Answer
$\mathcal{V}_1=515\ ft/s$
$\mathcal{V}_2=941\ ft/s$
Work Step by Step
From the energy balance:
$\dot{m}(h_1+\mathcal{V}_1^2/2)=\dot{m}(h_2+\mathcal{V}_2^2/2)$
$2c_p(T_2-T_1)=\mathcal{V}_1^2-\mathcal{V}_2^2$
From the mass balance:
$\dot{m}=\dfrac{P_1A_1\mathcal{V}_1}{RT_1}=\dfrac{P_2A_2\mathcal{V}_2}{RT_2}$
$\mathcal{V}_1=\mathcal{V}_2\dfrac{T_1P_2}{T_2P_1}$
Given $T_1=580°R,\ T_2=530°R,\ P_1=100\ psia,\ P_2=50\ psia,\ c_p=0.248\ Btu/lbm.°R$
and solving for $\mathcal{V}_2=941\ ft/s$,
hence $\mathcal{V}_1=515\ ft/s$
