Ventilation 101...Let's make this a sticky!

Ou8aCracker2

Well-Known Member
Another method that's worked for me is 3.14cfm x Watts/Tdiff

Example...

Ambient air is 68°F and wattage is 400w and I want a 10°F tdiff (temerature difference,lights on)...

3.14cfm x 400 = 1256cfm

1256cfm / 10°F = 125.6cfm.(after all pressure losses.
 

rocho

Well-Known Member
Hello,
Maybe it has already been explained right there but I am not very skilled with your language and even less skilled with the theory concerned the air physic.

If I reduce the speed of a small 187m3 / h extractor by a dimmer could I know how many W i may give to get ish 80m3 / h?

I only know that 50% of power DOES NOT correspond to 50% of airflow because I asked that to the Vents customer care(model code is TT100)

thanks in advance to those who will help me
 
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