Thermodynamic Property Methods
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Transcript Thermodynamic Property Methods
Ref.1: Ikoku, Natural Gas Production Engineering, John Wiley &
Sons, 1984, Chapter 7.
Ref.2: Menon, Gas Pipeline Hydraulic, Taylor & Francis, 2005,
Chapter 3.
1
Single Phase Gas Flow
Series Pipeline with Elevation Change
P4
P2
P5
P1
P3
5
2
Si
2
.
527
10
z
f
T
(
q
/
E
)
0.0375 g Z i
e
1
g
av
m
av
g
Si 2
2
i
i
i
sc
Li
, Si
Pi e Pi 1
5
di
Si
z avi Tavi
Si
z
f
T
e
1
av
m
av
Si 2
2
5
2
i
i
i
Pi e Pi 1 K
L
,
K
2
.
527
10
(
q
/
E
)
g
g sc
i S
di5
i
2
Single Phase Gas Flow
Series Pipeline with Elevation Change
P e P K
2
1
S1
2
2
zav1 f m1Tav1
d15
P e P K
zav2 f m2 Tav2
P32 e S3 P42 K
zav3 f m3 Tav3
P e P K
zav4 f m4 Tav4
2
2
2
4
S2
S4
2
3
2
5
d 25
d35
d 45
e S1 1
Le1 , Le1 L1
S1
e S2 1
Le2 , Le2 L2
S2
e S3 1
Le3 , Le3 L3
S3
e S4 1
Le4 , Le4 L4
S4
3
Single Phase Gas Flow
Series Pipeline with Elevation Change
z f T
S3 S 2 S1 zav4 f m4 Tav4
S 2 S1 av3 m3 av3
e
Le4 e
Le3
5
5
d4
d3
P12 e S1 S 2 S3 S 4 P52 K
z f T
zav1 f m1Tav1
S1 av2 m2 av2
e
Le2
Le1
5
5
d2
d1
n S j z
Si
f
T
avi mi avi
P12 e i1 Pn21 K e j1
L
ei
5
di
i 1
n
i 1
5
n S j z T
Si
f
f
j1 avi avi mi db
mb
2
2
i 1
P1 e Pn1 K zavTav 5 e
L
5 ei
db i 1
zavTav f mb di
n
i 1
4
Single Phase Gas Flow
Series Pipeline with Elevation Change
5
n S j z T
Si
f
f
j1 avi avi mi db
mb
2
2
i 1
P1 e Pn1 K zavTav 5 e
L
5 ei
db i 1
zavTav f mb di
zavi
Tavi
Assume :
1,
1, d b d1 , f mb f m 1
zav
Tav
i 1
n
P12 e S Pn21
2.527 105 g (qg sc / E ) 2 zavTav f m 1
5
1
d
LeSeries
n
Sj
5
n
Si
f
d
m
i
1
i 1
e j1
L
,
S
e
e
5
i
f
d
i 1
m1
i
i 1
Where : LeSeries
5
Single Phase Gas Flow
Series Pipeline with Elevation Change
Assume : f mi
P12 e S Pn21
0.032
1/ 3 (Weymouth)
di
8.0864 107 g (qg sc / E ) 2 zavTav
16 / 3
1
d
LeSeries
16 / 3
n S j
j1 d1
e
L
16 / 3 ei
di
i 1
i 1
Where : LeSeries
6
Single Phase Gas Flow
Parallel Pipeline with Elevation Change
LeA , dA , qA
LeB , dB , qB
P1
P2
LeC , dC , qC
q g sc
( P12 e S P22 )d 5
198.94 E
z
T
f
L
g av av m e
0.5
d
K
f m Le
5
0.5
( P12 e S P22 )
, Where : K 198.94 E
z
T
g av av
qtotal q A qB qC
0.5
d 5 0.5 d 5 0.5 d 5 0.5
5
db
C
A
B
K
K
f m Le
f m Le
f m Le
f m A Le A
B B
C C
b Parallel 7
0.5
Single Phase Gas Flow
Parallel Pipeline with Elevation Change
LeParallel
d b5
fm
b
d 5
A
f m A Le A
0.5
d B5
f m Le
B B
0.5
d C5
f m Le
C C
0.5 2
d b5
fm
b
C d 5 0.5
j
j A f m j Le j
2
Assume : d b d A , Le A LeB LeC and Weymouth equation
LeParallel
Le A
d 8 / 3 d 8 / 3
1 B C
d A
d A
2
8
Single Phase Gas Flow
Looped Pipeline with Elevation Change
LeB , dB , qB
LeA , dA , qA
P1
Px
LeC , dC , qC
P2
Assume : Le A LeB , d A d C , and Weymouth equation
LeParallel
q g sc
Le A
d
1 B
d A
8/3 2
( P e P )d
1112.1E
g z avTav LeLoop
2
1
S
2
2
16 / 3
A
LeSeries LeLoop
0.5
Le A
d
1 B
d A
8/3 2
e S A LeC
0.5
d
( P12 e S P22 )
, Where : K 1112.1E
K
Le
z
T
g av av
Loop
16 / 3
A
9
0.5
Single Phase Gas Flow
Looped Pipeline For Increasing Capacity
LeB , dB , qB
LeA , dA , qA
P1
Px
LeC , dC , qC
P2
Assume : Le A LeB , d A d C , and Weymouth equation
