Question

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 m²). The drying curve in the falling rate period is linear. If the drying area is 5 m², the time required for drying ___________ h (rounded off to 1 decimal place).

Hint:

For drying time calculation, consider both the constant rate period and the falling rate period, and add their respective times.

Answer 1

Given:

• Initial mass of wet solid, \( W_1 = 100 \, \text{kg} \)
• Initial moisture content (dry basis), \( X_1 = 0.30 \)
• Final moisture content, \( X_2 = 0.02 \)
• Critical moisture content, \( X_c = 0.10 \)
• Equilibrium moisture content, \( X_e = 0.01 \)
• Drying rate in constant rate period, \( R_c = 10 \, \text{kg/hr/m}^2 \)
• Drying area, \( A = 5 \, \text{m}^2 \)

Step 1: Dry Solid Mass (\( M_s \))
\[ W_1 = M_s + X_1 M_s = M_s(1 + 0.30) = 1.30M_s \Rightarrow M_s = \frac{100}{1.30} = 76.92 \, \text{kg} \] Step 2: Total Drying Rate (\( N_c \))
\[ N_c = R_c \times A = 10 \times 5 = 50 \, \text{kg/hr} \] Step 3: Time in Constant Rate Period (\( t_1 \))
\[ \Delta W_1 = M_s (X_1 - X_c) = 76.92 \times (0.30 - 0.10) = 15.38 \, \text{kg} \] \[ t_1 = \frac{15.38}{50} = 0.3076 \, \text{hr} \] Step 4: Time in Falling Rate Period (\( t_2 \))
\[ t_2 = \frac{M_s (X_c - X_e)}{N_c} \ln\left( \frac{X_c - X_e}{X_2 - X_e} \right) \] \[ t_2 = \frac{76.92 \times (0.10 - 0.01)}{50} \ln\left( \frac{0.09}{0.01} \right) = \frac{6.9228}{50} \times \ln(9) = 0.1385 \times 2.1972 = 0.3043 \, \text{hr} \] Step 5: Total Drying Time
\[ t_{\text{total}} = t_1 + t_2 = 0.3076 + 0.3043 = 0.6119 \, \text{hr} \] Final Answer: \( \boxed{0.6} \, \text{hr} \)