Answer
$\mathcal{V}_2=764.3\ m/s$
Work Step by Step
From tables A-5 and A-6:
Inlet ($P_1=6\ MPa, x_1=1$): $s_1=5.8902\ kJ/kg.K,\ h_1=2784.6\ kJ/kg$
Outlet ($P_2=1.2\ MPa, s_2=s_1$): $x_2=0.8533,\ h_2=2492.5\ kJ/kg$
From the energy balance:
$\dot{m}(h_1+\mathcal{V}_1^2/2)=\dot{m}(h_2+\mathcal{V}_2^2/2)$
$h_1+\mathcal{V}_1^2/2=h_2+\mathcal{V}_2^2/2$
Given $\mathcal{V}_1\approx 0\ m/s$ and solving for the outlet velocity:
$\mathcal{V}_2=764.3\ m/s$
