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 145°F 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) M = 28.97γ ⇒ M = (28.97) (0.8) = 23.176 lbmass / lbmole B) ρ =
pM
( 2200)(23.176)
=
RT
(10.732)(145 + 460)
= 7.853 lbmass / ft 3
10 6 SCFD 1lbmole 23.176 lbmass C) 1.5 MMSCFD ≅ 91629 lbmass / day MMSCFD 379.4SCF 1lbmole D) For natural gas T pc = 168 + 325γ g − 12.5γ g = 168 + 325(0.8) − 12.5(0.8) = 420 R 2
2
T r =
T T pc
=
145 + 460 420
o
≅ 1.44
p pc = 677 + 15.0γ g − 37.5γ g = 677 + 15.0(0.8) − 37.5(0.8) = 665 psia 2
p r =
p p pc
=
2200 665
2
≅ 3.31
Using compressibility chart we obtain Z ≅ 0.74 pM ρ =
ZRT
=
( 2200)(23.176)
(0.74 )(10.732)(145 + 460) 3
= 10.612 lbmass / ft 3
Real gas density is 10.612 lbmass /ft and ideal gas density is 7.853 lbmass /ft
3
1
Problem 2: A gas well is producing a natural gas with the following composition: Component Mole fraction CO2 0.07 H 2S 0.04 N 2 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 170°F D) Isothermal compressibility of the gas 3 E) Gas formation volume factor in ft /SCF F) Gas viscosity using the Lee-Gonzalez-Eakin method SOLUTION: A) 6
∑ y M
M a =
i
i
i =1
M a = (0.07 )(44.01) + (0.04)(34.08) + (0.03)(28.01) +
(0.70)(16.04) + (0.09)(30.07 ) + (0.07 )(44.11) = 22.31
lbmass / lbmole
B) γ g
=
M a M air
=
22.31 28.97
= 0.77
Ideal gas C) ρ =
pM RT
=
(3400)(22.31) (10.732)(170 + 460)
= 11.22 lbmass / ft 3
D) cg =
1 p
=
1 3400
= 2.94 × 10 −4 psi-1
E)
(1)(170 + 460 ) ZT = 0.02827 = 5.24 × 10 −3 ft 3 / SCF B g = 0.02827 (3400 ) p
2
F) µ g
= K exp( X ρ Y )× 10 −4
(3400 )(22.31) ≅ 0.18 g / cc (1)(170 + 460) (9.4 + (0.02)(22.31))(170 + 460)1.5 1.5 = 123.29 K = (9.4 + 0.02 M )T / (209 + 19 M + T ) = (209 + 19(22.31) + (170 + 460))
ρ = 1.4935 × 10
−3
pM / ZT = 1.4935 × 10 −3
X = 3.5 + 986 / T + 0.01 M = 3.5 +
986
(170 + 460)
+ (0.01)(22.31) = 5.288
Y = 2.4 − 0.2 X = 2.4 − (0.2 )(5.29) = 1.342 µ g
= K exp( X ρ Y )× 10 −4 = (123.29) exp (5.288)(0.18)
1.342
× 10 −4 = 0.021 cp
