# Wavelength of a Power Wave

Discussion in 'Physics' started by rahulpsharma, Apr 15, 2012.

1. ### rahulpsharma Thread Starter New Member

Sep 5, 2010
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I donno if my question will make sense but still cant contain my curiosity to ask... I came across this on net and am unable to understand the physical significance of the following:

Consider a 230VAC wave having a Frequency 50Hz... Speed of Wave (since it is an electromagnetic wave) is c... Therefore, wavelength of Voltage wave is SoL/Freq = 6000 Km...

Now wavelength is the distance between two peaks or troughs... right...?? So the first Peak occurs at 1500 Kms... Zero crossing will occur at 3000 Kms.... First trough will occur at 4500Kms and the last Zero Crossing will occur at 6000 Kms...

Now what does this physically mean.... Suppose the source of power - an alternator, for instance - is at a location A.... A Load is at a distance of say 200 meters.... So does it mean that in its journey from the alternator to the load, the voltage would never peak at all (because we saw that the first peak would arrive at a distance of 1500 Kms...!!!)..

I know I am getting it all wrong but would request posters to pls help me understand it...

Thanks

2. ### steveb Senior Member

Jul 3, 2008
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It sounds like you are asking a question about the difference between a standing wave and a traveling wave. A standing wave (like a ringing note on a guitar string) has fixed nodes (a node is zero point in the wave) in space where the wave amplitude is zero. A traveling wave has nodes that travel along the path.

In any transmission line, if you have properly terminated ends that do not reflect the wave, the wave will be a traveling wave. For anyone standing at a fixed point, they would see the wave oscillate at 50 Hz, with full amplitude, as the wave travels by.

It is possible to set up a standing wave on a transmission line, by shorting out the ends. This causes a reflections and the forward and backward waves add up to make a standing wave. The same thing happens on a guitar string. The fixed endpoints are like shorts that force the wave to be zero at the endpoints. A string without fixed endpoints can have traveling waves. The following page from the AAC ebook explains this better than most other books.

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3. ### Papabravo Expert

Feb 24, 2006
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Your picture was one where the wave is only a function of position. A so called standing wave. Waves are actually functions of both position AND time. What you need to get your head around is what that means.

4. ### Wendy Moderator

Mar 24, 2008
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An electromagnetic wave is photons, RF or light. If it is in a wire it is not really electromagnetic, though a wire can convert one into the other, usually involving resonance. The pattern of the wave bouncing in the wire can create standing waves, which is also a form of resonance.

http://en.wikipedia.org/wiki/Standing_wave_ratio

It can also apply with dealing with a transmission medium like a coax cable. It has a lot to do with mismatched transmission lines and AC signal generation (which can apply to everything from power lines to digital signal transmission to RF transmitter). The transmission line, load, and source all need to be matched to the same impedance.

5. ### steveb Senior Member

Jul 3, 2008
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It's not clear what you are trying to say here. Can you clarify what you mean? You say a wire can convert one into to other: what one entity into what other entity? What is the wire converting to and from? If the wave is not electromagnetic, then what is it a waving in? If it has frequency, wavelength and speed, it must be a wave of some type, and it seems that only an electromagnetic wave is reasonable as the explanation.

Perhaps you are referring to an antenna? So, a wire can be an antenna which converts free space waves to guided waves (receiving antenna) and guided waves to free space waves (transmitting antenna). Electromagnetic waves that are guided are still electromagnetic waves, and the fact that they are guided by conductive or dielectrics structures does not change the fundamental nature.

Apr 13, 2012
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7. ### Wendy Moderator

Mar 24, 2008
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A electric signal inside a wire is not electromagnetism. This is a fundamental definition issue. It is electromagnetic when it becomes a photon, but an AC current in a wire does not qualify as an electromagnetic wave. This is the point I'm trying to make to the OP. Do you disagree with the above?

8. ### Papabravo Expert

Feb 24, 2006
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I do most emphatically. The medium does not change the physics only the properties. In this case the velocity of propagation. Maxwell's equations are independent of the medium. Currents produce magnetic fields. Alternating currents produce changing magnetic fields and changing magnetic fields produce electric fields. Your comments above are just a bit outside the mainstream. Maybe time for a refresher.

