Stupid question of the day....

I'm afraid you've got a pretty limited notion as to what "nil" would be. Remember, with AC, one of the big concerns is the transmission of significant amounts of power over long distances - have you thought about how large those sorts of conductors ARE?

Bob M.

Wrong again. You seem to think that the current is uniform down to the "skin depth," and THEN it somehow starts to fall off. As John already pointed out, with seemingly unwarranted patience, that ain't so.

Once again: "do the math." And this time, go beyond just using the skin-depth calculator on your favorite web site, and actually figure out what the EFFECTS would be (in terms of resistive loss, heating, whatever) of the skin depth at 60 Hz in a conductor otherwise seemingly-properly-sized for the 800A service that John mentioned as an example.

You might be surprised by the result.

Bob M.

On Tue, 02 Aug 2005 18:40:17 GMT, "Bob Myers" Gave us:

No. What the figure tells one is where the current is near zero, which is that area beneath the skin depth, all the way to the center of the wire. The way the current passes through said skin depth area doesn't matter.

Try being less stupid. THAT is what is unwarranted here. Unless, of course, it just comes naturally for you.

Whatever.

More stupidity. That was merely one location that I pointed out. It explains it quite well, however, and much better than your insulting ass does.

Pure aluminum or pure copper runs will see no difference. The reason that skin effect affects power transmission lines is due to the lower conductivity cores that are typically used.

You might get along with folks, if you stop with the bullshit insults. Sorry if YOU don't see your remarks that way, but I know better. Both about the remarks, and the topic.

The Navy seems to think there's a significant problem with gold over copper? Do tell....

Bob M.

On Tue, 02 Aug 2005 18:43:52 GMT, "Bob Myers" Gave us:

Look up Galvanic reaction in ship hulls, and you will find that all Navy ships have provisions to reduce it.

Note again that my reference is to the effect, not the remarks about specific elements. Learn to read.

Current begins to fall off monotonically from the very surface for any wire size at any AC frequency. There's no hard "skin boundary", and the 1/e density is just a handy if arbitrary measurement point.

I don't see why this needs arguing over. In a given situation, you just calculate the effects and decide how they affect things. Sometimes a 200% increase in resistance doesn't matter, and sometimes a 1% increase does. But skin effect does often matter in real situations at 60 Hz, and shouldn't be always/automatically discounted.

John

High currents do not increase skin effect, that is true. But the variation in conductor admittance *caused* by skin effect is a larger problem with high current conductors than it is with low current applications.

Not so. I could show you several switchyards within a short drive that use many hollow tube conductors all over the yard.

daestrom

On Tue, 02 Aug 2005 13:48:47 -0500, John Fields Gave us:

BTW nearly any metal in contact with "pure water" makes it pure no more.

Nonsense. High voltage DC has about the same corona problems as high voltage AC. The amount of corona discharge is a function of the electric field gradient and has nothing to do with skin effect. Like I said before, you've mixed up two different phenomenon that are completely unrelated.

ACRS cables have steel wires, but they are not all bundled in the center. They are distributed in a circle about 1/3 of the way out from the center. Dead center is Al strands, as well as the outer periphery. The reason for the steel is *not* skin effect, nor have anything to do with corona discharge. It is strength reinforcement, pure and simple. Nothing more. The elasticity of an all AL conductor would cause too much 'stretch' in the conductor, and too much rise/fall with temperature change.

daestrom

That depends on one's definition of 'nil' I guess.

Not quite. If by 'heat the wire evenly', you mean heat is generated equally in each unit of cross-section, yes. Since the resistivity of the material is a constant, and if the current density is uniform throughout, then the amount of I^2R losses in each unit cross-section is the same. But the material in the center will be a higher *temperature* than that around the periphery. It's simple really, the heat generated in the center must be conducted to the circle of material surrounding it. The heat from the center, combined with the heat generated in the circle of material must now be conducted to the next circle of material surrounding that. And so on... So the material just under the surface has heat generated directly in it,

So the *temperature* profile throughout the conductor is far from 'even'. If the material has a positive temperature coefficient of resistivity (as do both copper and Al), then the resistence of the central core is higher than the outer surface. The exact amount of temperature difference is a function of the electrical resistivity and thermal conductance of the material.

daestrom

The Navy knows it's a problem, but then naval ships are in seawater. One must have an electrolyte to complete the 'circuit'. This is one reason why commercial work with Al conductors often requires the application of special 'grease' to seal the connection from moisture intrusion.

daestrom

Not at all. You're apparently using a very interesting, albeit incorrect, definition of "skin depth." As has already been pointed out numerous times, the "skin depth" figure that results from the calculations you've been using is where the current density is down to about 37% of its "surface" value (not 37% of the conductance or loss or any other nonsensical notion that you seemed to think in a previous post). There is clearly still current farther from the surface than the "skin depth," and it is also clear that the density above that value is non-uniform. This IS important, and again I would suggest you check the values through an actual loss calculation to see just how big the effect can be.

That comment is particularly ironic, along with:

given the following:

Talk about the pot complaining about the complexion of the kettle...

Further nonsense:

Translation: you didn't bother to run the numbers, or you wouldn't be saying something so obviously incorrect. Next time, show your work.

Bob M.

Ratio of AC resistance at 60 Hz to DC resistance for 17 mm diameter copper wire:

Going by "High Frequency Resistance", pages 3323-3325 of the 43rd edition of the "CRC Handbook":

They give a formula X=pi*d*SQR((2*u*f)/rho)*SQR(1000)

u is magnetic permeability, unity for copper.

rho is resistivity in microohm-cm.

They simplify this for copper, to x=10*d*.01071SQR(f)

d is diameter in centimeters, and f is frequency in Hz.

So, for 17 mm diameter copper wire at 60 Hz this "x" is 1.41.

Next is a table that gives ratio of AC resistance to DC resistance as a function of this "x".

This table has an entry for 1.4, giving AC resistance 1.020 times DC resistance.

- Don Klipstein ( snipped-for-privacy@misty.com)

But that wasn't the question. You were responding to a comment made in the specific context of gold-on-copper, to the effect that "galvanic reaction" was the reason that such a combination wasn't a good idea. Sorry, but the "galvanic reaction" of dissimilar metals has absolutely nothing to do with the subject at hand.

There actually very often IS another layer (commonly, nickel) placed between a copper conductor and a top protective layer of gold, but this has nothing whatsoever to do with a "galvanic reaction" between these two metals. (If it did, following the original incorrect response on this subject, the problem would then become WORSE due to the fact that there would now be two such interfaces rather than one. Remember, if you can, that the original comment along these lines said that a "galvanic reaction" was a problem between ANY two metals.) The reason that an intermediate layer of nickel is often used in this case has to do with the fact that, left to themselves, gold and copper will tend to diffuse into one another. This causes a problem in electrical applications (where gold-plating copper conductors is being done to prevent corrosion) primarily on the gold side of things, as the copper diffusing up through the gold layer will eventually reach the surface and create the very same corrosion problem that the gold was supposed to be preventing. Nickel doesn't diffuse into gold like copper does, hence its use here.

My, again with the personal attacks; I suppose in the absence of practical knowledge, that's about all one is left with.

Bob M.