9 process synthesis.pdf
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Floudas: Nonlinear and Mixed-Integer OptimizationChapter 7
Process Synthesis
Cheng-Liang ChenPSELABORATORY
Department of Chemical EngineeringNational TAIWAN University
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The Chemical Plant
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Key Questions
Q1: Which process units should be used in the processowsheet ?
Q2: How should the process units be interconnected ?
Q3: What are the optimal operating conditions andsizes of the selected process units ?
Multi-objective mixed discrete-continuous optimizationproblem
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Trade-offs in Process Synthesis
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Difficulties/Challenges in Process Synthesis
Q1: Combinatorial nature How can we deal with the large combinatorialproblem effectively ?
Q2: Nonlinear (nonconvex) characteristics How can we cope with the highly nonlinear modelw.r.t. the quality of its solution (i.e., local vs. global) ?
h
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Approaches in Process Synthesis
Heuristics and evolutionary search
Targets, physical insights
Optimization
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Optimization Approach in ProcessSynthesis
Step 1: Representation of AlternativesA superstructure is postulated in which all process aternativesstructures of interest are embedded and hence are candidates forfeasible or optimal process owsheets
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Step 2: Mathematical Model
minx ,y f (x , y )
s.t. h (x , y ) = 0
g (x , y ) 0
x X Rn
y Y = {0, 1} (?)
Step 3: Algorithmic Development
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Mathematical Model of SuperstructureModeling with 0 1 Variables Only
Propositional Logic Expressions:
OR denoted as AND denoted as
EXCLUSIVE OR denoted as NEGATION denoted as
P 1 P 2 is equivalent to P 1 P 2
P 1 P 2 P 1 P 2 P 1 P 2 P 1 P 2 P 1 P 1 P 2 P 1 P 21 1 1 1 1 0 1 01 0 0 1 0 0 0 10 1 0 1 1 1 1 10 0 0 0 1 1 1 0
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P i yi P i 1 yi
yi = 1 if clause P i is true0 if clause P i is false
Proposition Mathematical Representation1. P 1 P 2 P 3 y1 + y2 + y3 1
2. P 1 P 2 P 3
y1 1
y2 1y3 1
3. P 1 P 2 (P 1 P 2 ) 1 y1 + y2 1 (y1 y2 0)
4.(P 1 P 2 ) (P 2 P 1 )(P 1 P 2 )(P 1 if and only if P 2 )
y1 y2 0
y2 y1 0or y1 = y2
5. P 1 P 2 P 3 (exactly one) y1 + y2 + y3 = 1
6. (P 1 P 3 ) (P 2 P 3 )
(P 1 P 2 P 3 )y1 y3 0y2 y3 0
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Procedure to Obtain a Conjunctive Normal Form:
Step 1: Replace the implication by its equivalent disjunction
P 1 P 2 P 1 P 2
Step 2: Apply DeMorgans Theorem to put the negation inward
(P 1 P 2 ) P 1 P 2(P 1 P 2 ) P 1 P 2
Step 3: Distribute the logical OR over the logical ANDrecursively using the equivalent of
(P 1 P 2 ) P 3 = (P 1 P 3 ) (P 2 P 3 )
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Illustration:
(P 1 P 2 ) P 3 P 4 P 5
[(P 1 P 2 ) P 3 ] (P 4 P 5 )
[(P 1 P 2 ) P 3 ] (P 4 P 5 )[(P 1 P 2 ) P 3 ] (P 4 P 5 )
[(P 1 P 2 ) (P 4 P 5 )] [P 3 (P 4 P 5 )]
[P 1 P 2 P 4 P 5 ] [P 3 P 4 P 5 ]
1 y1 + 1 y2 + y4 + y5 1
