Answer
a) $T_2=376.6°C$
b) $A_1/A_2=6.46$
Work Step by Step
The energy balance:
$\dot{m}(h_1+\nu_1^2/2)=\dot{m}(h_2+\nu_2^2/2)$
From table A-6 (3 MPa, 400°C): $v_1=0.09938\ m³/kg,\ h_1=3231.7\ kJ/kg$
Given $\nu_1=40\ m/s,\ \nu_2=300\ m/s$
and solving for the outlet enthalpy: $h_2=3187.5\ kJ/kg$
From table A-6 (2.5 MPa, $h_2$): $v_2=0.11533\ m³/kg,\ T_2=376.6°C$
From the material balance:
$\frac{A_1\nu_1}{v_1}=\frac{A_2\nu_2}{v_2}$
Solving for the area ratio: $A_1/A_2=6.46$
