contoh soal "vertical oil well performance"
TRANSCRIPT
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Example 7-1
A productivity test was conducted on a well. The test results indicate that
the well is capable of producing at a stabilized flow rate of 110 STB/day
and a bottom-hole flowing pressure of 900 psi. After shutting the well for
24 hours, the bottom-hole pressure reached a static value of 1300 psi.Calculate:
Productivity index
AOF
Oil flow rate at a bottom-hole flowing pressure of 600 psi
Wellbore flowing pressure required to produce 250 STB/day
Diket :
q = 110 STB/day
Pwf = 900 psi
Pr = 1300 psi
Dit :a. PI / J
b. AOF / Qomax
c. Q @ 600 psi
d. Pwf @ q = 250 STB/day
Jawab : Pr > Pwf Estimate :
Pwf
a. 1300
1000
800
600
400
2000
J = 0.275 STB/ psi-day
Nilai Tetap
b. Pwf = 0
AOF = 357.5 STB/day
c. Pwf = 600 psi
q = 192.5 STB/day
d. q = 250 STB/day
)(Pr Pwf
qJ
daypsiSTBJ
/
)9001300(
110
Pr.JAOF
PwfJq Pr.
0, 1300
200
400
600
800
1000
1200
1400
P(Pressure,psi)
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1300 - Pwf = 909.0909
Pwf = 390.9091 psi
Example 7-2
A well is producing from a saturated reservoir with an average reservoir
pressure of 2500 psig. Stabilized production test data indicated that
the stabilized rate and wellbore pressure are 350 STB/day and 2000 psig,
respectively. Calculate:
Oil flow rate at pwf = 1850 psig
Calculate oil flow rate assuming constant J
Construct the IPR by using Vogels method and the constant productivity
index approach.
Diket :
Pr = 2500 psig
Qo = 350 STB/day
Pwf = 2000 psig
Dit :
a. Qo @ Pwf = 1850 psig
b. Qo @ constant J
c. Buat Grafik IPR - Vogel's Method
Jawab :
Saturated Oil Reservoir Pr = Pb
a. Vogel's Method
Qo(max) = 1067.07317 STB/day Nilai Tetap
Pwf = 1850 psig
J
Pwf Pr275.0
1300 Pwf
00
2
Pr8.0
Pr2.01
(max)
PwfPwf
Qo
Qo
2
2500
20008.0
2500
20002.01
350(max)
Qo
2
Pr8.0
Pr2.01(max)
PwfPwfQoQo
2
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Qo = 441.682927 STB/day
b.
J = 0.7 STB/ psig-day
Nilai Tetap
Qo = 455 STB/day
c. Estimate :
Pwf Vogel (Qo) Linear (Qo)
psig STB/day STB/day
2500 0 0
2200 218.2 210
2000 350.0 350
1500 631.7 700
1000 845.1 1050
500 990.2 14000 1067.1 1750
Example 7-3
An oil well is producing from an undersaturated reservoir that is characterized
by a bubble-point pressure of 2130 psig. The current average
reservoir pressure is 3000 psig. Available flow test data shows that the
well produced 250 STB/day at a stabilized pwf of 2500 psig. Construct
the IPR data.
Diket :Pb = 2130 psig
Pr = 3000 psig
Qo = 250 STB/day
Pwf = 2500 psig
Dit :
- Konsep data IPR
2500.
2500..
)(Pr Pwf
QoJ
20002500350
J
PwfJQo Pr.
185025007.0 Qo
0
500
1000
1500
2000
2500
3000
0 500
Pwf,Psig
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Jawab :
Undersaturated Reservoir Pr > Pb Pwf > Pb
J = 0.5 STB/day-psig
Nilai Tetap
Qob = 435 STB/dayEstimate :
Pwf Qo
psig STB/day
3000 0
2500 250 Pwf > Pb
2130 435 Pb
2000 498 Pwf < Pb
1500 709
1000 867
500 973
0 1027
= 591.6667
Example 7-4
The well described in Example 7-3 was retested and the following
results obtained:
Pwf = 1700 psig, Qo = 630.7 STB/day
Generate the IPR data using the new test data.
