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Power lines also radiate em waves:

enter image description here

And the radiation should cause some energy loss.

How can we deduce logically whether the radiation loss for 50/60Hz power transmission lines are negligible? I know in low frequencies the loss is less but in this case the lines can be too long upto 2km. Is there a way to simplfy this and formulate the radiation loss roughly? Should we start modelling it like a 60Hz 2km dipole antenna?

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  • \$\begingroup\$ The loss is not negligible. People go to great lengths to reduce power loss (higher voltage & higher towers), but by no means is it negligible. Likely, if there was no loss, you electric bill would be cut in half. \$\endgroup\$ – st2000 Oct 24 '17 at 13:28
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    \$\begingroup\$ @st2000 You're talking about resistive losses, which are a completely different matter. \$\endgroup\$ – Dave Tweed Oct 24 '17 at 13:29
  • \$\begingroup\$ The problem with the losses is that they greatly depend on what they couple into magnetically. Essentially you could build a transformer with just some wires on the ground and the loss would differ by how much you draw from that transformer. Which is a reason why in most countries you are not allowed to power light bulbs by simply air wiring them when you live under transmission lines. \$\endgroup\$ – PlasmaHH Oct 24 '17 at 13:31
  • \$\begingroup\$ @PlasmaHH "air wiring light bulbs"? what is "air wiring"? wiring to where? sorry i didnt get what you meant. \$\endgroup\$ – user16307 Oct 24 '17 at 13:33
  • \$\begingroup\$ @user16307: take a neon tube and stand under a high voltage transmission line. It will glow. Optimize and extend that idea to get more out of it. \$\endgroup\$ – PlasmaHH Oct 24 '17 at 13:43
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There are many kinds of power loss associated with power transmission lines:

  • Resistive - due to the current in the wires heating them up, and also induced currents in nearby conductors.

  • Leakage - current flowing in places that it shouldn't, such as contaminated insulators, corona discharge, etc.

  • Radiation - electromagnetic waves carrying energy away from the wires.

The last one is the one you're asking about, and it is by far the least significant of the three.

No, you can't treat the wire as a "dipole". You have to treat it as an actual transmission line, because the spacing among the conductors and between the conductors and ground is very much less than a wavelength. This is what limits the ability to couple electromagnetic energy to "free space" — the electrical and magnetic fields are almost entirely canceled out right at the source.

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  • \$\begingroup\$ Ok so you mean we cannot relate the radiation losses or quantify it with a simple model. Btw the losses due to skin effect in your first category? \$\endgroup\$ – user16307 Oct 24 '17 at 13:44
  • \$\begingroup\$ "the electrical and magnetic fields are almost entirely canceled out right at the source." are you talking about the near field? \$\endgroup\$ – user16307 Oct 24 '17 at 13:45
  • \$\begingroup\$ the "almost" is part of the reason why you have transposition towers \$\endgroup\$ – PlasmaHH Oct 24 '17 at 13:46
  • \$\begingroup\$ Yes, skin effect increases the effective resistance of the wire, increasing resistive losses. Thermal losses in equipment such as transformers (winding resistance, eddy currents) also fall into this category. And yes, I'm talking about "near field", since the "far field" is thousands of kilometers away. \$\endgroup\$ – Dave Tweed Oct 24 '17 at 13:50

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