Chapter 14 - Gas-Vapor Mixtures and Air-Conditioning - Problems - Page 751: 14-66E

$q_{in}=16.9\text{ Btu/lbm dry air}$

The amount of moisture in the air remains constant ($\omega_1 =\omega_2$) as it flows through the heating section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is $40$ psia. The properties of the air at the inlet and exit states are determined to be $$ \begin{aligned} P_{v 1} & =\phi_1 P_{g 1}=\phi_1 P_{\text {sat } @ 50^{\circ} \mathrm{F}}=(0.90)(0.17812 \mathrm{psia})=0.16031\ \mathrm{psia} \\ h_{g 1} & =h_{g @ 50^{\circ}\mathrm{F}}=1083.1\ \mathrm{Btu} / \mathrm{lbm} \\ \omega_1 & =\frac{0.622 P_{v 1}}{P_1-P_{v 1}}=\frac{0.622(0.16031 \mathrm{psia})}{(40-0.16031) \mathrm{psia}} \\ & =0.002503\ \mathrm{lbm~H}{ }_2 \mathrm{O} / \mathrm{lbm~} \mathrm{dry} \text { air } \\ h_1 & =c_p T_1+\omega_1 h_{\mathrm{g} 1} \\ & =\left(0.240 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\left(50^{\circ} \mathrm{F}\right)+(0.002503)(1083.1 \mathrm{Btu} / \mathrm{lbm}) \\ & =14.71\ \mathrm{Btu} / \mathrm{lbm~dry~} \mathrm{air} \\ P_{v 2} & =P_{v 1}=0.16031\ \mathrm{psia} \\ P_{g 2} & =P_{\text {sat } @ 120 \mathrm{F}}=1.6951 \mathrm{psia} \\ \phi_2 & =\frac{P_{v 2}}{P_{g 2}}=\frac{0.16031 \mathrm{psia}}{1.6951 \mathrm{psia}}=0.0946=\mathbf{9 . 4 6} \% \\ h_{g 2} & =h_g @ 120 \mathrm{F}=1113.2 \mathrm{Btu} / \mathrm{lbm} \\ \omega_2 & =\omega_1 \\ h_2 & =c_p T_2+\omega_2 h_{\mathrm{g} 2} \\ & =\left(0.240 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\left(120^{\circ} \mathrm{F}\right)+(0.002503)(1113.2 \mathrm{Btu} / \mathrm{lbm}) \\ & =31.59\ \mathrm{Btu} / \mathrm{lbm~dry} \text { air } \end{aligned} $$ From the energy balance on air in the heating section, $$ q_{\text {in }}=h_2-h_1=31.59-14.71=16.9 \text{ Btu/lbm dry air } $$