enpe 523 assignment _2 solution

8/11/2019 ENPE 523 Assignment _2 Solution

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UNIVERSITY OF CALGARY

INRTODUCTION TO RESERVOIR ENGINEERING

ENPE 523

Fall 2010

Solution of Assignment #2

Date: January 18, 2010 Due Date: January 25, 2010

Problem 1:A gas well is producing gas with a specific gravity of 0.8 at a rate of 1.5 MMSCF/day.

The average reservoir pressure and temperature are 2200 psi and 145F respectively.Assume ideal gas behaviour and calculate:

A) Apparent molecular weight of the gas

B)

Density of the gas at reservoir conditions

C) Flow rate in lb/day

D) Calculate density of the gas by assuming a real gas behaviour

SOLUTION:

A) lbmolelbmass/23.176(0.8)(28.97)M28.97M ===

B) 3/853.7)460145)(732.10(

)176.23)(2200(ftlb

RT

pMmass=

+==

C) daylblbmole

lb

SCF

lbmole

MMSCFD

SCFDMMSCFD mass

mass /916291

176.23

4.379

1105.1

6

D)

For natural gas

( ) ( ) RT ggpco4208.05.128.03251685.12325168

22 =+=+=

44.1420

460145

+==

pc

rT

TT

( ) ( ) psiap ggpc 6658.05.378.00.156775.370.1567722 =+=+=

31.3665

2200==

pc

rp

pp

Using compressibility chart we obtain

74.0Z

( )3/612.10

)460145)(732.10(74.0

)176.23)(2200(ftlb

ZRT

pMmass=

+==

Real gas density is 10.612 lbmass/ft3and ideal gas density is 7.853 lbmass/ft

3


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Problem 2:A gas well is producing a natural gas with the following composition:

Component Mole fractionCO2 0.07

H2S 0.04N2 0.03

C1 0.70

C2 0.09

C3 0.07

Calculate

A) Apparent molecular weight

B) Specific gravity

C) Gas density at 3400 psia and 170F

D) Isothermal compressibility of the gas

E) Gas formation volume factor in ft3/SCF

F)

Gas viscosity using the Lee-Gonzalez-Eakin method

SOLUTION:

A)

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( ) lbmolelbmass

M

MyM

a

i

i

ia

/31.2211.4407.007.3009.004.1670.0

01.2803.008.3404.001.4407.0

6

1

=++

+++=

==

B)

77.097.2831.22 ===

air

ag

MM

Ideal gas

C)

3/22.11)460170)(732.10(

)31.22)(3400(ftlb

RT

pMmass=

+==

D)

41094.23400

11 ===p

cg psi-1

E)

( )( )

( ) SCFft

p

ZTBg /1024.5

3400

460170102827.002827.0 33=

+==


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F)

( )( )( )

( )( )

( ) ( ) ( )( )( )( )( ) ( )( )

( ) ( )( )

( )( ) 342.129.52.04.22.04.2

288.531.2201.0460170

9865.301.0/9865.3

29.12346017031.221920946017031.2202.04.919209/02.04.9

/18.04601701

31.223400104935.1/104935.1

10exp

5.1

5.1

33

4

===

=++

+=++=

=+++++=+++=

+

==

=

XY

MTX

TMTMK

ccgZTpM

XK Y

g

( ) ( ) ( )( ) cpXK Yg 021.01018.0288.5exp29.12310exp4342.14 ===

Real gas

C)

( ) ( ) RT oggpc 84.41077.05.1277.03251685.1232516822 =+=+=

53.184.410

460170=

+==

pc

rT

TT

( ) ( ) psiap ggpc 32.66677.05.3777.00.156775.370.1567722 =+=+=

1.5

32.666

3400===

pc

r

p

pp


83.0Z

( )3/52.13

)460170)(732.10(83.0

)31.22)(3400(ftlb

ZRT

pMmass=

+==

D)

dp

dZ

Zpcg

11=

From compressibility chart we have:

1510632.666

04.004.004.0

04.0====

=

=

psipdp

dZ

dp

dZp

p

pp

dpdZ

c

c

pc

r

r

145 1022.210683.0

1

3400

111 === psidp

dZ

Zpcg


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4

E)

( )( )

( ) SCFft

p

ZTBg /1035.4

3400

46017083.002827.002827.0 33=

+==

F)

