Gas Pipeline II
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Transcript of Gas Pipeline II
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8/22/2019 Gas Pipeline II
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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
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Single Phase Gas Flow
Series Pipeline with Elevation Change
P2
P1
P3
P4
P5
ii
i
sciiii
avav
ig
i
i
S
i
i
gavmavg
i
S
i
Tz
ZS
S
eL
d
EqTfzPeP
0375.0,
1)/(10527.25
25
2
1
2
25
5
2
1
2 )/(10527.2,1
EqKS
eL
d
TfzKPeP
sc
i
iiii
gg
i
S
i
i
avmav
i
S
i
2
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8/22/2019 Gas Pipeline II
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Single Phase Gas Flow
Series Pipeline with Elevation Change
1
15
1
2
2
2
1
1,
1
11
1111
S
eLLL
d
TfzKPeP
S
ee
avmavS
2
25
2
2
3
2
2
1,
2
22
2222
S
eLLL
d
TfzKPeP
S
ee
avmavS
3
35
3
2
4
2
3
1,
3
33
3333
S
eLLL
d
TfzKPeP
S
ee
avmavS
4
45
4
2
5
2
4
1
,
4
44
4444
S
e
LLLd
Tfz
KPeP
S
ee
avmavS
3
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Single Phase Gas Flow
Series Pipeline with Elevation Change
1
111
2
2221
3
33312
4
444123
4321
5
1
5
2
5
3
5
42
5
2
1
eavmav
eavmavS
e
avmavSS
e
avmavSSS
SSSS
Ld
TfzLd
Tfze
Ld
TfzeL
d
Tfze
KPeP
n
i
e
i
avmavS
n
S
i
iii
i
j
j
n
i
i
L
d
TfzeKPeP
1
5
2
1
2
1
1
11
n
i
e
i
b
m
m
avav
avavS
b
m
avavn
S
i
b
iii
i
j
j
b
n
i
i
Ld
d
f
f
Tz
Tze
d
fTzKPeP
15
5
5
2
1
2
1
1
11
4
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Single Phase Gas Flow
Series Pipeline with Elevation Change
n
i
e
i
b
m
m
avav
avavS
b
m
avavn
S
i
b
iii
i
j
j
b
n
i
i
Ld
d
f
f
Tz
Tze
d
fTzKPeP
15
5
5
2
1
2
1
1
11
1,,1,1: 1 mmb
av
av
av
avffdd
T
T
z
zAssume
b
ii
Series
sc
e
mavavgg
n
S Ld
fTzEqPeP5
1
25
2
1
2
1
1)/(10527.2
n
i
i
i
i
i
j
j
Series
Sn
i
e
im
mS
e eSLd
d
f
feLWhere 1
1
1
1 ,:1 5
5
1
5
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8/22/2019 Gas Pipeline II
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Single Phase Gas Flow
Series Pipeline with Elevation Change
Series
sc
e
avavgg
n
S Ld
TzEqPeP
3/16
1
27
2
1
2
1
)/(100864.8
n
i
e
i
S
e i
i
j
j
SeriesL
d
deLWhere
13/16
3/16
1
1
1:
)(032.0
:3/1
Weymouthd
fAssume
i
mi
6
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Single Phase Gas Flow
Parallel Pipeline with Elevation Change
P1 P2
LeA , dA , qA
LeB , dB , qB
LeC, dC, qC
5.05
5.05
5.05
5.05
Paral le lbCCBBAA em
b
em
C
em
B
em
A
CBAtotal
Lf
dK
Lf
d
Lf
d
Lf
dK
qqqq
5.02
2
2
1
5.05
5.052
2
2
1 )(94.198:,)(
94.198
avavg
S
ememavavg
S
gTz
PePEKWhere
Lf
dK
LfTz
dPePEq
sc
7
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Single Phase Gas Flow
Parallel Pipeline with Elevation Change
25.0
5
5
25.0
5
5.0
5
5.0
5
5
C
Aj em
j
m
b
em
C
em
B
em
A
m
b
e
jj
b
CCBBAA
b
Paral le l
Lf
d
f
d
Lf
d
Lf
d
Lf
d
f
d
L
23/83/8
1
,:
A
C
A
B
e
e
eeeAb
d
d
d
d
LL
equationWeymouthandLLLddAssume
A
Paral le l
CBA
8
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Single Phase Gas Flow
Looped Pipeline with Elevation Change
P2P1LeA , dA , qA
LeB , dB , qB
LeC, dC, qCPx
C
AA
LoopSeries
A
Parallel e
S
A
B
e
ee
A
B
e
e Le
dd
LLL
dd
LL
2
3/82
3/8
11
equationWeymouthandddLLAssume CAee BA ,,:
5.02
2
2
1
5.03/16
5.03/162
2
2
1 )(1.1112:,)(
1.1112
avavg
S
e
A
eavavg
A
S
g
Tz