1 y3 + y4 + y5 1
or y1 + y2 y4 y5 1
y3 y4 y5 0
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Select Among Process Units i P U :
iP U
yi = 1 select only one unit
iP U
yi 1 select at most one unit
iP U
yi 1 select at least one unit
Select Unit i If Unit j Is Selected:
If unit j is selected (P j is true or yj = 1)then unit i should be selected too (P i is true or yi = 1)
Others: not denedP
j P
i P
jP
i (1 y
j) + y
i 1 y
j y
i 0
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Mathematical Model of SuperstructureModeling with Continuous and Linear 0 1 Variables
Activation and Deactivation of ContinuousVariables:
F Li yi F i F U i yi
F i = 0 for yi = 0
F Li F i F U i for yi = 1
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Activation and Relaxation of Constraints:If unit i does not exist (yi = 0) , relax equality/inequality constraintsIf unit i exists (yi = 1) , both equality/inequality constraints should
be activated
h (x ) + s +1 s1 = 0
g(x ) s2
s+1 + s
1 U 1 (1 yi )
s 2 U 2 (1 yi )s +1 , s
1 , s 2 0
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Nodes with Several Inputs:
Alternative 1:m
j =1
F j Uyi 0
F j 0, j = 1 , m
Alternative 2: F j Uyi 0F j 0, j = 1 , m
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Logical Constraints in Multiperiod Problems:If unit i is not selected (zi = 0),then yti = 0 for all periods of operation
Alternative 1:T
t =1
yti T zi 0
Alternative 2: yti zi 0 i = 1 , . . . , T
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Either-Or Constraints:
Either f 1 (x ) 0 or f 2 (x ) 0
f 1 (x ) U (1 y1 ) 0f 2 (x ) Uy1 0
y1 = 1f 1 (x ) 0f 2 (x ) U
ory1 = 0
f 1 (x ) U f 2 (x ) 0
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Constraint Functions in Logical Expressions:Illustration
If f (x ) 0, then g(x ) 0
P 1 , f (x ) 0 : L1 y1 + f (x ) U 1 (1 y1 )P 2 , g(x ) 0 : L2 (1 y2 ) g(x ) U 2 y2
P 1 P 2 = P 1 P 21 y1 + y2 1 (y1 y2 0)
y1 y2 0L 1 y1 + f (x ) U 1 (1 y1 ),
L 2 (1 y2 ) g(x ) U 2 y2
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Constraint Functions in Logical Expressions:Illustration
If f (x ) 0
P 1
and h(x ) = 0
P 2
, then g(x ) 0
P 3
P 1 P 2 P 3 (P 1 P 2 ) P 3 P 1 P 2 P 3 1 y1 + 1 y2 + y3 1 (y1 + y2 y3 1)
y1 + y2 y3 1L 1 y1 + f (x ) U 1 (1 y1 )L 2 (1 y3 ) g(x ) U 3 y3
h (x ) + s+1 s
1 = 0
s +1 + s1 U 2 (1 y2 )
s +1 , s1 0
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Constraint Functions in Logical Expressions:Illustration
(x ) = max {0, f (x )} = f (x ) if f (x ) 00 if f (x ) < 0
L 1 (1 y1 ) f (x ) U 1 y1L 2 (1 y1 ) (x ) f (x ) U 2 (1 y1 )
L 3 y1 (x ) U 3 y1
(x ) 0, (x ) f (x ) 0 L2 = L3 = 0
L 1 (1 y1 ) f (x ) U 1 y1 0 (x ) f (x ) U 2 (1 y1 )0 (x ) U 3 y1
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Mathematical Model of SuperstructureModeling with Bilinear Products of Continuous and
0 1 Variables Illustration:min
i j
x ij yij
Let hij = xij yij i, jx ij U (1 yij ) hij xij L (1 yij )
Ly ij hij Uyij
If yij = 1x ij hij xij
L hij U
If yij = 0x ij U h ij xij L
0 h ij 0
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Illustration:
L 0 (w) x U 0 (w) when y = 0L 1 (w) x U 1 (w) when y = 1
L 0 (w) + [ L 1 (w) L 0 (w)]y x U 0 (w) + [ U 1 (w) U 0 (w)]y (nonlinear product!)
