosity by using
1. Carr – Kobayashi- Burrows Method
2. Lee-Gonzales-Eakin Method
35. Given the following gas composition
Component
CO2
N2
C1
C2
C3
n-C4
n-C5
yi
0.06
0.03
0.75
0.07
0.04
0.03
0.02
If the reservoir pressure and temperature are 2500 psia and 175oF respectively ,
calculate
a. Gas density , by accounting for the presence of non-hydrocarbon component
by using Carr-Kobayashi-Burrows method.
b. Isothermal gas compressibility coefficient
c. Gas viscosity by using
1. Carr-Kobayashi-Burrows method
2. Lee-Gonzales-Eakin Method
36. A gas is producing at a rate of 1.22 mm scf/day . the specifying gravity of the
producing gas is 0.74 . if the average reservoir pressure and reservoir temperature
are 2000 psia and 125oF , calculate
a. Gas flow rate in ft3/day
b. Gas viscosity under reservoir condition
[153]
Dr.Jawad R.R. Alassal Kirkuk U.
37. A high – molecular – weight natural gas has the following composition :
Component
C1
C2
C3
C4
C5
C6
C7+
Yi
0.73
0.10
0.05
0.03
0.03
0.02
0.04
If the molecular weight and specifying gravity of C7+ are 135 and 0.81 , calculate
a. Specific gravity of the gas mixture
b. Density of the gas at 2000 psia and 120oF
38. a gas well is producing a hydrocarbon gas at a bottom hole pressure of 3100 psia
,the reservoir pressure and temperature are 3420 psia and 160oF. Given the
following additional well and gas properties data ,
Gas gravity = 0.700
k = 29 md
rw = 0.25 ft
h=12 ft
re= 660 ft
calculate the gas flow rate by using
a. The real gas potential approach
b. The pressure – squared method
[154]
Dr.Jawad R.R. Alassal Kirkuk U.
Chapter -4RESERVOIR FLUID PROPERTIES
Hydrocarbon systems found in petroleum reservoirs are known to display multiphase
behavior over wide ranges of pressures and temperatures. The most important phases that
occur in petroleum reservoirs are a liquid phase, such as crude oils or condensates, and a
gas phase, such as natural gases.
The conditions under which these phases exist are a matter of considerable practical
importance. The experimental or the mathematical determinations of these conditions are
conveniently expressed in different types of diagrams, commonly called phase diagrams.
The objective of this subject is to review the basic principles of hydrocarbon phase
behavior and illustrate the use of phase diagrams in describing and characterizing the
volumetric behavior of single-component, two-component, and multicomponent systems.
Single-Component Systems:
The simplest type of hydrocarbon system to consider is that containing one component.
The word component refers to the number of molecular or atomic species present in the
substance.
A single – component system is composed entirely of one kind of atom or molecule. We
often use the word pure to describe a single – component system.
The qualitative understanding of the relationship between temperature T, pressure p, and
volume V of pure components can provide an excellent basis for understanding the phase
behavior of complex petroleum mixtures. This relationship is conveniently introduced in
terms of experimental measurements conducted on a pure component as the component is
subjected to changes in pressure and volume at a constant temperature. The effects of
making these changes on the behavior of pure components are discussed next.
Suppose a fixed quantity of a pure component is placed in a cylinder fitted with a
frictionless piston at a fixed temperature T1. Furthermore, consider the initial pressure
exerted on the system to be low enough that the entire system is in the vapor state. This
initial condition is represented by point E on the pressure/volume phase diagram (p-V
diagram) as shown in Figure 1. Consider the following sequential experimental steps
taking place on the pure component:
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Dr.Jawad R.R. Alassal Kirkuk U.
1. The pressure is increased isothermally by forcing the piston into the cylinder.
Consequently, the gas volume decreases until it reaches point F on the diagram, where the
liquid begins to condense. The corresponding pressure is known as the dew-point
pressure, pd , and defined as the pressure at which the first droplet of liquid is formed.
2. The piston is moved further into the cylinder as more liquid condenses. This
condensation process is characterized by a constant pressure and represented by the
horizontal line FG. At point G, traces of gas remain and the corresponding pressure is
called the bubble-point pressure, pb, and defined as the pressure at which the first sign of
FIGURE 1 Typical pressure/volume diagram for a pure component.
Gas formation is detected. A characteristics of a single – component, at given temperature,
the dew- point pressure and the bubble – point pressure are equal.
3. As the piston is forced slightly into the cylinder, a sharp increase in the pressure (point
H ) is noted without an appreciable decrease in the liquid volume. That behavior evidently
reflects the low compressibility of the liquid phase.
By repeating these steps at progressively increasing temperatures, a family of curves of
equal temperatures (isotherms) is constructed as shown in Figure 1. The dashed curve
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connecting the dew points, called the dew-point curve (line FC), represents the states of
the ―saturated gas.‖ The dashed curve connecting the bubble points, called the bubblepoint curve (line GC), similarly represents the ―saturated liquid.‖ These two curves meet a
point C, which is known as the critical point. The corresponding pressure and volume are
called the critical pressure, pc, and critical volume, Vc, respectively. Note that, as the
temperature increases, the length of the straight line portion of the isotherm decreases
until it eventually vanishes and the isotherm merely has a horizontal tangent and inflection
point at the critical point.
This isotherm temperature is called the critical temperature, Tc, of that single component.
This observation can be expressed mathematically by the following relationship:
(
)
(
)
Referring to Figure 1, the area enclosed by the area AFCGB is called the two-phase region
or the phase envelope. Within this defined region, vapor and liquid can coexist in
equilibrium. Outside the phase envelope, only one phase can exist.
The critical point (point C) describes the critical state of the pure component and
represents the limiting state for the existence of two phases, that is, liquid and gas. In
other words, for a single-component system, the critical point is defined as the highest
value of pressure and temperature at which two phases can coexist. A more generalized
definition of the critical point, which is applicable to a single- or multicomponent system,
is this: The critical point is the point at which all intensive properties of the gas and liquid
phases are equal.
An intensive property is one that has the same value for any part of a homogeneous
system as it does for the whole system, that is, a property independent of the quantity of
the system. Pressure, temperature, density, composition, and viscosity are examples of
intensive properties.
Many characteristic properties of pure substances have been measured and compiled over
the years. These properties provide vital information for calculating the thermodynamic
properties of pure components as well as their mixtures. The most important of these
properties include :
 Critical pressure, pc .
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Dr.Jawad R.R. Alassal Kirkuk U.
 Critical temperature, Tc .
 Critical volume, Vc .
 Boiling point temperature, Tb .
 Acentric factor, ..
 Molecular weight, M.
 Specific gravity,
Those physical properties needed for hydrocarbon phase behavior calculations are
presented in Table 1 for a number of hydrocarbon and nonhydrocarbon components.
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Dr.Jawad R.R. Alassal Kirkuk U.
.
Another means of presenting the results of this experiment is shown in Figure 2, in which
the pressure and temperature of the system are the independent parameters. Figure 2
shows a typical pressure/temperature diagram ( p/T diagram) of a single-component
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Dr.Jawad R.R. Alassal Kirkuk U.
system with solids lines that clearly represent three different phase boundaries: vaporliquid, vapor-solid, and liquid-solid phase separation boundaries. As shown in the
illustration, line AC terminates at the critical point (point C) and can be thought of as the
dividing line between the areas where liquid and vapor exist. The curve is commonly
called the vapor-pressure curve or the boiling-point curve. The corresponding pressure at
any point on the curve is called the vapor pressure, pv, with a corresponding temperature
termed the boiling-point temperature.
FIGURE 2 Typical pressure/temperature diagram for a single-component system
The vapor-pressure curve represents the conditions of pressure and temperature at which
two phases, vapor and liquid, can coexist in equilibrium. Systems represented by a point
located below the vapor-pressure curve are composed only of the vapor phase. Similarly,
points above the curve represent systems that exist in the liquid phase. These remarks can
be conveniently summarized by the following expressions:
if p < Pv. the system is entirely in the vapor phase;
If p > pv . the system is entirely in the liquid phase;
If p = pv . the vapor and liquid coexist in equilibrium;
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Dr.Jawad R.R. Alassal Kirkuk U.
where p is the pressure exerted on the pure substance. It should be pointed out that these
expressions are valid only if the system temperature is below the critical temperature Tc
of the substance
EXAMPLE
A pure propane is held in a laboratory cell at 80oF and 200 psia. Determine the ―existence
state‖ (i.e., as a gas or liquid) of the substance.
Solution
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Dr.Jawad R.R. Alassal Kirkuk U.
component
Figure (3): Vapor pressure chart for hydrocarbon
From a Cox chart, the vapor pressure of propane is read as pv = 150 psi, and because the
laboratory cell pressure is 200 psi (i.e., p > pv), this means that the laboratory cell
contains a liquefied propane.
Two-Component Systems
A distinguishing feature of the single-component system is that, at a fixed temperature,
two phases (vapor and liquid) can exist in equilibrium at only one pressure; this is the
vapor pressure. For a binary system, two phases can exist in equilibrium at various
pressures at the same temperature. The following discussion concerning the description of
the phase behavior of a two-component system involves many concepts that apply to the
more complex multicomponent mixtures of oils and gases.
An important characteristic of binary systems is the variation of their thermodynamic and
physical properties with the composition. Therefore, it is necessary to specify the
composition of the mixture in terms of mole or weight fractions. It is customary to
designate one of the components as the more volatile component and the other the less
volatile component, depending on their relative vapor pressure at a given temperature.
Suppose that the examples previously described for a pure component are repeated, but
this time we introduce into the cylinder a binary mixture of a known overall composition
Consider that the initial pressure p1 exerted on the system, at a fixed temperature of T1, is
low enough that the entire system exists in the vapor state. This initial condition of
pressure and temperature acting on the mixture is represented by point 1 on the p/V
diagram of Figure
4. As the pressure is increased isothermally, it reaches point 2, at which an infinitesimal
amount of liquid is condensed. The pressure at this point is called the dew-point pressure,
pd, of the mixture. It should be noted that, at the dew-point pressure, the composition of
the vapor phase is equal to the overall composition of the binary mixture. As the total
volume is decreased by forcing the piston inside the cylinder, a noticeable increase in the
pressure is observed as more and more liquid is condensed. This condensation process is
continued until the pressure reaches point 3, at which traces of gas remain. At point 3, the
corresponding pressure is called the bubble-point pressure, pb. Because, at the bubble
point, the gas phase is only of infinitesimal volume, the composition of the liquid phase
therefore is identical with that of the whole system. As the piston is forced further into the
cylinder, the pressure rises steeply to point 4 with a corresponding decreasing volume.
