37

CHAPTER 9

Process Furnace Design Case

Study

9.1 PROBLEM STATEMENT AND SOLUTION

An economic process model was developed [1, 2] for an industrial metallurgical application. The

annual cost for a furnace operation in which the slag-metal reaction is a critical factor of the process

was considered and a modiﬁed version of the problem is presented. The objective was to minimize

the annual cost and the primal equation representing the model was:

Y = C

1

/(L

2

∗ D ∗ T

2

) + C

2

∗ L ∗ D + C

3

∗ L ∗ D ∗ T

4

. (9.1)

The model was subject to the constraint that:

D ≤ L.

The constraint must be set in geometric programming for which would be:

(D/L) ≤ 1 (9.2)

where

D = Depth of the furnace (ft)

L = Characteristic Length of the furnace (ft)

T = Furnace Temperature (K) .

For the speciﬁc example problem, the values of the constants were:

C

1

= 10

13

($ − ft

2

− K

2

)

C

2

= 100($/ft

2

)

C

3

= 5 ∗ 10

−11

(ft

−2

− K

−4

).

From the coefﬁcients and signs, the signum values for the dual are:

σ

01

= 1

σ

02

= 1

σ

03

= 1

σ

11

= 1

σ

1

= 1 .

38 9. PROCESS FURNACE DESIGN CASE STUDY

D

L

T=?

Figure 9.1: Process furnace.

The dual problem formulation is:

Objective Function ω

01

+ ω

02

+ ω

03

= 1 (9.3)

L terms −2ω

01

+ ω

02

+ ω

03

− ω

11

= 0 (9.4)

D terms −ω

01

+ ω

02

+ ω

03

+ ω

11

= 0 (9.5)

T terms −2ω

01

+ 4ω

03

= 0 . (9.6)

Using Equations (9.3)to(9.6), the values of the dual variables were found to be:

ω

01

= 0.4

ω

02

= 0.4

ω

03

= 0.2

ω

11

=−0.2 .

The dual variables cannot be negative and the negative value implies that the constraint is

not binding, that is it is a loose constraint. Thus, the problem must be reformulated without the

constraint and the dual variable is forced to zero, that is, ω

11

= 0 and the equations resolved. This

means that the constraint D ≤ L will be loose, that is D will be less than L in the solution. The new

dual becomes:

Objective Function ω

01

+ ω

02

+ ω

03

= 1 (9.7)

L terms −2ω

01

+ ω

02

+ ω

03

= 0 (9.8)

D terms −ω

01

+ ω

02

+ ω

03

= 0 (9.9)

T terms −2ω

01

+ 4ω

03

= 0 . (9.10)

Now the problem is that it has 4 equations to solve for three variables. If one examines

Equations (9.8) and (9.9), one observes that Equation (9.8) is dominant over Equation (9.9) and

thus Equation (9.9) will be removed from the dual formulation. The new dual formulation is:

9.1. PROBLEM STATEMENT AND SOLUTION 39

Objective Function ω

01

+ ω

02

+ ω

03

= 1 (9.11)

L terms −2ω

01

+ ω

02

+ ω

03

= 0 (9.12)

T terms −2ω

01

+ 4ω

03

= 0 . (9.13)

The new solution for the dual becomes:

ω

01

= 1/3

ω

02

= 1/2

ω

03

= 1/6

and by deﬁnition

ω

00

= 1.0 .

The dual variables indicate that the second term is the most important, followed by the ﬁrst

term and then the third term. The degrees of difﬁculty are now equal to:

D = T − (N + 1) = 3 − (2 + 1) = 0 .

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

(9.14)

= 1[[{(C

1

∗ 1/(1/3)1)}

(1/3∗1)

][{(C

2

∗ 1/(1/2)}

(1/2∗1)

][{(C

3

/(1/6))}

(1/6∗1)

]]

1

= 1[[{(1 ∗ 10

13

∗ 1/(1/3)1)}

(1/3∗1)

][{(100 ∗ 1/(1/2)}

(1/2∗1)

][{(5 ∗ 10

−11

/(1/6))}

(1/6∗1)

]]

1

= $11, 370 .

This can be expressed in a general form in terms of the constants as:

Y = (3C

1

)

1/3

(2C

2

)

1/2

(6C

3

)

1/6

. (9.15)

The values for the primal variables can be determined from the relationships between the

primal and dual which are:

C

1

∗ L

−2

∗ D

−1

∗ T

−2

= ω

01

Y (9.16)

C

2

∗ L ∗ D = ω

02

Y (9.17)

and C

3

∗ L ∗ D ∗ T

4

= ω

03

Y. (9.18)

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