Working backwards it seems to show that you have taken another hit from
your bong and come up with a microscopic cooling load of 1.472 Btu/hr
in a futile attempt to prove something.

Outdoor swamp cooler, 80% effective on the microscopic load, supply air
temp 73, airflow 1472/(1.08x7)=195 CFM

Water required at 80% effective, 4.5 x 195 x (79.5-28.4)/7000= 6.4
pounds per hour. Your flawed scheme on paper tries to approach what a
typical outdoor system does. I have said this two times before. Maybe
3rd time is a charm.

When you put it under a real load, you make the house a mixing box with
either the flooded floor or using an indoor evaporative cooler. Just
another inherent flaw.

The occupants get to experience the hot drafts infiltrating in, unless
of course you duct the make up air right to the swamp cooler. As I have
also been trying to hammer into your head, once you realize the
advantage of ducting the make up air directly to the indoor evaporative
cooler, a light bulb should illuminate and you will soon find that it
is best just to run it on straight outside air.

At 100% effective, an outdoor unit under the microscopic load

1472/(1.08x 15)= 91 CFM

Water required 4.5 x 91 x 64.1/7000= 3.75 pounds per hour

To put 500 CFM in context then, it is inefficient as there is not a lot
of temperature differential between the supply and room air.

Temperature differential between room and supply temp = 1472/(1.08 x
500)= 2.73 degrees. therefore you need a supply dry bulb of 80-2.73=
77.27.

W @ 77.27 db 65 wb =72.6 Grains

Water required 4.5 x 500 x 44.2/7000=14.2 pounds per hour.

NIck, please, go read some of the references I have recommended.