What size vent fan do I need and intake area size?

videoman

Here’s how you can calculate the amount of ventilation you need. If for instance you wanted to keep you grow room temperature from getting any more than 5° warmer than the intake air temperature, and you were using 400 watts of power, you’d make the calculation below in blue. The chart is based on the following formula. It is a well-established heat transfer formula.


(3.2 × 400) ÷ 5 = 256






---------------------------Calculating the passive intake.-------------------------

The Home Ventilating Institute recommends one square foot of open air inlet per 300 CFM of ventilation fan capacity.

If you were going to use 256 CFM, you’d want 256/300 square feet of intake area, which is 122.88 square inches.

Here are some options for the intake area for a 256 CFM ventilation fan:

1 hole - 12.5 inches in diameter.
2 holes – 8.84 inches in diameter.
3 holes – 7.22 inches in diameter.
4 holes – 6.25 inches in diameter.
5 holes – 5.59 inches in diameter.
6 holes – 5.11 inches in diameter.


Here is how to calculate the hole sizes:

1. Take the total area in square inches needed, in this case 122.88 square inches, and divide by the number of holes you want.
2. Then divide by Pi (3.14).
3. Take the square root of that value.
4. Then multiply by 2.

The answer is the diameter that each hole would need to be to make up the total area needed for intake.

A large number of small holes will create more backpressure than one large hole of equivalent area. This would be negligible unless you’re using a huge number of holes or you’re using ducting to supply the air to each intake hole. If you’re just cutting them in a wall you should be fine using 8 or less holes without having to take into account the extra backpressure.

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Hobo

wow video, you are posting away today, good stuff!

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greengreen

i know you just explained all that video. but i suck at math and lets say my FOAF had a rubbermaid setup 5 feetaway. Right now theres 2 1.5 inch holes for passive intakes. running 5 "100w" 26w CFL's Temp is at 81 with the window open and its like 70 outside. It has been a temp battle for me with 5 n 6 CFL's in the sockets. running a big exhaust with a rather big hole too.

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i am a pathological liar, and have illusions of grandeur. I post pictures of pot plants i found on the internet and i have never nor will i ever consider the cultivation of marijuana, as it is illegal.

Squirt

I just double checked the numbers and the formula seems to be off a bit for my setup. I flipped your formula though. I have a 400w HID and a 120cfm bathroom fan for my exhaust with a 4" output and about 15' of ducting on the on it.

So 3.2 x 400w = 1280 / 120cfm = about 10.6666667°F temp change. Correct?

In my room with the light off the temp is about 20-21°c (68-69.8°F), when the light has been on for a while the temps are about 24-25°c (75.2-77°F). I have never seen it change more then 4 or 5°c (9°F). I also have a 'active' air intake rather then a 'passive' air intake, would this be why the formula doesn't work for me?

Thanks.

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green_nobody

No, as far as i see this is about passive intake here and not active - since you then being blowing air forcefully into your grow this has to be as much as you evacuate from the room too, that is all you got to check with your setup then.

as far as i get video this is about how much air do you need to cool a bulb/grow room down - the second about how big your passive intake has to be.

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videoman

Green dude, is there a question in there somewhere?

Squirt, most probaly yes. I have been using this formula for a long time and it's always worked for me. The formula does no take into account a active intake, its assuming passive.

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greengreen

nevermind i figured it out ;P

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i am a pathological liar, and have illusions of grandeur. I post pictures of pot plants i found on the internet and i have never nor will i ever consider the cultivation of marijuana, as it is illegal.

videoman

okay let me spell it out for you all, if you have a 400 watt light, multiply that times 3.2 which equals 1280 than divide by the cfm of your fan, lets assume a 500 cfm fan for this scenario, so we have 1280 divided by 500 equals 2.56

This means that using a 400 watt light and a 500 cfm fan your temp will rise 2.56 degrees above ambient temperature.
Peace

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