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● Steady-state IPR: Assume constant outer boundary pressure, pe. Draw IPR curves for skin effects equal to 0, 5, 10 and
Posted: Thu May 05, 2022 5:52 pm
by answerhappygod
- Steady State Ipr Assume Constant Outer Boundary Pressure Pe Draw Ipr Curves For Skin Effects Equal To 0 5 10 And 1 (256.15 KiB) Viewed 32 times
● Steady-state IPR: Assume constant outer boundary pressure, pe. Draw IPR curves for skin effects equal to 0, 5, 10 and 50, respectively. Use a drainage radius of 2980 ft (A=640 acres). kμ = 8.2 md ky = 0.9 md h = 53 ft Pe - Pwf = 141.2 q Bu re kh In +5 rw Pi = 5651 psi When skin factor is 5, becomes: Pb = 1323 psi Co=1.4 x 10-5 psi-¹ Cw = 3 x 10-6 psi-1 Cf 2.8 x 10-6 Similarly, the multiplier of the rate for 0, 10 and 50 skin effects are 5.54, 11.6, and 35.9, respectively. psi-1 C₂ = 1.29 x 10-5 psi-¹ Pwf = 5651 - 8.589 6000 μ = 1.7 cp B 1.1 res bbl/STB 5000 R, 150 SCF/STB 0.19 4000 Sw = 0.34 API° = 28 3000 rw = 0.328 ft (7 7/8 well) 2000 1000 S = 10 S=5 S=0 0 1000 1200 Pwt (psi) 0 S = 50 200 400 600 q (STB/d) 800