2010 Case 6 PM
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18. A shell-and-tube heat exchanger uses saturated steam at 143°C to heat the water inside the tubes, raising its temperature from 15°C to 45°C. It is known that the heat transfer coefficient for steam condensation is 1.1×104 W/(㎡℃), while the convective heat transfer coefficient of water inside the tube is 3500 W/(㎡℃). When the thermal resistance of the tube walls, the thermal resistance due to fouling, and heat losses are ignored, and the tube walls can be treated as flat surfaces, what are the values for the overall heat transfer coefficient (W/(m²·°C)) and the average temperature of the heat transfer surface (°C)? (A) 2655, 127.2 (B) 3921, 141.3 (C) 3313, 140.3 (D) 2655, 115.7 Solution: K = 1/(1/α1 + 1/α2) = 1/(1/11000 + 1/3500) = 2655. △tm = (45 – 15)/ln = 112.3. t1 = 143; t2 = 143 – 112.3 = 30.7. Let the temperature of the flat wall be tQ = α1S(143 – t) = α2S(t – 30.7). Then 11000(143 – t) = 3500(t – 30.7), and thus t = 115.7 °C. Answer: D. How should the highlighted part be understood?
This post was last edited by kang2012 on 2016-8-28 at 11:21. This explanation is a bit complicated; let’s simplify it. Q = KS△tm = a1S(143 – average wall temperature) = a2S(t_average wall temperature – t_average cold temperature), where t_average cold temperature is equal to the average temperature of the heating medium minus △tm. In this case, the average temperature of the heating medium is 143, so the average cold temperature is 143 – 30.7. (In fact, the average cold temperature can be approximated as (15 + 45)/2, but this would result in a slight discrepancy in the calculation results for the exam.) If it’s difficult to understand, simply use the formula Q = KS△tm = a1S(143 – average wall temperature) to solve it.
Regarding the wall temperature, it can be calculated directly using the formula: (143 – tw) / (1/α1) = (tw – (15 + 45)/2) / (1/α2)