Hot water (0.01 m³ /min) enters the tube side of a counter current shell and tube heat exchanger at 80°C and leaves at 50°C. Cold oil (0.05 m³ /min) of density 800 kg/m³ and specific heat of 2 kJ/kg.K enters at 20°C. The log mean temperature difference in °C is approximately

Question

Quizplus · Accepted Answer

The log mean temperature difference (LMTD) is calculated using the formula: LMTD = (ΔT1 - ΔT2) / ln(ΔT1/ΔT2), where ΔT1 and ΔT2 are the temperature differences at the two ends of the heat exchanger. Here, ΔT1 = 80°C - 20°C = 60°C and ΔT2 = 50°C - 20°C = 30°C. Thus, LMTD = (60 - 30) / ln(60/30) = 30 / ln(2) ≈ 32°C.

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