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.
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
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).
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)].
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.
Latexman, I am not sure how you got 0.63 cu./lb because i did the calculations and I got about half of you value; please see my second .JPG attachment for comment.
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.
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.
