1 the nature and organization of optimization...
TRANSCRIPT
8
Chapter 1
The Nature and Organization of
Optimization Problems
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WHY OPTIMIZE?
1. Improved yields, reduced pollutants
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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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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Figure E1.4-3
Optimal Reflux for Different Fuel Costs
Flooding
constraint
Min reflux to
achieve separation
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Material Balance Reconciliation
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Least squares solution:
2
1
)(minii B
P
i
CA mmm
opt. mA is the “average” value
any constraints on mA?
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THREE INGREDIENTS IN OPTIMIZATION PROBLEM
1. Objective function economic model
2. Equality ConstraintsProcess model
3. Inequality Constraints
nx1
1
2
1. min f(x) x
2. subject to h( ) 0 (m )
3. g( ) 0 (m )
2(feasible region :)
3
dependent variables
independent variables
x
x
1 2relate to m and perhaps m
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TABLE 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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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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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
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
inventory, storage cost = k1D
production cost = k2 + k3 D
per run (set up operating
cost) cost per unit
(could be nonlinear)
QkD
QkDkC
D
Qn
DkknDkC
321
321 )(
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Step 3
evaluate constraints
continuous
integern
D>0
Step 4
simplification – none necessary
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Step 5
computation of the optimum
analytical vs. numerical solution
1
2
2
21 0
k
QkD
D
Qkk
dD
dC
opt
02
minimum? ifcheck
answer good
000,70000,30
optimumflat 622,31
10
0.4000,100.1
3
2
2
2
5
321
D
Qk
dD
Cd
D
D
Q
kkk
opt
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solution? analytical
02
run per cost suppose
2/3
4
2
21
2/1
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D
Qk
D
Qkk
dD
dC
Dkk
Step 6
Sensitivity of the optimum
subst Dopt into C
316.4
000,100
162.3
620,31
2
321
3
2
1
2
1
2
1
321
kQ
kk
Q
C
Qk
C
k
Qk
k
C
k
Qk
k
C
QkQkkC
opt
opt
opt
opt
opt
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000,1000.4000,100.1
158.02
1
00
581.12
1
810,152
1
321
1
2
3
21
2
2
11
2
1
1
2
Qkkk
Qk
Qk
Q
D
k
D
kk
Qk
k
D
kk
Qk
k
D
k
QkD
opt
opt
opt
opt
opt
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RELATIVE SENSITIVITY (Percentage change)
213
321
1
1
1
2
1
1
111
Con sens. abs.
Don sens. abs.
0683.0240,463
)0.1(31620
0932.0
5.0863.0
5.00683.0
5.00683.0
620,31
0.1240,463
ln
ln
/
/
1
3
3
22
11
1
kQkk
kQkk
C
k
k
CS
SS
SS
SS
SS
k
Qk
k
C
kC
k
C
kk
CCS
opt
optC
k
D
k
C
Q
D
Q
C
k
D
k
C
k
D
k
C
k
opt
opt
optoptoptC
k
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PIPELINE PROBLEM
variables parameters
V
p
f L
Re m
D pipe cost
electricity cost
#operating days/yr
pump efficiency
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Equality Constraints
2
2
0.2
4
Re /
2
.046 Re
Dv m
Dv
Lp v f
D
f
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min (Coper + Cinv.)
subject to equality constraints
22 vD
Lfp
need analytical formula for f
tubessmoothf 2.0Re046.
pump power cost
o
pC m
2
mass flow rate 4
D
m v
substituting for ∆p,
)(5.1
1
0.28.22.08.4
annualizedDCC
mDCC
inv
ooper
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5.1
1
8.48.222.0 TC cost Total DCDmCo
(constraint eliminated by substitution)
6.3 0.2 2 2.8
1
0.16
.32 .45 .03
1
opt
2
( )0 necessary condition for a minimum
solving,
( )
( )
opt velocity V
4
(sensitivity analysis)
opt o
opt o
opt
d TC
dD
CD m
C
CD m
C
m
D
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optimum velocity
non-viscous liquids 3 to 6 ft/sec.
gases (effect of ρ) 30 to 60 ft/sec.
at higher pressure, need to use different
constraint (isothermal)
1
2 211 1
1 2
1 1
ln2
.32324
upstream velocity
or use Weymouth equation
ppp fL
p S Vp p D
S gV
for large L, ln ( ) can be neglected
exceptions: elevation changes, slurries (settling),
extremely viscous oils (laminar flow,
f different)
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Heat Exchanger Variables
1. heat transfer area
2. heat duty
3. flow rates (shell, tube)
4. no. passes (shell, tube)
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
(given flow rate of one
fluid, inlet
temperatures, one
outlet temp., phys.
props.)