Repeating the previous examples at progressively increasing temperatures, a complete set
of isotherms is obtained on the p/V diagram of Figure 5 for a binary system consisting of
n-pentane and n-heptane. The bubble-point curve, as represented by line AC, represents
the locus of the points of pressure and volume at which the first bubble of gas is formed.
The dew-point curve (line BC) describes the locus of the points of pressure and volume at
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Dr.Jawad R.R. Alassal Kirkuk U.
which the first droplet of liquid is formed. The two curves meet at the critical point (point
C). The critical pressure, temperature, and volume are given by pc, Tc, and Vc,
respectively. Any point within the phase envelope (line ACB) represents a system
Figure (4): Pressure/Volume isotherm for a two – component system
consisting of two phases. Outside the phase envelope, only one phase can exist
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Dr.Jawad R.R. Alassal Kirkuk U.
Figure (5): pressure/ Volume diagram for the n- pentane and n-heptane system containing 52.4 wt % nheptane
If the bubble-point pressure and dew-point pressure for the various isotherms on a p/V
diagram are plotted as a function of temperature, a p/T diagram similar to that shown in
Figure 6 is obtained. Figure 6 indicates that the pressure/temperature relationships no
longer can be represented by a simple vapor pressure curve, as in the case of a singlecomponent system, but take on the form illustrated in the figure by the phase envelope
ACB.
The dashed lines within the phase envelope are called quality lines; they describe the
pressure and temperature conditions of equal volumes of liquid. Obviously, the bubblepoint curve and the dew-point curve represent 100% and 0% liquid, respectively.
Figure (6): Typical temperature / pressure diagram for a two- component system
Figure 7 demonstrates the effect of changing the composition of the binary system on the
shape and location of the phase envelope. Two of the lines shown in the figure represent
the vapor-pressure curves for methane and ethane, which terminate at the critical point.
Ten phase boundary curves (phase envelopes) for various mixtures of methane and ethane
also are shown. These curves pass continuously from the vapor-pressure curve of the one
pure component to that of the other as the composition is varied. The points labeled 1–10
represent the critical points of the mixtures as defined in the legend of Figure 7. The
dashed curve illustrates the locus of critical points for the binary system
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Dr.Jawad R.R. Alassal Kirkuk U.
It should be noted by examining Figure 7 that, when one of the constituents becomes
predominant, the binary mixture tends to exhibit a relatively narrow phase envelope and
displays critical properties close to the predominant component. The size of the phase
envelope enlarges noticeably as the composition of the mixture becomes evenly
distributed between the two components.
Figure (8) shows the critical loci for a number of common binary systems. Obviously, the
critical pressure of mixtures is considerably higher than the critical pressure of the
components in the mixtures. The greater the difference in the boiling point of the two
substances, the higher the critical pressure of the mixture.
Figure (7) :Phase diagram of a methane / ethane mixture
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Dr.Jawad R.R. Alassal Kirkuk U.
[166]
Figure (8) : Convergence pressures for binary
systems.
Three-Component Systems
The phase behavior of mixtures containing three components (ternary systems) is
conveniently represented in a triangular diagram, such as that shown in Figure (9). Such
diagrams are based on the property of equilateral triangles that the sum of the
perpendicular distances from any point to each side of the diagram is a constant and equal
to the length on any of the sides. Thus, the composition xi of the ternary system as
represented by point
A in the interior of the triangle of Figure (9) is
Where
Figure (9): Properties of the three-component diagram.
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Dr.Jawad R.R. Alassal Kirkuk U.
Typical features of a ternary phase diagram for a system that exists in the two-phase
region at fixed pressure and temperature are shown in Figure 10. Any mixture with an
overall composition that lies inside the binodal curve (phase envelope) will split into
liquid and vapor phases. The line that connects the composition of liquid and vapor phases
that are in equilibrium is called the tie line. Any other mixture with an overall composition
that lies on that tie line will split into the same liquid and vapor compositions. Only the
amounts of liquid and gas change as the overall mixture composition changes from the
liquid side (bubble-point curve) on the binodal curve to the vapor side (dew-point curve).
If the mole fractions of component i in the liquid, vapor, and overall mixture are x i , yi ,
and zi , the fraction of the total number of moles in the liquid phase nl is given by
This expression is another lever rule, similar to that described for binary diagrams. The
liquid and vapor portions of the binodal curve (phase envelope) meet at the plait point
critical point), where the liquid and vapor phases are identical
FIGURE (10): Three-component phase diagram at a constant temperature and pressure for a system that
forms a liquid and a vapor
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Dr.Jawad R.R. Alassal Kirkuk U.
Multicomponent Systems
The phase behavior of multicomponent hydrocarbon systems in the two-phase region, that
is, the liquid-vapor region, is very similar to that of binary systems. However, as the
system becomes more complex with a greater number of different components, the
pressure and temperature ranges in which two phases lie increase significantly.
The conditions under which these phases exist are a matter of considerable practical
importance. The experimental or the mathematical determinations of these conditions are
conveniently expressed in different types of diagrams, commonly called phase diagrams.
One such diagram is called the pressure-temperature diagram. Figure (11) shows a typical
pressure/temperature diagram (p/T diagram) of a multi- component system with a specific
overall composition. Although a different hydrocarbon system would have a different
phase diagram, the general configuration is similar.
These multicomponent p/T diagrams are essentially used to classify reservoirs, specify the
naturally occurring hydrocarbon systems, and describe the phase behavior of the reservoir
fluid. To fully understand the significance of the p/T diagrams, it is necessary to identify
and define the following key points on the p/T diagram:
• Cricondentherm (Tct): The cricondentherm is the maximum temperature above which
liquid cannot be formed regardless of pressure (point E). The corresponding pressure is
termed the cricondentherm pressure, p is termed the cricondentherm pressure, pct.
• Cricondenbar (pcb): The cricondenbar is the maximum pressure above which no gas
can be formed regardless of temperature (point D). The corresponding temperature is
called the cricondenbar temperature, Tcb.
• Critical point: The critical point for a multicomponent mixture is referred to as the
state of pressure and temperature at which all intensive properties of the gas and liquid
phases are equal (point C). At the critical point, the corresponding pressure and
temperature are called the critical pressure, pc, and critical temperature, Tc , of the
mixture. Phase envelope (two-phase region) The region enclosed by the bubble-point
curve and the dew-point curve (line BCA), where gas and liquid coexist in equilibrium, is
identified
• Quality lines: The dashed lines within the phase diagram are called quality lines. They
as the phase envelope of the hydrocarbon system.
describe the pressure and temperature conditions for equal volumes of liquids. Note that
the quality lines converge at the critical point (point C).
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Dr.Jawad R.R. Alassal Kirkuk U.
• Bubble-point curve: The bubble-point curve (line BC) is defined as the line
separating the liquid phase region from the two-phase region.
• Dew-point curve: The dew-point curve (line AC) is defined as the line separating the
vapor phase region from the two-phase region.
Figure (11): Typical P/T diagram of a multicomponent System.
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Dr.Jawad R.R. Alassal Kirkuk U.
Classification of Reservoirs and Reservoir Fluids
Petroleum reservoirs are broadly classified as oil or gas reservoirs. These broad
classifications are further subdivided depending on
1. The composition of the reservoir hydrocarbon mixture.
2. Initial reservoir pressure and temperature.
3. Pressure and temperature of the surface production.
4. Location of the reservoir temperature with respect to the critical temperature and the
cricondentherm.
In general, reservoirs are conveniently classified on the basis of the location of the In
general, reservoirs are conveniently classified on the basis of the location of the point
representing the initial reservoir pressure pi and temperature T with respect to the p/T
diagram of the reservoir fluid. Accordingly, reservoirs can be classified into basically two
types:
• Oil reservoirs If the reservoir temperature, T, is less than the critical temperature, Tc,
of the reservoir fluid, the reservoir is classified as an oil reservoir
. • Gas reservoirs If the reservoir temperature is greater than the critical temperature of
the hydrocarbon fluid, the reservoir is considered a gas reservoir.
Oil Reservoirs
Depending on initial reservoir pressure, pi, oil reservoirs can be subclassified into the
following categories:
1. Undersaturated oil reservoir If the initial reservoir pressure, pi (as represented by point
1 on Figure (11), is greater than the bubble-point pressure, pb, of the reservoir fluid, the
reservoir is an undersaturated oil reservoir.
2. Saturated oil reservoir When the initial reservoir pressure is equal to the bubble-point
pressure of the reservoir fluid, as shown on Figure 11 by point 2, the reservoir is a
saturated oil reservoir.
3- Gas-cap reservoir If the initial reservoir pressure is below the bubble-point pressure of
the reservoir fluid, as indicated by point 3 on Figure 11 the reservoir is a gas-cap or twophase reservoir, in which an oil phase underlies the gas or vapor phase. Crude oils cover a
wide range in physical properties and chemical compositions, and it is often important to
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Dr.Jawad R.R. Alassal Kirkuk U.
be able to group them into broad categories of related oils. In general, crude oils are
commonly classified into the following types:
 Ordinary black oil.
 Low-shrinkage crude oil.
 High-shrinkage (volatile) crude oil.
 Near-critical crude oil.
This classification essentially is based on the properties exhibited by the crude oil,
including:
• Physical properties, such as API gravity of the stock-tank liquid.
• Composition.
• Initial producing gas/oil ratio (GOR).
• Appearance, such as color of the stock-tank liquid.
• Pressure-temperature phase diagram.
Three of the above properties generally are available: initial GOR, API gravity, and color
of the separated liquid.
