Subtopic: Concentration and drying of liquid foods – evaporators, tray, drum and spray dryers
Question Statement:
A single effect evaporator is used to concentrate 5000 kg h⁻¹ of a 1.5 wt% sugar solution entering at 50 °C to a concentration of 2 wt% at 101.325 kPa. Steam supplied is saturated at 169.06 kPa (115 °C). The overall heat transfer coefficient is 1550 W m⁻² K⁻¹. The boiling point of solution is the same as that of water. The specific heat of the feed is 4.21 kJ kg⁻¹ K⁻¹. The latent heat of water at 100 °C is 2257.06 kJ kg⁻¹ and the latent heat of steam at 115 °C is 2216.52 kJ kg⁻¹. The required surface area for heat transfer is
Options:
(A) 6.9 m²
(B) 10.7 m²
(C) 13.9 m²
(D) 46.3 m²
Official Answer Key:
D
Step-by-Step Solution & Explanation
Mass balance (solids conserved): F·x_F = L·x_L ⟹ L = 5000×0.015/0.02 = 3750 kg/h; Vapor evaporated V = F−L = 1250 kg/h. Heat duty: Q = F·c_p(T_boil−T_feed) + V·λ_water(100°C) = 5000×4.21×(100−50) + 1250×2257.06 = 1,052,500 + 2,821,325 = 3,873,825 kJ/h = 1,076,063 W. ΔT = T_steam − T_boil = 115−100 = 15 K. A = Q/(U·ΔT) = 1,076,063/(1550×15) = ≈46.3 m².
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