Answer:
the specific volume will be ρ = 1.791 kg/m³
Explanation:
From conservation of mass:
m₃ out =m₁ in + m₂ in
m₃ out =m₁ in + m₂ in =0.08 kg/s + 0.15 kg/s = 0.23 kg/s
the volumetric flow rate q at the outlet will be
q = v*A , where v= velocity at the outlet and A= area of the outlet
q= 12.84 m/s * 0.01 m² = 0.1284 m³/s
the specific volume of the outlet fluid will be
ρ = m₃ out / q = 0.23 kg/s / 0.1284 m³/s = 1.791 kg/m³
ρ = 1.791 kg/m³
