8. fracture gradients

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 1 PETE 661 Drilling Engineering Lesson 8 Fracture Gradients

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8. Fracture Gradients

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Page 1: 8. Fracture Gradients

8. Fracture Gradients PETE 661 Drilling Engineering Slide 1

PETE 661Drilling Engineering

Lesson 8 Fracture Gradients

Page 2: 8. Fracture Gradients

8. Fracture Gradients PETE 661 Drilling Engineering Slide 2

Prediction of Fracture GradientsWell PlanningTheoretical Fracture Gradient Determination

Hubbert & WillisMatthews & KellyBen EatonComparison of Results

Experimental Frac. Grad. DeterminationLeak-off TestsLost Circulation

Gas Cut Mud

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 3

Read:Applied Drilling Engineering, Ch. 6

HW # 5Casing Design

due 10-8-04

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 4

Well PlanningSafe drilling practices require that the

following be considered when planning a well:

Pore pressure determination Fracture gradient determination Casing setting depth selection Casing design H2S considerations Contingency planning

Page 5: 8. Fracture Gradients

8. Fracture Gradients PETE 661 Drilling Engineering Slide 5

Fig. 7.21

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 6

fracture

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 7

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

Page 8: 8. Fracture Gradients

8. Fracture Gradients PETE 661 Drilling Engineering Slide 8

Formation Pressure and Matrix Stress

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

S = P +

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 9

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.

Page 10: 8. Fracture Gradients

8. Fracture Gradients PETE 661 Drilling Engineering Slide 10

Formation Pressure and Matrix Stress

Calculations:3. Matrix stress gradient,

psi

psi/ft

/ D = 0.522 psi/ft

PS

DDP

DSor

ft/psi478.0000.1DP

DS

D.,e.i

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 11

Formation Pressure and Matrix Stress

Calculations:

4. Matrix stress at 14,000 ft

= 0.522 psi/ft * 14,000 ft

= 7,308 psi

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 12

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.

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 13

Fracture Gradient Determination

1. Hubbert & Willis:

where F = fracture gradient, psi/ft

= pore pressure gradient, psi/ftDP

DP21

31Fmin

DP1

21Fmax

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 14

Fracture Gradient Determination

2. Matthews & Kelly:

where Ki = matrix stress coefficient

= vertical matrix stress, psi

DP

DKF i

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 15

Fracture Gradient Determination

3. Ben Eaton:

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

DP

1*

DPSF

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 16

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.

Page 17: 8. Fracture Gradients

8. Fracture Gradients PETE 661 Drilling Engineering Slide 17

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

Page 18: 8. Fracture Gradients

8. Fracture Gradients PETE 661 Drilling Engineering Slide 18

Also,

lb/galpsi/ft0.052

psi/ft0.823Fmin

lb/gal 15.83Fmin

Example - Hubbert and Willis

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 19

Example - Hubbert and Willis

DP1

21Fmax 735.01

21

= 0.8675 psi/ft

Fmax = 16.68 lb/gal

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 20

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 i

iK

Example

)given(ft/psi735.0DP

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 21

Example - Matthews and Kelly

(ii) Next, calculate the matrix stress.

ft ,depthDpsi ,pressure porePpsi ,stress matrix

psi ,overburdenS

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

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 22

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

ft449,5535.0915,2Di

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 23

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

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 24

Example - Matthews and Kelly

(v) Now calculate F:DP

DKF i

735.0000,11

915,2*685.0F

ft/psi9165.0

gal/lb63.17052.0

9165.0F

Page 25: 8. Fracture Gradients

8. Fracture Gradients PETE 661 Drilling Engineering Slide 25

0.685

5,449

Ki

Dep

th, D

i

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 26

Example

Ben Eaton:

DP

1*

DPSF

??DS

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 27

Variable Overburden Stress by Eaton

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

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 28

Fig. 5-5

At 11,000 ft = 0.46

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 29

Example - Ben Eaton

From above graphs, at 11,000 ft.:

DP

1DP

DSF

46.0;ft/psi96.0DS

735.046.01

46.0735.096.0F

F = 0.9267 psi/ft = 17.82 lb/gal

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 30

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

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 31

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 ).

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 32

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.

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 33

Similarities

Ben Eaton:

DP

1*

DPSF

Matthews and Kelly:

DP

DKF i

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 34

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 35

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

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 36

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 37

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 38

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 39

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 40

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 41

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?

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 42

Example

Leak-off pressure = PS + PHYD

= 1,000 + 0.052 * 9 * 4,000= 2,872 psi

Fracture gradient = 0.718 psi/ft

EMW = ?

ftpsi

000,4872,2

DP OFFLEAK

13.8 lb/gal

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 43

What is Gas Cut Mud?

After drilling through a formation containing gas, this “drilled gas” will show up in the mud returns at the surface.

Gas cut mud is mud containing some gas - from any source.

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 44

Gas Cut Mud

Effect of Drilling Rate Effect of Circulation Rate Mud/Gas Ratio at the bottom of the Hole Mud/Gas Ratio at the Surface Density of Gas Cut Mud Reduction of Bottom Hole Pressure due to Gas Cut Mud Safe Drilling Practices

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 45

How Critical is Gas Cut Mud?

Example Problem

Well depth = 15,000 ftHole size = 7 7/8”Drill pipe size = 4 1/2”Mud weight = 15 ppgDrilling Rate = 20 ft/hrCirc. rate = 7.0 bbl/min

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 46

How Critical is Gas Cut Mud?

Formation Properties

F100 T

F250 T

1.35 Z1 Z

25% Porosity Sand70% saturation gas Sand

S

B

B

S

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 47

Bottom-Hole Ratio of Mud Volume to Gas Volume:

This indicates there are 1990 volumes of mud to 1 volume of gas at the bottom of the hole.

