8. fracture gradients
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8. Fracture GradientsTRANSCRIPT
8. Fracture Gradients PETE 661 Drilling Engineering Slide 1
PETE 661Drilling Engineering
Lesson 8 Fracture Gradients
8. Fracture Gradients PETE 661 Drilling Engineering Slide 2
Prediction of Fracture GradientsWell PlanningTheoretical Fracture Gradient Determination
Hubbert & WillisMatthews & KellyBen EatonComparison of Results
Experimental Frac. Grad. DeterminationLeak-off TestsLost Circulation
Gas Cut Mud
8. Fracture Gradients PETE 661 Drilling Engineering Slide 3
Read:Applied Drilling Engineering, Ch. 6
HW # 5Casing Design
due 10-8-04
8. Fracture Gradients PETE 661 Drilling Engineering Slide 4
Well PlanningSafe drilling practices require that the
following be considered when planning a well:
Pore pressure determination Fracture gradient determination Casing setting depth selection Casing design H2S considerations Contingency planning
8. Fracture Gradients PETE 661 Drilling Engineering Slide 5
Fig. 7.21
8. Fracture Gradients PETE 661 Drilling Engineering Slide 6
fracture
8. Fracture Gradients PETE 661 Drilling Engineering Slide 7
Formation Pressure and Matrix Stress
Given: Well depth is 14,000 ft. Formation pore pressure expressed in equivalent mud weight is 9.2 lb/gal. Overburden stress is 1.00 psi/ft.
Calculate: 1. Pore pressure, psi/ft , at 14,000 ft2. Pore pressure, psi, at 14,000 ft3. Matrix stress, psi/ft
4. Matrix stress, psi
8. Fracture Gradients PETE 661 Drilling Engineering Slide 8
Formation Pressure and Matrix Stress
PSoverburden pore matrix stress = pressure + stress (psi) (psi) (psi)
S = P +
8. Fracture Gradients PETE 661 Drilling Engineering Slide 9
Formation Pressure and Matrix Stress
Calculations:
1. Pore pressure gradient= 0.433 psi/ft * 9.2/8.33 = 0.052 * 9.2= 0.478 psi/ft
2. Pore pressure at 14,000 ft= 0.478 psi/ft * 14,000 ft= 6,692 psig
Depth = 14,000 ft. Pore Pressure = 9.2 lb/gal equivalent Overburden stress = 1.00 psi/ft.
8. Fracture Gradients PETE 661 Drilling Engineering Slide 10
Formation Pressure and Matrix Stress
Calculations:3. Matrix stress gradient,
psi
psi/ft
/ D = 0.522 psi/ft
PS
DDP
DSor
ft/psi478.0000.1DP
DS
D.,e.i
8. Fracture Gradients PETE 661 Drilling Engineering Slide 11
Formation Pressure and Matrix Stress
Calculations:
4. Matrix stress at 14,000 ft
= 0.522 psi/ft * 14,000 ft
= 7,308 psi
8. Fracture Gradients PETE 661 Drilling Engineering Slide 12
Fracture Gradient Determination
In order to avoid lost circulation while drilling it is important to know the variation of fracture gradient with depth.
Leak-off tests represent an experimental approach to fracture gradient determination. Below are listed and discussed three approaches to calculating the fracture gradient.
8. Fracture Gradients PETE 661 Drilling Engineering Slide 13
Fracture Gradient Determination
1. Hubbert & Willis:
where F = fracture gradient, psi/ft
= pore pressure gradient, psi/ftDP
DP21
31Fmin
DP1
21Fmax
8. Fracture Gradients PETE 661 Drilling Engineering Slide 14
Fracture Gradient Determination
2. Matthews & Kelly:
where Ki = matrix stress coefficient
= vertical matrix stress, psi
DP
DKF i
8. Fracture Gradients PETE 661 Drilling Engineering Slide 15
Fracture Gradient Determination
3. Ben Eaton:
where S = overburden stress, psi = Poisson’s ratio
DP
1*
DPSF
8. Fracture Gradients PETE 661 Drilling Engineering Slide 16
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 each of the above three methods.
Summarize the results in tabular form, showing answers, in units of lb/gal and also in psi/ft.
