TAMU - PemexWell Control

Lesson 9B

Fracture Gradient Determination

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Fracture Gradient Determination

Hubbert and Willis

Matthews and Kelly

Ben Eaton

Christman

Prentice

Leak-Off Test (experimental)

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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

Casing design

H2S considerations

Contingency planning

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The Hubbert & Willis Equation

Provides the basis of fracture theory and prediction used today.

Assumes elastic behavior.

Assumes the maximum effective stress exceeds the minimum by a factor of 3.

min_emax_e *3

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Cohesion, c = 0Angle of Internal Friction, = 30 deg.

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The Hubbert & Willis Equation

If the overburden is the maximum stress, the assumed horizontal stress is:

H = 1/3(ob - pp) + pp

Equating fracture propagation pressure to minimum stress gives

pfp = 1/3(ob - pp) + pp

min_emax_e *3

obemin_e *3

1

pobpmin p*3

1p

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The Hubbert & Willis Equation

pfp = 1/3(ob - pp) + pp

pfp = 1/3(ob + 2pp) (minimum)

pfp = 1/2(ob + pp) (maximum)

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Matthews and Kelly

Developed the concept of variable ratio between the effective horizontal and vertical stresses, not a constant 1/3 as in H & W.

Stress ratios increase according to the degree of compaction

He = KMKve

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Matthews and Kelly

eH = KMKev

KMK = matrix stress coefficient

Including pore pressure,

H = KMK(ob - pp) + pp

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Matthews and Kelly

Equating fracture initiation pressure to the minimum in situ horizontal

stress gives

pfi = KMK(ob - pp) + pp

and

gfi = KMK(gob - gp) + gp

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Example 3.8

Given: Table 3.4 (Offshore LA)

Estimate fracture initiation gradients at 8,110’ and 15,050’ using Matthews and Kelly correlation

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TABLE 3.4

psi/ftft

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Example 3.8

KMK = 0.69

At 8,110 ft, KMK = 0.69:

gfi = 0.69(1 - 0.465) + 0.465

gfi = 0.834 psi/ft

KMK = 0.61

For the undercompacted interval at 15,050 ft, the equivalent depth is determined by Eq. 3.68:

=[15,050-(0.815*15,050)]/0.535

= 5,204 ftHere KMK = 0.61

Fig. 3.38

nob

pob

nob

Veeq gg

p

ggD

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Example 3.8 At 15,050 ft, KMK = 0.61:

gfi = 0.61*(1-0.815)+0.815 = 0.928 psi/ft

0.928 / 0.052 = 17.8 lb/gal!

Note: Overburden gradient was assumed to be 1.0 psi/ft

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Pennebaker’s Gulf Coast

gfi = Kp(gob - gp) + gp

where Kp is Pennebaker’s effective stress ratio

gfi = Kp(gob - gp) + gp

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Pennebaker’s overburden

gradient from Gulf Coast

regionDepth where

t = 100 sec/ftWel

l D

epth

, f

t

Overburden Gradient, psi/ft

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Pennebaker’s Effective Stress Ratio

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Example 3.9

Re-work Example 3.8

using Pennebaker’s correlations

where the travel time of 100 sec/ft

is at 10,000 ft

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8,110

15,050

100 sec at 10,000 ft

0.77 0.940.945 0.984

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Example 3.9

At 8,110’

gfi = 0.77(0.945 - 0.465) + 0.465

gfi = 0.835 psi/ft

At 15,050’

gfi = 0.94(0.984 - 0.815) + 0.815

gfi = 0.974 psi/ft (18.7 ppg)

gfi = Kp(gob - gp) + gp

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Eaton’s Gulf Coast Correlation

Based on offshore LA in moderate water depths

ppobE

Efi ggg

1g

ratio stress effective an is

term ratio sPoisson' bracketed the Note

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From Eaton

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Mitchell’s approximation for Eaton’s Overburden Relationship for the Gulf Coast:

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ob 000,1

D0006.0

000,1

D01494.084753.0g

a9.2...............000,1

D10*199.1

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2ob 11.80006.011.801494.084753.0g

35 11.810*199.1

ft/psi936.0gob

Example 3.10 Estimate the fracture gradient at 8,110 ft using Eaton’s Method

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Mitchell’s approximation for Eaton’s Poisson’s Ratio for the Gulf Coast

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E 000,1

D00668.0

000,1

D05945.023743.0

70.3.....000,1

D10*71.6

000,1

D00035.0

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3

438.0E

2E 11.800668.011.805945.023743.0

463 11.810*71.611.800035.0

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Example 3.10 – cont’d

69.3............ggg1

g ppobE

Efi

465.0465.0936.0438.01

438.0gfi

ft/psi832.0gfi

At 15,050 ft, Michell’s approximation yields:

gob = 0.977 psi/ft, E = 0.468 and gfi = 0.958 psi/ft.

