Ventilation 101

DDub

Well-Known Member
I have noticed many people on this site who have questions about Ventilation. Specifically how many CFM's are required to Exhaust their grow area. This is a very important thing to take into consideration when planning your grow room.

Many people don't know that there is a proven heat transfer formula to figure out your required CFM's.
**This Formula DOES NOT APPLY TO CFL LIGHTING**

First you must know 3 thing's before using this Formula:

1. We will call this (W) The amount of Light you will be using in Watts.

2. We will call this (V) The amount of Variance you would maximum like to have between your Normal Room Temperature and Growing Area again in Fahrenheit. (So if your Normal Room Temp. or Intake Temp. is 68 degrees Fahrenheit and you would like to have a maximum of 75 degrees Fahrenheit, you would use 7)

3. Of course to know #2 or (V) you must know the Temperature in Fahrenheit of your Normal Room Temperature or Intake Temperature.

The Formula is as follows:


3.2 x (W)
------------- = CFM
(V)

So if you have 1200 Watts of light, an Intake Temp. of 65 and would like to keep the Grow Area at or below 75. (W) is 1200 and (V) is 10.

3.2 x (1200) = 3840

3840/10 = 384 CFM

So 384 CFM's would be required to Exhaust this grow area and keep it at 75 degrees Fahrenheit max.

So now you have your Exhaust set up. What about the Intake. If you are going to use a passive intake there is another Formula to figure out the size of Intake you will need.

For every 300 CFM of Ventilation strength you will need One Square Foot of Open Air Inlet for a Passive Intake.

Here is the Formula, you must know (A) and (H)

A/H/Pi=Z Sq Root of Z x 2 = The Diameter each hole would need to be for Intake Area Needed.

(A) will be your Area in Square INCHES, this is how you calculate your sq inches needed from CFM used. You need to know the Sq Inches for the Formula. We will use the number used above of 384 CFM.
So 384/300 = 1.28 this is the sq ft needed
Take 1.28 and Multiply by 144.
1.28 x 144 = 184.32 square inches needed

(H) will be the number of Equal Circle Holes wanted

(Pi) is 3.14

A/H/Pi=Z Sq Root of Z x 2 = The Diameter each hole would need to be for Intake Area Needed.

This one is complicated so we will do a few.

So like above say we have 384 CFM of Exhaust. And we only want one hole.

First we must calculate the sq inches of ventilation required. I did this already for 384 cfm which is 184.32 sq inches needed. Now we divide by the number of holes wanted and we only want one, so the number stays the same. We take 184.32 and Divide by Pi or 3.14.
So 184.32/3.14 = 58.70
Now we take the Sq Root of 58.70 which is 7.66
Now we multiply 7.66 x 2 = 15.32

So we would need One 15.32 inch hole to act as a Passive Intake for our 384 cfm Exhaust Fan.

Now we'll do another say we have an Exhaust of 747 CFM. And would like 4 equal size Holes.

747/300 = 2.49
2.49 X 144 = 358.56 This is the amount of Sq Inches

Now 358.56/4 (Since we want 4 Holes) = 89.64
89.64/ Pi (3.14) = 28.54
The Square Root of 28.54 is 5.34
5.34 x 2 = 10.68

So we would need 4 holes of 10.68 inches.
 

7th1der

Well-Known Member
Plan on making a fan purchase today. Mind if I post my equation for you to grade, professor? I'm sure I will be stoned before I attempt to calculate all this! Lol
 

Hotwired

Well-Known Member
I believe these formulas would only account for a room that doesn't have air cooled lighting and uses one fan to exhaust the whole room including the heat from the lights.

There would have to be a significantly different formula to account for the use of 2 exhaust fans doing separate jobs. But I don't think you can calculate the heat output of an aircooled light. It would depend a lot on how many lights are in the line and so on.....

Many people use a fan for their filter and a fan for their lights. This would get weird after a while.
 

DDub

Well-Known Member
I believe these formulas would only account for a room that doesn't have air cooled lighting and uses one fan to exhaust the whole room including the heat from the lights.

There would have to be a significantly different formula to account for the use of 2 exhaust fans doing separate jobs. But I don't think you can calculate the heat output of an aircooled light. It would depend a lot on how many lights are in the line and so on.....

Many people use a fan for their filter and a fan for their lights. This would get weird after a while.
This is correct this formula can only be used in refernce to a sealed area other than an Exhuast and Intake (In the first formula). If only one air cooled hood is used this formula will be pretty accurate.

But Hotwired is correct adding more and more length to ducting, having multiple air cooled hoods, the shape in which hoods are arranged, and having multiple open end hoods. Will all make this formula incorrect. But many of the people in the situations I mentioned just above all would have sufficent experience (I would assume) to know how to set up a proper ventilation system. It is those who are not so experienced who can use this.
 

Skeksis

Well-Known Member
Looks like I should have gone with a 6" Vortex instead of the 4".

Oh well, I'll just use the 4" to cool the light separately and get another 6" to
exhaust the DR120.
 

tjd22

Member
i am using 2 1000 watt bulbs with aircooled hoods and insulated ducting. i have a 160 cfm inline duct fan to pull fresh air inside, with the insulated ducting attached to the lights (which are in a row.) On the other end of the lights i have a 400 cfm blower fan exhausting air to the attic. Will this be ok to exhaust the lights without many issues. this exhaust is only for the lights not the room.
 

researchkitty

Well-Known Member
here's the third vote in a row for a sticky...................... just click report post and 'make sticky plz', eventually the mods will ge tthe hint =) =)
 
screw the formula nothing will beat ol tried n true experimentation. purchase a thermo/hygrometer should give u an idea if heat vent is working. took me three configs ina 4x6x9 flobox 600w overhead coolhood/horizontal cool tube 400hps combo to get things <78 degrees tops.
 

researchkitty

Well-Known Member
screw the formula nothing will beat ol tried n true experimentation. purchase a thermo/hygrometer should give u an idea if heat vent is working. took me three configs ina 4x6x9 flobox 600w overhead coolhood/horizontal cool tube 400hps combo to get things <78 degrees tops.
Says the new guy trying to look cool bumping posts! :)
 
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