GATE AG 2017 • Q49 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Heat and Mass Transfer

Subtopic: Material and energy balances in food processing systems

Question Statement:

Hot water at 95 °C is sent through a countercurrent tube-in-tube heat exchanger with cold water entering at 25 °C. Hot/cold water specific heat capacity is 4.2 kJ kg⁻¹ K⁻¹. Flow rates of hot and cold water are 2.7 and 4.1 kg min⁻¹, respectively. Overall heat transfer coefficient is 55 W m⁻² K⁻¹ and the area of heat transfer is 5 m². Cold water outlet temperature from the heat exchanger in °C will be __________.

Official Answer Key:

55 to 56

Step-by-Step Solution & Explanation

• Using the ε-NTU method for a counter-flow exchanger. Chot = 2.7×4.2 = 11.34 kJ/min·K = 189 W/K. Ccold = 4.1×4.2 = 17.22 kJ/min·K = 287 W/K. • Cmin = 189 W/K (hot stream), Cmax = 287 W/K, Cr = Cmin/Cmax = 0.6585. • UA = 55×5 = 275 W/K. NTU = UA/Cmin = 275/189 = 1.455. • ε (counter-flow) = [1–exp(–NTU(1–Cr))]/[1–Cr·exp(–NTU(1–Cr))]. • NTU(1–Cr) = 1.455×0.3415 = 0.497. exp(–0.497) = 0.6084. • ε = (1–0.6084)/(1–0.6585×0.6084) = 0.3916/0.5993 ≈ 0.6535. • Qmax = Cmin(Th,in–Tc,in) = 189×70 = 13,230 W. Qactual = ε×Qmax = 0.6535×13,230 ≈ 8647 W. • Tc,out = Tc,in + Qactual/Ccold = 25 + 8647/287 ≈ 25 + 30.1 ≈ 55.1 °C, inside the accepted range 55–56.

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