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Fan amps and watts

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32 replies to this topic

#26
jaclaz

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I did a lot of stupid things with like stopping them barehand when spinning (you just need to press on the center).

Does this mean that you hadn't the guts to inset a finger near the outer edge? :unsure:

:whistle:

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#27
bphlpt

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As you pointed out, it takes no guts at all to push ones finger anywhere on the output side of (practically) any size computer case fan. It is all but impossible to cut ones self in such a manner. This is also true with common house "box" type fans with either plastic or metal blades, again on the output or exhaust side. I did it many times as a kid. However, while you essentially can't cut yourself that way, your fingers can be bent or pinched between the blades and either the housing or the grill so care should be exercised around small children.

Cheers and Regards

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#28
jaclaz

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Casually ;) a device that is engineered to actually cut fingers (not really, but similar enough) actually exists, it has a finely sharpened steel blade and it is powered by 4 AA batteries, so logically is in the same power range of less than 10 W.

And (SCOOP! :w00t: ) we have a movie of it! :yes:


Watch attentively the movie and time the rate at which cuts are performed, and note ( around 0:59 ) the number of gears needed to transform the relatively high speed (and very low torque) of the motor to enough (slow) torque actually needed to cut the (boneless) "finger".

Just in case:
http://science.howst...ipment/gear.htm

jaclaz

#29
bphlpt

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Apparently, it uses gears, as you say, "to transform the relatively high speed (and very low torque) of the motor to enough (slow) torque actually needed". The final gear then has a post attached which acts against what looks like a spring loaded blade, cutting the "finger" with a guillotine type action while the "finger" is pinched between the blade and the housing.

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#30
Ponch

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The specific device (fan) can run continuously at 12.8 V but it is designed to be powered at 12V (i.e. it has a target of being operated at 12 V though it can accept as low as 6V and as much as 12.8V).

Power absorption is another matter, if you attach a device rated for a higher power absorption (but within the right Voltage specs) to a power outlet, the device will attempt to get more power (up to it's own specifications).

As a matter of fact when you connect a load to a voltage output, the voltage level might drop a little, but NEVER increase, if your motherboard ouputs 12V (as a matter of fact, if you had a mutimeter, you would measure on it - as said - most probably 12.2 or 12.3 V with no load connected to it and exactly 12 V or more likely 11.9 V with a "heavy" load connected) you will NEVER be able to get 12.8V from it.

Examples (simplified, and in order to let you understand) in case of BOTH an unprotected/unlimited power supply and of an unprotected/unlimited device:
Power supply specs:
12V 1A hence 12W

  • If you attach to it a device rated 6V 1A (6W) continuous, the device will burn in no time (because the voltage operating range of the device is greatly exceeded) .
  • If you attach to it a device rated 12V 1A (12W) continuous, both the device and power supply will work for years without issues (this device may well have peaks or "MAX" of up to 40% more than standard absorption - or even higher "transient" current spikes without any consequences to the device or to the power supply)
  • If you attach to it a device rated 6V 2A (12W) continuous the power supply OR the device will burn in no time.(because the voltage operating range of the device is greatly exceeded OR the current capabilities of the power supply are greatly exceeded)
  • If you attach to it a device rated 24V 1A (24W) continuous the power supply is very likely to burn (because the device will attempt to draw more current to compensate for the lower voltage)

Additionally voltage and current are linked together by Ohm's Law:
http://en.wikipedia.org/wiki/Ohm's_law
that is the same, written differently, that gives you V x A = W (when talking of DC, NOT AC)
http://www.csgnetwor...m/ohmslaw2.html
the MAX current that device will draw is 0.70A at 12V, if you supply it with a higher voltage, let's say 12.8V it will draw LESS current, i.e.
12 x 0.70 = 12.8 x X -> X=~0.66 A

If you attach to an outlet (BTW surely protected from overcurrents) rated for 12 V 0.74A that particular fan, it will run normally at 12V (and NOT at any higher voltage) absorbing normally 0.51A with the possibility of peaks up to 0.70A.

The outlet can provide UP TO 0.74A at 12V (8.88W), the device will nornally use 0.51A at 12V (6.12W) with peaks up to 0.70A at 12V (8.40W).

You are well within the specs.

jaclaz

I know the thread IS about fans, but I intentionally separated that big chunk from the post because it starts with "specific device (fan)" then goes on with things like "a load" and "a device". I'd like to point out that those consideration are in fact all for that "specific device (fan)" (or a DC motor) as for instance Ohm's law would say a big "well no" to the 4th example in case of a resistor. An AA battery would power a 220v radiator for few hours without burning (in fact you'd die of cold before it burns) :D .

#31
jaclaz

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I know the thread IS about fans, but I intentionally separated that big chunk from the post because it starts with "specific device (fan)" then goes on with things like "a load" and "a device". I'd like to point out that those consideration are in fact all for that "specific device (fan)" (or a DC motor) as for instance Ohm's law would say a big "well no" to the 4th example in case of a resistor. An AA battery would power a 220v radiator for few hours without burning (in fact you'd die of cold before it burns) :D .

Well, noone talked of a battery, we were talking after having expressly said "(simplified, and in order to let you understand)" of a PC power supply connected to mains and about a brushless DC motor, which is both a resistive load and an inductive one (actually essentially the second), and unlike the "will burn", "very likely to burn" was used instead for example #4.

Ohm's Law was cited AFTER the 4 examples and as an additional point, to exemplify how powering (within a bearable range) a device (still talking of an electric DC motor) connected to a PC power supply) with higher voltage will reduce the amount of current needed and how current absorption is linked to the voltage level supplied.

In the case of an electric radiator suited for 220 V , actually norms are about 230 V with some slightly different tolerances:
http://en.wikipedia....Standardization
labeled 1000 W, we can use:
W = V x I to get I=1000/230=~4.35 A
and then use
I = V / R to get that the resistance of the thingy is about 53 Ohms as 230/53=~4.34
But now, when we use the AA battery 1.5 V we have:
I=1.5/4.34 =~0.35 A or 350 mA
Since a normal AA battery (alkaline) has a capacity of roughly 2200 mAh, and without taking into consideration a number of factors, it will last 2200/350=~ 6h30 which, while NOT being accurate, it is still more accurate than "a few hours". :whistle:

jaclaz

#32
Ponch

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You can't expect everyone to write 100 lines just to add a little precision. I had done a short calculation. The battery/power supply will not burn, nor is it very likely to. That was my point. 230v, 220v, 2500W, 1000W, alkaline 1.5v, rechargeable 1.2v, 2200mAh, 2500mAh, 6h30, 40h, a few hours, whatever, it will not burn.

Edited by Ponch, 22 December 2012 - 11:50 AM.


#33
jaclaz

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You can't expect everyone to write 100 lines just to add a little precision. I had done a short calculation. The battery/power supply will not burn, nor is it very likely to. That was my point. 230v, 220v, 2500W, 1000W, alkaline 1.5v, rechargeable 1.2v, 2200mAh, 2500mAh, 6h30, 40h, a few hours, whatever, it will not burn.

Sure it won't :).

jaclaz




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