A vessel of 0.3m3 capacity contains a mixture of air and steam which is 0.75 dry. If the pressure is 7 bar and temperature is 116.9 degree Celsius, find: Mass of water present, mass of dry saturated vapour, mass of air.
Answer key: 0.102kg, 0.307kg, 1.394kg
2. Relevant equations
Relative humidity = mv/mg = Pv/Pg (pressure of vapour/ pressure of dry saturated vapour)
humidity ratio = mv/ma
Moist Volume V = total volume/mass of dry air
3. The attempt at a solution
I was able to determine the mass of dry saturated vapour. The moist volume (V) of dry saturated vapour from the steam table is 0.9774m3/kg and hence:
0.9974 = volume/mass of dry saturated vapour
0.9974 = 0.3/mg
mg = 0.3069kg
I have absolutely no idea however, on how to find the other two masses. I don’t understand the term 0.75 dry, I don’t know how to find the mass of water (is this mass of water vapour???) without relative humidity. I’d appreciate any help. Thank you very much.