Linear programming is to be used to optimize the operation of the solvent splitter column shown in Fig. E19.6. The feed is naphtha, which has a value of $40/bbl in its alternate use as a gasoline blending stock. The light ends sell at $50/bbl, while the bottoms are passed through a second

Question

Linear programming is to be used to optimize the operation of the solvent splitter column shown in Fig. E19.6. The feed is naphtha, which has a value of [latex]\$ 40 / bbl[/latex] in its alternate use as a gasoline blending stock. The light ends sell at [latex]\$ 50 / bbl[/latex], while the bottoms are passed through a second distillation column to yield two solvents. A medium solvent comprising 50 to [latex]70 \%[/latex] of the bottoms can be sold for [latex]\$ 70 / bbl[/latex], while the remaining heavy solvent ( 30 to [latex]50 \%[/latex] of the bottoms) can be sold for [latex]\$ 40 / bbl[/latex].

Another part of the plant requires [latex]200 bbl /[/latex] day of medium solvent; an additional [latex]200 bbl / day[/latex] can be sold to an external market. The maximum feed that can be processed in column 1 is [latex]2,000 bbl / day[/latex]. The operational cost (i.e., utilities) associated with each distillation column is [latex]\$ 2.00 / bbl[/latex] feed. The operating range for column 2 is given as the percentage split of medium and heavy solvent. Solve the linear programming problem to determine the maximum revenue and percentages of output streams in column [latex]2 .[/latex]

Accepted Answer

Objective function is to maximize the revenue,

[latex]\max R=-40 x_{1}+50 x_{3}+70 x_{4}+40 x_{5}-2 x_{1}-2 x_{2}[/latex] (1)

*Balance on column 2

[latex]x_{2}=x_{4}+x_{5}[/latex] (2)

* From column 1,

[latex]\begin{aligned}x_{1} &=\frac{1.0}{0.60} x_{2}=1.667\left(x_{4}+x_{5} ight) (3) \x_{3} &=\frac{0.4}{0.60} x_{2}=0.667\left(x_{4}+x_{5} ight) (4)\end{aligned}[/latex]

Inequality constraints are

[latex]\begin{aligned}&x_{4} \geq 200 (5) \&x_{4} \leq 400 (6)\&x_{1} \leq 2000 (7)\&x_{4} \geq 0 \quad x_{5} \geq 0 (8)\end{aligned}[/latex]

The restricted operating range for column 2 imposes additional inequality constraints. Medium solvent is 50 to [latex]70 \%[/latex] of the bottoms; that is

[latex]0.5 \leq \frac{x_{4}}{x_{2}} \leq 0.7 ext { or } \quad 0.5 \leq \frac{x_{4}}{x_{4}+x_{5}} \leq 0.7[/latex]

Rewriting in linear form,

[latex]0.5 x_{2} \leq x_{4} \leq 0.7 ext { or } 0.5\left(x_{4}+x_{5} ight) \leq x_{4} \leq 0.7\left(x_{4}+x_{5} ight)[/latex]

Simplifying,

[latex]\begin{aligned}&x_{4}-x_{5} \geq 0 (9) \&0.3 x_{4}-0.7 x_{5} \leq 0 (10)\end{aligned}[/latex]

No additional constraint is needed for the heavy solvent. That the heavy solvent will be 30 to [latex]50 \%[/latex] of the bottoms is ensured by the restriction on the medium solvent and the overall balance on column 2 .

By using Excel-Solver,

	[latex]x_{1}[/latex]	[latex]x_{2}[/latex]	[latex]x_{3}[/latex]	[latex]x_{4}[/latex]	[latex]x_{5}[/latex]
Initial values	1	1	1	1	1
Final values	1333.6	800	533.6	400	400

max R = 13068.8

Constraints
[latex]X_{2}-X_{4}-X_{5}[/latex]	0
[latex]X_{1}-1.667 X_{2}[/latex]	7.467E-10
[latex]X_{3}-0.667 X_{2}[/latex]	-1.402E-10
[latex]X_{4}[/latex]	400
[latex]X_{4}[/latex]	400
[latex]X_{1}-1.667 X_{2}[/latex]	1333.6
[latex]X_{4}-X_{5}[/latex]	0
[latex]0.3 X_{4}-0.7 X_{4}[/latex]	-160

Table S19.6. Excel solution

Thus the optimum point is [latex]x_{1}=1333.6, x_{2}=800 ; x_{3}=533.6, x_{4}=400[/latex] and [latex]x_{5}=400[/latex].

Substituting into (5), the maximum revenue is [latex]13,068 \$[/latex] /day, and the percentage of output streams in column 2 is [latex]50 \%[/latex] for each stream.

Question 19.6: Linear programming is to be used to optimize the operation o...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions