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C. Fracturing Fluids and Proppants
C. Fracturing Fluids and Proppants
EXAMPLE C-1
From Table 5- 1 the consistency index for pipe and slot can be calcylated:
Determination of Rheological Properties of Power Law Fluids A hydraulic fracturing fluid has exhibited the experimentally determined behavior shown in Fig. C-1. These'data were measured in a concentric cylinder viscometer. Calculate the power law constants, determine the consistency for a pipe and a slot, and plot the apparent viscosity in a pipe for shear rates from 1 to I 0-3sec-I. The ratio of the radius of the cup to that of the bob in the viscometer was equal to 2.
Solution (Ref. Sections 5-2.2 and 5-2.3) From Fig. C- I , the slope is equal to 0.36, and thus n = n ' = 0.36. The consistency index K,, = 0.21 lbf sec"'lft', which is obtained from the intercept at y = 1. From the top equation in Table 5- 1 (and noting that p = 2),
and
Finally, the apparent viscosity (again from Table 5-1) is given for a pipe by
Po =
47,880 (0.143) - -6847 y o .64. Y'-o.36
(C-5)
A graph of the apparent viscosity vs. shear rate is shown in Fig. C-2.
Then, the power law for this fluid is from Eq. 5-4: T~~ =
0.125.jo.36.
(C-2)
c- 1
PRACTICAL COMPANION TO RESERVOIR STIMULATION
, i I
I
10
!=k c 1
II 10
0.1
1
100
Shear Rate, f (sec-’)
Figure C-1-Experimental
c- 2
data for Example C-1.
1000
FRACTURING FLUIDS A N D PROPPANTS
10000
1000
100
10
1
10
I00
1000
Shear Rate, f (sec-I)
Figure C-2-Apparent viscosity in a pipe for the fluid in Example C-1 .
c- 3
PRACTICAL COMPANION TO RESERVOIR STIMULATION
EXAMPLE C-2
Calculation of Total Leakoff Coefficient Assuming that the filter-cake leakoff coefficient is Cw,= 0.001 f t / 6 and using the data in Table C-1, calculate the total leakoff coefficient for both the gas-well and the oilwell data.
Solution (Ref. Section 5-7) From Eq. 5- 16, the viscosity control coefficient can be calculated (this is the same for both the oil and the gas well): C,, = 0.0469
=
{T
0.01 ft I &n.
C,. = (0.0374) (4000) (0.001)(0.12) (1.2 x
= 0.1 13.
0.025
C,. = (0.0374) (4000) = 0.005.
c-4
{
1
CL
1
-
0.001
+ - +1 -
0.01
I 0.113’
(C-9)
for the gas well. For the oil well a leading. to CL= 9 x similar calculation leads to CL = 7.7 x
I
k
=
I
lrnd
~
~
Ap = 4000psi
From Eq. 5- 17, the compressibility control coefficient can be calculated. For the gas well,
For the oil well,
As can be seen, the compressibility coefficient for an oil well can be a very effective means of leakoff control; for a gas well it is not. The total leakoff coefficient can be considered as a series of resistances, and thus
(C-7)
I I
f$
=
0.12
ct
=
1.2 x 10-4 psi (gas well)
c,
=
10-5 psi (oil well)
,u
=
0.025cp (gas well)
p
=
1 cp (oil well)
fiL
=
1ocp
Table C-1-Reservoir
(0.001) (0.12) 1
and fluid data for Example C-2.
I I
FRACTURING FLUIDS AND PROPPANTS
EXAMPLE C-3
Leakoff Control Comparisons Compare the total leakoff coefficient and control of it for Example C-2 assuming that the wall coefficient is halved ( 5 x or doubled (2 x Repeat the calculation by changing the leakoff viscosity from 10 cp to 1 cp and 100 cp. Use the data for the oil well only.
Solution (Ref. Section 5-7) The total leakoff coefficient, using the results from Eqs. C-6 and C-8 and calculating it as in Eq. C-9, is equal to 4.7 x (for C, = 5 x and 1.6 x (for C , = 2 x This is a substantial change from the leabff coefficient calculated in Example C-2. Essentially, the total leakoff
coefficient is halved if C, is halved and is doubled if C,, is doubled. If the leakoff viscosity is 1 cp (0.1 of the value used in Example C-2),then the viscosity control coefficient is equal to 0.032 (as in Eq. C-6) and the total leakoff coefficient is equal to 9.6 x 10-4(a 7% increase from the value in Example c-2). If the leakoff viscosity is 100 cp (10 times the value in Example C-2), then the viscosity control coefficient is equal to 0.0032 and the total leakoff coefficient is equal to 7.6 X 10-4(.a 15% decrease from the value in Example C-2). Thus, leakoff control is far more effective using a wall filter cake than by any other mechanism.
