Answer
$w_{1}=0.01377\text{ kg H}_2\text{O/kg dry air}$
$ϕ_{1}=64.4\%$
$T_{dp}=18.8^{∘}C$
Work Step by Step
(a) We obtain the properties of water vapor from EES. 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 $$
\omega_2=\frac{0.622 P_{\mathrm{g} 2}}{P_2-P_{\mathrm{g} 2}}=\frac{(0.622)(2.488 \mathrm{kPa})}{(100-2.488) \mathrm{kPa}}=0.01587 \mathrm{~kg~} \mathrm{H}_2 \mathrm{O} / \mathrm{kg} \text { dry air }
$$ Thus $$
\omega_1=\frac{\left(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)(21-26)^{\circ} \mathrm{C}+(0.01587)(2451.2 \mathrm{~kJ} / \mathrm{kg})}{(2548.3-88.10) \mathrm{kJ} / \mathrm{kg}}=0.01377\ \mathbf{k g ~ } \mathbf{H}_2 \mathrm{O} / \mathbf{k g ~ d r y ~ a i r}
$$ (b) The relative humidity $\phi_1$ is determined from $$
\phi_1=\frac{\omega_1 P_1}{\left(0.622+\omega_1\right) P_{g 1}}=\frac{(0.01377)(100 \mathrm{kPa})}{(0.622+0.01377)(3.3638 \mathrm{kPa})}=0.644 \text { or } \mathbf{6 4 . 4 \%}
$$ (c) The vapor pressure at the inlet conditions is $$
P_{v 1}=\phi_1 P_{g 1}=\phi_1 P_{\text {sat }} @ 266^{\circ \mathrm{C}}=(0.644)(3.3638 \mathrm{kPa})=2.166\ \mathrm{kPa}
$$ Thus the dew-point temperature of the air is $$
T_{\mathrm{dp}}=T_{\text {sat } @ P_r}=T_{\text {satt } @ 2.166 \mathrm{kPa}}=18.8^{\circ} \mathrm{C}
$$
