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 is __________ 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

Correct Answer: 0.5

Given: Initial mass of wet solid, \( W_1 = 100 \, {kg} \) Initial moisture content (dry basis), \( X_1 = 30% = 0.30 \, {kg water per kg dry solid} \) Final moisture content, \( X_2 = 2% = 0.02 \, {kg water per kg dry solid} \) Critical moisture content, \( X_c = 10% = 0.10 \, {kg water per kg dry solid} \) Equilibrium moisture content, \( X_e = 1% = 0.01 \, {kg water per kg dry solid} \) Drying rate in constant period, \( R_c = 10 \, {kg per hour per square meter} \) Drying area, \( A = 5 \, {m}^2 \) Step 1: Calculate Dry Solid Mass ($M_s$)
The initial wet solid contains 30% moisture (dry basis): \[ X_1 = \frac{{Mass of water}}{{Mass of dry solid}} = 0.30 \] Total initial mass: \[ W_1 = M_s + 0.30M_s = 1.30M_s \] \[ M_s = \frac{100}{1.30} = 76.92 \, {kg (dry solid)} \] Step 2: Moisture Contents
All moisture contents are on dry basis: \begin{align*} X_1 &= 0.30\ {kg water per kg dry solid}
X_c &= 0.10\ {kg water per kg dry solid}
X_2 &= 0.02\ {kg water per kg dry solid}
X_e &= 0.01\ {kg water per kg dry solid} \end{align*} Step 3: Drying Rate in Constant Rate Period
Total drying rate: \[ N_c = R_c \times A = 10 \times 5 = 50 \, {kg water per hour} \] Step 4: Time for Constant Rate Period (\(t_1\))
Water removed in constant rate period: \[ \Delta W_1 = M_s(X_1 - X_c) = 76.92(0.30 - 0.10) = 15.38 \, {kg} \] \[ t_1 = \frac{\Delta W_1}{N_c} = \frac{15.38}{50} = 0.3076 \, {hours} \] Step 5: Time for Falling Rate Period (\(t_2\))
Water removed in falling rate period: \[ \Delta W_2 = M_s(X_c - X_2) = 76.92(0.10 - 0.02) = 6.15 \, {kg} \] For linear falling rate: \[ t_2 = \frac{M_s(X_c - X_e)}{N_c} \ln\left(\frac{X_c - X_e}{X_2 - X_e}\right) \] \[ t_2 = \frac{76.92(0.10 - 0.01)}{50} \ln\left(\frac{0.09}{0.01}\right) \] \[ t_2 = \frac{6.9228}{50} \ln(9) = 0.1385 \times 2.1972 = 0.3043 \, {hours} \] Step 6: Total Drying Time
\[ t_{{total}} = t_1 + t_2 = 0.3076 + 0.3043 = 0.6119 \, {hours} \] Rounded to 1 decimal place: \[ t_{{total}} = \boxed{0.6} \, {hours} \]