Question

Steam enters a control volume operating at steady state at 3 bar and 160 ◦ C with a volumetric flow rate of 0.5 m3 /s. Saturated liquid leaves the control volume through exit #1 with a mass flow rate of 0.1 kg/s, and saturated vapor leaves through exit #2 at 1 bar with a velocity of 5 m/s. Determine the area of exit #2, in m2 .

Answer 1

superheated water vapor tables at 3Bar and 160 degrees:

Volume: V1: 0.482281 m^3/kg

superheated water vapor tables at 1Bar and 160 degrees:

Volume: V1: 0.880277 m^3/kg

Mass flow rate = Volumetric flow rate / volume = (0.5 m3 /s) / (0.48228 m^3/kg) = 1.024 kg / s

Of this mass 0.1 Kg/s is lost throughg saturated liquid

so, Mass flow fate coming out of exit 2 = 1.024 - 0.1 = 0.924 Kg/s

Again, Mass flow rate = Volumetric flow rate / volume = (Area * velocity) / volume

==> 0.924 Kg / s = Area * (5 m/s) / (0.880277 m^3/kg )

==> Area = 0.162 m^2