Download - Optimization of Process Flowsheets S,S&L Chapter 24 T&S Chapter 12 Terry A. Ring CHEN 5253
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• df/dx1= 3rd order polynomial• df/dx2= 2nd order polynomial• df/dx3= 4th order polynomial• Ns=24
v
v
v
N
i ii=1
i V
N
ij j i Ej=1
N
ij j i Ij=1
MinimizeJ x f xd
Subject to (s.t.) x 0,i 1, ,N
a x b,i 1, ,N
c x d,i 1, ,N
LINEAR PROGRAMING (LP)
equality constraints
inequality constraints
objective function
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 12,000The 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 maximizemaximize profit: 1 2J x 2.00x 1.40x
Step 3. Develop models for process and constraints. Only constraints on the three products are given:
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.
1 2
1 2
1 2
0.70x 0.31x 6,000
0.06x 0.09x 2,400
0.24x 0.60x 12,000
EXAMPLE LP –SOLUTION (Cont’d)
Step 5. Compute optimum. a) Inequality constraints define feasible space.
1 20.70x 0.31x 6,000
1 20.06x 0.09x 2,400
1 20.24x 0.60x 12,000Feasible
Space
EXAMPLE LP –SOLUTION (Cont’d)Step 5. Compute optimum. b) Constant J contours are positioned to find
optimum.
J = 10,000
J = 20,000
J = 27,097
x1 = 0, x2 = 19,355 bbl/day
1 20.70x 0.31x 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 12,000The 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
Minimize f{x} w.r.t d Subject to: c{x} = 0 g{x} 0 xL x xU
SUCCESSIVE QUADRATIC PROGRAMMING
The NLP to be solved is:
1. Definition of slack variables: mizxg ii ,,1,02
2. Formation of Lagrangian: 2,,, zxgxcxfzxL TT
Lagrange multipliers
Kuhn-Tucker multipliers
SUCCESSIVE QUADRATIC PROGRAMMING
2. Formation of Lagrangian: 2,,, zxgxcxfzxL TT
3. At the minimum: 0L
0
,,1,002
n)(definitio 0
0
0
2
migzL
zxgL
xcL
xgxcxfL
iiiiz
XT
XT
XX
i
Complementary slackness equations: either gi = 0 (constraint active) or i = 0 (gi < 0, constraint slack)
Jacobian matrices
OPTIMIZATION ALGORITHM x* w{d, x*}
Tear equations: h{d , x*} = x* - w{d , x*} = 0, w(d,x*) is a Tear Stream
Minimize f{x, d} w.r.t d
Subject to: h{x*, d} = x* - w{x*, d} = 0
c{x, d} = 0
g{x} 0
xL x xU
OPTIMIZATION ALGORITHM
equality constraints
inequality constraints
objective function
design variables
tear equations
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 (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
Tear equations converged loosely for each master iteration
Wegstein’s method
Simple Methods of Flow Sheet Optimization
• Golden Section Method• τ=0.61803
Golden Section Problem
• Replace CW HX and Fired Heater
• 1 Heat Exchanger
• Optimize w.r.t TLGO,out
• PV=(S-C)+i*CTCI
• S=0, C=$3.00/MMBTU in Fired Heater
• CTCI= f(HX Area)
XCHG-101
XCHG-102
1 2-Light Gas Oil
3
4-CW
5-CW
DVDR-1
1-Light Gas Oil
Q-1
6-Crude Oil
7-Crude Oil
4-Light Gas Oil
Fired Heater2
Golden Section Result
• Min Annual Cost of HX– CA=Cs(Q)+imCTCI(A(ΔTapp))
Aspen Optimization
• 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