Venting requirements for HIDs in a growroom

[rosscopico0]

Hi all,

I found this on another website and have simply rehashed (pardon the pun) most of it and removed some un-necessary information.

 

So I’ve just got some basic physics here to get a good idea of the likely temperature rise and/or ventilation requirements in growspaces. For anyone afraid of a bit of math, don’t be, read on its dead simple; if I can do it anyone can.

 

Keep in mind that this is a complex issue and there are still lots of factors that will influence things but this will help remove some of the guesswork involved.

 

First off as far as ventilation goes everyone is probably aware that ventilation (fresh air really) is required for the plants to be healthy and the commonly used rule of thumb is to provide at least 20 room air changes per hour. So you just calculate your room volume, multiply it by 20 and this gives you the amount of air you need to put through your room over an hour, fan ratings are often in m3/hour or cubic feet per min so make sure to use the correct units when calculating this.

 

The problem though with this is that the minimum 20 air changes per hour is only enough fresh air for your plants and it wont necessarily prevent overheating (or even humidity problems). This is particularly true in closet setups and where growers are trying to get the maximum amount of light possible into a space, such as in scrog setups where up to 100W per square foot is required.

 

Therefore although we know what our plants need we are often still left with the following common questions:

 

1. What is the most lighting I can put into my growspace?

2. What temperature am I likely to get in my growspace?

3. How much ventilation am I likely to need?

 

So how do we answer these questions? Well below is the basic formula we need to use:

 

Q = V x P x C x dT

 

Where:

Q = Amount of lighting (kW)

V = Volume of air being ventilated (m3/s)

P = Density of air (assume 1.2 kg/m3)

C = Specific heat capacity of air (assume 1.02 kJ/kgK)

dT = Temperature difference between ambient and growroom air

 

Here’s how you would use it for each question:

 

Q.1 What is the most lighting I can put into my growspace?

Assume: Ventilation = 240m3/hr (0.067m3/s), temperature of air entering room = 21degC, temperature of growspace to be no more that 26degC

 

Q = V x P x C x dT

Q = 0.067 x 1.2 x 1.02 x (26-21)

Q = 0.41 kW i.e. 400 W

 

Q.1 What temperature am I likely to get in my growspace?

Assume: Lighting = 400W (0.4kW), ventilation = 240m3/hr (0.067m3/s) and temperature of air entering room = 21degC

 

Q = V x P x C x dT

=> dT = Q / (V x P x C)

=> dT = 0.4 / (0.067 x 1.2 x 1.02)

=> dT = 4.87, i.e. 21 + 5 = 26degC in growspace

 

Q.2 How much ventilation am I likely to need?

Assume: Lighting = 400W (0.4kW), temperature of air entering room = 21degC and temperature of growspace to be no more that 26degC

 

Q = V x P x C x dT

=> V = Q / (P x C x dT)

=> V = 0.4 / (1.2 x 1.02 x (26-21))

=> V = 0.065 m3/s i.e. 240 m3/hr

 

So that’s it, once you get used to using it its very simple really, Just stuff the figures in a spreadsheet and let it do the work or if you know how to juggle equations then your set.

 

Now, this is still a pretty simplistic approach to such a complex bit of heat transfer and there are a few points you should keep in mind that will change things

 

* Consider the ability of the walls of your growspace to absorb heat. If for example they are concrete then they can act as a ‘heat sink’ but if they were lightweight they wont be able to absorb much heat at all. This is the reason people commonly notice temperature rises in their growspace as soon as Mylar is added. Instead of heat being absorbed by the walls (even lightweight ones) its reflected back into the room and increases your air temperature.

 

* Consider what your walls are connected to, if for example its outside then in colder climates this will also act as a heat sink but in warmer climates this could add to your heat problem..

 

* Consider the volume of your growspace, the smaller the space the more heat is a problem as large spaces can absorb much more heat.

 

* Fans blowing on the light fitting itself will turn ‘radiant’ heat into convective heat that can be removed by the ventilation that way you can get the light closer to plants.

 

* If you don’t have remote ballasts then add on at least another 25% to your lighting gain and if possible also put a fan blowing on the ballast for similar effect to that above.

 

* When selecting an extract fan remember to consider the effect of duct resistance or back-pressure. For example if you have long duct runs, too many duct bends or are drawing air through small holes then the fan will not pull as much air through as it says it should. Try to get the most out of your fan by minimising the fan resistance and oversize your fan by any from 25-50% to allow for this if you think its significant.

 

Hope some of you find this useful.

Edited June 18, 2004 by rosscopico0

[Skunkxxx]

Good One.I just changed my little grow area.I have 1200 watts of MH and 270 watts of Flou in a Closed area inside my room.It is About 160mm High and 140mm long and 60 or 70 deep The heat is running 28-31*C AND THE OUTSIDE TEMP IS HIGHER THAN THAT.I have a Cage fan that moves alot of air not sure how much. It came from a old furnace.Enough to heat a whole house.Great info Thanks

[rosscopico0]

No worries mate. I found it very useful when setting up my cupbooard, so figured it would be a great addition to the great aussie grow guide.

[Gazza2001au]

originaly from Flopeye, The heat is running 28-31*C AND THE OUTSIDE TEMP IS HIGHER THAN THAT

 

i dont really get that from using just a fan your temps should be 3-4 deg higher than the out side of the chamber temps unless your using some type of cooling device i.e aircon,fridge,venting from under your house etc...?

[rosscopico0]

Hey all,

I found another formula to pretty much do the same thing

Q = ( 1.76 W ) / Tc

 

Where Q is the ventilation in CFM

W is the wattage of heat being dissapated

Tc is the difference in temp between ambient & growroom air in degrees Celcius

 

For those of you who like using imperial measurements

Q = ( 3.16 W ) / Tf

 

Where Tf is the difference in temp between ambient & growroom air in degrees Farenheit

 

If your venting measurements are in m^3/ min then multiply by about 35.327 to convert to Cubic feet per min ( CFM )

 

Heres an example - If were to use a 150mm inline fan ( 4m^3/min ) to vent a 250w hps. Whats temp rise will I get through the area??

 

So to convert m^3/min - CFM, multiply by the above coefficient.

4*35.327= 141.3 CFM

 

Q = ( 1.76 W ) / Tc

Tc = ( 1.76 W ) / Q

= ( 1.76 * 250 ) /141.3

= 440/141.3

= 3.11* C rise

 

Another example - what is the biggest light I can use with a whisper fan ( 42m^3/min ) and maintain a max 3^*C rise?

 

42m^3/min x 35.327 = CFM = 1483.7 CFM

 

Q = ( 1.76 W ) / Tc

W = ( Q Tc ) / 1.76

= ( 1483.7 * 3 ) / 1.76

= 4451.2 / 1.76

= 2529 Watts!!!

 

Another example - if I have a 1000w light in my area & want a max 2*C rise.

 

Q = ( 1.76 W ) / Tc

= ( 1.76 * 1000 ) / 2

= 1760 / 2

= 880 CFM = 24.9 m^3 /min

 

Ive worked through about twenty examples to compare against the other formula & find the answers are always similar - I think it might have been my DIY calculating the cubic feet in a cubic meter ( with a tape measure!! ) I found a conversion thing which reckons 1m^3 = 35.31 cubic feet :scratchin

Oh well close enough!!! :reallyexcited:

Edited October 13, 2004 by rosscopico0

[Chev]

:thumbsup Nice one mate, you've been busy with ya homework

[rosscopico0]

Cheers Chev

 

If I can help someone to grow their own & avoid the black market, then I reckon a little positive karma has been sent around. The more positive karma out there, the better!! :yingyang