C-5
PRACTICAL COMPANION TO RESERVOIR STIMULATION
EXAMPLE C-4
and in oilfield units,
Friction Pressure Drop Calculation Calculate the pressure drop per 1000 ft of vertical tubing, with a 3.5-in. outside diameter, for a polymer solution at 100°F. The rheological properties, n' and K;;,,., are 0.39 and 4.5 x lo-* Ibf-sec"'/ft*,respectively. Perform this calculation for a range of injection rates up to 100 BPM. The fluid density is 60 Ib/ft3. Plot the results on log-log paper.
Solution (Ref. Section 5-2.3 and a fluid mechanics textbook) To calculate the friction pressure drop for any type flow, the Reynold's numbers must first be calculated. For a power law fluid this is given by
APf =
5.2 x IO-'ffpLv
2
d
(C- 16)
where the Ap, is the pressure in pounds per square inch. If N R e is greater than 2100, the flow is turbulent and the Fanning friction factor is given by (C-17) where
c=
log n'
+ 2.5
50
(C- 18)
and b = While Eq. C-10 is for consistent units, for oilfield units it becomes NRe
=
0.249pv""'d"'
(
961l'K '1I)";-
'
(C-11)
where p is in Ib/ft', v is in ft/sec, d is in in., and K' is in lbf-sec"'/ft*. The velocity is given by the obvious: (C- 12)
v = 17.17-, d2
20 v = 17.17 - = 3.81 f t / ~ e ~ . 3=
(C- 13)
=
16 ->
NRe
(C- 14)
leading to the classic pressure drop resulting from friction
Ap+ =
C- 6
2fr.PLVL d
(C- 15)
(C- 19)
(C-20)
From Eq. C- 1 I , the Reynold's number is then calculated: NRe
=
(0.249) (60) (38.1)1'61(3"'39) (96°,39)(4.5 x lo-*) ( 1.39)0.39 (C-21)
= 2.6 x lo4.
where q is barrels per minute. If the N R e is less than 2100, laminar flow is in effect, and in such case the Fanning friction factor is
fr.
7
The calculation of the pressure drop is then given by Eq. C-16. A sample calculation is performed with q = 20 BPM. From Eq. C- 13 (and remembering that for OD = 3.5 in. the ID z 3 in.), the velocity is
and in oilfield units, 4
1.4 - log n'
Therefore, the flow is turbulent. From Eqs. C- 17, C- 18 and C- 19, the friction factor is equal to 0.003, and from Eq. C- 16,
APf =
(5.2 x
= 453 psi.
(0.003) (60) (1000) (38.1)' 3
(C-22)
Figure C-3 is a plot of the expected pressure drops. Clearly, the laminar and turbulent flow regimes are indicated by the sharp bend in the results. These values are likely to be significantly lower in an actual operation because of the liberal use of friction reducing agents.
FRACTURING FLUIDS AND PROPPANTS
10000
f c 1000
c
100
f 10
100
1 Injection Rate (BPM)
Figure C-3-Friction
pressure drops for Example C-4.
c-I
PRACTICAL COMPANION TO RESERVOIR STIMULATION
EXAMPLE C-5
From Fig. 5-4, the shear rate is 50 sec-' at 100 ft and 60 sec-' at 250 ft. Thus,
Shear and Temperature Effects on Apparent Viscosity During Fracturing Given the rheological properties in Table C-2 and the simulated shear rate and temperature properties shown in Figs. 5-4 and 5-5, determine the apparent viscosities at 100and 250 ft along the fracture after 25% and 75% of the fluid has been pumped.
