Arts of Electronics 3rd Edition Exercise 1.6 About mains voltages

[Minor Tom]

Hello friends, my first post in this board!

This is the exercise:

Optional Exercise: New York CIty requires about 10^10 W of electrical power, at 115volts (this is plausible: 10 million people averaging 1 kilowatt each). A heavy power cable might be an inch in diameter. Let's calculate what will happen if we try to supply the power through a cable 1 foot in diameter made of pure copper. Its resistance is 0.05 microOhms (5 x 10^-8 Ohms) per foot. Calculate:
a) the power lost per foot from I^2R losses
b) the length of cable over which you will lose all 10^10 watts, and
c) how hot the cable will get, if you know the physics involved (sigma = 6 x 10^-12 W / K^4 cm^2).
If you have done your computations correctly, the results should seem preposterous. What is the solution to this puzzle?

I have calculated a) to be approx. 378 MegaWatts which surely will evaporate the cable, no need to calculate c) (And I don't know the formula anyway).
I assume the answer to the puzzle is 1) use much higher voltage to reduce the current  2) use multiple power plants with multiple cables.
What about AC versus DC (except that AC is preferred because it is easier to transform)?

As it would seem simple to calculate b), I am not so sure if it would be correct (even if the cable wouldn't vanish). As energy is lost in the cable, the voltage also drops. No cable of less than endless length is able to eat away all available power. Am I correct?

[AG6QR]

As for AC vs DC, it makes little difference.  AC is easy to transform with a conventional transformer, but AC also can suffer from issues due to power factor, which can require greater current, thus greater losses in the lines, to deliver the same power to the load.

If the line consumes a given amount of power per unit length at a given current, then there most certainly is a length that will consume any fixed amount of power.  If you actually tried to use that length of line, with a generator just barely capable of that power output, the generating station would just barely be able to keep the specified current flowing through the line with a dead short as a load.  If you then tried to use a nonzero resistance as the load, the voltage and current would have to drop, or something else in the generating station would hit a limit.

[T3sl4co1l]

Technically, AC vs. DC does matter to the problem -- we must account for skin effect, because we haven't been told to ignore any part of real physics, and we have a real physical setup here, including material properties (the cable is specified as copper!).

Effectively, it's a pipe with a wall thickness of maybe half an inch, at AC!  This increases a) and decreases b) by about tenfold.

The result for c) will change in an interesting way: the outer layer becomes vaporized, while the core remains cold.  It will "burn" from the outside in.



Actually, an even better question that follows: how long will it take to blow out this "fuse wire"?  Can it even reach full current (assuming DC now) before the conductor is completely exploded?

The reason is this:
Any path has inductance.  If we apply 115V DC to a conductor, say in a few microseconds (easily enough achieved, say with an explosively actuated switch -- since to be fair, we have to move a lot of metal quickly, to get a connection that's as solid as the cable it goes between), then the current will ramp up linearly after that time.  Say the conductor is 10 meters long.  The inductance of free space is ~1.257 uH/m, and there is a geometry factor that depends on arrangement, so for simplicity let's say it's 10uH total.  This gives a ramp rate of dI/dt = V/L = 115V / 10uH = 11.5 A/us == MA/s.  So it will take one second just to reach 115 x 10^6 watts (drawn from the supply).

Speaking of skin depth -- in that time, current will still be diffusing into the conductor (at a depth of a few inches), so the conductor will be heating up quite rapidly indeed!

The same situation at AC, becomes much less interesting: 10uH at 60Hz is 3.7 milliohms, or a paltry 30kA.  Such currents are in fact common in real world applications: aluminum smelting.  They use arm- or leg-sized busbar, and, I would assume, rectify it as soon as possible after the transformer.

Tim

[Minor Tom]

Many thanks for your answers, @AG6QR and @T3sl4co1l !
A very interesting aside, Teslacoil!

Perhaps the difficulties I have stem from the wild mix of ideal and real cable models in this exercise.

The calculation of power loss for 1 foot of cable seems to be valid (to me) only when this foot of cable is inserted into an ideal cable (which also connects some undefined load).

If you assume an ideal cable all the way and no voltage drops, loss of all power can simply be achieved by inserting a longer "real" cable.  It is calculated simply by dividing the power available by the power lost at 1 foot. Which is 26,45 foot.

To illustrate my difficulties:
Assume 50 foot of real cable, by using P(Loss) = I²*R = (P/V)²*R (which is the same formula I used to calculate the loss for 1 foot):
When you calculate this, more power is lost than what is available, which cannot be. There must be a logical failure somewhere - and I think I see what it is. By giving voltage and power, a resistance is defined. This resistance is reached for 26,45 foot of cable. So we simply calculated what length of real cable is required to achieve this load. If you inserted more loads (more resistance), the given power couldn't be achieved any more by the given voltage. With some consumer loads inserted as a start, the defined resistance would be reached even earlier (with less cable).
Which is essentially what AG6QR said, but I didn't understand initially. Correct?

[Maxlor]

To illustrate my difficulties:
Assume 50 foot of real cable, by using P(Loss) = I²*R = (P/V)²*R (which is the same formula I used to calculate the loss for 1 foot):
When you calculate this, more power is lost than what is available, which cannot be. There must be a logical failure somewhere - and I think I see what it is. By giving voltage and power, a resistance is defined. This resistance is reached for 26,45 foot of cable. So we simply calculated what length of real cable is required to achieve this load. If you inserted more loads (more resistance), the given power couldn't be achieved any more by the given voltage. With some consumer loads inserted as a start, the defined resistance would be reached even earlier (with less cable).
Which is essentially what AG6QR said, but I didn't understand initially. Correct?

Yes I think you got it. And to reaffirm it: it helps to just always go back to the bare basics, in this case, Ohms law. If U and I are fixed, R is fixed too! And if R is defined as R = R_cable + R_load, either one of R_cable and R_load can at most be as high as R. And if you still insist on changing R... one of U or I has to change too.

[AG6QR]

When you calculate this, more power is lost than what is available, which cannot be. There must be a logical failure somewhere - and I think I see what it is. By giving voltage and power, a resistance is defined. This resistance is reached for 26,45 foot of cable. So we simply calculated what length of real cable is required to achieve this load. If you inserted more loads (more resistance), the given power couldn't be achieved any more by the given voltage. With some consumer loads inserted as a start, the defined resistance would be reached even earlier (with less cable).
Which is essentially what AG6QR said, but I didn't understand initially. Correct?

Yes, I also think you've got it.

The point of the exercise is to show how impractical it would be to distribute power to a city at 120V, and to illustrate why higher voltages are used for distribution.

The point could have been made by asking a slightly different question:  If we must deliver this many watts to the load, at what cable length will the power dissipated in the cable equal the power delivered to the load, thus requiring us to double the size of our power plant and waste half of its capacity in transmission line losses?  Corollary questions: what would the voltage be at the load?  What would the voltage be at the power plant?

The hypotheticals are interesting to think about, but I hope the exercise brings home why we go to the trouble of using high voltage lines, and stepping voltage up and down for transmission.  Furthermore, the same principle is why we don't distribute significant amounts of power around our houses at 5V, even though many small electronic devices these days use 5V (USB) power.

[Minor Tom]

Thanks for your comments, Maxlor and AG6QR!
And I support your suggestions, AG6QR.

Yeah, I got what that Exercise is supposed to show.

Thanks to the formula P_loss = I²*R, one can see that a reducing the current by a factor of 100, the power loss is reduced by a factor of 10,000!