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ANSWER: 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} The convective heat transfer coefficient he is to be constart rate dorging at u043eu0441u0441u0438u044f, Calculated first to find out the It is assumed that heat transfer and drying top open surface As , the radiation 0 ondy (ie, no conduction Pressure I atm of for ain using the ideal law, humid volume gas Ya - 0.02 is RTG 1 y + 18 A ( 18 P u0420 u0721u0722u0718 3 - 1.02 -0.288 kg le Mass - 46.31) ( 0821) (393 u667a In = 1.148 m/ling ang air Density of air, sa 12/1.148 flow rate of the gas, G'aula - (4.5)(360)(188) = 14, 386 kg /m2 h he = 0.0204 (14386) 43.3 W/m2.K. Temperature of the solid in the constant role adiabtic Saturation temperature of air (at Ta = 120u20ac; humidity. Ya =0,02); 0.8 nows : 0.0204 0204 ( 6 0.8 Period Tw = 41.5u00b0C kg Ang day wet - bulb temperature of aigh, Ts Saturation hun at 41.5c, Yo =0.0545 is assumed constant ain) The temperature of the air hc (Ta-tw) Constant during rate, Ne u0722u071du0710 ( 43.3) (120 - 41 17 41.5 ) (2400 (10) - 1.412x10 kg / m2 s Given moisture contents : X; = 0.28 ; Xc = 0.12; = 0.005 , u6728 A X - 35 solid bading, bg/m2 0; XP Wsla Unbound moisture to be removed (ws/a) ( x; )(- x) - 35 (0.98 - 0.12 ) = 5.6 kg/mu00b2 time, to = 5.6 kg/mu00b2 Constant rate drying Nc 5.6 ty Im? Im ) 5.08 8 kg = 1. 1o2 hors Falling rate time tu0119 drying xo u0445 Xc - Ws X In Xp a +* 0.12-0 (2) u0645u062f (2) (u062f) - 0.12 - - (35). 5.08 0.005-0 = 2.627 hors total = tette doying time 1.102 +2.627 = 3.75 h ...