Answer
a) $\mathcal{V}_2=185\ m/s$
b) $A_1/A_2=0.887$
Work Step by Step
From table A-18:
Inlet (7°C): $\bar{h}_1=8141\ kJ/kmol$
Inlet (27°C): $\bar{h}_2=8723\ kJ/kmol$
The energy balance in molar basis:
$\dot{n}(\bar{h}_1+M.\mathcal{V}_1^2/2)=\dot{n}(\bar{h}_2+M.\mathcal{V}_2^2/2)$
Given $M=28\ kg/kmol,\ \mathcal{V}_1=275\ m/s$
and solving for the outlet velocity: $\mathcal{V}_2=185\ m/s$
The material balance for ideal gases:
$\frac{P_1A_1\mathcal{V}_1}{RT_1}=\frac{P_2A_2\mathcal{V}_2}{RT_2}$
Given $P_1=60\ kPa,\ P_2=85\ kPa,$ and
solving for the area ratio: $A_1/A_2=0.887$
