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In the specifications of a 10 KW BLDC motor I've seen

Motor Wire ------ 2 - Strand 1.9mm
Max Continuous Current ----- 200 Amps
Max Continuous Power ----- 10000 Watts

The details are as follows, it's a Scorpion HK-7050-340KV BLDC motor http://www.scorpionsystem.com/catalog/helicopter/motors_4/hk-70/HK_7050_340/

Now my question is, the max continuous current is a huge 200 amps but the winding used in the motor is just 2 strands of 1.9 mm diameter wire.

Why doesn't the coil burn out at such huge amps.

Thank you.

P.S.  If you like take a look at this https://www.youtube.com/watch?v=77uK19KxMuI        just to see the power of this monster in action.

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    \$\begingroup\$ In normal use, would the motor be in a stream of high velocity air? This could provide some cooling. Also, the motor output may be exaggerated a bit. \$\endgroup\$ – mkeith Jun 5 '17 at 6:06
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    \$\begingroup\$ Yeah seems exaggerated .... but believe it or not this monster really pulls 200 amps at full throttle. \$\endgroup\$ – Knight Jun 5 '17 at 6:36
  • \$\begingroup\$ Are you sure the windings aren't made of a whole load of wound coils all connected in parallel? That would make more sense if it were the case. \$\endgroup\$ – TonyM Jun 5 '17 at 6:56
  • \$\begingroup\$ Sorry but that isn't the case .... just two strands of 1.9 mm magnet wire is somehow magically withstanding such huge amps. \$\endgroup\$ – Knight Jun 5 '17 at 7:17
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    \$\begingroup\$ The other way to approach this question is to consider the temperature rating of the insulation on the wire. The link in the question says it is rated for 180C, which is around 350F. So, the wire has to get very hot before the insulation will fail. The tables you find online for maximum current in a wire are based on much lower temperature insulation (I believe in the range of 60-90C). \$\endgroup\$ – mkeith Jun 6 '17 at 4:31
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Ignoring the effect on the waveforms of motor winding inductance (Which you really cannot do, but ignoring it makes the math much less messy).

If we assume a standard three phase machine which at any given point has one winding directly across the supply and two windings effectively in series also across the supply, then for 200A of DC input one winding sees 133A and the other two see 67A each, further the 133A winding is only conducting for 1/3rd of each cycle while conducting 67A for the other 2/3rds of a cycle.

If these waveforms were square (which they are NOT), you would get average (Per winding):

((133^2)/3 + (67^2)*2/3) * 0.006 = 54W (IRMS = 94A) for a total copper loss of ~162W distributed across the three phases.

Assuming a motor Tcase of say 100 degrees C, and winding temperatures of say 200C (220C magnet wire is an off the shelf thing), then you can tolerate nearly 1.8degrees C per watt between the winding and the case, which should not be a showstopper (This ignores the heat from the iron losses which is likely significant).

Of course you still have to get rid of the 162W from the motor housing without letting its case get too hot (Big low speed machines often have auxiliary blowers to handle this).

While I have my doubts that that machine actually meets that spec, it is not completely impossible at first glance.

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  • \$\begingroup\$ Thank you very much for at least making a few things clear. I'm ready to accept the answer, but I just wanna wait for some time. \$\endgroup\$ – Knight Jun 5 '17 at 12:10
  • \$\begingroup\$ I got one thing for sure that when dealing with things related to heat and temperature (thermodynamics stuff) it is quite difficult to be assertive as there are a lot of variables and complex physics involved which decide the end result.We need a lot of data, careful analysis and thermodynamic simulations to be sure.So without sufficient data it is not that easy to answer. \$\endgroup\$ – Knight Jun 6 '17 at 11:58
  • \$\begingroup\$ To simplify, Heat loss is directly proportional to the temperature .So I believe that at some temperature below the maximum permitted core temperature which is around 180°C in this case, Heat loss is equal to heat produced thereby stabilizing the temperature. \$\endgroup\$ – Knight Jun 6 '17 at 11:59
  • \$\begingroup\$ 2 copper magnet wires in parallel each with a diameter of 1.9 mm is going to need about 400 amps of current at room temperature and with no air flow over it to melt since the melting temperature of copper is about 1083°C (1981°F) . In the case of this motor the max current flow is 200 amps and moreover with a good airflow to eliminate the heat .So even though this seems impossible I now believe that it isn't. Which is exactly what you and few others stated in the comments.Unfortunately I could only accept one answer. So I appreciate your insight and Thanking you all. \$\endgroup\$ – Knight Jun 6 '17 at 12:03

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