Gas Pipeline II

8/22/2019 Gas Pipeline II

1/18

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


2/18

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


3/18



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


4/18



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


5/18



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


6/18



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


7/18


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


8/18


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


9/18


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


10/18


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


11/18


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


12/18


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


13/18


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


14/18


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


15/18


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


16/18


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.


17/18


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


18/18


Example 2: Solution)(:costInstalled 5.15.1 CAfBbaseBBBbaseB LLxdCCLdCC

)3.68(1.69691.0 kmkmLx Bf

Gas Pipeline II

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