9 Process Synthesis.pdf

8/11/2019 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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Chen CL 3

The Chemical Plant


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Chen CL 4

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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Chen CL 5

Trade-offs in Process Synthesis


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Chen CL 6

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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Chen CL 7

Approaches in Process Synthesis

Heuristics and evolutionary search

Targets, physical insights

Optimization

Ch CL 8


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Chen CL 8

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

Ch CL 9


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Chen CL 9

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

Ch CL 10


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Chen CL 10

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

Chen CL 11


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Chen CL 11

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

Chen CL 12


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Chen CL 12

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 )

Chen CL 13


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Chen CL 13

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

Chen CL 14


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Chen CL 14

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

Chen CL 15


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Chen CL 15

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

Chen CL 16


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Chen CL 16

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

Chen CL 17


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Chen CL 17

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

Chen CL 18


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Chen CL 18

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

Chen CL 19


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Chen CL 19

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

Chen CL 20


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Chen CL 20

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

Chen CL 21


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Chen CL 21


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

Chen CL 22


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Chen CL 22


(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

Chen CL 23


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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

Chen CL 24


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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)

Chen CL 25


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Mathematical Model of SuperstructureModeling Nonlinearities of Continuous Variables

Chen CL 26


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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



y1 + y2 + y3 = 1 1 y1 , 2 y1 + y2

3 y2 + y3 , 4 y3

Chen CL 27


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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

Chen CL 28


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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

Chen CL 29


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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 )

Chen CL 30


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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

Chen CL 31


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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

Chen CL 32


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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

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