Question B1: A granular wet solid is taken on a tray bidali (Imx0-6m) and dried in a stream of hot aire air (120°C, humidity=0.02 kg per kg dry 多) air; velocity, u= 4.5m/s). The initial &c. moisture content of 28% (dry basis) is to be reduced to 0.5%. From laboratory tests it is known that the critical moisture content is 12% and the equilibrium moisture content is Hl negligible. The falling rate of drying is linear in the moisture content. If the solid loading (dry basis) is 35kg/m”, calculate the drying time. Assume that the air flow is large and its temperature drop across the tray is small. 15

(1)solutionFrom the gigen dataNow, calculating the drying time isAir temperature T_{\varphi}=120^{\circ} \mathrm{C} \Rightarrow 120+273\begin{aligned}\text { pressure } & =1 \mathrm{~atm} \\u & =4.5 \mathrm{~m} / \mathrm{sec}\end{aligned}humidity \gamma_{G}=0.02By using an ideal gas law,\begin{aligned}\nu_{H} & =\frac{R T_{G}}{P}\left(\frac{1}{29}+\frac{1}{18}\right) \\& =\frac{0.0821 \times 393}{1}\left(\frac{1}{29}+\frac{1}{18}\right) \\& =0.0821 \times 393\left(\frac{29418}{29 * 18}\right) \\\gamma_{H} & =1.148 \mathrm{~m}^{3} / \mathrm{kg} \text { dry air } \\C_{\text {density }} \text { of air } & =1.02 / 1.148 \\& =0.888 \mathrm{~kg} / \mathrm{m}^{3} \\\text { mass Flow rate } Q^{\prime} & =\mathrm{Cl} \rho \mathrm{g} \\& =4.5 \times 3600 \times 0.888 \\& =14.386 \mathrm{~kg} / \mathrm{m}^{2} . \mathrm{h}\end{aligned}Then,\begin{aligned}h_{C} & =0.0204\left(G^{\prime}\right)^{0.8} \\& =0.0204(14386)^{0.8}\end{aligned}h c=43 \cdot 3 \mathrm{w} / \mathrm{m}^{2} \cdot kTemperature of the solid is the Constant rate period = adiabatic saturation temperature of air\text { Q. F } T_{\varphi}=120^{\circ} \mathrm{C}, \quad \nu_{\varphi}=0.02Wet bulb temperature T_{\omega}=T_{s}=41.5^{\circ} \mathrm{C} at 41.5^{\circ} \mathrm{C} \quad \gamma_{\mathrm{s}}=0.0545 \mathrm{~kg}IF the temperature of the air is assumed constant.\text { Constant drying rate } \begin{aligned}N_{c} & =\frac{h_{c}\left(T_{G}-T_{\omega}\right)}{1 \omega} \\& =\frac{43.3 \times(120-41.5)}{2400 \times 1000} \\& =\frac{43.3 \times 78.5}{2400000} \\& =1.412 \times 10^{-3} \mathrm{~kg} / \mathrm{m}^{\gamma} \cdot \mathrm{sec}\end{aligned}moisture Content x_{i}=0.28, x_{c}=0.12x^{*}=0, \quad x_{f}=0.005solid loading ws / a=35 \mathrm{~kg} / \mathrm{m}^{2}\begin{aligned}\text { unbound moisture to be removed } & =\left(w_{s} / a\right)\left(x_{i}-x_{c}\right) \\& =35 \times(0.28-0.12) \\& =5.6 \mathrm{~kg} / \mathrm{m}^{2}\end{aligned}\text { constant rate drying fine } \begin{aligned}t_{c} & =\frac{5.6 \mathrm{~kg} / \mathrm{M}^{\gamma}}{N c} \\& =\frac{5.6}{5.08} \\& =1.102 \mathrm{hrs}\end{aligned}Faccing rate drying time is\begin{aligned}\hbar_{F} & =\frac{\omega_{s}}{a} \frac{x_{c}-x^{*}}{\mathrm{~N}} \ln \left(\frac{x_{c}-x^{*}}{x_{f}-x^{*}}\right) \\& =35\left(\frac{0.12-0}{5.08}\right) \ln \left(\frac{0.12-0}{0.005-0}\right) \\& =35\left(\frac{0.12}{5.08}\right) \ln \left(\frac{0.12}{0.005}\right) \\t_{F} & =2.627 \mathrm{hrs}\end{aligned}\begin{aligned}\text { rotal drying time } & =t_{c}+t_{f} \\& =1.102+2.627 \\& =3.75 \mathrm{hrg}\end{aligned} ...