Question

A wet solid containing 20% (w/w) moisture (based on mass of bone-dry solid) is dried in a tray-dryer. The critical moisture content of the solid is 10% (w/w). The drying rate (kg m$^{-2}$ s$^{-1}$) is constant for the first 4 hours, and then decreases linearly to half the initial value in the next 1 hour. At the end of 5 hours of drying, the percentage moisture content of the solid is \(\underline{\hspace{2cm}}\) % (w/w) (rounded off to one decimal place).

Hint:

Always split drying problems into constant-rate and falling-rate periods and analyze each separately.

Answer 1

Correct Answer: 8 - 8.4

Moisture contents (dry basis):
Initial: $X_0 = 0.20$
Critical: $X_c = 0.10$
Drying rate constant for first 4 h → constant-rate drying.
Moisture removed in constant-rate period:
$\Delta X = X_0 - X_c = 0.20 - 0.10 = 0.10$
After 4 hours, drying enters falling-rate period.
Drying rate decreases linearly to half over 1 hour.
Average drying rate in falling-rate period:
$R_{avg} = \frac{R_0 + 0.5R_0}{2} = 0.75R_0$
Moisture removed in falling-rate period relative to constant-rate equivalent:
Effective drying time = $0.75$ hour of constant-rate removal.
Thus moisture removed in falling period:
$0.75 \times (0.10 / 4) = 0.01875$
Final moisture:
$X_f = 0.10 - 0.01875 = 0.08125$
Convert to percentage:
$X_f = 8.1%$ (w/w)
Rounded: $8.1%$