$title Structural Optimization of Process Flowsheets (PROCSEL,SEQ=116)
$onText
The goal is the profitable production of chemical C.
which can be produced from chemical B where B may be
the raw material that can be purchased from the
external market or an intermediate that is produced
from raw material A. There are two alternative paths
of producing B from A. A mixed-integer nonlinear
formulation is presented to solve the optimal
production and capacity expansion problem.
Kocis, G R, and Grossmann, I E, Relaxation Strategy for the Structural
Optimization of Process Flow Sheets. Independent Engineering Chemical
Research 26, 9 (1987), 1869-1880.
Morari, M, and Grossmann, I E, Eds, Chemical Engineering Optimization
Models with GAMS. Computer Aids for Chemical Engineering Corporation,
1991.
Process flowsheet
A2 +-----+ B2 BP
+----->| 2 |----->+ |
A | +-----+ | | B1 +-----+ C1
---->| +----+------->| 1 |-------->
| +-----+ | +-----+
+----->| 3 |----->+
A3 +-----+ B3
Keywords: mixed integer nonlinear programming, process flow, chemical
engineering
$offText
$eolCom //
Positive Variable
a2 'consumption of chemical a in process 2'
a3 'consumption of chemical a in process 3'
b2 'production capacity of chemical b in process 2'
b3 'production capacity of chemical b in process 3'
bp 'amount of chemical b purchased in external market'
b1 'consumption of chemical b in process 1'
c1 'production capacity of chemical c in process 1';
Binary Variable
y1 'denotes potential existence of process 1'
y2 'denotes potential existence of process 2'
y3 'denotes potential existence of process 3';
Variable
pr 'total profit in million $ per year';
Equation
inout1 'input-output for process 1'
inout2 'input-output for process 2'
inout3 'input-output for process 3'
mbalb 'mass balance for chemical b'
log1 'logical constraint for process 1'
log2 'logical constraint for process 2'
log3 'logical constraint for process 3'
obj 'profit objective function';
* The original constraint for inout2 is b2 = log(1 + a2)
* but this has been convexified to the form used below.
* The same is true for inout3. so b2 and b3 are the
* output variables from units 2 and 3 respectively.
inout1.. c1 =e= 0.9*b1;
inout2.. exp(b2) - 1 =e= a2;
inout3.. exp(b3/1.2) - 1 =e= a3;
mbalb.. b1 =e= b2 + b3 + bp;
log1.. c1 =l= 2*y1;
log2.. b2 =l= 4*y2;
log3.. b3 =l= 5*y3;
obj.. pr =e= 11*c1 // sales revenue
- 3.5*y1 - y2 - 1.5*y3 // fixed investment cost
- b2 - 1.2*b3 // operating cost
- 1.8*(a2+a3) - 7*bp; // purchases
* demand constraint on chemical c based on market requirements
c1.up = 1;
Model process / all /;
solve process maximizing pr using minlp;