9 Process Synthesis.pdf

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Process Synthesis Floudas: Nonlinea Nonlinearr and Mixed-Integer Optimization Chapter 7

Process Synthesis

Given the specifications of the inputs (e.g., feed streams) that may corresponds to raw materia materials, ls, and the specificat specification ionss of the outputs outputs (e. (e.g., g., pr produc oducts, ts, by by-products) in a process system

Develop systematically process flowsheet(s) that transform the available raw materials into the desired products and which meet the specified performance criteria of Maximum profit or minimum cost Energy efficiency ☞ Good operability with respect to ✏ Flexibility ✏ Controllability ✏ Reliability ✏ Safety ✏ Environmental regulations ☞

Cheng-Liang Chen

☞

PSE

LABORATORY

Department of Chemical Engineering National TAIWAN University

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An Overall Process System

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The Chemical Plant

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Key Questions

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Trade-offs in Process Synthesis

Q1: Which process units should be used in the process flowsheet ? Q2: How should the process units be interconnected ? Q3: What are the optimal operating conditions and sizes of the selected process units ?

⇒ Multi-objective mixed discrete-continuous optimization problem

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Approaches in Process Synthesis

Difficulties/Challenges in Process Synthesis Q1: Combinatorial nature How can we deal with the large combinatorial problem effectively ?

➢

Heuristics and evolutionary search

➢

Targets, physical insights

Q2: Nonlinear (nonconvex) characteristics How can we cope with the highly nonlinear model w.r.t. the quality of its solution (local vs. global) ?

➢

Optimization

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Optimization Approach in Process Synthesis ➢

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Optimization Approach in Process Synthesis ➢

Step 1: Representation of Alternatives

Step 2: Mathematical Model

A superstructure is postulated in which ALL process alternatives

min f (x, y ) x,y

structures of interest are embedded and hence are candidates for

s.t. h(x, y) = 0

feasible or optimal process flowsheets

g (x, y ) ≤

0

x ∈ X ⊆ Rn y ∈ Y = {0, 1} ➢

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(?)

Step 3: Algorithmic Development

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Superstructure Modeling with 0 − 1 Variables


Application: select among process units i ∈ P U

Application: select among process units i ∈ P U

Let yi =

 

1 if unit i is

selected

i ∈ P U

0 if unit i is NOT selected

Select one & only one unit ⇒

  

yi = 1

i∈P U

Select at most one unit

⇒

Suppose yi = 1 if unit i ∈ P U is selected, = 0 otherwise yj = 1 if unit j ∈ P U is selected, = 0 otherwise

Question:

If unit j is selected (yj = 1) then unit i should be selected too (yi = 1)

Others: not defined yi ≤ 1

⇒ yj − y i ≤ 0

(check it)

i∈P U

Select at least one unit

⇒

i∈P U

Derivation: yi = 1 ⇔ P i is true; yi ≥ 1

yj = 1 ⇔ P j is true

P j ⇒ P i ⇒ ¬P j ∨ P i (?)

⇒ (1 − yj ) + yi ≥ 1 (?) ⇒ y

y ≤0

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Propositional Logic Expressions

Propositional Logic Expressions P i ≡ yi

denoted as ∨ denoted as ∧ denoted as ⊕ denoted as ¬

OR AND EXCLUSIVE OR NEGATION

⇒ yi

P 1 ⇒ P 2 is equivalent to ¬P 1 ∨ P 2

⇔ ¬P i ≡ 1 − yi 1 if clause P i is true = 0 if clause P i is false



Proposition

Mathematical Representation

1.

P 1 ∨ P 2 ∨ P 3

y1 + y2 + y3 ≥ 1 y1 ≥ 1

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 2

2.

P 1 ∧ P 2 ∧ P 3

1 1 0 0

1 0 1 0

1 0 0 0

1 1 1 0

1 0 1 1

0 0 1 1

1 0 1 1

0 1 1 0

3.

¬P 1 ∨ P 2 (P 1 ⇒ P 2) (¬P 1 ∨ P 2) ∧ (¬P 2 ∨ P 1) (P 1 ⇔ P 2) (P 1 if and only if P 2) P 1 ⊕ P 2 ⊕ P 3 (exactly one) (¬P 1 ∨ P 3) ∧ (¬P 2 ∨ P 3) (P 1 ∨ P 2 ⇒ P 3)

y3 ≥ 1

4. 5. 6.

