Discussion
A single effect evaporator is used to concentrate orange fruit juice with 2.7:1 concentration ratio of product solid to feed solid. The juice is fed to the evaporator at 100 kg.h−1 and 30 °C, where the product moisture evaporates at 60 °C. The specific heat of juice is 3.9 kJ.kg−1.°C−1. The latent heat of vaporisation at 60 °C is 2450 kJ.kg−1. Heat to accomplish the evaporation, in MJ.h−1, is nearest to
Answer: A
The heat required is about 165.96 MJ.h−1.
The solids are concentrated 2.7 times, so the product mass is the feed divided by 2.7, and the rest is evaporated:
\[ \begin{aligned} m_{p} &= \frac{100}{2.7} = 37.04\ \text{kg/h} \\ m_{v} &= 100 - 37.04 = 62.96\ \text{kg/h} \end{aligned} \]
Heat is needed to warm the whole feed from 30 °C to the boiling temperature of 60 °C, and then to evaporate the water:
\[ \begin{aligned} Q_{s} &= 100(3.9)(60 - 30) = 11\,700\ \text{kJ/h} \\ Q_{L} &= 62.96(2450) = 154\,259\ \text{kJ/h} \\ Q &= 165\,959\ \text{kJ/h} \approx 165.96\ \text{MJ/h} \end{aligned} \]
154.26 MJ/h is the latent heat alone, leaving out the sensible heating of the feed. 117.06 and 142.56 do not fit the mass balance: the feed must lose about 63 kg of water per hour, and that alone needs over 154 MJ.
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