Answer:
[tex]v=\frac{0.04149}{\pi D^2}[/tex]
Explanation:
-The mass flow rate at the outlet is equal to the sum of the mass flow rates of the constituents:
[tex]\dot m_3=\dot m_1+\dot m_2\\\\=11.5 \ kg/s[/tex]
#Outlet velocity is determined from the outlet's mass flow rate, the outlet area and the specific volume from table A-4
-Let the outlet pipe's diameter be [tex]D[/tex] meters.
-We apply the saturated liquid approximation:
[tex]v_o=\frac{4\dot m_3\alpha_3}{D_3^2\pi}\\\\=\frac{4\times 11.5\times 0.001017}{D^2\pi}\\\\=\frac{0.04149}{\pi D^2}[/tex]
Hence, the outlet's velocity is [tex]v=\frac{0.04149}{\pi D^2}[/tex]
