PROPOSED SOLUTION - ASSIGNMENT 4

Norwegian University of Science and Technology Course: Process Integration Department of Energy and Process Engineering Number: TEP 4215 Trondheim, 30.12.07, T. Gundersen Part: Heat Exchanger Networks PROPOSED SOLUTION - ASSIGNMENT 4 Task 1: Optimal Stream Split Ratio a) There are several factors that influence the sum of costs for heat exchangers I and II, and thus also the optimal split ratio (α and β). The main principle in this local optimization is to adjust the split ratio of stream H2 in such a way that the temperature driving forces are distributed in an economically optimal way. The most important factors (properties) in this respect are the following: Heat Transfer Coefficients: U I vs. U II Temperature Driving Forces: ΔT LM,I vs. ΔT LM,II Duties (Heat Effects): Q I vs. Q II Cost Laws for the two Heat Exchangers In this task, it is assumed that the duties for the two heat exchangers involved have been determined in the design phase (Q I = 2000 kw and Q II = 700 kw) and that these duties should not be changed during optimization. The fact that the duty of exchanger I is considerably larger than exchanger II could indicate that exchanger I should have better (more) driving forces than exchanger II. When considering the heat transfer equation for a pure counter-current heat exchanger: A = Q / ( U ΔT LM ) it is obvious that it is the product of the heat transfer coefficient (U) and the driving forces (ΔT LM ) that is important for the heat transfer area. This means that the driving forces should be distributed among the two heat exchangers (by manipulating the split ratio α / β and thereby the two temperatures T α and T β ) in a way that compensates for high or low heat transfer coefficient. A heat exchanger with a low U-value should be given larger driving forces (ΔT LM ) than a heat exchanger with good heat transfer conditions if other factors are the same. In this example it is given that the two heat exchangers have equal heat transfer coefficients (U = 0.05 kw/m 2 C). When it comes to driving forces, the two heat exchangers I and II have identical ΔT in the hot end of the exchangers (ΔT = 10 C = ΔT min ). The inlet temperatures for the cold streams, however, are different in the two heat exchangers, since stream C1 has a temperature of 40 C entering exchanger I, while stream C2 after heat exchanger III has a temperature of 66.67 C entering exchanger II. This indicates that T β should be larger than T α. Page 1 of 5

If the cost equations for the two exchangers are different it would make sense to try to reduce the heat transfer area for the most expensive heat exchanger while the area of the other heat exchanger then would be increased. b) Since the duty for the two exchangers are given (and constant), one alternative could have been to select a split ratio that gave isothermal mixing where the two branches of stream H2 meet (T α = T β = T t,h2 = 70 C). This would give the following values: α = Q I / (100 - T α ) = 2000 / (100 70) = 66.67 kw/ C β = Q II / (100 - T β ) = 700 / (100 70) = 23.33 kw/ C The sum of α and β is of course equal to 90 kw/ C. A better solution that is based on qualitative considerations about area requirements for the heat exchangers is to try to achieve similar profiles in the two exchangers, and thereby a better distribution of the available driving forces. This means to select split ratio as follows: α / β = mcp C1 / mcp C2 = 40 / 30 while α + β = 90 kw/ C This gives the following values: α = 51.43 kw/ C and β = 38.57 kw/ C with corresponding temperatures: T α = 61.11 C and T β = 81.85 C As one can see from these calculations, the two alternatives (isothermal mixing vs. good distribution of driving forces) give significantly different values for α and β. c) The optimal split ratio can be found by iteration (in this case simple trial and error ). The result from (b) is of course used as a (good) starting point. k = 1: α = 51.43 (and thereby β = 38.57) T α = 61.1 C A I = 2690 m 2 C I = 2,616,603 NOK T β = 81.9 C A II = 1128 m 2 C II = 1,469,709 NOK C tot = 4,086,312 NOK k = 2: α = 50 (and thereby β = 40) T α = 60.0 C A I = 2773 m 2 C I = 2,670,513 NOK T β = 82.5 C A II = 1103 m 2 C II = 1,448,353 NOK C tot = 4,118,866 NOK k = 3: α = 55 (and thereby β = 35) T α = 63.6 C A I = 2523 m 2 C I = 2,506,429 NOK T β = 80.0 C A II = 1208 m 2 C II = 1,537,319 NOK C tot = 4,043,748 NOK k = 4: α = 56 (and thereby β = 34) Page 2 of 5

