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The document describes optimizing the operation of a thermal cracker using linear programming. The objective is to maximize profit while satisfying furnace and process constraints. Variables are feed rates of various hydrocarbon inputs and recycled streams. The objective function expresses profit as revenue from products minus feed costs. Constraints ensure product outputs meet demands and the cracker, heat, and fuel requirements are met. Linear programming is used to determine the input rates that optimize the objective function while satisfying all constraints.

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11 views3 pages

File 000360

The document describes optimizing the operation of a thermal cracker using linear programming. The objective is to maximize profit while satisfying furnace and process constraints. Variables are feed rates of various hydrocarbon inputs and recycled streams. The objective function expresses profit as revenue from products minus feed costs. Constraints ensure product outputs meet demands and the cracker, heat, and fuel requirements are met. Linear programming is used to determine the input rates that optimize the objective function while satisfying all constraints.

Uploaded by

abir.morsi2021
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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REPUBLIQUE ALGERIENNE DEMOCRATIQUE ET POPULAIRE

Ministère de l’enseignement supérieur et


de la recherche scientifique ‫وزارة التعليــم العالــي والبحــث العلمـي‬
Ecole nationale polytechnique de ‫المدرسة الوطنية المتعددة التقنيات‬
Constantine ‫قسنطينة‬

Département Génie des procédés

Optimization

Optimization of a thermal cracker via linear programming

morsi abir katar nada


5ème année GP
2020/2021
The problem

Reactor systems that can be described by a “yield matrix” are potential candidates for the
application of linear programming. In these situations, each reactant is known to produce a
certain distribution of products. When multiple reactants are employed, it is desirable to
optimize the amounts of each reactant so that the products satisfy flow and demand
constraints. Linear programming has become widely adopted in scheduling production in
olefin units and catalytic crackers. In this example, we illustrate the use of linear
programming to optimize the operation of a thermal cracker sketched in this figure:

Figure: Flow diagram of thermal cracker.

Resolution of the problem

1. Setting up the objective function and constraints

Our objective is to maximize profit while operating within furnace and downstream
process equipment constraints.

Fuel x1
DGN x2
Gas oil x3
Propane x4
Ethane x5
Recycled propane x6
Recycled ethane x7

The objective function is formed as below:

1
Fobj = product value – feed cost

a- Product value:

Ethylene: E = 17.75*(0.25*x2 +0.2*x3+0.35*x4+0.5*x5+0.35*x6+0.5*x7)

Propylene: P = 13.79*(0.18*x2+0.15*x3 +0.15*x4 +0.01*x5 +0.15*x6 +0.01*x7)

Butadiene: B = 26.64*(0.05*x2+0.04*x3 +0.02*x4 +0.01*x5 +0.02*x6 +0.01*x7)

Gasoline: G = 9.93(0.3*x2+0.25*x3 +0.07*x4 +0.01*x5 +0.7*x6 +0.01*x7)

Product value = E+P+B+G = 11.23*x2+9.17*x3 +9.51*x4 +9.39*x5 +9.51*x6 +9.39*x7

b- Feed cost:

Feed cost = 10.14*x2+12.5*x3 +9.73*x4 +6.55*x5

So we obtain:

Fobj = 1.09*x2-3.33*x3 -0.22*x4 +2.84*x5 +9.51*x6 +9.39*x7

2. Setting up the constraints

Ethylene: E’= 0.25*x2+0.2*x3 +0.35*x4 +0.5*x5 +0.35*x6 +0.5*x7 ≤ 5000

Propylene: P’ = 0.18*x2+0.15*x3 +0.05*x4 +0.01*x5 +0.15*x6 +0.01*x7 ≤ 2000

Recycled propane: x6 = 0.01*x2+0.01*x3 +0.1*x4 +0.1*x6

0.01*x2+0.01*x3 +0.1*x4 -0.9*x6 = 0

Recycled ethane: x7 = 0.05*x2+0.04*x3 +0.06*x4 +0.4*x5 +0.06*x6 +0.4*x7

0.05*x2+0.04*x3 +0.06*x4 +0.4x5 +0.06x6 -0.6*x7 = 0

Cracker capacity: C = 1*x2+0.9*x3 +0.9*x4 +01.1*x5 +0.9*x6 +1.1*x7 ≤ 200000

Heat product: H = 18000*(0.21*x3+0.01*x2) +21520*(0.15*x2+0.1*x3 +0.25*x4 +0.07*x5


+0.25*x6 +0.07*x7) +21520*x1

H = 21.520*x1 +3408*x2+5932*x3 +5380*x4 +1506.4*x5 +5380*x6 +1506.4*x7 …(1)

Fuel requirement: Fr = 4553*x2+3900*x3 +5016*x4 +8364*x5 +5016*x6 +8364*x7 …(2)

We know that: (1) – (2) = 20000000 Btu/h

21.520*x1 -1145*x2+2032*x3 +364*x4 -6857.6*x5 +364*x6 -6857.6*x7 = 20000000

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