Process Optimization
Tier III: Optimization Design
Problems
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
Streams
H1
H2
H3
H4
H5
H6
H7
Hot Streams
Target
Supply
T1 (K)
T2 (K)
270
310
275
340
330
370
380
410
420
460
450
500
510
550
FCp
(kW/K)
350
250
400
275
450
500
350
Cold
Streams
C1
C2
C3
C4
C5
C6
Cold Streams
Supply
Target
t1 (K)
t2 (K)
250
300
290
340
310
360
380
460
450
480
490
530
FCp
(kW/K)
350
250
450
300
400
300
HEN Problem
Using the stream data given and a DTmin 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) Based on the optimum utility requirements,
come up with a possible heat exchange
network.
HEN Solution
Attempt to solve this problem before
proceeding to the solution.
Temperature Interval Diagram
Intervals
Cold Streams
Hot Streams
H7
10
11
12
420
410
410
400
390
380
370
380
370
360
350
340
330
320
310
300
340
330
320
310
300
290
275
270
260
265
260
250
C4
C3
C2
C1
FCp = 450
460
450
440
FCp = 250
470
460
450
C5
FCp = 350
20
21
H1
FCp = 350
19
H2
FCp = 250
14
15
16
17
18
H3
FCp = 400
13
H4
FCp = 275
9
500
490
480
FCp = 300
8
H5
FCp = 450
6
7
510
500
490
C6
FCp = 400
5
H6
FCp = 500
3
4
t
540
530
FCp = 300
2
FCp = 350
1
T
550
540
Table of Exchangeable Heat
Loads
Table of Exchangeable Loads - Hot Streams
Interval
i
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
H1,i
kW
3500
8750
1750
-
H2,i
kW
2500
2500
2500
2500
6250
-
H3,i
kW
8000
4000
4000
-
H4,i
kW
5500
2750
-
H5,i
kW
-
H6,i
H7,i
kW
kW
3500
10500
5000
10000
5000
4500
5000
13500
Total cooling required (kW)
Total, HHi
kW
3500
10500
0
5000
10000
5000
9500
13500
0
5500
2750
0
8000
4000
6500
2500
2500
6000
15000
1750
0
111500
Table of Exchangeable Heat
Loads
Table of Exchangeable Loads - Cold Streams
Interval
i
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
C1,i
kW
3500
8750
1750
3500
C2,i
kW
2500
2500
2500
2500
2500
-
C3,i
kW
9000
4500
4500
4500
-
C4,i
C5,i
C6,i
kW
kW
kW
9000
3000
8000
3000
4000
3000
9000
3000
6000
Total heating required (kW)
Total, HCi
kW
0
9000
3000
0
8000
7000
3000
9000
3000
6000
0
0
9000
7000
7000
7000
2500
6000
8750
1750
3500
100500
Cascade Diagram
QH,m in = 0
3500
1
17250
0
12
0
3500
10500
2
17250
9000
8000
13
5000
0
3
3000
4
4000
14
5
6500
15
8000
6
2500
16
7000
7
2500
17
3000
8
6000
18
9000
9
15000
19
3000
6000
10
14500
2750
0
11
17250
8750
14500
1750
20
15000
5500
6000
8250
18000
0
2500
8250
13500
13500
7000
8250
7000
9500
7000
12750
9000
5000
7000
13250
0
7000
10000
9000
16250
2000
5000
0
1750
14500
0
21
QC,m in =
11000
3500
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
120000
100000
D H (kW)
80000
60000
QC = 111500 kW
40000
20000
0
230
280
330
380
430
T (K)
480
530
580
Cold Composite Stream
Cold Streams
120000
100000
D H (kW)
80000
60000
QH = 100500 kW
40000
20000
0
230
280
330
380
430
T (K)
480
530
580
Optimized
120000
QH,m in = 0
100000
Cold composite
stream
D H (kW)
80000
60000
Hot
composite stream
40000
20000
QC,m in = 11000 kW
0
230
280
330
380 T (K) 430
480
530
580
220
250 270
320
370
470
520
570
t = T - DT
420
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 Problem
Five 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
$/tonne•km.
Transportation Problem
The distances between plants and warehouses, in km, are as
follows:
P1
P2
P3
P4
P5
W1
155
125
150
130
120
W2
140
110
135
115
100
W3
145
120
140
125
110
Plant 1 has a production capacity of 1300 tonnes, plant 2 can
make 1200 tonnes, plant 3 can make 1700 tonnes, plant 4
can make 1400 tonnes, and plant 5 can make 1600 tonnes.
Furthermore, market research suggests that the amount sold
at each warehouse is limited. Warehouse 1 can receive 2400
tonnes, warehouse 2 can receive 2000 tonnes, and
warehouse 3 can receive 2500 tonnes.
