Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relative Humidity and Dew Point after compression

  1. Sep 27, 2013 #1
    Hello all,

    Would appreciate some guidance here.

    I have an air compressor compressing ambient air to 3000 psi into a 425cu-in tank. I have the volumetric flow rate at the exit. If I was looking for relative humidity, dew point, and moisture content/accumulation at the exit, how would I go about doing so.

    The air would then be expanded from 3000 psi to 600 psi, how would I get the same information again?

    I'm trying to evaluate the size of the moisture separator and heat exchanger needed to keep the system out of moisture by getting those values.

    Thank you in advance.
     
  2. jcsd
  3. Sep 28, 2013 #2
    Hi,
    First arrive at the air temperature and Relative humidity at the compressor inlet. Also arrive at the pressure dew point required for your application.
    From the saturated water vapour property table, note down the partial pressure of water vapour at the air temperature and Rh
    As the air is compressed, the partial pressure of water vapour increases by compression ratio and temperature of air increases. Because of increase in partial pressure of water vapour, the vapour condenses.
    Cool the air to the temperature, at the line pressure which is the pressure dew point. Air will be saturated athis temperature.
    As the air expands to the application pressure, partial pressure of water vapour decreases and thus relative humidity decreases.

    I have enclosed a file for reference
     

    Attached Files:

  4. Sep 29, 2013 #3
    Sakthi1267,

    Thank you for the response. I've pulling hair for something simple(I think). But, I wanted to add that the compressor is a three stage compressor, which spits out the air at 225 deg C. Is it safe assume then that at exit the humidity is 100%?
     
  5. Sep 29, 2013 #4
    Hail,

    The compression ration for the compressor is 3014.7/14.7 Psi( 1 atm = 14.7 psi) = 205. Let us assume air at the inlet is 37 degree c And 95% Rh.

    Partial pressure of water vapour at 37 C , 95 % Rh = 5.8 Kpa,
    The partial pressure of water vapour increase by compression ratio times = 205 * 5.8 = 1189 Kpa.

    The saturation temperature of water vapour for 1189 kpa = 187 C

    In your case the air temperature at outlet is 225 C. As air temperature is higher than saturation temperature, there will not be any condensation happening at compressor outlet.

    As the air expands from 3014 psi to 600 Psi(application pressure), the partial pressure of water vapour decreases. by expansion ratio which is = 3014/600 = 5.02

    Partial pressure of water vapour at 600 Psi air pressure = 1189/5 = 237. Kpa
    saturation temperature corresponding to 237 kpa= 125 C

    If your application temperature is less than 125 C, condensation will happen.
     
  6. Sep 29, 2013 #5
    Sakthi,

    That was very well put. I understand the system a little better. To add though, I had mistyped the temp, it's in fact 225F and not 225C. So, condensate will occur after the compression. I'm trying to determine the actual condensate quantity of accumulation or at least the rate.

    Can I just obtain the absolute humidity in g/m^3 at the exit temperature(225F) and multiply by flow rate, Since RH is 100% at exit.

    And, if correct, would similar path be warranted for expansion?
     
  7. Sep 30, 2013 #6
    Hail,

    The product of absolute humidity and flow rate will provide you water vapor content of air at 225 F. It will not provide you the actual condensate value.
    Here is a methodology which can be followed to calculate the condensate value.
    Specific humidity of air at input condition (37 C, Rh 95%)= 0.66 Pw/(P-Pw) – Kg of water vapour / Kg of dry air.
    Pw - Partial pressure of water vapor
    P – Air pressure.
    = 0.66 * (5861.5)/(100000-5861.5)
    =41 g/kg of air – Note 1

    At the compressor outlet, (107 C,Rh 100%) presssur 3000 Psi= 0.66 Pw/(P-Pw) – Kg of water vapour / Kg of dry air
    Pw = 143.3 Kpa
    P = 3000 Psi = 20684 Kpa
    = 0.66 (143.3/(20684-143.3)
    = 4.6 g/kg of air. – Note 2

    Total condensate = Note 1- Note 2
    = 41 - 4.6
    = 36.4 g/Kg of dry air

    Multiply this value with mass flow rate of dry air.
     
  8. Oct 18, 2013 #7
    dew pt temperature

    hail,

    How did you calculate the dew point temperature, you have used the formula, i am not able to understand.

    But the calculation on condensate mass is correct.

    Generally dew point temperature will be derived from saturation property table if we know the saturation pressure.
     
  9. Oct 18, 2013 #8
    Sakthi,

    Thanks for getting back. The formula used is conducted by W. Wagner and A. Prub called "The IAPWS Formulation 1995 for the Thermodynamic Properties of Ordinary Water Substance for General and Scientific Use" in the Journal of Physical and Chemical Reference Data. I ran a few data points, and they were sufficiently accurate. This allows me to input certain ambient conditions, and now have to worry about running through the tables repeatedly.
     
  10. Oct 21, 2013 #9
    Hail,

    What is your requirement for dryness value at the application.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Relative Humidity and Dew Point after compression
  1. Too much Humidity (Replies: 6)

Loading...