The initial producing GOR perhaps is the most important indicator of fluid type. Color has
not been a reliable means of differentiating clearly between gas condensates and volatile
oils, but in general, dark colors indicate the presence of heavy hydrocarbons. No sharp
dividing lines separate these categories of hydrocarbon systems, only laboratory studies
could provide the proper classification. In general, reservoir temperature and composition
of the hydrocarbon system greatly influence the behavior of the system.
1. Ordinary black oil: A typical p/T phase diagram for ordinary black oil is shown in
Figure 12. Note that quality lines that are approximately equally spaced characterize this
black oil phase diagram. Following the pressure reduction path, as indicated by the
vertical line EF in Figure 12, the liquid shrinkage curve, shown in Figure 13, is prepared
by plotting the liquid volume percent as a function of pressure. The liquid shrinkage curve
approximates a straight line except at very low pressures. When
produced, ordinary black oils usually yield gas/oil ratios between 200 and 700 scf/STB
and oil gravities of 15 to 40 API. The stock-tank oil usually is brown to dark green in
color.
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Figure (12): P-T diagram for ordinary black oil
Figure (13): Liquid shrinkage curve for black oil
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2. Low-shrinkage oil: A typical p-T phase diagram for low-shrinkage oil is shown in
Figure 14. The diagram is characterized by quality lines that are closely spaced near the
dew-point curve. The liquid shrinkage curve, given in Figure 15, shows the shrinkage
characteristics of this category of crude oils. The other associated properties of this type of
crude oil are:
 Oil formation volume factor less than 1.2 bbl/STB.
 Gas-oil ratio less than 200 scf/STB.
 Oil gravity less than 35o API.
 Black or deeply colored.
 Substantial liquid recovery at separator conditions as indicated by point G on the
85% quality line of Figure 14
Figure (14): Typical phase diagram for low shrinkage oil
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Dr.Jawad R.R. Alassal Kirkuk U.
Figure (15): Shrinkage curve for low shrinkage oil
3. Volatile crude oil: The phase diagram for a volatile (high-shrinkage) crude oil is
given in Figure 16. Note that the quality lines are close together near the bubble point, and
at lower pressures, they are more widely spaced. This type of crude oil is commonly
characterized by a high liquid shrinkage immediately below the bubble point, shown in
Figure 17. The other characteristic properties of this oil include:
• Oil formation volume factor greater than 1.5 bbl/STB.
• Gas-oil ratios between 2000 and 3000 scf/STB.
• Oil gravities between 45° and 55o API.
• Lower liquid recovery of separator conditions, as indicated by point G on Figure 16
• Greenish to orange in color.
Solution gas released from a volatile oil contains significant quantities of stock-tank liquid
(condensate) when the solution gas is produced at the surface. Solution gas from black
oils usually is considered ―dry,‖ yielding insignificant stock-tank liquid when produced to
surface conditions. For engineering calculations, the liquid content of released solution
gas perhaps is the most important distinction between volatile oils and black oils. Anther
characteristic of volatile oil reservoirs is that the API gravity of the stock tank liquid
increases in the later life of the reservoirs.
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Dr.Jawad R.R. Alassal Kirkuk U.
Figure(16): P-T diagram for a volatile crude oil.
Figure (17): Typical shrinkage curve for a volatile crude oil
4. Near-critical crude oil: If the reservoir temperature, T, is near the critical
temperature, Tc, of the hydrocarbon system, as shown in Figure 18, the hydrocarbon
mixture Tc, of the hydrocarbon system, as shown in Figure 18, the hydrocarbon mixture is
identified as a near-critical crude oil. Because all the quality lines converge at the critical
point, an isothermal pressure drop (as shown by the vertical line EF in Figure (19) may
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Dr.Jawad R.R. Alassal Kirkuk U.
shrink the crude oil from 100% of the hydrocarbon pore volume at the bubble point to
55% or less at a pressure 10 to 50 psi below the bubble point. The
shrinkage characteristic behavior of the near-critical crude oil is shown in Figure (20) This
high shrinkage creates high gas saturation in the pore space and because of the gas-oil
relative permeability characteristics of most reservoir rocks; free gas achieves high
mobility almost immediately below the bubble-point pressure.
The near-critical crude oil is characterized by a high GOR, in excess of 3000 scf/STB,
with an oil formation volume factor of 2.0 bbl/STB or higher. The compositions of nearcritical oils usually are characterized by 12.5 to 20 mol% heptanesplus, 35% or more of
ethane through hexanes, and the remainder methane. It should be pointed out that nearcritical oil systems essentially are considered the borderline to very rich gas condensates
on the phase diagram
Figure (18): phase diagram for a near – critical crude oil
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Dr.Jawad R.R. Alassal Kirkuk U.
Figure (19): Typical liquid shrinkage curve for a near – critical crude oil
Figure (20): liquid shinkage curves for crude oil systems
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PROPERTIES OF CRUDE OIL SYSTEMS
Petroleum (an equivalent term is crude oil) is a complex mixture consisting predominantly
of hydrocarbons and containing sulfur, nitrogen, oxygen, and helium as minor
constituents. The physical and chemical properties of crude oils vary considerably and are
dependent on the concentration of the various types of hydrocarbons and minor
constituents present. An accurate description of physical properties of crude oils is of a
considerable importance in the fields of both applied and theoretical science and
especially in the solution of petroleum reservoir engineering problems.
Physical properties of primary interest in petroleum engineering studies include:
 Fluid gravity
 Specific gravity of the solution gas
 Gas solubility
 Bubble-point pressure
 Oil formation volume factor
 Isothermal compressibility coefficient of undersaturated crude oils
 Oil density
 Total formation volume factor
 Crude oil viscosity
 Surface tension
Data on most of these fluid properties are usually determined by laboratory
experiments performed on samples of actual reservoir fluids. In the absence of
experimentally measured properties of crude oils, it is necessary for the petroleum
engineer to determine the properties from empirically derived correlations.
Crude Oil Gravity
The crude oil density is defined as the mass of a unit volume of the
crude at a specified pressure and temperature. It is usually expressed in
pounds per cubic foot. The specific gravity of a crude oil is defined as the
ratio of the density of the oil to that of water. Both densities are measured
at 60°F and atmospheric pressure:
ϒo =
(1)
Where:
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Dr.Jawad R.R. Alassal Kirkuk U.
ϒo = specific gravity of the oil
= density of the oil, lb/ft3.
= density of the water, lb/ft3.
It should be pointed out that the liquid specific gravity is dimensionless,
but traditionally is given the units 60°/60° to emphasize the fact that
both densities are measured at standard conditions. The density of the
water is approximately 62.4 lb/ft3.
Although the density and specific gravity are used extensively in the
petroleum industry, the API gravity is the preferred gravity scale. This
gravity scale is precisely related to the specific gravity by the following
expression:
o
API =
(2)
The API gravities of crude oils usually range from 47° API for the lighter crude oils to 10°
API for the heavier asphaltic crude oils.
Example:
Calculate the specific gravity and the API gravity of a crude oil system with a measured
density of 53 lb/ft3 at standard conditions.
Solution:
Step 1. Calculate the specific gravity from Equation 1:
ϒo =
= 53./62.4 = 0.849
Step 2. Solve for the API gravity:
o
API =
= 35.2
Specific Gravity of the Solution Gas:
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Dr.Jawad R.R. Alassal Kirkuk U.
The specific gravity of the solution gas ϒg is described by the weighted average of the
specific gravities of the separated gas from each separator.
This weighted-average approach is based on the separator gas-oil ratio,
or:
(3)
Where
n = number of separators
Rsep = separator gas-oil ratio, scf/STB
ϒsep = separator gas gravity
Rst = gas-oil ratio from the stock tank, scf/ STB
ϒst = gas gravity from the stock tank
Example
Separator tests were conducted on a crude oil sample. Results of the test in terms of the
separator gas-oil ration and specific gravity of the separated gas are given below:
Separator
#
Primary
Intermediate
Stock tank
Pressure psig
Temperature
660
75
0
150
110
60
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Dr.Jawad R.R. Alassal Kirkuk U.
Gas-Oil Ratio
scf /STB
724
202
58
Gas Specific
Gravity
0.743
0.956
1.296
Calculate the specific gravity of the separated gas.
Solution:
Estimate the specific gravity of the solution by using Equation 3
ϒg =
(
)(
) (
)(
) (
)(
)
= 0.819
Gas Solubility:
The gas solubility Rs is defined as the number of standard cubic feet of gas which will
dissolve in one stock-tank barrel of crude oil at certain pressure and temperature. The
solubility of a natural gas in a crude oil is a strong function of the pressure, temperature,
API gravity, and gas gravity.
For a particular gas and crude oil to exist at a constant temperature, the solubility
increases with pressure until the saturation pressure is reached.
At the saturation pressure (bubble-point pressure) all the available gases are dissolved in
the oil and the gas solubility reaches its maximum value. Rather than measuring the
amount of gas that will dissolve in a given stock-tank crude oil as the pressure is
increased, it is customary to determine the amount of gas that will come out of a sample of
reservoir crude oil as pressure decreases.
A typical gas solubility curve, as a function of pressure for an undersaturated crude oil, is
shown in Figure 21. As the pressure is reduced from the initial reservoir pressure pi, to the
bubble-point pressure pb, no gas evolves from the oil and consequently the gas solubility
remains constant at its maximum value of Rsb. Below the bubble-point pressure, the
solution gas is liberated and the value of Rs decreases with pressure. The following five
empirical correlations for estimating the gas solubility are given below:
 Standing’s correlation
 The Vasquez-Beggs correlation
 Glaso’s correlation
 Marhoun’s correlation
 The Petrosky-Farshad correlation
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Figure 21: Gas-solubility pressure diagram.
Standing’s Correlation
Standing (1947) proposed a graphical correlation for determining the gas solubility as a
function of pressure, gas specific gravity, API gravity, and system temperature. The
correlation was developed from a total of 105 experimentally determined data points on
22 hydrocarbon mixtures from California crude oils and natural gases. The proposed
correlation has an average error of 4.8%. Standing (1981) expressed his proposed
graphical
*(
)
+
correlation in the following more convenient mathematical form
With:
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Dr.Jawad R.R. Alassal Kirkuk U.