8. Fracture Gradients PETE 661 Drilling Engineering Slide 17
1. Hubbert & Willis:
The pore pressure gradient,
F 13
1 2 *0.735 0.823 psiftmin
D2P1
31Fmin
PD
0.735 psift
Example - Hubbert and Willis
8. Fracture Gradients PETE 661 Drilling Engineering Slide 18
Also,
lb/galpsi/ft0.052
psi/ft0.823Fmin
lb/gal 15.83Fmin
Example - Hubbert and Willis
8. Fracture Gradients PETE 661 Drilling Engineering Slide 19
Example - Hubbert and Willis
DP1
21Fmax 735.01
21
= 0.8675 psi/ft
Fmax = 16.68 lb/gal
8. Fracture Gradients PETE 661 Drilling Engineering Slide 20
2. Matthews & Kelly
In this case P and D are known, may be calculated, and is determined graphically.
(i) First, determine the pore pressure gradient.
DK
DPF i
iK
Example
)given(ft/psi735.0DP
8. Fracture Gradients PETE 661 Drilling Engineering Slide 21
Example - Matthews and Kelly
(ii) Next, calculate the matrix stress.
ft ,depthDpsi ,pressure porePpsi ,stress matrix
psi ,overburdenS
S = P + = S - P = 1.00 * D - 0.735 * D = 0.265 * D = 0.265 * 11,000 = 2,915 psi
8. Fracture Gradients PETE 661 Drilling Engineering Slide 22
Example - Matthews and Kelly
(iii) Now determine the depth, , where, under normally pressured conditions, the rock matrix stress, would be 2,915 psi.
iD
Sn = Pn +n n = “normal”1.00 * Di = 0.465 * Di + 2,915
Di * (1 - 0.465) = 2,915
ft449,5535.0915,2Di
8. Fracture Gradients PETE 661 Drilling Engineering Slide 23
Example - Matthews and
Kelly
(iv) Find Ki from the plot on the right, for
For a south Texas Gulf Coast well,
Di = 5,449 ft
Ki = 0.685
8. Fracture Gradients PETE 661 Drilling Engineering Slide 24
Example - Matthews and Kelly
(v) Now calculate F:DP
DKF i
735.0000,11
915,2*685.0F
ft/psi9165.0
gal/lb63.17052.0
9165.0F
8. Fracture Gradients PETE 661 Drilling Engineering Slide 25
0.685
5,449
Ki
Dep
th, D
i
8. Fracture Gradients PETE 661 Drilling Engineering Slide 26
Example
Ben Eaton:
DP
1*
DPSF
??DS
8. Fracture Gradients PETE 661 Drilling Engineering Slide 27
Variable Overburden Stress by Eaton
At 11,000 ftS/D = 0.96 psi/ft
8. Fracture Gradients PETE 661 Drilling Engineering Slide 28
Fig. 5-5
At 11,000 ft = 0.46
8. Fracture Gradients PETE 661 Drilling Engineering Slide 29
Example - Ben Eaton
From above graphs, at 11,000 ft.:
DP
1DP
DSF
46.0;ft/psi96.0DS
735.046.01
46.0735.096.0F
F = 0.9267 psi/ft = 17.82 lb/gal
8. Fracture Gradients PETE 661 Drilling Engineering Slide 30
Summary of Results
Fracture Gradient psi.ft lb/gal
Hubbert & Willis minimum: 0.823 15.83
Hubbert & Willis maximum: 0.868 16.68
Mathews & Kelly: 0.917 17.63
Ben Eaton: 0.927 17.82
8. Fracture Gradients PETE 661 Drilling Engineering Slide 31
Summary of Results
Note that all the methods take into consideration the pore pressure gradient. As the pore pressure increases, so does the fracture gradient.
In the above equations, Hubbert & Willis apparently consider only the variation in pore pressure gradient. Matthews & Kelly also consider the changes in rock matrix stress coefficient, and in the matrix stress ( Ki and i ).
8. Fracture Gradients PETE 661 Drilling Engineering Slide 32
Summary of Results
Ben Eaton considers variation in pore pressure gradient, overburden stress and Poisson’s ratio,
and is probably the most accurate of the three methods. The last two methods are actually quite similar, and usually yield similar results.
8. Fracture Gradients PETE 661 Drilling Engineering Slide 33
Similarities
Ben Eaton:
DP
1*
DPSF
Matthews and Kelly:
DP
DKF i
8. Fracture Gradients PETE 661 Drilling Engineering Slide 34
8. Fracture Gradients PETE 661 Drilling Engineering Slide 35
Experimental Determination of Fracture Gradient
The leak-off test
Run and cement casing Drill out ~ 10 ft
below the casing seat Close the BOPs Pump slowly and
monitor the pressure
8. Fracture Gradients PETE 661 Drilling Engineering Slide 36
8. Fracture Gradients PETE 661 Drilling Engineering Slide 37
4580105120120120120120120 40 20
8. Fracture Gradients PETE 661 Drilling Engineering Slide 38
8. Fracture Gradients PETE 661 Drilling Engineering Slide 39
8. Fracture Gradients PETE 661 Drilling Engineering Slide 40
8. Fracture Gradients PETE 661 Drilling Engineering Slide 41
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?
