Transcript L11-Optimization of Process Flowsheets.ppt

Optimization of Process
Flowsheets
S,S&L Chapter 24
T&S Chapter 12
Terry A. Ring
CHEN 5253
OBJECTIVES
On completion of this course unit, you are
expected to be able to:
– Formulate and solve a linear program (LP)
– Formulate a nonlinear program (NLP) to
optimize a process using equality and
inequality constraints
– Be able to optimize a process using
Aspen/ProMax beginning with the results
of a steady-state simulation
• Data/ModelAnalysisTools/Optimization
• Calculators/SimpleSolver or Advanced Solver/
Degrees of Freedom
• Over Specified Problem
– Fitting Data
– Nvariables>>Nequations
• Equally Specified Problem
– Units in Flow sheet
– Nvariables=Nequations
• Under Specified Problem
– Optimization
– Nvariables<<Nequations
Optimization
• Number of Decision Variables
– ND=Nvariables-Nequations
• Objective Function is optimized with respect to ND
Variables
– Minimize Cost
– Maximize Investor Rate of Return
• Subject To Constraints
– Equality Constraints
• Mole fractions add to 1
– Inequality Constraints
• Reflux ratio is larger than Rmin
– Upper and Lower Bounds
• Mole fraction is larger than zero and smaller than 1
PRACTICAL ASPECTS
• Design variables, need to be identified and
kept free for manipulation by optimizer
– e.g., in a distillation column, reflux ratio
specification and distillate flow specification are
degrees of freedom, rather than their actual
values themselves
• Design variables should be selected AFTER
ensuring that the objective function is
sensitive to their values
– e.g., the capital cost of a given column may be
insensitive to the column feed temperature
• Do not use discrete-valued variables in
gradient-based optimization as they lead to
discontinuities in f(d)
Optimization
• Feasible Region
– Unconstrained Optimization
• No constraints
– Uni-modal
– Multi-modal
– Constrained Optimization
• Constraints
– Slack
– Binding
Modality
• Multimodal
• Unimodal
– (X1& X2<0)
Stationary Points
• Maximum number of solutions
– Ns= πNDegree of partial differential Equation
• Local Extrema
– Maxima
– Minima
• Saddle points
• Extrema at infinity
– Example
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df/dx1= 3rd order polynomial
df/dx2= 2nd order polynomial
df/dx3= 4th order polynomial
Ns=24
LINEAR PROGRAMING (LP)
objective function
Nv
Minimize J x   fi xi
d
i=1
Subject to (s.t.) xi  0,i  1,
Nv
a x
j=1
ij
j
Nv
c x
j=1
ij
j
, NV
 b,i
 1,
i
, NE
equality constraints
 d,i
 1,
i
, NI
inequality constraints
w.r.t. design variables
The ND design variables, d, are
adjusted to minimize f{x} while
satisfying the constraints
EXAMPLE LP – GRAPHICAL SOLUTION
A refinery uses two crude oils, with yields as below.
Volumetric Yields
Max. Production
Crude #1
Crude #2
(bbl/day)
Gasoline
70
31
6,000
Kerosene
6
9
2,400
Fuel Oil
24
60
12,000
The profit on processing each crude is:
$2/bbl for Crude #1 and $1.4/bbl for Crude #2.
a) What is the optimum daily processing rate for
each grade?
b) What is the optimum if 6,000 bbl/day of gasoline
is needed?
EXAMPLE LP –SOLUTION (Cont’d)
Step 1. Identify the variables. Let x1 and x2 be the daily
production rates of Crude #1 and Crude #2.
Step 2. Select objective function. We need to
maximize profit: J x  2.00x1  1.40x2
Step 3. Develop models for process and constraints.
Only constraints on the three products are given:
0.70x1  0.31x2  6, 000
0.06x1  0.09x2  2, 400
0.24x1  0.60x2  12, 000
Step 4. Simplification of model and objective function.
Equality constraints are used to reduce the number of
independent variables (ND = NV – NE). Here NE = 0.
EXAMPLE LP –SOLUTION (Cont’d)
Step 5. Compute optimum.
a) Inequality constraints define feasible space.
0.06x1  0.09x2  2, 400
Feasible Space
0.24x1  0.60x2  12, 000
0.70x1  0.31x2  6, 000
EXAMPLE LP –SOLUTION (Cont’d)
Step 5. Compute optimum.
b) Constant J contours are positioned to find optimum.
x1 = 0, x2 = 19,355 bbl/day
J = 27,097
J = 20,000
J = 10,000
0.70x1  0.31x2  6, 000
EXAMPLE LP – GRAPHICAL SOLUTION
A refinery uses two crude oils, with yields as below.
Volumetric Yields
Max. Production
Crude #1
Crude #2
(bbl/day)
Gasoline
70
31
6,000
Kerosene
6
9
2,400
Fuel Oil
24
60
12,000
The profit on processing each crude is:
$2/bbl for Crude #1 and $1.4/bbl for Crude #2.
a) What is the optimum daily processing rate for
each grade?
19,355 bbl/d
b) What is the optimum if 6,000 bbl/day of gasoline
is needed?
0.7*x1+0.31*x2=6,000, equality constraint added
0.31*19,355=6,000
Solving for a Recycle Loop
• Newton-Raphson
– Solving for a root
– F(xi)=0
• Optimization
– Minimize/Maximize w.r.t. ND variables (d) s.t.
constraints
– F(xi) = 0, G(xi) < 0, H(xi) > 0
SUCCESSIVE QUADRATIC PROGRAMMING
The NLP to be solved is:
Minimize f{x}
w.r.t d
Subject to: c{x} = 0
g{x}  0
xL  x  xU
1. Definition of slack variables: gi x   zi2  0, i  1, , m
2. Formation of Lagrangian:
Lx , , , z   f x   T c x   T g x   z 2

