Petroleum Production Engineering 343e17
This document was ed by and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this report form. Report 3b7i
Overview 3e4r5l
& View Petroleum Production Engineering as PDF for free.
More details w3441
- Words: 1,802
- Pages: 20
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 . 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 . 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 . 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 () 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)
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 . 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 . 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 . 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 () 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)
Related Documents 3m3m1z
Petroleum Production Engineering 343e17
December 2019 66
Petroleum Production Engineering Boyun Guo 5r633t
December 2019 61
Petroleum Production Engineering.pdf 195g6x
December 2019 59
Petroleum Production Systems 1z2l24
August 2020 0
Petroleum Engineering Sample Questions 4dw2t
February 2022 0
Basic Petroleum Engineering Practices.pdf 4w1t2w
December 2019 32More Documents from "hawar abdul-khaliq" 1j1v59
10-chapter-electrochemistry-mcqs.pdf 1q2gl
October 2019 38
Operations Research Ppt.ppt 1rby
November 2019 72
3 Improved Oil Recovery.pdf 2h725f
December 2019 36