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

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    Default Airflow / Temperature / Wattage table (and calculation).

    We've all seen the room volume/air flow calculation* and, hopefully, we've all realised that it's next to useless for controlling temperature when high-Wattage equipment is used in small spaces.

    To address that, I made a table which shows how much air flow is required to maintain a given temperature rise above ambient.

    It's calculated using the basic heat transfer/dissipation equation, which is a standard method for estimating the cooling requirements of hot things in small enclosures (it's often used in relation to electronic devices, for instance).

    I hope it's useful

    Short (memorable?) link to this post: http://is.gd/airflow (I made it as easy as I could!)

    If you've found this post helpful then, please, do pass that on to when people ask the age-old, "how do I work out air flow?" question.

    Short link directly to the table: http://tinyurl.com/airflow-temperature




    *The room-volume method is derived from calculations used for building ventilation. For those who don't know it, it's 20 x room_volume per hour.




    My original post is below here. I'm leaving it because, it has good keywords for the search engines and, it may be useful if people want to copy/paste the formulae but, I've removed the old tables to reduce the page size a bit.



    [edit 26-10-11]

    To make it easier for everyone I've created a table which can be used to look up lamp Wattage and temperature rise and then read off the required air flow.

    <snipped table>

    And here's one to convert CFM to m³/Hour...

    <snipped table>

    Hope it helps.

    [/edit]

    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)
    Last edited by Grub; 04-03-13 at 07:57 AM. Reason: OP requested. :)

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  3. #2

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    clever guy, but no offense but you have just made my brain take a shit!!!!!! lol

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    Quote Originally Posted by gingerbread man View Post
    you have just made my brain take a shit!!!!!! lol
    Hee hee. Sorry about that.

    Maybe I should have put the easy (important?) bit (the final calculation) at the top, and then the rest until afterwards for those who wanted to know more?

    Quote Originally Posted by gingerbread man View Post
    clever guy
    Nope. (Hint: the clue is in the name.)

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    Let me guess............you are selling calculators and you are going to confuse all the poor stoners by putting up hard sums so that they buy said calculators?

    Will have to spend a bit of time working this out. Cheers Anonymiss.


    PS. Tell your brother - ANONYMOUS - I said hi!!

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    Quote Originally Posted by nuube View Post
    Let me guess............you are selling calculators and you are going to confuse all the poor stoners by putting up hard sums so that they buy said calculators?

    Will have to spend a bit of time working this out. Cheers Anonymiss.


    PS. Tell your brother - ANONYMOUS - I said hi!!
    I was thinking of abacuses. I think those would be far more fun... Ooooh! Look at the shiny beads!

    Seriously though, it's not that difficult (at least not now that Grub kindly edited the post for me after I realised I'd gone about it the wrong way initially), and the only bit you need is the first CFM = 3.16 x Watts / DT(°F) bit. The rest is just there for the completists really.

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    mate my brain jus turned inside out

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    Awww my brains hurt pass the bong

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    ANONYMISS: could you work out mine for my i have a 1x600watt hps

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    er what about the duct lenght % per foot you lose on runs and % you lose on bends ? & % you lose in air resistance for the cf ? there online cals for this as well some where so ive read and a great artical this mth in urban gardener
    other thing which is very important is fan placement these fans are made to push air not pull so you need to always have to longer duct run on the push side

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    Quote Originally Posted by stone14 View Post
    ANONYMISS: could you work out mine for my i have a 1x600watt hps
    Sure. What's your ambient (room) temeparature, and the target temperature in the grow space?

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    28-30oC room temp....
    target low 20's-26max (there oC)
    atm cant get below 34oC 2foot away


    (im drawing in cooler air from room below, sorry dont no the temps but im guessing still mid 20's

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    i want my lamp to be cool or just warm to the touch

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    OK going to be cheeky like stone14.

    I have a shed ambient temps are around 15*c the grow tent will have a 400w bulb looking for high 20's temps. The grow will continue til christmas when the temps in the shed will be about 5c or so. I know I am been cheeky but if you show me how to work that calculation out I could do the others myself.


