# Thermodynamics: Air and Steam Mixtures

1. Apr 12, 2014

### nightingale

1. The problem statement, all variables and given/known data
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.

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.

Last edited: Apr 12, 2014
2. Apr 12, 2014

### maajdl

What does that mean? :

"... a mixture of air and steam which is 0.75 dry ..."

would that mean that 75% of the water is vapor and 25% is liquid ?
If this is the case, you know that the steam is saturated and you can find its partial pressure Pv.
Then you know also the partial pressure of air: Ptot - Pv = 7 - Pv .

The rest of the problem is then a simple conversion of quantities from partial pressure to masses.

3. Apr 21, 2014

### nightingale

Which means that the air is 100% saturated in order for the saturated vapour (Pg) to be equal to Pv. Am I correct?
Because if from the steam table I would only get Pg, not Pv. But since you said I can find Pv, that means Pg is equal to Pv?

Thank you.

4. Apr 21, 2014

### Staff: Mentor

Yes.

Chet

5. Apr 21, 2014

### nightingale

Thank you, everyone!