DRILLING
ENGINEERING
Vahid Salimi
Textbook
Applied Drilling Engineering,
by :Adam T. Bourgoyne Jr.,
Martin E. Chenevert,
Keith K. Millheim
F.S. Young Jr.,.
Contents:
pore pressure and fracture pressure
drilling hydraulics
casing design
under balanced drilling
directional drilling
Chapter 1
pore pressure and fracture pressure
Hydrostatic Pressure
• Hydrostatic pressure is defined as the pressure exerted
by a column of fluid.
• The pressure is a function of the average fluid density
and the vertical height or depth of the fluid column.
• Mathematically, hydrostatic pressure is expressed as:
HP (psi) = 0.052 x ρf (ppg) x D (ft)
where:
HP = hydrostatic pressure
ρf = average fluid density
D = true vertical depth or height of the column
Hydrostatic Pressure(cont’d)
Hydrostatic pressures can easily be converted to
equivalent mud weights and pressure gradients.
Hydrostatic pressure gradient is given by:
HG = HP / D … (psi/ft)
Example
Calculate the hydrostatic pressure for the
following wells:
• a. mud weight = 9 ppg, hole depth = 10100 ft MD (measured
depth), 9900 ft TVD (true vertical depth)
• b. mud gradient = 0.468 psi / ft, hole depth = 10100 ft MD
(measured depth), 9900 ft TVD (true vertical depth)
solution
a.
HP (psi) = 0.052 x ρf (ppg) x D (ft) = 0.052 x 9 x 9900 = 4632 psi
b.
Hydrostatic pressure = fluid gradient (psi / ft) x depth
(ft)..........psi
= 0.468 (psi /ft) x 9900(ft) = 4633 psi
 Mud Weight (MW) should be kept heavy enough so that hydrostatic
head of mud column is higher than formation pressure at any depth.
Usually 150 psi
 Need to know formation pressure in order to determine MW
Pf + 150 = 0.052 MW D
Pf
MW
D
150
Formation Pressure, psi
Mud Weight, ppg
True Vertical Depth, ft
Safety, psi
Example
 You are drilling with 7.9 ppg
oil base mud. If the formation
pressure is predicted 5,000 psi
at 9,000 ft true vertical depth,
what is the required MW in
order to have 150 psi
overpressure ?
5,000 + 150 = 0.052 MW 9,000
MW = 11 ppg
OVERBURDEN PRESSURE
 The overburden pressure is defined as the pressure exerted
by the total weight of overlying
formations above the point of interest.
 The total weight is the combined weight of both the formation solids
(rock matrix) and formation fluids in the pore space.
 The overburden pressure can therefore be expressed as the hydrostatic
pressure exerted by all materials overlying the depth of interest:
σov = 0.052 x ρb x D
where
σov = overburden pressure (psi)
ρb = formation bulk density (ppg)
D = true vertical depth (ft)
OVERBURDEN PRESSURE(cont’d)
 Overburden gradient under field conditions of
varying lithological and pore fluid density is given by:
σovg= 0.433[(1 – φ)ρma +(φxρf)]
where
σovg= overburden gradient, psi/ft
φ= porosity expressed as a fraction
ρf= formation fluid density
ρma= matrix density
matrix and fluid densities
Substance
Sandstone
Limestone
Dolomite
Anhydrite
Halite
Gypsum
Clay
Freshwater
Seawater
Oil
Gas
Density (gm/cc)
2.65
2.71
2.87
2.98
2.03
2.35
2.7 - 2.8
1.0
1.03 - 1.06
0.6 - 0.7
0.15
 To convert densities from gm/cc to gradients in psi/ft use:
Gradient (psi/ft) = 0.433 x (gm /cc)
 To convert from psi/ft to ppg, use:
Density (ppg) = gradient (psi/ft) / 0.052
Pore pressure
 The magnitude of pressure in the pore of formation known as the pore pressure
 Pore pressure
= formation pressure
=formation fluid pressure
=reservoir pressure
=pressure in fluid contained in the pore spaces of the rock
Example
 Determine the pore pressure of a
normally pressured formation in
the Gulf of Mexico at 9,000’ depth.
Solution
p = 0.465 psi/ft * 9,000 ft = 4,185 psig
Homework: Pore Pressure Profiles
 The following pore pressure information has been supplied
for the well you are about to drill.
 a. Plot the following pore pressure/depth information on a
P-Z diagram :
 b. Calculate the pore pressure gradients in the formations
from surface; to 8000ft; to 8500ft; and to 9500ft. Plot the
overburden gradient (1 psi/ft) on the above plot.
 Determine the mud weight required to drill the hole section:
down to 8000ft; down to 8500ft; and down to 9500ft.
 Assume that 200 psi overbalance on the formation pore
pressure is required.
 c. If the mudweight used to drill down to 8000ft were used to
drill into the formation pressures at 8500ft what would be the
over/underbalance on the formation pore pressure at this
depth?
 d. Assuming that the correct mudweight is used for drilling at
8500ft but that the fluid level in the annulus dropped to 500 ft
below drillfloor, due to inadequate hole fill up during tripping.
What would be the effect on bottom hole pressure at 8500ft ?
 e. What type of fluid is contained in the formations below
8500ft.
Normal Pore Pressure
 Pressure of a column of water extending from the formation to the surface
 The magnitude of normal pore pressure varies with the concentration of
dissolved salts, type of fluid, gases present and temperature gradient.
=0.433 psi/ft for fresh water
=0.465 psi/ft for seawater
Subnormal Formation Pressure
 Subnormal pore pressure is defined as any formation pressure that
is less than the corresponding fluid hydrostatic pressure at a given
depth.
 Subnormal formation pressure can cause lost circulation of water as
the drilling fluid.
ABNORMAL PORE PRESSURE
Abnormal pore pressure is defined as any pore pressure
that is greater than the hydrostatic pressure of the
formation water occupying the pore space.
 Abnormal pressure is sometimes called overpressure or
geopressure.
Abnormal pressure can be thought of as being made up
of a normal hydrostatic component plus an extra amount
of pressure.
This excess pressure is the reason why surface control
equipment (e.g. BOPs) are required when drilling oil and
gas wells.
ABNORMAL PORE PRESSURE(cont’d)
 Abnormal formation pressure can
cause a kick with water as the
