Answer
a) $\dot{V}_1=0.283\ m³/s$
b) $\dot{W}_s=68.8\ kW$
Work Step by Step
From table A-20:
Inlet ($T_1=300K$): $\bar{h}_1=9431\ kJ/kmol$
Outlet ($T_2=450K$): $\bar{h}_2=15483\ kJ/kg$
From the material balance:
$\dot{m}=\frac{P_1\dot{V}_1}{RT_1}$
Given $P_1=100\ kPa,\ R=0.1889\ kJ/kg.K,\ T_1=300\ K,\ \dot{m}=0.5\ kg/s$:
$\dot{V}_1=0.283\ m³/s$
From the energy balance, with $M=44\ kg/kmol$:
$\dot{W}_s+\dot{m}h_1=\dot{m}h_2$
$\dot{W}_s=\frac{\dot{m}}{M}(\bar{h}_2-\bar{h}_1)$
$\dot{W}_s=68.8\ kW$
