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SOLUTION : The convective heat transter coefficient he is to be calculated first to find aut the constant rate drying. It is assumed that heat transfer and drying occur. at the top open surface only (i.e, 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}{9}+\frac{1}{18}\right)=\left(\frac{(0.0821)(393)}{1.0}\right)\left(\frac{1}{99}+\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(\sigma^{\prime}\right)^{0.8} \\& =0.0204(14386)^{0.8} \\& =43.3 \mathrm{w} / \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, \left.\gamma_{G}=0.02\right);wet-bulb temperature of air, T_{\omega}=41.5^{\circ} \mathrm{C}=T_{\delta},Saturation humidity at 41.5^{\circ} \mathrm{C}, Y_{S}=0.0545 \mathrm{tg} / \mathrm{kg} dryThe temperature of the air is assumed constart airConstConstant during rate,\begin{aligned}N_{c} & =\frac{h_{c}\left(T_{G}-t_{\omega}\right)}{\lambda_{\omega}} \\& =\frac{(43.3)(120-41.5)}{(2400)(1000)} \\& =1.412 \times 10^{-3} \mathrm{~kg} / \mathrm{m}^{2 \mathrm{~s}}\end{aligned}Given moisture contents: x_{i}=0.28 ; x_{c}=0.12;x^{*}=0 ; \quad x_{f}=0.005 ;Solid bading, w_{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} \cdot \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^{*}}{w} \lambda_{n}\left(\frac{x_{c}-x^{*}}{x_{f}-x^{*}}\right) \\=(35) \cdot\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}\begin{aligned}\text { total drying time }=t_{c}+t_{f} & =1.102+2.627 \\& =3.75 \mathrm{~h} .\end{aligned}     "I hope this helps! Best of luck with the rest of your coursework". Please provide your valuable feedback by a thumbsup. Your thumbsup is a result of my efforts. Your LIKES are very important for me so please LIKEu2026. Thanks in advance!ud83dude0a. ...