Answer
a) $\nu_1=60.8\ m/s$
b) $T_2=685.8\ K$
Work Step by Step
For ideal gases:
$\dot{m}=\frac{P_1A_1\nu_1}{RT_1}$
Given $\dot{m}=6000\ kg/h,\ P_1=1\ MPa,\ T_1=773\ K,\ R=0.1889\ kJ/kg.K,\ A_1=0.004\ m²$,
and solving for the velocity: $\nu_1=60.8\ m/s$
From the material balance in molar base:
$\dot{n}(\bar{h}_1+M.\nu_1^2/2)=\dot{n}(\bar{h}_2+M.\nu_2^2/2)$
From Table A-20 (500°C) $\bar{h}_1=30797\ kJ/kmol$, given $M=44\ kg/kmol,\ \nu_2=450\ m/s$ and
solving for the outlet enthalpy: $\bar{h}_2=26423\ kJ/kmol$
Back to Table A-20: $T_2=685.8\ K$
