Subtopic: Steady state heat transfer in conduction, convection and radiation
Question Statement:
Heat and Mass Transfer: Steady state heat transfer in conduction, convection and radiation; transient heat transfer in simple geometry; working principles of heat exchangers; diffusive and convective mass transfer; simultaneous heat and mass transfer in agricultural processing operations; material and energy balances in food processing systems; water activity, sorption and desorption isotherms.
A tube-in-tube counter-flow heat exchanger is heating oil from 35°C to 77°C by circulating hot water at 100°C. The outlet temperature of water is 70°C. The log-mean-temperature difference (LMTD) is
Options:
(A) exactly equal to the mean arithmetic temperature difference
(B) significantly greater than the mean arithmetic temperature difference
(C) significantly smaller than the mean arithmetic temperature difference
(D) very nearly equal to the mean arithmetic temperature difference
Official Answer Key:
D
Step-by-Step Solution & Explanation
For counter-flow: ΔT1 (hot-in end) = 100 − 77 = 23°C; ΔT2 (hot-out end) = 70 − 35 = 35°C.
Arithmetic mean ΔT = (23+35)/2 = 29.0°C.
LMTD = (ΔT1−ΔT2)/ln(ΔT1/ΔT2) = (23−35)/ln(23/35) = (−12)/ln(0.6571) = −12/(−0.4198) = 28.6°C.
Since the ratio ΔT1/ΔT2 = 23/35 = 0.657 is not far from 1, LMTD (28.6°C) is very close to the arithmetic mean (29.0°C) — as is generally true whenever the two terminal temperature differences are of comparable magnitude (ratio less than ~2). Answer: (D) very nearly equal to the mean arithmetic temperature difference.
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