Answer
a) $T_2=436.5\ K$
b) $P_2=331\ kPa$
Work Step by Step
The energy balance for the system:
$\dot{E}_{in}=\dot{E}_{out}$
$\dot{m}(h_1+\nu_1^2)=\dot{m}(h_2+\nu_2^2)$
Given $h_1=503.02 kJ/kg\ @(600kPa,500K),\ \nu_1=120\ m/s,\ \nu_2=380\ m/s$:
and solving for the outlet enthalpy: $h_2=438.02\ kJ/kg$
From Table A-17: $T_2=436.5\ K$
For ideal gases:
$P\dot{V}=\dot{m}RT\rightarrow \frac{P_1}{T_1}A_1\nu_1= \frac{P_2}{T_2}A_2\nu_2$
With $A_1/A_2=2$ and solving for the outlet pressure: $P_2=331\ kPa$
