Question

A 4 m tank contains wet steam at 1 MPa. What is the volume occupied by the vapour phase if the quality is 60%?

Answer 1

At 1 MPa the steam table the specific volume of liquid (v_f) is 0.0011 and specific volume of vapour (v_g) is 1.8  

given quality x = 0.6

so specific volume v = v_f + x*(v_g-v_f) = 0.0011 + (0.6*(1.8-0.0011) = 1.08 m³/kg

so given volume is 4 m³ tank

volume = specific volume * mass

mass = volume / specific volume = 4/1.08 = 3.7 kg

quality is defined as % of vapor present in total mixture so here we can take 3.7*0.6 = 2.22 kg of vapor is present in total mixture

So volume of vapor = specific volume of vapor*mass of vapor = 1.8*2.2 = 3.96 m³