9. ### steveb Senior Member

Jul 3, 2008
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I disagree with some of the above. I agree that electromagnetism and photons are basically the same thing. But, I disagree that AC current in a wire does not qualify as electromagnetism because it is the very source of electromagnetism. AC current goes hand in hand with AC magnetic field, and AC magnetic field goes hand in hand with AC electric fields, and the interaction of the AC electric field and the AC magnetic field is the electromagnetic wave.

Why can't we have photons with frequency of 50 Hz? I go back to my previous question. If we have a propagation speed, a propagation wavelength and a propagation frequency, then we have a wave. What is the nature of the wave, if not electromagnetic?

Granted, the wavelength is very very long, and we may not want to consider it in analyses, but that does not change the nature of the physics. Over a long enough distance, the transmission is a wave-guided transmission with energy swapping between magnetic fields and electric fields (called an electromagnetic wave).

We may not normally consider the electromagnetic character in all 50/60 Hz power transmission situations, but the OPs question is specifically about the electromagnetic nature of a very long transmission line, of many thousands of kilometers.

10. ### Wendy Moderator

Mar 24, 2008
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While it is in the wire it is a current. When it leaves the wire it is a photon. You have standing waves from the electricity bouncing back and forth, but I have never heard this referred to as a electromagnetic wave. If it is in an antenna that is a transducer, which converts one form of energy into another, in this case an electricity into photons (AKA EM radiation). Like many transducers, antenna's are bidirectional.

SWR is caused by impedance mismatch, this is where SWR comes from. It is the currents alternating with the voltages that actually create the standing waves.

I am not a physics wiz, but I remember the basics of my college specialty.

I remember a story about when coax was new and a HAM wanting to try this new technology out (right after WWII). He had a weak concept of matching impedance, so treated the coax as a conventional wire for a relatively high power transmitter. Needless to say, the standing waves in this set up was horrible. Where the currents maxed out it exceeded the rating of the coax wire, so it melted the internal insulation from the heat. You could see the current waves as "goose eggs" in the coax where the insulation melted and expanded.

I mention this story because I think it relates to the OP's question.

It is important to note that if the impedances are matched in a transmission line there are no standing waves. They generally are a bad thing, except in antennas.

Standing waves in antenna's are the mechanism by which they work, and the antenna need not be perfect, though you actually get gain by designing it for ideal performance.

11. ### Papabravo Expert

Feb 24, 2006
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When an electromagnetic wave radiates from an antenna it is most definitely NOT a photon. It is just a wave.

12. ### steveb Senior Member

Jul 3, 2008
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In the classical theory of electromagnetism, we don't think about or talk about photons at all. They are just waves, as you say. However, we know from quantum physics (QED particularly) that electromagnetic radiation of all types are photons. This is part of the well-known wave-particle duality, of all things. So, gamma-rays, X-rays, ultraviolet, visible, infrared, microwave, radio, short waves, and below is all photons, and this includes 50/60 Hz wave propagation over a long transmission line also.

13. ### steveb Senior Member

Jul 3, 2008
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I'm still not following your underlying point. I'm not arguing with your practical knowledge, which is extensive. I'm just pointing out that fundamentally transmission lines are wave guides, and the waves they transmit are electromagnetic waves. The OP started with a discussion of wavelength, which implies a wave must exist. I ask again, what is the wave nature if not electromagnetic?

Your above statement is very confusing and imprecise. You say, "while IT is in the wire ..." and "When IT leaves the wire ...". What is the IT you are referring to? It seems to me that the IT is electromagnetic fields. The thing is that when a transmission line is long enough, it becomes the medium of propagation. So everything you are saying about free space waves (i.e those waves coming off the antenna) applies to a transmission line too. The waves are now guided, but they have all the characteristics of electromagnetic waves, including the fact that they are photons (if you prefer to think about that viewpoint).