X = 0.0125 API - 0.00091 (T – 460)
Where:
T = temperature, oR
P = system pressure, psia
ϒg = solution gas specific gravity.
It should be noted that Standing’s equation is valid for applications at and below the
bubble-point pressure of the crude oil.
Example:
The following experimental PVT data on six different crude oil systems are available.
Results are based on two-stage surface separation.
Oil #
T
API
P > Pb
1
250
2377
715
1.528
38.13
150
60
47.1
0.851
2
220
2620
768
1.474
40.95
100
75
40.7
0.855
3
260
2051
693
1.529
37.37
100
72
48.6
0.911
4
237
2884
968
1.619
38.92
60
120
40.5
0.898
5
218
3045
943
1.570
37.70
200
60
44.2
0.781
6
180
4239
807
1.385
46.79
85
173
27.3
0.848
Where:
T = reservoir temperature, °F
pb = bubble-point pressure, psig
Bo = oil formation volume factor, bbl/STB
psep = separator pressure, psig
Tsep = separator temperature, °F
co = isothermal compressibility coefficient of the oil at a specified pressure, psi -1
Using Standing’s correlation, estimate the gas solubility at the bubble point pressure and
compare with the experimental value in terms of the absolute average error (AAE).
[184]
Dr.Jawad R.R. Alassal Kirkuk U.
Solution:
Apply Equation 4 (2-70) to determine the gas solubility. Results of the calculations are
given in the following tabulated form:
Oil #
X
1
2
3
4
5
6
0.361
0.309
0.371
0.312
0.322
0.177
Predicted Rs
Equation 270
838
817
774
969
1012
998
2.297
2.035
2.349
2.049
2.097
1.505
Measured Rs
%
Error
751
768
693
968
943
807
11.6
6.3
11.7
0.108
7.3
23.7
AAE = 10.1 %
.
The Vasquez-Beggs Correlation:
Vasquez and Beggs (1980) pr
esented an improved empirical correlation for estimating Rs. The correlation was obtained
by regression analysis using 5,008 measured gas solubility data points. Based on oil
gravity, the measured data were divided into two groups. This division was made at a
value of oil gravity of 30°API. The proposed equation has the following
form:
*
(
)+
Values for the coefficients are as follows:
Cofficient
C1
C2
C3
API > 30
0.0178
1.1870
23.931
0.0362
1.0937
25.7240
The authors proposed the following relationship for
adjustment of the gas gravity gg to the reference separator pressure:
[185]
Dr.Jawad R.R. Alassal Kirkuk U.
(
[
)(
)(
)
(
)]
where
ϒgs = gas gravity at the reference separator pressure
ϒg = gas gravity at the actual separator conditions of psep and Tsep
psep = actual separator pressure, psia
Tsep = actual separator temperature, °R
Example:
Using the PVT of the six crude oil systems of Example 1, solve for the gas solubility.
Solution:
Oil #
1
2
3
4
5
6
Predicted Rs Measured Rs
Equation 271
0.8731
0.855
0.911
0.850
0.814
0.834
779
733
702
820
947
841
751
768
693
968
943
807
%
Error
3.76
-4.58
1.36
15.2
0.43
4.30
AAE = 4.9%
Glaso’s Correlation:
Glaso (1980) proposed a correlation for estimating the gas solubility as a function of the
API gravity, pressure, temperature, and gas specific gravity.
The correlation was developed from studying 45 North Sea crude oil samples. Glaso
reported an average error of 1.28% with a standard deviation of 6.98%. The proposed
relationship has the following form:
[186]
Dr.Jawad R.R. Alassal Kirkuk U.
*(
Where
(
)(
)
)+
is a correlating number and is defined by the following expression :-
With
[
( )]
Marhoun’s Correlation
[
]
where
ϒg =gas specific gravity
ϒ0=stock-tank oil gravity
T =temperature, °R
a–e =coefficients of the above equation having these values:
a =185.843208
b =1.877840
c =-3.1437
d =-1.32657
e =1.398441
The Petrosky-Farshad Correlation
*(
)
[187]
Dr.Jawad R.R. Alassal Kirkuk U.
+
x = 7.916 (10-4) (API)1.5410 - 4.561(10-5 ) (T - 460)1.3911
where:
p = pressure, psia
T = temperature, °R
Bubble-Point Pressure:
The bubble-point pressure pb of a hydrocarbon system is defined as the
highest pressure at which a bubble of gas is first liberated from the oil.
This important property can be measured experimentally for a crude oil
system by conducting a constant-composition expansion test.
In the absence of the experimentally measured bubble-point pressure, it
is necessary for the engineer to make an estimate of this crude oil property
from the readily available measured producing parameters. Several graphical
and mathematical correlations for determining pb have been proposed
during the last four decades. These correlations are essentially based on
the assumption that the bubble-point pressure is a strong function of gas
solubility Rs, gas gravity ϒg, oil gravity API, and temperature T, or:
pb = f (RS, ϒg, API, T)
Several ways of combining the above parameters in a graphical form or
a mathematical expression are proposed by numerous authors, including:
 Standing
 Vasquez and Beggs
 Glaso
 Marhoun
 Petrosky and Farshad
[188]
Dr.Jawad R.R. Alassal Kirkuk U.
Standing’s Correlation
[(
(
)
)
]
With:
a = 0.00091 (T -460) -0.0125 (API)
Where:
pb = bubble-point pressure, psia
T = system temperature, °R
Standing’s correlation should be used with caution if nonhydrocarbon
components are known to be present in the system.
The Vasquez-Beggs Correlation:
*(
)(
) +
With:
a =-C3 API/T
Cofficient
C1
C2
C3
API > 30
27.624
0.914328
11.172
56.18
0.84246
10.393
Glaso’s Correlation:
(
)
(
)
[
Where:
p*b is a correlating number and defined by the following equation:
[189]
Dr.Jawad R.R. Alassal Kirkuk U.
(
)]
(
) ( ) (
)
where
Rs =gas solubility, scf/STB
t =system temperature, °F
ϒg =average specific gravity of the total surface gases
a, b, c =coefficients of the above equation having the following values:
a =0.816
b =0.172
c =-0.989
For volatile oils, Glaso recommends that the temperature exponent b
be slightly changed, to the value of 0.130.
Marhoun’s Correlation:
where
T =temperature, °R
ϒo=stock-tank oil specific gravity
ϒg=gas specific gravity
a–e =coefficients of the correlation having the following values:
a =5.38088 x10-3 b =0.715082
c =-1.87784
d =3.1437
e =1.32657
[190]
Dr.Jawad R.R. Alassal Kirkuk U.
The reported average absolute relative error for the correlation is 3.66% when compared
with the experimental data used to develop the correlation.
The Petrosky-Farshad Correlation:
*
(
)
+
x = 7.916 (10-4) (API)1.5410 - 4.561(10-5 ) (T - 460)1.3911
Oil Formation Volume Factor:
The oil formation volume factor, Bo, is defined as the ratio of the volume of oil (plus the
gas in solution) at the prevailing reservoir temperature and pressure to the volume of oil at
standard conditions. Bo is always greater than or equal to unity. The oil formation volume
factor can be expressed mathematically as:
Bo =
(
)
(
)
Where:
Bo =oil formation volume factor, bbl/STB
(Vo)p,T =volume of oil under reservoir pressure p and temperature T, bbl
(Vo)sc =volume of oil is measured under standard conditions, STB
A typical oil formation factor curve, as a function of pressure for an undersaturated crude
oil (pi > pb), is shown in Figure 22. As the pressure is reduced below the initial reservoir
pressure pi, the oil volume increases due to the oil expansion. This behavior results in an
increase in the oil formation volume factor and will continue until the bubble-point
pressure is reached. At pb, the oil reaches its maximum expansion and consequently
attains a maximum value of Bob for the oil formation volume factor. As the pressure is
reduced below pb, volume of the oil and Bo are decreased as the solution gas is liberated.
When the pressure is reduced to atmospheric pressure and the temperature to 60°F, the
value of Bo is equal to one.Most of the published empirical Bo correlations utilize the
following generalized relationship:
(
)
[191]
Dr.Jawad R.R. Alassal Kirkuk U.
Figure 22: Oil formation volume factor versus pressure.
Six different methods of predicting the oil formation volume factor are
presented below:
 Standing’s correlation
 The Vasquez-Beggs correlation
 Glaso’s correlation
 Marhoun’s correlation
 The Petrosky-Farshad correlation
 Other correlations
It should be noted that all the correlations could be used for any pressure
equal to or below the bubble-point pressure.
Standing’s Correlation:
*
( )
Where:
T = temperature, °R
ϒo = specific gravity of the stock-tank oil
[192]
Dr.Jawad R.R. Alassal Kirkuk U.
(
)+
ϒg = specific gravity of the solution gas.
The Vasquez-Beggs Correlation:
(
)[
)(
]
Where:
R =gas solubility, scf/STB
T =temperature, °R
ϒgs=gas specific gravity as defined by Equation 2-72
Values for the coefficients C1, C2 and C3 are given below:
Cofficient
API > 30
C1
C2
C3
Vasquez and Beggs reported an average error of 4.7% for the proposed correlation.
Glaso’s Correlation:
Bo =1.0 +10A
Where:
A =-6.58511 +2.91329 log B*ob -0.27683 (log B*ob)2
B*ob is a correlating number and is defined by the following equation:
(
( )
[193]
Dr.Jawad R.R. Alassal Kirkuk U.
)
where
T =temperature, °R
ϒo=specific gravity of the stock-tank oil
Marhoun’s Correlation:
with the correlating parameter F as defined by the following equation:
The coefficients a, b and c have the following values:
a =0.742390
b =0.323294
c =-1.202040
where T is the system temperature in °R.
The Petrosky-Farshad Correlation:
(
)*
(
Where
T = temperature , Ro
= Specific gravity of the stock tank oil
[194]
Dr.Jawad R.R. Alassal Kirkuk U.