Diket :
Pwf = 1700 psig
Qo = 630.7 STB/day
Jawab :
)(Pr Pwf
QoJ
25003000250
J
PbJQob Pr
213030005.0 Qob
2
8.02.018.1 Pb
Pwf
Pb
PwfPbJQobQo
8.1
maxPbJ
QoQo
2
8.02.018.1 Pb
Pwf
Pb
PwfPbJQobQo
PbJQob Pr
0
500
1000
1500
2000
2500
3000
3500
0 2
Pwf,psig
Pb
Pr
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J = 0.50 STB/day-psig Nilai tetap
Qob = 435 STB/day
Generate the IPR :
Pwf Qo
psig STB/day
3000 0
2500 250 Pwf > Pb
2130 435 Pb
2000 498 Pwf < Pb
1500 709
1000 867500 973
0 1027
Example 7-5
Using the data given in Example 7-2, predict the IPR where the average
reservoir pressure declines from 2500 psig to 2200 psig.
Diket :
(Pr)p = 2500 psig (Pr)f = 2200 psig
Qo = 350 STB/day
Pwf = 2000 psig(Qo max)p = 1067.1 STB/day
Ditanya :
- Prediksi IPR pd Pr = 2200 psig
Jawab :
(Qo max)f = 849 STB/day Nilai Tetap
Pwf Qo
psig STB/day
2200 0
2000 133
1500 417
1000 631
500 775
2
8.02.018.1
PrPb
Pwf
Pb
PwfPbPb
J
3000
0
500
1000
1500
2000
2500
3000
3500
0 2
Pwf,psig
Pb
p
f
p
fpQofQo
Pr
Pr8.02.0
Pr
Prmaxmax
2
Pr8.0
Pr2.01(max)
PwfPwfQoQo
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0 849
Example 7-6
The information given in Examples 7-2 and 7-5 is repeated here for
convenience:
Current average pressure = 2500 psig Stabilized oil flow rate = 350 STB/day
Stabilized wellbore pressure = 2000 psig
Generate the current IPR data and predict future IPR when the reservoir
pressure declines from 2500 to 2000 psig by using Wiggins method.
Diket :
(Pr)p = 2500 psig (Pr)f = 2200 psig
Qo = 350 STB/day
Pwf = 2000 psig
Ditanya : a. Data IPR saat ini
b. Data IPR yang akan datang
Jawab :
a. Menghitung Qo max Saat Ini / Current dgn menggunakan Wiggins method
(Qo max)p = 1264 STB/day Nilai Tetap
Generate the current IPR data by using Wiggins method and
compare the results with those of Vogels.
Pwf, psig
Wiggins Vogels
2500 0 0 Vogel's Method
2200 216 218
2000 350 350
1500 651 6321000 904 845
500 1109 990
0 1264 1067
b. Menghitung Qo max yang akan datang / Future dgn menggunakan Wiggins method
Qo, STB/day
2
Pr48.0
Pr52.01max
PwfPwfQoQo
2
Pr
48.0
Pr
52.01
max)(PwfPwf
QopQo
.0Pr2.01(max)
PwfQoQo
5
10
15
20
25
30
Pwf,Psig
2
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(Qo max)f = 989 STB/day Nilai Tetap
Pwf, psig Qo, STB/dayWiggins
2200 0
2000 129
1500 418
1000 657
500 848
0 989
Example 7-7
A well is producing from a saturated oil reservoir that exists at its saturation
pressure of 4000 psig. The well is flowing at a stabilized rate of
600 STB/day and a pwf of 3200 psig. Material balance calculations provide
the following current and future predictions for oil saturation and
PVT properties.
