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 06-25-2010, 07:24 PM #10 (permalink) gone-ot ...beats walking...   Join Date: Jul 2009 Location: . Posts: 6,190 Thanks: 179 Thanked 1,522 Times in 1,123 Posts ...Swamp Cooler Theory 101: eqn 1: E(s) = ( DBD / WBD ) = (T1 - T2) / (T1 - T3) ...where: E(s) = Saturation Efficiency, actual change divided by maximum possible change. DBD = Dry-Bulb Depression, air temperature change from HOT to COOL, actual. WBD = Wet-Bulb Depression, air temperature change, HOT to COOL, maximum possible. T1 = Outside (or inlet) HOT air temperature. T2 = Inside (or discharge) COOL air temperature. T3 = Wetbulb temperature, maximum adiabetic air cooling possible. ...for example, if T1 = 105°F and T2 = 73.9°F and T3 = 66.1°F, then the saturation efficiency E(s) would be 0.80, or 80%: E(s) = (105.0-73.9)/(105.0-66.1) = 0.80 x 100% = 80% ...unfortunately, that E(s) value is for Aspen Excelsior, not a FRAM™ paper automotive air filter, which probably has an E(s)-value of ~ 0.5 or less. ...never the less, let's backsolve the above equation to see how "cool" the air going into the engine (ie: T2) would be with E(s) = 0.5: eqn 2: T2 = T1 - E(s)*(T1 - T3) = 105.0 - 0.5*(105.0 - 66.1) = (105.0 - 19.4) = 85.6°F ...a (theoretical) temperature reduction of 19.4°F. ...if you know the relative humidity (RH%) values of T1, T2 and T3, Eqn 1 becomes: eqn 3: E(s) = SQRT[ (rh2 - rh1)/(rh3 - rh1) ] ...where: rh1 = relative humidity at T1 of the inlet (outside) air. rh2 = relative humidity at T2 of the discharge (after the wet filter) air. rh3 = relative humidity at T3; theoretically 100%, but practically less than 80%. ...so, because of the HUGE volume of air drawn by the engine through the air filter, a large amount of water is required to raise rh2. For instance, in the first example above, the outside air would be: T1 = 105°F @ rh1 = 12% and the inside air (going into engine) would be: T2 = 73.9°F @ rh2 = 68%. ...unfortunately, THAT's gonna take a LOT of water, especially if you're driving at freeway speeds (higher RPMs) for any appreciable time...literally, gallons-per-hour! ...and, here's the equation for calculating "how many" gallons-per-hour (GPH) of water you'd need: eqn 4: GPH = [CFM*WBD*E(s)] / 8700 ...where, 8700 is a conversion factor based upon (a) 8.34 lb. water-per-gallon and (b) 1043 BTU-per-pound of water. CFM is a function of engine displacement (CID), volumetric efficiency (~ 80%) and RPM: eqn 5: CFM = (CID*RPM*0.8) / 3456 Last edited by gone-ot; 06-30-2010 at 12:06 PM.. Reason: added GPH and CFM equations.