Question 25, Morning of Day 2, 2012
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25. A brewery wishes to cool 100 m3 of hot wort at 97°C to 8°C within 60 minutes using a plate heat exchanger. The inlet and outlet temperatures of the cooling water are 3°C and 80°C respectively. The average density of the hot wort is 1040 kg/m3, with an average specific heat capacity of 4.10 kJ/(kg·K); the average specific heat capacity of the cooling water is 4.186 kJ/(kg·K), and its density is 1000 kg/m3. The heat transfer efficiency of the heat exchanger is 95%. What is the value for cooling water consumption (m3/h)? A 118 B 131 C 138 D 124. I checked previous solutions but am not satisfied; please advise, experts.
Calculate normally, and then divide by the heat transfer efficiency
100x1040x4.10x(97-8)=0.95x1000x4.186x(80-3)xD, D=123.9
Why is 0.95 multiplied by the cooling water side? Using this calculation, the heat required to raise the temperature of the cooling water exceeds the heat released by the hot liquid; it’s a bit confusing. It should be 0.95 multiplied by the hot side instead, right?
The purpose is cooling, and the cooling medium only functions at 95% efficiency; there must be cooling losses
There is a detailed introduction to the TianDa version P171!