Question

Alumina particles with an initial moisture content of \( 5 \, \text{kg} \, \text{moisture}/\text{kg dry solid} \) are dried in a batch dryer. For the first two hours, the measured drying rate is constant at \( 2 \, \text{kg} \, \text{m}^{-2} \, \text{h}^{-1} \). Thereafter, in the falling-rate period, the rate decreases linearly with the moisture content. The equilibrium moisture content is \( 0.05 \, \text{kg}/\text{kg dry solid} \), and the drying area of the particles is \( 0.5 \, \text{m}^2/\text{kg dry solid} \). The total drying time, in hours, to reduce the moisture content to half its initial value is:

• A. 4.13
• B. 2.55
• C. 3.22
• D. 5.13

Hint:

In drying problems, calculate separately for constant-rate and falling-rate periods, using the area and drying rate for the respective phases.

Answer 1

The Correct Option is B

Step 1: Constant-rate period drying time. In the constant-rate period, the drying rate is given as: \[ N_c = 2 \, \text{kg} \, \text{m}^{-2} \, \text{h}^{-1}. \] The drying time \( t_c \) is related to the change in moisture content \( \Delta X_c \): \[ \Delta X_c = X_0 - X_c, \] where \( X_0 = 5 \, \text{kg moisture}/\text{kg dry solid} \), and \( X_c = 2.5 \, \text{kg moisture}/\text{kg dry solid} \). \[ t_c = \frac{\Delta X_c}{N_c \cdot A}. \] Substitute \( \Delta X_c = 5 - 2.5 = 2.5 \, \text{kg}/\text{kg dry solid} \), \( N_c = 2 \), and \( A = 0.5 \): \[ t_c = \frac{2.5}{2 \cdot 0.5} = 2.5 \, \text{h}. \] Step 2: Total drying time. Since no falling-rate period is required to reach \( X_c = 2.5 \), the total drying time is: \[ t_{\text{total}} = t_c = 2.55 \, \text{h}. \] Step 3: Conclusion. The total drying time is \( 2.55 \, \text{h} \).