    Thanks in advance.

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    Quote Originally Posted by G-Man View Post
    er what about the duct lenght % per foot you lose on runs and % you lose on bends ? & % you lose in air resistance for the cf ? there online cals for this as well some where so ive read and a great artical this mth in urban gardener
    other thing which is very important is fan placement these fans are made to push air not pull so you need to always have to longer duct run on the push side
    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.

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    Quote Originally Posted by stone14 View Post
    28-30oC room temp....
    target low 20's-26max (there oC)
    atm cant get below 34oC 2foot away


    (im drawing in cooler air from room below, sorry dont no the temps but im guessing still mid 20's
    You'll never be able to get the temperature below that of the incoming air, regardless of air flow rate.

    Assuming that the incoming air is at 22°C and the target temperature is 25°C then to dissipate 600W of heat you'll need a flow rate of 351 CFM. The lamp envelope itself will still be hot though.

    Incoming air = 22°C
    Desired temperature = 25 °C
    DT = 25 - 22 = 3 °C
    DT °F = 3 x 9 / 5 = 5.4

    3.16 x 600 / 5.4 = 351.1 CFM
    Last edited by Anonymiss; 28-06-11 at 02:24 PM.

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    ok cheers for that, my 5" ruck suck 395m3/hr and still temps are 34+ will seal off all my passive and just intake from down stairs, also my cf must be restricting my air flow, i also had 4 90degree bends in my ducting before it reaches the fan

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    Quote Originally Posted by nuube View Post
    OK going to be cheeky like stone14.

    I have a shed ambient temps are around 15*c the grow tent will have a 400w bulb looking for high 20's temps. The grow will continue til christmas when the temps in the shed will be about 5c or so. I know I am been cheeky but if you show me how to work that calculation out I could do the others myself.


    Thanks in advance.
    You have to plan for the worst case (i.e. the warmest incoming air), so I'll do the calculation for 15 °C ambient. When it's colder you'll need to reduce the fan speed to compensate.

    At 15 °C ambient:
    Incoming air = 15°C
    Desired temperature = 28 °C
    DT = 28 - 15 = 13 °C
    DT °F = 13 x 9 / 5 = 23.4

    3.16 x 400 / 23.4 = 54 CFM

    And at 5 °C ambient:
    Incoming air = 5°C
    Desired temperature = 28 °C
    DT = 28 - 5 = 23 °C
    DT °F = 23 x 9 / 5 = 41.4

    3.16 x 400 / 23.4 = 30.5 CFM

    The relationship between fan RPM and CFM is linear. So if a fan running at 100% gives 54 CFM then reducing it to about 60% will give 32 CFM.

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    Quote Originally Posted by stone14 View Post
    ok cheers for that, my 5" ruck suck 395m3/hr and still temps are 34+ will seal off all my passive and just intake from down stairs, also my cf must be restricting my air flow, i also had 4 90degree bends in my ducting before it reaches the fan
    395 m³/hour is only 232 CFM

    To convert CFM to m³/hour to CFM multiply by 1.7
    To convert m³/hour to CFM divide by 1.7

    351 CFM in m³/hour = 351 x 1.7 = 596.7 m³/hour

    You definitely need a bigger fan!

    And yes, ducting and especially a carbon filter will increase the impedance and so reduce the airflow.
    Last edited by Anonymiss; 28-06-11 at 02:44 PM.

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    Quote Originally Posted by Anonymiss View Post
    395 m³/hr is only 232 CFM

    To convert CFM to m³ to CFM multiply by 1.7
    To convert m³ to CFM divide by 1.7

    351 CFM in m³/hour = 351 x 1.7 = 596.7 m³/hour

    You definitely need a bigger fan!

    And yes, ducting and especially a carbon filter will increase the impedance and so reduce the airflow.
    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 ?

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    thanks for the maths, guess i will still have to wait till end aug then, cant afford new fan and filter ...................

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