Well, perhaps a bit more clarification. The evaporation of
non-atomized water is not, strictly speaking, the only mechanism
for moist air enthalpy change.

The enthalpy (or energy content) of moist air at pressure Pb,
temperature t (�C) and mixing ratio r (g H20 / kg dry air) is
defined by:

 h [kJ / kg moist] = 1.00464  t + 0.001846  r x t + 2.5  r

Note: by convention, the enthalpy of dry air ( r = 0 ) at 0�C is
equal to zero. Negative values of enthalpy are actually possible
and indicate that the energy content of the air / vapor mixture is
less than the energy content of dry air at 0�C.

From this it can clearly be seen that as the water content (truly
gaseous water) increases, the enthalpy of the resulting mixture
increases. That is, it contains (or can contain) a greater amount
of heat. So, *for any given pressure*, the higher the humidity, the
higher the heat transfer capacity will be. However, as has been
said, the higher the humidity, the less dense the air. So, there
are a couple of competing things happening here;

1. As the humidity increases, the enthalpy, and heat transfer
efficiency, increases for any given pressure.

2. As the humidity increases, the density (and hence pressure)
decreases.

The enthalpy changes can be readily obtained from the equation
above, or using a standard psychrometric chart. The pressure change
is not as easy to calculate, since unfortunately

PV=nRT applies to ideal gases only. And, while saturated steam can
*almost* be considered ideal, water vapor cannot. Primarily due to
the issues Frank raised about non-atomized clusters, and
combinability issues.

Keith Hughes
'86 Westy Tiico "Marvin"

> Date:    Fri, 28 Jun 2002 23:32:29 +0100
> From:    Clive Smith <clive.harman-smith@NTLWORLD.COM>
> Subject: Re: Humid Air - Some Comments
>
> Absolutely wonderful
>
> And talking of wonderment, I wonder if its the last word?