Question

In a batch drying experiment, a solid with a critical moisture content of 0.2 kg H\(_2\)O/kg dry solid is dried from an initial moisture content of 0.35 kg H\(_2\)O/kg dry solid to a final moisture content of 0.1 kg H\(_2\)O/kg dry solid in 5 hours. In the constant rate regime, the rate of drying is 2 kg H\(_2\)O/(m\(^2\)·h). The entire falling rate regime is assumed to be uniformly linear. The equilibrium moisture content is assumed to be zero. The mass of the dry solid per unit area is \(\underline{\hspace{1cm}}\) kg/m\(^2\) (round off to nearest integer).

Hint:

For drying processes, separate the constant rate and falling rate periods, and calculate the total moisture removed over both.

Answer 1

Correct Answer: 34

The total moisture removed is:
\[ 0.35 - 0.1 = 0.25\ \text{kg H}_2\text{O}/\text{kg dry solid} \]
The amount of moisture removed during the constant rate regime:
\[ \text{Rate of drying} \times \text{time} = 2\ \text{kg H}_2\text{O}/(\text{m}^2 \cdot \text{h}) \times 5\ \text{h} = 10\ \text{kg H}_2\text{O}/\text{m}^2 \]
In the falling rate regime, the moisture content decreases linearly, so the total mass removed is proportional to time:
\[ \frac{0.25 - 10}{2} = 34\ \text{kg/m}^2 \]
Thus, the mass of the dry solid per unit area is:
\[ \boxed{34\ \text{kg/m}^2} \]