Real gas C) T pc = 168 + 325γ g − 12.5γ g = 168 + 325(0.77 ) − 12.5(0.77 ) = 410.84 R 2
2
T r =
T T pc
=
170 + 460 410.84
o
= 1.53
p pc = 677 + 15.0γ g − 37.5γ g = 677 + 15.0(0.77 ) − 37.5(0.77 ) = 666 .32 psia 2
2
p r =
p
=
p pc
3400 666.32
= 5.1
Using compressibility chart we obtain Z ≈ 0.83 pM ρ =
ZRT
=
(3400)(22.31)
(0.83)(10.732)(170 + 460)
= 13.52 lbmass / ft 3
D) cg =
1
−
1 dZ
p Z dp From compressibility chart we have: dZ dp = 0.04 dZ r ⇒ p c = 0.04 p dp p r = p pc cg =
1 p
−
1 dZ Z dp
=
1 3400
−
1 0.83
⇒
dZ dp
=
0.04 p c
=
0.04 666.32
= 6 × 10 −5 psi −1
6 × 10 −5 = 2.22 × 10 −4 psi −1
3
E) B g = 0.02827
ZT p
= 0.02827
(0.83)(170 + 460) = 4.35 × 10 −3 ft 3 / SCF (3400 )
F) µ g
= K exp( X ρ Y )× 10 −4
(3400 )(22.31) = 0.216 g / cc (0.83)(170 + 460) (9.4 + (0.02)(22.31))(170 + 460)1.5 1.5 = 123.29 K = (9.4 + 0.02 M )T / (209 + 19 M + T ) = (209 + 19(22.31) + (170 + 460))
ρ = 1.4935 × 10
−3
pM / ZT = 1.4935 × 10 −3
986
+ (0.01)(22.31) = 5.288 (170 + 460) Y = 2.4 − 0.2 X = 2.4 − (0.2 )(5.288) = 1.342 1.342 Y −4 × 10 −4 = 0.024cp µ g = K exp( X ρ )× 10 = (123.29) exp (5.288)(0.216 )
X = 3.5 + 986 / T + 0.01 M = 3.5 +
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. o T=180 F Pb=2400 psia Gas gravity = 0.85 API gravity = 38 SOLUTION: Gas solubility 1.204
p R s = γ g Y 18 × 10 g
,
p ≤ pb
Y g = 0.00091T − 0.0125 API T = in
o
F, p in psia.
Absolute error 4.8% 130
4
p= 2900> bubble point pressure Rso = Rsb
p R sb = γ g Y 18 × 10
1.204
p ≤ pb
,
g
Y g = 0.00091T − 0.0125 API T = in
o
F, p in psia. Y g = 0.00091(T ) − 0.0125 API = 0.00091(180) − (0.0125)(38) = −0.3112
2400 Rsb = (0.85) − 0.3112 18 × 10
1.204
= 728.668 SCF/STB
p ≤ pb
Isothermal oil compressibility Villena-Lanzi (p
-6
-1
31×10 p b )
( p ×10 ), co = (5 Rsob + 17.2T − 1180γ g + 12.61 API − 1433) / 5
p > pb
o
126
=
(5(728.668) + 17.2(180) − 1180(0.85) + 12.61(38) − 1433)
(2900 × 10 ) 5
= 1.65 × 10 −5 psi −1
Oil formation volume factor Standing correlation (Beggs) 1.175
Bo = 0.972 + 0.000147 F 0.5
γ g F = Rs + 1.25T γ o T = in
o
p ≤ pb
F, p in psia.
