Transcript Slide 1

Chapter 1
•USES OF OPTIMIZATION
•FORMULATION OF OPTIMIZATION PROBLEMS
•OVERVIEW OF COURSE
1
Chapter 1
OPTIMIZATION OF CHEMICAL PROCESSES
T.F. EDGAR, D.M. HIMMELBLAU, and L.S. LASDON
UNIVERSITY OF TEXAS
MCGRAW-HILL – 2001 (2nd ed.)
PART I – PROBLEM FORMULATION
II – OPTIMIZATION THEORY AND METHODS
III – APPLICATIONS OF OPTIMIZATION
APPENDICES (MATRIX OPERATIONS)
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PHILOSOPHY OF BOOK
Chapter 1
•Most undergraduates learn by seeing how a method
is applied
•Practicing professionals need to be able to recognize
when optimization should be applied (Problem formulation)
•Optimization algorithms for reasonably-sized problems
are now fairly mature
•Focus on a few good techniques rather than encyclopedic
coverage of algorithms
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Chapter 1
Chapter 1
The Nature and Organization of
Optimization Problems
4
WHY OPTIMIZE?
1. Improved yields, reduced pollutants
Chapter 1
2. Reduced energy consumption
3. Higher processing rates
4. Reduced maintenance, fewer shutdowns
5. Better understanding of process (simulation)
But there are always positive and negative factors to be
weighed
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6
Chapter 1
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Chapter 1
Chapter 1
OPTIMIZATION
• Interdisciplinary Field
Max Profit
Min Cost
Max Efficiency
• Requires
1. Critical analysis of process
2. Definition of performance objective
3. Prior experience (engr. judgment)
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9
Chapter 1
10
Chapter 1
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Chapter 1
Chapter 1
Min reflux to
achieve separation
Figure E1.4-3
Flooding
constraint
Optimal Reflux for Different Fuel Costs
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13
Chapter 1
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Chapter 1
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Chapter 1
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Chapter 1
Chapter 1
Material Balance Reconciliation
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Least squares solution:
P
min
 (m
Chapter 1
i 1
A
 m C i  m Bi )
2
opt. mA is the “average” value
any constraints on mA?
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Chapter 1
THREE INGREDIENTS IN OPTIMIZATION PROBLEM
1 . O b je c tiv e fu n c tio n
e c o n o m ic m o d e l


3 . In e q u a lity C o n s tra in ts 
P ro c e s s m o d e l
Chapter 1
2 . E q u a lity C o n s tra in ts
1 . m in f( x )
x n x1
2 . su b je ct to h( x )  0
(m 1 )
3 . g( x )  0
(m 2 )
(fe a sib le
2
re g io n  :)
3


in d e p e n d e n t v a ria b le s 
d e p e n d e n t v a ria b le s
re la te to m 1 a n d p e rh a p s m 2
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Chapter 1
TABLE 1
Chapter 1
THE SIX STEPS USED TO SOLVE OPTIMIZATION PROBLEMS
1. Analyze the process itself so that the process variables
and specific characteristics of interest are defined, i.e.,
make a list of all of the variables.
2. Determine the criterion for optimization and specify
the objective function in terms of the above variables
together with coefficients. This step provides the
performance model (sometimes called the economic
model when appropriate).
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Chapter 1
3. Develop via mathematical expressions a valid process
or equipment model that relates the input-output variables
of the process and associated coefficients. Include both
equality and inequality constraints. Use well-known
physical principles (mass balances, energy balances),
empirical relations, implicit concepts, and external
restrictions. Identify the independent and dependent
variables (number of degrees of freedom).
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Chapter 1
4. If the problem formulation is too large in scope:
(A)Break it up into manageable parts and/or
(B)Simplify the objective function
5. Apply a suitable optimization technique to the
mathematical statement of the problem.
6. Check the answers and examine the sensitivity of the
result to changes in the coefficients in the problem and
the assumptions.
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EXAMPLES – SIX STEPS OF OPTIMIZATION
Chapter 1
specialty chemical
100,000 bbl/yr.
2 costs inventory (carrying) or storage, production cost >
how many bbl produced per run?
Step 1
define variables
Q = total # bbl produced/yr (100,000)
D = # bbl produced per run
n = # runs/yr
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Step 2
develop objective function
Chapter 1
inventory, storage cost = k1D
production cost
= k2
per run
(set up
cost)
+
k3
D
operating
cost per unit
(could be nonlinear)
C  k1 D  n ( k 2  k 3 D )
n
Q
D
C  k1 D  k 2
Q
D
 k 3Q
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Step 3
evaluate constraints
n
integer
Chapter 1
continuous
D>0
Step 4
simplification – none necessary
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Step 5
computation of the optimum
analytical vs. numerical solution
Chapter 1
dC
 k1 
dD
D
k 2Q
0
k 2Q

opt
D
2
k1
k1  1 .0
Q  10
D
opt
k 2  10 , 000
k 3  4 .0
5
 31 , 622
flat optimum
30 , 000  D  70 , 000
 good answer
check if minimum?
2
d C
dD
2

