UNIVERSITY OF KIRKUK College of Engineering - Pet. Eng. Department
PETROLEUM PRODUCTION ENGINEERING BOYUN GUO – WILLIAM C. LYONS – ALI GHALAMBOR
CHPTER -3RESERVOIR DELIVERABILITY
Solution of Problems page 3/43 By: Hawar AbdulKhaliq Hamma Gul
1
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
2
)(
)
)
( )
Calculated data points are: qo = J* (Pe – pwf)
Pwf (psi) 0 1000 2000 3000 4000 5000
qo (stb/day) 757.6075 606.086 454.5645 303.043 151.5215 0
6000
5000
3000
2000
1000
0 800
700
600
500
400
300
qo (stb/day)
200
"Transient IPR curve"
3
100
0
pwf (psi)
4000
2. For steady state flow,
J*=
* (
)
+
(
=
(
)(
)(
)
)* (
)
+
J* = 0.13938 bbl/day/psi Calculated points are:
Pwf (psi) 0 1000 2000 3000 4000 5000
qo (stb/day) 696.9 557.52 418.14 278.76 139.38 0 6000
4000
3000 2000 1000 0 800
700
600
500
400
300
200
100
qo (stb/day)
"Steady state IPR curve" 4
0
pwf (psi)
5000
3. For pseudo steady state,
J*=
(
(
=
)
(
)(
)( )
) (
)
J*= 0.147202 bbl/day/psi
Calculated points are: Pwf (psi) 0 1000 2000 3000 4000 5000
qo (stb/day) 736.01 588.808 441.8 294.404 147.202 0 6000
5000
3000
2000
1000
0 800
700
600
500
400
300
200
100
qo (stb/day)
"Pseudo-steady state IPR curve" 5
0
pwf (psi)
4000
3.2 Construct IPR of a vertical well in an
saturated oil reservoir
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.
6
)( )
) (
)
Calculated points by Vogel's equation:
q = q max [
(
Pwf (psi) 0 1000 2000 3000 4000 4500
)
(
) ]
qo (stb/day) 3796.2 3477.507 2858.867 1940.28 721.7467 0 5000 4500 4000
3000 2500
2000 1500 1000 500 0 4000
3500
3000
2500
2000
1500
1000
qo (stb)
"IPR curve"
7
500
0
pwf (psi)
3500
3.3
Construct IPR of a vertical well in an unsaturated oil reservoir
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.
8
(
) )
Calculated points by: [
q = J* (Pe – Pb) +
Pwf (psi) 0 500 1000 1500 2000 2500 3000 5000
(
)
(
) ]
qo (stb/day) 4769.2 4648.767 4431.987 4118.86 3709.387 3203.567 2601.4 0
6000
4000 3000
2000 1000 0 6000
5000
4000
3000
2000
qo (stb)
9
1000
0
pwf (psi)
5000
3.4
Construct IPR of a vertical well in an unsaturated oil reservoir
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) +
[
(
10
)
(
) ]
Pwf (psi) 0 500 1000 1500 2000 2500 3000 3500 5000
qo (stb/day) 1052 1028.716 988.498 931.3461 857.2604 766.2408 658.2873 533.4 0 6000 5000
3000 2000 1000 0 1200
1000
800
600
400
qo (stb/day)
11
200
0
pwf (psi)
4000
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 +
Pwf (psi) 0 500 1000 1500 2000 2500 3000 3500 5000
.92
[
(
[
(
)
qo (stb/day) 1227.12 1199.96 1153.048 1086.384 999.9664 893.7967 767.8746 622.2 0
12
(
)
) ]
(
) ]
6000 5000
3000 2000 1000 0
1400
1200
1000
800
600
qo (stb/day)
13
400
200
0
pwf (psi)
4000
3.5 Construct IPR of a well in saturated oil reservoir using both 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) 0 500 1000 1500 2000 2500 3000 3500
qo (stb/day) 602.46 575.4108 528.6894 462.2958 376.2301 270.4922 145.0822 0
14
)
Vogel's equation
Fetkovich's equation: (
n= c=
)
(
(
(
)
)
)
(
)
(
)
calculated data points are: q= 0.0023 (
)
Pwf (psi) 0 500 1000 1500 2000 2500 3000 3500
qo (stb/day) 1077.052 1059.431 1006.12 915.6465 785.0386 608.4835 372.5929 0 4000 3500 3000 2500 2000 1500 1000 500 0
1200
1000
800
600
400
200
qo (stb/day)
15
0
pwf(fetkovich's model) pwf(vogel's model)
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: Reservoir pressure Average pressure (psi) Productivity index J* (stb/day-psi) Oil viscosity (cp) Oil formation volume factor (rb/stb) Relative permeability to oil
present 2200 1.25 3.55 1.20 0.82
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:
16
) ] (
) +
future 1500 ------3.85 1.15 0.65
Reservoir press. = 2200 psi Pwf (psi) q (stb/day) 2200 1980 1760 1540 1320 1100 880 660 440 220 0
Reservoir press. = 1500 psi Pwf (psi) q (stb/day)
0 262.7782 501.1118 715.001 904.4458 1069.446 1210.002 1326.113 1417.78 1485.002 1527.78
1500 1350 1200 1050 900 750 600 450 300 150 0
0 136.654 260.596 371.826 470.344 556.15 629.244 689.626 737.296 772.254 794.5
2500
2000
1500
pwf (present) 1000
500
0 2000
1500
1000
500
0
qo (stb/day)
*in the above table, col. 2 calculated from eq.2 & col.4 from eq.1
17
pwf (future)
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 (
)
(
)
18
)
Pe = 3000 psi Pe = 2500 psi Pe = 2000 psi pwf (psi) q(stb/day) pwf (psi) q(stb/day) pwf (psi) q(stb/day) 3000 2700 2400 2100 1800 1500 1200 900 600 300 0
0 684 1296 1836 2304 2700 3024 3276 3456 3564 3600
2500 2250 2000 1750 1500 1250 1000 750 500 250 0
0 570 1080 1530 1920 2250 2520 2730 2880 2970 3000
2000 1800 1600 1400 1200 1000 800 600 400 200 0
0 456 864 1224 1536 1800 2016 2184 2304 2376 2400
Pe = 1500 psi Pe = 1000 psi pwf (psi) q(stb/day) pwf (psi) q(stb/day) 1500 1350 1200 1050 900 750 600 450 300 150 0
0 342 648 918 1152 1350 1512 1638 1728 1782 1800
19
1000 900 800 700 600 500 400 300 200 100 0
0 228 432 612 768 900 1008 1092 1152 1188 1200
3500
3000
2000
1500
pwf (psi)
2500
(pe=3000 psi) (pe=2500 psi) (pe-1000 psi) (pe=2000 psi)
1000
500
0 4000
3500
3000
2500
2000
1500
1000
500
0
qo (stb/day)
"IPR curve , problem 3.7"
20
(pe=1500 psi)