Transportation Problem
What combination of shipments will
maximize the profit that this company can
earn, and what is that profit? Use Lingo to
solve this.
Attempt to solve this problem before
proceeding to the solution.
Transportation Problem
Solution
Before Lingo can be used, this problem must be broken
down into components:
Profit = Revenue – Expenses
What is revenue?
Revenue = S(selling price)*(quantity sold)
  SPi X ij
j
i
= SP1(Sx1j) + SP2(Sx2j) + SP3(Sx3j)
(i refers to a warehouse property, while j refers to a plant
property)
Transportation Problem
Solution
What are the expenses? The cost of production
and the cost of shipping.
Expenses = Production cost + Shipping cost
The costs of shipping from each plant to each
warehouse are given below.
P1
P2
P3
P4
P5
Shipping Costs
Cost ($/tonne-km) W1 ($/tonne) W1 ($/tonne)
0.30
46.50
42.00
0.35
43.75
38.50
0.31
46.50
41.85
0.34
44.20
39.10
0.29
34.80
29.00
W1 ($/tonne)
43.50
42.00
43.40
42.50
31.90
Transportation Problem
Solution
Production cost = S(cost per unit)*(quantity
produced)
  C j X ij
j
i
=SCj*x1j + SCj*x2j + SCj*x3j
Shipping cost = S(quantity shipped)*(shipping
price)
  X ij S ij
j
i
=Sx1j*S1j + Sx2j*S2j + Sx3j*S3j
Transportation Problem
Solution
The objective function is now:
SPi X ij   C j X ij   X ij S ij
maximize 
j
i
j
i
j
i
Transportation Problem
Solution
The constraints:
Sx1j = 2400
Sx2j = 2000
Sx3j = 2500
Sxi1 <= 1300
Sxi2 <= 1200
Sxi3 <= 1700
Sxi4 <= 1400
Sxi5 <= 1600
Production and Destination Details
i
j
P1
P2
P3
P4
P5
W1
W2
W3
X11
X12
X13
X14
X15
X21
X22
X23
X24
X25
X31
X32
X33
X34
X35
Lingo Solution
Lingo Solution
Question #3
Optimizing Minimum Approach
Temperature
DTmin 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.
DTmin 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 $/year•m2. If
the heat exchanger is expected to have a heat
exchange coefficient of U = 1.2 kW/m2,
investigate where the optimum minimum
approach temperature lies. Hint: It is between
DTmin = 5 K and 20 K.
DTmin Optimization
What is the optimum minimum
approach temperature in this case?
Use DTmin = 5 K, 10 K, and 20 K to develop
your solution.
Attempt to solve this problem before
proceeding to the solution.
Optimum DTmin Solution
T2 (K)
400
FCp (kW/K)
200
DH (kW)
hot stream
T1 (K)
300
t2 (K)
430
FCp (kW/K)
150
DH (kW)
cold stream
t1 (K)
330
20000
15000
Using the algebraic method, the utility
requirements and exchanged heat are
calculated for each DTmin.
DTmin (K) Qexchanged (kW) QC,min (kW)
5
9750
10250
10
9000
11000
20
7500
12500
QH,min (kW)
5250
6000
7500
QH+C (kW)
15500
17000
20000
Optimum DTmin Solution
Next, for each case the inlet and outlet
temperatures of the heat exchanger are
calculated so that the log mean temperature
difference can be calculated.
Qexchanged  FCp(T2  T1 )  FCp(t 2  t1 )
DTLM
(T2  t 2 )  (T1  t1 )

 T2  t 2 

ln 
 T1  t1 
Optimum DTmin Solution
Then the area of each
heat exchanger is
calculated.
Qex  UADTLM
Q
A
UDTLM
2
DTmin (K) Qexchanged (kW) DTLM (K) Area (m )
5
9750
11.1
732.0
10
9000
16.4
457.3
20
7500
25.7
243.2
Optimum DTmin Solution
Finally, the annual utilities cost, heat exchanger
cost, and total cost are calculated and plotted as
a function of DTmin.
Utilities  90  QH ,min  40  QC ,min
Exchanger  600  Area
DTmin
(K)
QC,min
(kW)
QH,min
(kW)
Area
(m2)
Utilities cost
($/year)
Exchanger cost
($/year)
Total cost
($/year)
5
10
20
10250
11000
12500
5250
6000
7500
732.0
457.3
243.2
882500
980000
1175000
439189
274390
145914
1321689
1254390
1320914
Optimum DTmin Solution
Annualized Cost as a Function of DTm in
1400000
1200000
Cost ($/year)
1000000
The optimum minimum
approach temperature
is near 10 K
800000
600000
Utilities
Exchanger
Total
400000
200000
0
5
10
15
DTmin (K)
20
The End
This is the end of the Process Optimization
module.