)
(
)
+
Material Balance Equation:
where ρo = density of the oil at the specified pressure and temperature, lb/ft3.
Example:
The following experimental PVT data on six different crude oil systems are available.
Results are based on two-stage surface separation.
Oil
#
1
2
3
4
5
6
T
Pb
Rs
Bo
250
220
260
237
218
180
2377
2620
2051
2884
3065
4239
751
768
693
968
943
807
1.528
1.474
1.529
1.619
1.570
1.385
38.13
40.95
37.37
38.92
37.70
46.79
Psep
Tsep
API
150
100
100
60
200
85
60
75
72
120
60
173
47.1
40.7
48.6
40.5
44.2
27.3
0.851
0.855
0.911
0.898
0.781
0.848
Calculate the oil formation volume factor at the bubble-point pressure by using the six
different correlations. Compare the results with the experimental values and calculate the
absolute average error (AAE).
Solution:
Crude Oil
1
2
3
4
5
6
%AAE
Exp.
Bo
1.528
1.474
1.529
1.619
1.570
1.385
-
Method
1
1.506
1.487
1.495
1.618
1.571
1.461
1.7
Method
2
1.474
1.450
1.451
1.542
1.546
1.389
2.8
Method
3
1.473
1.459
1.461
1.589
1.541
1.438
2.8
[195]
Dr.Jawad R.R. Alassal Kirkuk U.
Method
4
1.516
1.477
1.511
1.575
1.554
1.414
1.3
Method
5
1.552
1.508
1.556
1.632
1.584
1.433
1.8
Method
6
1.525
1.470
1.542
1.623
1.599
1.387
0.6
Where:
Method 1 =Standing’s correlation.
Method 2 =Vasquez-Beggs correlation.
Method 3 =Glaso’s correlation.
Method 4 =Marhoun’s correlation.
Method 5 =Petrosky-Farshad correlation.
Method 6 =Material balance equation.
Isothermal Compressibility Coefficient of Crude Oil:
Isothermal compressibility coefficients are required in solving many reservoir engineering
problems, including transient fluid flow problems, and they are also required in the
determination of the physical properties of the undersaturated crude oil.
By definition, the isothermal compressibility of a substance is defined mathematically by
the following expression:
(
)
For a crude oil system, the isothermal compressibility coefficient of the oil phase co is
defined for pressures above the bubble-point by one of the following equivalent
expressions:
( )(
)
(
)(
)
(
)(
)
Where:
[196]
Dr.Jawad R.R. Alassal Kirkuk U.
co = isothermal compressibility, psi-1
ρo = oil density lb/ft3
Bo = oil formation volume factor, bbl/STB
At pressures below the bubble-point pressure, the oil compressibility
is defined as:
where Bg = gas formation volume factor, bbl/scf
There are several correlations that are developed to estimate the oil compressibility at
pressures above the bubble-point pressure, i.e., undersaturated crude oil system. Three of
these correlations are presented below:
 The Vasquez-Beggs correlation.
 The Petrosky-Farshad correlation.
 McCain’s correlation.
The Vasquez-Beggs Correlation:
(
)
Where:
T = temperature, °R
p = pressure above the bubble-point pressure, psia
Rsb = gas solubility at the bubble-point pressure
ϒgs = corrected gas gravity.
The Petrosky-Farshad Correlation:
(
[197]
Dr.Jawad R.R. Alassal Kirkuk U.
)
Where:
T =temperature, °R
Rsb =gas solubility at the bubble-point pressure, scf/STB
Oil Formation Volume Factor for Undersaturated Oils
[
(
)]
Where:
Bo = oil formation volume factor at the pressure of interest, bbl/STB
Bob = oil formation volume factor at the bubble-point pressure, bbl/STB
p = pressure of interest, psia
pb = bubble-point pressure, psia
Crude Oil Density:
The above equation may be used to calculate the density of the oil at pressure below or
equal to the bubble-point pressure
Standing (1981) proposed an empirical correlation:
[
( )
(
)]
Density of the oil at pressures above the bubble-point pressure can be calculated
with:
[198]
Dr.Jawad R.R. Alassal Kirkuk U.
[ (
)]
Crude Oil Viscosity:
Crude oil viscosity is an important physical property that controls and influences the flow
of oil through porous media and pipes. The viscosity, in general, is defined as the internal
resistance of the fluid to flow. The oil viscosity is a strong function of the temperature,
pressure, oil gravity, gas gravity, and gas solubility. Whenever possible, oil viscosity
should be determined by laboratory measurements at reservoir temperature and pressure.
The viscosity is usually reported in standard PVT analyses. If such laboratory data are not
available, engineers may refer to published correlations, which usually vary in complexity
and accuracy depending upon the available data on the crude oil. According to the
pressure, the viscosity of crude oils can be classified into three categories:
• Dead-Oil Viscosity
The dead-oil viscosity is defined as the viscosity of crude oil at atmospheric
pressure (no gas in solution) and system temperature.
• Saturated-Oil Viscosity
The saturated (bubble-point)-oil viscosity is defined as the viscosity of
the crude oil at the bubble-point pressure and reservoir temperature.
• Undersaturated-Oil Viscosity
The undersaturated-oil viscosity is defined as the viscosity of the crude oil at a pressure
above the bubble-point and reservoir temperature.
Estimation of the oil viscosity at pressures equal to or below the bubble-point pressure is a
two-step procedure:
Step 1. Calculate the viscosity of the oil without dissolved gas (dead oil), mob, at the
reservoir temperature.
Step 2. Adjust the dead-oil viscosity to account for the effect of the gas solubility at the
pressure of interest.
[199]
Dr.Jawad R.R. Alassal Kirkuk U.
At pressures greater than the bubble-point pressure of the crude oil, another adjustment
step, i.e. Step 3, should be made to the bubble-point oil viscosity, mob, to account for the
compression and the degree of under-saturation in the reservoir. A brief description of
several correlations that are widely used in estimating the oil viscosity in the above three
steps is given below.
METHODS OF CALCULATING VISCOSITY
OF THE DEAD OIL:
Several empirical methods are proposed to estimate the viscosity of
the dead oil, including:
 Beal’s correlation
 The Beggs-Robinson correlation
 Glaso’s correlation
These three methods are presented below.
Beal’s Correlation:
From a total of 753 values for dead-oil viscosity at and above 100°F, Beal (1946)
developed a graphical correlation for determining the viscosity of the dead oil as a
function of temperature and the API gravity of the crude. Standing (1981) expressed the
proposed graphical correlation in a mathematical relationship as follows:
(
(
)
)(
)
With
(
)
where
=viscosity of the dead oil as measured at 14.7 psia and reservoir
temperature , cp
T = temperature , oR
The Beggs-Robinson Correlation:
[200]
Dr.Jawad R.R. Alassal Kirkuk U.
Beggs and Robinson (1975) developed an empirical correlation for determining the
viscosity of the dead oil. The correlation originated from analyzing 460 dead-oil viscosity
measurements. The proposed relationship is expressed mathematically as follows:
Where
(
)
An average error of -0.64% with a standard deviation of 13.53% was reported for the
correlation when tested against the data used for its development. Sutton and Farshad
(1980) reported an error of 114.3% when the correlation was tested against 93 cases from
the literature.
Glaso’s Correlation:
Glaso (1980) proposed a generalized mathematical relationship for
computing the dead-oil viscosity. The relationship was developed from
experimental measurements on 26 crude oil samples. The correlation has
the following form:
[
(
)](
)
where the coefficient a is given by:
a = 10.313 [log(T -460)] -36.447
[201]
Dr.Jawad R.R. Alassal Kirkuk U.
[
(
)]
METHODS OF CALCULATING THE SATURATED OIL VISCOSITY:
Several empirical methods are proposed to estimate the viscosity of
the saturated oil, including:
 The Chew-Connally correlation
 The Beggs-Robinson correlation
These two correlations are presented below:
The Chew-Connally Correlation:
(
With
) (
)
[
(
(
(
)
(
)]
)
)
(
)
Where:
µob = viscosity of the oil at the bubble-point pressure, cp
µod = viscosity of the dead oil at 14.7 psia and reservoir
temperature, cp
The experimental data used by Chew and Connally to develop their
correlation encompassed the following ranges of values for the independent
variables:
Pressure, psia: 132–5,645
Temperature, °F: 72–292
Gas solubility, scf/STB: 51–3,544
[202]
Dr.Jawad R.R. Alassal Kirkuk U.
Dead oil viscosity, cp: 0.377–50
The Beggs-Robinson Correlation:
From 2,073 saturated oil viscosity measurements, Beggs and Robinson (1975) proposed
an empirical correlation for estimating the saturated-oil viscosity. The proposed
mathematical expression has the following form:
(
Where
(
)
)
(
)
The reported accuracy of the correlation is -1.83% with a standard deviation of 27.25%.
The ranges of the data used to develop Beggs and Robinson’s equation are:
Pressure, psia: 132–5,265
Temperature, °F: 70–295
API gravity: 16–58
Gas solubility, scf/STB: 20–2,070
METHODS OF CALCULATING THE VISCOSITY OF THE
UNDERSATURATED OIL
Oil viscosity at pressures above the bubble point is estimated by first calculating the oil
viscosity at its bubble-point pressure and adjusting the bubble-point viscosity to higher
pressures. Vasquez and Beggs proposed a simple mathematical expression for estimating
the viscosity of the oil above the bubble-point pressure. This method is discussed below.
The Vasquez-Beggs Correlation:
(
Where
With
[203]
Dr.Jawad R.R. Alassal Kirkuk U.
)
(
)
The data used in developing the above correlation have the following
ranges:
Pressure, psia: 141–9,151
Gas solubility, scf/STB: 9.3–2,199
Viscosity, cp: 0.117–148
Gas gravity: 0.511–1.351
API gravity: 15.3–59.5
The average error of the viscosity correlation is reported as -7.54%.