( )( )( )

( )( )

( ) ( ) ( )( )( )( )

( ) ( )( )

( ) ( )( )

( )( ) 342.1288.52.04.22.04.2

288.531.2201.0460170

9865.301.0/9865.3

29.12346017031.2219209

46017031.2202.04.919209/02.04.9

/216.046017083.0

31.223400104935.1/104935.1

10exp

5.1

5.1

33

4

===

=++

+=++=

=+++

++=+++=

=+

==

=

XY

MTX

TMTMK

ccgZTpM

XK Y

g

( ) ( ) ( )( ) cpXK Yg 024.010216.0288.5exp29.12310exp4342.14 ===

Problem 3:Use correlations to estimate the gas solubility, formation volume factor, oil viscosity and

isothermal compressibility of oil by considering following reservoir conditions and fluids

properties:

P=2900 psia.

T=180oF

Pb=2400 psiaGas gravity = 0.85

API gravity = 38

SOLUTION:

Gas solubility

psia.inpF,in

0125.000091.0

pp,1018

o

b

204.1

=

=

=

T

APITY

pR

g

Ygs g

Absolute error 4.8%

130


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p= 2900> bubble point pressure

Rso= Rsb

psia.inpF,in

0125.000091.0

pp,1018

o

b

204.1

=

=

=

T

APITY

pR

g

Ygsb g

( ) ( ) ( )( )

( ) b

204.1

3112.0ppSCF/STB668.728

1018

240085.0

3112.0380125.018000091.00125.000091.0

=

=

===

sb

g

R

APITY

Isothermal oil compressibilityVillena-Lanzi (p


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6

835.05.13138

5.141

5.131

5.1415.131

5.141=

+=

+==

APIAPI o

o

( ) 18.96018025.1835.0

85.0668.728251

5.05.0

=+

=+

= T.RF

o

g

s

( ) STBbblFBob /441.118.960000147.0972.0000147.0972.0 175.1175.1 =+=+=

( )[ ] ( )[ ] STBbblppcBB boobo /453.1290024001065.1exp441.1exppbpfor

5 ===

>

Oil viscosity

( )[ ] ( ) ( ) 36.0180log5644.038025086.08653.11loglog ==+od At bubble point pressure

( ) cpobdobdobd 73.1110101436.0101log436.0436.036.0 ===+==+

B

obdob A =

( ) 336.0100668.728715.10 515.0 =+= A

( ) 55.0150668.72844.5 338.0 =+= B

( ) cpA Bobdob 454.073.1336.055.0===

( ) ( )( )[ ] 258.029001098.8513.11exp29006.2 5187.1 == m ( ) ( ) cppp mbobo 477.02400/2900454.0/

258.0===


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Practice Problem 1:Use correlations to estimate the gas solubility, formation volume factor, viscosity and

isothermal compressibility of reservoir brine at pressures of 3000 psia.

T=150oF

Salinity = 15% by weightGas gravity = 0.75

Brine density = 69.3 lb/ft3

Solution Gas water Ratio:

( )285854.00840655.010= ST

swp

sw

R

R

( )( )( )

5.0

10285854.0

150150840655.0

=

=

swp

sw

R

R

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) (( 6

936247

3102752

37242

10-2

11037049.21501034122.21501053425.8150130237.002505.910

0.00482

1501094883.21501005553.31501044241.71001021.1

2.556

150101654.21501091663.11501012265.615839.8

=

++=

=

+=

=

+=

C

B

A

( ) ( ) 22.513000102300000482.0556.2 26 =+= swpR

( ) STBSCFRsw /61.722.155.0 ==


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Water formation volume factor:

ppTppTV

TTV

wp

wt

1072139

2742

1025341.21058922.31072834.11095301.1

1050654.51033391.11000010.1

=

++=

( ) ( )

( )( ) ( ) ( ) ( )

( )

-0.00422

30001025341.2

30001058922.315030001072834.115030001095301.1

0.022397

1501050654.51501033391.11000010.1

210

72139

2742

=

=

=

++=

wp

wt

V

V

( )( ) bbl/STB1.018100422.01022397.01Bw =+=

Water isothermal compressibility (p


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710-2.665

0.00482

2.556

=

=

=

C

B

A

( )( ) 0.003221300010665.2200482.0 7 =+=

T

swp

p

R

Gas gravity = 0.75

( ) ( )

