Exercise: Develop an IPR curve for the well described below. The drainage radius is 1490 ft and the skin effect is zero. 1.6 kH = 13 md (from cores) koh p[1 – 0.2(Puf/ P) – 0.8(Puf/P)²] 254.2B.HO[In(0.472re/rw) + s] 1.5 h = 115 ft 1.4 Pi = 4350 psi Рь 3 4350 psi 1.3 When p = 4350 psi, 1.2 Co = 1.2 × 10-5 Cu = 3 x 106 psi- 1.1 1.0 !- 0,2 Pws 0.8 1000 2000 3000 4000 5000 Cf = 3.1 × 10-6 Psi-1 Pressure (psi) 90 = 1411. Pwf C; = 1.25 × 10-5 psi- 0.020 0.016 Ho = 1.7 cp 5000 0.012 Yo = 32° API 0.008 Ys = 0.71 4000 T = 180°F 0.004 Tpc = 395°R Тре 0.000 3000 1000 2000 3000 4000 5000 D= 4350 Pressure (psi) Ppc = 667 psi Sy = 0.3 2000 1000.0 $ = 0.21 800.0 %3D 1000 rw = 0.406 ft 600.0 400.0 200.0 500 1000 1500 Flowrate, q, (STB/d) 0.0 The IPR curve is no longer straight line! 1000 2000 3000 Pressure (psi) 4000 5000 Rs(SCF/STB) 3, (res t/sCF) , (res BBI/STB) Flowing Bottomhole Pressure, Pw (psi)
Exercise: Develop an IPR curve for the well described below. The drainage radius is 1490 ft and the skin effect is zero. 1.6 kH = 13 md (from cores) koh p[1 – 0.2(Puf/ P) – 0.8(Puf/P)²] 254.2B.HO[In(0.472re/rw) + s] 1.5 h = 115 ft 1.4 Pi = 4350 psi Рь 3 4350 psi 1.3 When p = 4350 psi, 1.2 Co = 1.2 × 10-5 Cu = 3 x 106 psi- 1.1 1.0 !- 0,2 Pws 0.8 1000 2000 3000 4000 5000 Cf = 3.1 × 10-6 Psi-1 Pressure (psi) 90 = 1411. Pwf C; = 1.25 × 10-5 psi- 0.020 0.016 Ho = 1.7 cp 5000 0.012 Yo = 32° API 0.008 Ys = 0.71 4000 T = 180°F 0.004 Tpc = 395°R Тре 0.000 3000 1000 2000 3000 4000 5000 D= 4350 Pressure (psi) Ppc = 667 psi Sy = 0.3 2000 1000.0 $ = 0.21 800.0 %3D 1000 rw = 0.406 ft 600.0 400.0 200.0 500 1000 1500 Flowrate, q, (STB/d) 0.0 The IPR curve is no longer straight line! 1000 2000 3000 Pressure (psi) 4000 5000 Rs(SCF/STB) 3, (res t/sCF) , (res BBI/STB) Flowing Bottomhole Pressure, Pw (psi)
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![Exercise: Develop an IPR curve for the well described below. The drainage radius is 1490 ft
and the skin effect is zero.
1.6
k = 13 md (from cores)
1.5
köhỹ[1 – 0.2(Pwƒ/p) – 0.8(Pwƒ/P)²]
254.2 Boo[In(0.472re/rw)+s]
90 ==
h = 115 ft
1.4
Pi = 4350 psi
Pb = 4350 psi
1.3
1.2
1.1
When p = 4350 psi,
Co
1.2 x 10-5
Cw =
3 x 10-6
1.0
0
4000
5000
3000
Pressure (psi)
Psi-1
Psi-1
Cf = 3.1 x 10-6
90 = 1411. 1-0.2 - 0.8
1411 [1
- 0.8 (Por)
P
p
c₁ = 1.25 × 10-5 psi-¹
x
F = 1.7 cp
Ho
5000
Yo = 32° API
Yg = 0.71
4000
T = 180°F
3000
5000
Tpc = 395°R
Ppc = 667 psi
2000 3000
Pressure (psi)
2000
Sw = 0.3
1000
0
5000
2000
3000 4000
Pressure (psi)
$ = 0.21
rw = 0.406 ft
Bo (res BBI/STB)
R$ (SCF/STB)
0.020
0.016
0.012
0.008
0.004
0.000
1000.0
B. (res ft³/SCF)
800.0
600.0
400.0
200.0
0.0
0
0
1000
1000
1000
2000
4000
Flowing Bottomhole Pressure. p (psi)
p=4350
0
500
1500
1000
Flowrate, q (STB/d)
The IPR curve is no longer straight line!](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb492b9f4-e556-4f7b-a8d5-7076ae35298b%2F74a8cccf-9808-4673-97c0-82a40828f532%2Fhdh6r3q_processed.png&w=3840&q=75)
Transcribed Image Text:Exercise: Develop an IPR curve for the well described below. The drainage radius is 1490 ft
and the skin effect is zero.
1.6
k = 13 md (from cores)
1.5
köhỹ[1 – 0.2(Pwƒ/p) – 0.8(Pwƒ/P)²]
254.2 Boo[In(0.472re/rw)+s]
90 ==
h = 115 ft
1.4
Pi = 4350 psi
Pb = 4350 psi
1.3
1.2
1.1
When p = 4350 psi,
Co
1.2 x 10-5
Cw =
3 x 10-6
1.0
0
4000
5000
3000
Pressure (psi)
Psi-1
Psi-1
Cf = 3.1 x 10-6
90 = 1411. 1-0.2 - 0.8
1411 [1
- 0.8 (Por)
P
p
c₁ = 1.25 × 10-5 psi-¹
x
F = 1.7 cp
Ho
5000
Yo = 32° API
Yg = 0.71
4000
T = 180°F
3000
5000
Tpc = 395°R
Ppc = 667 psi
2000 3000
Pressure (psi)
2000
Sw = 0.3
1000
0
5000
2000
3000 4000
Pressure (psi)
$ = 0.21
rw = 0.406 ft
Bo (res BBI/STB)
R$ (SCF/STB)
0.020
0.016
0.012
0.008
0.004
0.000
1000.0
B. (res ft³/SCF)
800.0
600.0
400.0
200.0
0.0
0
0
1000
1000
1000
2000
4000
Flowing Bottomhole Pressure. p (psi)
p=4350
0
500
1500
1000
Flowrate, q (STB/d)
The IPR curve is no longer straight line!
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