8. Fracture Gradients

transcript

8. Fracture Gradients PETE 661 Drilling Engineering Slide 1

PETE 661Drilling Engineering

Lesson 8 Fracture Gradients

Prediction of Fracture GradientsWell PlanningTheoretical Fracture Gradient Determination

Hubbert & WillisMatthews & KellyBen EatonComparison of Results

Experimental Frac. Grad. DeterminationLeak-off TestsLost Circulation

Gas Cut Mud

Read:Applied Drilling Engineering, Ch. 6

HW # 5Casing Design

due 10-8-04

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

Fig. 7.21

fracture

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

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

S = P +

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.

Calculations:3. Matrix stress gradient,

psi/ft

/ D = 0.522 psi/ft

ft/psi478.0000.1DP

D.,e.i

Calculations:

4. Matrix stress at 14,000 ft

= 0.522 psi/ft * 14,000 ft

= 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 experimental approach to fracture gradient determination. Below are listed and discussed three approaches to calculating the fracture gradient.

1. Hubbert & Willis:

where F = fracture gradient, psi/ft

= pore pressure gradient, psi/ftDP

31Fmin

21Fmax

2. Matthews & Kelly:

where Ki = matrix stress coefficient

= vertical matrix stress, psi

3. Ben Eaton:

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

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.

1. Hubbert & Willis:

The pore pressure gradient,

1 2 *0.735 0.823 psiftmin

31Fmin

0.735 psift

Example - Hubbert and Willis

lb/galpsi/ft0.052

psi/ft0.823Fmin

lb/gal 15.83Fmin

21Fmax 735.01

= 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.

Example

)given(ft/psi735.0DP

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

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

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

Di * (1 - 0.465) = 2,915

ft449,5535.0915,2Di

Example - Matthews and

(iv) Find Ki from the plot on the right, for

For a south Texas Gulf Coast well,

Di = 5,449 ft

Ki = 0.685

(v) Now calculate F:DP

735.0000,11

915,2*685.0F

ft/psi9165.0

gal/lb63.17052.0

9165.0F

Example

Ben Eaton:

Variable Overburden Stress by Eaton

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

Fig. 5-5

At 11,000 ft = 0.46

Example - Ben Eaton

From above graphs, at 11,000 ft.:

46.0;ft/psi96.0DS

735.046.01

46.0735.096.0F

F = 0.9267 psi/ft = 17.82 lb/gal

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

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 ).

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.

Similarities

Ben Eaton:

Matthews and Kelly:

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

4580105120120120120120120 40 20

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 = ?

000,4872,2

DP OFFLEAK

13.8 lb/gal

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.

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

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

How Critical is Gas Cut Mud?

Formation Properties

F100 T

F250 T

1.35 Z1 Z

25% Porosity Sand70% saturation gas Sand

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.

hrbbl0.2110

hrbbl420

gs 0.7*porosity 25.0*cu.ft 61.5

bbl*hr

ft 20in/ft 12

in 877

hrmin 60*

minbbl 7

GasMud

MudGas

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

(PV = ZnRT)

Mud/gas Volume Ratio at the Surface:

279.4465

1990 VolumeGas VolumeMud:surface At

990,1VolumeGasVolumeMud:BottomAt

465BottomatGasSurfaceatGas:Expansion

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(

Mudsurf

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:

AASred.gas P

PPln TC)Z(100

TZCPΔP

Mud Density at the Surface:

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

AASred.gas P

PPln TC)Z(100

TZCPΔP

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

Average T and Z

175.12

35.11Z

560710T

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

AASred.gas P

PPln TC)Z(100

TZCPΔP

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

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.

Summary of Gas-Cut Mud Problem

At bottom:

Gas expansion:

990,1rate generation gasraten circulatio mud

465bottomat volumesurfaceat volume

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

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)

psi 31

TZ)C100(TZCPp

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

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

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...

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

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

(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.

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

(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.

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

8. Fracture Gradients

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