Old Pipeline : qold
2
qnew
qold
d
d
, qnew K
K
Le e Le
Le
C
A
Loop
/3
d 16
A
LeLoop
/3
d 16
A
Le A e S A LeC
16 / 3
A
SA
0.5
Le A e S A LeC
Le A
SA
e
LeC
8/3 2
d
1 B
d A
0.5
16 / 3
A
10
Single Phase Gas Flow
Looped Pipeline For Increasing Capacity
LeB , dB , qB
LeA , dA , qA
P1
Assume : x f e
Le A
Le A e LeC
2
qnew
qold
Px
SA
1 x fe
1
x fe
d
1 B
d A
8/3
2
1 x fe
LeC , dC , qC
P2
e S A LeC
Le A e S A LeC
q 2
1 old
qnew
x fe
1
1
8/3 2
d
1 B
d A
11
Single Phase Gas Flow
Optimum Diameter of Looped Pipeline
LeB , dB , qB
LeA , dA , qA
P1
Px
LeC , dC , qC
P2
Assume that the cost of a pipe with one inch diameter and one foot length is Cbase
Installed cost : CB Cbased B1.5 LB CB Cbased B1.5 x f ( LA LC )
q 2
For horizontal pipeline : C B Cbase ( LA LC ) 1 old
qnew 1
For d BOptimum :
dC B
0 (Solve this problem graphicall y)
dd B
d B1.5
1
d
1 B
d A
8/3 2
Single Phase Gas Flow
Example 1: Description
P4
P2
P5
P1
P3
For the above pipeline, calculate the exit pressure of each segments (P2, P3, P4,
P5) based on Weymouth equation and Hysys software.
Feed specifications:
T1=50 oF, P1=1000 psia, qsc=200 MMscfd, C1=90%, C2=10% (mole)
Pipeline specifications:
L1=20 km, ΔZ1=2000 m, L2=30 km, ΔZ2=-500, L3=15 km, ΔZ3=600, L4=50 km,
ΔZ4=-800, d1= d2= d3= d4= 20 in, T5=35 oF, E=0.95(in Weymouth)
13
Single Phase Gas Flow
Example 1: Solution
Tav 502.5 R, M g yi M i 17.45 g
o
Mg
M Air
0.602
2 P13 P53
816.7 psia
Assume P5 600 psia Pav 2
2
3 P1 P5
Ppc 674.3 psia , Tpc 355.5 oR Tpr 1.414, Ppr 1.211 zav 0.85
S1
0.0375 g Z1
zavTav
0.3468, S 2 0.0867, S3 0.1041, S 4 0.1387
S 0.2255, LeSeries Le1 Le2 e S1 Le3 e S1 S2 Le4 e S1 S2 S3 499488 ft
Hysys
7
2
8
.
0864
10
(
q
/
E
)
zavTav
g
g sc
2
S
2
P5 e P1
LeSeries P5 612(655) psia
16 / 3
d1
Single Phase Gas Flow
Example 1: Solution
S 0.3642, LeSeries Le1 Le2 e S1 Le3 e S1 S2 279072 ft
Hysys
7
2
8
.
0864
10
(
q
/
E
)
zavTav
g
g sc
2
S
2
P4 e P1
LeSeries P4 700(711) psia
16 / 3
d1
S 0.3642, LeSeries Le1 Le2 e S1 211803 ft
Hysys
7
2
8
.
0864
10
(
q
/
E
)
zavTav
g
g sc
2
S
2
P3 e P1
LeSeries P3 773(782) psia
16 / 3
d1
P e
2
2
S1
Hysys
2 8.0864 10 g (qg sc / E ) zavTav
Le1 P2 805(800) psia
P1
16 / 3
d1
15
7
2
Single Phase Gas Flow
Example 2: Description
LB , dB
P1
LA , dA
Px
LC , dC
An old pipeline has the following specification:
Feed specifications:
T1=50 oC, P1=1000 psia, qold =1500 MMscfd, C1=90%, C2=10% (mole)
Pipeline specifications:
LA+LC =100 km, ΔZ=0, dA= dC= 42 in, T2=42 oC, E=0.95(in Weymouth)
a) Calculate the outlet pressure (P2)
b) The pipeline capacity is increased by 25% (qnew=1875 MMscfd). Calculate
the optimum length and diameter of the looped pipeline.
P2
Single Phase Gas Flow
Example 2: Solution
Tav 575 R, M g yi M i 17.45 g
o
Mg
M Air
0.602
2 P13 P23
876 psia
Assume P2 740 psia Pav 2
2
3 P1 P2
Ppc 674.3 psia, Tpc 355.5 oR Tpr 1.62, Ppr 1.30 zav 0.905
L 328084 ft,
7
2
8
.
0864
10
(
q
/
E
)
z avTav
g
old
2
2
P2 P1
16 / 3
d
A
L P2 738 psia (738)
Single Phase Gas Flow
Example 2: Solution
Installed cost : CB Cbased B1.5 LB CB Cbased B1.5 x f ( LA LC )
x f 0.691 LB 69.1 km (68.3 km)