How many lights will 1 fan vent?

[Geeorge]

If I have four 600w lights with air cooled reflectors (the hood type) and each of those reflectors has a 6" diameter duct flange, would one 6" extractor fan be enough to vent the heat? The fan I have in mind moves 660 m3 (RVK150 L1).

Last year I had a 6" cool tube for a single 600w bulb and used a 6" tent intake fan to vent the heat, but it didn't do shit, probably reduced the temperature by 1 degree and almost melted my cheap B&Q ducting. Is there a rule of thumb I can use re number of lights : m3 per hour of air needed to vent?

[M_C]

number of lights : m3 per hour of air needed to vent[/B]?

Welllll... if you understand math well....

Calculating the airflow required to maintain a stable temperature, and so determine the size of fan needed, is surprisingly simple.

It boils down to CFM = 3.16 x Watts / DT(°F)

Watts is the lighting power (and any other 'hot' things in the grow space) in Watts.
DT is the allowable temperature rise within the enclosure (i.e. desired temperature minus ambient temperature) in °F.

So, to work out the size of fan required (in CFM) simply plug your lighting Wattage into the equation along with the DT value.

An example is:
Ambient temerature = 20 °C
Target temperature of the enclosure = 25 °C
DT in °C = 25 - 20 = 5
DT in °F = 5 x (9 / 5) = 9
Lighting Wattage = 250 Watts

Plugging these values into the equation gives:

CFM = 3.16 x 250 / 9 = 87.77

This represents the actual throughput required but it doesn't take account of the static pressure necessary to overcome the system impedance (how hard the fan has to suck or blow). But for a free-air system with no ducting or filters it should be fairly accurate.


For those who want a bit more info on how to work it out, here it is.

First, you'll need to know the amount of heat that needs to be dissipated.

The general equation for heat transfer is:

q = Cp x W x DT

where:
q = amount of heat transferred
Cp = specific heat of air
DT = allowable temperature rise within the enclosure
W = mass flow

Mass flow is defined as:

W = CFM x Density

DT is the difference between the ambient air (room) temperature and the target temperature inside the grow space in °F.

At sea level the density of air is 1.2041 kg/m3 (at 20°C) and the specific heat capacity (under typical room conditions) is 1.006 kJ/kgC. After doing some substitution and conversion this gives:

CFM = 3.16 x Watts / DT(°F)

Quoted from one of missy's posts... means bugger all to me though

[MrChedda.]

If I have four 600w lights with air cooled reflectors (the hood type) and each of those reflectors has a 6" diameter duct flange, would one 6" extractor fan be enough to vent the heat? The fan I have in mind moves 660 m3 (RVK150 L1).

Last year I had a 6" cool tube for a single 600w bulb and used a 6" tent intake fan to vent the heat, but it didn't do shit, probably reduced the temperature by 1 degree and almost melted my cheap B&Q ducting. Is there a rule of thumb I can use re number of lights : m3 per hour of air needed to vent?

answer is no, i ran the same setup 4 coolshade hoods via y splitter via rvk l1 it doesnt evenly distribute the air out so two lights was always baking hot mate, better solution get another 6" rvk for the other 2 and get some auto steppers and hook them upto that to keep noise down and get better air flow evenly out of the shades

[Geeorge]

Cheers fellas. Wow that's a complex calculation, kudos to whoever worked it all out. Basically you hit the nail on the head MrChedda 2 fans = 1320 CFM and the calculation says ~1500 for 2400 watts.

[Steveboy09]

I use x3 600watt cool tubes on a 250mm pole star fan that's 1125cfm I think seems legit