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Homework Help: Thermodynamics Gas Mixture - Dew Point Temperature

  1. Apr 3, 2017 #1
    1. The problem statement, all variables and given/known data
    View attachment 116989

    Only Number 1, not number 2

    2. Relevant equations
    Dew Point Temperature T = Saturated Temperated at Vapor Pressure

    Partial Pressure = (mole fraction) x (Mixture Pressure)

    3. The attempt at a solution

    The dew point temperature is only dependent on the pressure of the water vapor in the mixture. This pressure is equal to

    P (vapor) = 0.15 (100 kPa) = 15 kPa. Using the saturation tables, the temperature corresponding to this pressure is 53.97 degrees Celsius.

    The 2nd part is where I am having difficulty. The temperature cools down to 43.97 degrees Celsius. Now I have to find the mass flow rate of the water to determine how much time it will take to collect 10kg of water. This means I need to determine the mass flow rate of the vapor too? I know the mole fractions add up to 100%
  2. jcsd
  3. Apr 3, 2017 #2


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    What is the vapour pressure of water at 43.97°C? So what fraction of the water condenses in the heat exchanger? You know the flow rate of water vapour into the HE, so that gives you the rate of collection of liquid water.
  4. Apr 3, 2017 #3
    The saturation pressure at that temperature is 9.14 kPa using linear interpolation with the saturation tables. Ok, so the rate given was for the mixture which is the same for all components. For some reason, I was thinking each component might have a different mass flow rate.

    Then the part I'm confused is how to determine the fraction of water condensing. I know the fraction of water vapor and so the fraction of vapor and fraction of liquid water should add up to 0.15, correct?

    Edit: The new vapor pressure divided by the mixture pressure will give me a different mole fraction, wouldn't it? So is the mole fraction no longer 0.15?
  5. Apr 3, 2017 #4
    Am I supposed to use the humidity ratio to solve this? I thought the 0.1 kg/s was the mass flow rate of dry air

    ω = 0.622 x (Saturated Pressure)/ (Pressure mixture - Saturated Pressure)
  6. Apr 4, 2017 #5


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    As I read it, 0.1 kg/s at 100 kPa is the mass flow rate of the CO2-H2O-N2 mixture. If it isn't, there's not enough information given to answer part 1. This rate is not "the same for all components"; it is the sum of the rates for the individual components, which are proportional to their mass fractions (note what you are given is the volume fractions).
    So the entry pressure is 15 kPa, and the equilibrium pressure is 9.14 kPa. What fraction of the water vapour condenses? How much mass is that per second?
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