Is it possible to get a dry cold air mixture to be equal in weight with a wet hot air mixture?
I was looking through Mollier diagrams but they all talk about energy so now i do not know what to search for.
Lets assume same pressure.
Mats, if your question relates to aero (downforce or drag) then air density is important. Note that Specific Volume on the psychrometric chart does not include the mass of the water vapour so that will have to be added in.
If your question relates to engine performance, it is oxygen per unit volume that is important. This is inversely proportional to specific volume on the psychrometric chart since this is expressed in m^3/(kg of dry air) so moisture is not included in the mass term, but is included in the volume term.
Moisture content is also useful if the engine is detonation limited (allows more ign adv or boost) but is generally detrimental because of the oxygen it displaces.
To illustrate all this, look at 3 points on the chart, starting with 25*C - 20% relative humidity. From the chart, the specific volume is 0.85 m3/kg and water vapor content is 4g/kg. Dry air density is the inverse of SV ie 1/0.85 = 1.176. Actual air density is 1.176 x (1000 + 4)/1000 = 1.18
If we move to 100% RH at the same temp, we see that SV is now about 0.87 so engine power will be down by about 2.3% just because of the humidity. Water content is up to 20g/kg. Dry air density is 1/0.87 = 1.149. Actual air density is 1.149 x (1000 + 20)/1000 = 1.172
If we look at a cold day - say 8*C but still 20% RH, the SV is now 0.8, so engine power will now be up by 6.25% from point 1 and 8.75% from point two.
Edited by gruntguru, 23 January 2014 - 02:55.