I disagree with Ernest slightly.  In our application there should be plenty
of excess heat at or near the rad to create a rapid vaporization.  Our
question is where the evaporation would be most effective

I suggest the following.
In a perfect world we would weep the water out of the surface of the rad and
the water would take the heat with it as it evaporates.  This is by far the
most efficient.  Heat is absorbed raising the temp of the water to the
vaporization point and then the heat would all.  Very little would be blown
away as there is plenty of excess heat and air flow for near instant
evaporation.

So I suggest that a nozzle squirting water directly on the opening of the
rad would be most effective.

But don't take my word for it.  The turbo racing crowd uses these technique
all the time and have products that do just this.  Google something like
"turbo intercooling water spray" to find them.

> > Don't know for certain, but I would think that since the temp of the air
> > stream before the radiators is way below the evaporation temperature for
> > water,  you wouldn't have the heat energy to change much of the water
into
> > vapor.  So not certain just how much cooling of the air you would get.
Just
> > changing it into a fine mist alone does not do the job, of course, it
must
> > be changed into a vapor in order to benefit form the latent heat of
> > vaporization.

> Even ice evaporates.  It's called sublimation. 8*)
>
> A pot of water sitting in a cold room will eventually evaporate
> completely, it just takes a long time.  Spread the water out to expose
> more surface area, and it evaporates quickly (but still very slowly).
> The evaporation can only occur because the ambient air adds a small
> amount of energy to the water in the form of heat. And the heat transfer
> only occurs at the boundary layer.  Add heat faster, and the evaporation
> happens faster.  The evaporation temp, often called the boiling point,
> is just the temperature where the liquid converts to gas at such a rate
> that the process uses all the energy added to the liquid for changing
> into gas.

> Misting a liquid increases it's surface area dramatically.  Every tiny
> droplet is a sphere exposed to air on all sides.  The finer the mist,
> the greater the surface area.  With such a large area exposed, the mist
> will 'scavenge' the air for latent heat.  Make the mist fine enough, and
> it's possible to pull the temp down to the temp of the water almost
> immediately.  Add any heat to the system (like from a radiator) and the
> water will also scavenge that quickly.
>
> If the water was just squirted onto the radiator, there will be a huge
> reduction in surface area.  Most likely, most of the water will be blown
> out the exit duct before there's enough time to vaporize or even add
> heat to it.
>
> It basically goes on what you said before:
>
>    3 gallons of water (assuming it all evaporated- which of course, it
>    would not, as some would be blown through the core unevaporated - 50%?
>    just don'tknow) would provide 22,345 BTU of heat dissipation.  So 3
>    gallons would
>
> The mist decreases that 50% figure to a useful number.
>
> --
> http://www.ernest.isa-geek.org/
> "Ignorance is mankinds normal state,
>    alleviated by information and experience."
>                                    Veeduber
>
>
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