tier iii: optimization design problems derek mccormack section 1: sample problems
TRANSCRIPT
Process Optimization
Tier III: Optimization Design Problems
Derek McCormack
Section 1:
Sample Problems
Introduction
Three sample problems have been given here to work on. The first is a heat exchange network optimization problem. The second is a transportation optimization problem to be solved with Lingo. The third problem deals with optimizing a heat exchanger’s minimum approach temperature.
Question #1
Optimization of a Heat Exchange Network by Thermal Pinch Analysis
Optimization of a Heat Exchange Network
A plant has the following stream data:
Hot Target Supply FCpStreams T1 (K) T2 (K) (kW/K)
H1 270 310 350H2 275 340 250H3 330 370 400H4 380 410 275H5 420 460 450H6 450 500 500H7 510 550 350
Hot StreamsCold Supply Target FCp
Streams t1 (K) t2 (K) (kW/K)C1 250 300 350C2 290 340 250C3 310 360 450C4 380 460 300C5 450 480 400C6 490 530 300
Cold Streams
HEN Problem
Using the stream data given and a Tmin of 10 K, do the following :
a) Determine the optimum heating and cooling utilities required by using the algebraic thermal pinch analysis method. Do you notice anything special with this example?
b) Now solve this problem using the graphical method, keeping in mind the results obtained above.
c) Create a possible heat exchange network for this situation based on the optimized conditions.
HEN Solution
Attempt to solve this problem before proceeding to the solution.
Temperature Interval Diagram
550 540540 530
510 500
490 480500 490
470 460460 450450 440
420 410410 400
390 380380 370
275 265270 260260 250
300 290310 300320 310
370 360
340 330350 340
330 320
Hot Streams Cold StreamsIntervals
21
14
19
18171615
13
1211
10
9
8
20
76
5
43
2
1
FC
p =
350
H1
H7FC
p =
350
H6
FC
p =
500 H5FC
p =
450 H4FC
p =
275
H3FC
p =
400
H2F
Cp
= 250
C1
FC
p =
350
C2
FC
p =
250
C3
FC
p =
450C5
FC
p =
400C6
FC
p =
300
FC
p =
300
C4
T t
Table of Exchangeable Heat Loads
Interval H1,i H2,i H3,i H4,i H5,i H6,i H7,i Total, HHi
i kW kW kW kW kW kW kW kW1 - - - - - - 3500 35002 - - - - - - 10500 105003 - - - - - - - 04 - - - - - 5000 - 50005 - - - - - 10000 - 100006 - - - - 5000 - 50007 - - - - 4500 5000 - 95008 - - - - 13500 - - 135009 - - - - - - - 0
10 - - - 5500 - - - 550011 - - - 2750 - - - 275012 - - - - - - - 013 - - 8000 - - - - 800014 - - 4000 - - - - 400015 - 2500 4000 - - - - 650016 - 2500 - - - - - 250017 - 2500 - - - - - 250018 3500 2500 - - - - - 600019 8750 6250 - - - - - 1500020 1750 - - - - - - 175021 - - - - - - - 0
Total cooling required (kW) 111500
Table of Exchangeable Loads - Hot Streams
Table of Exchangeable Heat Loads
Interval C1,i C2,i C3,i C4,i C5,i C6,i Total, HCi
i kW kW kW kW kW kW kW1 - - - - - - 02 - - - - - 9000 90003 - - - - - 3000 30004 - - - - - - 05 - - - - 8000 - 80006 - - - 3000 4000 - 70007 - - - 3000 - - 30008 - - - 9000 - - 90009 - - - 3000 - - 3000
10 - - - 6000 - - 600011 - - - - - - 012 - - - - - - 013 - - 9000 - - - 900014 - 2500 4500 - - - 700015 - 2500 4500 - - - 700016 - 2500 4500 - - - 700017 - 2500 - - - - 250018 3500 2500 - - - - 600019 8750 - - - - - 875020 1750 - - - - - 175021 3500 - - - - - 3500
Total heating required (kW) 100500
Table of Exchangeable Loads - Cold Streams
Cascade DiagramQH,min =
QC,min =
17250
17250
12
13
14
15
16
17
18
19
20
8000
2500
4000
6500
2500
0
1750
15000
6000
0
9000
7000
7000
7000
2500
6000
8750
1750
16250
13250
12750
8250
8250
8250
14500
14500210 3500
11000
3500
1
2
3
4
5
6
7
8
9
10500
5000
0
5000
10000
3500
0
13500
9500
0
9000
3000
0
8000
7000
3000
9000
3000
5000
2000
7000
9000
7000
13500
18000
15000
0
10
112750
5500
14500
6000
0
17250
No Pinch Point?