Consider fiber optic cable, which is a waveguide for light. A coax cable is a waveguide for RF, and a power transmission line is a waveguide for 50/60 Hz waves.

I understand your practical point about the frequency being so low that over short distance, you can think in terms of circuit theory or network theory, but fundamentally, over 1000s of kilometers, the concept of electromagnetic waves is valid, and that viewpoint is the only way to get at the OPs question.

14. ### Wendy Moderator

Mar 24, 2008
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I suspect we will have to disagree. A twin set of wires is not a wave guide, neither is coax. A wave guide is not a standard transmission line. It is a different animal, used because using wires in any form is not practical at those frequencies. One transfers EM, the other electricity. "It" is pretty obvious, being I have mentioned EM ratidation. It is the same reason waveguide exists, as standard wire will radiate energy in the form of EM radiation.

I have quite a bit of experience from fiber optics too, but it doesn't apply here.

I also strongly disagree with EM radiating from a wire not being photons. What do you thing the mechanism is? Radio waves, which is the usual form of EM radiation, is also photons.

It is the fundamental concept of the duality of light, as both photons and as waves. EMR is the wave side of the concept, but they are still talking RF, light and other radiation. So when you flip the coin you are also talking about photons.

But this is off track. I agree with Steve about the basics of line frequencies. They typically do not worry about impedance matching because of the really long wavelengths involved. I got lost in the minutia of standing waves, which is indeed an electrical phenomena that is closely related to EM radiation.

There are losses for power companies, mostly resistive, but some radiated power. The latter is fairly minor, though I have no numbers on it. Theft of service laws exist because it is possible to couple into power lines with a simple coil. This is usually detected by power companies (they monitor their product very carefully) and dealt with.

Last edited: Apr 16, 2012
15. ### steveb Senior Member

Jul 3, 2008
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Well, yes, I definitely disagree with your definition of a waveguide. Here is a reference that talks about single wire, twin wire and coaxial waveguides. The terms "transmission line" and "waveguide" are basically synonomous.

http://www.st-andrews.ac.uk/~jcgl/Scots_Guide/audio/part6/page3.html

Anyway, terminology is just terminology, and I'm not worried about that. However, waveguides, transmission lines, or wires, ... they all can carry electromagnetic field energy and guide it along a path.

Last edited: Apr 16, 2012
16. ### Papabravo Expert

Feb 24, 2006
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To understand transmission lines and propagation it is not necessary to delve into quantum mechanics. Maxwell's equations do not require it. I know enough quantum mechanics to know that whatever insight it offers at very small distances must approach the classical limit at macro distances. Since we are dealing with power systems and very long wavelengths I fail to see how a quantum approach is either useful or valid.

17. ### steveb Senior Member

Jul 3, 2008
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It may not be useful, but it is certainly valid.

If we only care about what is useful, then we have to conceed that Bill is correct. For most practical cases, the wavelength is so long at 50 Hz, that you don't need to consider the electromagnetic effects. Circuit theory and network theory suffices.

I was only concerned about the OPs question. He is the one who brought up wavelength. So the question itself is about what is valid, and not so much about what is useful.

18. ### Wendy Moderator

Mar 24, 2008
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Useful, perhaps not, valid, definitely. It is always important to remember they are the same thing, it defines a lot of their properties.

We will have to agree to disagree on some points. The last sentence for example. I say electricity, you say electromagnetic field energy. Electromagnetic energy and electricity are not the same thing, just closely related.

My argument for the traveling wave guides is weak I have to admit, but it is closer to fiber optic than it is to coax. All of them have an impedance, which is not usually worried about in power lines, though the power company is definitely interested in maximum power transfer. Perhaps paying attention to power transfer and the impedance issue takes care of itself?

Last edited: Apr 16, 2012
19. ### Papabravo Expert

Feb 24, 2006
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So what exactly is the use or the validity of considering the transmission of AC power from a quantum mechanical perspective. I think you can understand everything you need to know from the perspective of classical physics.

20. ### Wendy Moderator

Mar 24, 2008
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To the OP, Have I confused the issue?