1990

hrbbl0.2110

hrbbl420

gs 0.7*porosity 25.0*cu.ft 61.5

bbl*hr

ft 20in/ft 12

in 877

4

hrmin 60*

minbbl 7

GasMud

2

MudGas

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 48

Ratio of surface volume of gas to bottom-hole volume of gas:

This shows there are 465 volumes of gas at the surface per volume of gas at the bottom of the hole

465 )R(710psi)(1.35) 7.14()R0psi)(1)(56 (11,700

law) (gas TT

ZZ

PP

VV

B

S

B

S

S

B

B

S

(PV = ZnRT)

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 49

Mud/gas Volume Ratio at the Surface:

279.4465

1990 VolumeGas VolumeMud:surface At

990,1VolumeGasVolumeMud:BottomAt

465BottomatGasSurfaceatGas:Expansion

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 50

Mud Density at the Surface:

So the mud weight has been cut 2.84 ppg (from 15 to 12.16) ppg

ppg 16.121279.4

ppg0)*1(ppg 15*(4.279)

Volume Density) udsurface)(M @ vol vol/gas(

Total

Mudsurf

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 51

It should be noted that in actual situations the mud cut would probably be less because we have assumed all gas stays in the mud-gas mixture. A certain amount of gas will break out.

The effects of gas cut mud on the hydrostatic head:

S

SB

SS

AASred.gas P

PPln TC)Z(100

TZCPΔP

Mud Density at the Surface:

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 52

R re, temperatuSurface - T

factorility compressib Surface - ZR re, temperatuAverage - T

factorility compressib Average - Zpsi pressure, Surface - P

surface at the fluid totalof % Gas - C wellof bottomat pressure cHydrostati - P

S

S

A

A

S

B

S

SB

SS

AASred.gas P

PPln TC)Z(100

TZCPΔP

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 53

18.94%4.2791100%*1

mud of vol.gas of Vol.

100%*gas of Vol.C

psi 11,700ft 15000*ppg 15*0.052PB

Hydrostatic Pressure and C

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 54

Average T and Z

175.12

35.11Z

6352

560710T

A

A

R

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 55

Reduction in BHP

psi 30.57 ΔP

14.714.711,700ln

560)18.94)(1)((100(635).7)(1.175)(18.94)(14ΔP

red.gas

red.gas

S

SB

SS

AASred.gas P

PPln TC)Z(100

TZCPΔP

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 56

The resulting bottom hole pressure will be

p = 11,700 - 30.57

BHP = 11,669 psi

This means the gas reduced the hydrostatic head by only 30.57 psi!

Reduction in BHP

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 57

Conclusion

It can be seen that the surface gas cut of approx. 3 PPG resulted in a bottom hole pressure reduction of only 30.57 psi.

There is one other factor that reduces the effect of gas cut mud even further and that is the effect of drilled solids in the mud. Drilled solids will tend to raise the overall density of the mud.

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 58

Summary of Gas-Cut Mud Problem

At bottom:

Gas expansion:

990,1rate generation gasraten circulatio mud

465bottomat volumesurfaceat volume

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 59

Summary of Gas-Cut Mud Problem

At surface:

i.e. At the surface, the mud mix contains one part of gas (by volume) for each 4.279 parts of good mud.

279.4465990,1

raten circulatio gasncirculatio mud

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 60

Summary of Gas-Cut Mud Problem

Density of mix

1279.4)0*1()15*279.4(

volume total weighttotal

Density of Mud at surface = 12.16 #/gal

(-2.84 lb/gal)

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 61

Summary of Gas-Cut Mud Problem

psi 31

PPPln

TZ)C100(TZCPp

S

SB

SS

AAS

A reduction in the mud density at the surface by 2.84 lb/gal resulted in a reduction in BHP of:

Page 62: 8. Fracture Gradients

8. Fracture Gradients PETE 661 Drilling Engineering Slide 62

Note:

It is very important in any drilling operation:

To recognize the symptoms of increasing pore pressure

To be able to estimate the magnitude of the pore pressure

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 63

Note cont’d:

To know the fracture gradients of the exposed formations

To maintain the drilling practices within controllable limits

To keep in mind that any one symptom of increasing pore pressure may not be sufficient to provide the basis for precise conclusions

Look at all the indicators...

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 64

ROP F.L.Temp Cl - MUD t d Gas Units SH YP

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 65

What should be done when gas cut mud is encountered?

(1) Establish if there is any fire hazard. If there is a fire hazard, divert flow through mud-gas separation facilities.

(a) Notify any welder in area(b) Notify all rig personnel of the

pending danger

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 66

What should be done when gas cut mud is encountered?

(2) Determine where the gas came from. If the casing seat fracture gradient is being approached, and there is some concern about raising the mud weight:

Stop drilling and circulate, and observe the gas response. If source is drilled gas, the gas rate will decrease.

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 67

What should be done when gas cut mud is encountered?

(a) If the gas units completely return to the original background gas, it would probably be safe to resume drilling.

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 68

What should be done when gas cut mud is encountered?

(b) If there has been ample circulation time and the gas units do not drop back to the original background level, but stay at a higher value, this indicates that the mud weight is approaching the pore pressure and consideration should be given to increasing the mud weight.

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8. Fracture Gradients PETE 661 Drilling Engineering Slide 69

What should be done when gas cut mud is encountered?

Establish Where did the gas come from?

(a) Drilled gas - no increase in mud weight is required

(b) Increasing pore pressure - (abnormal pore pressure) - May have to increase mud weight