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

/ω

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

/ω

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 =[(ω

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

,ω

01

= 0.26087,ω

02

= 0.17391,ω

03

= 0.44952,ω

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 =[(ω

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 .

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