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The Powerline Ekibastuz–Kokshetau in Kazakhstan holds the record for having the highest operating transmission voltage in the world, running at over 1 megavolt. Why did they choose to deliver energy this way?

EDIT:

If higher voltage means thinner wire can be used for transmission, why doesn't the rest of the developed world operate at transmissions this high?

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While I believe a question asking "why transmit power at high voltage?" is too simple for this site, the answer to "why does the Ekibastuz–Kokshetau powerline run at a higher voltage than other HV lines?" is interesting and not straightforward. Perhaps the latter is indeed the intended question. –  tyblu Mar 11 '12 at 0:17
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4 Answers

up vote 15 down vote accepted

The design of powerlines is a complex matter, in which many decisions overlay.

The Powerline Ekibastuz–Kokshetau is a relatively recent build, finished 1985. There were two more lines spawning off it, one towards Moscow which is now driven with 500 kV, the other was dismantled.

It is connected to a large power plant that was built at about the same time.

It runs a long distance through a relatively empty area.

One can assume it was the prototype project for the idea of electricity distribution on scarcely populated areas in the soviet influence sphere.

What would influence an electricity provider to build a 1MV power line?

  • Build a huge power plant (not happening often)

  • In an area with low population density (not many people complaining about the build)

  • Not having a distribution network in place (only happening in the so called 2nd world)

  • Needing the power elsewhere (Ekibastus Plant is 4GW, power line is 5 GVA)

Simply put, anyone else who might need a 1MV power line, had something else built before it was economically feasible to build 1MV lines. Seeing the Moscow branch of this particular line being run at 500 kV despite being designed for 1MV says something about that.

So, if a 1MV power line gets built again, it might first be in Argentina or Brazil. But only if they decide to build huge power plants in places where most of the electricity is needed elsewhere.

Also, a lot changed in power plant technology in the 20 years since then. Smaller plants are more feasible, solar and wind technologies are finding their place. Today, a town like Kokshetau would get a medium-sized plant, and be done. Megaprojects to transport electricity are not needed much anymore.

I suppose the power line is the quirk of a 5-year-plan, really. If so, it was meant to be the beginning of a massive power distribution system for the rural parts of the influence sphere. But before more could be built, the system collapsed.

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Power is current times voltage (\$I\times V\$). The power loss in a wire is \$I^2\times R\$, so, if you increase the voltage by a factor of, say, 4, the current required will be reduced by a factor of 4 and the loss in wire will be reduced by a factor of 16.

I assume the power line is really long, so using higher voltage means thinner wire can be used. This is one of the main reasons why AC won the current wars - back then there was no easy way of stepping up/down DC voltage.

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Right, I have clarified the question above. I was really wanting to know why the rest of the world doesn't operate at this voltage as high as in Kazakhstan. Sorry for not being clear. –  Gabriel Fair Mar 11 '12 at 6:10
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Basically there're two factors. As voltage gets higher the current gets lower and losses get lower and that allows for thinner wires. On the other hand as voltage gets higher better insulation is required everywhere - posts have to be higher (so that no discharge into the ground happens), distance between the wires needs to be greater, and much better insulation is needed in the transformers at the ends of the line. So raising the voltage reduces transmission losses and wires cross-section, but induces lots of problems with high voltage itself. That's why the actual used voltage is a tradeoff - high enough to not lose too much energy as heat and not tool high so that the system can be manufactured and run.

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Understanding why there is such a voltage is simple, if one takes care to what we are speaking of.

An answer

An answer could be : Power loss is \$I^2\times R_{wire}\$. So for a constant transmitted power \$V\times I\$, by increasing voltage we decrease \$I\$, thereby diminishing loss.

Ok, but did we not just hide our \$V\$ variable ? Indeed, by Ohm's Law \$V = I\times R\$, so the power loss is also \$V^2\over R\$.

So did we actually did worse by increasing the voltage ?


What does it mean to dissipate power \$I^2\times R\$ ?

  • It first means that the cable, by its nature, resist to the flow of electrons. its electrons like to be in a state of equilibrium and don't like to be pushed by new entrants
  • It also says that if, quite naturally, we say that the dissipated power is = the amount flowing, \$I\$ * some force \$F\$ to be overcome, then that force is proportional to the amount flowing itself : the more there is, the more the force will be strong. Of course, we can give that force a name, and that is precisely the voltage between the cable extremities.

When you think of it, it is not surprising that the dissipated power is quadratic. If you have very big cable, then it would make sense that dissipated power is linear. You pay a constant price for each electron that comes in. In a smaller cable, the cable becomes saturated and its capacity to accept new electron diminishes.


Putting it all together

Having said all that, it is quite clear what the error of the naive reasoning is : we were using the voltage between the ground and the first extremity of the cable. but the only quantity that makes sense is the voltage across endpoints of the cable.

Another view on this, is that every time you speak of a voltage, you have to know not only the amount of Volt it has but also the 2 points it refers to. They are part of the definition. In itself, a tension of 10 Volts has no physical meaning. A tension of 10 Volts between point A and point B, on the contrary, does have a meaning.

Going back to the problem, by increasing the voltage between the ground and the 1st extremity of the cable, we need a lower intensity to transmit the same amount of energy to someone else, who will take take this current and consume it at ground level voltage.

Conclusion

That lower intensity yields to a lower dissipated energy \$I^2\times R = I\times V_2\$ in the cable of resistance \$R\$, where \$V_2 = I\times R\$ is the voltage drop across the cable.

An equivalent way of seeing this is that it will induce a lower voltage drop between the central and the consumer.

The limit is that you need to have special equipment. At one extreme if the tension so too high, the electron of the air itself will be pushed around, and electrical discharge (aka a "plasma") will be created.

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Although it's not practically useful in most contexts, one can meaningfully describe an absolute voltage of a point, referring to the potential difference between that point and a point an infinite distance away in the void of space. Such a measure would be equivalent to expressing height as distance from the center of the earth. If the top of a person's head were 41,852,012'7" from the center of the earth, and the feet were 41,852,006'9" feet from the center, the person would be 5'10" tall. Of course, while one could determine a person's stature that way, relative measurement is easier. –  supercat Mar 16 '12 at 15:12
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I think that the confusion in the first paragraph is due to the fact that you use the same R for two different things: one is the resistance of the isolation, the other is the resistance of the wire. –  clabacchio Mar 23 '12 at 9:42
    
@supercat good point –  nicolas Oct 3 '12 at 12:50
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