Lagrange multipliers
Kuhn-Tucker multipliers

SUCCESSIVE QUADRATIC PROGRAMMING
2. Formation of Lagrangian:
Lx , , , z   f x   T c x   T g x   z 2


3. At the minimum: L  0 
X L  X f x   T X c x   T X g x   0
 L  c x   0
 L  g x   z 2  0 (definitio n)
Jacobian matrices
zi L  2i zi  0  gi i  0, i  1, , m
0
Complementary slackness equations:
either gi = 0 (constraint active)
or i = 0 (gi < 0, constraint slack)
OPTIMIZATION ALGORITHM
x*
w{d, x*}
Tear equations: h{d , x*} = x* - w{d , x*} = 0, w(d,x*) is a Tear Stream
OPTIMIZATION ALGORITHM
objective function
Minimize f{x, d}
w.r.t d
tear equations
Subject to: h{x*, d} = x* - w{x*, d} = 0
c{x, d} = 0
g{x}  0
xL
 x
xU
design variables
equality constraints
inequality constraints
inequality constraints
REPEATED SIMULATION
Minimize f{x, d}
w.r.t d
S.t. h{x*, d} = x* - w{x*, d} = 0
c{x, d} = 0
g{x}  0
xL  x  xU
Sequential iteration
of w and d (tear
equations are
converged each
master iteration).
INFEASIBLE PATH APPROACH (SQP)
Minimize f{x, d}
w.r.t. d
S.t. h{x*, d} = x* - w{x*, d} = 0
c{x, d} = 0
g{x}  0
xL  x  xU
Both w and d are
adjusted simultaneously,
with normally only one
iteration of the tear
equations.
COMPROMISE APPROACH
Minimize f{x, d}
w.r.t. d
S.t. h{x*, d} = x* - w{x*, d} = 0
c{x, d} = 0
g{x}  0
xL  x  xU
Wegstein’s method
Tear equations
converged loosely for
each master iteration
Simple Methods of Flow Sheet
Optimization
• Golden Section Method
• τ=0.61803
Golden Section Problem
• Replace CW HX
4-CW
1
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2-Light Gas Oil
XCHG-102
5-CW
1 Heat Exchanger
1 Fired Heater
Optimize VP w.r.t TLGO,out
VP=(S-C)+i*CTCI
S=0, C=$3.00/MMBTU in Fired Heater
Q=80,000lb/h*0.5BTU/lb*(440F-T
CTCI= f(HX Area)
1-Light Gas Oil
DVDR-1
6-Crude Oil
Q-1
3
4-Light Gas Oil
XCHG-101
7-Crude
Oil
Fired 2Heater
LGO,out)
Q=500,000lb/h*0.45BTU/lb(TCO,out-240F)
Q=UAΔTLM
Golden Section Result
• Min Annual Cost of HX
– CA=-Cs(Q)+imCTCI(A(ΔTLM))
Aspen Optimization
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Use Design I Aspen File
MeOH Distillation-4.bkp
Optimize DSTWU column
V=D*(R+1)
Minimize V
w.r.t. R
s.t. R≥Rmin