Rated current of wires

[Anonymous No.1648319]

We have a power line with a load that is almost completely inductive.

If I've understood it right it's just the reactive component of the current drawn by the load what turns into heat in the wire's equivalent resistor. Then, if the load draws let's say 100 amps why do we need a wire with a size rated for more than 100 amps if it's not going to heat up?

[Anonymous No.1648321]

idk surge

[Anonymous No.1648325]

>>1648319

You need wire rated for 100 amps because there will be 100 amps flowing through it and it will heat up

[Anonymous No.1648333]

>>1648319
>If I've understood it right it's just the reactive component of the current drawn by the load what turns into heat in the wire's equivalent resistor.

All the current through a resistor heats it. You don't understand what reactive means, and you probably don't understand anything else you said. How the fuck were you tasked with this problem?

[Anonymous No.1648365]

>>1648333

If you think I don't understand what reactive power or anything else I said is just test me you arrogant prick.

What I have problems with is with your first senctence "All the current through a resistor heats it" because I have a fucking book in front of me that says the opposite. If you spoke my language and I had any camera with me I'd take a picture of it and post it.

[Anonymous No.1648366]

>>1648333
>>1648325
wrong.
reactive power the voltage and current are out of phase, heat is produced as power, out pf phase means when current is max, voltage is 0, voltage is max, current is 0.
therefore power is 0.
q.e.d no power in the cable therefore no heat.

this is what tesla was working on before they killed him, why do you think tesla coils are all inductors? wireless power doesn't need cables.

[Anonymous No.1648376]

>>1648366
wrong.
the question is about the current ampacity of the wire not reactive power. current flows through the wire and it has resistance. so the heat loss is I^2xR. think about how a transformer works.

[Anonymous No.1648377]

>>1648366

CURRENT AND VOLTAGE IN A RESISTOR CANNOT BE OUT OF PHASE.

goddamn where did you learn your shit?

[Anonymous No.1648379]

>>1648365
if it has pictures and formulas, the language doesn't matter. post it.

[Anonymous No.1648382]

>>1648333
>tasked
hope not. that sounds like homework right?
he just put it as
>We have
I hope that doesn't mean he is actually building that lol

[Anonymous No.1648402]

>>1648319
>"This metal melts at 100 Celsius, why do we need to keep the heat much less than 100 Celsius"

Boy are you one special guy

[Anonymous No.1648412]

>>1648365
>If you think I don't understand what reactive power or anything else I said is just test me you arrogant prick.

let me see if I follow: you can pass 1000000 reactive amps through a 100 ohm 1/8 watt resistor? cool.

[Anonymous No.1648420]

>>1648377
woah this is incredible, so any inductive or reactive loads with shitty power factor, we can just add a resistor and it will return the power factor to 1?
good to know good to know. oh, wires are already resistive? so they can never be out of phase is what you are saying. good to know good to know.
dumbass.

[Anonymous No.1648423]

>>1648420

I did not say that. Go back to EE 101 and learn how current and voltage are in a resistor. Here's a clue: it's called Ohm's law, and the current through a resistor will always be in phase with the voltage across that resistor.

jeez.

[Anonymous No.1648434]

>>1648366

I concede, you win. You can use a 0.1mm^2 wire for that 20kVA pure inductive load of yours.


Wrap the wire around your neck a few dosen times before turning on the power, the "reactive current" is definitely not going to blow your head off.

[Anonymous No.1648437]

>>1648319
Euro style schematic is better, than ANSI.

[Anonymous No.1648448]

>>1648420
>woah this is incredible, so any inductive or reactive loads with shitty power factor, we can just add a resistor and it will return the power factor to 1?
What you say is basically true. If you add a series resistor the power factor will approach 1 as you increase the resistance.

[Anonymous No.1648450]

>>1648448
>the power factor will approach 1 as you increase the resistance.
because you're measuring the power factor of the resistor, not the inductive/reactive component.

[Anonymous No.1648452]

>>1648450
The power factor of the resistor is always 1
Where the fuck are you from?

[Anonymous No.1648472]

>>1648402
>melts at 100 Celsius

[Anonymous No.1648478]

Sure, the load is inductive, but the cable still has a resistive component and will heat up.

[Anonymous No.1648480]

As I understand the inductor only moves the phase of the current.
>Then, if the load draws let's say 100 amps
>why do we need a wire with a size rated for more than 100 amps if it's not going to heat up?
Because the load literally draws 100A. There's 100A going through the wires.
>If I've understood it right it's just the reactive component of the current drawn by the load what turns into heat
I think you misunderstood something. I think it's supposed to mean reactive component as in resistor/resistance only heads up. The current is still real.

I forgot most of this shit so I can't give you a good answer.

[Anonymous No.1648526]

I Iove these threads.

[Anonymous No.1648603]

>>1648319
>inductive
The fact that it's inductive has no bearing on any of it. If there is current in the wire, it will heat up based on its resistance. Thicker wire has less resistance, therefore it will heat up less.

All power distribution systems have derating to account for safety margins. In your typical home, branch circuit will have a 15A breaker, but in commercial settings, if a 15A appliance is expected to run continuously (e.g. a large blender in some commercial kitchen) derating factors will cause you to install wire that could otherwise typically carry 25 amps instead of 15 (while still using a 15A circuit breaker).
14 AWG wire (15A) will become warm if you run it continuously at 15A and as such, the insulation will degrade prematurely (probably in the realm of years instead of decades) potentially causing shorts.

[Anonymous No.1648650]

The cable is a resistor. It maybe be a 0.10 ohm resistor but it's still a resistor and resistors draw more power the hotter they are. So if you push 20 amps through a wire rated for 2 at 75c it'll start to heat up faster and faster until it melts.

[Anonymous No.1648690]

>>1648650
>resistors draw more power the hotter they are
what are negative temp coefficient thermistors for $500

[Anonymous No.1648699]

>>1648319
The current passing through the wire heats up the wire.

[Anonymous No.1648770]

>>1648319
>>1648365
>>1648366

DON'T GIVE UP YET! Ignore all the haters who actually went to class.

[Anonymous No.1648794]

>>1648319
When calculating the heating of a resistive element you use a DC equivalent voltage (rms). When you're talking about 120v ac, the 120v is the DC equivalent voltage.

An inductor should be thought of as something that stores power then releases it when the magnetic field collapses, it's why the phase is shifted. The current is still flowing it's just delayed, therefore you still need a wire rated for 100amps.

[Anonymous No.1648796]

>>1648794
I should add that there is a thing called back emf that will resist current flow.
This happens in motors which are inductive loads.

[Anonymous No.1649004]

>>1648366
>the absolute state of /diy/

[Anonymous No.1649159]

>wrong

Hearty kek

[Anonymous No.1649160]

>>1648319
Imagine the power

[Anonymous No.1649168]

Interestingly enough I just ran into a similar problem at work. The M.E.s sized out a 10.5kW generator for a customer with a PF of 1. Now that they sent it out into the field the customer is complaining that they are getting a lot of voltage drop. They told me to check out the problem and after a lot of poking around about wire sizes and what kinda load the customer is putting on the generator it finally hit me that all of the load are inductive and that the power factor is most likely totally fucked.

[Anonymous No.1649170]

>>1649168
http://dansdata.com/gz039.htm

[Anonymous No.1649426]

Funny thread. Seriously though, I read that if you have too much reactive power as a consumer, your utility company may charge you more so you need something called "power correction" like a bunch of capacitors. I don't get this. Do you steal power by being reactive?? It is just stored and returned back, right? And your company says nice try, but you used that power to heat your oven or whatever, pay up? But if I return it back why should I pay for it? A restaurant down the road can re-use the same power for their oven and return it as well, right?

[Anonymous No.1649458]

>>1649426
I can't tell if you are joking.

[Anonymous No.1649664]

>>1649426
Ding ding ding! This thread has officially hit rock bottom now.

No the company charges you for Kvas. If it takes you more amps to perform the same amount of work because your pf is fucked and you don't wanna buy caps you get charged for the extra Kvas. It also heats up the cables and trips breakers when you hit the ampacity threshold of the circuits.

Here's a good analogy.

You go to the bar and you buy a mug of beer. The mug is the cable, the contents are the amps. The liquid is usable amps. The foam is amps that contribute to reactive power (as an example) . You correct your pf by adding caps, or in this case, reducing the head in the mug with a pinch of salt (or a salty finger) , for more beer. You always pay for the mug, but you are better satied with a full mug of beer. If all you get are half beer/half foam, you need 2 mugs to satisfy the same thirst. Hence why it's more expensive.

