Contoh Soal "Vertical Oil Well Performance"

8/3/2019 Contoh Soal "Vertical Oil Well Performance"

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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

18501850


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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


7/31

(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

Contoh Soal "Vertical Oil Well Performance"

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