Message-ID: <3FD1F8BD.7010704@nc.rr.com>
Date: Sat, 06 Dec 2003 10:41:49 -0500
From: Ernest Christley <echristley@nc.rr.com>
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To: Rotary motors in aircraft <flyrotary@lancaironline.net>
Subject: Re: [FlyRotary] Re: spray bars
References: <list-2830003@logan.com>
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Ed Anderson wrote:

> 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