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I seek an equation for water/oxygen in an airmass

  1. Sep 11, 2012 #1
    I seek an equation to determine the ratio of water molecules / O2 molecules in a mass of air ------- given density, temperature and relative humidity. (Moles H2O / Moles O2)

    The volume can be any size, a cubic meter is fine. I’m just interested in the ratio.
    Density can be expressed in any measure.

    I would prefer temperature in F but C is fine.

    Because water displaces other elements in an airmass I believe this could be complicated. Depending on how much Oxygen is displaced or how predictable its displacement is.

    I seek to do this over a range of densities that match those from the psychrometric tables for %RH calculation.

    = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

    Trivia:
    I need some accuracy to then see if this has any correlation with wildland fire behavior.
    Perhaps paralleling Celsius Crossover (Canadian Extreme Fire Behavior) or Fahrenheit Crossover (Southern US moisture of extinction).

    Thanks
     
  2. jcsd
  3. Sep 11, 2012 #2
    (1) The components of air all move together, in spite of the different masses of the molecules, so that a single figure of 20.95% for oxygen in dry air is serviceable in nearly all situations -- more of that later.

    (2) The amount of water vapour in air is determined by the saturation vapour pressure at a particular temperature multiplied by the relative humidity. The SVP is accurately tabulated; there are algebraic expressions for SVP as a function of temperature, but I am not sure that any of them will be accurate enough for your intended purposes.

    (3) When a particular loading of water vapour is introduced into a parcel of dry air, all components of the air will be equally "displaced", so that with y% water vapour, the oxygen content of standard air will be 20.95% * 100 / (100+y)

    (4) Although air circulates and mixes evenly in principle, there is in practice quite a significant difference between, say, rural air and urban air, or indoor air and outdoor air, in oxygen content. If you are concerned with bushfire behaviour, particularly after a fire has started, the air that you are wanting to model may well have a greatly reduced oxygen content.

    The standard dry atmosphere is a typical, ideal, or average concept.
     
  4. Sep 12, 2012 #3

    Borek

    User Avatar

    Staff: Mentor

    What JohnRC wrote, plus amount of CO2 is not constant (just like oxygen), although concentration of CO2 is relatively small, so its effects can be below accuracy of your calculations.
     
  5. Sep 12, 2012 #4
    Last edited: Sep 12, 2012
  6. Sep 13, 2012 #5
    For proportion of water vapour in air

    p(H2O) atm = RH%/100 * 176160 * exp (-4057/(T°C + 236))

    p(O2) atm = 20.95 * p(local)/{1 + p(H2O)/p(local)}

    ratio water vapour: oxygen = p(H2O)/p(O2)

    Be very careful with "standard pressure". I have taken it to be 101.3 kPa. Note that most aneroid barometers (needle on dial) have been adjusted to read local values of "mean sea level pressure", and the pressure indicated on the dial will be nowhere near the actual local pressure if you are at any significant altitude. You will need a pressure from a mercury barometer, or from a pressure reading multiplied by an altitude correction factor. If you have an accurate local reading for mean sea level pressure, the actual atmospheric pressure is given by

    p(local) = p(MSL) * exp (-0.00013627 * h metre)
     
  7. Sep 13, 2012 #6
    Get the red ink pen out.

    I don’t know how to utilize the exponential function so start with that correction.
    Thanks JohnRC for your help.


    I’m in an atmosphere with a station pressure of 25 inHg (84.6595 kPa),
    a dry bulb temperature of 60 degrees F (15.5555 C)
    and a RH of 19%

    p(H2O) atm = RH%/100 * 176160 * exp (-4057/(T°C + 236))

    p(H2O) atm = .19 * 176160 * exp (-4057/(15.5555°C + 236))

    p(H2O) atm = 33470.4 * (16.1276537)

    p(H2O) atm = 539,799.02

    - - - - - - - - -

    p(O2) atm = 20.95 * p(local)/{1 + p(H2O)/p(local)}

    p(O2) atm = 20.95 * 84.6595 / {1 + 539,799.02 / 84.6595}

    p(O2) atm = 20.95 * .0132

    p(O2) atm = .27



    again appreciated
     
  8. Sep 13, 2012 #7
    oh dear! I suspect a typo or two on my part. But perhaps not.

    Conditions: 19% RH, 15.55°C

    p(H2O) = 0.19 * 176160 * exp (-4057/251.55)
    = 0.19 * 176160 * exp(–16.128)
    = 33470 * 9.901 E -8
    =0.003314 atm

    The exponential function simply means the transcendental number e = 2.71828... raised to that power. If you are not using a spreadsheet or calculator that has the function inbuilt, then exp(-x) = e^(-x) = 10^(-x/2.3026...) and you can look it up in your "antilogarithm" tables.

    To proceed any further, we need to know height above sea level and p(MSL) which will be the local "pressure" as recorded at a nearby weather station.

    Suppose that the weather station has said 991 hPa for local MSL pressure and that the altitude is 1400 ft = 417 m

    991 hPa = 991/1013.25 atm = 0.978 atm

    p(local) = 0.978 * exp (-0.00013627 * 417) = 0.978 * exp (-0.0568) = 0.978 * 0.945
    = 0.924 atm

    p(O2) = 0.924 * 0.2095 / (1 + 0.003314/0.924) = 0.1936 / 1.00359
    = 0.1929 atm

    ratio = 0.003314/0.1929 = 1.72% or 1 in 58.2
     
  9. Sep 13, 2012 #8
    OK;

    Just to be sure before we go any further.

    This
    ratio = 0.003314/0.1929 = 1.72% or 1 in 58.2
    is saying there is 1 molecule of water for every 58.2 molecules of O2 at that elevation (station pressure - the actual air pressure around you), temp and RH.
     
    Last edited: Sep 13, 2012
  10. Sep 13, 2012 #9
    Whoops! I have made a mistake. 1 in 58.2 would mean 1 molecule of water in 58.2 molecules of (oxygen or water)

    I should have said 1 to 58.2,

    which correctly describes the number of oxygen molecules that go with each water molecule.

    (that would mean 1 in 59.2)
     
  11. Sep 13, 2012 #10
    You have sharpened your ink pen.
    well done and thanks.

    Now to dig out excel for this IMac.
     
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