1.11 fracture gradients
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Fracture Gradients 1.11- 1
1.11
Fracture Gradients
Fracture Gradients 1.11- 2
Prediction of Fracture Gradients4Well Planning4Theoretical Fracture Gradient Determination
4Hubbert & Willis4Matthews & Kelly4Ben Eaton4Comparison of Results
4Experimental Frac. Grad. Determination4Leak-off Tests
Fracture Gradients 1.11- 3
Well Planning4Safe drilling practices require that the
following be considered when planning a well:
4 Pore pressure determination4 Fracture gradient determination4 Casing setting depth selection4 Casing design
Fracture Gradients 1.11- 4
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
Fracture Gradients 1.11- 5
Formation Pressure and Matrix Stress
PSoverburden pore matrix stress = pressure + stress (psi) (psi) (psi)
S = P +
Fracture Gradients 1.11- 6
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.
Fracture Gradients 1.11- 7
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
Fracture Gradients 1.11- 8
Formation Pressure and Matrix Stress
Calculations:
4. Matrix stress at 14,000 ft
= 0.522 psi/ft * 14,000 ft
= 7,308 psi
Fracture Gradients 1.11- 9
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.
Fracture Gradients 1.11- 10
Fracture Gradient Determination
1. Hubbert & Willis:
where F = fracture gradient, psi/ft
= pore pressure gradient, psi/ftDP
DP21
31Fmin
DP1
21Fmax
Fracture Gradients 1.11- 11
Fracture Gradient Determination
2. Matthews & Kelly:
where Ki = matrix stress coefficient
= vertical matrix stress, psi
DP
DKF i
Fracture Gradients 1.11- 12
Fracture Gradient Determination
3. Ben Eaton:
where S = overburden stress, psi g = Poisson’s ratio
DP
1*
DPSF
g
g
Fracture Gradients 1.11- 13
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.
Fracture Gradients 1.11- 14
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
Fracture Gradients 1.11- 15
Also,
lb/galpsi/ft0.052
psi/ft0.823Fmin
lb/gal 15.83Fmin
Example - Hubbert and Willis
Fracture Gradients 1.11- 16
Example - Hubbert and Willis
DP1
21Fmax 735.01
21
= 0.8675 psi/ft
Fmax = 16.68 lb/gal
Fracture Gradients 1.11- 17
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
Fracture Gradients 1.11- 18
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
Fracture Gradients 1.11- 19
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
Fracture Gradients 1.11- 20
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
Fracture Gradients 1.11- 21
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
Fracture Gradients 1.11- 22
0.685
5,449
Ki
Dep
th, D
i
Fracture Gradients 1.11- 23
Example
Ben Eaton:
DP
1*
DPSF
g
g
??DS
g
Fracture Gradients 1.11- 24
Variable Overburden Stress by Eaton
At 11,000 ftS/D = 0.96 psi/ft
Fracture Gradients 1.11- 25
Fig. 5-5
At 11,000 ftg = 0.46
Fracture Gradients 1.11- 26
Example - Ben Eaton
From above graphs, at 11,000 ft.:
DP
1DP
DSF
gg
46.0;ft/psi96.0DS
g
735.046.01
46.0735.096.0F
F = 0.9267 psi/ft = 17.82 lb/gal
Fracture Gradients 1.11- 27
Summary of Results
Fracture Gradient psi.ft lb/gal
Hubbert & Willis minimum: 0.823 15.83Hubbert & Willis maximum: 0.868 16.68Mathews & Kelly: 0.917 17.63Ben Eaton: 0.927 17.82
Fracture Gradients 1.11- 28
Summary of Results
4 Note that all the methods take into consideration the pore pressure gradient. As the pore pressure increases, so does the fracture gradient.
4 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 ).
Fracture Gradients 1.11- 29
Summary of Results
4 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.
Fracture Gradients 1.11- 30
Similarities
Ben Eaton:
DP
1*
DPSF
g
g
Matthews and Kelly:
DP
DKF i
Fracture Gradients 1.11- 31
Fracture Gradients 1.11- 32
Experimental Determination of Fracture Gradient
The leak-off test
4 Run and cement casing4 Drill out ~ 10 ft
below the casing seat4 Close the BOPs4 Pump slowly and
monitor the pressure
Fracture Gradients 1.11- 33
Fracture Gradients 1.11- 34
Fracture Gradients 1.11- 35
Fracture Gradients 1.11- 36
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