Why high voltage for power transmission?

As you increase the voltage, more people get fried trying to steal power. This decreases your loses.

Luhan

If the entire power distribution system operated at 120/240V the conductors would have to be several feet thick!!!

P = I*V and I = V/R V = IR

So

P = I^2*R

or

P = V^2/R

So the second power dissipation equation is reduced by a factor of R while the first is multiplied by it. This helps you with the idea if you want to send X volts or X amps along the line with some fixed R. Always choose X volts because you will be wasted less power. i.e.

P1 = X^2*R P2 = X^2/R

Then P1 = R^2*P2

i.e., you will be wasting R^2 times the amount of power by choosing to send X amps instead of X volts. for R > 1 this is huge.

e.g.,

Say you want to transfer power of 100Watts. You can do this by choosing any I and V such that I*V = 100. Or I = 100/V.

choosing a real large number for V, say, 1000, gives I = 100/1000 = 1/10 A and the power we are trying to send is 100W(obviously, because this is what we started with). Or we could choose I = 1000A while V = 1/10V.

Now, we have a dissipation in the line so what is our actual power at the other end, say, of a 10Ohm line?

100W - 1/10A*10 = 99W

VS

100W - 1000A*10 = -99900W

i.e., in the first case of using 1000V we only lost 1W and in the second case we lost all of our power.

Why does this happen though? You have to look into why ohms law exists. It has to do with how electrons flow in wire. As electrons flow they are bumping into atoms of the wire.

An analogy is like a freeway. If you just have a few cars going really fast then its usually not a problem... but what happens if you have a lot of cars going slow? Current is like the number of cars and voltage is there speed. (and just note that they are inversely proportional) If you want a lot of cars then they all must go slow and going slow makes people mad(power dissipation). If you just allow a few fast cars then no one is mad. The total power behind the cars is like momentum. If you have 10 cars going 1 MPH or 1 car going 10 MPG then you still have the same momentum. But we loose a little momentum from people getting off the freeway from being pissed because its to slow... i.e., electrons hitting atoms and loosing energy to that atom creating heat.

The idea is simple though. Sending one electron through a wire at a high speed is much less likely to collide with an atom. Sending a large amount of electrons through a wire, even at a very slow speed will have many more collisions. The reason has to do with the interaction between the electrons themselfs too. Its kinda like a whole crowd of people trying to walk over a man hole but avoid falling in. Its much harder to do the more people you have. If you walking towards it and you think you can jump over it but get bumped in the process you might fall in.

It's scales with the square of the voltage *dropped across the transmission line*, which is (hopefully) quite small compared to the voltage dropped across the load. To avoid this confusion, people normally talk about "I-squared R" losses in transmission lines, since the same current flows through the transmission line as through the load.

As such, if you double the input voltage, the current required by the load for the same delivered power is halved. Therefore, the power lost in the line -- I^2*Rline -- has been quartered.

This is why high voltages are used for power distribution.

---Joel Kolstad

That's spelled "losses."

--- Power goes as the square of the current.

View in Courier:

100V / 1KV / 10A--> / +--------+ +-----+ / 1A--> R1 +-----+ | | | ~|--------[0.5R]---------| |-----+ | | | | | [10R] | ~|--------[0.5R]---------| |-----+ | +-----+ R2 +-----+ | | GEN XFMR +--------+ LOAD

R1 and R2 represent the transmission line impedance, so with a 1kV source pushing 1 ampere through the line the loss is:

P = I²(R1+R2) = 1A² * 1R = 1 watt

With a 10kV source pushing 100mA through the line, however:

100V / 10kV / 10A--> / +--------+ +-----+ /0.1A--> R1 +-----+ | | | ~|--------[0.5R]---------| |-----+ | | | | | [10R] | ~|--------[0.5R]---------| |-----+ | +-----+ R2 +-----+ | | GEN XFMR +--------+ LOAD

P = I²(R1+R2) = 0.1A² * 1R = 0.01 watt,

Which is a hundred-fold decrease in loss for a ten-fold increase in line voltage.

-- John Fields Professional Circuit Designer

No. Your losses are primarily through resistive causes, and they are proportional to the CURRENT (not voltage) flowing through the wire. So you want to minimize CURRENT in the transmission line. Power is the product of voltage times current. Use a transformer to step voltage UP at the generating station. This allows the amount of CURRENT pumped down the wire to be LOWER, by the same ratio that you multiplied voltage. Now move your electric power several hundred miles. At the receiving end, reverse the voltage change, and voila! You've got power ready for toasters, computers, etc.