drilling fluid.
Normal and Abnormal Pore Pressure
Normal Pressure Gradients
West Texas: 0.433 psi/ft
Gulf Coast: 0.465 psi/ft
Depth, ft
Abnormal
Pressure
Gradients
10,000’
Pore Pressure, psig
Pore Pressure vs. Depth
0
0.433 psi/ft
0.465 psi/ft
8.33 lb/gal
9.0 lb/gal
5,000
Normal
10,000
Abormal
15,000
20,000
8
9
10
11 12
13
14
15
Pore Pressure Equivalent, lb/gal
16
{ Density of mud required to control this pore pressure }
Fracture Gradient
Pore Pressure
Gradient
Transition zone
The upper portion of the region of abnormal
pressure is called the transition zone
Causes Of Abnormal Pore Pressure
Compaction Effects
Diagenetic Effects
Differentional Density Effects
Fluid Migration Effects
Diagenetic Effects
 With increasing pressure and temperature, sediments
undergo a process of chemical and physical changes
collectively known as diagenesis.
 Diagenesis is the alteration of sediments and their
constituent minerals during post depositional compaction.
 Diagenetic processes include the formation of new
minerals, recrystallisation and lithification.
 Diagenesis may result in volume changes and water
generation which if occurring in a seabed environment
may lead to both abnormal or sub-normal pore pressure.
Clay Diagenesis
 Clay Diagenesis (Conversion of Smectite to Illite)
 If the water released in this process cannot escape during compaction, then the
pore fluid will support an increased portion of the overburden and will thus be
abnormally pressured.
 Diagenesis of Sulphate Formations
 Anhydrite (CaSO4) is diagenetically formed from the dehydration of gypsum
(CaSO4.2H2O).
 During the process large volumes of water are released and this is accompanied by
a 30-40% reduction in formation volume
HIGH PRESSURE
NORMAL PRESSURE
7. Abnormal Pressure
661. Drilling Engineering
Slide 33
Homework
When crossing faults it is possible to go from normal pressure
to abnormally high pressure in a short interval.
7. Abnormal Pressure
661. Drilling Engineering
Slide 37
Well “A” found only Normal Pressure ...
7. Abnormal Pressure
661. Drilling Engineering
Slide 38
Methods of estimating pore pressure
Direct measurement
It is possible only when the formation has been drilled
It is expensive
Indirect measurement
The main parameter is the variation of porosity with depth (porosity
dependent parameter)
If pore pressure is normal, porosity-dependent parameter (x) have an easily
recognized trend because of the decreased porosity with increased depth of
burial and compaction.
A departure from the normal pressure trend signals a probable transition
zone.
Detection of the depth at which this departure occurs is critical because
casing must be set in the well before excessively pressured permeable zones
can be drilled safely.
Prediction and Detection of Abnormal
Pressure Zones
1. Before drilling
 Shallow seismic surveys
 Deep seismic surveys
 Comparison with nearby wells
7. Abnormal Pressure
661. Drilling Engineering
Slide 40
Prediction and Detection of Abnormal
Pressure Zones
2. While drilling
 Drilling rate, gas in mud, etc. etc.
 D - Exponent
 DC - Exponent
 MWD - LWD
 Density of shale (cuttings)
7. Abnormal Pressure
661. Drilling Engineering
Slide 41
Prediction and Detection of Abnormal
Pressure Zones
3. After drilling
 Resistivity log
 Conductivity log
 Sonic log
 Density log
7. Abnormal Pressure
661. Drilling Engineering
Slide 42
Compaction Theory of Abnormal Pressure
During deposition, sediments are compacted by
the overburden load and are subjected to greater
temperatures with increasing burial depth.
Porosity is reduced as water is forced out.
Hydrostatic equilibrium within the compacted
layers is retained as long as the expelled water is
free to escape
If water cannot escape, abnormal pressures
occur
Compaction Theory
 In Porous formation the overburden stress is supported by rock matrix
stress and pore pressure
 Bulk Density = ρm (1-Ф) + ρf Ф
 The average porosity in sediments ,generally decreases with increasing depth
- due to the increasing overburden
 This results in an increasing bulk density with increasing depth, and
increasing rock strength
 Average Porosity Ф = ρm - ρb / ρm – ρf
 Plot Ф Vs. Depth on similog graph.
Example
Calculate the overburden stress at a depth of
7,200 ft in the Santa Barbara Channel.
Assume
φo = 0.37
ρma = 2.6 gm/cc
kφ = 0.0001609 ft-1
ρf = 1.044 gm/cc
Solution
Homeworks
Homeworks (cont’d)
Homeworks (cont’d)
Pore pressure prediction methods
 Measure the porosity indicator (e.g.density) in normally
pressured, clean shales to establish a normal trend line.
 When the indicator suggests porosity values that are higher than
the trend, then abnormal pressures are suspected to be present.
 The magnitude of the deviation from the normal trend line is
used to quantify the abnormal pressure.
Equivalent Matrix Stress Method
Example
 Estimate the pore pressure at 10,200’ if the equivalent depth
is 9,100’. The normal pore pressure gradient is 0.433 psi/ft.
The overburden gradient is 1.0 psi/ft.
ASSUME:
 At 9,100’, pne = 0.433 * 9,100 = 3,940 psig
 At 9,100’, σobe = 1.00 * 9,100 = 9,100 psig
 At 10,200’, σob = 1.00*10,200 = 10,200 psig
Solution
Approach 2: Empirical correlation
 More accurate
 Need numerous data
 Uses (Xo-Xn) or (Xo/Xn) to predict the
magnitude of the abnormal pressure
Prediction of pore pressure by seismic data
Homework
Pore pressure indications while drilling
Drilling rate (ROP)
 gas in mud
Pit level
Flowline temperature
Rate Of Penetration(ROP)
 Drill bits break the rock by a combination of several processes
including: Compression (weight-on-bit), shearing (rpm) and sometimes
jetting action of the drilling fluid.
 The speed of drilling is described as the rate of penetration (ROP) and is
measured in ft/hr.
 The rate of penetration is affected by numerous parameters namely:










Weight On Bit (WOB)
Revolutions Per Minute (RPM)
bit type
bit wear
hydraulic efficiency
degree of overbalance
drilling fluid properties
hydrostatic pressure and
hole size
Formation properties
TABLE 2.8 -
Note, that many factors can influence the drilling rate, and some of these
factors are outside the control of the operator.
75
Effect of bit weight and hydraulics on penetration
rate
Inadequate
hydraulics or
excessive imbedding
of the bit teeth in the
rock
0
76
Drilling rate
increases more or
less linearly with
increasing bit
weight.
A significant
deviation from this
trend may be
caused by poor
bottom hole
cleaning
Effect of Differential Pressure on Drilling Rate
Differential pressure is the
difference between
wellbore pressure and
pore fluid pressure
Decrease can be due to:
• The chip hold down effect
• The effect of wellbore
pressure on rock strength
77
 If all parameters affecting ROP are held constant whilst drilling a
uniform shale sequence then ROP should decrease with depth. This
is due to the natural increased compaction with depth reflecting a
decrease in porosity and increased shale density and increased
shale (compressive) strength.
 When entering an abnormally pressured shale, the drill bit sees a
shale section which is undercompacted. The increased porosity and
decreased density of the undercompacted section results in the
formation becoming more ‘drillable’ as there is less rock matrix to
remove. Consequently ROP increases, assuming all drilling
parameters were kept constant.
Drilling underbalanced
can further increase the
drilling rate.
79
Drilling Rate as a Pore Pressure Predictor
Penetration rate depends on a number of different parameters.
R = K(P1)a1 (P2)a2 (P3)a3… (Pn)an
Modified d-exponent
81