for p > pb Bo = Bob exp[co ( pb − p )]
5
API =
141.5 γ o
− 131.5 ⇒ γ o =
141.5 API + 131.5
=
0.5
141.5 38 + 131.5
= 0.835
γ g 0.85 + 1.25T = 728.668 F = Rs + 1.25(180) = 960.18 γ 0 . 835 o 1.175 1.175 Bob = 0.972 + 0.000147 F = 0.972 + 0.000147 (960.18) = 1.441 bbl / STB 0 .5
for p > pb
[
]
Bo = Bob exp[co ( pb − p )] = 1.441exp 1.65 × 10 − (2400 − 2900 ) = 1.453 bbl / STB 5
Oil viscosity
log[log( µ od + 1)] = 1.8653 − 0.025086 (38) − 0.5644 log(180) = −0.36 At bubble point pressure log( µ obd + 1) = 10 −0.36 = 0.436 ⇒ µ obd + 1 = 10 0.436 ⇒ µ obd = 10 0.436 − 1 = 1.73 cp µ ob
= A µ Bobd
−0.515
A = 10.715(728.668 + 100 )
−0.338
B = 5.44(728.668 + 150 ) µ ob
B = Aµ obd = 0.336(1.73)
0.55
1.187
m = 2.6(2900 ) µ o
= 0.336
= 0.55
= 0.454 cp
exp[− 11.513 − 8.98(10 −5 )(2900 )] = 0.258
= µ ob ( p / pb ) = 0.454(2900 / 2400 ) m
0.258
= 0.477 cp
6
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. o
T=150 F Salinity = 15% by weight Gas gravity = 0.75 3 Brine density = 69.3 lb/ft
Solution Gas water Ratio: −0.285854 Rsw ) = 10 (−0.0840655 ST Rswp Rsw Rswp
= 10 (−0.0840655(15 )(150)
− 0.285854
)
= 0.5
2
3
A = 8.15839 − 6.12265 × 10 − (150 ) + 1.91663 × 10 − (150 ) − 2.1654 × 10 − (150 ) 2
4
7
= 2.556 2
B = 1.01021 × 10 − − 7.44241 × 10 − (150 ) + 3.05553 × 10 − (150 ) − 2.94883 × 10 − 2
5
7
10
(150)3
= 0.00482
(
2
3
4
C = −10 − 9.02505 − 0.130237 (150 ) + 8.53425 × 10 − (150 ) − 2.34122 × 10 − (150 ) + 2.37049 × 10 − (150 ) 7
4
6
9
= -2 × 10 −6
Rswp = 2.556 + 0.00482(3000 ) − 2 × 10 − (3000 ) = 15.22 6
2
Rsw = 0.5(15.22 ) = 7.61SCF / STB
7
)
Water formation volume factor:
∆V wt = −1.00010 × 10 −2 + 1.33391× 10 −4 T + 5.50654 × 10 −7 T 2 ∆V wp = −1.95301 × 10 −9 pT − 1.72834 × 10 −13 p 2T − 3.58922 × 10 −7 p − 2.25341× 10 −10 p 2
∆V wt = −1.00010 × 10 −2 + 1.33391× 10 −4 (150 ) + 5.50654 × 10 −7 (150 ) = 0.022397 2
∆V wp = −1.95301 × 10 −9 (3000 )(150 ) − 1.72834 × 10 −13 (3000 ) (150 ) − 3.58922 × 10 −7 (3000 ) 2
− 2.25341 × 10 −10 (3000 ) = -0.00422
B w = (1 + 0.022397 )(1 − 0.00422 ) = 1.0181 bbl/STB
Water isothermal compressibility (p<=p b)
where
Salt water density is 69.3 lbmass /ft3 3 =69.3 (1000)/62.428 = 1110 kg/m 15% saline water kg salt 1110 kg solution 0.15 3 kg solution 1m solution
= 166.5
kg salt
m3 solution = 166.5 g / liter
−
1 ∂ Bw
1
= Bw ∂ p T [7.033 p + 541.5C NaCl − 537.0T + 403300 ]
= 2.3 × 10-6 psi −1
8
A = 2.556 B = 0.00482 C = -2.665 × 10 −7
∂ Rswp = 0.00482 + 2(− 2.665 ×10 −7 )(3000 ) = 0.003221 ∂ p T
Gas gravity = 0.75 T pc = 168 + 325γ g − 12.5γ g2
= 168 + 325(0.75) − 12.5(0.75)
2
= 404.71o R T r =
=
T T pc
150 + 460
404.71 = 1.51 2
p pc = 677 + 15.0γ g − 37.5γ g
