[hotrod]

Well I used a 3 and a 4 gal/hr nozzle on my pre-compressor WI setup on a turbo that was rated with a max flow of 360 CFM. The 4 gal/hr was a bit too much, and the 3 seemed to be about right. Since I was pushing that turbo very hard a lot of that water went to cool the compressor discharge to sane levels.

In your case you have:
6.6 L displacement @ 30 psi, lets assume 4000 rpm --- engine needs about --

6.6 L = 403 CID x 2000/1728 = 466 CFM, at 30 psi equals a pressure ratio of `approx 2:1 . Your engine should need about 932 CFM at 4000 rpm or 64.3 lb/min of air flow. (this is air flow into the compressor inlet at atmospheric pressure).

Not sure what the VE for a diesel engine is so this is at 100% VE, for a gasonline engine you'd mulitply by about .85 to get a real world air flow.

If we do a quick on the napkin computation of the specific heat of that air flow we get:

64.3 lb/min x .024 btu/lb x 100 deg F delta T = 154.3 BTU/min deg F

Latent heat of water = 970 BTU/lb deg F at atmospheric pressure. (it will be a little lower at high pressure but this is close enough for a ball park computation).

To absorbe that heat flow you, would need to evaporate about 0.16 lb of water /min or 73 ml/min of water. That would be just over a 1 gal/hr water flow, or about the same as an Aquamist 0.4 mm nozzle at 40 psi line pressure.

Not trying to be condescending by working it out, but just wanted to show where the numbers were coming from, so everyone could follow the reasoning.

Larry