R

T

o

ggpc

71.404

75.05.1275.0325168

5.12325168

2

2

=

+=

+=

51.1

71.404

460150

=

+=

=pc

rT

TT

( ) ( )

psia

p ggpc

16.667

75.05.3775.00.15677

5.370.15677

2

2

=

+=

+=

5.4

667

3000

=

=

=pc

rp

pp


79.0Z

Gas formation volume factor


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10

( )( )( )

SCFft

ftbblSCFft

p

ZTBg

/000809.0

615.51/1054.4

3000

46015079.002827.0

02827.0

3

3

33

=

=

+=

=

( )

16

6

1086.4

003221.00181.1

000809.0103.2

=

+=

psi

Water viscosityMcCain correlation

B

w AT=1

( ) ( ) ( )

984.0

)15(1055586.1)15(1047119.5)15(1079461.6)15(1063951.212166.1

42.83

151072213.815313314.01540564.8574.109

4635242

332

=

++=

=

++=

B

A

cp

Tw

603.0

42.83 984.01

=

=

( ) ( )( )( )cp

w

658.0

603.03000101062.33000100295.4944.0295

=

++=


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Practice Problem 2:A dry gas reservoir is initially at an average pressure of 6000 psia and temperature of 160oF. The gas has a specific gravity of 0.65. What will the average reservoir pressure be

when one-half of the original gas in SCF has been produced? Assume the volume

occupied by the gas in the reservoir remains constant. If the reservoir originally contained

1MMft

3

of reservoir gas, how much gas has been produced at a final pressure of 500psia?

SOLUTION

psia6000=ip

R620F160 == oiT

65.0=

Part 1Calculate critical pressure and temperature using gas gravity

Gas gravity = 0.65

( ) ( )

R

T

o

ggpc

97.373

65.05.1265.0325168

5.12325168

2

2

=

+=

+=

66.1

97.373

460160

=

+=

=pc

rT

TT

( ) ( )

psia

p ggpc

91.670

65.05.3765.00.15677

5.370.15677

2

2

=

+=

+=

94.8

91.670

6000

=

=

=pc

rp

pp

Using compressibility chart we obtain06.1Z

21 VV =

2

12

nn = ,


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12

1

1111

V

RTnZp =

2

222

2V

RTnZp =

1

2

11

22

1

2

21

ZZ

nZnZ

pp ==

1

1

22

2

1p

Z

Zp

=

( )600006.12

1 22

=

Zp

p2 can be obtained by trial and error as given in the following.

First trial: Z2=1

( ) psiap 28306000

06.1

1

2

12 =

=

Calculate Z2

22.4

91.670

2830

2830

=

=

=pc

rp

p

Using Tr= 1.66 and Pr=4.22 gives Z2=0.84 1Assume Z2=0.84

( ) psiap 2377600006.1

84.0

2

1

2

=

=

Calculate Z2

54.3

91.670

2377

2377

=

=

=pc

rp

p

Using Tr= 1.66 and Pr=3.54 gives Z20.84 (converged solution)

( ) psiap 23776000

06.1

84.0

2

12 =

=


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Part 2Initial moles of gas in the reservoir

( )( )( )( )( )

850693

460160732.1006.1

106000 6

=

+=

=RTZ

Vpn

i

iii

in SCF = 850693(379.4)= 323 MMSCF

Using

75.0

91.670

500

=

=

=pc

rp

pp

Using Tr= 1.66 and Pr=0.75 gives Z30.94 , the remaining moles in the reservoir are

( )( )( )( )( )

79941

460160732.1094.0

10500 63

333

=

+=

=RTZ

Vpn

Remaining in SCF = 79941(379.4)=30MMSCF

Produced gas =ni-n3=323-30=293 MMSCF

Another solution method for Part 1

( )600006.12

1 22

=

Zp

( )

22.491.670

2830

2830

2830600006.1

1

2

1

2

22

2

==

=

=

==

c

r

rc

pZ

p

Z

pp

Z

p

Using Chart gives Z2= 0.84

( )

( )

psia

Zp

2377

600006.1

84.0

2

1

600006.12

1 22

=

=

=


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enpe 523 assignment _2 solution

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