Solution (Ref. Section 5-3) A. 25% offluidpumped From Fig. 5-5 the temperature at 100 ft is 185°F and at 250 ft is 200°F (the maximum temperature). Fromkble C-2, the n' and K'are 0.392 and 0.258 (interpolation) and 0.4133 and 0.1 169, respectively. From Fig. 5-4, the shear rate is 130 sec-' at 100 ft and 300 sec-' at 250 ft. The apparent viscosity is given by
p ( , = 47,880 K'v"
-',
p u = (47,880) (0.350) (50-0.597)= 1621 cp
(C-26)
at 100 ft, and
p , = (47,880) (0.1169) (60-0"s7) = 506 cp
(C-27)
at 250 ft. Temp (OF)
Time (hr)
n'
K'
0.4145
0.3384
0.4133
0.1 169
0.5686
0.0339
0.6895
0.0139
(C-23)
and thus at 100 ft,
p,, = (47,880) (0.258) (130-"
602)
= 660 cp;
(C-24)
= 197 CP.
(c-2-51
whereas at 250 ft,
p ( , = (47,880) (0.1 169) (300-"
5x7)
This is a substantial reduction in the viscosity, and it has a major effect on the expected role of the pad fluid, which is to generate the desired fracture widths.
B. 75% of fluid pumped Again, from Fig. 5-5 the temperatures are 155°F and 200"F, respectively, for the 100-ft and 250-ft locations. From Table C-2 the rheological properties for 155°F are 0.350 and 0.403, respectively.
Table C-2-Flheology of a borate crosslinked fluid containing 40-lb/lOOO gal hydroxypropyl guar.
C- 8
FRACTURING FLUIDS AND PROPPANTS
EXAMPLE C-6
Calculation of Settling Velocity Using the data in Table C-3, calculate the settling velocity for a particle with a diameter equal to 2.5 x m.
Solution (Ref. Section 5-6) From Eq. 5- 12, which is the modified Stokes law, u, =
(9.78) (2650 - 1000)(2.5 x 18 (2)
+
(9.78) (2650
-
I
n'
=
0.37
K'
=
4Pa-s"'
P O
=
2Pa-s
g
=
9.78 m/sec2
Table C-3-Data
1000) (2.5 x 10-4)2
i
for Example C-6.
18 (4) 0.63
4
(2.5 x
(C-28) 10-413
and u , = 2.8 x lo-'
leading to u, = 3.18 x
+
2.6 x 10-3uf'63,
(C-29)
ds.
c-9
PRACTICAL COMPANION T O RESERVOIR STIMULATION
EXAMPLE C-7
Calculation of Stress-Induced Proppant-Pack Permeability Impairment: Long-Term Stresses Use the data in Fig. 5-27 for 20140 sand to calculate the stressinduced permeability reduction for 2000 and 4000 psi drawdown. Will the stress on the proppant pack increase or decrease with time? Assume that the reservoir is 10,000 ft deep with an average density equal to 165 lb/ft3, the Poisson’s ratio is 0.25, the initial reservoir pressure is 6500 psi, and the poroelastic constant is 0.7.
Solution (Ref. Section 5-9) The effective minimum horizontal stress can be‘estimated from Eq. 2-4 1: CT’H,mitr
=
0.25 ( 165) ( 10,000) 0.75 ~
[
= 2303 psi.
144
0
k
I l l I ?I
4
- 0.25 (165) (l0,OOO) - (0.7) - 0.75 144
+ (0.7) (5000)
(~ooo)]
= 6152 psi.
(C-3 1)
With the 4500-psi bottomhole pressure,
- (0.7) (6500)]
c ~ H , = ~ 6152 ~ ~ ; -~ (0.7) ~
(C-30)
This is the stress that the proppant will experience without drawdown. Thus, the proppant-pack permeability will be about 300,000 md (from Fig. 5-27).
c-I0
With a 2000-psi drawdown ( p,,f = 4500 psi) experienced near the wellbore, the instantaneous effective stress will increase by (0.7) (2000) = 1400 (o’= 3700 psi); with a 4000psi drawdown (pH.,= 2500 psi), the effective stress will increase by (0.7) (4000) = 2800 (0= 5100 psi) and result in permeability values equal to 200,000 and 150,000 md, respectively. In the long term, the reservoir pressure will decline. As a result, the absolute minimum horizontal stress will decrease (e.g., fdr p = 5000 psi) as given by
(4500) = 3000 psi;
(C-32)
and with the 2500-psi bottomhole pressure, I T ’ H ~ ~ , ~ ;= ~ ~6152 -
(0.7) (2500) = 4400 psi.
