GATE 2025 CH – Question 65
A wet solid of 100 kg containing 30 wt% moisture is to be dried to 2 wt% moisture in a tray dryer. The critical moisture content is 10 wt% and the equilibrium moisture content is 1 wt%. The drying rate during the constant rate period is 10 kg/(h m$^2$). The drying curve in the falling rate period is linear. If the drying area is 5 m$^2$, the time required for drying is ____ h (rounded off to 1 decimal place).
Note: All the moisture values given are on dry-solid basis.
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Correct answer: 0.59 to 0.61
Explanation
Since the moisture values are on a dry-solid basis, $X_1 = 0.30$, $X_c = 0.10$, $X^* = 0.01$ and $X_2 = 0.02$ kg/kg dry solid. The dry solid in the 100 kg is $S_s = \frac{100}{1.30} = 76.92$ kg. The constant rate drying rate for the whole area is $10 \times 5 = 50$ kg/h. Constant rate period: $t_c = \frac{S_s(X_1 - X_c)}{R_cA} = \frac{76.92 \times 0.20}{50} = 0.308$ h. Falling rate period (rate linear in $X - X^*$): $t_f = \frac{S_s(X_c - X^*)}{R_cA}\ln\frac{X_c - X^*}{X_2 - X^*} = \frac{76.92 \times 0.09}{50}\ln 9 = 0.1385 \times 2.197 = 0.304$ h. The total is $0.308 + 0.304 = 0.61 \approx 0.6$ h.