Present Future
Pr 4000 3000
o, cp 2.4 2.2
Bo, bbl/STB 1.2 1.15
kro 1 0.66
Generate the future IPR for the well at 3000 psig by using Standings
method.
Diket :
Qo = 600 STB/day
Pwf = 3200 Psig
Ditanya :
- Future IPR @ Pr 3000 psig
Jawab :
(Qo)max = 1829 STB/day
Jp* = 0.823 STB/day / psi
Pr
r84.0
Pr
r15.0maxmax)(
pppQofQo
2
Pr48.0
Pr52.01max
PwfPwfQoQo
Pr8.01Pr1max
PwfPwf
Qo
Qo
Pr
max8.1*Qo
Jp
1500
2000
2500
3000
3500
Pwf,psig
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Jf* = 0.618 STB/day / psi
Generate the future IPR
Pwf, psig Qo, STB/day
3000 0
2500 286
2000 527
1500 722
1000 870
500 9730 1031
Example 7-8
A well is producing from an undersaturated-oil reservoir that exists at
an average reservoir pressure of 3000 psi. The bubble-point pressure is
recorded as 1500 psi at 150F. The following additional data are available:
stabilized flow rate = 280 STB/day
stabilized wellbore pressure = 2200 psi
h = 20 rw = 0.3 re = 660 s = -0.5
k = 65 md
mo at 2600 psi = 2.4 cp
Bo at 2600 psi = 1.4 bbl/STB
Calculate the productivity index by using both the reservoir properties
and flow test data
Diket :
Undersaturated-Oil Reservoir
Pr = 3000 psi k = 65 md
Pb = 1500 psi o @ 2600 psi = 2.4 cp
Tb = 150 F o @ 2600 psi = 1.4 bbl/STB
Qo = 280 STB/day
Pwf = 2200 psi
h = 20
rw = 0.3re = 660
s = -0.5
Ditanya :
- Productivity Index
- Flow Test Data
Jawab :
p
f
oo
Kro
oo
Kro
JpJf
**
2
Pr8.0
Pr2.01
8.1
Pr*
ff
f PwfPwfJfQo
0
500
1000
0 200
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Fetkovich's Method
J = 0.42 STB/day / psi
From Production Data :
J = 0.35 STB/day / psi
Example 7-9
A four-point stabilized flow test was conducted on a well producing
from a saturated reservoir that exists at an average pressure of 3600 psi.
Qo, STB/day Pwf, psig263 3170
383 2890
497 2440
640 2150
a. Construct a complete IPR by using Fetkovichs method.
b. Construct the IPR when the reservoir pressure declines to 2000 psi.
Diket :
Pr = 3600 psi
Ditanya :
a. IPR - Fetkovich's Methodb. IPR @ Pr = 2000 psi
Jawab :
a. Qo, STB/day Pwf, psig (Pr2
- Pwf2)10
-6psi
2
263 3170 2.9111
383 2890 4.6079
497 2440 7.0064
640 2150 8.3375
3600 12.96
n = 0.854
AOF = 940 STB/day
Srw
reoo
khJ
75.0ln
00708.0
)(Pr Pwf
qJ
10
100
(Pr2-
Pwf2)x1
0-6psi2
36002
67
10log10log
105log750log
n
nPwfQo
C22Pr
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Cp = 0.0007924
IPR using Fetkovich Method
Pwf, psig
3600 0
3000 3402500 534
2000 684
1500 796
1000 875
500 922
0 937
b. Diket :
(Pr)f = 2000 psi
Cf = 0.0004402
Pwf, psi Qo, STB/day
2000 0
1500 94
1000 150
500 181
0 191
Example 7-10
Using the data given in Example 7-9, generate the future IPR data
when the reservoir pressure drops to 3200 psi.
