stimulation 11

Fractured Well Performance

How do you know your fracture design was successful?

• Pumped all fluids and proppants without any mechanical

problems or screenout or additional costs?

…………….Yes for service company, completions engineer

• Achieved a predicted production goal? …………….Yes for reservoir and completions engineer

• Achieved economic success? …………….Yes for manager, production company

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Fracture Conductivity Models

1. Infinite Conductivity Model

• Negligible pressure loss in the fracture

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rw rwa xf

pwf2

fx

war

Sew

rwa

r

Pwf



2. Uniform flux Model

• Slight pressure gradient corresponding to a uniformly distributed flux

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rw rwa xf

pwf

qe

fx

war

Pwf q



3. Finite conductivity Model

• Constant, limited fracture conductivity

• Dimensionless Fracture Conductivity

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rw rwa xf

pwf

Fcd=50

Fcd

Fcd=10

fkx

wf

k

cdF

McGuire – Sikora Empirical Model (1960), Modified by Holditch (1975)

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Basis:

Pseudosteady state flow

Constant rate production

Square drainage shape

Compressible fluid

Entire interval is propped

Both xf and kfw are important

1. Length is dominant variable Constraining stimulation 2. kfw sufficiently high

Well spacing/shape

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xf

2ye

2xe

xf

re

Limit penetration to 80% of the reservoir boundary,

%80

ex

fx


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Remarks: • For a given Lf/Le ratio, the solution approaches a maximum asymptotic value • Thus at large relative conductivity (low-k), productivity can be increased by increasing Lf and not the conductivity. • For a given Lf, there exists an optimal fracture conductivity • Theoretical maximum increase in productivity ratio is 13.6


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A

B

C

D Point A – length limited, increase treatment size to Point B – conductivity limited, change proppant type to Point C – length limited, increase treatment size to Point D – reached drainage radius

Tinsley et al. empirical model (1969) Effect of fracture conductivity and fracture height on well performance Rwa = f (xf, Fcd)

Basis: • Steady state flow • Constant rate production • Cylindrical reservoir geometry • Incompressible fluid flow

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Fracture Characterization Parameter, X

Tinsley et al. empirical model (1969)

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Empirical Equations

a. For 0.1 < X < 3

DDCer

fxCBFE

))((28.425.183.1tan785.0/

b. X > 3

C

ZZYFFE

1)tan(tan

where

668.9

334.0334.3

XB

92.008.0

fh

hC

h

fhD 75.01

38.240.6

2

84.4

XXF

er

fx

XY

32.127.2

84.064.1

2

24.1

XXZ

Empirical Equations

a. For 0.1 < X < 3

DDCer

fxCBFE

))((28.425.183.1tan785.0/

b. X > 3

C

ZZYFFE

1)tan(tan

where

668.9

334.0334.3

XB

92.008.0

fh

hC

h

fhD 75.01

38.240.6

2

84.4

XXF

er

fx

XY

32.127.2

84.064.1

2

24.1

XXZ

Example

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A given well currently produces at a constant bottomhole flowing pressure of 1200 psia. It was determined this well would be a good candidate for hydraulic fracturing. Two fracture treatments are proposed by the service company: Treatment A with xf = 100 ft and FcD = 200 Treatment B with xf = 250 ft and FcD = 10. To assess the results of the fracture treatments, rate decline is used to forecast

performance. Assume hf = h Reservoir data for a typical well pi = 2600 psia moi = 0.2 cp Boi = 1.642 bbl/stb h = 66 ft rw = 0.33 ft cti = 30 x 10-6 psi-1 f = 0.117 Sw = 0.32 re = 744 ft k = 0.25 md

Example

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• Time to start of pss

tpss = 77 days

where tDApss = 0.1 (center of a circle) and A = p re2

• Dimensionless time

tD = 2,258 t/rwa2

DApsst

k

AT

c3790

psst

fm

2

war

tc

kt000264.0

Dt

fm

Example

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12320033.

