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Homework Help: Thermodynamics: Air and Steam Mixtures

  1. Apr 12, 2014 #1
    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.
    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.
    Last edited: Apr 12, 2014
  2. jcsd
  3. Apr 12, 2014 #2


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    Gold Member

    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.
  4. Apr 21, 2014 #3
    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.
  5. Apr 21, 2014 #4

  6. Apr 21, 2014 #5
    Thank you, everyone!
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