1. Optimization of Process Flowsheets
Sieder Chapter 21
Terry A. Ring
CHEN 5253

2. On completion of this course unit, you are expected to be able to:
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/

3. Degrees of Freedom Over Specified Problem Equally Specified Problem
Fitting Data
Nvariables>>Nequations
Equally Specified Problem
Units in Flow sheet
Nvariables=Nequations
Under Specified Problem
Optimization
Nvariables<<Nequations

4. 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

5. 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)

6. Optimization Feasible Region Unconstrained Optimization
No constraints
Uni-modal
Multi-modal
Constrained Optimization
Constraints
Slack
Binding

7. Modality Multimodal Unimodal Min/Max Local Slope=df/dx1+df/dx2 = 0
(|X1| & |X2|<6)
Unimodal
(X1& X2<0)
Min/Max
Local Slope=df/dx1+df/dx2 = 0

8. Stationary Points Maximum number of Extrema
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

9. LINEAR PROGRAMING (LP)
objective function
equality constraints
inequality constraints
w.r.t. design variables
The ND design variables, d, are adjusted to minimize f{x} while satisfying the constraints

10. 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.
What is the optimum daily processing rate for each grade?
What is the optimum if 6,000 bbl/day of gasoline is needed?

11. EXAMPLE LP –SOLUTION (Cont’d)
Step 1. Identify the variables. Let x1 and x2 be the daily production rates [bbl/day] of Crude #1 and Crude #2.
Step 2. Select objective function. We need to maximize profit:
Step 3. Develop models for process and constraints. Only constraints on the three products are given:
Gasoline
Kerosene
Fuel Oil
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.

12. EXAMPLE LP –SOLUTION (Cont’d)
Step 5. Compute optimum.
Inequality constraints define feasible space.
Kerosene
Feasible Space
Fuel Oil
Gasoline

13. EXAMPLE LP –SOLUTION (Cont’d)
Step 5. Compute optimum.
Constant J contours are positioned to find optimum.
x1 = 0, x2 = 19,355 bbl/day
J = 27,097
J = 20,000
J = 10,000

14. 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.
What is the optimum daily processing rate for each grade? 19,355 bbl/day
What is the optimum if 6,000 bbl/day of gasoline is needed?
0.7*x1+0.31*x2=6,000, equality constraint added *19,355=6,000

15. 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

16. 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:
2. Formation of Lagrangian:
Lagrange multipliers
Kuhn-Tucker multipliers

17. SUCCESSIVE QUADRATIC PROGRAMMING
2. Formation of Lagrangian:
3. At the minimum:
Jacobian matrices
Complementary slackness equations:
either gi = 0 (constraint active)
or i = 0 (gi < 0, constraint slack)

18. OPTIMIZATION ALGORITHMx*
w{d, x*}
Tear equations: h{d , x*} = x* - w{d , x*} = 0, w(d,x*) is a Tear Stream

19. OPTIMIZATION ALGORITHM
objective function
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
tear equations
equality constraints
inequality constraints
inequality constraints
design variables

20. 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).

21. 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.

22. 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
Tear equations converged loosely for each master iteration
Wegstein’s method

23. Simple Methods of Flow Sheet Optimization
Golden Section Method τ=

24. Golden Section Problem
Replace CW HX
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
CTCI= f(HX Area)
440°F
240°F
Q=80,000lb/h*0.5BTU/lb*(440F-TLGO,out)
Q=500,000lb/h*0.45BTU/lb(TCO,out-240F)
Q=UAΔTLM

25. Golden Section Result Min Annual Cost of HX CA=-Cs(Q)+imCTCI(A(ΔTLM))
Min Found a 7 evaluations

26. Aspen Optimization Use Design I Aspen File MeOH Distillation-4.apw
Optimize DISTL column
V=D*(R+1)
Minimize V
w.r.t. R
s.t. R≥Rmin,
xD>0.99 (methanol),
xB<0.01 (methanol)