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7/30/2019 inflow performance "IPR"
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University of Kirkuk
College of Engineering / Pet. Eng. Department
Production Engineering
"Inflow Performance"
- Home work -
Hawar Abdul-khaliq
4th stage
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7/30/2019 inflow performance "IPR"
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H.W.
construct the IPR of a vertical well in an oil reservoir
considers:
1.Transient flow at 1 month .2.Steady state flow.
3.Pseudo steady state flow.
The following data are given:
K=8.2 md. , =19% ,h=53 ft , Pe=5651 psi , Pb=50 psi , Bo=1.1
=1.7 cp. , A=640 acre. , re=2980 ft , rw= 0.328 ft , S=0
Ct=1.29*10
-5
psi
-1
Solution:
time = 30 day = 720 hr.
1.q= ( )
= 1161.98 bbl/day
J= = 0.20745 bbl/day/psi
2.q= * + = 1011.46 bbl/day
J=
= 0.1806 bbl/day/psi
3.q= = 1102.2 bbl/day
J=
= 0.19678 bbl/day/psi
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pwf (psi) q)1 q)2 q)3
0 1466.051 1020.492 1111.995200 1414.164 984.3749 1072.639
400 1362.278 948.2577 1033.284600 1310.391 912.1405 993.928800
1258.505 876.0233 954.57231000 1206.619 839.906 915.21671200
1154.732 803.7888 875.8611400 1102.846 767.6716 836.50531600
1050.959 731.5544 797.14961800 999.0729 695.4372 757.79392000
947.1865 659.3199 718.43822200 895.3 623.2027 679.08252400 843.4136
587.0855 639.72682600 791.5272 550.9683 600.37112800 739.6408
514.851 561.01543000 687.7544 478.7338 521.65973200 635.868
442.6166 482.3043400 583.9816 406.4994 442.94833600 532.0952
370.3821 403.59263800 480.2087 334.2649 364.23694000 428.3223
298.1477 324.88124200 376.4359 262.0305 285.52564400 324.5495
225.9132 246.16994600 272.6631 189.796 206.81424800 220.7767
153.6788 167.4585
5000168.8903 117.5616 128.1028
5200 117.0039 81.44434 88.747095400 65.11745 45.32712
49.39145600 13.23104 9.209892 10.03575651 0 0 0
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Transient flow
Steady state flow
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0500100015002000
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q
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020040060080010001200
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Pseudo steady state flow
* according the productivity index J and IPR , the best flow
in
this reservoir is transient flow after this flow the
pseudo-steady
state is the best , and steady-state flow .
* the formula of the eq. of the transient flow depends on time
(as
time increases J decreases ).
* when the pressure pwf below bubble point pb, the IPR appearas
a curve and J constant because the reservoirat the pressure
below pb is two-phase flow .
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020040060080010001200
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