(c-33)
Thus, the effective stress on the proppant decreases with time.
EXAMPLE C-1
From Table 5- 1 the consistency index for pipe and slot can be calcylated:
Determination of Rheological Properties of Power Law Fluids A hydraulic fracturing fluid has exhibited the experimentally determined behavior shown in Fig. C-1. These'data were measured in a concentric cylinder viscometer. Calculate the power law constants, determine the consistency for a pipe and a slot, and plot the apparent viscosity in a pipe for shear rates from 1 to I 0-3sec-I. The ratio of the radius of the cup to that of the bob in the viscometer was equal to 2.
Solution (Ref. Sections 5-2.2 and 5-2.3) From Fig. C- I , the slope is equal to 0.36, and thus n = n ' = 0.36. The consistency index K,, = 0.21 lbf sec"'lft', which is obtained from the intercept at y = 1. From the top equation in Table 5- 1 (and noting that p = 2),
and
Finally, the apparent viscosity (again from Table 5-1) is given for a pipe by
Po =
47,880 (0.143) - -6847 y o .64. Y'-o.36
(C-5)
A graph of the apparent viscosity vs. shear rate is shown in Fig. C-2.
Then, the power law for this fluid is from Eq. 5-4: T~~ =
0.125.jo.36.
(C-2)
c- 1
PRACTICAL COMPANION TO RESERVOIR STIMULATION
, i I
I
10
!=k c 1
II 10
0.1
1
100
Shear Rate, f (sec-’)
Figure C-1-Experimental
c- 2
data for Example C-1.
1000
FRACTURING FLUIDS A N D PROPPANTS
10000
1000
100
10
1
10
I00
1000
Shear Rate, f (sec-I)
Figure C-2-Apparent viscosity in a pipe for the fluid in Example C-1 .
c- 3
PRACTICAL COMPANION TO RESERVOIR STIMULATION
EXAMPLE C-2
Calculation of Total Leakoff Coefficient Assuming that the filter-cake leakoff coefficient is Cw,= 0.001 f t / 6 and using the data in Table C-1, calculate the total leakoff coefficient for both the gas-well and the oilwell data.
Solution (Ref. Section 5-7) From Eq. 5- 16, the viscosity control coefficient can be calculated (this is the same for both the oil and the gas well): C,, = 0.0469
=
{T
0.01 ft I &n.
C,. = (0.0374) (4000) (0.001)(0.12) (1.2 x
= 0.1 13.
0.025
C,. = (0.0374) (4000) = 0.005.
c-4
{
1
CL
1
-
0.001
+ - +1 -
0.01
I 0.113’
(C-9)
for the gas well. For the oil well a leading. to CL= 9 x similar calculation leads to CL = 7.7 x
I
k
=
I
lrnd
~
~
Ap = 4000psi
From Eq. 5- 17, the compressibility control coefficient can be calculated. For the gas well,
For the oil well,
As can be seen, the compressibility coefficient for an oil well can be a very effective means of leakoff control; for a gas well it is not. The total leakoff coefficient can be considered as a series of resistances, and thus
(C-7)
I I
f$
=
0.12
ct
=
1.2 x 10-4 psi (gas well)
c,
=
10-5 psi (oil well)
,u
=
0.025cp (gas well)
p
=
1 cp (oil well)
fiL
=
1ocp
Table C-1-Reservoir
(0.001) (0.12) 1
and fluid data for Example C-2.
I I
FRACTURING FLUIDS AND PROPPANTS
EXAMPLE C-3
Leakoff Control Comparisons Compare the total leakoff coefficient and control of it for Example C-2 assuming that the wall coefficient is halved ( 5 x or doubled (2 x Repeat the calculation by changing the leakoff viscosity from 10 cp to 1 cp and 100 cp. Use the data for the oil well only.
Solution (Ref. Section 5-7) The total leakoff coefficient, using the results from Eqs. C-6 and C-8 and calculating it as in Eq. C-9, is equal to 4.7 x (for C, = 5 x and 1.6 x (for C , = 2 x This is a substantial change from the leabff coefficient calculated in Example C-2. Essentially, the total leakoff
coefficient is halved if C, is halved and is doubled if C,, is doubled. If the leakoff viscosity is 1 cp (0.1 of the value used in Example C-2),then the viscosity control coefficient is equal to 0.032 (as in Eq. C-6) and the total leakoff coefficient is equal to 9.6 x 10-4(a 7% increase from the value in Example c-2). If the leakoff viscosity is 100 cp (10 times the value in Example C-2), then the viscosity control coefficient is equal to 0.0032 and the total leakoff coefficient is equal to 7.6 X 10-4(.a 15% decrease from the value in Example C-2). Thus, leakoff control is far more effective using a wall filter cake than by any other mechanism.
