PETE 411
Well Drilling
Lesson 22
Prediction of
Fracture Gradients
1
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
 Lost Circulation
2
Read:
Applied Drilling Engineering, Ch. 6
HW #12
Casing Design
due Nov. 1, 2002
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NOTE:
On all HW and Quizzes please put:
* PETE 411/501 (or 411/502)
* Name, written legibly
* Number of HW or Quiz
(on the outside)
Thank you!
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Well Planning
 Safe drilling practices require that the
following be considered when
planning a well:
 Pore pressure determination
 Fracture gradient determination
 Casing setting depth selection
 Casing design
 Mud Design, H2S considerations
 Contingency planning
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Fig. 7.21
6
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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 ft
2. Pore pressure, psi, at 14,000 ft
3. Matrix stress, psi/ft
4. Matrix stress, psi
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Formation Pressure and Matrix Stress
S =S  PP + 
overburden
stress
(psi)
pore
= pressure
(psi)
+
matrix
stress
(psi)
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Formation Pressure and Matrix Stress
Calculations:
Depth = 14,000 ft.
Pore Pressure = 9.2 lb/gal equivalent
Overburden stress = 1.00 psi/ft.
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
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Formation Pressure and Matrix Stress
Calculations:
3. Matrix stress gradient,
S P
psi
S P 
or
 
psi/ft
D D D
 S P
i.e.,
 
 1.000  0.478 psi / ft
D D D
 / D = 0.522 psi/ft
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Formation Pressure and Matrix Stress
Calculations:
4. Matrix stress (in psi) at 14,000 ft
= 0.522 psi/ft * 14,000 ft
 = 7,308 psi
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 four
approaches to calculating the fracture
gradient.
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Fracture Gradient Determination
1. Hubbert & Willis:
where
Fmin
1  2P 
 1 

3
D 
Fmax
1  P
 1  
2  D
F = fracture gradient, psi/ft
P
= pore pressure gradient, psi/ft
D
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Fracture Gradient Determination
2. Matthews & Kelly:
K i
P
F 

D
D
where
Ki = matrix stress coefficient
 = vertical matrix stress, psi
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Fracture Gradient Determination
3. Ben Eaton:
P
S P  g 
 
F  
 * 
D
 D   1 g 
where
S = overburden stress, psi
g = Poisson’s ratio
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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 four methods.
Summarize the results in tabular form, showing
answers, in units of lb/gal and also in psi/ft.
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Example - Hubbert and Willis
1. Hubbert & Willis:
Fmin
1  2P 
 1 

3
D 
P
psi
The pore pressure gradient,
 0.735
D
ft
Fmin
1
psi
 1  2 *0.735  0.823
3
ft
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Example - Hubbert and Willis
Also,
Fmin 
0.823 psi / ft
 psi / ft 
0.052 

 lb / gal 
Fmin  15.83 lb / gal
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Example - Hubbert and Willis
Fmax
1  P
 1  
2  D
1
 1  0.735 
2
= 0.8675 psi/ft
Fmax = 16.68 lb/gal
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Example
P K i
2. Matthews & Kelly F 

D
D
In this case P and D are known,  may be
calculated, and K i is determined graphically.
(i) First, determine the pore pressure gradient.
P
 0.735 psi / ft
D
(given )
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Example - Matthews and Kelly
(ii) Next, calculate the matrix stress.
S=P+
=S-P
= 1.00 * D - 0.735 * D
= 0.265 * D
= 0.265 * 11,000
 = 2,915 psi
S  overburden , psi 
  matrix stress , psi 




P  pore pressure , psi 


D  depth , ft

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Example - Matthews and Kelly
(iii) Now determine the depth, D i , where,
under normally pressured conditions, the
rock matrix stress,  would be 2,915 psi.
Sn = Pn + n
n = “normal”
1.00 * Di = 0.465 * Di + 2,915
Di * (1 - 0.465) = 2,915
2,915
Di 
 5,449 ft
0.535
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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
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Example - Matthews and Kelly
(v) Now calculate F:
K i
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
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Example
Ben Eaton:
P
S P  g 
 
F  
 * 
D
 D   1 g 
S
?
D
g?
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Variable Overburden Stress by
Eaton
At 11,000 ft
S/D = 0.96 psi/ft
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Fig. 5-5
At 11,000 ft
g = 0.46
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Example - Ben Eaton
From above graphs,
at 11,000 ft.:
S
 0.96
D
psi / ft;
 S P  g
F    
 D D  1  g
 P
 
 D
g  0.46
 0.46 
F  0.96  0.735  
  0.735
 1  0.46 
F = 0.9267 psi/ft
= 17.82 lb/gal
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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
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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 ).
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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 four methods. The last two
methods are actually quite similar, and
usually yield similar results.
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Similarities
Ben Eaton:
P
S P  g 
 
F  
 * 
D
 D   1 g 
 Ki
P
F 

D
D
Matthews and Kelly:
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9 10
11 12
Pore Pressures
14
16
18
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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
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80
105
120
120
120
120
120
120
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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?
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Example
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 = ?
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