8. Fracture Gradients PETE 661 Drilling Engineering Slide 42
Example
Leak-off pressure = PS + PHYD
= 1,000 + 0.052 * 9 * 4,000= 2,872 psi
Fracture gradient = 0.718 psi/ft
EMW = ?
ftpsi
000,4872,2
DP OFFLEAK
13.8 lb/gal
8. Fracture Gradients PETE 661 Drilling Engineering Slide 43
What is Gas Cut Mud?
After drilling through a formation containing gas, this “drilled gas” will show up in the mud returns at the surface.
Gas cut mud is mud containing some gas - from any source.
8. Fracture Gradients PETE 661 Drilling Engineering Slide 44
Gas Cut Mud
Effect of Drilling Rate Effect of Circulation Rate Mud/Gas Ratio at the bottom of the Hole Mud/Gas Ratio at the Surface Density of Gas Cut Mud Reduction of Bottom Hole Pressure due to Gas Cut Mud Safe Drilling Practices
8. Fracture Gradients PETE 661 Drilling Engineering Slide 45
How Critical is Gas Cut Mud?
Example Problem
Well depth = 15,000 ftHole size = 7 7/8”Drill pipe size = 4 1/2”Mud weight = 15 ppgDrilling Rate = 20 ft/hrCirc. rate = 7.0 bbl/min
8. Fracture Gradients PETE 661 Drilling Engineering Slide 46
How Critical is Gas Cut Mud?
Formation Properties
F100 T
F250 T
1.35 Z1 Z
25% Porosity Sand70% saturation gas Sand
S
B
B
S
8. Fracture Gradients PETE 661 Drilling Engineering Slide 47
Bottom-Hole Ratio of Mud Volume to Gas Volume:
This indicates there are 1990 volumes of mud to 1 volume of gas at the bottom of the hole.
1990
hrbbl0.2110
hrbbl420
gs 0.7*porosity 25.0*cu.ft 61.5
bbl*hr
ft 20in/ft 12
in 877
4
hrmin 60*
minbbl 7
GasMud
2
MudGas
8. Fracture Gradients PETE 661 Drilling Engineering Slide 48
Ratio of surface volume of gas to bottom-hole volume of gas:
This shows there are 465 volumes of gas at the surface per volume of gas at the bottom of the hole
465 )R(710psi)(1.35) 7.14()R0psi)(1)(56 (11,700
law) (gas TT
ZZ
PP
VV
B
S
B
S
S
B
B
S
(PV = ZnRT)
8. Fracture Gradients PETE 661 Drilling Engineering Slide 49
Mud/gas Volume Ratio at the Surface:
279.4465
1990 VolumeGas VolumeMud:surface At
990,1VolumeGasVolumeMud:BottomAt
465BottomatGasSurfaceatGas:Expansion
8. Fracture Gradients PETE 661 Drilling Engineering Slide 50
Mud Density at the Surface:
So the mud weight has been cut 2.84 ppg (from 15 to 12.16) ppg
ppg 16.121279.4
ppg0)*1(ppg 15*(4.279)
Volume Density) udsurface)(M @ vol vol/gas(
Total
Mudsurf
8. Fracture Gradients PETE 661 Drilling Engineering Slide 51
It should be noted that in actual situations the mud cut would probably be less because we have assumed all gas stays in the mud-gas mixture. A certain amount of gas will break out.
The effects of gas cut mud on the hydrostatic head:
S
SB
SS
AASred.gas P
PPln TC)Z(100
TZCPΔP
Mud Density at the Surface:
8. Fracture Gradients PETE 661 Drilling Engineering Slide 52
R re, temperatuSurface - T
factorility compressib Surface - ZR re, temperatuAverage - T
factorility compressib Average - Zpsi pressure, Surface - P
surface at the fluid totalof % Gas - C wellof bottomat pressure cHydrostati - P
S
S
A
A
S
B
S
SB
SS
AASred.gas P
PPln TC)Z(100
TZCPΔP
8. Fracture Gradients PETE 661 Drilling Engineering Slide 53
18.94%4.2791100%*1
mud of vol.gas of Vol.
