Community Answer

Honor CodeSolved 1 Answer

See More Answers for FREE

Enhance your learning with StudyX

Receive support from our dedicated community users and experts

See up to 20 answers per week for free

Experience reliable customer service

Get Started

(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} ...