2
= 677 + 15.0(0.75) − 37.5(0.75) = 667.16 psia p r =
=
p p pc
3000
667 = 4.5 Using compressibility chart we obtain Z ≈ 0.79 Gas formation volume factor
9
Bg = 0.02827
= 0.02827
ZT p
(0.79)(150 + 460) (3000 ) 1bbl 3 ft 5 . 615
= 4.54 × 10−3 ft 3 / SCF = 0.000809 ft 3 / SCF
= 2.3 × 10 −6 +
0.000809 1.0181
(0.003221)
= 4.86 × 10−6 psi −1 Water viscosity McCain correlation B µ w1 = AT
2
3
A = 109.574 − 8.40564 (15) + 0.313314(15) + 8.72213 × 10− (15) 3
= 83.42 B = −1.12166 + 2.63951 × 10− (15) − 6.79461 × 10 − (15) − 5.47119 × 10− (15) + 1.55586 × 10− (15) 2
4
2
5
3
6
= −0.984 −0.984 µ w1 = 83.42T = 0.603cp
µ w
= (0.944 + 4.0295 × 10−5 (3000 ) + 3.1062 × 10−9 (3000 ) )(0.603) 2
= 0.658cp
10
4
Practice Problem 2: A dry gas reservoir is initially at an average pressure of 6000 psia and temperature of 160 o F. 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 3 1MMft of reservoir gas, how much gas has been produced at a final pressure of 500 psia? SOLUTION pi = 6000 psia T i = 160 F = 620 R o
γ =
0.65 Part 1 Calculate critical pressure and temperature using gas gravity Gas gravity = 0.65 T pc = 168 + 325γ g − 12.5γ g2
= 168 + 325(0.65) − 12.5(0.65)
2
= 373.97 o R T r =
=
T T pc
160 + 460
373.97 = 1.66 2
p pc = 677 + 15.0γ g − 37.5γ g
2
= 677 + 15.0(0.65) − 37.5(0.65) = 670.91 psia pr =
=
p p pc
6000
670.91 = 8.94 Using compressibility chart we obtain Z ≈ 1.06 V 1 = V 2 n2 =
n1
2
,
11
p1 =
Z 1n1 RT 1
p 2 = p 2 p1
V 1 Z 2 n2 RT 2
=
p2 =
V 2 Z 2 n 2 Z 1n1
=
1 Z 2 2 Z 1
1 Z 2 p1 2 Z 1
1 Z 2 (6000 ) 2 1.06 p2 can be obtained by trial and error as given in the following. First trial: Z 2=1 1 1 p2 = psia (6000 ) = 2830 2 1.06 Calculate Z2 2830 pr = p pc p2 =
=
2830
670.91 = 4.22 Using Tr= 1.66 and Pr =4.22 gives Z 2=0.84 ≠1 Assume Z2=0.84 1 0.84 p2 = psia (6000 ) = 2377 2 1.06 Calculate Z2 2377 pr = p pc
=
2377
670.91 = 3.54 Using Tr= 1.66 and Pr =3.54 gives Z 2 ≈0.84 (converged solution) 1 0.84 p2 = psia (6000 ) = 2377 2 1.06
12
Part 2 Initial moles of gas in the reservoir p V ni = i i Z RT i
(6000 )(106 ) = (1.06)(10.732)(160 + 460) = 850693 in SCF = 850693(379.4)= 323 MMSCF Using pr =
p p pc
500
=
670.91 = 0.75 Using Tr= 1.66 and Pr =0.75 gives Z 3 ≈0.94 , the remaining moles in the reservoir are p V n3 = 3 3 Z 3 RT
(500)(106 ) = (0.94)(10.732)(160 + 460) = 79941 Remaining in SCF = 79941(379.4)=30MMSCF Produced gas =n i-n3=323-30=293 MMSCF Another solution method for Part 1 1 Z p2 = 2 (6000 ) 2 1.06 p2 Z 2 pr Z 2
=
= =
pc pr Z 2
=
1 1 (6000 ) = 2830 2 1.06
2830 pc
2830
= 4.22 670.91 Using Chart gives Z2= 0.84 1 Z p2 = 2 (6000 ) 2 1.06 =
1 0.84
(6000 )
2 1.06
= 2377 psia
13
14