Diket :
nb = 0.854 Pr = 3200 psi
Cb = 0.0007924
Pb = 3600 psi
n/nb = Dimensionless Flow Exponent
nb = Flow exponent Bubble Point Pressure
Qo, STB/day = 0.00079 (Pr2
- Pwf2)0.854
1100
0
500
1000
1500
2000
2500
3000
3500
4000
0 200
Pressure(Pwf),psig
p
fCpCf
Pr
Pr
854.022Pr PwfCfQo
32Pr
1503.0Pr
12459.0Pr
10577.01
PbPbPbnb
n
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n/nb = 1.0041
C/Cb = 0.6592
Solve for future values of nf and Cf from
nf = 0.8575
Cf = 0.00052
Flow rate is expressed as :
(Qo)max = 535.94 STB/day
Pwf, Psi Qo, STB/day
3200 0
2000 350.48
1500 433.23
1000 490.74
500 524.70
0 535.94
Example 7-11
The following reservoir and flow-test data are available on an oil well:
Pressure data :
Pr = 4000 psi ; Pb = 3200 psi
Flow test data: pwf = 3600 psi Qo = 280 STB/day
Generate the IPR data of the well.
32Pr
13066.2Pr
17981.4Pr
15718.31
PbPbPbCb
C
nb
nnbnf
Cb
C
CbCf
fnCfQo 2Prmax
fnPwfCfQo 22Pr 0
500
1000
1500
2000
2500
3000
3500
0 10
Pressure(Pwf),
Psi
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Diket :
Pr = 4000 psi Pr > Pb
Pb = 3200 psi Pr > Pwf
Pwf = 3600 psi
Qo = 280 STB/day
Ditanya :- Data IPR Sumur
Jawab :
J = 0.7 STB/day/psi Nilai Tetap
Jika Pwf > Pb menggunakan Rumus :
Jika Pwf < Pb menggunakan Rumus :
Pwf, psi Qo, STB/day
4000 0
3800 140
3600 2803400 420
3200 560
3000 696
2800 823
2600 941
2200 1151
2000 1243
1000 1571
500 1653
0 1680
)(Pr Pwf
qJ
)36004000(
280
J
PwfJQo Pr
222
Pr PwfPbPb
JPbJQo
0
500
1000
1500
2000
2500
3000
3500
4000
4500
0 200 400 600 800 1000
Pwf,psi
Qo, STB/day
WELL PERFORM
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Q
0
82.5
137.5
192.5
247.5
302.5357.5
82.5, 1000
137.5, 800
192.5, 600
247.5, 400
302.5, 200
IPR
IPR
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357.5, 0100 200 300 400
Q (Flow Rate, STB/day)
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1000 1500 2000
Qo, STB/day
IPR
Vogel's Method
Linear
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STB/day
IPR
0 400 600 800 1000 1200
Qo, STB/day
IPR
Pwf > Pb
Pwf < Pb
Qob
Pwf > Pb
Pwf < Pb
Qo max
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2500
21302000
1500
1000
500
000 400 600 800 1000 1200
Qo, STB/day
IPR
Pwf > Pb
Pwf < Pb
Qob
2200
2000
1500
1000
500
00
500
1000
1500
2000
2500
0 200 400 600 800 1000
Pwf,Psig
Qo, STB/day
IPR
IP
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2
Pr8Pwf
0
00
00
00
00
00
00
0 500 1000 1500
Qo, STB/day
Current IPR
Wiggin's Method
Vogel's Method
2500
Future IPR
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,
129, 2000
418, 1500
657, 1000
848, 500
989, 00
500
1000
1500
2000
0 200 400 600 800 1000 1200
Pw
f,Psig
Qo, STB/day
Wiggin's M
Future IPR
Standing's Method
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400 600 800 1000 1200
Qo, STB/day
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Plot (Pr2 - Pwf2) VS Qo
Plot
Linear (Plot)
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1000
Qo, STB/day
AOF = 940750
400 600 800 1000
Flow Rate (Qo), STB/day
IPR using Fetkovich method
Series1
Series2
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200 300 400 500 600
Flow Rate (Qo), STB/day
IPR
Series1
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1200 1400 1600 1800
NCE
IPR
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R
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ethod