PNG 410: Applied Reservoir Engineering

Home work assignment #3 Solution Problem 1: We have a single phase gas reservoir with the following field data:

Contour number

Contour area (acre)

Thickness (ft)

Pressure psia

Water saturation

Porosity

1 551 5 4200 0.34 0.22 2 410 6 4185 0.25 0.22 3 290 5 4102 0.20 0.22 4 142 4 4055 0.20 0.22 5 80 5 4044 0.20 0.22 6 0 - - - -

1.1 Please calculate the reservoir volume using trapezoidal rule and pyramidal rule. What

is difference you get using the two different integration rules? a. Calculation of reservoir volume using trapezoidal rule:

3 51 2 41 2 2 3 3 4 4 5 5 6

31 21 2 1 2 2 3 2 3

( ) ( ) ( ) ( ) ( )2 2 2 2 2

5 6 5 4 5 = (551 410) (410 290) (290 142) (142 80) (80 0)

2 2 2 2 2 = 6226.5 acre-ft

( ) ( )3 3

trapzoidal

pyramidal

Z ZZ Z ZV A A A A A A A A A

ZZ ZV A A A A A A A A

A

3 4 3 4

544 5 4 5 5 6 5 6

( )3

( ) ( )3 3

= 6113.1 acre-ft

A A A A

ZZA A A A A A A A

1.2 Please calculate the area-averaged and volume-average pressure of the reservoir.

,

,

551 4200 410 4185 ... 80 40444154.08 psia

551 410 ... 142 80

551 5 4200 410 6 4185 ... 80 5 40444157.58

551 5 410 6 290 5 142 4 80 5

i ii

avg area

ii

i ii

avg area

ii

A pp

A

V pp psia

V

1.3 Please calculate the original gas in place, using the trapezoidal rule for the reservoir volume.

5

i 1

i wi

(1 ) 998.767 acre-ft , here V is the volume of reservoir portion

calculated from trapzoidal rule, and and S are the corresponding porosity and water saturation.

gas i i wii

V V S

1.4 If the gas formation volume factor at the average initial pressure is 0.00453 cu ft/SCF,

what is the original gas in place in units of SCF?

gi,

43,560 cu ft/acre-ft 998.767 acre-ft = 43.5 MM CF

convert to SCF under standard condition using B

/ 9.60 MMM SCF

gas

gas g

V

G V B

Problem 2: A small gas reservoir has an initial pressure of 3200 psia and temperature 220 deg F. The pressure-production history and gas volume factors are as follows: Pressure (psia)

Gp (MM SCF)

Bg

(cu ft/SCF)

Z factor p/z slope Estimated G (MMSCF) assuming volumetric reservoir)

3200 0 0.0052622 0.875337 3655.7332925 79 0.0057004 0.866741 3374.711 -3.55725 1027.6852525 221 0.0065311 0.857427 2945.476 -3.02278 1116.4272125 452 0.0077360 0.854542 2486.711 -1.98599 1483.124 2.1 Calculate the initial gas in place using production data at the end of each production intervals, assuming volumetric behavior. For each pressure, calculate the z factor using

gg

i

p i

B0.02829 , z = 0.0609B

0.02829 (220 460)

use the above z formula to calculate z at each p, and p/z values.

p( ) pzFor each interval, p/z vs Gp plot has slope of =- .G z

G can be estimated

rrg r

r

pzTB p

p

G

for each interval, as seen in the table.

2.2 Is this reservoir a volumetric reservoir or water drive reservoir? Explain. 2.2 G value is not a constant, indicating water drive reservoir. 2.3 Plot p/z vs cumulative production.

p/z vs Gp

0

500

1000

1500

2000

2500

3000

3500

0 100 200 300 400 500

Gp (MM SCF)

p/z

2.4 If the initial volume of gas in place is 1018 MM SCF, and if there is no appreciable water production, what is the volume of water invaded at the end of each period?

g

p e g

e1

Using the following equation: G(B )

Because W is negligible, W (B )

at each interval, there is , , , and G values.

First interval; W 79 MM SCF 0.0057004 cu ft/SCF - 1

gi e p g w p

p g gi

p g gi

B W G B B W

G B G B

G B B

e2

e3

018 MM SCF (0.0057004-0.0052622)

= 0.004244 MM SCF

= 4244 SCF = 4244 SCF/(5.615 cu ft/bbl)= 755.8 STB

W 27004.97 STB

W 174237.5 STB

Problem 3: The production of a gas field is given below:

p/z (psia) Gp (MMM SCF) 6553 0.393 6468 1.642 6393 3.226 6329 4.260 6246 5.504 6136 7.538 6080 8.749

3.1 draw p/z vs Gp. Is this a volumetric reservoir?

p/z vs Gpp/z = -56.889Gp + 6569.5

R2 = 0.998

6,000

6,100

6,200

6,300

6,400

6,500

6,600

0 2 4 6 8 10

Gp (MMM SCF)

p/z

(p

sia)

This is a volumetric reservoir, because the p/z vs Gp figure is linear. 3.2 what is the initial gas in place? Initial gas in place is the Gp when p/z is zero. This give about 6569.5/56.889 = 115.5 MMM SCF. 3.3 What percentage of the initial gas in place will be recovered at a p/z of 900? At p/z = 900 psia, Gp = (900 – 6569.5) /(- 56.889) = 99.66 MMM SCF Recovery factor at that point = 99.66 / 115.5 * 100% = 86.3% Problem 4: Calculate the daily gas production including the condensate and water gas equivalents for a reservoir with the following daily production: Separator gas production = 6 MM SCF Condensate production = 100 STB Stock tank gas production = 21 M SCF Fresh water production = 10 bbl Initial reservoir pressure = 6000 psia Current reservoir pressure = 2000 psia Reservoir temperature = 225F Water vapor content of 6000 psia and 225 deg F = 0.86 bbl/MM SCF Condensate gravity = 50API Solution:

water

p

50

141.50.78

131.542.43

144.841.008

133000 716.51 SCF

GE 7390 SCF

G

= 6000000+21000+716.51 100+7390 10

= 6,166,551

o

owo

o

ogasCondensate

wo

ps st gasCondensate water

API

API

M

GEM

G G GE GE

SCF