Stupid question of the day....

[TokaMundo]

On Tue, 02 Aug 2005 10:54:26 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> Gave us:

Before any difference could even be noted, the wired diameter would
have to be over 16 mm.

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.

 

[Bob Myers]

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:v4cve15bd0ar0d02f0da10fbm87640sr98@4ax.com...

For AC at this frequency there is nil skin effect.


Not nil. Do the math.


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]

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:i2dve15rqf7183m844bjidvu0v98rn9cjf@4ax.com...

On Tue, 02 Aug 2005 10:54:26 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> Gave us:

Before any difference could even be noted, the wired diameter would
have to be over 16 mm.

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.


0.85cm 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.

 

[Bob Myers]

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:mhbve1p34rfo2cpma8te55gjbgclte2a0j@4ax.com...

If you connect Au to Cu and put a Current through it, for best results
AC,
the Cu starts corroding at the transistion from Cu to Au. This is always
the
case when putting to metals together, the greater the difference between
the
metals the faster the corroding will be.

---
That's not true.

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]

On Tue, 02 Aug 2005 18:35:31 GMT, "Bob Myers"
<nospamplease@address.invalid> Gave us:

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:v4cve15bd0ar0d02f0da10fbm87640sr98@4ax.com...
For AC at this frequency there is nil skin effect.


Not nil. Do the math.


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?

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.

 

[TokaMundo]

On Tue, 02 Aug 2005 18:40:17 GMT, "Bob Myers"
<nospamplease@address.invalid> Gave us:

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:i2dve15rqf7183m844bjidvu0v98rn9cjf@4ax.com...
On Tue, 02 Aug 2005 10:54:26 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> Gave us:

Before any difference could even be noted, the wired diameter would
have to be over 16 mm.

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.


0.85cm 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.

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 M.

 

[TokaMundo]

On Tue, 02 Aug 2005 18:43:52 GMT, "Bob Myers"
<nospamplease@address.invalid> Gave us:

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:mhbve1p34rfo2cpma8te55gjbgclte2a0j@4ax.com...

If you connect Au to Cu and put a Current through it, for best results
AC,
the Cu starts corroding at the transistion from Cu to Au. This is always
the
case when putting to metals together, the greater the difference between
the
metals the faster the corroding will be.

---
That's not true.

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....

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]

On Tue, 2 Aug 2005 10:40:37 -0500, "DBLEXPOSURE"
<celstuff@hotmail.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:4ofue1lca6tpqat9iqvepk6dhplv1v9335@4ax.com...
On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
electricdummy@hotmail.com> wrote:


If you connect Au to Cu and put a Current through it, for best results AC,
the Cu starts corroding at the transistion from Cu to Au. This is always
the
case when putting to metals together, the greater the difference between
the
metals the faster the corroding will be.

---
That's not true.

--
John Fields
Professional Circuit Designer

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.\

--
John Fields
Professional Circuit Designer

 

[DBLEXPOSURE]

"John Fields" <jfields@austininstruments.com> wrote in message
news:ebfve1d9njtughtckfk1i3oqithcl1j1u1@4ax.com...

On Tue, 2 Aug 2005 10:40:37 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> wrote:


"John Fields" <jfields@austininstruments.com> wrote in message
news:4ofue1lca6tpqat9iqvepk6dhplv1v9335@4ax.com...
On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
electricdummy@hotmail.com> wrote:


If you connect Au to Cu and put a Current through it, for best results
AC,
the Cu starts corroding at the transistion from Cu to Au. This is always
the
case when putting to metals together, the greater the difference between
the
metals the faster the corroding will be.

---
That's not true.

--
John Fields
Professional Circuit Designer

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.\

--
John Fields
Professional Circuit Designer

---
Less "noble", or more anodic.
---
Same difference
---
not a dielectric as you have stated
----

With the preface, "If I recal correctly", Correction noted.

---
(but there is no metal named "coper",

so i'll assume you meant "copper")
---

BFD

 

[John Fields]

On Tue, 2 Aug 2005 18:59:37 +0200, "Alexander"
<electricdummy@hotmail.com> wrote:

, hakte DBLEXPOSURE op ons in met:

"John Fields" <jfields@austininstruments.com> wrote in message
news:4ofue1lca6tpqat9iqvepk6dhplv1v9335@4ax.com...
On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
electricdummy@hotmail.com> wrote:


If you connect Au to Cu and put a Current through it, for best
results AC, the Cu starts corroding at the transistion from Cu to
Au. This is always the
case when putting to metals together, the greater the difference
between the
metals the faster the corroding will be.

---
That's not true.

--
John Fields
Professional Circuit Designer

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?
---

But at least ThanX for confirming my statement and not saying its not true
without giving a reason as someone else did.

---
Whether I gave a reason or not is unimportant, what matters is that
a factual error got corrected.


--
John Fields
Professional Circuit Designer

 

[John Larkin]

On Tue, 02 Aug 2005 18:28:32 GMT, TokaMundo <TokaMundo@weedizgood.org>
wrote:

On Tue, 02 Aug 2005 10:54:26 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> Gave us:

Before any difference could even be noted, the wired diameter would
have to be over 16 mm.

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.

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]

On Tue, 02 Aug 2005 17:31:30 GMT, TokaMundo
<TokaMundo@weedizgood.org> wrote:

On Tue, 02 Aug 2005 04:41:38 -0500, John Fields
jfields@austininstruments.com> Gave us:

On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
electricdummy@hotmail.com> wrote:


If you connect Au to Cu and put a Current through it, for best results AC,
the Cu starts corroding at the transistion from Cu to Au. This is always the
case when putting to metals together, the greater the difference between the
metals the faster the corroding will be.

---
That's not true.

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
Professional Circuit Designer

 

[John Fields]

On Tue, 2 Aug 2005 14:09:31 -0500, "DBLEXPOSURE"
<celstuff@hotmail.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:ebfve1d9njtughtckfk1i3oqithcl1j1u1@4ax.com...
On Tue, 2 Aug 2005 10:40:37 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> wrote:


"John Fields" <jfields@austininstruments.com> wrote in message
news:4ofue1lca6tpqat9iqvepk6dhplv1v9335@4ax.com...
On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
electricdummy@hotmail.com> wrote:


If you connect Au to Cu and put a Current through it, for best results
AC,
the Cu starts corroding at the transistion from Cu to Au. This is always
the
case when putting to metals together, the greater the difference between
the
metals the faster the corroding will be.

---
That's not true.

--
John Fields
Professional Circuit Designer

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.\

--
John Fields
Professional Circuit Designer

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

---
Hardly. "Nobel" was the inventor of dynamite, while "noble", in the
context which makes sense in this thread, refers to chemical
inactivity.
---

---
not a dielectric as you have stated
----

With the preface, "If I recal correctly", Correction noted.

---
(but there is no metal named "coper",
so i'll assume you meant "copper")
---
BFD

---
Yup; wrong is wrong.

--
John Fields
Professional Circuit Designer

 

[daestrom]

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:vdote1tb8oed91mpoiqsa0pag4s4lclrnm@4ax.com...

On Tue, 02 Aug 2005 00:45:43 GMT, "daestrom"
daestrom@NO_SPAM_HEREtwcny.rr.com> