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I work in aviation. I was told today that one of the systems of the aircraft had drawn more power than the generator could provide and that had caused it to fail, to get too hot. I don't understand how this could happen. How can a load draw more power than a source can provide? That doesn't make sense to me.

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  • \$\begingroup\$ VTC - Questions regarding the use of electrical/electronic devices are off topic here. \$\endgroup\$ – Michael Karas Apr 6 '17 at 9:29
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    \$\begingroup\$ @MichaelKaras really not a usage question, by no means. \$\endgroup\$ – Marcus Müller Apr 6 '17 at 9:39
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    \$\begingroup\$ @K Jason .This gen burn out seems bad .How could it get overloaded if there are circuit breakers that are sized under the generators capability .Are things getting a bit tetchey on the tarmac? I think that you could also post on Aviation Stack . \$\endgroup\$ – Autistic Apr 6 '17 at 10:58
  • \$\begingroup\$ @MarcusMüller This is a troubleshoot question lacking necessary investigation. \$\endgroup\$ – Dmitry Grigoryev Apr 6 '17 at 19:14
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Assuming circuits are good and the MCB protecting the generator works I could suggest a short between the generator and MCB or failure of the windings insulation causing an internal short, both subject to failure due to age, vibration, or manufacturing fault..

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It should have said safely provide.

But also, there's often a confusion between drawing power and drawing current.

For example, you can get the maximum power out of any voltage source if your load has exactly the resistance of the internal resistance of that source.

You can of course also plug in a much, much lower (practically, a "short") resistance load. The power flowing into that load will then be lower than the maximum power of the source, but the current will be much higher than what's flowing at the max power point (at the expense of the voltage dropping very significantly).

As an example, take an AA battery. (THOUGHT experiment. Don't do this at home, kids.) Let's say that has an internal resistance of 0.2 Ω. Now, when you take a 0.2 Ω resistor (which would probably be something like a piece of thin steel wire), that resistor would get very hot – the maximum amount of power the battery can supply gets converted to heat.

Next, really short the battery with a thick piece of copper and very good contacts (that's where you usually get into trouble - 0.2Ω isn't all that much, and making a contact way better than that is hard) and short-circuit your battery. What will now get hot?

Right, the battery, and not so much the copper wire. The power converted to heat in the copper wire isn't very high, and most of the power is "wasted" across the internal resistance of the battery.

I'd recommend not shorting batteries, by the way. It's dangerous. They can get extremely hot, depending on the type catch fire, or even explode.

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  • \$\begingroup\$ I don't think typical AA and AAA batteries are that dangerous. I have shortcircuited a bunch of (old) AAA batteries and they didn't even get hot. \$\endgroup\$ – Oskar Skog Apr 6 '17 at 9:20
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    \$\begingroup\$ @OskarSkog true, AA (alkaline/zinc) type batteries are relatively safe (they can source surprisingly high currents, but are pretty big and heat-resistant, so that the energy they can supply until they run flat can't do much harm). But someone's going to do this with a LiIOn battery somewhere, or look into the spark while trying to figure out how hot a truck battery gets when shorting it... \$\endgroup\$ – Marcus Müller Apr 6 '17 at 9:22
  • \$\begingroup\$ That was a very understandable answer. And thank you for it. I wonder the generator's behavior depends on power demands somewhat and a larger demand for power can tell the generator to work harder. \$\endgroup\$ – K Jason Apr 6 '17 at 9:24
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    \$\begingroup\$ @KJason that's exactly how power generation happens. For example, assume the generator in question just supplies the cabin heating. If the heating's turned off, there will be very little mechanical power being consumed from the motor by the generator. Turn on the heating, and the generator will, from the perspective of the motor, act like a big break that the motor has to work against. \$\endgroup\$ – Marcus Müller Apr 6 '17 at 9:31
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    \$\begingroup\$ @OskarSkog Dropping it in salt water is literally the recommended way of disposing of an immediately dangerous lithium polymer battery in an emergency. The water will absorb most of the energy and so is more likely to suppress rather than encourage fireworks. \$\endgroup\$ – Muzer Apr 7 '17 at 14:25
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The aircraft generator should be protected by a series of fuses and/or circuit breakers to constrain both the maximum current in each sub-circuit and the maximum total current.

Think of the wiring in your home- there is a main breaker and a bunch of breakers in a panel to protect each circuit. For example, there might be a 300A fuse protecting a main 28V bus, and a series of 30-100A aircraft-rated magnetic/thermal breakers protecting various sub-circuits. As well as protecting the generator itself, it is necessary to protect the wiring in the aircraft so that a fire cannot easily occur as a result of a simple overload.

If the (very expensive) generator failed due to overload then there may be an engineering issue with the electrical system. Those in charge of maintenance will probably want a DAR (Designated Airworthiness Representative) or other qualified person to have a look at it.

The way it could happen is simple- imagine you have a 15A home circuit and you plug two 12A kettles into it. The 15A breaker should blow protecting the wiring, but if it is faulty (or deliberately bypassed) the wiring could overheat. In your case, both the wiring and the generator itself could overheat. This could be dangerous and will probably be expensive.

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  • \$\begingroup\$ I understand from this how wires could get hot. That was clear. I could picture insulation melting and windings touching. That would be enough for failure right there right? \$\endgroup\$ – K Jason Apr 6 '17 at 21:24
  • \$\begingroup\$ Yes, something like that would probably be the failure mode. If you get shorted turns the failure can cascade because the internal current in the windings will increase. \$\endgroup\$ – Spehro Pefhany Apr 7 '17 at 2:41
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The reason you get confused, is that the statement is incomplete. "...the aircraft had drawn more power than the generator could provide (without overheating) and caused it to fail."
The statement in parenthesis is what is missing. The reason it is missing, is that it is commonly understood that the given power rating is for continuous use, without overheating the generator.
However, the generator is capable of delivering more power (at the cost of overheating) for a short period of time (the time it takes to burn the generator's winding).
You should now be able to see that a generator can provide more power (called max power), even if it is just for a short time, than the "normally quoted" power (called continuous power).

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