33
CHAPTER 8
Gas Transmission Pipeline Case
Study
8.1 PROBLEM STATEMENT AND SOLUTION
The energy crisis is with us today, and one of the problems is in the transmission of energy. A gas
transmission model was developed [20, 14] to minimize the total transmission cost of gas in a new
gas transmission pipeline. The problem is more difficult than the previous case studies as several of
the exponents are not integers. The primal expression for the cost developed was:
C = C
1
L
1/2
V /(F
0.387
D
2/3
) + C
2
D
V + C
3
/(L
F)+ C
4
F/L . (8.1)
Subject to:
(V /L) F.
The constraint must be restated in the geometric form as:
(V /(LF )) ≤−1 , (8.2)
where
L = Pipe length between compressors (feet)
D = Diameter of Pipe (in)
V = Volume Flow Rate (ft
3
/sec)
F = Compressor Pressure Ratio Factor.
Figure 8.1 is a sketch of the problem indicating the variables and is not drawn to scale.
L
D
F
F
CompressorCompressor
V
Figure 8.1: Gas transmission pipeline.
34 CHAPTER 8. GAS TRANSMISSION PIPELINE CASE STUDY
For the specific problem, the values of the constants were:
C
1
= 4.55
10
5
C
2
= 3.69
10
4
C
3
= 6.57
10
5
C
4
= 7.72
10
5
.
From the coefficients and signs, the signum values for the dual are:
σ
01
= 1
σ
02
= 1
σ
03
= 1
σ
04
= 1
σ
11
=−1
σ
1
=−1 .
The dual problem formulation is:
Objective Function ω
01
+ ω
02
+ ω
03
+ ω
04
= 1 (8.3)
L terms 0.5ω
01
ω
03
ω
04
+ ω
11
= 0 (8.4)
F terms 0.387ω
01
ω
03
+ ω
04
+ ω
11
= 0 (8.5)
V terms ω
01
+ ω
02
ω
11
= 0 (8.6)
D terms 0.667ω
01
+ ω
02
= 0 . (8.7)
Using Equations (8.3) to (8.7), the values of the dual variables were found to be:
ω
01
= 0.26087
ω
02
= 0.17391
ω
03
= 0.44952
ω
04
= 0.11570
ω
11
= 0.43478
and by definition
ω
00
= 1,
and
ω
10
= ω
mt
= σ
m
σ
mt
ω
mt
= (1)
(1
0.43478) = 0.43478 where m = 1 and t = 1 .
8.1. PROBLEM STATEMENT AND SOLUTION 35
The objective function can be found using the dual expression:
Y = d(ω) = σ
M
m=0
T
m
t=1
(C
mt
ω
mo
mt
)
σ
mt
ω
mt
σ
(8.8)
= 1[[{(4.55
10
5
1/0.26087)}
(1
0.26087)
]
[{(3.69
10
4
1/0.17391)}
(1
0.17391)
]
[{(6.57
10
5
1/0.44952)}
(1
0.44952)
]
[{(7.72
10
5
1/0.11570)}
(1
0.11570)
]
[{(1
0.43478/0.43478)}
(1
0.43478)
]]
1
= $1.3043
10
6
/ yr .
The degrees of difficulty are equal to:
D = T (N + 1) = 5 (4 + 1) = 0 . (8.9)
The values for the primal variables can be determined from the relationships between the
primal and dual which are:
C
1
L
1/2
V /(F
0.387
D
2/3
) = ω
01
Y (8.10)
C
2
D
V = ω
02
Y (8.11)
C
3
/(L
F) = ω
03
Y (8.12)
C
4
F/L = ω
04
Y (8.13)
V /(F
L) = ω
11
10
= 1 . (8.14)
The fully general expressions are somewhat difficult, but the variables can be expressed in
terms of the constants and objective function as:
F =[(C
3
ω
04
)/(C
4
ω
03
)]
1/2
(8.15)
V = C
3
/(ω
03
Y) (8.16)
L =[(C
3
C
4
)/(ω
03
ω
04
)]/Y (8.17)
D =[
02
ω
03
)/(C
2
C3)]
Y
2
. (8.18)
Using the values of Y = 1.3043
10
6
,C
1
= 4.55
10
5
,C
2
= 3.69
10
4
,C
3
= 6.57
10
5
,C
4
=
7.72
10
5
01
= 0.26087
02
= 0.17391
03
= 0.44952
04
= 0.11570, and ω
11
= 0.43478
one obtains:
F =[(C
3
ω
04
)/(C
4
ω
03
)]
1/2
=[(6.57
10
5
0.11570)/(7.72
10
5
0.44952)]
1/2
= 0.468
V = C
3
/(ω
03
Y) = 6.57
10
5
/(0.44952 1.3043
10
6
) = 1.1205 ft
3
/sec
L =[(C
3
C
4
)/(ω
03
ω
04
)]
1/2
/Y =[(6.57
10
5
7.72
10
5
)/(0.44952
0.1157)]
1/2
/1.3043
10
6
= 2.3943 ft
D =[
02
ω
03
)/(C
2
C
3
)]
Y
2
=[(0.17391
0.44952)(3.69
10
4
6.57
10
5
)]
(1.3043
10
6
)
2
= 5.4857 in .

Get Geometric Programming for Design and Cost Optimization now with O’Reilly online learning.

O’Reilly members experience live online training, plus books, videos, and digital content from 200+ publishers.