Need to know relative humidity (RH) of compressed air 1

[GasDistEng]

Hi-

Air at atmospheric pressure (14.7 PSIA) enters the inlet of a simple air compressor at 50 Deg F. drybulb and 50% RH. The air exits the compressor at 80 Deg F. drybulb at 90 PSIG. How do you calculate the RH of the compressed air?

Any help would be greatly appreciated!

Thank you,

TB

 

[CESSNA1]

GASDISTENG: There are a couple of ways. One is to find a psychromatic chart that is for 90 psi. The second is to buy some software that will do it. The third is to combine the appropriate equations (I do not know what they are offhand).

If there is an aftercooler after compression to obtain the 90 deg f discharge air; and the moisture is removed by a moisture separator then the RH of the discharge air will be around 90%-99%. The RH changes of a given amount of air changes with temoerature and pressure.

Regards
Dave

 

[25362]

From tables obtain the vapor pressure of water at 80 deg F: 0.5068 psia. The total pressure is: 90+14.7 = 104.7 psia at the compressor's exit.

Convert to specific humidity: 0.622x0.5068/(104.7-0.5068)=
~ 0.003 lb water/lb dry air.
0.622 is the ratio of molecular weights for water and air.

Since you started with about 0.004 lb water/lb dry air, as read from a psychrometric chart, some water would condense out and would probably have to be removed.
The compressed air would thus be saturated with a RH=100%.

If you cool this compressed air, it will continue to be saturated, ie, RH=100% at the new conditions, even as more moisture condenses out.

Remember that % RH is not an absolute quantity but a dimensionless ratio (x 100) of the actual vapor pressure of moisture in air to the (tabulated-saturation) vapor pressure of water at the given dry bulb temperature.

It may help visiting:

thread391-88862, thread798-76900 and, for example,

http://www.hankisonintl.com/Air/answer2.htm

 

[GasDistEng]

25362-

Thanks a million. I followed you 100% did the calcs out and came up with the same. However, assume same conditions except say 1% rh to start? This would give you 0.00007533882 lbs of H20 / lb of dry air at initial (atmospheric conditions and 0.003 lbs of H20 / lb of dry air at after you compress to 90 PSIG. How would you find RH?

Thanks you have been very helpful.

TB

 

[25362]

I didn't explain myself with sufficient clarity. If you start with 0.0000753 lb water/lb dry air you will not increase this value to 0.003 lb/lb upon compression.

The latter is the dew point moisture concentration or maximum specific humidity at the compressor's outlet.

Since you are below that value to start with, the initial specific humidity remains constant.

RH is the ratio between the actual VP and the VP of water at the new dry bulb temperature.

At 80 deg F, VP of water = 0.5068 psia (from tables).
The "new" VP of water vapor after compression would be: (0.0000753/0.622)x(90+14.7) = 0.0127 psia.

By definition the new RH = 100*0.0127/0.5068 = 2.5%

 

[MintJulep]

Relative humidity isn't commonly used when dealing with compressed air systems because it's relative.

With all the temperature changes associated with compression, intercoolers, aftercoolers and such, RH values keep changing even though the amount of water in the air may remain the same.

Specific humidity or dew point are generally more useful.

 

[GasDistEng]

25362-

Thanks again. I will post a link of my final calculation under this post. You helped me a lot.

Thanks again