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

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/

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

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 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

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

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

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?

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.

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

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

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

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)

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 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

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 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 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

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

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)