# Superconductivity - Current returned to power station

Discussion in 'Physics' started by Skeebopstop, Feb 17, 2009.

1. ### Skeebopstop Thread Starter Active Member

Jan 9, 2009
358
3
I was laying in bed last night and an interesting question popped into my head.

"What would happen if I could give current a lossless return path to the power station:

1. Be charged a kilowatt hour rate? I cannot imagine how given I have not dissipated any power, however perhaps the power company only measures your current throughput and assumes the power rating to be at 220VAC*Irms.

2, The only way I could return the current without dissipating it as an equivalent power, would be with a superconductive wire. This would, if it were possible, essentially 'short circuit' a phase on the generator and, if my thinking is correct, force one of the phases (I am assuming to do this on a single phase) to 0V and as such, bring that phase out of operation for the rest of the network."

O how I love laying in bed thinking of this stuff. Leave me alone mind!

Last edited by a moderator: Feb 17, 2009
2. ### b.shahvir Active Member

Jan 6, 2009
444
0
Dear James,

I am pleased to know we share some similarity. Most of my questions/technical thoughts or solutions to them pop-up in my head when I am in bed ( occassionally, I also benefit from time spent in the restroom)

Coming to your thought, theoretically lossless line would mean zero resistance....resulting in infinite current....resulting in infinite magnetic field! Practically though, the magnetic field around a superconductor would be finite but massive. This would destroy your energy meter as well as the utility company switchgears/equipments.

The thought being theoretical, a theoretical Energy meter would not function since current thru voltage coil or circuit would be zero due to zero voltage across it (short circuit loop); the current coil alone cannot contribute to measurement of the electrical energy consumed. Hence, it would be impossible for the utility company to charge you for electrical energy consumption (whether theoretical or practical) as it cannot be measured!

Kind Regards,
Shahvir

3. ### Skeebopstop Thread Starter Active Member

Jan 9, 2009
358
3
O the woes of the restroom. Luckily I keep it well stocked with interesting magazines so I needn't think of such things as this during my 10 minutes of bliss!

So to the other question, if all current is returned to the generator, it would in effect 'short' that phase of the power network. However, lets say the distances are long enough such that by the time it takes to return through the superconductive neutral line to the generator, it would be somewhat out of phase.

How can one reason through the implications of returning all of the current back through the neutral line with no losses (i.e. voltage drop).

4. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
What about power dissipated by the load?

5. ### BillO Distinguished Member

Nov 24, 2008
985
136
Yeah, my question exactly. Are you just shorting this thing out or is there a load on the other end, like a city, or something?

BTW, it does not matter whether your distribution lines are lossless or not, or you have a load or not, all the current that goes out comes back in. You are aware of this, right?

Last edited: Feb 18, 2009
6. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Yep, if your meter spins the rotor, you get a bill.

7. ### Skeebopstop Thread Starter Active Member

Jan 9, 2009
358
3
There is a load in that the rest of the city, however on 'your' setup, you do a lossless short circuit.

8. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Then the step down transformer feeding your home explodes, unless it's equipped with a circuit breaker.

9. ### b.shahvir Active Member

Jan 6, 2009
444
0
I there is load (city).... meter spin and you pay for juice you consume.....if no load and line considered lossless, meter not spin due to 'zero' volts across pressure coils.... bid adieu to utility company switchgear & your electrical installations .

Sometimes its good to realize concepts in theory only!

10. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Actually, it would be the other way around. No meter spin indicates no current flow. Voltage would still be there, as with any open circuit.

11. ### b.shahvir Active Member

Jan 6, 2009
444
0
In absence of voltage coil amps, current coil alone cannot maintain meter spin, whatever be the strength of magnetic field by current coil.... meter disc may crawl but not spin indicating much less power consumption.

