Abby Normal wrote:

"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 in
each 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...

Nothing new about it. You merely said "maintain it at 80F inside."

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.

This makes no sense to me. Care to explain more?

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

This makes no sense to me. Care to explain more?

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

So?

Ignoring fan heat

But the outdoor unit has a big blower, vs a small exhaust fan...

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.

I've been saying that a swamp cooler with appropriate controls would
achieve the same performance as an indoor scheme.

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).

But "indoor units" are 100% effective.

Take a stab at how much air and water the indoor scheme ends up using.

I've done that in great detail, several times.

Be a lot more than what a comparable outdoor unit would use.

Wrong.

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.

Wrong.

The problem is you refuse to factor in the importance of the wet bulb
temperature.

It seems to me that the problems are your arrogance and ignorance. If you
could set aside the arrogance, you might cure your ignorance by learning
more about the 300-year-old physics you talk about with no understanding.

Nick