How to calcualte how much power an undervolted fan will draw
I bought 4 Papst 6424H 6.75" fans
specs here @ 24 volts they draw 26 Watts how much power would they draw at 12 Volts? 13 Watts? at 6 Volts... 6.5 Watts? That calculation would work on a resister but on coils? They are 90Deg out of phase arn't they?more then one coil... more 90Deg out of phase calcualtions? I would like to use a Orbitlal Matrix MX212 to control them. Since the MX212 only has 3 fan connectors would having 2 fans on one connector be to much of a power draw? Would having 1 fan on a connector be to much? I'm hopeing someone has the knowledge because I don't want to test it and find out I burned out my MX212 Thanks |
Assuming you don't stall them, power drawn is propertional to the cube of the speed of the fan. Speed is proportional to voltage.
So at 12V, you draw 0.5^3 as much power as at 24, which is 3.25W. At 6V the fan will have stalled. In fact given 12V is the minimum voltage, you'll not really be able to adjust them much off a standard PC supply as they'll stall at around 10-11V. According to specs the maximum you can pull off a MX212 fan header is 750mA, at 12V that's 9W. So you can run 2 fans off one header, though given you won't be able to adjust them, you won't really see much benefit over just hooking them directly to the 12V. |
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With the MX212 I have run other fans at 10% of it's rotations speed and it actualy powers down for a second then starts back up... So I think the MX212 tracks rotation and raises voltage enough to start the fans spinning again... I'll have to test it more in the future. Thanks for the info :) |
MH, your first calc assumes constant current, which is not how a resistor behaves. Cut the voltage in half, and the current drops in half also, resulting in 1/4 power.
Aerodynamic power is the cube of speed, but that doesn't seem to match these switching fans. I just measured a Yate Loon 120. Feeding it 6V instead of 12 reduced the current by half. So that matches the resistor model nicely. Not matching the aero power calc is not intuitive for me. Do you need a scope to measure these switching motors properly? |
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so if my fans follow the same principal as yours the power drawn would be about 13 watts. about 1 watt over thee recomended feed. :cry: but there are 3 outputs on the MX212 and i have 4 fans to control. looks like i may not be able to use the MX212 for the fans :mad: I don't have acccess to a scope. i suppose i could go buy an amp meter. but after i get these measurments i would probably not uese it again...( i have hade a Multimeter for voltage checks and continuity for a few years and never needed anything more untill now) One option wich i will now try is email teh manufacture and C if they can give me the values acurately... but from past emxperience that sometimes takes weeks before getting a reply. I guess one option would be to use mCubed T-Balancer 4 fan headers @ 15 Watts each... |
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(2) Is new to me. Butcher's Law? |
Taken from the behaviour of brudhless DC motors Les. And confirmed by fan air flow info on comair rotron - specially, air volume moved is proportional to fan speed. Halving the voltage halves the free air flow, implying it has halved the speed of the fan.
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Just my casual non-scientific observations:
In unrestricted free-flow, the fan's power draw (Watts) is roughly proportional to the voltage squared down to the stall voltage. Fan power draw is related to the back-pressure. Fans will draw more power in free-air, than when acting against a restriction. Panasonic gives PQ curves for their fans at different voltages, which can be seen here: http://www.panasonic.com/industrial/...pdf/fba12g.pdf Pretty sure that Panasonic actually measured such. From this we can make some further observations: Fan peak pressure is roughly proportional to the voltage squared. Fan pressure - while peaking at proportional to the voltage squared at "dead head", is not proportional to the voltage squared once air-flow occurs. Pressure seems to be proportional to the voltage squared divided by the ratio of the new and old air-flow rates against a resistance. Fan peak flow is roughly directly proportional to voltage In fact axial fans can be observed to act very similarly to (DC) water pumps. Both of these observations from Panasonic's axial fan PQ curves also seem to match my own observations of DC water-cooling pumps, however the peak flow of water pumps does not seem to be as linear with voltage as axial fans, but I suspect this is related to fitting resistance effects. None of the above contradicts the Vent-Axia fan laws. A doubling of voltage will result in a quadrupling of motor power, but will not result in a doubling of air-flow against resistance. To do so requires roughly sqrt(8) or ~2.8x the original voltage. The actual proportional values of this will depend on whether various sections of the "system" are in laminar or turbulent flow. |
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So to recap this then,
If doubling the voltage quadrups the power output. then it is safe to say dividing the voltage in half would drop the output power by a quarter? since @ 24 volts the fan draws 26Watts so @ 12Volts the fan would draw 6.5 Watts and with 4 fans and only 3 fan headers on the MX212 that would mean i'd still would have to double up. putting it to 13Watts which is 1 Watt over the spec. will it burn it up... do I want to spend $100 to find out...? have to ponder about that... |
Yes, 13W of power will fry a fan header. 12W is rather generous for most headers too. I've fried mobo fan headers with just 8W of power draw. Personally I would never put more than 6W on a mobo fan header.
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it is for a Orbital Matrix 212 listed maximum output power at 1000mA. (on page 14) I have had 2 120mm evercool fans hooked up to one header in the past... they have a draw of 4.56W each. |
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