 log R 
60N 
d 

12W 
 log 6 
10 d b 

R  ft /hr
N  rpm
d  d  exponent
W  Bit W eight, lbf
d b  Bit Diamet er,in
Or, in its most used form:
 The D Exponent basically attempts to correct the ROP for changes in RPM,
weight on bit and hole size.
 The D exponent increases linearly with depth, reflecting increased rock
strength with depth. For abnormally pressured shales, the D exponent
deviates from the normal trend and actually decreases with depth.
82
dc-exponent
Mud weight also affects R
An adjustment to d may be made:
dc = d (rn /rc)
where
dc = exponent corrected for mud density
rn = normal pore pressure gradient
rc = effective mud density in use
83
d-exponent
 The d-exponent normalizes R for any
variations in W, db and N
 Under normal compaction, R should
decrease with depth. This would cause d
to increase with depth.
 Any deviation from the trend could be caused
by abnormal pressure.
84
Example
• While drilling in a Gulf Coast shale,
R = 50 ft/hr
W = 20,000 lbf
N = 100 RPM
ECD = 10.1 ppg
(Equivalent Circulating Density)
db = 8.5 in
• Calculate d and dc
85
Solution
 50 
log 

 2.079
60 * 100 

d

 12 * 20,000   1.554
log 

6
 10 * 8.5 
d  1.34
 0.465 
dc  1.34 

 0.052 * 10.1
dc  1.19
86

 log R
60N
d

12 W
log

10 6 d b







 rn 
d c  d  
 rc 
Example
solution
Ratio Method
 The ratio method is much simpler and does not
require values of overburden. To calculate
pore pressure, use the following formula:
Homework
• Using the Eaton Method, calculate the pore
pressure at depth 12000 ft given:
 dcn (from normal trend) = 1.5 d-units
 dco (from new trend) = 1.1 d-units
 Overburden gradient = 19 ppg
 Normal pore pressure in area = 9 ppg
Fracture Pressure
Prediction of Fracture Gradients
Well Planning
Theoretical Fracture Gradient Determination
Hubbert & Willis
Matthews & Kelly
Ben Eaton
Comparison of Results
Experimental Frac. Grad. Determination
Leak-off Tests
Fracture Gradients
1.11- 101
In-Situ Earth Stresses
Example
Fracture Gradient Determination
2. Matthews & Kelly:
K is
P
F 