C-5
PRACTICAL COMPANION TO RESERVOIR STIMULATION
EXAMPLE C-4
and in oilfield units,
Friction Pressure Drop Calculation Calculate the pressure drop per 1000 ft of vertical tubing, with a 3.5-in. outside diameter, for a polymer solution at 100°F. The rheological properties, n' and K;;,,., are 0.39 and 4.5 x lo-* Ibf-sec"'/ft*,respectively. Perform this calculation for a range of injection rates up to 100 BPM. The fluid density is 60 Ib/ft3. Plot the results on log-log paper.
Solution (Ref. Section 5-2.3 and a fluid mechanics textbook) To calculate the friction pressure drop for any type flow, the Reynold's numbers must first be calculated. For a power law fluid this is given by
APf =
5.2 x IO-'ffpLv
2
d
(C- 16)
where the Ap, is the pressure in pounds per square inch. If N R e is greater than 2100, the flow is turbulent and the Fanning friction factor is given by (C-17) where
c=
log n'
+ 2.5
50
(C- 18)
and b = While Eq. C-10 is for consistent units, for oilfield units it becomes NRe
=
0.249pv""'d"'
(
961l'K '1I)";-
'
(C-11)
where p is in Ib/ft', v is in ft/sec, d is in in., and K' is in lbf-sec"'/ft*. The velocity is given by the obvious: (C- 12)
v = 17.17-, d2
20 v = 17.17 - = 3.81 f t / ~ e ~ . 3=
(C- 13)
=
16 ->
NRe
(C- 14)
leading to the classic pressure drop resulting from friction
Ap+ =
C- 6
2fr.PLVL d
(C- 15)
(C- 19)
(C-20)
From Eq. C- 1 I , the Reynold's number is then calculated: NRe
=
(0.249) (60) (38.1)1'61(3"'39) (96°,39)(4.5 x lo-*) ( 1.39)0.39 (C-21)
= 2.6 x lo4.
where q is barrels per minute. If the N R e is less than 2100, laminar flow is in effect, and in such case the Fanning friction factor is
fr.
7
The calculation of the pressure drop is then given by Eq. C-16. A sample calculation is performed with q = 20 BPM. From Eq. C- 13 (and remembering that for OD = 3.5 in. the ID z 3 in.), the velocity is
and in oilfield units, 4
1.4 - log n'
Therefore, the flow is turbulent. From Eqs. C- 17, C- 18 and C- 19, the friction factor is equal to 0.003, and from Eq. C- 16,
APf =
(5.2 x
= 453 psi.
(0.003) (60) (1000) (38.1)' 3
(C-22)
Figure C-3 is a plot of the expected pressure drops. Clearly, the laminar and turbulent flow regimes are indicated by the sharp bend in the results. These values are likely to be significantly lower in an actual operation because of the liberal use of friction reducing agents.
FRACTURING FLUIDS AND PROPPANTS
10000
f c 1000
c
100
f 10
100
1 Injection Rate (BPM)
Figure C-3-Friction
pressure drops for Example C-4.
c-I
PRACTICAL COMPANION TO RESERVOIR STIMULATION
EXAMPLE C-5
From Fig. 5-4, the shear rate is 50 sec-' at 100 ft and 60 sec-' at 250 ft. Thus,
Shear and Temperature Effects on Apparent Viscosity During Fracturing Given the rheological properties in Table C-2 and the simulated shear rate and temperature properties shown in Figs. 5-4 and 5-5, determine the apparent viscosities at 100and 250 ft along the fracture after 25% and 75% of the fluid has been pumped.
