I left out skin effect. Which may or may not be significant.
Sorry - off on the other boards for a bit.
gazpacho No, I am not implying that there are no resistive losses in an AC transmission line, but that they appear as current losses, not voltage losses. Current is not lost in DC systems because there are no shunt components in a point-to-point transmission line. But the parallel capacitance of the tranmission line allows AC current to return short of the terminating element.
Crafter_man, still looking for cite. In my attempt to explain transmission lines without using jargon like resonance and phase margins, I may have oversimplified to the point of confusion to an EE.
If you read to the end of the link I provided, it describes how most models omit completely the resistive element of the transmission line, because for a properly tuned and terminated line, the reactive components dominate completely. It’s a suitable approximation for all but the longest transmission lines.
And even in those, the inclusion of a resistance element is not the same value as the linear resistance of a DC transmission wire. I recall this being glossed over as the “simple answer” in my undergrad transmission theory course. This hasn’t been so easy to find a cite for, but the web is a big place.
I didn’t mean to imply that there was any overall gains using AC, just that at the time the power distribution systems were invented, the voltage losses were much less than using DC, allowing much further transmission over practical distances using practical voltages. This isn’t the only tradeoff point, but it’s a very important one, and in the histories I’ve read, the one that shot DC down in the early battle of AC vs DC. (cite)
A lot of the cites I’m finding point to multiple point generation and distribution more difficult with DC due to the lack of a DC equivalent to a transformer, which was basically gazpacho’s’s point, and some cite both. But Edison was confident he could overcome this scheme by building local generators to serve neighborhood-scale markets.
The efficiency that makes a practical difference, in the end, is cost efficiency, and the ability to centrally generate power and distribute it cost effectively to rural and suburban customers won out in the end.
Not necessarily. If energy is lost, the available power is reduced at the load end. This will appear as a voltage loss in a single-mesh DC loop, due to Kirchoff’s Current Law. But in an AC circuit, the loss can appear as a phase shift between the current and voltage, or as a reduction in voltage, or in current, depending on the specifics of the source, load, and transmission line.
You cannot apply the simple rules of DC circuits to AC transmission lines. It just don’t work… and it’s not nearly as intuitive.
If you imagine DC electricity as water thru a pipe you gain some useful insight, but AC is more like sound transmission thru your plumbing. It’s nothing like the flow of water in your tap, and more like the weird noises your pipes make when the taps are running.
A current loss? If there is current flowing in the conductors there will be a voltage drop across the line. There is a current flowing in AC power lines. Your explanation of losses due to the capacitance only make AC power lines worse than DC power lines.
bughunter, I’m having difficulty accepting your explanation of voltage relationships as having any impact on the choice of AC over DC. The Ferranti effect wasn’t even known about until after a fire at Ferranti’s Deptford power station.
The Ferranti effect doesn’t rely on using a distributed parameter model, either. It works out just fine using a lumped parameter model.
Say you have a transmission line of length x. It has a series inductance of L per unit length, and a shunt capacitance of C per unit length. When the line is not supplying any load current, the receiving end voltage is given by:
V[sub]R[/sub] = V[sub]S[/sub] / (1 - 0.5 [symbol]w[/symbol][sup]2[/sup] L C x[sup]2[/sup])
Note that V[sub]R[/sub] is necessarily greater than V[sub]S[/sub]. As the load current is increased above zero, the receiving end voltage falls because of the voltage drop across R and [symbol]w[/symbol]L. For high voltage systems, [symbol]w[/symbol]L is about ten times larger than R, meaning that voltage regulation is much worse on AC systems than on DC systems.
At some particular level of load current, the receiving end voltage will equal the sending end voltage, and this is the situation you are describing. However, transmission lines are operated over a wide range of loading levels, and it would be pure coincidence if I[sup]2[/sup][symbol]w[/symbol]L equalled V[sup]2[/sup]/[symbol]w[/symbol]C at any particular time.
What really happens is that as the loading level on the line changes, shunt inductors and shunt capacitors are switched in and out of service to counteract the piss-poor voltage regulation of the AC line.
All I know is that the biggest west coast electric grid backbone uses several DC lines. The 845 mileInter-Tie lCelilo (OR)-Sylmar (CA) line, uses 800-kV DC.
I have a slight tendency to believe that the EEs who figured out that sending all that juice was best done with DC weren’t idiots. (I was good friends with two sons of the then BPA head in college. They were smart as could-be. Not so sure about the father.)
In short: DC is better if you have the tech but it costs more so the longer the link the better.
The Ferranti Effect? Never heard of it. You’re the man, Desmostylus. 
So it would appear that, if certain conditions are met, the load voltage can actually be higher than the source voltage. Fascinating.
But this does not necessarily mean more power will be available to the load. All else being equal, is it still be true that a DC transmission line is less lossy (i.e. more efficient) than an AC transmission line? I would think that this would be true even with the Ferranti Effect in play…
Yes. There’s no charging current, so more load current can be carried for a given conductor size. No charging current also means less I[sup]2[/sup]R losses for a given load current.
There’s no losses cause by induced currents in parallel conductors (e.g. earth wires, fences, telcoms cables).
No EM-radiation losses, either.
You also have smaller towers and narrower easements (right-of-ways), because the peak voltage is the same as the rms voltage.
It doesn’t say that at all.
Don’t worry about cite for this.
Crafter_Man has already mentioned skin effect, which serves to increase R.
There’s also another factor. To accurately model a long transmission line using lumped equivalents, it’s not sufficient to simply take the R, L and C per unit length, and multiply by the total length x.
The R and L also have to be multiplied by a correction factor:
sinh([symbol]w[/symbol]x√(LC))/([symbol]w[/symbol]x√(LC)),
whilst C has to be multiplied by a correction factor of
tanh(0.5[symbol]w[/symbol]x√(LC))/(0.5[symbol]w[/symbol]x√(LC)).
But, unless x is very large, those correction factors are very, very close to 1, meaning you can safely ignore them.
What’s “very large” in this instance? Transmission lines can run into the hundreds of miles…
In this case, “very large” happens to be hundreds of miles.
The wavelength in air at 60 Hz is 5,000 km. 500 km is 1/10 of a wavelength, and at this point, the effect becomes noticeable.
Let’s try an example.
[symbol]w[/symbol]L = 0.3 [symbol]W[/symbol]km[sup]-1[/sup]
[symbol]w[/symbol]C = 3 x 10[sup]-6[/sup] Skm[sup]-1[/sup]
The factor sinh([symbol]w[/symbol]x√(LC))/([symbol]w[/symbol]x√(LC)) for different values of x is tabulated below:
x (km) factor
50 1.000
100 1.002
150 1.003
200 1.006
250 1.009
300 1.014
350 1.018
400 1.024
450 1.031
500 1.038
550 1.046
600 1.055
650 1.065
700 1.075
750 1.087
800 1.099
850 1.112
900 1.126
950 1.141
1000 1.157