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Homework Help: Calculating the effect of humidity volume and temp of compressed air

  1. Jan 20, 2014 #1
    This isn't homework, I'm a hobbyist, not a student. Posting here because questions I've previously posted were moved to this section. My intent is to better understand humidity and it's role in compressed air for an automotive application. All help and direction is greatly appreciated. Thank you in advance.

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

    A turbocharger compresses air which is then cooled, what is the new temperature?

    How do I calculate the temperature of compressed air with different humidity values?
    How do I calculate the volume or density of compressed air with different humidity values?

    Atmospheric pressure - 14.7
    Atmospheric air temperature - 77°F
    Relative Humidity - 77%
    Compressor efficiency - 60%
    Compressed air cooler efficiency - 80%
    Compressed air cooler media (water) temperature- 70°F
    Air compressed to - 14.7psig/29.4psia/2bar

    2. Relevant equations

    I haven't any idea where to start.

    3. The attempt at a solution
    Pretty sure I've figured out how to accurately calculate for dry air, where using the following method and values above result in a temperature of 110.62°F and a volume of 0.53 it's original with 0% humidity:

    Solving for temperature


    (k) 1.4 -1 = 0.4 / 1.4 = 0.286
    (T1) 77°F + 459.69 = 536.69°Ra
    (P1) 14.7psia
    (P2) 29.4psia

    T2= 654.36°Ra - 459.69 = 194.67°F


    CE - Compressor Efficiency = 60%

    T3 = (T2 - T1) / CE + T1
    (654.36-536.69) / 0.60 + 536.69 = 732.81°Ra or T3 = 273.1°F


    HE - Cooler efficiency - 80%
    WT - Cooler media (water) temperature - 70°F (529.69°Ra)

    T4 = T3 - ((T3 - WT) * HE)
    732.81 - ((732.81-529.69) * 0.80)
    732.81 - 162.50 = 570.31°Ra T4 = 110.62°F


    Solving for volume or compression ratio

    CR = T1 / T2 x PR

    Using T1, T4 and pressure ratio of 2 from above...

    536.69 / 570.31 x 2 = 1.88:1 Compression Ratio

    1 / 1.88 = 0.53

    So, what was 1 cubic foot of 77°F air should now be 0.53 cubic feet and 110.62°F

    Any links or insight on how to account for humidity and the effects of latent heat is very much appreciated.

    Thank you again.
    Last edited: Jan 20, 2014
  2. jcsd
  3. Jan 21, 2014 #2
  4. Jan 21, 2014 #3
    You pretty much do it the same thing as for dry air, except you use a slightly different value of k. If you know the relative humidity and the temperature, then you know the mole fraction of water vapor in the gas phase, correct. If you know the mole fraction of water vapor x, then you can calculate the weighted average molar heat capacity at constant volume Cv (weighted by the mole fractions of air and water vapor). The weighted average heat capacity at constant pressure is obtained by adding R. Then you can calculate k for the moist air. If at any point you need to use the ideal gas law, you need to use a molar weighted average of the molecular weight.
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