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Summary

Note that all the methods take into consideration the pore pressure gradient.

As the pore pressure increases, so does the fracture gradient.

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Summary

Hubbert and Willis apparently consider only the variation in pore pressure gradient…

Matthews and Kelly also consider the changes in rock matrix stress

coefficient and the matrix stress.

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Summary

Ben Eaton considers variation in pore pressure gradient, overburden stress, and Poisson’s ratio.

It is probably the most accurate of the three.

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Summary

The last two are quite similar and yield similar results.

None of the above methods considers the effect of water depth.

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Christman’s approach

Christman took into consideration the effect of water depth on overburden stress.

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Example 3.11

Estimate the fracture gradient for a normally pressured formation located 1,490’ BML. Water depth = 768 ftAir gap = 75 ftSea Water Gradient = 0.44 psi/ft

Assume Eaton’s overburden for the Santa Barbara Channel.

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Example 3.11 - Solution

From Fig. 3.42, next page

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1,490

0.451

Fig. 3.42- Christman’s Correlation for Santa Barbara Channel

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Example 3.12

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Fig. 3.45 - Procedure used to determine the effective stress ratio in Example 3.12.

Effective stress ratio

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From Barker

and Wood

AndEaton and

Eaton

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Experimental Determination

Leak-off test, LOT, - pressure test in which we determine the amount of pressure required to initiate a fracture

Pressure Integrity Test, PIT, pressure test in which we only want to determine if a formation can withstand a certain amount of pressure without fracturing.

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

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?

Example

Leak-off pressure = PS + PHYD

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

= 2,872 psi

Fracture gradient = 0.718 psi/ft

EMW = ?

ft

psi

000,4

872,2

D

P OFFLEAK

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PIT

4,000’

10.0 ppg

??

How much surface pressure will be required to test the casing seat to 14.0 ppg equivalent?

ps = 0.052 * (EMW - MW) * TVDshoe

ps = 0.052 * (14.0 - 10.0) * 4,000

ps = 832 psi

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LOT

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Leak-off

Rupture

Propagation

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Example 3.21

Interpret the leak-off test.

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Solution

pfi = 1,730 + 0.483 * 5,500 - 50 1,730 psi = leak off pressure 0.483 psi/ft = mud gradient in well 5,500’ depth of casing seat 50 psi = pump pressure to break

circulation pfi = 4,337 psi = 0.789 psi/ft

= 15.17 ppg

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Poor Cement Job

What could cause this?

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Example

Surface hole is drilled to 1,500’ and pipe is set. About 20’ of new hole is drilled after cementing. The shoe needs to hold 14.0 ppg equivalent on a leak off test. Mud in the hole has a density of 9.5 ppg.

9.5 ppg

1,500’

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Example

What surface pressure do we need to test to a 14.0 ppg equivalent?

(14.0 - 9.5) * 0.052 * 1,500 = 351 psi

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Example

The casing seat is tested to a leak off pressure of 367 psi. What EMW

did the shoe actually hold?

367/(0.052*1,500) + 9.5

EMW = 14.2 ppg

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Example

After drilling for some time, TD is now 4,500’ and the mud weight is 10.2 ppg. What is the maximum casing pressure that the casing seat can withstand without fracturing?

10.2 ppg

1,500’

4,500’

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Example

Max. CP = (EMW - MW) * 0.052 * TVDshoe

Max. CP = (14.2 - 10.2) * .052 * 1500

Max. CP = 312 psi

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Example

Now we are at a TD of 7,500 with a mud weight of 13.7 ppg. What is the maximum CP that the shoe can withstand?

Max. CP = (14.2 - 13.7) * 0.052 * 1,500

Max. CP = 39 psi