Sensible heat gain of 10,000 Btu/hr -excludes make up air/infiltration

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.

Now suppose the house leaks 200 cfm of air (about average in the US.)
In the indoor scheme, the fan would only move 1160 cfm, and the cooler
might reduce its airflow to 1160 cfm (real vs inflated cooler cfm :-),
so the cooler and the indoor scheme would have equivalent performance.

But what can the cooler do if the house leaks more air or we need less
cooling? Suppose we only need 200 cfm of outdoor air? It can't reduce
the airflow to zero and still evaporate water, so it will have to move
excess outdoor air through the house and use excess water, ie the indoor
scheme will use less air and water in this case.

Working backwards, if C = 200, P = 0.03236C = 6.472 pounds per hour,
and 1000P = 6472 Btu/h = (105-80)200 + Q makes cooling load Q = 1472
Btu/h. If the swamp cooler needs (say) 500 cfm min to evaporate water,
(105-80)500 + 1472 = 1000P makes P = 13.92 pounds per hour, over twice
the water required by the indoor scheme.

Nick