100%*gas of Vol.C
psi 11,700ft 15000*ppg 15*0.052PB
Hydrostatic Pressure and C
8. Fracture Gradients PETE 661 Drilling Engineering Slide 54
Average T and Z
175.12
35.11Z
6352
560710T
A
A
R
8. Fracture Gradients PETE 661 Drilling Engineering Slide 55
Reduction in BHP
psi 30.57 ΔP
14.714.711,700ln
560)18.94)(1)((100(635).7)(1.175)(18.94)(14ΔP
red.gas
red.gas
S
SB
SS
AASred.gas P
PPln TC)Z(100
TZCPΔP
8. Fracture Gradients PETE 661 Drilling Engineering Slide 56
The resulting bottom hole pressure will be
p = 11,700 - 30.57
BHP = 11,669 psi
This means the gas reduced the hydrostatic head by only 30.57 psi!
Reduction in BHP
8. Fracture Gradients PETE 661 Drilling Engineering Slide 57
Conclusion
It can be seen that the surface gas cut of approx. 3 PPG resulted in a bottom hole pressure reduction of only 30.57 psi.
There is one other factor that reduces the effect of gas cut mud even further and that is the effect of drilled solids in the mud. Drilled solids will tend to raise the overall density of the mud.
8. Fracture Gradients PETE 661 Drilling Engineering Slide 58
Summary of Gas-Cut Mud Problem
At bottom:
Gas expansion:
990,1rate generation gasraten circulatio mud
465bottomat volumesurfaceat volume
8. Fracture Gradients PETE 661 Drilling Engineering Slide 59
Summary of Gas-Cut Mud Problem
At surface:
i.e. At the surface, the mud mix contains one part of gas (by volume) for each 4.279 parts of good mud.
279.4465990,1
raten circulatio gasncirculatio mud
8. Fracture Gradients PETE 661 Drilling Engineering Slide 60
Summary of Gas-Cut Mud Problem
Density of mix
1279.4)0*1()15*279.4(
volume total weighttotal
Density of Mud at surface = 12.16 #/gal
(-2.84 lb/gal)
8. Fracture Gradients PETE 661 Drilling Engineering Slide 61
Summary of Gas-Cut Mud Problem
psi 31
PPPln
TZ)C100(TZCPp
S
SB
SS
AAS
A reduction in the mud density at the surface by 2.84 lb/gal resulted in a reduction in BHP of:
8. Fracture Gradients PETE 661 Drilling Engineering Slide 62
Note:
It is very important in any drilling operation:
To recognize the symptoms of increasing pore pressure
To be able to estimate the magnitude of the pore pressure
8. Fracture Gradients PETE 661 Drilling Engineering Slide 63
Note cont’d:
To know the fracture gradients of the exposed formations
To maintain the drilling practices within controllable limits
To keep in mind that any one symptom of increasing pore pressure may not be sufficient to provide the basis for precise conclusions
Look at all the indicators...
8. Fracture Gradients PETE 661 Drilling Engineering Slide 64
ROP F.L.Temp Cl - MUD t d Gas Units SH YP
8. Fracture Gradients PETE 661 Drilling Engineering Slide 65
What should be done when gas cut mud is encountered?
(1) Establish if there is any fire hazard. If there is a fire hazard, divert flow through mud-gas separation facilities.
(a) Notify any welder in area(b) Notify all rig personnel of the
pending danger
8. Fracture Gradients PETE 661 Drilling Engineering Slide 66
What should be done when gas cut mud is encountered?
(2) Determine where the gas came from. If the casing seat fracture gradient is being approached, and there is some concern about raising the mud weight:
Stop drilling and circulate, and observe the gas response. If source is drilled gas, the gas rate will decrease.
8. Fracture Gradients PETE 661 Drilling Engineering Slide 67
What should be done when gas cut mud is encountered?
(a) If the gas units completely return to the original background gas, it would probably be safe to resume drilling.
8. Fracture Gradients PETE 661 Drilling Engineering Slide 68
What should be done when gas cut mud is encountered?
(b) If there has been ample circulation time and the gas units do not drop back to the original background level, but stay at a higher value, this indicates that the mud weight is approaching the pore pressure and consideration should be given to increasing the mud weight.
8. Fracture Gradients PETE 661 Drilling Engineering Slide 69
What should be done when gas cut mud is encountered?
Establish Where did the gas come from?
(a) Drilled gas - no increase in mud weight is required
(b) Increasing pore pressure - (abnormal pore pressure) - May have to increase mud weight