Stupid question of the day....

[TokaMundo]

Not so. At 0.85 cm depth, current density is down to 1/e (ie, only
0.37 of) the surface density. That's pretty significant.

0.85 cm is pretty thick. 8.5 mm in fact. Double that to get 17mm.

Unless the wire is larger than 17mm at 60Hz, the entire wire will
carry current. VERY simple math.

 

[TokaMundo]

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?

As I noted, the wire would have to be greater than 17mm in diameter.

A steel strand of 3/4 an inch covered with aluminum strands to a
finished thickness of an inch or so would be quite ideal. The entire
depth of the aluminum would carry the energy, and the steel would see
near none of it.

Coming back down to the consumer level, using any standard household
wire, the effect IS most certainly NIL. Even the 25 or 40kW
transformer hanging out on the pole has no need for any such
considerations.

If the wire feeding you house is over 17mm in diameter, there MIGHT
be a small difference in the ohmic resistance of the line. A very
small difference. Certainly not the 37% that was suggested.

 

[Bob Myers]

Very much so as close to nil as it gets. Review the math.

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.

 

[Bob Myers]

0.85cm is 8.5 mm. That means that the wire has to be bigger than
that number as a radius before the current flow anywhere else besides
the entire wire.

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.

 

[TokaMundo]

Wrong again. You seem to think that the current is uniform
down to the "skin depth," and THEN it somehow starts to
fall off.

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.

As John already pointed out, with seemingly unwarranted
patience,

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

that ain't so.
Whatever.

Once again: "do the math." And this time, go beyond just using
the skin-depth calculator on your favorite web site,

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.

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.

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 be surprised by the result.

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.

 

[Bob Myers]

It's called galvanic reaction.

The Navy seems to think it's real. Does that make you an idiot?

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


Bob M.

 

[TokaMundo]

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

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.

 

[John Fields]

Firstly, Aluminium is Al not Au. Au is gold. You are speaking of
aluminium and coper?

---
I doesn't make any difference, (but there is no metal named "coper",
so i'll assume you meant "copper") there won't be any corrosion
unless the dissimilar metals are in contact with each other in the
presence of an electrolyte, not a dielectric as you have stated.

Galvanic Corrosion Is possible when Al and Cu are in contact with one and
other. If I recal correctly a dialectric such as water needs to be present.
Cathodic protection, (electric current) can be used to slow or stop this
proccess. I Imagine reversing the polarity may speed it up. Aluminium is
the "Less Nobel" of the two metals so I would imagine that it would be the
one to corrode.

---
Less "noble", or more anodic.

If he truly meant a gold-copper couple, the copper, being more
anodic than gold, would corrode.

BTW, pure water _is_ a dielectric and dissimilar metals in contact
with each other and pure water would not corrode.\

 

[DBLEXPOSURE]

---
I doesn't make any difference, (but there is no metal named "coper",
so i'll assume you meant "copper") there won't be any corrosion
unless the dissimilar metals are in contact with each other in the
presence of an electrolyte, not a dielectric as you have stated.


---
Less "noble", or more anodic.

If he truly meant a gold-copper couple, the copper, being more
anodic than gold, would corrode.

BTW, pure water _is_ a dielectric and dissimilar metals in contact
with each other and pure water would not corrode.\

 

[John Fields]

Op [GMT+1=CET], hakte DBLEXPOSURE op ons in met:

Firstly, Aluminium is Al not Au. Au is gold. You are speaking of
aluminium and coper?

Galvanic Corrosion Is possible when Al and Cu are in contact with one
and other. If I recal correctly a dialectric such as water needs to
be present. Cathodic protection, (electric current) can be used to
slow or stop this proccess. I Imagine reversing the polarity may
speed it up. Aluminium is the "Less Nobel" of the two metals so I
would imagine that it would be the one to corrode.

Correct I also added the remark of the diëlectricum to the discussion.

---
No you added the remark about the _electrolyte_, which was correct.
---

And your remark about Aluminium is correct, however as stated in some
applications I have seen an Copper core and an Gold (aurum) shell. And since
the combination gold-copper is worse then the well known combination
aluminium-copper.

---
In what way is it worse?

Looking at:

http://www.ocean.udel.edu/seagrant/publications/corrosion.html

It seems that the distance between gold and copper (0.52V) is the
same as the distance between copper and aluminum, so why would the
rate of corrosion be worse for a gold-copper couple than for
copper-aluminum?
---

 

[John Larkin]

0.85 cm is pretty thick. 8.5 mm in fact. Double that to get 17mm.

Unless the wire is larger than 17mm at 60Hz, the entire wire will
carry current. VERY simple math.

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

 

[John Fields]

It's called galvanic reaction.

The Navy seems to think it's real. Does that make you an idiot?

---
Back looking for some more lumps, bonehead? OK, I'm happy to
oblige...

First, it's called "galvanic corrosion" and, second, if you knew
anything about it and had somehow managed to pull your head out of
your ass before commenting, you might have noticed that the poster
made no mention of the electrolyte required for the corrosion to
occur. That's why what he said wasn't true.

 

[John Fields]

---
I doesn't make any difference, (but there is no metal named "coper",
so i'll assume you meant "copper") there won't be any corrosion
unless the dissimilar metals are in contact with each other in the
presence of an electrolyte, not a dielectric as you have stated.


---
Less "noble", or more anodic.

If he truly meant a gold-copper couple, the copper, being more
anodic than gold, would corrode.

BTW, pure water _is_ a dielectric and dissimilar metals in contact
with each other and pure water would not corrode.\

 

[daestrom]

It becomes more prevalent as frequency goes up, not current.

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.

VERY high frequency. NOT AC line frequencies.

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

daestrom

 

[TokaMundo]

BTW, pure water _is_ a dielectric and dissimilar metals in contact
with each other and pure water would not corrode.\

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

 

[TokaMundo]

You're an idiot.

First, it's called "galvanic corrosion"

Wrong. The result is corrosion. The activity is called "reaction".

Your favorite web site which you posted a reference to speaks about
the end result.

The moniker I gave speaks about the process itself.

You're a fucking jackass. Everybody speaks about that.

and, second, if you knew
anything about it and had somehow managed to pull your head out of
your ass

Two more reasons you should be on everyone's filtered list.

before commenting,

I commented on how much of an asshole you are. When I say
something, you come back demanding proofs, yet you get to make a
jackjawed remark like "not true" and think you won't see anything said
about how much of an ass you are? Sorry, CHUMP! You don't get that.

you might have noticed that the poster
made no mention of the electrolyte required for the corrosion to
occur.

Oh boy!

That's why what he said wasn't true.

And THAT is also what you should have said in your post, dumbfuck.

 

[TokaMundo]

Yup; wrong is wrong.

And that be you. Your method is wrong, and wrong is wrong.

You're fucked John. No way out.

 

[daestrom]

Yes, but also in voltages >=15 kV there's a signifigant skin effect,
that's
why all transmission conductors are constructed with a steel *core* and an
*aluminium* outer sheath, because the current tends to flow on the skin of
the conductor.I mentioned corona discharge, to bring into evidence the
very
strong electric field around the conductor in very high voltages.

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

 

[daestrom]

For AC at this frequency there is nil skin effect.

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

Current in a wire will heat the wire evenly if it is of one
material.

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,
*PLUS* all the heat generated in interior material conducted into it. For
uniform heat generation throughout the material, it is simple integration to
show that the temperature profile is a parabolic with the apex at the
centerline and temperature falling off as one moves further from the center
to the outer surface.

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

 

[daestrom]

It's called galvanic reaction.

The Navy seems to think it's real. Does that make you an idiot?

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