Pf is the concept that an inductive or capacitive load will shift the voltage and sinusoids out of phase with each other (more foam). U correct the pf to use the least amount of amps to perform the same work (pinch of salt, more room for beer) . There is such as thing as too much as well (too much salt or too much caps makes for overvoltage conditions). Most residences don't have to worry about this. Industrial sites with big services do because bean counters and high demand applications.

Poor troll/10 for making me respond. I was really enjoying watching jimmies rustle on this concept. But you took it to a whole new level with your reuse power bs. Fuck you.

[Anonymous No.1649667]

>>1649426
Ding ding ding! This thread has officially hit rock bottom now.

No the company charges you for Kvas. If it takes you more amps to perform the same amount of work because your pf is fucked and you don't wanna buy caps you get charged for the extra Kvas. It also heats up the cables and trips breakers when you hit the ampacity threshold of the circuits.

Here's a good analogy.

You go to the bar and you buy a mug of beer. The mug is the cable, the contents are the amps. The liquid is usable amps. The foam is amps that contribute to reactive power (as an example) . You correct your pf by adding caps, or in this case, reducing the head in the mug with a pinch of salt (or a salty finger) , for more beer. You always pay for the mug, but you are better satied with a full mug of beer. If all you get are half beer/half foam, you need 2 mugs to satisfy the same thirst. Hence why it's more expensive.

Pf is the concept that an inductive or capacitive load will shift the voltage and amperage sinusoids out of phase with each other (more foam). U correct the pf to use the least amount of amps to perform the same work (pinch of salt, more room for beer) . There is such as thing as too much as well (too much salt or too much caps makes for overvoltage conditions). Most residences don't have to worry about this. Industrial sites with big services do because bean counters and high demand applications.

Poor troll/10 for making me respond. I was really enjoying watching jimmies rustle on this concept. But you took it to a whole new level with your reuse power bs. Fuck you.

[Anonymous No.1649789]

>>1648452
If there is leading current or voltage and it passes through a resistor, it doesn't magically make them have the same phase... That's all

[Anonymous No.1649795]

>>1648448
yes that's how power correction is done: a large resistor in series with the inductive load. (large meaning on the order of megohms). or maybe in parallel, not sure.

[Anonymous No.1649799]

>>1649789
>it doesn't magically make them have the same phase... That's all

It isn't magic, it's reality. In a pure resistor the voltage and current cannot be anything but in phase. You quite simply do not fully understand impdedance and AC circuit analysis, which is why it seems like magic to you.

[Anonymous No.1649804]

>>1649795
>yes that's how power correction is done: a large resistor in series with the inductive load. (large meaning on the order of megohms). or maybe in parallel, not sure.

that might work in a tiny lab bench circuit, but when you have a factory full of motors you do not burn a million watts in a resistor, you use capacitors.

[Anonymous No.1649815]

>>1649667
>Kvas
It's kVA.

[Anonymous No.1649854]

>>1649815
I'll admit I was phone posting.

[Anonymous No.1649991]

>>1649799
I'm a different anon. Perhaps you should re-read the comment you replied too.
He didn't say pure resistance. He just said, that if there already is leading current or voltage, the resistor is not going to change that.

[Anonymous No.1650003]

>>1649991
Cont.
I was wrong, as is the another anon who posted. Learn something new everyday.

Polite sage.

[Anonymous No.1650301]

>>1649667
>No the company charges you for Kvas
No, they charge you for wattage not kva, as the meter doesn't register the reactive power. So if you have a poor power factor, they have to provide more power than they charge you for. But of course they can detect that and charge you a power factor penalty. Also if you have to use the beer head analogy than you need to pee in the empty mug, so your piss will be the reactive power you are returning back. that's considering beer was also piss as you are returning the original 'substance'. So if you think about it, technically the utility company doesn't provide more power, since the reactive power is returned back to the grid. But reactive loads create 'ebb and flow' surges that put higher strain on the distribution systems. So it is basically not about extra power but more about current surges.

[Anonymous No.1650317]

>>1650301
I receive 10 industrial high voltage bills at work. Believe me we get charged for kva if pf is low. Yes the reactive power stresses the lines but pf is not a 1:1 transfer back out the main transformer. It takes alot to get to that point for us. The utility here cares more about harmonics as our lines are beefier than our main transformers. I can show you cases where we could run 25% more equipment with cap banks because of the current so you're not wrong. Our service entrances are minimum 100MVA to 250 so overall pf is kind of moot at the entrance. In any case reactive power does noone any good and the beer analogy is for non QEWs, and an attempt to help this thread.

If you look at most commercial and industrial bills yes the rate is by kWh but the first 50kVA is free. Then comes the pf clause and the pf at on peak and off peak...

[Anonymous No.1650333]

>>1650317
>we get charged for kva if pf is low
So you probably have a reactive power meter installed.
>harmonics
well that's the different kind of the power factor that i don't really understand well. I was considering a linear inductive load that has no harmonics, just a current lag. In this case the reactive power is just additional current that is circulating back and forth without doing any work but strictly speaking the power is returned back to the grids, right? For example, I have this huge toroidal transformer that I use to melt wrenches when I am bored. It has a huge Bmax so I was able reduce the number of windings on the primary so it can draw up to 1 amp without saturating. Now imagine I have a 100 of these transformers each drawing 1 amp but with no load. It is just magnetizing current circulating back and forth as the magnetic field collapses. Now imaging I have 100 neighbors and all of them also have 100 transformers each drawing 1A. So we'd we drawing 10KA in our neighborhood and none of the meters will be able to register it!!! How cool is that.

[Anonymous No.1650575]

>>1650333
>So you probably have a reactive power meter installed.
Its industrial so ya.
>Harmonics
Harmonics are signals that come out as odd multiples of the base frequency due to switching noise from say vfds. Causes hell on the gear because skin effect. We filter up to the 13th. It's not pf.

BTW i dunno why you guys mix up import/export with pf. It's a 4 quadrant thing, not 2. Just because I'm experiencing shitty pf don't mean I'm exporting. If I'm exporting I'd be getting a nasty phonecall because we are leading, or another way to say it, the cap banks don't disengage on low load. Pic related.

With your example you'll only overload the lines and xformers. The meters will notice. The utility will be pissed. Don't piss off the utility.

Its been a while, and I'm half asleep. Correct me if I'm wrong.

[Anonymous No.1650717]

>>1648319
I'm an electrical engineer. What you asked makes no sense.

[Anonymous No.1650819]

>>1650575
I don't know, it seems like if you are leading you have an over corrected capacitive load? I am talking about a simple case of an uncorrected inductive load like a motor. I don't fully understand your diagram, but in the case of an inductive load, the reactive power is 'imported' I guess, then given back the next cycle. Does it mean it is both imported and exported? An uncorrected inductive load results in a higher current being drawn from the mains but part of that current is reactive so it flows back and forth without doing any work. So you may say an inductive load temporary 'steals' reactive power from the grid. But it wouldn't be registered by a conventional meter so I don't know how would the utility know? I guess their equipment may get overheated. That's one way to find out for sure. But they don't install reactive meters at residential drops. Speaking of the transmission lines, it gets trickier since an unloaded line acts like a capacitor, and as more current flows through it becomes more of an inductor. So I guess one may say that the transmission line 'provides' reactive power and consumers with poor pf consume it?

[Anonymous No.1650833]

wow like i never thought 4cha would be like this. here i am shitposting on endomorphs on fit, hegel on lit and inductive capacitors with electricians on diy, its so crazy, I even manage to disregard direction on biz and K!!!

[Anonymous No.1650834]

>>1650833
fag

[Anonymous No.1650839]

>>1648437
Nah, the squggly lines indicate the electrons have to slow down through multiple corners whereas the block indicates a fork in the road leading to better electron traffic. Of course one could assume that the block end junction causes a lot of accidents with the electron traffic and hence slows down flow as well so I guess both are valid.

[Anonymous No.1650842]

>>1648319
maybe this helps

[Anonymous No.1650856]

>>1648319
you are misainterpreting your sauce.

what that means is that load itself only produces heat for it's resistive portion.

the amount of heat the wire generates to transport that much current will still be the same, the wire doesn't care if it's an a/c motor or a BAMF of a halogen bulb, or an switched mode power supply