Answer
$m_{a}=1.421 \text {kg/s}$
$v_{2}=0.8706 \text{ m}^3\text{/kg dry air}$
$V_{2}=17.5\text{ m/s}$
Work Step by Step
The amount of moisture in the air remains constant ($\omega_1=\omega_2$) as it flows through the cooling section since the
process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total
pressure is $1$ atm. The properties of the air at the inlet state are determined from the psychrometric chart to be
$h_1=76.14 \mathrm{~kJ} / \mathrm{kg}$ dry air
$\omega_1=0.01594 \mathrm{~kg} \mathrm{H}{ }_2 \mathrm{O} / \mathrm{kg}$ dry air $\left(=\omega_2\right)$
$\boldsymbol{v}_1=0.8953 \mathrm{~m}^3 / \mathrm{kg}$ dry air
The mass flow rate of dry air through the cooling section is $$
\begin{aligned}
\dot{m}_a & =\frac{1}{v_1} V_1 A_1 \\
& =\frac{1}{\left(0.8953 \mathrm{~m}^3 / \mathrm{kg}\right)}(18 \mathrm{~m} / \mathrm{s})\left(\pi \times 0.3^2 / 4 \mathrm{~m}^2\right) \\
& =1.421 \mathrm{~kg} / \mathrm{s}
\end{aligned}
$$ From the energy balance on air in the cooling section, $$
\begin{aligned}
-\dot{Q}_{\text {out }} & =\dot{m}_a\left(h_2-h_1\right) \\
-(750 / 60) \mathrm{kJ} / \mathrm{s} & =(1.421 \mathrm{~kg} / \mathrm{s})\left(h_2-76.14\right) \mathrm{kJ} / \mathrm{kg} \\
h_2 & =67.35 \mathrm{~kJ} / \mathrm{kg} \text { dry air }
\end{aligned}
$$ (b) The exit state of the air is fixed now since we know both $h_2$ and $\omega_2$. From the psychrometric chart at this state we read $$
\begin{aligned}
& T_2=\mathbf{2 6 . 5} \mathrm{C} \\
& \phi_2=\mathbf{7 3 . 1} \% \\
& \boldsymbol{v}_2=0.8706 \mathrm{~m}^3 / \mathrm{kg} \text { dry air }
\end{aligned}
$$ (c) The exit velocity is determined from the conservation of mass of dry air, $$
\begin{gathered}
\dot{m}_{a 1}=\dot{m}_{a 2} \longrightarrow \frac{\dot{V}_1}{v_1}=\frac{\dot{V}_2}{v_2} \longrightarrow \frac{V_1 A}{v_1}=\frac{V_2 A}{v_2} \\
V_2=\frac{v_2}{v_1} V_1=\frac{0.8706}{0.8953}(18 \mathrm{~m} / \mathrm{s})=17.5 \mathrm{~m} / \mathrm{s}
\end{gathered}
$$