2 k 2Q
D
3
0
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Chapter 1
suppose cost per run  k 2  k 4 D

dC
 k1 
dD
Chapter 1
analytical
k 2Q
D
2
k 4Q

2D
3/2
1/ 2
0
solution?
Step 6
Sensitivity of the optimum
subst Dopt into C
C
opt
C
 2 k1k 2 Q  k 3Q
opt
 k1
C
31 , 620

k 1Q
3 . 162
k2
opt
k3
C
k 2Q
k1
opt
k 2
C

Q
opt
Q

100 , 000
k1k 2
Q
 k3
4 . 316
30
D
opt
k 2Q

k1
D
opt
Chapter 1
 k1
D
opt
k 2
D

k1
2 k1
k 2Q
1
k1
2k2
 15 ,810
1 . 581
opt
k3
D

k 2Q  1
0
opt
Q

k1  1 .0
0
k 2Q
1
k1
2Q
k 2  10 , 000
0 . 158
k 3  4 .0
Q  100 , 000
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RELATIVE SENSITIVITY (Percentage change)
S k1 
C
C
opt
Chapter 1
C
C
opt
/C
opt

 k1 / k1
 463 , 240
opt
 k1

k 2Q
 ln C
opt
 ln k 1
k1  1 .0
 31 , 620
k1
S k 1  0 . 0683
S k1   0 . 5
S k 2  0 . 0683
S k 2  0 .5
S k 3  0 . 863
S Q  0 .5
S Q  0 . 932
S k3  0
C
D
C
D
C
D
C
S k1 
C
C
opt
D
k1
 k1 C
opt
abs. sens. on D
abs. sens. on C

31620 (1 . 0 )
 0 . 0683
463 , 240
k1  k 2  Q  k 3
k 3  k1  Q  k 2
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Chapter 1
PIPELINE PROBLEM
variables
param eters
V

p

f
L
Re
m
D
pipe cost
electricity cost
#operatin g days/yr
pum p efficiency
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Equality Constraints
Chapter 1

D
2
v  m
4
Re  Dv  / 
p  2 v
2
L
f
D
f  .0 4 6 R e
 0 .2
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Chapter 1
min (Coper + Cinv.)
subject to equality constraints
 p  2 f
L
v
2
Chapter 1
D
need analytical formula for f
f  . 046 Re
 0 .2
pum p pow er cost  C om
smooth tubes
p

m  m a ss flo w ra te 
D
4
2
v
substituting for ∆p,

C oper  C o D
C inv  C 1 D
1 .5
 4 .8

0 .2
2 .8
 2 .0
m 
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( annualized
)
 0 .2  2 2 .8  4 .8
1 .5
Total cost  TC  C o   m D
 C1D
(constraint eliminated by substitution)
d (T C )
0
n e ce ssa ry co n d itio n fo r a m in im u m
Chapter 1
dD
so lv in g ,
(D
opt
)
6 .3

Co

C1

0 .2
C  
opt
(D )   o 
 C1 


0 .1 6
o p t v e lo city V
opt
2
 m


 .3 2
2 .8
m
.4 5

.0 3
m

4
D opt 
2
(se n sitiv ity a n a lysis)
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optimum velocity
Chapter 1
non-viscous liquids
gases (effect of ρ)
3 to 6 ft/sec.
30 to 60 ft/sec.
at higher pressure, need to use different
constraint (isothermal)


 p1
ln

 fL
p2
2 p1 

p 

 .323 
p1  p 2 
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 D






2
S
V
 1 1








S 1   gV1  upstream velocity
or use W eym outh equation
for large L, ln ( ) can be neglected
exceptions: elevation changes, slurries (settling),
extremely viscous oils (laminar flow, 38
f different)
Chapter 1
Heat Exchanger Variables
(given flow rate of one
1. heat transfer area
fluid, inlet
2. heat duty
temperatures, one
3. flow rates (shell, tube)
outlet temp., phys.
4. no. passes (shell, tube)
props.)
5. baffle spacing
6. length
7. diam. of shell, tubes
8. approach temperature
9. fluid A (shell or tube, co-current or countercurrent)
10.tube pitch, no. tubes
11.velocity (shell, tube)
12.∆p (shell, tube)
13.heat transfer coeffs (shell, tube)
14.exchanger type (fins?)
15.material of construction
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