Ventilation <= driving me CFM (Certifiably Fu<King Mad)

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Superslow

Active Member
Hello Everyone, this is my first post (I think) so have a read and see if you can help! Cause I'm currently doing this ===> :wall:

I am in the middle of designing a 'Friends' indoor vegetable grow room! and to use the method regularly seen, recommended, and explained on RIU. One would simply work out the volume of the room divided by how many minutes you want the air of that area changed!

i.e CFM needed = Cubic Feet of the room divided by e.g 5. 5 being every 5 Minutes the air would be replaced!

NOTE: To convert CFM to CMH use the folowing formula CFM X 0.03 X 60 = CMH <== borrowed this from somewhere on this forum! :-P

My problem with this theory is that is doesn't take into account, lighting, vented lights, the amount of plants (co2 needed) and grow area! For an example my friend wants to use 3 x 600w HID lights and has poor access to ventilation but has alot of space available, his room is 10' x 10' with a 7'6" ceiling which equals 10*10*7.5= 750 cu ft, so if he wants to change his air every minute he would hypothetically need a fan of approximately 850+ CFM to allow for back pressure and flow restrictions. But if he increases his room by 200 cu ft, he would also need to increase his fan by the same, although the only thing that has changed is the volume of the room! My problem with this equation is, there are no more lights(heat), no more plants, (no more Co2 used), so I feel this formula is floored somewhat! I think the formula should be worked out per Light (if using HID, because of heat) and per plant in the case of CFL's. I Know this is a basic description of how many air changes would be required but, I hope you all get the idea of what I'm talking about. as another example if I was to grow in the corner of a large warehouse with Three 600w HID lights imagine what size the fan would need to be! Is anyone following me?
Ok so what Would someone recommend to be the minimum Cubic Feet per Minute required to maintain a reasonable environment, per 600w HPS light or (per watt or per 1000w if this could become a formula)
He currently has a 6" inlet and a 6" outlet and would like to think this would be ok! can YOU help Me and Him out please!!!


Thanks in Advance

SuperSlow
 
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macwmv

Well-Known Member
imo the formulas and all that is useless 3 600w are gona make a lot of heat i would def vent them and if u have money then get the biggest fan u can afford think of it this way either u get a fan that u need to run almost all the time or u get a huge fan that lowers the temps by 1 degree a min and run it a few times a day ...
 
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Superslow

Active Member
Hello again people, anyone have an idea, of a formula, or if i can get away with a 6" fan (lets say 6" can fan with carbon scrubber) for 3 600w HPS with digital ballasts?
 
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SheetMetalGod

Member
as far as a 6" to 8" taper from home depot goes that will not help much because you are still trying to move air through a 6" outlet a 6" round pipe will allow 350cfm at 1.0" of water column(static pressure) as compared to 200cfm at .3"of water column(static pressure). pipe size/fan motor(hp)/pressure in the system are all key to proper ventilation. when it is all said and done you are limited by the size of your pipe.
 
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macwmv

Well-Known Member
i dont no about all that cuz mine works so ..... 6in vent to 6to8 converter then 8inch piping running 2 600w vented hoods and i can get my temps as low as 70 if i want maintaining it between 72-78 is cake....
 
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