1.11 fracture gradients

Fracture Gradients 1.11- 1
1.11
Fracture Gradients

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

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

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

 PS
overburden pore matrix
stress = pressure + stress
(psi) (psi) (psi)
S = P + 

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.

Calculations:
3. Matrix stress gradient,
psi
psi/ft
 / D = 0.522 psi/ft
 PS
DD
P
D
S
or


  ft/psi478.0000.1
D
P
D
S
D
.,e.i 


Calculations:
4. Matrix stress at 14,000 ft
= 0.522 psi/ft * 14,000 ft
 = 7,308 psi

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.

1. Hubbert & Willis:
where F = fracture gradient, psi/ft
= pore pressure gradient, psi/ft
D
P







D
P2
1
3
1
Fmin







D
P
1
2
1
Fmax

2. Matthews & Kelly:
where Ki = matrix stress coefficient
 = vertical matrix stress, psi
D
P
D
K
F i




3. Ben Eaton:
where S = overburden stress, psi
g = Poisson’s ratio
D
P
1
*
D
PS
F 





g
g





 


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.

1. Hubbert & Willis:
The pore pressure gradient,
 F
1
3
1 2*0.735 0.823
psi
ft
min   







D
2P
1
3
1
Fmin
P
D
0.735
psi
ft

Example - Hubbert and Willis

Also,







lb/gal
psi/ft
0.052
psi/ft0.823
Fmin
lb/gal15.83Fmin 








D
P
1
2
1
Fmax  735.01
2
1

= 0.8675 psi/ft
Fmax = 16.68 lb/gal

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.
D
K
D
P
F i


iK
Example
)given(ft/psi735.0
D
P


Example - Matthews and Kelly
(ii) Next, calculate the matrix stress.


















ft,depthD
psi,pressureporeP
psi,stressmatrix
psi,overburdenS

S = P + 
 = S - P
= 1.00 * D - 0.735 * D
= 0.265 * D
= 0.265 * 11,000
 = 2,915 psi

(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
ft449,5
535.0
915,2
Di 

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

(v) Now calculate F:
D
P
D
K
F i



735.0
000,11
915,2*685.0
F 
ft/psi9165.0
gal/lb63.17
052.0
9165.0
F 

0.685
5,449
Ki
Depth,Di

Example
Ben Eaton:
D
P
1
*
D
PS
F 





g
g





 

??
D
S
g

Variable Overburden Stress by
Eaton
At 11,000 ft
S/D = 0.96 psi/ft

Fig. 5-5
At 11,000 ft
g = 0.46

Example - Ben Eaton
From above graphs,
at 11,000 ft.:
D
P
1D
P
D
S
F 













g
g
46.0;ft/psi96.0
D
S
g
  735.0
46.01
46.0
735.096.0F 







F = 0.9267 psi/ft
= 17.82 lb/gal

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

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 ).

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.

Similarities
Ben Eaton:
D
P
1
*
D
PS
F 





g
g





 

Matthews and Kelly:
D
P
D
K
F i




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

1.11 fracture gradients

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