Question A granular wet solid is taken on a tray (1 mx0.6 m) and dried in a stream of hot air (120°C; humidity = 0.02 kg per kg dry air; velocity = 4.5 m/s). The flow of hot air is perpendicular to the face of the wet solid on the tray. The initial 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 is negligible. The falling rate of drying is linear in the moisture content. If the solid loading (dry basis) is 35 kg/m², calculate the total (constant and falling rate) drying time. Assume that the air flow is large and its temperature drop across the tray is small'.

  The convective heat transfer coffficient he is to be calculated first to find at the constart rate drying. It is assumed that heat transfer and drying occur. at the top open surface only (ie, no conduction or radiation).Air temperature (assumed constart), T_{G}=120^{\circ} \mathrm{C}=393 \mathrm{~K}.Pressure =1 \mathrm{~atm}.using the ideal gas law, humid volume of air for Y_{G}=0.02 is\begin{array}{l}\gamma_{H}=\frac{R T_{G}}{P}\left(\frac{1}{2 g}+\frac{1}{18}\right)=\left(\frac{(0.0821)(393)}{1.0}\right)\left(\frac{1}{2 g}+\frac{1}{18}\right) \\\gamma_{M}=1.148 \mathrm{~m}^{3} /(\mathrm{kg} \text { dry air })\end{array}Density of air, S_{G}=1.02 / 1.148=0.888 \mathrm{~kg} / \mathrm{r}^{3}Mass flow rate of the gas, G^{\prime}=u \rho_{5}\begin{array}{l}=(4.5)(3600)(0.888) \\=14,386 \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{h}\end{array}Now,\begin{aligned}h_{c} & =0.0204\left(\hbar^{\prime}\right)^{0.8} \\& =0.0204(14386)^{0.8} \\& =43.3 \omega / \mathrm{m}^{2} . \mathrm{k} .\end{aligned}Temperature of the solid in the constant rate Period = adiabtic saturation temperature of air (at T_{G}=120^{\circ} \mathrm{C}; humidity, Y_{G}=0.02 );wet-bulb temperature of air, T_{\omega}=41.5^{\circ} \mathrm{C}=T{ }^{\omega}, Saturation humidity at 41.5^{\circ} \mathrm{C}, Y_{S}=0.0545 \mathrm{~kg}(\mathrm{~kg} dry The temperature of the air is assumed constant\begin{array}{l}=\frac{(43.3)(120-41.5)}{(2400)(1000)} \\=1.412 \times 10^{-3} \mathrm{~kg} / \mathrm{m}^{2 \mathrm{~s}}\end{array}Given moisture contents: x_{i}=0.28 ; x_{c}=0.12;x^{*}=0 ; \quad x_{f}=0.005Solid bading, \omega_{s} / a=35 \mathrm{~kg} / \mathrm{m}^{2}Unbound moisture to be removed =\left(w_{s} / a\right)\left(x_{i}-x_{c}\right)\begin{array}{l}=35(0.28-0.12) \\=5.6 \mathrm{~kg} / \mathrm{m}^{2}\end{array}Constant rate drying time, t_{c}=\frac{5.6 \mathrm{~kg} / \mathrm{m}^{2}}{n_{c}}\begin{array}{l}=\frac{5.6 \mathrm{~kg} / \mathrm{m}^{2}}{5.08 \mathrm{~kg} / \mathrm{m}^{2} \mathrm{~h}} \\=1.102 \mathrm{hrs} .\end{array}Falling rate drying time, t_{f}\begin{array}{l}=\frac{\omega_{s}}{a} \cdot \frac{x_{c}-x^{*}}{N} \lambda_{n}\left(\frac{x_{c}-x^{*}}{x_{f}-x^{*}}\right) \\=(35) .\left(\frac{0.12-0}{5.08}\right) \lambda_{n}\left(\frac{0.12-0}{0.005-0}\right) \\=2.627 \text { hrs }\end{array}\text { Total drying time } \begin{aligned}=t_{c}+t_{f} & =1.102+2.627 \\& =3.75 \mathrm{~h} .\end{aligned}   ...