pinch 70 C 70 C 4 We want to cool both the hot streams to the pinch temperature. The next step is to find the duty for the two heat exchangers:

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1 Ene Process Integration (3 ECTS credits) P Espoo 2016 EXERCISE 2 SOLUTIONS 1 MER heat exchanger network First we draw the stream-grid and calculate the enthalpy rate change, Q, above and below the pinch for each stream. pinch Q = cp*m = C Q = 70 Q = C 2 cp*m = C Q = 75 Q = C cp*m = C 3 Q = C Q = 450 cp*m = Q = above the pinch below the pinch According to the Euler method the minimum number of units: U min,mer 1) above + (N 1) under = (5 1) + (4 1) = 7 = (N above the pinch To ensure that the driving forces are correct close to the pinch the following requirement must be met for the cold and hot streams to be matched (cp ṁ) cold (cp ṁ) hot. The only way this can be met is to match stream number 1 and 4 and stream number 2 and 3. cp 4 ṁ 4 = = cp 1 ṁ 1 cp 3 ṁ 3 = = cp 2 ṁ 2 We want to cool both the hot streams to the pinch temperature. The next step is to find the duty for the two heat exchangers: Q above 1,4 = min{350, 450} = 350 kw Q above 2,3 = min{90, 90} = 90 kw Since there is not enough heat in stream 1 and 2 to heat stream 4 to the target temperature we need to add a heater. In order to calculate the duty of the heater we need the temperature of stream 4 after the heat exchanger: The duty of the heater is then: T 4 = = 140 C Q heater = 5.0 ( ) = 100kW

2 below the pinch Below the pinch the requirement to ensure correct driving forces close to the pinch is: (cp ṁ) hot (cp ṁ) cold The only match meeting this requirement is stream 1 and 3. The duty of the heat exchanger is then: Q below 1,3 = min{70, 50} = 50kW The exit temperature for stream 1 from the heat exchanger is then: T 1 = = 65.7 C We now have only hot streams left below the pinch and these need to be cooled with coolers. The duty of the coolers are then: Q cooler,1 = 3.5 ( ) = 20kW Q cooler,2 = 1.5 (80 30) = 75kW The heat exchanger network is shown in the drawing below. pinch C C 60 C Q = 20 kw 140 C 2 C 30 C Q = 75 kw 115 C 45 C C H Q = 90 kw 4 Q = 50 kw Q = 100 kw Q = 350 kw above the pinch below the pinch Note that the cooling and heating is the same as the targets found in Exercise 1). The reason why the number of units above the pinch is less than the expected 5 1 = 4 is that we have a perfect match between two streams. This results in two sub-networks above the pinch, so the maximum number of units needed is: U above = 5 2 = 3. 2 Problem table algorithm a) First we must adjust the start and target temperatures. This is done by subtracting 1 2 T min from the hot streams temperatures and adding 1 2 T min to the cold streams temperatures. T min adjusted stream # type cp ṁ T start [ C] T target [ C] T start [ C] T target [ C] 1 hot hot cold cold

3 b) We can then draw the streams and identify the intervals: 175 C C 5.0 kj/ks 135 C C 2.0 kj/ks 75 C 4 55 C 3.5 kj/ks 50 C 3 c) 25 C 1.5 kj/ks interval [ C] hot cp m cold cp m surplus/deficit [kw ] = = = = = Based on this information we can draw a heat cascade diagram C C C C C C C 35.0 kw 35.0 kw 45.0 kw kw 42.5 kw 37.5 kw 35.0 kw kw 45.0 kw kw 57.5 kw 37.5 kw 5.0 kw 95.0 kw

4 Heat can be transfered from a higher temperature to a lower, but not vice versa. A negative heat flow is not feasible since it would mean a heat transfer from a low temperature to a high. Looking at the cascade we can identify the largest thermodynamic in-feasibility, which in this case is 100 kw. We can make it thermodynamically feasible by adding this amount of heat to the top of the cascade. The minimum utility consumptions and the pinch temperature can be directly read from this cascade. In this case the minimum hot- and cold utility consumption is 100 kw and 95 kw respectively and the pinch temperature is 75 C. The pinch temperature is found where no heat is cascaded. 3 HINT a) T ( C) Q (kw) b) First we must remember to adjust for T min, which means that the high- and low pressure steams adjusted temperatures are 185 C and 125 C respectively. The total minimum hot utility consumption is 100 kw. We are able use 90 kw low pressure steam. This leaves 10 kw for high pressure steam. The grand composite curve is shown in the figure below.

5 190 high pressure steam low pressure steam T ( C) Q (kw) 4 Heat pump a) First we must adjust the temperatures for T min. Then we check how much heat that is available for the evaporator (at 65 C). We can get approximately 30 kw for the evaporator. This corresponds to 45 kw heat from the condenser; COP Q C Q C = COP Q C Q E COP 1 Q E Q C = 3 30 = This means that with help of the heat pump we can reduce the low pressure steam consumption with 45 kw and the cooling water consumption with 30 kw. The electricity consumption for the heat pump is 15 kw.

6 b) condenser 110 T ( C) evaporator Q (kw)