Discussion
650 g of a wet food containing 405 g water is dried in a tray dryer to a final moisture content of 6.8 % (dry basis). It is observed that the drying process occurs under constant rate period and it takes 8 h. The rate of drying (in kg/h) is
Answer: D
The rate of drying is 0.0485 kg/h.
Dry solids = 650 - 405 = 245 g. A final moisture of 6.8% on a dry basis means water left = 0.068 x 245 g:
\[ \begin{aligned} \text{water left} &= 0.068 \times 245 = 16.66\ \text{g} \\ \text{removed} &= 405 - 16.66 = 388.34\ \text{g} \\ \text{rate} &= \frac{0.38834\ \text{kg}}{8\ \text{h}} = 0.0485\ \text{kg/h} \end{aligned} \]
A sample of 650 g cannot lose anywhere near 78 to 129 kg in an hour, so the large options cannot be right whatever the working. Because the whole drying is in the constant-rate period, the average rate equals the constant rate.
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