PROPERTIES OF RESERVOIR WATER:
Water Formation Volume Factor (Bw)
The water formation volume factor (Bw) can be calculated by the following Mathematical
expression :
Bw = A1 + A2 p +A3 P2
Where the coefficients A1 – A3 are given by the following expression:
Ai =a1 +a2 (T – 460) +a3 (T -460)2
With a1 – a3 given for gas – free and gas saturated water:
Gas-Free Water
A1
A2
A3
a1
0.9947
-4.228(10-6)
1.3(10-10)
A1
A2
A3
a1
0.9911
-1.093(10-6)
-5.0(10-11)
a2
5.8(10-6)
1.8376(10-8)
-1.3855(10-12)
Gas-Saturated Water
a2
6.35(10-5)
-3.497(10-9)
6.429(10-13)
[204]
Dr.Jawad R.R. Alassal Kirkuk U.
a3
1.02(10-6)
-6.77(10-11)
4.285(10-15)
a3
8.5(10-7)
4.57(10-12)
-1.43(10-15)
Water Viscosity:
Meehen (1980) proposed water viscosity correlation that accounts the effect of pressure
and density:
µw =µwD [ 1+3.5 x10-2 P2 (T -40)]
with:
µwD = A +B/T
A = 4.518 x10-2 + 9.313 x10-7 Y -3.93 x 10-12 Y2
B = 70.634 +9.576 x10-10 Y2
Where:
µw = brine viscosity at P and T, cp
µwD = = brine viscosity at 14.7 and T, cp
P = pressure psia.
T = Temperature , oF
Y = water salinity, ppm.
Brill and Beggs (1978) presented a simpler equation which considers only temperature
effects:
µw = exp( 1.003 – 1.479 x10-2 T + 1.982 x 10-5T2)
where T is in o F and µw in cp
Pressure Volume Temperature (PVT)
PVT tests are designed to study and quantify the phase behavior and properties of
reservoir fluids at simulated recovery conditions.
[205]
Dr.Jawad R.R. Alassal Kirkuk U.
Accurate and reliable phase behavior and volumetric data are essential elements for proper
management of petroleum reservoirs.
Most reservoirs are produced by depletion in which the reservoir pressure declines as
fluids are recovered; the reservoir temperature stays practically constant in most recovery
methods, the main variable that determines the behavior of fluids under reservoir
conditions during depletion is therefore the reservoir pressure. Hence relatively simple
tests which simulate recovery processes are conducted by varying the fluid pressure.
The main emphasis is on the volumetric data at the reservoir and surface temperatures;
hence the name PRESSURE-VOLUME-TEMPERATURE (PVT) data becomes
applicable.
FLUID SAMPLING
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 properties values
to be used in the various design procedures. As important is the quality of the sample
collected to ensure that the fluids tested are representative of the reservoir crude oil.
PVT analysis of a reservoir fluid comprises the determination of: The relation between pressure and volume at reservoir temperature.
 Various physical constants required for reservoir engineering calculations such as
Viscosity, Density, and Compressibility… etc.
 The effect of separator pressure and temperature on Oil formation volume factor,
Gas-Oil ratio… etc.
 The chemical composition of the most volatile components.
The value to be attached to the laboratory determination depends on whether the sample
investigated is representative of the reservoir contents, 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 minimize errors caused by differences in relative
movements of oil and gas after solution gas has been liberated.
The obtaining of the samples can be accomplished by: Surface sampling
 Sub-surface sampling
Subsurface samples can only be representative of 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.
[206]
Dr.Jawad R.R. Alassal Kirkuk U.
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.
Surface samples of oil and gas are taken from the separator connected with the well; the
surface oil and gas are then recombined in the laboratory on the basis of the producing gas
oil ratio.
Particular care therefore must be exercised in the field to obtain reliable sample and
accurate measurement of the gas oil ratio and separator conditions.
PRESSURE-VOLUME-TEMPERATURE (PVT) TESTS
The main applications of the PVT data are: To provide data for reservoir calculations
 To provide physical property data for well flow calculations
 For surface facility design
In reservoir calculations the PVT tests and subsequent report provides the source of the
reservoir fluids properties necessary to describe the behavior of the reservoir over its
development and production. The tests conducted therefore have to take in to
consideration the process going on both above and below the saturation pressure.
The main PVT tests for oil system plus associated compositional analysis:1. The flash vaporization or relative volume tests
2. The differential vaporization test
3. The separator test
4. Viscosity measurement
5. Composition analysis
1.
Flash Vaporization (Relative volume test):-
In this test the sample is placed in an equilibrium cell at pressure equal or greater than the
reservoir pressure and at constant temperature (reservoir temperature), then the pressure
reduced incrementally by expanding the fluid volume, starting from high pressure to the
lowest possible pressure, the gas liberated below the point of saturation pressure
remaining in equilibrium with the oil throughout the experiment, that is the system overall
composition remains constant.Figure 22 shows the procedure of this vaporization.
[207]
Dr.Jawad R.R. Alassal Kirkuk U.
The flash vaporization test also known as constant composition expansion, and pressure
volume relation test gives the relation between pressure (P) and volume (V) of reservoir
fluid at constant temperature (T), by plotting the volume of the system versus pressure a
break is obtained in the slope this occurs at the bubble point pressure (figure 23).
Figure (22) Laboratory Flash Vaporization Procedur
Figure 23 :determination of bubble point pressure with data from Flash vaporization.
[208]
Dr.Jawad R.R. Alassal Kirkuk U.
The bubble point, or saturation pressure is that pressure below which gas is liberated.
Hence a two-phase system is formed where as above the bubble point pressure a one
phase system is present (under-saturated liquid).
To carry out a relative volume test, the PVT cell is setup as shown in figure 22.
The PVT cell is filled with a certain quantity of reservoir fluid at pressure above the
estimated bubble point pressure and room temperature, and then the temperature of the
cell is raised to the reservoir temperature. During heating it is necessary to maintain
pressure by increasing the content volume of the PVT cell, when the pressure remains
constant the temperature is reached the reservoir temperature. The thermal expansion
factor ( ) can then be calculated from the volume change and it is equal to
=
(
)
Where:V1 = Volume of oil at room temperature T1
V2 = Volume of oil at reservoir temperature T2
The thermal expansion factor is expressed in ○C-1 or, ○F-1
The compressibility of the oil phase above the bubble point pressure can now be
calculated.
=
(
)
Where:V1 = Volume of oil at pressure P1
V2 = Volume of oil at pressure P2
The compressibility is expressed in reciprocal of pressure units e.g. Psi -1, atm-1 .
The system volume is commonly reported by the relative volume (Vr) defines as the
ratio of the total volume to the initial bubble point volume.
Relative volume (Vr) = (V/Vb)
Where:
V = Volume of oil at any pressure
[209]
Dr.Jawad R.R. Alassal Kirkuk U.
Vb = Volume of oil at bubble point pressure
Example: The data from a flash vaporization on a black oil at 220oF are given below.
Determine the bubble point pressure and prepare a table of pressure and relative volume
for the reservoir fluid study.
[210]
Dr.Jawad R.R. Alassal Kirkuk U.
Solution:
First, plot pressure against total volume, determine the point at which the two line cross,
and read bubble – point pressure and volume at the bubble point.
Pb = 2620 psig, Vb 63.316 cc.
Second, divide all volumes in the data set by Vb and add Pb to the table
For instance , at 2516 psig
Relative volume = 64.291 cc/ 63.316 cc = 1.0154.
See figure 23.
2. Composition Test (Direct Flash Test):An important test on all reservoir fluid samples is the determination of the fluid
composition. The most common method of compositional analysis of high pressure fluids
is to flash a relatively large volume of the fluid sample to the atmospheric pressure to
form generally two stabilized phases of gas and liquid, the two phases are individually
analyzed and then numerically recombined using the ratio of the separated phases, the gas
and liquid phases are commonly analyzed by gas chromatography and distillation
respectively, the properties of oil such as gas oil ratio (GOR), oil formation volume factor
[211]
Dr.Jawad R.R. Alassal Kirkuk U.
(Bo) and density and gas phase properties such as gas gravity (γg), compressibility,
density can be calculated.
3. Differential Vaporization (or, Liberation)Test:The sample of reservoir liquid in the laboratory cell is brought to bubble point pressure,
and temperature is set at reservoir temperature.
Pressure is reduced by increasing cell volume, and the cell is agitated to ensure
equilibrium between the gas and liquid. Then, all the gas expelled from the cell while
pressure in the cell is held constant by reducing cell volume.
The gas volume is collected, and its quantity and specific gravity are measured. The
volume of liquid remaining in the cell, Vo is measured.
This process is shown in figure 24.
Figure 24: Laboratory Differential vaporization Procedure
The process is repeated in steps until atmospheric pressure is reached. Then temperature is
reduced to 60oF, and the volume of remaining liquid is measured. This called residual oil
from differential vaporization, or residual oil.
Each of the values of volume of cell liquid , Vo is divided by the volume of
[212]
Dr.Jawad R.R. Alassal Kirkuk U.
the residual oil. The results is called relative oil volume and is given by symbol BoD
The volume of gas removed during each step is measured both at cell conditions and at
standard conditions. The Z- factor is calculated as:
Z = VR PR TSc / (VSc PSc TR)
Where the subscript R refers to conditions in the cell. Formation volume factors of the gas
removed are calculated with z – factors using :
Bg = 0.0282 Z T/P Cuft/SCF
The total volume of gas removed during the entire process is the amount of gas in solution
at the bubble point. This total volume divide by the volume of residual oil, and the units
are converted to standard cubic feet per barrel of residual oil. The symbol RsDb represents
standard cubic feet of gas removed per barrel of residual oil.
The gas remaining in solution at any lower pressure is calculated by subtracting the sum
of the gas removed down to and including the pressure of interest from the total volume of
gas removed. The result is divided by the volume of residual oil, converted to SCF/
residual bbl, and reported as RSD
Relative total volume at any pressure is calculated as :
BtD = BoD + Bg (RSDb – RSD )
Table below show these calculations.
Example: The data from a differential vaporization on a black oil at 220 OF are given
below. Prepare a table of solution gas- oil ratios, relative oil volumes, and relative total
volumes by this differential process.
Also include Z- factors and formation volume factors of the increments of gas removed.
[213]
Dr.Jawad R.R. Alassal Kirkuk U.
