Humidity affect density?

The density of moist air is always less than that of dry air at constant dry bulb temperature. The more moisture the less density of the mixture.

The density of standard air is 0.075lb/cu.ft where as the density of saturated steam(or water vapor) at 60F is mere 0.0008292lb/cu.ft.

 

The numbers that quark gives don't look right. Water has mol. wt. of 18, wile O2 and N2 have wts. of 32 and 28. I suspect that at a given temp. and pressure, humid air would have a slightly lower density than dry air, but nowhere near that much

 

Saturated Steam @ 60°F:

Temperature (T) 15.56 C 60.00 F
Pressure (P) 0.0177 bar 0.256 psi
Density Saturated Liquid (f) 998.97 kg/m3 62.364 lb/ft3
Saturated Vapor (g) 0.013272 kg/m3 0.000829 lb/ft3
Specific Volume Saturated Liquid (vf) 0.0010010 m3/kg 0.016035 ft3/lb
Saturated Vapor (vg) 75.344 1206.9
Internal Energy Saturated Liquid (uf) 65.31 kJ/kg 28.08 Btu/lb
Evaporated (ufg) 2331.6 1002.4
Saturated Vapor (ug) 2396.7 1030.4
Enthalpy Saturated Liquid (hf) 65.31 kJ/kg 28.08 Btu/lb
Evaporated (hfg) 2464.6 1059.6
Saturated Vapor (hg) 2530.0 1087.7
Entropy Saturated Liquid (sf) 0.23258 kJ/kg-K
(mayer) 0.05555 Btu/lb-R
Evaporated (sfg) 8.5360 2.0388
Saturated Vapor (sg) 8.7684 2.0943

 

Mathman,

To be frank, I was indeed puzzled by the hint of your post, initially, for quite sometime. Good steam properties calculator like steamtab which is based on IAPWS formulae should never lie, so it became my responsibility to check for a possible solution. This is what comes to my mind.

The specific volume of any gas, at STP, can easily be calculated by Avagadro's law if we know the molecular weight of the gas. Molecular weight of water is 18, as you rightly said, and average molecular weight of air is 29.

So, one lb mole of water vapor occupies 379/18 = 21.05 cu.ft at STP. So the density is about 1/21.05 = 0.0475lb/cu.ft. One lb mole of air occupies 379/29 = 13.07 cu.ft at STP. So the density is about 1/13.07 = 0.0765lb/cu.ft. So what you said should be right.

But the key word here is STP(14.696psia and 60F - a general standard, though there are many variations to the definition of STP). The saturation pressure of water vapor, at 60F, is 0.256398psia. So, you have to expand the water vapor from 14.696psia to 0.256398psia(isothermally) and the ratio is about 57.31. So the density should be 0.0475/57.31 = 0.0008288lb/cu.ft. This is close to the value I specified above and I think my procedure is fair if not accurate.

Thanks for pointing out.

Regards,

 

I think the data you are discussiing is for steam, which is not the same as for 100% humidity air. I believe the original question was about the latter.

 

In a sense, you are both right.

The key is 'saturated' steam, or water vapor. At 60F the sat pressure is 0.2563 psia, compared to 14.7 psia, or 1 atm.

If one were to look at the property of 'saturated' steam at 1 atm (14.7 psia), which would have a temperature of 212F, then the density is 0.037 lbm/ft³.

The partial pressure of water vapor in air is very low - near the sat. pressure IIRC. It's been so long since I've looked at this stuff - my memory is a bit 'foggy'.

Here's a basic discussion on humidity - Getting a handle on humidity

 

There are two types of saturation when we deal with moist air. One is air saturation with water and the other is water vapor(or steam) saturation at the given partial pressure corresponding to the dry bulb temperature. Though air is not saturated with water vapor at any RH below 100%, water vapor is at saturated state as the partial pressure of water vapor is its saturation pressure at the given temperature. So the above process is valid and infact RH calculation based on the ratio of partial pressure of vapor at the given temperature to the partial pressure of vapor when air is fully saturated is done by the above method.