Rich wrote:

Still wondering how to push 5000 cfm through a 14" duct (at 54 mph :-)

That is probably to feed two or more rooms. The outlet of the cooler is
probably 18 x 18 inches (324 sq in). Run that into a plenum and then feed
two or more of the 14" ducts.

I also suspect that swamp cooler cfms are inflated, as fans used to be.
Air King's 9166 whole house fan is listed as 8908 cfm in Grainger's 394
catalog, but Air King says it only delivers 3560 under the current test
method, which measures the actual cfm flowing through the fan, without
adding the room air entrained by that airstream after it leaves the fan.

Most of the swamp coolers in Grainger's catalog are now listed with an
"industry standard rating" ("a numerical rating for comparing similar
models and sizing purposes") instead of "CFM." For instance, their chart
says a 5NV70 WisperCool model RW5000 cooler with an "industry standard
rating" of 5000 can cool a 1200 ft^2 house in Las Vegas.

Regardless there is still a significant amount of air being moved. That is
the whole principal of a swamp cooler. You have to provide adequate exhaust
too by keeping windows open...

Moving lots of air wastes water and electricity. A Las Vegas homeowner
might do better with Sam's portable 797895 Arctic Breeze cooler mounted
inside a house near an open low window and an exhaust fan in a higher
window with a one-way plastic film damper. Turn on the cooler when the
house temp reaches 80 F and turn on the exhaust fan when the RH reaches
60% to keep the house air at the upper right corner (80 F and w = 0.012)
of the ASHRAE 55-2004 comfort zone.

A 1200 ft^2 house with R30 walls and R40 ceiling and 96 ft^2 of R3 windows
and 0.2 ACH (I suppose nobody builds houses like that in Las Vegas) would
have a thermal conductance of 1200ft^2/R30 for the ceiling + 96/2 for the
windows + 1024/30 for the walls + 0.2x1200x8/60 for air leaks, totaling
128 Btu/h-F. July looks like the worst-case month for cooling, when it's
91.1 F over an average day, with an average low and high of 76.2 and 105.9
and humidity ratio w = 0.0066 pounds of water per pound of dry air.

Keeping that house 80 F while evaporating P lb/h of water into C cfm of
outdoor air means 1000P = (91.1-80)(128+C). P = 60C(0.075)(0.012-0.0066)
= 0.0243C makes C = 108 cfm and P = 2.62 lb/h, ie 7.6 gallons per day.
If the house has significant thermal mass (eg a floorslab), we can save
more water and energy by only running the cooler at night.

Why do we need 5000 cfm???

Nick