Hot refined oil at 120°C enters a concentric tube-in-tube heat exchanger at 20 kg/min; oil is cooled by water entering at 30°C from the other end at 50 kg/min. Specific heats: oil=1.9, water=4.2 kJ/kg°C. Effectiveness of the HE = 0.7. Assuming steady state, no heat loss, find the exit temperature of water (°C).
Options:
(A) 39.9
(B) 41.4
(C) 57.7
(D) 63.0
Official Answer Key:
B
Step-by-Step Solution & Explanation
Ch(oil) = 20×1.9 = 38 kJ/min°C (Cmin, since smaller); Cc(water) = 50×4.2 = 210 kJ/min°C (Cmax). q_max = Cmin×(Th,in−Tc,in) = 38×90 = 3420 kJ/min. q_actual = 0.7×3420 = 2394 kJ/min. Water temperature rise = q/Cc = 2394/210 = 11.4°C. Water exit temperature = 30+11.4 = 41.4°C.
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