Answer
$w_{1}=0.00963\text{ kg H}_2\text{O/kg dry air}$
$ϕ_{1}=45.7\%$
$h_{1}=49.65\text{ kJ/kg dry air}$
Work Step by Step
(a) We obtain the properties of water vapor from EES. The specific humidity $\omega_1$
$$
\omega_1=\frac{c_p\left(T_2-T_1\right)+\omega_2 h_{f g 2}}{h_{g 1}-h_{f 2}}
$$ where $T_2$ is the wet-bulb temperature, and $\omega_2$ is determined from $$
\omega_2=\frac{0.622 P_{\mathrm{g} 2}}{P_2-P_{g 2}}=\frac{(0.622)(1.938 \mathrm{kPa})}{(95-1.938) \mathrm{kPa}}=0.01295 \mathrm{~kg~} \mathrm{H}_2 \mathrm{O} / \mathrm{kg} \text { dry air }
$$ Thus, $$
\omega_1=\frac{\left(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)(17-25)^{\circ} \mathrm{C}+(0.01295)(2460.6 \mathrm{~kJ} / \mathrm{kg})}{(2546.5-71.36) \mathrm{kJ} / \mathrm{kg}}=0.00963 \mathrm{~kg~} \mathbf{H}_2 \mathbf{O} / \mathbf{k g ~ d r y ~ a i r ~}
$$ (b) The relative humidity $\phi_1$ is determined from $$
\phi_1=\frac{\omega_1 P_1}{\left(0.622+\omega_1\right) P_{\mathrm{g} 1}}=\frac{(0.00963)(95 \mathrm{kPa})}{(0.622+0.00963)(3.1698 \mathrm{kPa})}=0.457 \text { or } 45.7 \%
$$ (c) The enthalpy of air per unit mass of dry air is determined from
$$
\begin{aligned}
h_1 & =h_{a 1}+\omega_1 h_{v 1} \\
& \equiv c_p T_1+\omega_1 h_{g 1} \\
& =\left(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(25^{\circ} \mathrm{C}\right)+(0.00963)(2546.5 \mathrm{~kJ} / \mathrm{kg}) \\
& =49.65 \mathrm{~kJ} / \mathrm{kg} \text { dry air }
\end{aligned}
$$