Solution:
All calculation shown will be at 2100 psig.
First, calculate solution gas oil ratio
RSD = (
(
)
)(
)
)
684 SCF/ bbl
Second, calculate relative oil volume:
BoD =
= 1.515 res.bbl/ residual bbl
Third, calculate Z- factor
Z = VR PR TSc / (VSc PSc TR)
Z=
(
)(
(
)(
)(
)(
)(
)
)
= 0.851
Fourth, calculate the gas formation volume factor:
Bg = 0.0282 Z T/P Cuft/SCF
Bg = 0.0282 x 0.851 x 680/(2114.7) = 0.00771 cuft/ SCF
[214]
Dr.Jawad R.R. Alassal Kirkuk U.
Fifth, calculate relative total volume:
BtD = BoD + Bg (RSDb – RSD )
BtD = ( 1.515 res. Bbl/residual bbl ) + ( 0.00771 cuft)/(5.615 cuft/bbl) (854 -684
scf/residual bbl) = 1.748 res. bbl/ residual bbl
All the calculations are shown in table below.
[215]
Dr.Jawad R.R. Alassal Kirkuk U.
[216]
Dr.Jawad R.R. Alassal Kirkuk U.
Chapter -5The Material Balance Equation (MBE)
The MBE for any hydrocarbon system is simply a volume balance which equates the total
production to the difference between the initial volume of hydrocarbon in the reservoir
and the current volume.
Assumptions of the MBE
1. Uniform pressure throughout the reservoir
2. No variation in PVT properties throughout the reservoir at certain pressure
3. Constant reservoir volume
4. Constant temperature
The MBE is a zero dimensional model, meaning that it is evaluated at a point in the
reservoir.
MBE can be used for:
1. Determine volume of hydrocarbon in place (N,G)
2. Determine water in flux (We) and mechanism
3. Determine average reservoir pressure
4. Determine the size of gas cap (m) in oil reservoir
5. Predication of reservoir performance.
Conditions to apply MBE :
1. Needs good quality of pvt data.
2. Accurate production history.
3. Average reservoir pressure information.
4. Always apply from initial pressure to current pressure.
Gas Reservoir:
Reservoir contains free gas as the only hydrocarbon system. The gas is either dry, wet or
condensate.
 Dry Gas : composed of methane and non-hydrocarbon such N2, CO2. The gas
remains single phase from reservoir to the separator condition.
[217]
Dr.Jawad R.R. Alassal Kirkuk U.
 Wet Gas :is mainly composed of methane and other light components. Wet gas
will not drop out condensate in the reservoir, but produces some condensate at the
surface. Generally, if the reservoir temperature is greater than critical temperature
the reservoir is gas.
 Condensate Reservoir: composed of heavy hydrocarbon, and reservoir
temperature lies between critical point and circondentherm. The gas will drop out
liquid in the reservoir.
Gas reservoir may have water influx from an aquifer or no water influx called
volumetric reservoir.
Most calculation in gas engineering involves gas formation volume factor (Bg) and gas
expansion factor (Eg = 1/Bg). Bg is defined as the volume occupied by moles of gas (n) at
certain pressure (p), and temperature (T) to that occupied at standard conditions.
Determination of Initial Gas in Place
There are commonly two approaches:
1. Volumetric Method.
2. Material Balance Method.
Volumetric method:
(
)⁄
Where:
G = initial gas-in-place, Scf
A = area of the reservoir, acres, from map
h = average reservoir thickness, ft, from Log
[218]
Dr.Jawad R.R. Alassal Kirkuk U.
= average weighted porosity, from Log
Swi = irreducible water saturation, from Log
Equation 1 can be applied at the initial pressure (pi) and at a depletion pressure (p), in
order to calculate the cumulative gas production.
Gas production (Gp) = Initial gas in place – Remaining gas in- place
(
)⁄
(
)( ⁄
(
)⁄
Or
i=
⁄
)
= Initial pore volume
Example :
A gas reservoir has the following characteristics:
A = 3000 acres, h = 30 ft,
= 0.15, Swi = 0.2, T = 150o F, pi = 2000 psi
Zi = 0.82
P, psi
Z
2000
0.82
1000
0.88
400
0.92
Calculate the cumulative gas produced and recovery factor at 1000 psia and 400 psia
Solution:
1. PV = pore volume =
2. Bg
P
Z
Bg, cf/Scf
[219]
Dr.Jawad R.R. Alassal Kirkuk U.
2000
1000
400
3.
0.0054
0.0152
0.0397
)⁄
(
4.
0.82
0.88
0.92
)⁄
(
5.
(
)⁄
6. The recovery factor (RF) =
⁄
7.
8.
⁄
⁄
The ultimate (RF) for volumetric gas reservoir will range from 80% to 90%. If a strong
water drive is present, trapping of residual gas at higher pressure can reduce the RF to
the range of 50% to 80%. So, operate the field with maximum production.to reduce
reservoir pressure to a low value, thereby minimizing the actual gas volume occupying
residual pore space.
Method of Material Balance Equation:
The material balance equation (MBE) in simple form is:
Production = expansion +influx + injection
[220]
Dr.Jawad R.R. Alassal Kirkuk U.
The Production terms
Produced gas ( Gp), & water (Wp) at surface
At reservoir condition:
The Expansion terms
 Gas Expansion.
 Water Expansion.
 Rock Expansion.
1. Gas Expansion
2. Water Expansion (
)
⁄
Substituting Eq. 2 in Eq1
[221]
Dr.Jawad R.R. Alassal Kirkuk U.
(
)
(
Eq 3
)
⁄(
)
Eq4
Substituting Eq. 4 in Eq3
(
Eq 5
)
Where :
Eq 5 is the water expansion term
3.Rock Expansion
(
)
(
)
Sum the expansion terms
(
)
(
)
(
)
or
(
(
)
)
)
(
The Influx terms
Water influx (We), injection either water or gas or both
Therefore, the general material balance is
(
)
(
)
(
(
[222]
Dr.Jawad R.R. Alassal Kirkuk U.
)
)
+We
Eq(A)
Play with eq. (A) for many cases
Case1
1. Neglect the expansion terms of water and rock, water influx, no injection and no
water production
(
Substitute definition of
)
(
)
in Eq (6) the
(
)
Or
(
)
Multiply Eq.7 by (pi /zi )× (p/z)
(
)
(
)
Divide eq.(8) by G
Equation (9) is a straight line equation. Plot of
slope of -
, & when Gp = 0, then
, and When
[223]
Dr.Jawad R.R. Alassal Kirkuk U.
vs
with
= 0, then G =Gp
-
Eq. (9) is applied only for volumetric gas reservoir.
If there is water influx the plot of vs. Gp will be as shown in figure. below:
[224]
Dr.Jawad R.R. Alassal Kirkuk U.
Other Plots can be used to recognize the presence of water influx
1. Cole plot
2. Modified Cole plot
3. Energy plot
Cole Plot:
Cole neglects or ignores the water influx from MBE
(
)
Plot
(
)
vs. Gp results as shown in the figure below
Modified Cole Plot
(
)
[225]
Dr.Jawad R.R. Alassal Kirkuk U.
Where:
(
)
)
(
Comparison between Cole plot and Modified Cole plot
1. Reservoirs those are subject to weak aquifer and significant
both plots have negative slope.
. In this case,
2. Reservoirs with
is significant but there is no aquifer attached. in this
case, the original Cole plot will have a negative slope while the Modified plot
will be horizontal.
Generally a negative slope for a weak aquifer?
Because
The numerator in
(
)
(is small (We) grows slower than denominator (
Energy Plot : Which is the plot of Log (1 –
If:
Slope = 1 volumetric reservoir
Slope < 1 water influx
Slope >1 gas leaking from reservoir or bad data
Extrapolation to (1) ( P=0) then the intercept = G
[226]
Dr.Jawad R.R. Alassal Kirkuk U.
) vs. GP
)
1
G
The Material Balance Expressed As A Linear Equation
Now return to equation (A), which is :
)
(
)
using the nomenclature of Havlena and Odeh, equation (1) can be
following form:
(
(
)
(
)
equation 2 is a straight line equation
where:
=(
= (
)
)
(
Eq4
Eq5
)
[227]
Dr.Jawad R.R. Alassal Kirkuk U.
written in the
Therefore:
<<
and
is frequently neglected
to determine G, we have to select the proper model of (We) to give a straight line.
Apply VEH model: (I will not go in detail in calculating We)
∑
⁄
plot ⁄
vs. ∑
wil yield a straight line, provided that the ∑
accurately assumed. The intercept is G and the slope is B
Non linear plots will result if the aquifer is improperly characterized.
is
Example:
The volumetric estimate of the gas initially in place for a dry gas reservoir range from 1.3
to 1.65×1012 Scf. production, pressure and the pertinent gas expansion,
are
in the attached table.
Calculate original gas in place (G)
[228]
Dr.Jawad R.R. Alassal Kirkuk U.
Time
M
P
psia
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
2883
2881
2874
2866
2857
2849
2841
2826
2808
2794
2782
2767
2755
2741
2726
2712
2699
2688
2667
∑
(
)
0
4
18
34
52
68
85
116.5
154.5
185.5
212
246
273.5
305.6
340
373.5
405
432.5
455.5
(
√(
)
⁄
)
5.5340
24.5967
51.1776
76.9246
103.3184
131.5371
180.0178
240.7764
291.3014
336.6281
392.8592
441.3134
497.2907
556.1110
613.6513
672.5969
723.0868
771.4902
0.3536
0.4647
0.6487
0.7860
0.9306
1.0358
1.0315
1.0594
1.1485
1.2425
1.2905
1.3702
1.4219
1.4672
1.5714
1.5714
1.6332
1.7016
13835
1.3665
1.5052
1.4793
1.5194
1.5475
1.5452
1.5584
1.5708
1.5879
1.5979
1.6136
1.6278
1.6256
1.6430
1.6607
1.6719
1.6937
solution:
1. Assume a volumetric gas reservoir
2. Plot p/z vs. Gp or
3. Plot of
vs
vs. Gp
shows upward curvature as shown in figure below, indicating
water in flux.
vs. ∑
4. Assume a linear water influx, plot
√
⁄(
) as shown
in the figure below, G = 1.325×1012 Scf, and BL= 212.7×103 cu.ft/ psia-month.5
[229]
Dr.Jawad R.R. Alassal Kirkuk U.
