Most online guides provide the same formula: [interior house volume in cubic
feet] / 2 = recommended fan CFM. Accordingly, a 3200 sq. ft. house would
need a hefty 12,800 CFM fan.
I saw an online product ad stating that a 9,600 CFM fan (on high speed) is
sufficient up to 3200 sq. ft.
- What is the truth ?
- Is the formula an overkill or not ?
- Where do you find a consumer version of a 13,000 CFM whole house fan ?
What every works. :-)
Really there are many factors, including the weather where you live, the
insulation you have construction materials, the design of the home (single
vs. multi story) the number of people etc.
Depends on the weather, among other things. The example below worked well
with a 347 cfm fan and a 1024 square foot house.
...Gary might run a WHF in Billings in July, when the daily min temp is 58 F
and the daily max is 87 and w = 0.0080, so evaporative cooling can help.
With a F cfm fan and constant day and night temps, he might have something
like this, viewed in a fixed font:
58/87-----/ ---www----- T RC = 6K/400 = 15 hours, with the fan off.
| 1/400 | 70 = 87+(T-87)e^-(12/15) = 47.91+0.449T
-----www----| makes T = 49.2 F, so it looks like
| he needs a longer time constant
--- 6,000 Btu/F to keep the house 70 F max when
--- it's 87 F outdoors for 12 hours.
If 2" of concrete floor over hollow blocks adds 32'x32'(5+2/12x25) = 9387
Btu/F and raises RC to 38 hours with the fan off, 70 = 87+(T-87)e^-(12/38)
= 23.3+0.732T, and T = 63.8 F at dawn. If 63.8 = 58+(70-58)e^-(12/RCn),
RCn = -12/ln((63.8-58)/(70-58)) = 16.5 hours = 15387/(400+F), F = 532 cfm.
But why waste coolth by keeping the house close to 63.8 F just after dawn?
If the slab is T (F) at dawn and we keep the house air 70 F all day with
a cooling thermostat and a slow ceiling fan (without which cool air will
stay near the floor) and an occupancy sensor, the house will only need
12h(87-70)400 = 81,600 Btu of coolth, max, and 81600 = (70-T)x15387 makes
T = 64.7, so RCn = 20.6 hours and F = 347 cfm, or less, with indoor mist.
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