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Water Vapor

Water Vapor

Water Vapor

(OP)
I know my relative humidity, temperature, and pressure, and I am trying to find the specific volume. I do not know the quality. Any help is appreciated.

RE: Water Vapor

(OP)
It is asking me to upgrade to the professional engineering edition. Do you what this means.

RE: Water Vapor

No.

Good luck,
Latexman

RE: Water Vapor

As the first step you can calculate the specific humidity in the way shown below:

Specific humidity = 0.622*Pv/(Pa-Pv) kgH2O per kg of dry air

where Pa=atmospheric pressure mbar
      Pv=Phi*Pgd/100 mbar
      Phi=relative humidity %
      Pgd=saturation pressure (mbar from steam tables) at the dry bulb temperature

After this you calculate the molecular weight, the gas constant and then the density from PV=mRT

Regards,

athomas236

RE: Water Vapor

If your pressure is atmospheric, specific volume can be read directly from a psychrometric chart in volume per mass of dry air (or as I was taught per mass of bone dry air).

If your pressure is significantly different from the atmospheric pressure the psychrometric chart is based on, that complicates matters.  However, if the air pressure you are working with is greater than the chart's base pressure or it can be isothermally compressed to the base pressure of the psychrometric chart without going through the dewpoint (condensation), then you can look up the specific volume at the chart's base pressure and then correct it to the actual pressure using Boyle's law:

P1V1 = P2V2

If the air pressure you are working with is less than the chart's base pressure and an isothermal compression goes through the dewpoint, then follow the post by athomas236 (or get the psychrometric software to work).

Good luck,
Latexman

RE: Water Vapor

You would not have any quality value because you are in the superheated range except at 100%RH where x=100%

RE: Water Vapor

(OP)
The equation is:
PV=nRT
  =mRT/M
This translates to:
Pv=RT/M

where v is the specific volume and R is the universal gas constant. I do not know M or v.

I am given:
P=350 psi
T=160 F
Relative humidity = 40%

How do you calculate molecular weight or the specific volume?

RE: Water Vapor

A relative humidity of 40% means that the partial pressure of water vapor equals 4/10 of the vapor pressure of water at the system temperature.

At 160 F the vapor pressure of water is about 4.75 psia.  The partial pressure of water vapor is 1.9 psia.  The mole fraction of water vapor = 1.9/(350+14.7).  Notice I assumed it was 350 psig.  The mole fraction of air = 1-1.9/(350+14.7).

Average molecular weight = 18 x 1.9/(350+14.7) + 29 x [1-1.9/(350+14.7)].

v = RT/MP

Ta Da!


 

Good luck,
Latexman

RE: Water Vapor

The above responses are correct,however, formulae do not give a good idea into the fundamental. Hopefully you have knowledge of steam tables, Mollier and T-s diagrams, and of the equation of state for ideal gases and vapors.  See my .JPG attachment into a classic analysis found in thermodynamics about the answer that you seek.   

RE: Water Vapor

ajs83,

Are you after the specific volume of the mixture (0.63 ft3/lb) or the partial volume of water vapor?

Good luck,
Latexman

RE: Water Vapor

You divided the universal gas constant by the individual gas constant of air, not the molecular weight of air.  It threw you off by about a factor of 2.

Good luck,
Latexman

RE: Water Vapor

Ok! thanks

RE: Water Vapor

(OP)
I am looking for the dew point temperature.

RE: Water Vapor

At the dewpoint temperature, the vapor pressure of water = the partial pressure of water, i.e. 100% R. H.  The partial pressure of water in your case is about 1.9 psia.  The temperature at which the vapor pressure of water = 1.9 psia is about 124 degrees F.

Good luck,
Latexman

RE: Water Vapor

Look at my first sketch where the dew point is located.  I am assuming that you kow how to read steam tables or the Mollier diagram.

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