Fracture Gradient - Part II

7/31/2019 Fracture Gradient - Part II

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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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Fracture Gradient Determination

Read AWC Chapter 4 all


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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 andprediction used today.

Assumed elastic behavior.

Assumed effective stress exceeds the

minimum by a factor of 3.


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If the overburden is maximum, the assumedhorizontal stress is:

H = 1/3(ob - pp) + pp Equating fracture propagation pressure to

minimum stress gives

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


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

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


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Matthews and Kelly

Developed the concept of variable ratiobetween the effective horizontal and vertical

stresses, not a constant 1/3 as in H & W.

Stress ratios increase according to the

degree of compaction

eH = KMKev


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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 theminimum in situ horizontal stress gives

pfi = KMK(ob - pp) + pp and

gfi = KMK(gob - gp) + gp


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Example 4.8

Given: Table 4.4 (Offshore LA) Estimate fracture initiation gradient at 8110

and 15,050 using Matthews and Kelly

correlation


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Example 4.8

KMK = 0.69

For 8110

gfi = 0.69(1 - .465) + .465

gfi = 0.834 psi/ft

KMK = 0.61

For the undercompacted

interval at 15,050, the

equivalent depth is

determined by:

De=

[15050-(.815*15050)]/.535

= 5204


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Example 4.8

gfi = 0.61*(1-.815)+.815 = .928 psi/ft

Note: Overburden gradient was assumed to

be 1.0 psi/ft


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Penebakers Gulf Coast

gfi = Kp(gob - gp) + gp

where Kp

is Penebakers effective stress

ratio


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Penebakers overburden

gradient from Gulf

Coast region

Depth where

t = 100 sec/ft


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Penebakers Effective Stress Ratio


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Example 4.9

Re-work Example 4.8 using Penebakerscorrelations where the travel time of 100

sec/ft is at 10,000


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Example 4.9

At 8110 gfi = 0.77(0.945 - 0.465) + 0.465

gfi

= 0.835 psi/ft

At 15050

gfi = 0.94(0.984 - 0.815) + 0.815

gfi = 0.974 psi/ft


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Eatons Gulf Coast Correlation

Based on offshore LA in moderate waterdepths

( )

ratiostresseffectiveanis

termratiosPoisson'bracketedtheNote

1ppob

E

E

fi gggg +

=


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Mitchells approximation


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Example 4.10


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Example 4.10


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Summary

Note that all the methods take intoconsideration the pore pressure gradient.

As the pore pressure increases, so does the

fracture gradient


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Summary

Hubbert and Willis apparently consider onlythe 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 porepressure gradient, overburden stress, and

Poissons ratio.

It is probably the most accurate of the three.


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Summary

The last two are quite similar and yieldsimilar results.

None consider the effect of water depth.


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Christmans approach

Christman took into consideration the effectof water depth on overburden stress.


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Example 4.11

Estimate the fracture gradient for aformation located 1490 BML. Water depth

is 768, air gap is 75.

Repeat for water depth of 1500


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Example 4.11


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Example 4.11


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Christman

Christman also noted that anomalously lowfracture gradients seemed to be associated

with formation having low bulk densities

for the burial depth. He then developed thecorrelation in Fig 4.45


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Example 4.12

Re-work the first part of Example 4.11 ifthe logged bulk density at 1490 BML is

2.08 g/cc


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gfi = 0.6 * (0.73-0.452) + 0.452

gfi = 0.619 psi/ft


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Prentice method

Water depth of 1000

Total depth = 4000

Water gap = 200


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Prentice method

Convert the water depth to an equivalentsection of formation.

E.g. 1000 * 0.465 psi/ft = 465 psi

From Eatons overburden stress chart the

stress gradient at 4000 equals 0.89 psi/ft


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Prentice method

465 psi/0.89 psi/ft = 522 equivalent depth

Calculate and convert apparent fracture

gradient to actual fracture gradient

522 + 3000 = 3522 equivalent


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Prentice method

From Eatons fracture gradient chart, thegradient at 3522 = 13.92 ppg

or

Fracture pressure = 0.052 * 13.92 * 3522

= 2549 psi


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Prentice method

The effective fracture gradient from themud flow line at the drill ship deck to the

casing seat is:

2549 * 19.23/(200 + 1000 + 3000)

= 11.67 ppg

F = 2549/4200 = 0.607 psi/ft 0.607/.052 = 11.67 ppg


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Experimental Determination

Leak-off test, LOT, - pressure test in whichwe determine the amount of pressurerequired to initiate a fracture

Pressure Integrity Test, PIT, pressure test inwhich we only want to determine if a

formation can withstand a certain amount ofpressure without fracturing.


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PIT

4000

10.0 ppg

??

How much surface pressure will be

required to test the casing seat to a

14.0 ppg equivalent?

ps = (EMW - MW) * 0.052 * TVDshoe

ps = (14.0 - 10.0) * 0.052 * 4000

ps = 832 psi

LOT


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LOT


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Leak-off

Rupture

Propagation

Example 4 22 - 2


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Example 4.22 - 2Interpret the leak-off test.


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Solution

Pfi = 1730 + .483*5500 - 50 1730 psi = leak off pressure

0.483 psi/ft = mud gradient in well

5500 depth of casing seat

50 psi = pump pressure to break circulation

Pfi = 4337 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 drilledto 1500 and pipe isset. About 20 of newhole is drilled after

cementing. The shoeneeds to hole 14.0 ppgequivalent on a leakoff test. Mud in the

hole has a density of9.5 ppg.

9.5 ppg

1500


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Example

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

(14.0 - 9.5) * .052 * 1500 = 351 psi


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Example

The casing seat is tested to a leak offpressure of 367 psi. What EMW did the

shoe actually hole?

367/.052/1500 + 9.5 = 14.2 ppg EMW


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Example

After drilling for some

time, TD is now 4500

and the mud weight is

10.2 ppg. What is the

maximum casing

pressure that the

casing seat can

withstand withoutfracturing?

10.2 ppg

1500

4500


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Example

Max. CP = (EMW - MW) * .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 7500 with a mudweight of 13.7 ppg. What is the maximum

CP that the shoe can withstand?

Max. CP = (14.2 - 13.7) * .052 * 1500

Max. CP = 39 psi

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Fracture Gradient - Part II