Solution (Ref. Section 5-3) A. 25% offluidpumped From Fig. 5-5 the temperature at 100 ft is 185°F and at 250 ft is 200°F (the maximum temperature). Fromkble C-2, the n' and K'are 0.392 and 0.258 (interpolation) and 0.4133 and 0.1 169, respectively. From Fig. 5-4, the shear rate is 130 sec-' at 100 ft and 300 sec-' at 250 ft. The apparent viscosity is given by
p ( , = 47,880 K'v"
-',
p u = (47,880) (0.350) (50-0.597)= 1621 cp
(C-26)
at 100 ft, and
p , = (47,880) (0.1169) (60-0"s7) = 506 cp
(C-27)
at 250 ft. Temp (OF)
Time (hr)
n'
K'
0.4145
0.3384
0.4133
0.1 169
0.5686
0.0339
0.6895
0.0139
(C-23)
and thus at 100 ft,
p,, = (47,880) (0.258) (130-"
602)
= 660 cp;
(C-24)
= 197 CP.
(c-2-51
whereas at 250 ft,
p ( , = (47,880) (0.1 169) (300-"
5x7)
This is a substantial reduction in the viscosity, and it has a major effect on the expected role of the pad fluid, which is to generate the desired fracture widths.
B. 75% of fluid pumped Again, from Fig. 5-5 the temperatures are 155°F and 200"F, respectively, for the 100-ft and 250-ft locations. From Table C-2 the rheological properties for 155°F are 0.350 and 0.403, respectively.
Table C-2-Flheology of a borate crosslinked fluid containing 40-lb/lOOO gal hydroxypropyl guar.
C- 8
FRACTURING FLUIDS AND PROPPANTS
EXAMPLE C-6
Calculation of Settling Velocity Using the data in Table C-3, calculate the settling velocity for a particle with a diameter equal to 2.5 x m.
Solution (Ref. Section 5-6) From Eq. 5- 12, which is the modified Stokes law, u, =
(9.78) (2650 - 1000)(2.5 x 18 (2)
+
(9.78) (2650
-
I
n'
=
0.37
K'
=
4Pa-s"'
P O
=
2Pa-s
g
=
9.78 m/sec2
Table C-3-Data
1000) (2.5 x 10-4)2
i
for Example C-6.
18 (4) 0.63
4
(2.5 x
(C-28) 10-413
and u , = 2.8 x lo-'
leading to u, = 3.18 x
+
2.6 x 10-3uf'63,
(C-29)
ds.
c-9
PRACTICAL COMPANION T O RESERVOIR STIMULATION
EXAMPLE C-7
Calculation of Stress-Induced Proppant-Pack Permeability Impairment: Long-Term Stresses Use the data in Fig. 5-27 for 20140 sand to calculate the stressinduced permeability reduction for 2000 and 4000 psi drawdown. Will the stress on the proppant pack increase or decrease with time? Assume that the reservoir is 10,000 ft deep with an average density equal to 165 lb/ft3, the Poisson’s ratio is 0.25, the initial reservoir pressure is 6500 psi, and the poroelastic constant is 0.7.
Solution (Ref. Section 5-9) The effective minimum horizontal stress can be‘estimated from Eq. 2-4 1: CT’H,mitr
=
0.25 ( 165) ( 10,000) 0.75 ~
[
= 2303 psi.
144
0
k
I l l I ?I
4
- 0.25 (165) (l0,OOO) - (0.7) - 0.75 144
+ (0.7) (5000)
(~ooo)]
= 6152 psi.
(C-3 1)
With the 4500-psi bottomhole pressure,
- (0.7) (6500)]
c ~ H , = ~ 6152 ~ ~ ; -~ (0.7) ~
(C-30)
This is the stress that the proppant will experience without drawdown. Thus, the proppant-pack permeability will be about 300,000 md (from Fig. 5-27).
c-I0
With a 2000-psi drawdown ( p,,f = 4500 psi) experienced near the wellbore, the instantaneous effective stress will increase by (0.7) (2000) = 1400 (o’= 3700 psi); with a 4000psi drawdown (pH.,= 2500 psi), the effective stress will increase by (0.7) (4000) = 2800 (0= 5100 psi) and result in permeability values equal to 200,000 and 150,000 md, respectively. In the long term, the reservoir pressure will decline. As a result, the absolute minimum horizontal stress will decrease (e.g., fdr p = 5000 psi) as given by
(4500) = 3000 psi;
(C-32)
and with the 2500-psi bottomhole pressure, I T ’ H ~ ~ , ~ ;= ~ ~6152 -
(0.7) (2500) = 4400 psi.
(c-33)
Thus, the effective stress on the proppant decreases with time.