The Pine splintered when he pondered


"105F db 65F wb ambient



.... 100(Pw-Pa)/(105-65) = -1 (Bowen, 1926) makes the water vapor
pressures
Pw-Pa = 0.4 "Hg, and Pw = e^(17.863-9621/(460+65)) = 0.6296,
approximately,
using a Clausius-Clapeyron approximation, so Pa = 0.2296, and the
absolute
outdoor humidity ratio wo = 0.62198/(29.921/Pa-1) = 0.00481 pounds of
water
per pound of dry air.


maintain it at 80F inside.


.... (105-80)C+10K = 1000P makes P = 0.025C + 10 pounds per hour of
water,
since C cfm of airflow moves about C Btu/h-F and evaporating each pound

of water takes about 1000 Btu. With wi = 0.0120 (an efficient corner of

the ASHRAE 55-2004 comfort zone), P = 0.075x60C(wi-wo) = 0.03236C,
since
each cubic foot of air weighs about 0.075 pounds and there are 60
minutes
each an hour (want to argue about that? :-) so C = 1360 cfm and P = 44,

ie we evaporate 5.28 gallons per hour of water. "

Well I can look at it from your newly chosen conditions then, you are
going to maintain 80F and W=0.012 or 84 grains, so you would NOW be
setting the 'improved controls' to 55% RH then rather 60%. To minimze
air flow and water use, then an indoor cooler could saturate this air
and produce air saturated at 68.13F, 103.3 Grains.

Would need to saturate 2901 CFM of room air through the indoor unit,
exhaust 1007 CFM and use 35.99 pounds per hour, there an improvement on
your 44 pounds per hour and a little less exhaust.

A 100% effective outdoor unit would still be 617 CFM and 25 pounds of
water.

A realistic outdoor unit at about 80% effectiveness would produce air
at 73 db/65 wb, 79.5 grains

Ignoring fan heat

CFM = 10,000/(1.08 x (80-73)) =1322.8

Water used = 4.5 x 1322.8 x (79.5-28.4)=43.5 lbs per hour, a similar
amount of air and water as you were guessing an indoor unit would move.

An indoor unit of comparable 80% effectiveness could produce air at
80-(.8x(80-68.13)=70.5 F (70.5Fdb, 68.13 WB, 99.5 Grains). Take a stab
at how much air and water the indoor scheme ends up using. Be a lot
more than what a comparable outdoor unit would use.

The outdoor unit is inherently superior as it directly treats the heat
of the outside air, and can supply cooler air to the space than what an
indoor unit can. Less air, less water, automatically deals with the
sensible heat of outside air, and only needs one fan. Indoor scheme you
need two fans, more water, need to move more air. FLAWED.

The problem is you refuse to factor in the importance of the wet bulb
temperature. I have given you a couple references now to this but you
refuse to consider this. So it is pretty hard to argue with someone who
does not understand an important concept. Typical of an electrical EIT,
perhaps the world's oldest one

Flooding the floor or even a 'dampened slab' is a stupid idea, of
similar magnitude of stupidity as exhausting the air to the attic or to
an attached garage.

Spore cases have an enzyme coating and the spores themselves do not
immediately extract moisture from the ambient air to germninate. What
spores need is a wet food source and dampening a slab and in particular
any dirt on that floor or say where the floor meets a baseboard trim or
water wicking up that trim or the paper in the sheet rock at the
bottom of the walls and you have mold. The moisture in the food
dissolves the enzymes on the spore case and creates a 'nutrient broth'
that enters the spore by osmosis. Osmosis being the same principle by
which I am trying to get the concept of wet bulb to enter your mind. So
mold does not really care what the RH of the air is, but mold does care
about what the moisture content of the food is.

If you want to improve evaporative cooler, look at indirect evap
cooling, been around for a while, just new to you. Keep trying to
re-invent the wheel Rube just really think about it when you give
people advice to blow air to an attached garage.