Indication of Water Drive
Havlena – Odeh MBE Plot of the Above Example
HW1 :
(1) Find the gas in place from the given production data
(2) Draw Cole & Modified Cole plot.
[230]
Dr.Jawad R.R. Alassal Kirkuk U.
(3) Draw the Energy Plot.
(4) Choose any a suitable model of water influx.( assume any information you need)
cf =
6.x 10-6 1/psi
cw =
3.x 10-6 1/psi
Sw =
Ts =
Ps
0.15
60
14.7
F
Psia
T= 200oF
Production data
Time
Year
0
1
2
3
4
5
6
7
8
9
10
(psia)
p
6411
5947
5509
5093
4697
4319
3957
3610
3276
2953
2638
Z Factor
1.1192
1.089
1.0618
1.0374
1.0156
0.9966
0.9801
0.9663
0.9551
0.9467
0.9409
Gp
B SCF
0
5.475
10.950
16.425
21.900
27.375
32.850
38.325
43.800
49.275
54.750
Wp
STB
0
378
1,434
3,056
5,284
8,183
11,864
16,425
22,019
28,860
37,256
Drive Indices for Gas Reservoirs (DI)
(DI), indicate the relative magnitude of the various energy forces contributing to the
driving mechanism of the reservoir.
Divide the general (MBE) by
[231]
Dr.Jawad R.R. Alassal Kirkuk U.
(
)
Define the following:
1. Gas Drive Index (GDI) as
(
)
2. Compressibility Drive Index (CDI) as
(
)
3. Water Drive Index (WDI) as:
If the summation is not equal one a fluctuation above or below one means the quality of
the collected production data with time is not accurate.
HW
For the Production data given below :
1- Generate a (P/Z) Vs (Gp) plot and find G
2- Generate a Cole Plot and Modified Cole plot .Do these plot indicate water influx is
occurring ? find G
3- Generate a Havlena – Odeh (F Vs Et) and find G
4- Find the drive indices.
Cf =4 × 10-6 psi-1 , Cw =3.× 10-6 psi-1 , Sw = 0.2, T= 200oF
Pressure, psia
2876
2824
2755
2688
2570
2435
2226
Z- factor
0.907
0.905
0.903
0.902
0.901
0.900
0.901
[232]
Dr.Jawad R.R. Alassal Kirkuk U.
Gp, MMSCF
384
550
788
1002
1445
1899
2670
An Abnormally Pressured Gas Reservoirs:
These reservoirs sometimes called ―an overpressure‖ or ―geo-pressure‖ gas reservoirs.
These reservoir are defined as reservoirs with pressure gradient greater than a normal
pressure gradient, i.e. over 0.5 psi/ ft ,A typical p/z vs. Gp plot for an abnormally pressure
gas reservoirs will exhibit two straight line as shown in the figure below.
The first straight line corresponds to the ―apparent‖ gas reservoir behaviour with an
extrapolation that given the ―apparent‖gas in place‖.
The second straight line corresponds to the normal pressure behaviour with an
extrapolation given the actual initial gas in place (G)
MBE for Pot (Tank) Aquifer (Jawad MSc):
The general MBE is:
(
(
)
(
Use the following definitions:
[233]
Dr.Jawad R.R. Alassal Kirkuk U.
)
)
(
)
(
)
(
(
)
)
(
)
(
)
⁄
Eq2
Where = Ct = Cw Sw + cf
Substitute eq.2 in eq1 and arrange the MBE, the result is a straight line equation too:
(
)
*
(
)
+
A plot of Y vs X results in a straight line for pseudo steady state the G = (1/ slop) of that
line. From the intercept , estimate ( )
EXAMPLE :
Estimate rD and initial Gas in - Place
ø
0.05
Cw
3.50E-06
Cf
6.50E-06
Sw
0.44
T,oF
318
[234]
Dr.Jawad R.R. Alassal Kirkuk U.
P,psia
4275
Z
1.0145
Gp,BSCF
0
Wp,MBbl
0
Bw
1.08
4240
1.0126
3.83
30.0492
1.08
4120
1.0061
20.7
162.2653
1.08
4060
1.0041
52.5
412.2092
1.08
3983
1.0002
85.7
672.8247
1.08
3982
1.0002
108
847.8916
1.08
3918
0.9970
116
911.9707
1.08
3941
0.9981
143
1118.764
1.08
3894
0.9958
173
1355.079
1.09
3878
0.9950
180
1415.978
1.09
3848
0.9936
191
1501.955
1.09
3782
0.9916
231
1809.842
1.09
3583
0.9834
320
2510.039
1.09
3538
0.9810
378
2966.75
1.09
3525
0.9845
395
3103.751
1.09
3515
0.9811
407
3197.055
1.09
3504
0.9820
422
3310.341
1.09
3449
0.9797
466
3654.256
1.09
3425
0.9788
496
3890.602
1.09
3320
0.9748
561
4420.918
1.09
3268
0.9730
590
4643.533
1.09
3266
0.9757
654
5148.02
1.09
3178
0.9737
694
5464.427
1.09
2996
0.9679
797
6269.743
1.09
2909
0.9654
871
6853.17
1.09
2842
0.9635
933
7340.276
1.09
2770
0.9616
1006
7911.578
1.09
2695
0.9598
1070
8414.238
1.09
2627
0.9582
1123
8831.839
1.09
2562
0.9558
1191
9350.379
1.09
2454
0.9546
1274
10068.65
1.09
2384
0.9533
1331
10613.22
1.09
2310
0.9521
1409
11444.22
1.09
[235]
Dr.Jawad R.R. Alassal Kirkuk U.
2220
0.9507
1464
12170.84
1.09
2062
0.9487
1517
12912.06
1.09
TANK TYPE MODEL MBE
1.2E-7
G = 1/Slope = 2.94 TSCF
1.0E-7
8.0E-8
6.0E-8
Y
4.0E-8
2.0E-8
-1.9E-21
-2.0E-8
-4.0E-8
-6.0E-8
-8.0E-8
0.0E+0
5.0E+4
1.0E+5
1.5E+5
2.0E+5
2.5E+5
3.0E+5
3.5E+5
X
Intercept=
1.75E-08
slope= 2.9E-13
G=
Ra/Re=
3.45E+12
2.08
HW:
For the Production data given below:
1- Generate a (P/Z) Vs (Gp) plot and find G
2- Generate a Cole Plot and Modified Cole plot .Do these plot indicate water influx is
occurring? find G
3- Generate a Havlena – Odeh (F Vs Et) and find G
4- Find the drive indices.
[236]
Dr.Jawad R.R. Alassal Kirkuk U.
Time Year
0.00
5.58
6.75
7.33
8.83
9.50
10.41
11.41
12.24
13.08
15.17
17.00
19.00
19.75
20.58
21.75
22.66
23.83
24.75
25.75
27.00
Ø
Cw
Cf
Sw
1-Sw
P,psia
4265
3997
3923
3886
3837
3795
3762
3709
3672
3646
3461
3333
3215
3170
3113
3045
2987
2887
2824
2772
2685
0.05
3.50E-06
6.50E-06
0.44
0.56
Z
1.0150
1.0019
0.9987
0.9970
0.9951
0.9934
0.9920
0.9900
0.9888
0.9879
0.9812
0.9763
0.9725
0.9712
0.9699
0.9683
0.9667
0.9640
0.9625
0.9614
0.9596
T=
200 0F
Gp
Bscf
0
230.517
295.94
337.519
421.695
459.81
508.42
568.138
602.276
650.988
798.654
929.707
1071.868
1125.064
1185.288
1267.995
1330.54
1406.141
1454.759
1512.137
1584.244
Wp
Mbbl
0
1357.4
1742.2
1986.9
2482.8
2707.5
2993.6
3345.2
3545.8
3832
4702.3
5473.7
6310.7
6642
7020.9
7449.5
8044
8844.1
9492.7
10275.4
11430.3
Bw
bbl/bbl
1.08
1.08
1.08
1.08
1.08
1.08
1.08
1.08
1.09
1.09
1.09
1.09
1.09
1.09
1.09
1.09
1.09
1.09
1.09
1.09
1.09
Gp for MBE:
Gp = Dry gas sold + Equivalent gas of condensate produced + Original water vapour
in produced gas + Vent gas from condensate gas.
The gas equivalent (GE) of one stock tank barrel of condensate liquid is:
[237]
Dr.Jawad R.R. Alassal Kirkuk U.
Therefore,
(
(
)
)
Gas equivalent to water vapour
Example:
Calculate the total daily reservoir gas production (Gp) and water production (Wp) for
MBE calculation.
Separator gas production = 10MMScf/ D
Condensate Production = 150 STB/D
Stock tank gas production (vent gas)= 30 MScf /D
Fresh water production = 15 bbl/ D
Condensate gravity = 55 API
Formation water salinity = 150,000 ppm
Solution:
(
)
(
)
[238]
Dr.Jawad R.R. Alassal Kirkuk U.
(
(
)
)
(
(
)
)
Condensate gas = 150×782.9 = 0.117 MMScf /D
Water vapour in original gas =311.8 lb/ MMScf from chart
Volume of dissolved water in produced gas = (311.8 ×(10+.117))/350.8
= 9 bbl/D
Gas equivalent to these (9 bbl of water) is
Therefore, Gp for MBE = 10+ 0.117 +0.0066 +0.03 = 10.213 MMScf / D
⁄
Homework
Calculate the total daily reservoir gas production (Gp) and water production (Wp) for
MBE calculation.
Separator gas production = 22 MMScf/ D
Condensate Production = 170 STB/D
Stock tank gas production (vent gas) = 45 MScf /D
Fresh water production = 22 bbl/ D
Condensate gravity = 52 API
Pi =3500 psia Ti= 2150 F
[239]
Dr.Jawad R.R. Alassal Kirkuk U.
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