Help with question 21 from the morning session of Juhua 2016
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21. A feed liquid at 120°C is continuously added to the bottom of a stirred-tank jacketed reactor, and cooling water at 30°C is circulated in the jacket to cool the feed liquid. When stirring is stopped, the two fluids flow in parallel, with the outlet temperatures of the material stream and the cooling water being 90°C and 42°C respectively. If all other conditions in the reactor remain unchanged and stirring is continued to increase the heat transfer coefficient, what will be the outlet temperature of the liquid mixture (°C) in this case? (A) 86.3 (B) 74.1 (C) 67.8 (D) 62.5 I have no idea at all; are there missing conditions?
:dizzy: That’s exactly how it was in the original question
120----90 30-----42; the temperature difference when operating in parallel is 66.8. The total heat quantity Q remains constant, so K increases, and the temperature difference can only decrease. 62.5<66.8, therefore choose D
The original question was: If the heat transfer coefficient is increased by 3 times, can it be calculated? Could you give the calculation process?
The original question was: If the heat transfer coefficient is increased by 3 times, can it be calculated? Could you give the calculation process?
It’s asking about the temperature of the material at the exit, not the temperature difference