744ln

744

1001593.0

X

.723.3

ln

ln

ftwar

warer

wrer

FE

Calculate rwa Case A: From the figure this value of X indicates an infinite conductivity fracture; thus rwa = xf/2 = 50 ft. Case B: X = 15.4 xf/re = 0.336 From the figure or equations,

X=123 X=15.4

xf/re = 0.336 FE=3.3

Example

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1

10

100

1000

1 10 100 1000 10000

qo

,bo

pd

time,days

Case B

Case A

Unfrac

Example

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Depletion rate decline

Apply Fetkovich’s type curve

a. Calculate tDd

Dt*.5warerln

1*

12warer

2DBtDdt

b. Calculate qDd assuming exponential decline

DdteDdq

c. Calculate qD

Ddqwarer

DdAqDq *5.ln

1

d. Calculate qo

Transient rate decline

a. From Example 1 the tpss = 77 days, therefore

dimensionless time

2

war

tc

kt000264.0

Dt

fm

tD = 2,258 t/rwa2

b. Dimensionless rate is from the approximate

solutions corresponding to tD.

c. Calculate flow rate from

oq

)wf

pi

p(kh

oB

o2.141

Dq

m qo = 498 qD

Decline Curve Analysis Advanced Topics

Composite of analytic and empirical type curves (Fetkovich, 1980)

GRM-Engler-09


Example – Use type curve matching to analyze the given well data

GRM-Engler-09

0.01

0.1

1

10

0.001 0.01 0.1 1 10 100Dimensionless Time, tdD

Dim

en

sio

nle

ss F

low

Rate

, q

dD

1.00.80.60.40.2b=0

LinearreD=2.5

510

2050200

5000QdD=1-exp(-tdD)

b = 0.4, re/rwa = 1000


Example - Results

GRM-Engler-09

Estimated Properties Units Rate-Time Rate-Cum Cum-Time

Productivity Factor, PF Mscf/d/psi 0.176 0.176 0.171

Pore Volume, PV MMcf 14.463 14.026 14.026

Initial Gas-In-Place, IGIP Bscf 2.091 2.028 2.028

Drainage Acreage Acres 86.5 83.9 83.9

Equivalent Drainage Radius, r e ft 1094.9 1078.3 1078.3

Apparent Wellbore Radius, r wa ft 1.1 1.1 1.1

Skin Factor, S - -1.1 -1.1 -1.1

Flow Capacity, kh md-ft 2.628 2.628 2.548

Permeability, k md 0.08211 0.08211 0.07963

Current Recovery, % IGIP % 50.0 51.6 51.6

Current Recovery, Per Acre Mscf/acre 12097.6 12474.4 12474.4

91.10 Mscf/d

334533 Mscf

1380543 Mscf

67.39 %

16294.79 Mscf/acre Projected Recovery, Per Acre

Last Reported Rate, qlast

Remaining Reserves

Estimated Ultimate Recovery (EUR)

Projected Recovery, % IGIP


Example – Use type curve matching to analyze well data from example 1.8

GRM-Engler-09

0.01

0.1

1

10

0.001 0.01 0.1 1 10 100Dimensionless Time, tdD

Dim

en

sio

nle

ss F

low

Rate

, q

dD

1.00.80.60.40.2b=0

LinearreD=2.5

510

2050200

5000QdD=1-exp(-tdD)

b = 0.8, re/rwa = 5


Example - (results)

GRM-Engler-09 b = 0.8, re/rwa = 5

Estimated Properties Units Rate-Time

Productivity Factor, PF Mscf/d/psi 0.139

Pore Volume, PV MMcf 15.281

Initial Gas-In-Place, IGIP Bscf 0.718

Drainage Acreage Acres 79.7

Equivalent Drainage Radius, re ft 2636

Apparent Wellbore Radius, rwa ft 2636

Skin Factor, S - -9.4

Flow Capacity, kh md-ft 1.762

Permeability, k md 0.044

Current Recovery, % IGIP % 73.3

Current Recovery, Per Acre Mscf/acre 6604

Transient solution for finite-conductivity vertical fractures, (Agarwal, et al.,1979)

pkh

oBq2.141

Dq

m

2

fx

tc

kt0063.

xfD

t

fm

fkx

wf

k

CDF

Basis:

