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 i1 Pn21  K   e j1
L
ei 
5
di

i 1 


n
i 1

5
n  S j z T
Si

f
f


 j1 avi avi mi db

mb
2
2
i 1



P1  e Pn1  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


 j1 avi avi mi db

mb
2
2
i 1



P1  e Pn1  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 Pn21 
2.527 105  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 j1
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 Pn21 
0.032
 1/ 3 (Weymouth)
di
8.0864 107  g (qg sc / E ) 2 zavTav
16 / 3
1
d
LeSeries

16 / 3
n  S j
 j1 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)