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Thread: Airflow / Temperature / Wattage table (and calculation).

  1. #21

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    also that equasion you worked out for 1x600wt..... for 2x600wt would you just x2 it?

    so anything more than 2x600watts your going to need more than 1 ex-take fan+cf to keep it cool?

    because both 8" and 10" only goes upto 1150m3/hr................ and a big 12" 1300m3/hr

  2. #22

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    Quote Originally Posted by stone14 View Post
    just too much mathes for me to work out, lol but thanks for doing it for me. ahh so i do need a biger fan, so i will have to buy a new 8" 760m3/hr fan, but my cf has a max of 480m3/hr..... will still be ok or wil i have to buy new cf also ?
    No problem

    As for the CF rating, I really don't know. I'd be tempted to try the one you have and see what happens, or ask the manufacturer (Rhino?).

    The important factor with carbon filters is what's known as the 'dwell time'. This is how long the air spends in contact with the carbon as it passes through the media. Increasing the flow rate will reduce the dwell time because the air will take less time to pass through the filter and so won't have as much time to react with the media, which in turn will lead to less efficient scrubbing.

  3. #23

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    Quote Originally Posted by stone14 View Post
    also that equasion you worked out for 1x600wt..... for 2x600wt would you just x2 it?

    so anything more than 2x600watts your going to need more than 1 ex-take fan+cf to keep it cool?

    because both 8" and 10" only goes upto 1150m3/hr................ and a big 12" 1300m3/hr
    Yep. My mistake, sorry about that. It is double.

    3.16 x 1,200 / 5.4 = 702.2 CFM (or 1,194 m³/hour)

    If it was my space then I'd be tempted to use two extractors in opposing top corners, but I'd definitely check the noise levels first.

    If you use two identical fans then the total noise level will be the dB rating of one fan plus 3 dB, which will sound about 1.23 times as loud.

    Basically, if (one small fan's dB + 3) is less than (one large fan's dB) a dual-fan setup will be quieter.

    Is there any way you can draw the incoming air from a cooler place? Or maybe even use a chiller/air-conditioner?

  4. #24

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    i have an air con 12,000bu in the cuboard and a 5" tt fan thats pulling air up into my tent throu hole in ceiling, and its still giving me 34-35oC 2 foot below my lamp at 8pm at nyt but my intake vents arnt sealed on my tent, i will duck tape them 2moro and try again

  5. #25

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    Quote Originally Posted by Anonymiss View Post
    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)
    very nice can the cfm be converted into m/3,is it multiply by 0.589 or something.

  6. #26

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    Quote Originally Posted by high riser View Post
    very nice can the cfm be converted into m/3,is it multiply by 0.589 or something.
    m³/hour = CFM x 1.7

  7. #27

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    ahh thats it, dont know where i got that figure from??
    Thanks alot very helpfull info.
    Peace

  8. #28

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    Quote Originally Posted by Anonymiss View Post
    The calculation provided above will only tell you how much airflow you need to achieve in order to dissipate a given amount of heat.

    System impedance will always be a factor when it comes to fan selection, and to work it out in detail you need to know the impedance curve of the system in question and then relate that to the static-pressure:flow-rate curve of the fan.

    This graph shows some (imaginary) system-impedance/flow-rate/pressure curves. Find the desired flow rate on the bottom, go up until you hit the relevant impedance curve, and then go left to determine the static pressure required to achieve that flow rate. (This is for illustration purposes and so doesn't have any actual values attached.)



    From this, it's obvious that a higher impedance system will need a higher pressure to achieve a given flow rate.

    This next graph shows the pressure:flow-rate curve for a typical (but imaginary) fan. (Accurate data for specific fans should be available from the manufacturers of those fans.)



    This graph shows the throughput a given (imaginary) fan will achieve through systems of differing impedance. The intersections between the fan curve and the impedance curves will give the flow rate and pressure for this particular fan in those systems.




    Although it's possible to calculate the impedance (losses) caused by ductwork it is fairly complicated. It's definitely not impossibly difficulty though, but I haven't spent much time on it yet. If (when?) I do then I'll post a short how-to, or at least some rough correction factors.
    cheers for the reply i think when working out your needed fan size for the lights temps etc you do have to factor in ducting loses as well because it can be a big loss 50% plus is a easy loss! and that can ruin you whole set up and have to do it again (got the t shirt lol), if you have got a few bends in the and long runs and fans positioned wrongly
    i have a good formula some where that has worked very well on the three rooms ive now built, might be a good addition to this thread the site doesnt have a good extraction formula any where
    Anonymiss good thread a bit scientific for most of us i would think, but the pretty graphs are nice !
    its always good to have some help when working out fans sizes it can be confusing for sure!

    good informative thread bro

  9. #29

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    Quote Originally Posted by high riser View Post
    ahh thats it, dont know where i got that figure from??
    Thanks alot very helpfull info.
    Peace
    The 0.588 figure is for converting m³/hour (CMH) to CFM

    CFM = CMH / 1.7 = CMH x 0.588

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  11. #30

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    Quote Originally Posted by G-Man View Post
    cheers for the reply i think when working out your needed fan size for the lights temps etc you do have to factor in ducting loses as well because it can be a big loss 50% plus is a easy loss! and that can ruin you whole set up and have to do it again (got the t shirt lol), if you have got a few bends in the and long runs and fans positioned wrongly
    i have a good formula some where that has worked very well on the three rooms ive now built, might be a good addition to this thread the site doesnt have a good extraction formula any where
    Anonymiss good thread a bit scientific for most of us i would think, but the pretty graphs are nice !
    its always good to have some help when working out fans sizes it can be confusing for sure!

    good informative thread bro
    Absolutely. The effect of ducting and other restrictons certainly does have to be taken into account, but only after determining the air flow required to maintain a particular temperature rise. Once that's known, the impedance/losses can be factored in and used to specify a fan which will actually deliver that throughput in a particular installation.

    One thing that seems counter-intuitive at first is the fact that room size makes no difference, it's only the amount of heat being generated that matters. Of course, that assumes no loss of heat through other routes, such as the walls, but these are likely to be minimal in most situations (do tents get especially warm to the touch, or radiate much heat?).

    Yeah, it's all a bit 'sciency', but it's a pretty complex subject which gets into the relams of air velocity in the duct (in feet per second), friction losses between the moving air and the duct wall, head (pressure) loss through bends, etc. Even things like the way the air-density changes with pressure and temperature have an effect, but that may be going too far!

    To make things easier (and reduce the 'science'!) I've created a table of lamp Wattages and temperature rises so people can simply look up what they need rather than having to calculate anything. (Thanks to Grub for editing my initial post to add them there.)

    The original calculation (and the table) only determines the actual throughput required, and if that throughput is achieved then it should be accurate.

    I guess that what's needed now is a table of correction factors, or something like that. Along the lines of 'multiply by this much per metre of 4" ducting/6" filter/45° bend' or similar. That would definitely help, and would probably provide some reasonable estimates of the actual fan sizes necessary.
    Last edited by Anonymiss; 29-06-11 at 01:43 PM.

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