a) Use super-targeting to get the right minimum heat recovery approach
temperature (HRAT) for the data of problem 2. Use the following cost data.
Step 1: Choose HRAT =10
400
350
300
250
200
150
100
50
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Step 2: Cascade the heat surplus through the intervals, add heat so that no deficit is
cascaded. Find the pinch position, minimum heating utility and cooling utility.
T (degree C)
358 54616 Min Heat Ultility
1 -14049
335.9 40568
2 -22219
280 18348
3 -415
279.3 17934
4 -5776
268.1 12158
5 -369
262.7 11789
6 -5585
238.2 6204
7 -503
235.4 5701
8 226
232.7 5927
9 1758
224.7 7686
10 15893
198.9 23579
11 1350
190.9 24928
12 -1092
186.1 23836
13 -3031
176.2 20805
14 -484
175.2 20321
15 -11623
165.3 8698
16 -1238
163 7460
17 -1296
160 6164
18 -6164
134 0 Pinch
19 18061
115.2 18061
20 7845
108.2 25907
21 -1402
90 24504
22 -3570
68 20934
23 -2866
56 18068
24 -2792
49 15276
25 738
45 16014
26 680
40 16694 Min Cool Ultility
Pinch T = 134 C, Hot utility =54616 kW, Cold utility = 16694 kW
Step 3: Find number of exchangers
There are 13 streams above the pinch + 1 utility and 7 streams below the pinch+ one
utility:
The number of exchangers (N-1)above + (N-1)below = 13 + 7 = 20 exchangers.
4. Create composite curves:
The total area is estimated from the composite curve diagram.
A=Q/(U*LMTD)
Q Th1 Th2 Tc1 Tc2 ΔTml A
1 680 45 40 20 21 22.03 123.52
2 2029 56 45 21 23 28.25 287.27
3 4134 68 56 23 28 36.16 457.26
4 9264 90 68 28 39 45.03 822.83
5 587 91 90 39 40 50.93 46.10
6 8628 108 91 39 54 53.28 647.77
7 11929 115 108 54 74 47.37 1007.32
8 29028 134 115 74 131 15.22 7630.46
9 14612 176 134 131 149 11.53 5067.12
10 555 177 176 149 150 27.20 81.59
11 2334 182 177 150 153 27.98 333.75
12 2035 186 182 153 155 29.49 275.93
13 267 186 186 155 155 30.46 35.05
14 2894 191 186 155 157 32.05 361.22
15 7993 199 191 157 163 34.87 916.83
16 3896 202 199 163 165 36.33 429.00
17 830 202 202 165 166 36.18 91.82
18 32593 225 202 166 205 26.75 4874.29
19 8402 233 225 205 216 18.18 1849.17
20 2469 235 233 216 219 17.00 580.80
21 1823 238 235 219 221 17.17 424.73
22 14762 263 238 221 239 20.65 2859.43
23 4115 268 263 239 243 24.42 674.17
24 3525 279 268 243 248 27.97 504.12
25 13482 336 279 248 264 49.04 1099.70
26 5033 600 336 264 270 169.43 118.82
27 49583 600 600 270 348 289.25 685.68
Total area 32285.72316 m2
Step 5:
Utility cost per year:
= (hot utility)*(hot cost) + (cold utility)*(cold cost) ($)
= $ 5.71 million
Calculate annual capital cost: (assuming time span of 5 years)
={(# exchangers)*(cost per exchanger) + (area)*(cost per square meter)}/(time span)
=$ 5.70 million
Total= Utility + Capital Cost = $11.4 million
Step 6: Repeat the same procedure for different values of HRAT. HRAT for optimum
cost is 200C
a) Assume that hot oil at 390oC is available as utility. Determine the outlet
temperature of this oil, if its usage is minimum. Discuss the costs associated with the
usage of heating oil. Is it always advisable using the lowest possible flow rate? Why?
Create Grand Composite Curve for HRAT = 10
400
Grand Composite Curve
350
300
Temperature(celsius)
250
200
150
100
50
0
0 10000 20000 30000 40000 50000 60000 70000
Enthalpy(kW)
For minimum flow rate, we need the steepest possible line (and minimum return
temperature) . The minimum return temperature is approximately 250 C.
A cost of using heating oil is the energy required to heat it to the input temperature. Using
the minimum flow rate and exit temperature then requires the oil to be heated over a farther
temperature range to bring it back to the input temperature, so energy costs of this could
outweigh the advantage of using the minimum flow rate.
b) Assume that all hot utility has to be satisfied using a furnace. What are your
stack losses? What is the real utility consumption? Discuss what would be a suitable
flame temperature to use.
Assuming the furnace utility temp is 600 C (the utility temp given in the first part of
the problem), the minimum usage will be tangent to the composite curve:
Grand
600 Composite Curve
500
400
Temperature(celsius)
300
200
100
0
-40000 -20000 0 20000 40000 60000 80000
Enthalpy(kW)
Stack loss: 36000 Kw. The total real utility consumption is 60774+36000 Kw. Stack
losses can be reduced if the furnace temperature is increased above 600 oC. The
limit is given by material considerations.
