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My house has a 24" Dayton belt drive whole house fan that's at least 20 years old but looks exactly like the current model 2EAX5. It's nice and quiet but I want it to move more air. It has a 1/3hp two speed motor with a 2" pulley driving an 8" pulley on the fan. The motor doesn't have any markings to tell its speed, but I've found quite a few that run at 1140 & 1725 rpm. With the 8" pulley that works out to a fan speed of 285 & 430 rpm, about 75% of the currently advertised 370 & 560 rpm. I don't know how many cfm that comes out to but current ads with a 6" pulley say 2500-3700 cfm.

I recently acquired a Cool Attic / Ventamatic 30" belt drive whole house fan, model CX30BD2SPD. It also has a 1/3hp two speed motor, marked 1140 & 1725 rpm, with a 2" pulley, but driving a 6" pulley on the fan. This spins the fan at 380 & 575 rpm, very comparable to the current Dayton advertised speed. Because this fan is 6" larger and spins 25% faster it moves more than double the air the Dayton does, advertised 5400-7800 cfm.

To summarize...

Dayton 24": 6" pulley, fan speed: 370 / 560 rpm, air flow: 2500 / 3700 cfm (advertised)

Dayton 24": 8" pulley, fan speed: 285 / 430 rpm, air flow: ??? / ??? cfm (not enough)

Attic Cool 30": 6" pulley, fan speed: 380 / 575 rpm, air flow: 5400 / 7800 cfm (advertised, LOTS more than the 24")

Attic Cool 30": 8" pulley, fan speed: 285 / 430 rpm air flow: ??? / ??? cfm

Why not swap out for the larger fan with the 6" pulley? My attic doesn't have enough ventilation for it.

I've thought about swapping the pulleys, putting the 6" on the 24" fan and the 8" on the 30" fan. I'm reasonably certain that would get the 24" up to the advertised 2500-3700 cfm. How can I estimate how much air the slowed-down 30" fan would move? Is it a linear function of speed, blade area, or some combination of the two?

Once I know how much air is moved, I can figure out how much more ventilation my attic might need for either one.

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  • It's hard to tell what you want here. Changing the rough opening for the fan from 24" to 30" seems like a big job, so why not assume the CFM is linear and try the 6" pulley on the 24" fan? That seems very easy to test. I suspect calculating CFM based on RPM is non-trivial, but linear is probably close enough for your purposes.
    – JPhi1618
    Commented Nov 25, 2019 at 21:15
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    What problem are you you trying to solve ? A whole house fan cools the house , not necessarily the attic. I have a 2 speed 24 " ( don't have the other numbers handy) and it makes good air flow through a 2500 sq.ft. house on "low" and gets the house to the outside temperature in several minutes . On" high" it makes too much breeze. for us. Our louver opening is about 30 X 3o ".
    – blacksmith37
    Commented Nov 25, 2019 at 21:16
  • I'm trying to get more air flow. I agree that reframing the opening from 24" to 30" is a big job, as is adding potentially several square feet of attic ventilation. I want to estimate the air flow for each combination so I can figure out how much ventilation I may need to add in the attic to make them work.
    – Eric Simpson
    Commented Nov 25, 2019 at 21:20
  • Get one of those inexpensive air flow gauges and measure each one.
    – JACK
    Commented Nov 25, 2019 at 21:24
  • I have a 30 X 30 louvered shutter in the ceiling ; but I could set either a 24" or 30" fan on the rafters with very little carpentry.
    – blacksmith37
    Commented Nov 26, 2019 at 1:54

2 Answers 2

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unless the fan is shrouded or ducted and sealed to your ceiling it will be quite hard to tell how much air is moved because much of the air could just recirculate around the fan.

the simplest way is to

Area = pi * radius^2

B = fan pitch in inches / 12

C = fan rpm

1 rpm = air moved so many feet at 0% slippage or 100% efficiency

max theoretical cubic feet per min = A * B * C

i don't know what % slippage would occur, that would be based on how well the fan blade is designed. Then account for however much regurgitation of air around the fan due to no shroud gap and not sealed to ceiling. this also doesn't account for any varying blade pitch (i.e. more pitch near center less near tip).

the easiest and cheapest way might be to build a 1-2 foot duct on output side of fan and grab some cheap wind speed gauge to measure the output air speed from fan. then math would be Area * speed; with wind speed converted to feet per sec.

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  • +1 for sounding smart. I wish I was smart enough to double check your ideas and calculations.
    – Lee Sam
    Commented Nov 26, 2019 at 0:17
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TLDR... Based on the numbers below, and considering the effort to reframe the louver opening or add more attic ventilation, I'll try the following until I get enough air flow...

  • Change the 24" fan pulley to the original 6" with no ventilation change.
  • Add more attic ventilation.
  • Replace the 24" fan with the 30" and change the 30" fan pulley to 8".
  • Change the 30" fan pulley back to 6".

How did I get there?

There are 3 basic "fan laws".

  • First, the volume of air moved (CFM) is a linear function of change of fan speed; double the fan speed and you double the volume of air moved.
  • Second, the static pressure changes as the square of the amount of change of fan speed; double the fan speed, quadruple the pressure.
  • Third, the power needed changes as the cube of the change of fan speed; doubling the speed takes 8 times the power.

Be careful! These "laws" are ideal approximations only for the same type of fan. You can't use calcs for a 24" Dayton fan directly with a 30" Attic Cool fan or vice-versa.

The published specs show both fans using 1/3hp motors. Looking up motor speeds, 1/3hp 2-speed 115v 1ph motors often run at 1140/1725 rpm. Comparing that to published fan speeds for the two fans shows both of them using a 3:1 reduction, e.g. a 2" motor pulley to a 6" fan pulley, running the fans around 380/575 rpm.

The 24" Dayton currently has a 2" motor and 8" fan pulley, running it slower than designed - 285 / 430 rpm. By the 1st fan law, it will put out up to 3700 cfm * (430/575) = 2767 cfm. By the 2nd law, it puts out (430/575)^2, or 55.9% of its normal pressure at that speed. The published specs say it takes 6.7 amps or 335 Watts at that speed. By the 3rd law, it would take about (430/575)^3 = 41.8% of the published power, or about 2.8 amps.

Crunching the numbers for all 4 fan combos I get...

  • Dayton 24": 8" pulley (calculated), fan speed: 285 / 430 rpm, air moved: 1925 / 2767 cfm, current: 2.1 / 2.8A
  • Dayton 24": 6" pulley (published), fan speed: 370 / 560 rpm, air moved: 2500 / 3700 cfm, current: 4.6 / 6.7A
  • Attic Cool 30": 8" pulley (calculated), fan speed: 285 / 430 rpm, air moved: 4050 / 5830 cfm, current: 1.3 / 2.7A
  • Attic Cool 30": 6" pulley (published), fan speed: 380 / 575 rpm, air moved: 5400 / 7800 cfm, current: 3.0 / 6.5A

My attic has enough ventilation now for the 24" Dayton with an 8" fan pulley. It's marginal for the 24" Dayton with a 6" pulley. Adding enough ventilation for the 30" Attic Cool with 8" pulley is a substantial change that may be best done when I have the roof done next year. Adding enough ventilation for the 30" Attic Cool with a 6" pulley (10-12ft^2 recommended, even more with insect screening!) could mean major structural changes to the gables and/or porch, which is likely more than I want to do.

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