PROPOSED SOLUTION - ASSIGNMENT 3

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1 Norwegian University of Science and Technology Course: Process Integration Department of Energy and Process Engineering Number: TEP 45 Trondheim, 0..07, T. Gundersen Part: Heat Exchanger Networks PROPOSED SOLUTION - ASSIGNMENT Task : Minimum Energy Consumption and the Fewest Number of Units a) The process Heat Cascade is drawn by using all supply/target temperatures for the process streams (see the figure below). The required minimum driving forces are included in the heat cascade by a difference between hot and cold interval temperatures of ΔT min = 0ºC. ST Q H 70 C 60 C kw 0 kw kw 45 C R 5 C 4 65 kw 0 kw kw 0 kw C R 45 kw + 75 kw 50 C 40 C 60 C 50 C kw R R 4 Q C 0 C 0 C CW The following heat balances must be satisfied for the temperature intervals in the cascade: R = Q H + 60 or Q H = R 60 R = R +.5 or R = R.5 R = R 8.5 or R = R R 4 = R + 75 or R = R 4 75 Q C = R 4 5 or R 4 = Q C + 5 It should be obvious that to minimize external heating (Q H ) and external cooling (Q C ), is the same as to minimize the residual values R i for i =,4. The residuals are not allowed to be Page of 5

2 negative, since that would indicate an attempt to transfer heat towards a higher temperature, which is not feasible by using heat exchangers. Minimizing energy consumption (Q H and Q C ) therefore results in the situation where at least one of the residuals becomes equal to zero. Two solution strategies are possible: i) Try to identify which residual that is most likely to become zero, and then check that no other residuals are negative. In this case, R is such a candidate (since R = R 8.5). This gives the following values for the other residuals as well as the heating and cooling demands, Q H and Q C : R = 8.5, R = 80, R 4 = 75, Q H = 0, Q C = 60 ii) Try by first setting external heating (Q H ) equal to zero and then calculate the residuals in the heat cascade based on that assumption. In this case we obtain: R = 60, R = 6.5, R = 0, R 4 = 55, Q C = 40 Then we identify the (if such residuals exist) largest negative residual (or accumulated heat deficit), which in this case is R = 0. In order to make the heat cascade thermodynamically feasible, external heating must then be at least 0 kw. This means that all residuals as well as external cooling will have their values increased by 0 kw. In both cases, the conclusion is as follows: Q H,min = 0 kw, Q C,min =, T pinch = C / (R = 0) b) Minimum number of units when we aim for some, but not necessarily maximum heat recovery is: U min = (N ) global = (n H + n C + n util ) = ( + + ) = 5 Minimum number of units for maximum heat recovery is calculated by realizing that in this case we have to decompose at the Pinch. Then we must have two separate heat exchanger networks, one above and one below the Pinch point. On the basis of Pinch temperatures established in sub-task (a), we have: Streams above pinch: Streams below pinch:,,, 4, ST,,, CW This gives the following (MER is short for Maximum Energy Recovery): U min,mer = (N ) above + (N ) below = (5 ) + (4 ) = 7 Page of 5

3 Task : Heat Exchanger Network using the Pinch Design Method The stream Grid is drawn with data for mcp and enthalpy changes for the process streams above and below Pinch: ΔQ C Pinch C 60 C mcp.0 ΔQ 50 C C 0 C C 0 C C i) Design of network above Pinch: The requirements for number of streams and/or stream branches in order to bring all hot streams down to Pinch temperature is: n C n H With two hot streams and two cold streams, this requirement is satisfied. The requirements for pair of streams (i,j) that should be matched in so-called Pinch Exchangers is (to satisfy sufficient temperature driving forces): mcp Cj mcp Hi Of the two alternative choices for Pinch exchangers [(,) and (,4)] or [(,4) and (,)] only the last one satisfies the mcp requirements and therefore achieves proper driving forces for the two Pinch exchangers: mcp 4 = 4.0 >.0 = mcp and mcp =.0 >.5 = mcp Heat exchanger I between hot stream and cold stream 4 will have the following duty when we apply the tick-off rule (making the heat exchanger as large as possible in order to reduce the number of units as much as possible): Q I = min {40, 40} = 40 kw Correspondingly for exchanger II between hot stream and cold stream : Q II = min {, 0} = kw The temperature for cold stream after heat exchange with hot stream in exchanger II is: Page of 5

4 T = 80 + /.0 = 5 C The rest of the heating of cold stream must be done with external heating in the form of steam: or: Q H =.0 (5 5) = 0 kw Q H = 0 = 0 kw Checking the network above Pinch (see figure below) against Performance Targets from Task, we have: Q H = 0 kw = Q H,min (OK!) U = < 4 = U min,above (NB!!) The explanation why the number of heat exchangers above Pinch is one less than the (N ) rule indicates is the fact that we have a perfect match between hot stream and cold stream 4 that shall deliver and receive exactly 40 kw. This creates two sub-networks above Pinch: { and 4} and {, and ST} This is why we get: U above = (N S) = 5 =, rather than (N ) = 4 (q.e.d.) ii) Design of network below Pinch: The requirements for number of streams and/or stream branches in order to be able to bring all cold streams up to Pinch temperature is: n H n C With two hot streams and only one cold stream this requirement is satisfied. The requirement for pair of streams (i,j) that should be matched in so-called Pinch exchangers is (to ensure sufficient driving forces): mcp Hi mcp Cj Of the two alternatives for the single Pinch exchanger below Pinch (,) or (,), only the first one satisfies the mcp requirement and thereby achieves feasible driving forces: mcp =.0 >.0 = mcp Heat exchanger III between hot stream and cold stream will have the following duty: Q III = min {, 0} = kw The temperature of cold stream entering exchanger III (remember that we start our design at the Pinch and then work our way towards lower temperatures): T = 80 /.0 = 5 C Since we now have moved away from the Pinch for cold stream, it is allowed to exchange heat with hot stream even though the mcp rule is not satisfied. The reason is that we now Page 4 of 5

5 (contrary to the situation at the Pinch) have (much) more than minimum driving forces, and that some reduction in ΔT is acceptable, but the driving forces must all the time be checked against ΔT min. Heat exchanger IV between hot stream and cold stream will have the following duty: Q IV = min {, (0 )} = 0 kw The temperature of hot stream out of heat exchanger IV is: T = 0 /.5 = 70 C The rest of the cooling of hot stream must be done with external cooling in the form of cooling water: Q C =.5 (70 0) = or Q C = 0 = Checking the network below Pinch (see the figure below) against targets from Task, we have: Q C = = Q C,min (OK!) U = = U min,below (OK!) Drawing the complete MER Network then concludes this assignment: 70 C I Pinch III 60 C 50 C II IV 70 C 0 C 5 C H 5 II 80 III 5 IV 0 C 40 C 0 kw kw I 4 kw 0 kw 40 kw Page 5 of 5

PROPOSED SOLUTION - ASSIGNMENT 2

PROPOSED SOLUTION - ASSIGNMENT 2 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 Integration PROPOSED