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Old 06-25-2010, 07:24 PM   #10 (permalink)
gone-ot
...beats walking...
 
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...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 = 105F and T2 = 73.9F and T3 = 66.1F, 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) = 85.6F ...a (theoretical) temperature reduction of 19.4F.

...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 = 105F @ rh1 = 12% and the inside air (going into engine) would be: T2 = 73.9F @ 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.
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