Modeling and Optimization

CHEN 4470 – Process Design Practice

Dr. Mario Richard EdenDepartment of Chemical Engineering

Auburn University

Lecture No. 9 – Modeling Reactive Systems and Mathematical Optimization

February 20, 2007

• Background Material– Multimedia software

– Documents placed on webpage• Rate-controlled reactions from Aspen Help• Workshop document on reactive systems

Kinetic Reactors in Aspen 1:6

• Example – Ethylbenzene Production– Feed stream conditions

• Benzene 120 lbmol/hr• Ethylene 100 lbmol/hr• Toluene 5 lbmol/hr• Temperature 400C• Pressure 20 atm

– Reactor specifications• Reactor specified as adiabatic with a 5 psi pressure

drop• Reactor length: 100 ft• Reactor diameter 10 ft

Kinetic Reactors in Aspen 2:6

• Example – Ethylbenzene Production– Reaction Data

• r2j in kmol/(s*m3)

• Ci in kmol/m3

Kinetic Reactors in Aspen 3:6

216 6 2 4 6 5 2 5C H + C H C H C Hr

226 5 2 5 2 4 6 4 2 5 2C H C H + C H C H (C H )r

236 4 2 5 2 6 6 6 5 2 5C H (C H ) + C H 2C H C Hr

246 5 3 2 4 6 5 2 5 3 6C H CH + 2C H C H C H + C Hr

22,500 kcal/kmole

2 4 6 6

621 1.00 10 (e )RT

C H C Hr C C

22,500 kcal/kmole

2 4 6 5 2 5

522 6.00 10 (e )RT

C H C H C Hr C C

25,000 kcal/kmole

6 6 6 4 6 5 2

623 ( )7.80 10 (e )RT

C H C H C Hr C C

20,000 kcal/kmole

2 4 6 5 3

8 224 1.80 10 (e )RT

C H C H CHr C C

216 6 2 4 6 5 2 5C H + C H C H C Hr

226 5 2 5 2 4 6 4 2 5 2C H C H + C H C H (C H )r

236 4 2 5 2 6 6 6 5 2 5C H (C H ) + C H 2C H C Hr

246 5 3 2 4 6 5 2 5 3 6C H CH + 2C H C H C H + C Hr

22,500 kcal/kmole

2 4 6 6

621 1.00 10 (e )RT

C H C Hr C C

22,500 kcal/kmole

2 4 6 5 2 5

522 6.00 10 (e )RT

C H C H C Hr C C

25,000 kcal/kmole

6 6 6 4 6 5 2

623 ( )7.80 10 (e )RT

C H C H C Hr C C

20,000 kcal/kmole

2 4 6 5 3

8 224 1.80 10 (e )RT

C H C H CHr C C

• Example – Ethylbenzene Production (Cont’d)– Components renamed from Aspen standards

Kinetic Reactors in Aspen 4:6

• Example – Ethylbenzene Production (Cont’d)– Reaction set defined as “Powerlaw”

Kinetic Reactors in Aspen 5:6

Click here to specify the kinetics

• Example – Ethylbenzene Production (Cont’d)

Kinetic Reactors in Aspen 6:6

/ [ ] in E RTir kT e C

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 + 2.25B + 1000 – A – B

(1a) max J = 1000 + 2.55A + 1.25B

Optimization Example 1 5:7

• Solution (Continued)– Analagously, we can substitute equation (2a)

into equation (3) to obtain a revised constraint:

0.40A + 0.50B + 1000 – A – B 700

1000 – 0.60A – 0.50B 700

– 0.60A – 0.50B – 300

(3a) 0.60A + 0.50B 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)

0

100

200

300

400

500

600

0 100 200 300 400 500

Total Number of Type A Loaves

To

tal N

um

be

r o

f T

yp

e B

Lo

av

es

Point 1

Point 2

Point 3

B = 600 - 0.6A

B = 600 - 1.2A

A = 500

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 m2/hr*$25/m2 = $67.5/hr

• Cost B = $20/hr + (1 – 0.90)*35 m2/hr*$25/m2 = $107.5/hr

• Cost C = $15/hr + (1 – 0.85)*50 m2/hr*$25/m2 = $202.5/hr

• Inspection rate Type 1: 30 m2/hr * 12 hr/shift = 360 m2/shift

• Inspection rate Type 2: 35 m2/hr * 12 hr/shift = 420 m2/shift

• Inspection rate Type 3: 50 m2/hr * 12 hr/shift = 600 m2/shift

Optimization Example 2 4:4

• Solution (Continued)– Now we can formulate the optimization problem

• Next Lecture – March 1– Physical property prediction & computer aided

molecular design

• Plant Trip to Degussa– March 13 (Itinerary will be provided soon)

• Planning 10:30 – 2:30 PM activities at Degussa• Leaving AU around 6:45 AM, returning to AU around

6:00 PM

– Working on transportation possibilities• Alternative is to car pool (500 miles roundtrip)• Mileage reimbursement roughly $240 per car

– Signup sheet• Needed for lunch arrangements, hard hats etc.• Indicate also if you are interested in driving

Other Business