Answer
$w_{1}=0.01127\text{ lbm H}_2\text{O/lbm dry air}$
$ϕ_{1}=59.2\%$
$T_{dp}=59.8^{∘}$
Work Step by Step
(a) The specific humidity $\omega_1$ is determined from $$
\omega_1=\frac{c_p\left(T_2-T_1\right)+\omega_2 h_{f g 2}}{h_{g 1}-h_{f 2}}
$$ where $T_2$ is the wet-bulb temperature, and $\omega_2$ is determined from $$
\omega_2=\frac{0.622 P_{g 2}}{P_2-P_{g 2}}=\frac{(0.622)(0.30578 \text { psia })}{(14.3-0.30578) \text { psia }}=0.01359\ \mathrm{lbm~} \mathrm{H}_2 \mathrm{O} / \mathrm{lbm} \text { dry air }
$$ Thus, $$
\omega_1=\frac{\left(0.24 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)(65-75)^{\circ} \mathrm{F}+(0.01359)(1056.5 \mathrm{Btu} / \mathrm{lbm})}{(1093.9-33.08) \mathrm{Btu} / \mathrm{lbm}}=\mathbf{0 . 0 1 1 2 7}\ \mathbf{l b m ~ \mathbf { H } _ { 2 }} \mathbf{O} / \mathbf{l b m} \text { dry air }
$$ (b) The relative humidity $\phi_1$ is determined from $$
\phi_1=\frac{\omega_1 P_1}{\left(0.622+\omega_1\right) P_{g 1}}=\frac{(0.01127)(14.3 \mathrm{psia})}{(0.622+0.01127)(0.4302 \mathrm{psia})}=0.5918 \text { or } \mathbf{5 9 . 2} \%
$$ (c) The vapor pressure at the inlet conditions is $$
P_{v 1}=\phi_1 P_{g 1}=\phi_1 P_{\text {sat } @ 75^{\circ} \mathrm{F}}=(0.5918)(0.4302 \mathrm{psia})=0.2546\ \mathrm{psia}
$$ Thus the dew-point temperature of the air is $$
T_{\mathrm{dp}}=T_{\text {sat } @ P_\gamma}=T_{\text {sat } @ 0.2546 \mathrm{psia}}=\mathbf{5 9 . 8} ^{\circ} \quad \text { }
$$
