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Old 08-23-2010, 10:27 PM   #11 (permalink)
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Originally Posted by nayeliesuncle View Post
I would use the system at EOC, with the fans at full blast.
The water pump isn't turning during EOC.

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Old 08-23-2010, 11:51 PM   #12 (permalink)
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It can be, if you convert to an electric water pump (only available for some engines).
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Old 08-24-2010, 12:59 PM   #13 (permalink)
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Somebody sells a system similar to this for use on home a/c units. I believe the someone here tried it. It was reported that the a/c unit ran less than without it. The issue was there is mineral build up as the water evaporates. It used tap water (or whatever you push through a hose at it). This mineral build up would decrease heat transfer out of the radiator. I don't know if collected rain or distilled water would prevent this.

I wouldn't keep the water under the hood, I would keep it away from engine heat so there is a bigger delta T.

I would also consider a nozzle that makes more of a cone pattern that a washer nozzle does. This would more evenly cover the radiator.

Good luck.
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Old 08-25-2010, 01:19 PM   #14 (permalink)
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You would'nt be able to carry enough water to make much differance.

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Old 08-25-2010, 06:44 PM   #15 (permalink)
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Originally Posted by CrazyRick View Post
You would'nt be able to carry enough water to make much differance.
You don't need that much water for evaporation cooling.

Actually, a common error at work is to use far too much water.
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Old 08-25-2010, 09:20 PM   #16 (permalink)
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...Swamp Cooler Theory 101:

eqn 1: E(s) = ( DBD / WBD ) = (T1 - T2) / (T1 - T3)

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 adiabatic 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...but "mister spray" can approach 0.95.

...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) ]

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; 06-08-2013 at 12:00 AM..
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Old 05-23-2011, 03:44 PM   #17 (permalink)
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"...Swamp Cooler Theory 101:..."

"...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! ..."

what if you captured the H2O to re use it? Maybe a radiator in the bottom of a stainless steel box. Water vapor condenses on the walls and drips down to the radiator. Seems like trying to take advantage of the phase change from liguid to gas could yield some really impressive results. Some of you engineering/scientist types may be able to explain what I'm thinking of (or why it wouldn't work).

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Old 05-23-2011, 04:12 PM   #18 (permalink)
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You can't spray enough water on a radiator to cool it w/o a continuous flow of water.
You are beating a dead horse.

Install an electric fan in front and set on a push instead of a pull function. install a manual switch or a temp comtrolled switch.
put a shroud around the fan to keep the airmoving thru.

You can dance all around this water issue but it is impractical.

someone mentioned the sprayer on the a/c. remeber this system is runnung off a water hose that is 'on' providing a constsance source of water
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Old 05-24-2011, 08:12 AM   #19 (permalink)
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never seen a water spray bar on an intercooler before, though i have seen plenty of CO2 spray bars, same general concept, but the science behind it is completely different

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Old 05-24-2011, 09:56 AM   #20 (permalink)
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The idea was to use the sprayer for emergency cooling. Why doesn't anyone read the whole thread anymore?

"ʞɐǝɹɟ ɐ ǝɹ,noʎ uǝɥʍ 'ʇı ʇ,usı 'ʎlǝuol s,ʇı"

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