PePEKWhere
L
dK
LTz
dPePEq
LoopLoop
sc
9
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Single Phase Gas Flow
Looped Pipeline For Increasing Capacity
P2P1LeA , dA , qA
LeB , dB , qB
LeC, dC, qCPx
5.03/16
5.03/16
,:
LoopC
A
A e
Anew
e
S
e
Aold
L
dKq
LeL
dKqPipelineOld
equationWeymouthandddLLAssume CAee BA ,,:
C
AA
C
A
A
C
A
A
Loop
e
S
A
B
e
e
S
e
e
S
e
A
e
A
old
new
Le
d
d
L
LeL
LeL
d
Ld
q
q
23/8
3/16
3/16
2
1 10
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Single Phase Gas Flow
Looped Pipeline For Increasing Capacity
P2P1LeA , dA , qA
LeB , dB , qB
LeC, dC, qCPx
C
A
A
C
A
e
C
A
A
A
e
e
S
e
eS
f
e
S
e
e
fLeL
Lex
LeL
LxAssume
1:
23/8
2
23/8
2
1
11
1
1
1
1
A
B
new
old
f
f
A
B
fold
new
d
d
q
q
x
x
d
d
xq
qe
e
e
11
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Single Phase Gas Flow
Optimum Diameter of Looped Pipeline
P2P1LeA , dA , qA
LeB , dB , qB
LeC, dC, qCPx
23/8
5.12
1
11
1)(:
A
B
B
new
oldCAbaseB
d
d
dqqLLCCpipelinehorizontalFor
)(:costInstalled 5.15.1 CAfBbaseBBBbaseB LLxdCCLdCC
Assume that the cost of a pipe with one inch diameter and one foot length is Cbase
y)graphicallproblemthisSolve(0d
d:
B
BB
d
CdFor
Optimum
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Single Phase Gas Flow
Example 1: Description
P2
P1
P3
P4
P5
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=50oF,P1=1000 psia, qsc=200 MMscfd, C1=90%, C2=10% (mole)
Pipeline specifications:
L1=20 km, Z1=2000 m,L2=30 km, Z2=-500,L3=15 km,Z3=600,L4=50 km,
Z4=-800, d1= d2= d3= d4= 20 in, T5=35 oF,E=0.95(in Weymouth)13
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Single Phase Gas Flow
Example 1: Solution
602.045.17,5.502Air
g
giig
o
avM
MMyMRT
psiaPP
PP
PpsiaPAssume av 7.8163
2
600 2521
3
5
3
1
5
85.0211.1,414.15.355,3.674 avprpro
pcpc zPTRTpsiaP
1387.0,1041.0,0867.0,3468.0
0375.0
432
1
1
SSSTz
Z
S avav
g
psiaPLd
TzEqPeP
f teLeLeLLLS
Series
sc
Series
e
avavggS
SSS
e
SS
e
S
eee
)655(612)/(100864.8
499488,2255.0
53/161
27
2
1
2
5
321
4
21
3
1
21
Hysys
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Single Phase Gas Flow
Example 1: Solution
psiaPLd
TzEq
PeP
f teLeLLLS
Series
sc
Series
e
avavggS
SS
e
S
eee
)711(700
)/(100864.8
279072,3642.0
43/161
27
2
1
2
4
21
3
1
21
psiaPLd
TzEqPeP
f teLLLS
Series
sc
Series
eavavggS
S
eee
)782(773)/(100864.8
211803,3642.0
33/16
1
27
21
23
1
21
psiaPL
d
TzEqPeP e
avavggS sc )800(805)/(100864.8
23/16
1
27
2
1
2
2 1
1
Hysys
Hysys
Hysys
15
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Single Phase Gas Flow
Example 2: Description
P2P1LA , dA
LB , dB
LC, dCPx
An old pipeline has the following specification:
Feed specifications:
T1=50oC,P1=1000 psia, qold=1500 MMscfd, C1=90%, C2=10% (mole)
Pipeline specifications:
LA+LC=100 km, Z=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.
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Single Phase Gas Flow
Example 2: Solution
602.045.17,575Air
g
giig
o
avM
MMyMRT
psiaPPPPPpsiaPAssume av 876
32740 2
2
2
1
3
2
3
12
905.030.1,62.15.355,3.674 avprpro
pcpc zPTRTpsiaP
)738(738)/(100864.8
,328084
23/16
27
2
1
2
2 psiaPLd
TzEqPP
f tL
A
avavoldg
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Single Phase Gas Flow
Example 2: Solution)(:costInstalled 5.15.1 CAfBbaseBBBbaseB LLxdCCLdCC
)3.68(1.69691.0 kmkmLx Bf