L 0 L 0 (w) L 0
U 0 U 0 (w) U 0L 1 L 1 (w) L 1U 1 U 1 (w) U 1
L 0 (w) + [ L 1 L 0 ]y x U 0 (w) + [ U 1 U 0 ]yL 1 (w) + [ L 0 L 1 ](1 y) x U 1 (w) + [ U 0 U 1 ](1 y)
If y = 0L 0 (w) x U 0 (w)
L 1 (w) + L 0 L 1 x U 1 (w) + U 0 U 1
If y = 1L 0 (w) + L 1 L 0 x U 0 (w) + U 1 U 0
L 1 (w) x U 1 (w)
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Mathematical Model of SuperstructureModeling Nonlinearities of Continuous Variables
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Modeling Separable Concave Functions:
Separable Costs: C (x ) = C 1 (x) + C 2 (x) + C 3 (x)
A : [ 1 , C ( 1 )]; B : [ 2 , C ( 2 )]; C : [ 3 , C ( 3 )]; D : [ 4 , C ( 4 )];
x = 1 1 + 2 2 + 3 3 + 4 4
C (x ) = 1 C ( 1 ) + 2 C ( 2 )+ 3 C ( 3 ) + 4 C ( 4 )
1 + 2 + 3 + 4 = 1 1 , 2 , 3 , 4 0
Ex:
x = 2 2 + 3 3
C (x ) = 2 C ( 2 ) + 3 C ( 3 ) 2 + 3 = 1 2 , 3 0( 1 , 4 = 0)
y1 = 1 if 1 x 20 otherwise
y2 = 1 if 2 x 30 otherwise
y3 = 1 if 3 x 40 otherwise
y1 + y2 + y3 = 1 1 y1 , 2 y1 + y2
3 y2 + y3 , 4 y3
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3
=segments
C (x ) = 1 C ( 1 ) + + 4 C ( 4 )x = 1 1 + + 4 4 1 + 2 + 3 + 4 = 1 1 y1 2 y1 + y2 3 y2 + y3 4 y3y1 + y2 + y3 = 1 1 , 2 , 3 , 4 0
y1 , y2 , y3 {0, 1}3
K
=segments
C (x) =K +1
i =1
i C ( i )
x =K +1
i =1
i i
K +1
i =1
i = 1
1 y1 i yi 1 + yi , i = 2 , . . . , K K +1 yK
K
i =1
yi = 1
i 0, iyi {0, 1}K
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Convexication (of Nonlinear Fcn.s of Continuous Var.s) :
f (x ) =j
cj i x ii , cj 0, x i 0
zi = ln x i xi = ez i
f (z ) =j
cj i x ii =j
cj i e i z i =j
cj ew
w, ew
: convex f (z ) : convex
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Convexication: Illustration
f (x 1 , x 2 , x 3 ) = x1 x 2 + x 0 .61 + x 1x 3
z1 = ln x 1 z2 = ln x 2 z3 = ln x 3
f (z1 , z2 , z 3 ) = ez 1 ez 2 + e0 .6 z 1 + ez 1 z 3
= ez 1 + z 2 + e0 .6 z 1 + ez 1 z 3
(convex in z1 , z 2 , z3 )
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Convexication: Illustration
f (x 1 , x 2 , x 3 , x 4 ) = x1 x 2 x 3 x 4
z1 = ln x 1 , z2 = ln x 2 , z3 = ln x 3 , z4 = ln x 4
f (z1 , z 2 , z3 , z 4 ) = ez 1 + z 2 ez 3 + z 4
difference of two convex fcn.s
non-convex
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Partitions (1)
f =f 1 for x af 2 for x > a
f = f 1 y + f 2 (1 y)a (1 y) Uy + x ay + U (1 y)
y = 0 , 1
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Partitions (2)
f =
f 1 for x a
f 2 for a < x bf 3 for x > b
f = f 1 y1 + f 2 y2 + f 3 y3y1 + y2 + y3 = 1a (1 y1 ) Uy1 + x ay 1 + U (1 y1 )ay 2 U (1 y2 ) + x by2 + U (1 y2 )
by3 U (1 y3 ) + x b(1 y3 ) + Uy3y1 , y2 , y3 = 0 , 1