Modeling and Optimization CHEN 4470 – Process Design Practice Dr. Mario Richard Eden Department of Chemical Engineering Auburn University Lecture No. 9 – Modeling Reactive Systems and Mathematical Optimization February 7, 2006

Background Material –Multimedia software –Documents placed on webpage Rate-controlled reactions from Aspen Help Workshop document on reactive systems Kinetic Reactors in Aspen 1:5

Example – Ethylbenzene Production –Feed stream conditions Benzene120 lbmol/hr Ethylene100 lbmol/hr Toluene 5 lbmol/hr Temperature400  C Pressure20 atm –Reactor specifications Reactions and kinetics as given in project description Reactor specified as adiabatic with a 5 psi pressure drop Reactor length:100 ft Reactor diameter 10 ft Kinetic Reactors in Aspen 2:5

Example – Ethylbenzene Production (Cont’d) –Components renamed from Aspen standards Kinetic Reactors in Aspen 3:5

Example – Ethylbenzene Production (Cont’d) –Reaction set defined as “Powerlaw” Kinetic Reactors in Aspen 4:5 Click here to specify the kinetics

Example – Ethylbenzene Production (Cont’d) Kinetic Reactors in Aspen 5:5 IMPORTANT Must be specified in SI units, see document on website or Aspen help

Optimization Example 1 1:7

Optimization Example 1 2:7 Solution –Objective is to maximize profit by identifying production rates of three types of bread –Define variables A: Number of 1 kg loaves produced of bread type A B: Number of 1 kg loaves produced of bread type B C: Number of 1 kg loaves produced of bread type C –Define profit function Profit A = Sales revenue of Type A – Cost of producing A Profit B = Sales revenue of Type B – Cost of producing B Profit C = Sales revenue of Type C – Cost of producing C Total Profit J = Profit A + Profit B + Profit C

Optimization Example 1 3:7 Solution (Continued) –Using the information given in the problem statement, we can define the individual profits as: Profit A = ($5 – 0.40*$1– 0.30*$1.5 – 0.30*$2)*A = 3.55A Profit B = ($3.5 – 0.50*$1– 0.50*$1.5)*B = 2.25B Profit C = ($2 – 1*$1)*C = C

Optimization Example 1 4:7 Solution (Continued) –The hint given in the problem statement suggest to reformulate the equations to be functions of A and B. From equation (2) we can obtain an expression for C: (2a) C = 1000 – A – B (2b)A + B  1000 –Substituting equation (2a) into equation (1) gives: max J = 3.55A B – A – B  (1a)max J = A B

Optimization Example 1 5:7 Solution (Continued) –Analagously, we can substitute equation (2a) into equation (3) to obtain a revised constraint: 0.40A B – A – B  700  1000 – 0.60A – 0.50B  700  – 0.60A – 0.50B  – 300  (3a)0.60A B  300

Optimization Example 1 6:7 Solution (Continued) –This gives the revised optimization problem: –We are now able to plot the constraints.

Optimization Example 1 7:7 Solution (Continued) The optimal solution to a LP lies on a vertex of the feasibility region, i.e. Point 1,2 or 3. (A,B) = (0, 600) J = $1,750 (A,B) = (500, 300) J = $2,650 (A,B) = (500, 0) J = $2,275 Optimal solution: A = 500, B = 300, C = 200 J = $2,650

Optimization Example 2 1:4

Optimization Example 2 2:4 Solution –Objective is to minimize daily inspection cost by allocation of inspection teams –Define variables X: Number of type 1 inspection teams Y: Number of type 2 inspection teams Z: Number of type 3 inspection teams –Define inspection cost Cost A = Wages for Type 1 + Cost of Errors by Type 1 Cost B = Wages for Type 2 + Cost of Errors by Type 2 Cost C = Wages for Type 3 + Cost of Errors by Type 3 Total Cost J = X*Cost A + Y*Cost B + Z*Cost C

Optimization Example 2 3:4 Solution (Continued) –Using the information given in the problem statement, we can define the individual inspection cost and inspection rates as follows: Cost A = $30/hr + (1 – 0.95)*30 m 2 /hr*$25/m 2 = $67.5/hr Cost B = $20/hr + (1 – 0.90)*35 m 2 /hr*$25/m 2 = $107.5/hr Cost C = $15/hr + (1 – 0.85)*50 m 2 /hr*$25/m 2 = $202.5/hr Inspection rate Type 1: 30 m 2 /hr * 12 hr/shift = 360 m 2 /shift Inspection rate Type 2: 35 m 2 /hr * 12 hr/shift = 420 m 2 /shift Inspection rate Type 3: 50 m 2 /hr * 12 hr/shift = 600 m 2 /shift

Optimization Example 2 4:4 Solution (Continued) –Now we can formulate the optimization problem

Next Lecture – February 14 –Physical property prediction –Computer aided molecular design No Class or Meetings on February 9 –Medical Exam in Montgomery –Will be out of town February 10 –Progress report no. 1 can be turned in to my mailbox by 4:00 PM on Friday Other Business