Answer
$\mathcal{V}_2=2119\ ft/s$
Work Step by Step
From table A-17E:
Initial( $T_1=540°F=1000°R$): $s_1^{\circ}= 0.75042\ Btu/lbm.°R,\ h_1=240.98\ Btu/lbm$
Isentropic process therefore:
$\Delta s = 0 \rightarrow s_2^{\circ}-s_1^{\circ}-R\ln(P_2/P_1)=0$
Given $P_2=12\ psia,\ P_1=60\ psia,\ R=0.06855\ Btu/lbm.°R$
$s_2^{\circ}=0.6401\ Btu/lbm.°R$
Back to table A-17E:
Final ($s_2^{\circ}$): $h_2=152.15\ Btu/lbm$
From the energy balance:
$\dot{m}(h_1+\mathcal{V}_1^2/2)=\dot{m}(h_2+\mathcal{V}_2^2/2)$
Given $\mathcal{V}_1=200\ ft/s$:
$\mathcal{V}_2=2119\ ft/s$
