Answer
$P_{a}= 96.01\text{ kPa}$
$w=0.0129\text{ kgH}_2\text{O/kg dry air}$ $h=52.78\text{ kJ/kg dry air}$
Work Step by Step
Analysis (a) The partial pressure of dry air can be determined from $$
\begin{aligned}
& P_v=\phi P_g=\phi P_{\text {sat }}\left(a 0^{\circ} \mathrm{C}=(0.85)(2.3392 \mathrm{kPa})=1.988\ \mathrm{kPa}\right. \\
& P_a=P-P_v=98-1.988=96.01\ \mathbf{k P a}
\end{aligned}
$$ (b) The specific humidity of air is determined from $$
\omega=\frac{0.622 P_v}{P-P_v}=\frac{(0.622)(1.988 \mathrm{kPa})}{(98-1.988) \mathrm{kPa}}=0.0129 \mathrm{~kg} \mathrm{H_{2 }} \mathbf{O} / \mathrm{kg} \text{ dry} \text { air }
$$ (c) The enthalpy of air per unit mass of dry air is determined from $$
\begin{aligned}
h & =h_a+\omega h_v \equiv c_p T+\omega h_g \\
& =\left(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(20^{\circ} \mathrm{C}\right)+(0.0129)(2537.4 \mathrm{~kJ} / \mathrm{kg}) \\
& =\mathbf{5 2 . 7 8} \mathrm{kJ} / \mathbf{k g} \text { dry air }
\end{aligned}
$$