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y1 − y2 ≤ 0 y2 − y1 ≤ 0

or y 1 = y 2

y1 + y2 + y3 = 1 y1 − y3 ≤ 0 y2 − y3 ≤ 0

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Procedure to Obtain a Conjunctive Normal Form

Procedure to Obtain a Conjunctive Normal Form Illustration

Step 1: Replace the implication by its equivalent disjunction ⇐⇒ ¬P 1 ∨ P 2

Step 2: Apply DeMorgan’s Theorem to put the negation inward ¬(P 1 ∧ P 2) ⇐⇒ ¬P 1 ∨ ¬P 2 ¬(P 1 ∨ P 2) ⇐⇒ ¬P 1 ∧ ¬P 2

➢

1 − y1 + y2 ≥ 1 (y1 − y2 ≤ 0)


P 1 ⇒ P 2 ➢

y2 ≥ 1

Step 3: Distribute the logical OR over the logical AND recursively using the equivalent of (P 1 ∧ P 2) ∨ P 3 =⇒ (P 1 ∨ P 3) ∧ (P 2 ∨ P 3)

(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] ⇓ y1 + y2 − y4 − y5 ≤ 1 1 − y1 + 1 − y2 + y4 + y5 ≥ 1 or ≥1 ≤0 1

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Superstructure Modeling with Continuous and Linear 0 − 1 Variables


Activation and Deactivation of Continuous Variables

Activation and Relaxation of Constraints If unit i does not exist (yi = 0), relax equality/inequality constraints If unit i exists (yi = 1), both equality/inequality constraints should be activated

F iLyi ≤ F i ≤ F iU yi

⇓ F i = 0

− h(x) + s+ = 0 1 − s1

for yi = 0

g (x) ≤ s2

F iL ≤ F i ≤ F iU for yi = 1

+ 1

s + s− ≤ U 1 · (1 − yi) 1 s2 ≤ U 2 · (1 − yi) + 1

−

s , s1 , s2 ≥ 0

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Nodes with Several Inputs

Logical Constraints in Multiperiod Problems

m

Alternative 1:



If unit i is not selected (zi = 0), then yit = 0 for all periods of operation

F j − U yi ≤ 0

j =1

F j ≥ 0,

j = 1, m

T

Alternative 1: Alternative 2:

yit − T · zi ≤ 0

t=1

F j − U yi ≤ 0 F j ≥ 0,



j = 1, m

Alternative 2:

yit − zi ≤ 0 ∀i = 1, . . . , T

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Either-Or Constraints

Constraint Functions in Logical Expressions

Either f 1(x) ≤ 0 or f 2(x) ≤ 0

If f (x) ≤ 0, then g (x) ≥ 0

⇓

⇓

f 1(x) − U (1 − y1) ≤ 0

P 1, f (x) ≤ 0 :

f 2(x) − U y1 ≤ 0

P 2, g (x) ≥ 0 : L2(1 − y2) ≤ g (x) ≤ U 2y2 − 

⇓ y1 = 1

⇓P 1 ⇒ P 2 y1 = 0

or

f 1(x) ≤ 0 f 2(x) ≤ U

L1y1 +  ≤ f (x) ≤ U 1(1 − y1)

¬P 1 ∨ P 2⇓

=⇒

1 − y1 + y2 ≥ 1 (y1 − y2 ≤ 0)

f 1(x) ≤ U

⇓

f 2(x) ≤ 0

y1 −

≤ 0

y2

L1y1 +  ≤ f (x) ≤ U 1(1 − y1), L2(1 − y2) ≤ g (x) ≤ U 2y2 − 

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If f (x) ≤ 0 and h(x) = 0, then g (x) ≥ 0

         P 1

P 2

φ(x) = max{0, f (x)} =

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)

⇓

L1(1 − y1) ≤

L1y1 +  ≤ f (x) ≤ U 1(1 − y1) L2(1 − y3) ≤ g (x) ≤ U 3y3 −  − h(x) + s+ 1 − s1 = 0 − s+ 1 + s1 ≤ U 2 · (1 − y2 )

s+ s− ≥ 0

f (x) if f (x) ≥ 0

0

if f (x) < 0 ≤ U 1y1

f (x)

⇒ L2(1 − y1) ≤ φ(x) − f (x) ≤ U 2(1 − y1) L3y1 ≤

y1 + y2 − y3 ≤ 1

 

≤ U 3y1

φ(x)

φ(x) ≥ 0, ⇓ φ(x) − f (x) ≥ 0 ⇓ L1(1 − y1) ≤

f (x)

L2 = L 3 = 0

≤ U 1y1 − 

0 ≤ φ(x) − f (x) ≤ U 2(1 − y1) 0 ≤

φ(x)

≤ U 3y1

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Superstructure Modeling with Bilinear Products of Continuous and 0 − 1 Variables min

 i

Let ⇒

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L0(w ) ≤ x ≤ U 0(w ) when y = 0

xij yij

L1(w ) ≤ x ≤ U 1(w ) when y = 1

j

hij = xij yij

⇓

∀i, j

L0(w ) + [ L1(w ) − L0(w )]y ≤ x ≤ U 0(w) + [ U 1(w ) − U 0(w)]y (nonlinear product!)