D
D
where
Ki = matrix stress coefficient
s = vertical matrix stress, psi
Fracture Gradients
1.11- 106
Example
A Texas Gulf Coast well has a pore pressure gradient
of 0.735 psi/ft. Well depth = 11,000 ft.
Calculate the fracture gradient in units of lb/gal using
Matthews & Kelly method
Summarize the results in tabular form, showing
answers, in units of lb/gal and also in psi/ft.
Fracture Gradients
1.11- 107
Example
P K is
F 
D
D
2. Matthews & Kelly
s may be calculated, and K i is determined
In this case P and D are known,
graphically.
(i) First, determine the pore pressure gradient.
P
 0.735
D
Fracture Gradients
psi / ft
1.11- 108
(given )
Example - Matthews and Kelly
(ii) Next, calculate the matrix stress.
S=P+s
s=S-P
= 1.00 * D - 0.735 * D
= 0.265 * D
= 0.265 * 11,000
s = 2,915 psi
Fracture Gradients
1.11- 109
S  overburden, psi 
s  mat rixstress, psi 




P  porepressure, psi


D  depth, ft

Example - Matthews and Kelly
(iii) Now determine the depth, D,i where,
under normally pressured conditions, the rock
matrix stress, s would be 2,915 psi.
Sn = Pn + sn
n = “normal”
1.00 * Di = 0.465 * Di + 2,915
Di * (1 - 0.465) = 2,915
2,915
Di 
 5,449 ft
0.535
Example Matthews and
Kelly
(iv) Find Ki from the
plot on the right, for
Di = 5,449 ft
For a south Texas
Gulf Coast well,
Ki = 0.685
Fracture Gradients
1.11- 111
Example - Matthews and Kelly
(v) Now calculate F:
K is
P
F 

D
D
0.685 * 2,915
F 
 0.735
11,000
 0.9165
psi / ft
0.9165
F 
 17.63
0.052
lb / gal
Leak off Test
 A test carried out to the point where the formation leaks off
Fracture Gradients
1.11- 114
Experimental Determination of Fracture
Gradient
Example:
In a leak-off test below the casing
seat at 4,000 ft, leak-off was
found to occur when the
standpipe pressure was 1,000 psi.
MW = 9 lb/gal.
What is the fracture gradient?
Slide 115
Solution
Leak-off pressure = PS + DPHYD
= 1,000 + 0.052 * 9 * 4,000
= 2,872 psi
PLEAK OFF 2,872

D
4,000
psi
ft
Fracture gradient = 0.718 psi/ft
EMW = ?
13.8 lb/gal
Slide 116
Homework
 While performing a leak off test the surface pressure at leak
off was 940 psi. The casing shoe was at a true vertical depth of
5010 ft and a mud weight of 10.2 ppg was used to conduct
the test.
Calculate: the maximum allowable mud weight at this depth .
Homework
 A leakoff test was carried out just below a 13 3/8" casing shoe at 7000 ft.
TVD using 9.0 ppg mud. The results of the tests are shown below.
What is the maximum allowable mud weight for the 12 1/4" hole section ?
BBLS PUMPED
1
1.5
2
2.5
3
3.5
4
4.5
5
SURFACE PRESSURE
400
670
880
1100
1350
1600
1800
1900
1920
(psi)
Equivalent Circulating Density (ECD)
 When the drilling fluid is circulating through the drillstring, the
borehole pressure at the bottom of the annulus will be greater than
the hydrostatic pressure of the mud.
 The extra pressure is due to the frictional pressure required to pump
the fluid up the annulus. This frictional pressure must be added to
the pressure due to the hydrostatic pressure from the column of mud
to get a true representation of the pressure acting against the
formation a the bottom of the well.
 An equivalent circulating density (ECD) can then be calculated from
the sum of the hydrostatic and frictional pressure divided by the true
vertical depth of the well.
Homework
 If the circulating pressure losses in the annulus of the above
well is 300 psi when drilling at 7500ft with 9.5ppg mud, what
would be the ECD of the mud at 7500ft.