T α = 64.3 C A I = 2484 m 2 C I = 2,480,461 NOK T β = 79.4 C A II = 1237 m 2 C II = 1,561,221 NOK C tot = 4,041,682 NOK k = 5: α = 58 (and thereby β = 32) T α = 65.5 C A I = 2415 m 2 C I = 2,433,537 NOK T β = 78.1 C A II = 1307 m 2 C II = 1,618,475 NOK C tot = 4,052,012 NOK k = 6: α = 57 (and thereby β = 33) T α = 64.9 C A I = 2448 m 2 C I = 2,456,221 NOK T β = 78.8 C A II = 1270 m 2 C II = 1,588,069 NOK C tot = 4,044,290 NOK The diagram below shows graphically how the total investment cost for heat exchangers I and II varies with the split ratio (here represented by the mcp value for the branch of H2 that is referred to as α). Notice that the optimum in this case is relatively flat (the deep valley is a graphical fake and comes as a result of the small region of the y-axis that is included. The difference in total cost for α = 50 kw/ C and α = 56 kw/ C is 77,184 NOK, which represents only 1.9% of the minimum cost. The difference in cost between the minimum point obtained through optimization (α = 56 kw/ C) and the reasonable starting point based on driving force considerations (α = 51.43 kw/ C) is only 44,630 NOK or 1.1%. When it comes to isothermal mixing for stream H2 (corresponds to an α-value of 66.67 kw/ C), the total cost is 4,645,928 NOK, which is 15% above the minimum. The reason is that in this case the driving forces are very small in the cold end of heat exchanger II. C I + C II (NOK) 4 120 000 4 110 000 4 100 000 4 090 000 4 080 000 4 070 000 4 060 000 4 050 000 4 040 000 4 030 000 48 50 52 54 56 58 60 62 α (kw/ C) Page 3 of 5

Task 2: Heat Exchanger Network with Loops and Paths Number of heat exchangers in the given MER network is: U = 7 Fewest number of units without Pinch decomposition: U min = (N 1) = (2 + 2 + 2 1) = 5 Number of independent loops in the network is then: L = U U min = 7 5 = 2 Identifying these loops: A: 1 (I) 4 (II) 2 (IV) 3 (III) 1 B: 1 (III) 3 (IV) 2 (C2) CW (C1) 1 Loops A and B can be combined to a new loop C, but this loop will then not be independent: C: 1 (I) 4 (II) 2 (C2) CW (C1) 1 A reasonable strategy is to try to remove the smallest units first by using heat load loops in the network. This will cause the smallest damage in the network (driving forces) and will remove heat exchangers with high specific cost (NOK/kW recovered). Then, the driving forces (ΔT ΔT min ) are re-established (if possible) by using paths from hot to cold utility. In this network, none of the heat exchangers are really significantly smaller than the others; nevertheless we will try to remove the smallest unit, which is the cooler C2 with a duty of 1400 kw. However, this unit is required in order to reach the target temperature of C for hot stream 2. None of the cold streams can cool stream 2 down to its target temperature. The next heat exchanger in the network to be considered is then exchanger III with a duty of 1600 kw. This unit is part of both loops (A and B). If we use the principle of +X and X, we notice that exchanger III can not be removed by loop B, since that would give a new duty for cooler C2 of -200 kw, which of course does not make sense thermodynamically. Loop A is manipulated in the following way to remove heat exchanger III: Q I ' = Q I + 1600 = 4200 kw Q II ' = Q II 1600 = 200 kw Q IV ' = Q IV + 1600 = 5600 kw Q III ' = Q III 1600 = 0 kw The updated network without exchanger III is shown in the figure below, where new duties for the heat exchangers are given together with new temperatures. Despite the fact that we removed a rather large heat exchanger, there are no cross-over situations in the network. The requirement for minimum driving forces (ΔT min = C) is, however, not satisfied in the cold end of exchanger I (140 C versus 123.3 C) and in the hot end of heat exchanger IV (175.6 C versus 160 C). Next, we identify possible heat load paths from steam (H) to cooling water (C1 and C2): a: ST (H) 4 (I) 1 (C1) CW b: ST (H) 4 (II) 2 (C2) CW By using path (a) it is possible to increase the hot outlet temperature for heat exchanger I from 140 C to the required 123.3 C + ΔT min = 143.3 C. This can be achieved by increasing the duty of the steam heater (H) and one of the coolers (C1) by 99 kw, while the duty of heat exchanger I is reduced correspondingly with 99 kw. This action will, however, Page 4 of 5

not solve the problem for heat exchanger IV, thus we also need to consider manipulation of path (b). By increasing steam and cooling water consumption in units (H) and (C2) and reducing the duty of exchanger II, the driving forces will increase both in the cold end of exchanger I and in the hot end of exchanger IV. I 280 140 1 C1 2400 180 175.6 51.1 2 C2 1400 160 II IV 60 3 mcp [30] [45] [40] 5600 260 193.3 123.3 1 H 4 4000 4200 200 Actually, by using path (b), the driving forces are exactly restored by removing heat exchanger II, which can be done by the following manipulation of the network: Q H ' = Q H + 200 = 4200 kw, Q II '' = Q II ' 200 = 0 kw, Q C2 ' = Q C2 + 200 = 1600 kw The final network with only 5 units (2 units have been removed from the original MER design) is shown below. Notice that energy consumption has increased only by 200 kw while we have removed two heat exchangers of 1600 kw and 1800 kw respectively. The main reason for this low energy penalty is the fact that streams 2 and 3 have quite equal mcp values (45 and 40 kw/ C). I 280 140 1 C1 60 [60] mcp [30] 2400 180 55.6 2 C2 1600 160 5600 260 190 1 H 4 4200 4200 IV 3 [45] [40] [60] Page 5 of 5