Water balance on water cooler

[Mrepp]

I want to figure out how to determine what the air temperature would be out of a swamp cooler if the RH of incoming air is 70 and the temp is 90 F. Saturation efficiency of the cooler is 95%. I think it is accurate to read the dry bulb temp off the chart, but want to know how to do a water mass balance and what other specifying information would be needed to be given to a vendor, the given conditions.

 

[quark]

If the saturation efficency of the cooler is 95% then the final DBT will be (WBT + 0.95 x WBD), where WBT is initial wet bulb temperature and WBD is wet bulb depression (the difference between initial DBT and WBT).

Otherwise you can do this way.

One lb of water requires 1000 btu to evaporate and the process is isenthalpic. You know initial enthalpy. Check how much heat is required to evaporate x lb of water (you should know x). Use the equation Q = 1.08 x cfm x (T2-T1) to know the temperature.

Also do a search on evaporative air coolers.

Regards,

 

[Mrepp]

Thanks for the response quark. I did start to do 1000 BTU, # water difference at the two conditions, but didn't come up with a resonable number. What is the 1.08, BTU/ft3 F?

 

[quark]

The above equation is simplified version of mCpdT. For air, specific heat is 0.24 btu/lb deg.F and density of air is 0.075lbs/cu.ft. Multiply that with 60 to get hours.

So 0.24 x 0.075 x 60 = 1.08 (now we talk about the heat in volume flow terms)

This link will give you approximation about the final DBT based on initial DBT and RH.

http://www.climatecontrollers.co.uk/newpages/allyou-need-to-know/allplus.htm#a2

Best Regards,