Notice that in this case the cascade diagram has no residuals that fall below zero. In this case, all of the heating needs of the cold streams are met by the hot streams, with an excess of heat left over. No heating utility is required, and the minimum cooling utility is 11,000 kW.
Hot Composite Stream
Hot Streams
0
20000
40000
60000
80000
100000
120000
230 280 330 380 430 480 530 580
T (K)
H
(kW
)
QC = 111500 kW
Cold Composite Stream
Cold Streams
0
20000
40000
60000
80000
100000
120000
230 280 330 380 430 480 530 580
T (K)
H
(kW
)
QH = 100500 kW
Optimized
0
20000
40000
60000
80000
100000
120000
230 280 330 380 430 480 530 580T (K)
H
(kW
)
220 270 320 370 420 470 520 570
t = T - T250
QH,min = 0
QC,min = 11000 kW
Cold composite stream
Hot composite stream
No Pinch Point?
Here we can see that we do not get a typical pinch point. The head of the cold composite stream cannot be moved below the tail of the hot composite stream. In this case, all of the heating requirements can be met by the hot streams, but 11,000 kW of cooling utility are still needed.
Question #2
Optimization of Transportation
Route Problem
Transportation ProblemFive chemical plants produce a chemical to be shipped and sold at three different selling stations. Each plant has a different production cost and shipping cost, while each warehouse that receives the product sells it for a different price. Warehouse 1 sells for 95 $/tonne, warehouse 2 sells for 90 $/tonne, and warehouse 3 sells for 93 $/tonne. The cost of production at each of the plants are as follows: plant 1 costs 42 $/tonne, plant 2 costs 45 $/tonne, plant 3 costs 43 $/tonne, plant 4 costs 46 $/tonne, and plant 5 costs 55 $/tonne. To ship from plant 1 costs 0.30 $/tonne•km, from plant 2 costs 0.35 $/tonne•km, from plant 3 costs 0.31 $/tonne•km, from plant 4 costs 0.34 $/tonne•km, and plant 5 costs 0.29 $/tonne•km.
Transportation ProblemThe distances between plants and warehouses, in km, are as follows:
Plant 1 has a production capacity of 1300 tonnes, plant 2 can make 1200 tonnes, plant 3 can make 1700 tonnes, plant 4 can make 1400 tonnes, and plant 5 can make 1600 tonnes. Furthermore, market research suggests that the amount sold at each warehouse is limited. Warehouse 1 can receive 2400 tonnes, warehouse 2 can receive 2000 tonnes, and warehouse 3 can receive 2500 tonnes.
W1 W2 W3
P1 155 140 145P2 125 110 120P3 150 135 140P4 130 115 125P5 120 100 110
Transportation Problem
What combination of shipments will maximize the profit that this company can earn, and what is that profit? Use Lingo to
solve this.
Attempt to solve this problem before proceeding to the solution.
Transportation Problem Solution
Before Lingo can be used, this problem must be broken down into components:
Profit = Revenue – Expenses
What is revenue?
Revenue = (selling price)*(quantity sold)
= SP1(x1j) + SP2(x2j) + SP3(x3j)
(i refers to a warehouse property, while j refers to a plant property)
j i
ijiXSP
Transportation Problem Solution
What are the expenses? The cost of production and the cost of shipping.
Expenses = Production cost + Shipping cost
The costs of shipping from each plant to each warehouse are given below.