⇓

xij − U (1 − yij ) ≤ hij ≤ xij − L(1 − yij )

L0 ≤ L0(w ) ≤ L0

Lyij ≤ hij ≤ U yij

U 0 ≤ U 0(w ) ≤ U 0 L1 ≤ L1(w ) ≤ L1

⇓ If yij = 1

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Superstructure Modeling with Bilinear Products of Continuous and 0 − 1 Variables

U 1 ≤ U 1(w ) ≤ U 1

If yij = 0

L0(w ) + [ L1 − L0]y ≤ x ≤ U 0(w ) + [ U 1 − U 0]y

xij ≤ hij ≤ xij

xij − U ≤ hij ≤ xij − L

L1(w) + [ L0 − L1](1 − y) ≤ x ≤ U 1(w ) + [ U 0 − U 1](1 − y )

L ≤ hij ≤ U

0 ≤ hij ≤ 0

⇓ If y = 0 If y = 1 L0(w ) ≤ x ≤ U 0(w) L0(w ) + L1 − L0 ≤ x ≤ U 0(w ) + U 1 − U 0 L1(w ) + L0 − L1 ≤ x ≤ U 1(w) + U 0 − U 1 L1(w ) ≤ x ≤ U 1(w )

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Modeling Nonlinearities of Continuous Variables

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Modeling Nonlinearities of Continuous Variables Separable Concave Functions

y1 =



Separable Costs: C (x) = C 1(x) + C 2(x) + C 3(x) ⇓ A : [γ 1, C (γ 1)]; B : [γ 2, C (γ 2)]; C : [γ 3, C (γ 3)]; D : [γ 4, C (γ 4)]; ⇓ x = λ 1γ 1 + λ2γ 2 + λ3γ 3 + λ4γ 4 x = λ 2γ 2 + λ3γ 3 C (x) = λ 1C (γ 1) + λ2C (γ 2) C (x) = λ 2C (γ 2) + λ3C (γ 3) Ex: λ2 + λ3 = 1 + λ3C (γ 3) + λ4C (γ 4) λ1 + λ2 + λ3 + λ4 = 1 λ2, λ3 ≥ 0 λ1, λ2, λ3, λ4 ≥ 0 (λ1, λ4 = 0) ⇓ 1 if γ 1 ≤ x ≤ γ 2 1 if γ 2 ≤ x ≤ γ 3 1 if γ 3 ≤ x ≤ γ 4 y2 = y3 = 0 otherwise 0 otherwise 0 otherwise y1 + y2 + y3 = 1 λ1 ≤ y 1 , λ2 ≤ y1 + y2 λ3 ≤ y2 + y3, λ4 ≤ y3





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Separable Concave Functions

Convexification of Nonlinear Fcn.s of Continuous Var.s

              

f (x) =

K +1

C (x) =

3 =⇒ segments

       

C (x) = λ 1C (γ 1) + · · · + λ4C (γ 4)

i=1

λiγ i

f (z ) =

K +1

λ 1 ≤ y1

=⇒

λ3 ≤ y2 + y3

segments

λ 4 ≤ y3 y1 + y2 + y3 = 1 λ1, λ2, λ3, λ4 ≥ 0 y1, y2, y3 ∈ {0, 1}3



α

cj Πixi i =

j

λi = 1

K

λ2 ≤ y1 + y2

cj ≥ 0, xi ≥ 0

zi = ln xi ⇓ xi = e zi

i=1

λ1 + λ2 + λ3 + λ4 = 1

α

cj Πixi i ,

j

K +1

x =

x = λ 1γ 1 + · · · + λ4γ 4

λiC (γ i )



λ 1 ≤ y1

cj Πieαizi =

j

w, ew : convex

i=1

 ⇒



cj ew

j

f (z ) : convex

λi ≤ yi−1 + yi , i = 2, . . . , K λK +1 ≤ yK K

yi = 1

i=1

λi ≥ 0,

∀i

yi ∈ {0, 1}K

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Convexification: Illustration

Convexification: Illustration

f (x1, x2, x3) = x 1x2 + x01.6 + xx1 3

z1 = ln x1 ⇓ z2 = ln x2 ⇓ z3 = ln x3 f (z1, z2, z3) = ez1 ez2 + e0.6z1 + ez1−z3

= ez1+z2 + e0.6z1 + ez1−z3 (convex in z1, z2, z3)

f (x1, x2, x3, x4) = x 1x2 − x3x4 z1 = ln x1, z2 = ln x2,

⇓ z3 = ln x3, z4 = ln x4

f (z1, z2, z3, z4) = ez1+z2 − ez3+z4

difference of two convex fcn.s ⇒ non-convex

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Partitions (1) f =

⇓

 

Partitions (2)

f 1 for x ≤ a f =

f 2 for x > a

⇓

f = f 1y + f 2(1 − y )

 

f 1 for x ≤ a f 2 for a < x ≤ b f 3 for x > b

f = f 1y1 + f 2y2 + f 3y3

a(1 − y ) − U y +  ≤ x ≤ ay + U (1 − y )

y1 + y2 + y3 = 1

y = 0, 1

a(1 − y1) − U y1 +  ≤ x ≤ ay1 + U (1 − y1) ay2 − U (1 − y2) +  ≤ x ≤ by2 + U (1 − y2) by3 − U (1 − y3) +  ≤ x ≤ b(1 − y3) + U y3 y1, y2, y3 = 0, 1

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Thank You for Your Attention Questions Are Welcome

9 Process Synthesis.pdf

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