Answer
$P_{v}= 1.988\text{ kPa}$
$P_{a}= 83.01\text{kPa}$
$w=0.0149\text{ kgH}_2\text{O/kg dry air}$
$h=57.90\text{ kJ/kg dry air}$
Work Step by Step
(a) The partial pressure of dry air can be determined from $$
\begin{aligned}
& P_v=\phi P_g=\phi P_{\text {sat } @ 20{ }^{\circ} \mathrm{C}}=(0.85)(2.3392 \mathrm{kPa})=1.988 \mathrm{kPa} \\
& P_a=P-P_v=85-1.988=83.01\ \mathrm{kPa}
\end{aligned}
$$ (b) The specific humidity of air is determined from
$$
\omega=\frac{0.622 P_v}{P-P_v}=\frac{(0.622)(1.988 \mathrm{kPa})}{(85-1.988) \mathrm{kPa}}=0.0149 \mathrm{~kg} \mathrm{H_{2 }} \mathbf{O} / \mathbf{k g ~ d r y ~ a i r}
$$ (c) The enthalpy of air per unit mass of dry air is determined from $$
\begin{aligned}
h & =h_a+\omega h_v \equiv c_p T+\omega h_g \\
& =\left(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(20^{\circ} \mathrm{C}\right)+(0.0149)(2537.4 \mathrm{~kJ} / \mathrm{kg}) \\
& =\mathbf{5 7 . 9 0} \mathbf{k J} / \mathbf{k g} \text { dry air }
\end{aligned}
$$
