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Old 12-03-2017, 12:05 AM #1
irvas
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Ventilation help

Hi,

I am trying to figure out how to calculate so I get the right exhaust. Ive been reading ventilation 101 but i can't find it out.

My grow room is following 1.20x0.9x1.8 (meter).

1.20x0.9x1.8 = 1,9544 and then i take 1,9544 x 60 = 118.64 m3/h

But then i dont understand how to get further. How can I calculate that I will have a 600w lamp in there and also that it it will be resistance in the air cooled reflector?

Hope anyone of you guys can help or teach me how to get further with this.

Thank you all in advance!
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Old 12-03-2017, 12:21 AM #2
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With aircooled reflectors you really need to run a dedicated blower unless it's a 1 lamp system as it mkes no sence to use 2 with such confined space. You will be ok running a 5" RVK fan between the filter & the hood. Working it as a exhaust/blower. You will need to tape up the Hood & Glass & make sure it's done as they're notorious for growers getting busted those setups.

Fans have a max efficiency with ductwork in a straight line with no more then 3m in a run between airmovers. You should concider this vector with all ventilation application. 8" fans & above will be quieter then 6" but both will be overkill for a tent, a room however is different.

If you was to uprate the filter from the standard to a maximum flow this will further increase the efficiency.

Some folks say that double filterd fans work too hard but I don't believe that having grown constant for a decade now. The flow of air regardless of the figures needs to do it's job & if your able to get the flow just right with double filtered then do it.
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Old 12-03-2017, 12:36 AM #3
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Thank you for your fast reply.

The fan that you are refering to, 5" RVK is taking out 220m3/h, is that correct? Just in case they dont have that one, so I know how much it should extrat per hour.

I will only use 1 single 600 w hps during flower, so I don't have to think about calculatiing different.

So if the ventilation system is not in a straight line, I should get a stronger fan? It will not be more then 3 meters from the fan to where the air will go out.

I will grow in a tent, I forgot to mention that. So with tape the hood, you mean just take ductape or any other and just tape around the "door" of the hood to make sure no hot air is coming out?
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Old 12-03-2017, 01:13 AM #4
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Tape all the seams on the hood with the Ductwork metal tape & seal the glass. The 5" RVK is enough for the tent you describe, I'm not sure how much m3/h it puts out. I use a different brand, more powerfull, quieter & cheaper but the RVK are the industry standerd Scrubber, they're not a dedicated Blower the propellor/impellers are the wrong shape for it.

You just need to take into acount the bends so with the 5" overrated m3/h you will be fine.
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Old 12-03-2017, 01:21 AM #5
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I will seal it very good.

But if you know, or anyone else, could you please explain how to calculate when my tent is
4x3x6 and I want to use 600w. The formula you would use to calculate it.

And also, what do you guys think about if it is necessary with active intake or if passive is enough?
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Old 12-03-2017, 02:28 AM #6
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You work it out based on how much C02 concentration you need in the room, in the ideal world if temps/humidity are stable.


Calculate the volume of air into the grow chamber from the individual fan size, cmf into cfh so 1.2 x 1.3 x.90 4x3x6 = ________ x (Air Changes per hour) & ÷ that by 60. This gives you the air exchange of the 5" in your room.

Here is a guide to basic air exchange per hour.

https://www.contractingbusiness.com/...opy%20copy.jpg

L x W x H x 60 = m3

m3 x 60 = m3/hour

m3/hour x 1.33 = Reduction on fan performance


A 5" RVK moves 220m3/hour enough to do your space
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Old 12-03-2017, 02:46 AM #7
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Ok, I will just adjust it after how the temps will be. I did not know if to have a active intake was important as well.

4x3x6 = 72 x (Air Changes per hour 20?) & ÷ that by 60. The result i get is all wierd.

I think i will just buy the fan that you told me, i cant figure this out any other way hehe.
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Old 12-03-2017, 03:59 PM #8
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Now i sorted it out. I will buy a fan that has the capacity of 240 m3/h where the speed is adjustable, the fan VENTS SILENT-M 100MM - 240 M3/H.

Can i buy a carbon filter that has capacity of 500 m3/h. Will that be to big? Or what would be the capacity fitting better with that fan?
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Old 12-03-2017, 09:56 PM #9
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The oversize of the filters & fans make the air to be treated not stay in contact with the carbon long enough to remove all the scent. In high humidity Water soluble terpenes are escaping through moisture loss that it isn't scrubbed effectively. It's worse with a over size fan to filter but it depends on what side of the fan the filter is going. I mentioned about the 3 types of fan in the earlier posts. My opinion & experience is that with overkill the flow manages to make the specified requirements in small areas so you needn't have 3x 20mm ports with an opening for passive intake. See how it works for you. I have a 6 turbo filter with a 5" fan i push the air through a reducer on the filter.
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