Electric wires and loss of energy

When electricity is lost when it is transported through electric wires what forms of energy does the lost electricity take? Heat, magnetic field or both?

If it is both do superconductors lose less of one more than the other?

Originally posted by Metal Brain
When electricity is lost when it is transported through electric wires what forms of energy does the lost electricity take? Heat, magnetic field or both?

If it is both do superconductors lose less of one more than the other?

Wire loss or line loss is through heat. Line loss occurs because of the voltage divider effect. The formula for the voltage drop across the wire is

Vloss = Rwire/(Rwire + Rload) * Vsupply

where R is resistance. The power loss from the wire is

P = Vloss*Vloss/Rwire

Superconductor means R in the 'wire' is very close to 0 - so Vloss will also tend to go to 0 and thus very little power is lost/heat dissipated.

Originally posted by SwissGambit
Wire loss or line loss is through heat. Line loss occurs because of the voltage divider effect. The formula for the voltage drop across the wire is

Vloss = Rwire/(Rwire + Rload) * Vsupply

where R is resistance. The power loss from the wire is

P = Vloss*Vloss/Rwire

Superconductor means R in the 'wire' is very close to 0 - so Vloss will also tend to go to 0 and thus very little power is lost/heat dissipated.

Superconductor means R in the 'wire' is very close to 0

Not quite; it is not merely "very close" to 0 but 0.
Superconductor means a material that can conduct electricity with literally zero resistance at least up to a certain threshold current.
A conductor with R being very close to 0 would be merely a very good conductor.

There is presumably some loss due to magnetic interference, but that would be the same whether the wire is superconducting or not. I suspect there would be more loss with AC than with DC.
I guess there may even be some gain, but not consistently.
I believe solar storms result in major fluctuations on the power lines.

Originally posted by twhitehead
There is presumably some loss due to magnetic interference, but that would be the same whether the wire is superconducting or not. I suspect there would be more loss with AC than with DC.
I guess there may even be some gain, but not consistently.
I believe solar storms result in major fluctuations on the power lines.

It is theoretically possible to give a superconductor cable shielding from external electrical/magnetic interference just by surrounding the superconductor cable core with a shield made of tube of superconductor and put an insulator between the two ( to prevent current going from the core to the shield ) .
If this is coupled with using DC rather than AC to prevent loss via emission of electromagnetic radiation ( usually radio waves ) due to fluctuations in current that comes with AC, such a superconductor cable should be able to give zero loss at least in theory.

However, even with this set-up, there probably will still be some loses albeit minute loses and, in this case, the loses would be due to small amounts of background ionising radiation leaking into the superconductor core ( which would be very hard to block completely ) of the cable where it would break up some of the Cooper pairs of electrons in the superconductor thus slightly hindering the otherwise totally smooth uninterrupted continuous flow of electrons.
But such a lose would be so tiny as to be pretty much academic.

Originally posted by SwissGambit
Wire loss or line loss is through heat. Line loss occurs because of the voltage divider effect. The formula for the voltage drop across the wire is

Vloss = Rwire/(Rwire + Rload) * Vsupply

where R is resistance. The power loss from the wire is

P = Vloss*Vloss/Rwire

Superconductor means R in the 'wire' is very close to 0 - so Vloss will also tend to go to 0 and thus very little power is lost/heat dissipated.

Not quite. It does lose through heat but if you have a long transmission line running 50 or 60 hertz, that means it is an antenna and a good fraction of it is actually transmitted as 50 or 60 Hz radio waves. That is a radio wave with a wavelength of over 3000 miles, close to 5000 Km. If you have a wire length of 1500 miles, that represents close to 1/2 wavelength and if precautions are not taken ( you can engineer out most of that radiation) you would lose a lot more than the 10% now lost in long distance lines. If you wanted to radiate that energy you could transmit more than 90% of it as radio waves. Fortunately you can prevent most of that loss by having two lines close together, one the + and the other the - (AC but one will be the opposite polarity), then it becomes a transmission line and not a whole lot is lost but some still leaks out.

Some energy is also lost to corona discharges off dirty insulators which, at 1 million volts for the big guys, is quite a problem. Corona discharge is where the electrons spit themselves right off a conductor if the voltage is high enough and the radius of curvature of the conductor is small enough which in this case could be a damp piece of dirt sticking out of a high voltage insulator where the voltage gradient builds up higher than the insulator can hold back, huge local electric fields literally accelerate electrons right into the sky.

When that happens and you are nearby with an old AM radio you can hear it hissing and buzzing for miles around. They fix that when the find it by taking a helicopter and spraying the insulator with some cleaning liquid. A very dangerous job as you can imagine. The pilot is flying the thing very close to the lines and a long nozzle dangles out of the chopper and sprays the insulator to clean it.

Not a job I would lust after.....

Originally posted by sonhouse
Not quite. It does lose through heat but if you have a long transmission line running 50 or 60 hertz, that means it is an antenna and a good fraction of it is actually transmitted as 50 or 60 Hz radio waves. That is a radio wave with a wavelength of over 3000 miles, close to 5000 Km. If you have a wire length of 1500 miles, that represents close to 1/2 wave ...[text shortened]... es out of the chopper and sprays the insulator to clean it.

Not a job I would lust after.....

Heh, I knew that. At least, I knew the radiation bit. 😛

I once worked on some PCBs that had 1GHz clocks. The wavelength was about a foot, which meant we had transmission line effects to contend with. Good times. 😀