Cost ($/tonne-km) W1 ($/tonne) W1 ($/tonne) W1 ($/tonne)P1 0.30 46.50 42.00 43.50P2 0.35 43.75 38.50 42.00P3 0.31 46.50 41.85 43.40P4 0.34 44.20 39.10 42.50P5 0.29 34.80 29.00 31.90
Shipping Costs
Transportation Problem Solution
Production cost = (cost per unit)*(quantity produced)
=Cj*x1j + Cj*x2j + Cj*x3j
Shipping cost = (quantity shipped)*(shipping price)
=x1j*S1j + x2j*S2j + x3j*S3j
j i
ijijSX
j i
ijjXC
Transportation Problem Solution
The objective function is now:
maximize j i
ijiXSP j i
ijjXC j i
ijijSX
Transportation Problem Solution
The constraints:
x1j = 2400
x2j = 2000x3j = 2500
xi1 <= 1300
xi2 <= 1200
xi3 <= 1700
xi4 <= 1400
xi5 <= 1600
j i W1 W2 W3
P1 X11 X21 X31
P2 X12 X22 X32
P3 X13 X23 X33
P4 X14 X24 X34
P5 X15 X25 X35
Production and Destination Details
Lingo Solution
Lingo Solution
Question #3
Optimizing Minimum Approach Temperature
Tmin Optimization
A hot process stream coming out of a distillation tower has a specific heat flow rate, FCp, of 200 kW/K and must be cooled from 400 K to 300 K. Another process stream with an FCp of 150 kW/K must be heated from 330 K to 430 K before it enters a processing unit. A significant savings in utility costs can be realized by passing these streams through a heat exchanger.
Tmin Optimization
Heating utility is available at a cost of approximately 90 $/kW•year, while cooling utility is available at approximately 40 $/kW•year. Based on an expected useful life of 10 years, the heat exchanger is estimated to have an annualized fixed cost of about 600 $/year•m2. If the heat exchanger is expected to have a heat exchange coefficient of U = 1.2 kW/m2, investigate where the optimum minimum approach temperature lies. Hint: It is between Tmin = 5 K and 20 K.
Tmin Optimization
What is the optimum minimum
approach temperature in this case?
Use Tmin = 5 K, 10 K, and 20 K to develop your solution.
Attempt to solve this problem before proceeding to the solution.
Optimum Tmin Solution
Using the algebraic method, the utility requirements and exchanged heat are calculated for each Tmin.
Tmin (K) Qexchanged (kW) QC,min (kW) QH,min (kW) QH+C (kW)5 9750 10250 5250 15500
10 9000 11000 6000 1700020 7500 12500 7500 20000
T1 (K) T2 (K) FCp (kW/K) H (kW)
hot stream 300 400 200 20000
t1 (K) t2 (K) FCp (kW/K) H (kW)
cold stream 330 430 150 15000
Optimum Tmin Solution
Next, for each case the inlet and outlet temperatures of the heat exchanger are calculated so that the log mean temperature difference can be calculated.
)()( 1212 ttFCpTTFCpQexchanged
11
22
1122
ln
)()(
tTtT
tTtTTLM
Optimum Tmin Solution
Then the area of each heat exchanger is calculated.
LM
LMex
TU
QA
TUAQ
Tmin (K) Qexchanged (kW) TLM (K) Area (m2)
5 9750 11.1 732.010 9000 16.4 457.320 7500 25.7 243.2
Optimum Tmin Solution
Finally, the annual utilities cost, heat exchanger cost, and total cost are calculated and plotted as a function of Tmin.
Tmin
(K)
QC,min
(kW)
QH,min
(kW)
Area
(m2)Utilities cost
($/year)Exchanger cost
($/year)Total cost
($/year)
5 10250 5250 732.0 882500 439189 132168910 11000 6000 457.3 980000 274390 125439020 12500 7500 243.2 1175000 145914 1320914
AreaExchanger
QQUtilities CH
600
4090 min,min,
Optimum Tmin Solution
Annualized Cost as a Function of Tmin
0
200000
400000
600000
800000
1000000
1200000
1400000
5 10 15 20
Tmin (K)
Co
st (
$/ye
ar)
Utilities
Exchanger
Total
The optimum minimum approach temperature is near 10 K
The End
This is the end of the Process Optimization module.