Lacustral wrote:

I'd like to use an exhaust fan (fan from a ceiling grille to outside) to
run overnight in the summer, with the windows open, so that my house is cool
in the morning. I'm not sure how much CFM is needed.

That depends on where you live and the house thermal mass and insulation
and how cool you want it. Whole house fans can work almost as well as AC
for all but a few days per year near Philadelphia. NREL says June has
average daily min and max temps of 61.8 and 81.7 F. July is the warmest
month, with 67.2 and 86.1 and humidity ratio w = 0.0133 pounds of water
per pound of dry air. An indoor fan or a window AC for dehumidification
can help ensure comfort in July.

I don't want a big powerful whole house fan because I'd like it to be
quiet. Just a small fan that keeps running overnight.

But you may need a large fan. Lasko's 2155A 16" window box fan has
a thermostat and sliding mounts to fit windows and it's efficient,
moving 2470 cfm with 90 watts on the highest of 3 speeds, and it's
fairly quiet. About $50 at ACE Hardware stores.

A house with an open window or two upstairs is a natural cool air trap.
At night, warm house air will rise up out of the window and cooler night
air will slide in and replace it. During the day, air stops moving. More
air and heat can flow with a fan that runs at night, eg an upstairs window
with a 1-way damper outside (eg plastic film, hinged at the top edge) and
an downstairs window with a damper inside and a differential thermostat or
an outdoor 70 F thermostat that runs the fan when outdoor air is cooler.

... can you tell me how much CFM gets your house down to the temperature
of the outside air

It can never reach the daily min, but a larger fan can keep it closer.
A Q cfm fan is like a 1/Q F-h/Btu thermal resistor or a Q Btu/h-F conductor
between outdoor air and the house mass. This may be the bottleneck, if the
exposed mass surface is large, with a 1.5 Btu/h-F-ft^2 airfilm conductance.

(I could calculate the cubic feet of airspace in my house, divide by CFM
of a fan, and come up with a guess, but I'm sure it's not that simple -
the hot stuff in the house is heating up the air, fans aren't completely
efficient about clearing out the inside air, etc.)

If a house has infinite thermal mass (most don't :-) and G Btu/h-F of
conductance and a Q cfm fan and a constant indoor temp Ti and 8-hour night
and 16-hour day temps Tn and Td and the fan moves 8Q(Ti-Tn) Btu at night
and the house gains 16(Td-Ti)G Btu/h during the day and these heatflows
are equal, Ti = (QTn+2GTd)/(Q+2G), approximately.

For example, G = 800 and Q = 100 and Ti = 80 and Tn = 70 makes
Ti = (100x70+2x800x80)/(100+1600) = 79.4. G = 400 and Q = 2470 makes
Ti = (2470x70+2x400x80)/(2470+800) = 72.4.

We might model a house with C Btu/h of thermal mass and G Btu/h-F of
conductance and a Q cfm fan and Tmin and Tmax indoor temps like this,
viewed in a fixed font like Courier:

1/Q (closed at night)
---www--- \---
| | --------------
| 1/G | Tmax . . .
Tn/Td ---------www--------*--- Tmin/Tmax . . .
| Tmin . .
| ------- -------
--- C night day
---
|
|
-

RCn = C/(Q+G) and RCd = C/G and Tmin=Tn+(Tmax-Tn)e^(-8/RCn) and
Tmax = Td+(Tmin-Td)e^(-16/RCd), and Tmin = [Tn+(Td-Tn)e^(-8/RCn)
-Tde^(-16/RCd-8/RCn)]/(1-e^(-16/RCd-8/RCn), if I did that right.

For example, C = 6000 and G = 400 and Q = 2470 and Ti = 80 and
Tn = 70 makes RCn = 2.1 hours and RCd = 15 hours, so...

20 C=6000'house thermal mass (Btu/F)
30 G=400'house conductance (Btu/h-F)
40 Q=2470'fan cfm
50 RCN=C/(G+Q)'night time constant (hours)
60 RCD=C/G'day time constant (hours)
70 TN=70'night temp (F)
80 TD=80'day temp (F)
90 TMIN=(TN+(TD-TN)*EXP(-8/RCN)-TD*EXP(-16/RCD-8/RCN))/(1-EXP(-16/RCD-8/RCN))
100 TMAX=TD+(TMIN-TD)*EXP(-16/RCD)
110 PRINT TMIN,TMAX'min and max indoor temps (F)

70.14393 76.608

Nick