Answer
a) $\mathcal{V}_2=62.0\ m/s$
b) $P_2=91.1\ kPa$
Work Step by Step
From table A-17:
Inlet ($P_1=80\ kPa, T_1=27°C$): $h_1=300.19\ kJ/kg$
Outlet ($T_2=42°C$): $h_2=315.27\ kJ/kg$
From the energy balance:
$\dot{m}(h_1+\mathcal{V}_1^2/2)=\dot{m}(h_2+\mathcal{V}_2^2/2)+\dot{Q}$
Given $\dot{m} = 2.5\ kg/s,\ \mathcal{V}_1=220\ m/s,\ \dot{Q}=18\ kW$:
and solving for the outlet velocity: $\mathcal{V}_2=62.0\ m/s$
From the material balance:
$\dot{m}=\frac{P_2A_2\mathcal{V}_2}{RT_2}$
Given $A_2=0.04\ m²,\ R=0.287\ kJ/kg.K$:
solving for the pressure: $P_2=91.1\ kPa$
