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 –N variables >>N equations Equally Specified Problem –Units in Flow sheet –N variables =N equations Under Specified Problem –Optimization –N variables <<N equations

Optimization Number of Decision Variables –N D =N variables -N equations Objective Function is optimized with respect to N D 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 R min –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 –(X 1 & X 2 <0)

Stationary Points Maximum number of solutions –N s = πN Degree of partial differential Equation Local Extrema –Maxima –Minima Saddle points Extrema at infinity –Example df/dx 1 = 3 rd order polynomial df/dx 2 = 2 nd order polynomial df/dx 3 = 4 th order polynomial N s =24

LINEAR PROGRAMING (LP) equality constraints inequality constraints objective function w.r.t. design variables The N D 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 YieldsMax. Production Crude #1Crude #2(bbl/day) Gasoline70316,000 Kerosene692,400 Fuel Oil246012,000 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 x 1 and x 2 be the daily production rates of Crude #1 and Crude #2. maximize 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: Step 4. Simplification of model and objective function. Equality constraints are used to reduce the number of independent variables (N D = N V – N E ). Here N E = 0.

EXAMPLE LP –SOLUTION (Cont’d) Step 5. Compute optimum. a)Inequality constraints define feasible space. Feasible 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 x 1 = 0, x 2 = 19,355 bbl/day

EXAMPLE LP – GRAPHICAL SOLUTION A refinery uses two crude oils, with yields as below. Volumetric YieldsMax. Production Crude #1Crude #2(bbl/day) Gasoline70316,000 Kerosene692,400 Fuel Oil246012,000 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*x *x 2 =6,000, equality constraint added 0.31*19,355=6,000

Solving for a Recycle Loop Newton-Raphson –Solving for a root –F(x i )=0 Optimization –Minimize/Maximize w.r.t. N D variables (d) s.t. constraints –F(x i ) = 0, G(x i ) 0

Minimizef{x} w.r.t d Subject to:c{x} = 0 g{x}  0 x L  x  x U SUCCESSIVE QUADRATIC PROGRAMMING The NLP to be solved is: 1. Definition of slack variables: 2. Formation of Lagrangian: Lagrange multipliers Kuhn-Tucker multipliers

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

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

Minimizef{x, d} w.r.t d Subject to: h{x *, d} = x * - w{x *, d} = 0 c{x, d} = 0 g{x}  0 x L  x  x U OPTIMIZATION ALGORITHM equality constraints inequality constraints objective function design variables tear equations inequality constraints

REPEATED SIMULATION Minimizef{x, d} w.r.t d S.t.h{x *, d} = x * - w{x *, d} = 0 c{x, d} = 0 g{x}  0 x L  x  x U Sequential iteration of w and d (tear equations are converged each master iteration).

INFEASIBLE PATH APPROACH (SQP) Minimizef{x, d} w.r.t. d S.t.h{x *, d} = x * - w{x *, d} = 0 c{x, d} = 0 g{x}  0 x L  x  x U Both w and d are adjusted simultaneously, with normally only one iteration of the tear equations.

COMPROMISE APPROACH (SQP) Minimizef{x, d} w.r.t. d S.t.h{x *, d} = x * - w{x *, d} = 0 c{x, d} = 0 g{x}  0 x L  x  x U Tear equations converged loosely for each master iteration Wegstein’s method

Simple Methods of Flow Sheet Optimization Golden Section Method τ=

Golden Section Problem Replace CW HX and Fired Heater 1 Heat Exchanger Optimize w.r.t T LGO,out PV=(S-C)+i*C TCI S=0, C=$3.00 /MMBTU in Fired Heater C TCI = f( HX Area )

Golden Section Result Min Annual Cost of HX –C A =C s (Q)+i m C TCI (A(ΔT app ))

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≥R min