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# 1 PETE 411 Well Drilling Lesson 22 Prediction of Fracture Gradients.

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1 PETE 411 Well Drilling Lesson 22 Prediction of Fracture Gradients
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1

PETE 411

Well Drilling

Lesson 22

2

Well Planning Theoretical Fracture Gradient Determination

Hubbert & Willis Matthews & Kelly Ben Eaton Comparison of Results

Experimental Frac. Grad. Determination Leak-off Tests Lost Circulation

3

Applied Drilling Engineering, Ch. 6

HW #12Casing Designdue Nov. 1, 2002

4

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!

5

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

6

Fig. 7.21

7

8

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

9

Formation Pressure and Matrix Stress

PS

overburden pore matrix stress = pressure + stress (psi) (psi) (psi)

S = P +

10

Formation Pressure and Matrix Stress

Calculations:

= 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.

11

Formation Pressure and Matrix Stress

Calculations:

psi

psi/ft

/ D = 0.522 psi/ft

PS

DD

P

D

Sor

ft/psi478.0000.1D

P

D

S

D.,e.i

12

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

13

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.

14

1. Hubbert & Willis:

where F = fracture gradient, psi/ft

P

D

P21

3

1Fmin

D

P1

2

1Fmax

15

2. Matthews & Kelly:

where Ki = matrix stress coefficient

= vertical matrix stress, psi

D

P

D

KF i

16

3. Ben Eaton:

where S = overburden stress, psi

= Poisson’s ratio

D

P

1*

D

PSF

17

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.

18

1. Hubbert & Willis:

F1

31 2*0.735 0.823

psi

ftmin

D

2P1

3

1Fmin

P

D0.735

psi

ft

Example - Hubbert and Willis

19

Also,

F0.823 psi / ft

0.052psi / ftlb / gal

min

F 15.83 lb / galmin

Example - Hubbert and Willis

20

Example - Hubbert and Willis

D

P1

2

1Fmax 735.01

2

1

= 0.8675 psi/ft

Fmax = 16.68 lb/gal

21

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

PF i

iK

Example

)given(ft/psi735.0D

P

22

Example - Matthews and Kelly

(ii) Next, calculate the matrix stress.

ft ,depthD

psi ,pressure poreP

psi ,stress matrix

psi ,overburdenS

S = P +

= S - P = 1.00 * D - 0.735 * D = 0.265 * D = 0.265 * 11,000 = 2,915 psi

23

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

ft449,5535.0

915,2Di

24

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

25

Example - Matthews and Kelly

(v) Now calculate F:D

P

D

KF i

735.0000,11

915,2*685.0F

ft/psi9165.0

gal/lb63.17052.0

9165.0F

26

27

Example

Ben Eaton:

D

P

1*

D

PSF

??D

S

28

Variable Overburden Stress by Eaton

At 11,000 ft

S/D = 0.96 psi/ft

29

Fig. 5-5

At 11,000 ft

= 0.46

30

Example - Ben Eaton

From above graphs,

at 11,000 ft.:D

P

1D

P

D

SF

46.0;ft/psi96.0D

S

735.046.01

46.0735.096.0F

F = 0.9267 psi/ft = 17.82 lb/gal

31

Summary of Results

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

32

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 ).

33

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.

34

Similarities

Ben Eaton:

D

P

1*

D

PSF

Matthews and Kelly:

D

P

D

KF i

35

9 10 11 1214

1618

Pore Pressures

36

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

37

38

4580105120120120120120120 40 20

39

40

41

42

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.

43

Example

Leak-off pressure = PS + PHYD

= 1,000 + 0.052 * 9 * 4,000

= 2,872 psi

EMW = ?

ft

psi

000,4

872,2

D

P OFFLEAK

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