Of course, there are losses in these transformers. But you expect these to be small, compared to the power you saved along the transmission path.

Saves on COPPER (or aluminum) too, and that is a big deal.

I believe this AC transformation process was the undoing of Thomas Edison's foray into the electric power business. George Westinghouse promoted AC, while Edison promoted DC. So Westinghouse "saw the light," while Edison didn't, notwithstanding Edison's most famous invention. I'm sure some historian will set me straight, if I got this wrong.

--
That\'s only true if there\'s no transformer or a 1:1 transformer
between the line and the load.

Tank you.

Luhan

Wrong, since the number of electrons in a wire doesn't change. Wires are like pipes which are always full of water, and the "water" can flow at different speeds, but the number of water molecules in the pipe

remains the same.

TRANSMISSION LINE ANALOGY:

An electric transmission line is like a long circular drive belt wrapped around a pair of drive wheels. But this drive belt has high friction. (Imagine that the belt is rubbing on the ground.) If we spin the drive wheel fast, the belt heats up from all the rubbing, and lots of energy is wasted.

To solve the problem, we can put gear boxes at either end, and gear the belt down to low speed (but at higher belt tension.) Now we can transmit energy at the same rate along the belt as before, but the belt moves very slowly and with immense pressure.

Voltage is like pressure difference. Current is like speed. And power transformers are like gearboxes.

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Current is like the number of cars and voltage is there speed.

Thanks for the nice diagram John. I think that the fault in my reasoning was to assume that a higher transmission voltage meant a correspondingly higher line current. After all, if you increase the voltage into a fixed resistor, the current goes up.

So ... why doesn't the current go up? Is the impedance seen at the generator lower for a higher voltage, due to a different transformer?

Stephen

Ya.

It's convienient to think in terms of load power. Then, current is whatever the power is divided by the voltage you're running at. I mean, as the power company, your customers demand some amount of power, day in, day out, for the most part.

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

I think he actually meant "losers."

John

Too bad Mr. Rockefeller didn't get behind Mr. (Dr?) Tesla - we'd have had wireless power transmission a century ago. )-;

Thanks! Rich

huh. Do you know what current is? Current is the measure of the number of electrons passing through a cross section of wire per second. 1 Columb of current per second is 1 amp.

So the number of electrons in a wire does change depending w.r.t to current.

Wires are not like pipes which are always full. Even if they were you can compress water and hence change the density => changing the number of molecules of water in the pipe.

With your (wrong) logic, if I filled a pipe full of water on the moon then I'd have EXACTLY the same amount of water molecules in a pip that was filled under the atlantic ocean. This isn't true because the densities are different.

The fact is that you can sent a few electrons or many electrons in a wire and heat is generated from friction generated when electrons collide with atoms in the the wire. This reduces the energy of the electron by an amount gained from the atom. It increases the vibration energy of the atom and on a global scale we sense this as heat which is felt from radition given off by the atoms as they stop vibrating.

{snipped}

Now I understand transformer oil.

-- john

Hi John,

As far power into the lin goes, the loads can just be reflected back through whatever transformers exist so that sooner or later the *effective* load current equals the input current.

I purposely simplified the discussion since the OP didn't appear that experienced. Additional simplifications included ignoring the phase of everything that's going on...

The number of electrons in a wire depends on its voltage w.r.t. another wire. The number in the pair of wires stays the same, but the negative wire will have a slight excess of electrons over the positive one. This will depend on the voltage between the wires and their spacing, and will be independent of the current through them. If you have a pair of wires at 1 cm apart, charged to 100V, they will have an imbalance of electrons of about 34.7 billion per meter. For 12 ga. wire, the number of electrons in the wire pair is about 1.6E25 per meter, so the imbalance is really microscopic: 1 part in 5E14. This is true for a charged pair of wires, whether there is zero current through them or many amps. There'll be a change in number of electrons due to a current that causes a change in voltage, but only in the tiny amount needed to charge the wires' capacitance to the new voltage.

-- john

LOL! But what about ferro-resonant transformers? I don't see the ATF!!

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms