1
UNIVERSITY OF KIRKUK
College of Engineering - Pet. Eng. Department
PETROLEUM
PRODUCTION
ENGINEERING BOYUN GUO – WILLIAM C. LYONS – ALI GHALAMBOR
CHPTER -3-
RESERVOIR DELIVERABILITY
Solution of Problems page 3/43
By: Hawar AbdulKhaliq Hamma Gul
2
3.1 Construct IPR of a vertical well in an oil reservoir.
Consider (1) transient flow at 1 month , (2) steady – state flow , and (3) pseudo –
steady state flow. the following data are given:
Porosity , ϕ = 0.25
Effective horizontal permeability , k = 10 md
Pay zone thickness , h = 50 ft
Reservoir pressure , pe = 5000 psi
Bubble point pressure , pb = 100 psi
Fluid formation volume factor , Bo = 1.2
Fluid viscosity , µo = 1.5 cp.
Total compressibility , Ct = 0.0000125 psi-1
Drainage area , A = 640 acre (re =2.980 ft)
Wellbore radius , rw = 0.328 ft
Skin factor , S = 5
Solution :
time = 30 day = 24 x 30 = 720 hr.
1. For transient flow,
J*=
(
)
= ( )( )
( )( ) ( ) (
( )( )( )( ) ) ( )
J* = 0.1515215 bbl/day/psi
3
Calculated data points are:
qo = J* (Pe – pwf)
Pwf (psi) qo (stb/day)
0 757.6075
1000 606.086
2000 454.5645
3000 303.043
4000 151.5215
5000 0
"Transient IPR curve"
0
1000
2000
3000
4000
5000
6000
0100200300400500600700800
pw
f (p
si)
qo (stb/day)
4
2. For steady state flow,
J*=
* (
) +
= ( )( )
( )( ) * (
) +
J* = 0.13938 bbl/day/psi
Calculated points are:
Pwf (psi) qo (stb/day)
0 696.9
1000 557.52
2000 418.14
3000 278.76
4000 139.38
5000 0
"Steady state IPR curve"
0
1000
2000
3000
4000
5000
6000
0100200300400500600700800
pw
f (p
si)
qo (stb/day)
5
3. For pseudo steady state,
J*=
(
)
= ( )( )
( )( ) (
)
J
*= 0.147202 bbl/day/psi
Calculated points are:
Pwf (psi) qo (stb/day)
0 736.01
1000 588.808
2000 441.8
3000 294.404
4000 147.202
5000 0
Pseudo-steady state IPR curve""
0
1000
2000
3000
4000
5000
6000
0100200300400500600700800
pw
f (p
si)
qo (stb/day)
6
saturated oil reservoir Construct IPR of a vertical well in an .23
Using Vogel's equation .the following data are given:
Porosity , ϕ = 0.20
Effective horizontal permeability , k = 80 md
Pay zone thickness , h = 55 ft
Reservoir pressure , pe = 4500 psi
Bubble point pressure , pb = 4500 psi
Fluid formation volume factor , Bo = 1.1
Fluid viscosity , µo = 1.8 cp.
Total compressibility , Ct = 0.000013 psi-1
Drainage area , A = 640 acre (re =2.980 ft)
Wellbore radius , rw = 0.328 ft
Skin factor , S = 2
Solution :
Assume pseudo-steady state flow ,
J*=
(
)
= ( )( )
( )( ) (
)
1.518476 STB/day/psi=
q max =
= ( )( )
= 3796.2 STB/day.
7
Calculated points by Vogel's equation:
q = q max [ (
) (
) ]
Pwf (psi) qo (stb/day)
0 3796.2
1000 3477.507
2000 2858.867
3000 1940.28
4000 721.7467
4500 0
IPR curve""
0
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
05001000150020002500300035004000
pw
f (p
si)
qo (stb)
8
unsaturated oil reservoir Construct IPR of a vertical well in an .33
Using generalized Vogel's equation .the following data are given:
Porosity , ϕ = 0.25
Effective horizontal permeability , k = 100 md
Pay zone thickness , h = 55 ft
Reservoir pressure , pe = 5000 psi
Bubble point pressure , pb = 3000 psi
Fluid formation volume factor , Bo = 1.2
Fluid viscosity , µo = 1.8 cp.
Total compressibility , Ct = 0.000013 psi-1
Drainage area , A = 640 acre (re =2.980 ft)
Wellbore radius , rw = 0.328 ft
Skin factor , S = 5.5
Solution :
Assume pseudo-steady state flow ,
J*=
(
)
= ( )( )
( )( ) (
)
J* = 1.300687 STB/day/psi
qb = J* (Pe – Pb)
= (1.300687) (5000 – 3000) = 2601.4 STB/day.
qv =
= (1.300687) (3000) / 1.8 = 2167.8 STB/day.
9
Calculated points by:
q = J* (Pe – Pb) +
[ (
) (
) ]
Pwf (psi) qo (stb/day)
0 4769.2
500 4648.767
1000 4431.987
1500 4118.86
2000 3709.387
2500 3203.567
3000 2601.4
5000 0
0
1000
2000
3000
4000
5000
6000
0100020003000400050006000
pw
f (p
si)
qo (stb)
10
unsaturated oil reservoir Construct IPR of a vertical well in an .43
Using generalized Vogel's equation .the following data are given:
Reservoir pressure , pe = 5500 psi
Bubble point pressure , pb = 3500 psi
Tested following bottom-hole pressure in well A , pwf 1 = 4000 psi
Tested production rate from well A , q1 = 400 stb / day
Tested following bottom-hole pressure in well B, pwf 1 = 2000 psi
Tested production rate from well B , q1 = 1000 stb/day
Solution :
Well A : Pwf1 > pb
J* =
( )
= 400 / (5500 - 4000) = 0.2667 stb/day/psi.
qb = J* (Pe – Pb)
= 0.2667 (5500 – 3500) = 533.4 stb/day
qv =
= (0.2667) (3500) / 1.8 = 518.6 stb/day
Calculated points by:
q = J* (Pe – Pb) +
[ (
) (
) ]
11
Pwf (psi) qo (stb/day)
0 1052
500 1028.716
1000 988.498
1500 931.3461
2000 857.2604
2500 766.2408
3000 658.2873
3500 533.4
5000 0
0
1000
2000
3000
4000
5000
6000
020040060080010001200
pw
f (p
si)
qo (stb/day)
12
Well B : Pwf1 < pb
J* =
(( )
[ (
) (
) ])
=
(( )
[ (
) (
) ])
= 0.3111 stb/day/psi
qb = J* (Pe – Pb) = 0.3111 (5500 – 3500) = 622.2 stb/day
qv =
= (0.3111) (3500) / 1.8 = 604.92 stb/day
Calculated points by:
qo= J* (Pe – Pb) +
[ (
) (
) ]
qo = 622.2 + .92 [ (
) (
) ]
Pwf (psi) qo (stb/day)
0 1227.12
500 1199.96
1000 1153.048
1500 1086.384
2000 999.9664
2500 893.7967
3000 767.8746
3500 622.2
5000 0
13
0
1000
2000
3000
4000
5000
6000
0200400600800100012001400
pw
f (p
si)
qo (stb/day)
14
3.5 Construct IPR of a well in saturated oil reservoir using both Vogel's equation
and Fetkovich's equation .
The following data are given:
Reservoir pressure , pe = 3500 psi
Bubble point pressure , pb = 3500 psi
Tested following bottom-hole pressure , pwf 1 = 2500 psi
Tested production rate at pwf 1 , q1 = 600 stb / day
Tested following bottom-hole pressure , pwf 2 = 1500 psi
Tested production rate at pwf 2, q2 = 900 stb/day
Solution:
Vogel's equation: qomax.=
(
) (
)
qomax.=
(
) (
)
calculated data points are:
Pwf (psi) qo (stb/day)
0 602.46
500 575.4108
1000 528.6894
1500 462.2958
2000 376.2301
2500 270.4922
3000 145.0822
3500 0
15
Fetkovich's equation:
n = (
)
(
)
(
)
(
)
c =
( )
( )
calculated data points are:
q= 0.0023 ( )
Pwf (psi) qo (stb/day)
0 1077.052
500 1059.431
1000 1006.12
1500 915.6465
2000 785.0386
2500 608.4835
3000 372.5929
3500 0
0
500
1000
1500
2000
2500
3000
3500
4000
020040060080010001200
qo (stb/day)
pwf(fetkovich's model)
pwf(vogel's model)
16
3.6 Determine the IPR for a well at the time when the average reservoir pressure
will be 1500 psig. The following data are obtained from laboratory tests of well fluid
samples:
future present Reservoir pressure
1500 2200 Average pressure (psi)
------- 1.25 Productivity index J* (stb/day-psi)
3.85 3.55 Oil viscosity (cp)
1.15 1.20 Oil formation volume factor (rb/stb)
0.65 0.82 Relative permeability to oil
Solution:
J*f =
(
)
(
)
(
)
(
)
9534 stb/day-psi
Vogel's equation for future IPR :
q = ( )( )
[ (
) (
) ]
( )( )
* (
) (
)
+
Vogel's equation for present IPR :
q = ( )( )
[ (
) (
) ]
( )( )
* (
) (
)
+
Calculated data points are:
17
Reservoir press. = 1500 psi Reservoir press. = 2200 psi
q (stb/day) Pwf (psi) q (stb/day) Pwf (psi) 0 1500 0 2200
136.654 1350 262.7782 1980
260.596 1200 501.1118 1760
371.826 1050 715.001 1540
470.344 900 904.4458 1320
556.15 750 1069.446 1100
629.244 600 1210.002 880
689.626 450 1326.113 660
737.296 300 1417.78 440
772.254 150 1485.002 220
794.5 0 1527.78 0
*in the above table, col. 2 calculated from eq.2 & col.4 from eq.1
0
500
1000
1500
2000
2500
0500100015002000
qo (stb/day)
pwf (present)
pwf (future)
18
3.7 Using Fetcovich's method , plot the IPR curve for a well in which
pi is 3000 psia and J'o = 4×10
-4 stb/day-psi . Predict the IPRs of the
well at well shut-in static pressures of 2500 psia, 2000 psia, 1500 psia,
and 1000 psia.
Solution:
The value of J'O at 2500 psia is:
J'O = J
'i
( )
J'O = 4*10
-4 (
) ( )
The value of J'O at 2000 psia is:
J'O = 4*10
-4 (
) ( )
and the value of J'O at 1500 psia is:
J'O = 4*10
-4 (
) ( )
and the value of J'O at 1000 psia is:
J'O = 4*10
-4 (
) ( )
Calculated data points are:
qo = J'O ( ) ( )
19
Pe = 2000 psi Pe = 2500 psi Pe = 3000 psi
q(stb/day) pwf (psi) q(stb/day) pwf (psi) q(stb/day) pwf (psi) 0 2000 0 2500 0 3000
456 1800 570 2250 684 2700
864 1600 1080 2000 1296 2400
1224 1400 1530 1750 1836 2100
1536 1200 1920 1500 2304 1800
1800 1000 2250 1250 2700 1500
2016 800 2520 1000 3024 1200
2184 600 2730 750 3276 900
2304 400 2880 500 3456 600
2376 200 2970 250 3564 300
2400 0 3000 0 3600 0
Pe = 1000 psi Pe = 1500 psi
q(stb/day) pwf (psi) q(stb/day) pwf (psi) 0 1000 0 1500
228 900 342 1350
432 800 648 1200
612 700 918 1050
768 600 1152 900
900 500 1350 750
1008 400 1512 600
1092 300 1638 450
1152 200 1728 300
1188 100 1782 150
1200 0 1800 0
20
"IPR curve , problem 3.7"
0
500
1000
1500
2000
2500
3000
3500
05001000150020002500300035004000
pw
f (p
si)
qo (stb/day)
(pe=3000 psi)
(pe=2500 psi)
(pe-1000 psi)
(pe=2000 psi)
(pe=1500 psi)
