Fan amps and watts
Posted 17 December 2012 - 06:21 PM
Posted 18 December 2012 - 05:26 AM
if you have a computer with a case fan or a PSU with a 120 mm fan, observe it accurately (with a torch since not only the PC is off but - for safety reasons - you have cut out electricity in your neighbourhood ).
That fan is mounted to EXTRACT air from the case.
The hypothetically finger cutting blades are on the INNER side.
A typical fan (like the one we were talking about) will have 7 (seven) blades.
Roughly 2/3 of the circular corona between the central motor and the outer of the fan is "blades" and 1/3 is "air".
At the very outer part of the fan, let's assume 110 mm you have a circumference of 110*Pi.
Your "target" when you will attempt to insert a finger in it is at the most 1/7*1/3*110*Pi=~16.45 mm wide.
The thickness of the fan is 38 mm (i.e. the actual hypothetically cutting edges of the blades are some 35 mm from the exterior).
But the blade is rotating by definition at 3.000 RPM, this means that it's speed at 110 mm outer diameter is 110*3000/60=5,500 mm/s
Your target is "covered" in 16.45/5,500=~0,00299 seconds, let's say 3/1000 of a second.
You finger has to travel linearly "inside" the target for 35 mm in 3/1000 of a second to reach the cutting edge of the blade, i.e. it has to travel at around 11,706 mm/s i.e. 11.7 m/s, i.e. roughly 42 km/h.
Presuming that your hand starts from still with the point of your finger at 10 cm from the fan, the tip of your finger will need to accelerate from 0 to 42 km/h in 10 cm.
Assuming that acceleration is linear, the 10 cm will be covered at an average speed of 5.85 m/s (or 21 km/h) and will thus take .10/5.85=0,017094 seconds, i.e. roughly 2/100.
So we have 11.7/0.02=585 m/s2 or 59.65 G.
I guess that if your finger (and hand and arm) can stand that kind of acceleration, it will survive the fan-through experience alright
Posted 18 December 2012 - 01:46 PM
I did a lot of stupid things with like stopping them barehand when spinning (you just need to press on the center). No computer fan except perhaps one modded with with razor blades could cut a finger.
Posted 19 December 2012 - 04:55 AM
Posted 19 December 2012 - 12:16 PM
Cheers and Regards
Posted 19 December 2012 - 01:22 PM
And (SCOOP! ) we have a movie of it!
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:
Posted 19 December 2012 - 02:20 PM
Cheers and Regards
Posted 22 December 2012 - 04:29 AM
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:
that is the same, written differently, that gives you V x A = W (when talking of DC, NOT AC)
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.
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) .
Posted 22 December 2012 - 08:37 AM
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:
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".
Posted 22 December 2012 - 11:40 AM
This post has been edited by Ponch: 22 December 2012 - 11:50 AM
Posted 22 December 2012 - 12:00 PM
Sure it won't .