12. ### Skeebopstop Thread Starter Active Member

Jan 9, 2009
358
3
Lets focus on the main question than. What happens to the generator, ignoring transformers in the way etc.., if I am able to short circuit its line voltage via superconductivity. Lets imagine it being over long enough distances that it won't be 'instantaneous' return of the current.

Consider only one of the three phases for now.

I visualize a massive bus brake, but will this bring the generator to a halt? Or will it just neatralize any voltage on that phase, so the rest of the network now sees 0V on that phase?

13. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
We are talking about an open circuit, yes? There is voltage across an open circuit, but no current flows in an open circuit. The meter disk will not turn.

14. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Do you mean shorting the load? Presumably, an over-current protection device will do it's job.

15. ### Skeebopstop Thread Starter Active Member

Jan 9, 2009
358
3
Anyone have any input to this?

16. ### b.shahvir Active Member

Jan 6, 2009
444
0
I think your concept of electrical current is something which comes and goes like a railway train. This is not the case...electrons are basic constituents of electrical current and they are present everywhere.

Please keep in mind when i say electrical current, it does not mean that electrons shoot out from the negative terminal of the generator.... then travel all around the world and then comes to the positive terminal of the generator after some time delay. this visualization is incorrect.

All the electrons being present everywhere move at the same time and hence practically there is no time delay between electrons shooting out of negative terminal or the electrons being absorbed by the positive terminal. Only thing which involves time is the feeling of voltage or EMF wave if it starts from say point A of the transmission line.... and if the line is sufficiently long reaches point B after a finite amount of time delay. the Emf wave travels at the speed of light and the time delay maybe due to line inductances (non-ideal conditions).

Coming to your point of its effect on generator; practically, voltage drop will occur only if an impedance is present in the line. but the line being lossless (theoretically zero impedance, hence rest of the network would see 0 volts), infinite current will flow, infinite magnetic field will develop, which will destroy the generator right away due to massive electro-mechanical stresses! (forget braking of generator).
I hope this helps, matey.

Last edited: Feb 20, 2009
17. ### BillO Distinguished Member

Nov 24, 2008
985
136
It would act to load the generator and more torque would be required to turn it. If we are talikng about a real generator that for some reason had no over-current protection, the short circuit would probably exceed the current capacity of the windings causing them to melt.

edit: BTW, in any normal configuration the other phases would be affected. There are only 3 lines used in normal power transmission. If you'd like to think of it this way, each of the phases must carry portions of the supply and return and are 120 degrees out of phase. This relationship would be broken.

Last edited: Feb 20, 2009
18. ### Skeebopstop Thread Starter Active Member

Jan 9, 2009
358
3
yes, for simplicity I only considered a single phase of the 3 phase transmission network.

Shahvir: I feel it good to remind myself of this. I was aware but hadn't let it re-enter my mind for quite sometime and perhaps should keep it more permanently.

The question then becomes what happens first. I am going to assume that the load to the generator is 'limited' to a point where it will never 'burn' out the windings. Thus if you will, the generator windings are also superconductive.

I realize it seems like I keep superimposing new rules on this, I guess it is difficult to probe for what I am looking for!

To re-iterate my question, assuming non-realistic conditions, if I did manage to sustain a short without blowing up the generator, would I stall the generator? I know shorting the 'phases' acts to brake, but what if you short just 'one phase' to neutral?

19. ### b.shahvir Active Member

Jan 6, 2009
444
0
Assuming non-realistic conditions.if the short does sustain without blowing up the generator, the generator will come to an extremely violent STOP (brake) even if you consider one phase to neutral short of a 3 phase generator.

The reason being the one phase winding will produce massive AC magnetic field (though pulsating in nature and not revolving) which is sufficient to magnetically brake the generator to an abrupt but destructive halt!
If you need a more detailed explanation please let me know.

20. ### Skeebopstop Thread Starter Active Member

Jan 9, 2009
358
3
So how would this correspond to load imbalances on a 3-phase generator?

Lets say for example, the L1 phase is loaded 4x as heavily as L2 and L3.