• The fracture has finite conductivity that is uniform throughout the fracture

• The fracture has two equal wing lengths

• The reservoir is infinite acting


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Low FCD case

• Start with point A

• Double xf

• Reduces FCD by ½

• and reduces tDxf by ¼

Result:

An increase in fracture length

results in no difference in flow

rate

A A’

fkx

wf

k

CDF

2

fx

tc

kt0063.

xfD

t

fm


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Low FCD case

• Start with point A

• Double xf and

• kfw is increased by

a factor of 4, then

• FCD would double

and tDxf reduces by ¼

Result:


And fracture conductivity results in an increase in flow rate

A

A’’

fkx

wf

k

CDF

2

fx

tc

kt0063.

xfD

t

fm


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High FCD case

• Start with point B

• Double xf and

• Reduces FCD by ½

• and reduces tDxf by ¼

Result:


results in an increase in flow rate

B B’

fkx

wf

k

CDF

2

fx

tc

kt0063.

xfD

t

fm


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FCD ≤ 3

Cannot be improved significantly by increasing xf

with same fracture conductivity

FCD ≥ 30

Increasing xf will be more beneficial than increasing conductivity

fkx

wf

k

CDF

2

fx

tc

kt0063.

xfD

t

fm

Guidelines


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Example

A low-permeability gas well required a massive hydraulic fracture (MHF) to become productive.

The following data was acquired for analysis.

pi 2394 psia mgi 0.0176 cp k 0.0081 md

Tf 260F h 32 ft

rw 0.33 ft cti 2.34 x 10-4

psi-1

f 0.107 A 640 Acres

pwf 1600 psia z 0.93

The well has produced for a little less than a year, with the performance data shown below.

Time,days q, mscfd

20 625

35 476

50 408

100 308

150 250

250 208

300 192

The objectives are to compute the fracture length and fracture flow capacity, and to predict future

performance.


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Example Example Transient Type Curve Matching for

Finite Conductivity Fractures

Match Point

q = 100 mcfd t = 100 days

qD = 2 tDxf = .0038

10

100

1000

1 10 100 1000

q,m

sc

fd

time,days

Solution

Type curve match with the finite conductivity

fracture solution

mDq

q

2

wfp

2

iph

zTgi

1422k

m (q/qD)m = 50

Since kh were determined from an earlier well test,

the y-axis match is fixed.

mDt

t

tc

g

k000264.02

fx

fm xf

2= 4.853*(t/tD)m

xf = 357 ft.

From the type curve, Fcd = 500; thus the fracture

flow capacity can be computed by,

ftmdfkxCDFwfk 1446)(*


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Example

Find the time to the beginning of pss

DApsst

k

AT

c3790

psst

fm tpss = 23,951 days (66 yrs)

where tDApss = 0.1 (center of a circle) and A = p re2

Future performance

a. Assume a time

b. Calculate tdxf

c. From type curve find corresponding qD

d. Calculate q


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Example

A Mesaverde gas well in the San Juan Basin was stimulated with a hydraulic fracture treatment.

The reported fracture parameters are a fracture half-length of 1030 ft and a dimensionless

fracture conductivity of 500.

pi 1175 psia mgi 0.0143 cp k 0.04 md

Tf 173F h 29 ft

rw 0.33 ft cti 5.2 x 10-4

psi-1

f 0.08 A 320 Acres re 2106 ft

pwf 600 psia z 0.87

Objective

Calculate the rate-time forecast for the well assuming an infinite-conductivity fracture.


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Example Solution

Time to start of pss

DApsst

k

AT

c3790

psst

fm tpss = 3,272 days (9 years)

Calculate apparent wellbore radius

Rwa = xf/2 = 515 ft.

sewrwar S = -7.4

Calculate Dimensionless time

2

war

tc

kt000264.0

Dt

fm

tD = 1.61 x 10-3

t {days}

Calculate dimensionless rate

oq

)wf

pi

p(kh

oB

o2.141

Dq

m qo = 0.00946 qD {mscfd}

For transient rate decline, find qD from (1)uniform flux solution – approximate equations,

and (2) infinite conductivity solution from type curve.

For depletion rate decline use Fetkovich type curve.

stimulation 11

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