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  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}   ...