fracture gradient - part i

Fracture Gradients

By Dr. Eissa shokir

Prediction of Fracture Gradients4Well Planning4Theoretical Fracture Gradient Determination4Hubbert & Willis4Matthews & Kelly4Ben Eaton4Comparison of Results

4Experimental Frac. Grad. Determination4Leak-off Tests

Well Planning

4Safe 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

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

Formation Pressure and Matrix Stress

σ+= PSoverburden pore matrix

stress = pressure + stress(psi) (psi) (psi)

S = P + σ

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.

Formation Pressure and Matrix Stress

Calculations:3. Matrix stress gradient,

psi

psi/ft

σ / D = 0.522 psi/ft

σ+= PS

DDP

DSor σ

+=

( ) ft/psi478.0000.1DP

DS

D.,e.i −=−=

σ

Formation Pressure and Matrix Stress

Calculations:

4. Matrix stress at 14,000 ft

= 0.522 psi/ft * 14,000 ft

σ = 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 experimentalapproach to fracture gradient determination. Below are listed and discussed three approaches to calculating the fracture gradient.

Fracture Gradient Determination

1. Hubbert & Willis:

where F = fracture gradient, psi/ft

= pore pressure gradient, psi/ftDP

⎟⎠⎞

⎜⎝⎛ +=

DP21

31Fmin

⎟⎠⎞

⎜⎝⎛ +=

DP1

21Fmax

Fracture Gradient Determination

2. Matthews & Kelly:

where Ki = matrix stress coefficientσ = vertical matrix stress, psi

DP

DKF i +

σ=

Fracture Gradient Determination

3. Ben Eaton:

where S = overburden stress, psiγ = Poisson’s ratio

DP

1*

DPSF +⎟⎟

⎠

⎞⎜⎜⎝

⎛γ−

γ⎟⎠⎞

⎜⎝⎛ −

=

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/galusing 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 13

1 2 *0.735 0.823 psiftmin = + =

⎟⎠⎞

⎜⎝⎛ +=

D2P1

31Fmin

PD

0.735 psift

=

Example - Hubbert and Willis

Also,

⎟⎟⎠

⎞⎜⎜⎝

⎛=

lb/galpsi/ft0.052

psi/ft0.823Fmin

lb/gal 15.83Fmin =

Example - Hubbert and Willis

Example - Hubbert and Willis

⎟⎠⎞

⎜⎝⎛ +=

DP1

21Fmax ( )735.01

21

+=

= 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.

DK

DPF iσ+=

σiK

Example

)given(ft/psi735.0DP

=

Example - Matthews and Kelly

(ii) Next, calculate the matrix stress.

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

====

ft ,depthDpsi ,pressure porePpsi ,stress matrix

psi ,overburdenSσ

S = P + σ

σ = S - P= 1.00 * D - 0.735 * D= 0.265 * D= 0.265 * 11,000

σ = 2,915 psi

Example - Matthews and Kelly

(iii) Now determine the depth, , where, under normally pressured conditions, the rock matrix stress, σ would be 2,915 psi.

iD

Sn = Pn + σn n = “normal”1.00 * Di = 0.465 * Di + 2,915

Di * (1 - 0.465) = 2,915

ft449,5535.0915,2Di ==

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

Example - Matthews and Kelly

(v) Now calculate F:DP

DKF i +

σ=

735.0000,11

915,2*685.0F +=

ft/psi9165.0=

gal/lb63.17052.0

9165.0F ==

0.685

5,449

Ki

Dep

th, D

i

Example

Ben Eaton:

DP

1*

DPSF +⎟⎟

⎠

⎞⎜⎜⎝

⎛γ−

γ⎟⎠⎞

⎜⎝⎛ −

=

??DS

=γ=

Variable Overburden Stress by Eaton

At 11,000 ftS/D = 0.96 psi/ft

Fig. 5-5

At 11,000 ftγ = 0.46

Example - Ben Eaton

From above graphs,at 11,000 ft.:

DP

1DP

DSF +⎟⎟

⎠

⎞⎜⎜⎝

⎛−

⎟⎠⎞

⎜⎝⎛ −=

γγ

46.0;ft/psi96.0DS

=γ=

( ) 735.046.01

46.0735.096.0F +⎟⎠⎞

⎜⎝⎛

−−=

F = 0.9267 psi/ft= 17.82 lb/gal

Summary of Results

Fracture Gradientpsi.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

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 σi ).

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.

Similarities

Ben Eaton:

DP

1*

DPSF +⎟⎟

⎠

⎞⎜⎜⎝

⎛γ−

γ⎟⎠⎞

⎜⎝⎛ −

=

Matthews and Kelly:

DP

DKF i +

σ=

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