Exercise: Develop an IPR curve for the well described below. The drainage radius is 1490 ftand the skin effect is zero.1.6k = 13 md (from cores)1.5köhỹ[1 – 0.2(Pwƒ/p) – 0.8(Pwƒ/P)²]254.2 Boo[In(0.472re/rw)+s]90 ==h = 115 ft1.4Pi = 4350 psiPb = 4350 psi1.31.21.1When p = 4350 psi,Co1.2 x 10-5Cw =3 x 10-61.00400050003000Pressure (psi)Psi-1Psi-1Cf = 3.1 x 10-690 = 1411. 1-0.2 - 0.81411 [1- 0.8 (Por)Ppc₁ = 1.25 × 10-5 psi-¹xF = 1.7 cpHo5000Yo = 32° APIYg = 0.714000T = 180°F30005000Tpc = 395°RPpc = 667 psi2000 3000Pressure (psi)2000Sw = 0.310000500020003000 4000Pressure (psi)$ = 0.21rw = 0.406 ftBo (res BBI/STB)R$ (SCF/STB)0.0200.0160.0120.0080.0040.0001000.0B. (res ft³/SCF)800.0600.0400.0200.00.00010001000100020004000Flowing Bottomhole Pressure. p (psi)p=4350050015001000Flowrate, q (STB/d)The IPR curve is no longer straight line!

Interproximal reduction (IPR) (also known as interproximal enamel reduction (IER) / slendering, air…

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)

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter10: Heat Exchangers

Section: Chapter Questions

Problem 10.19P

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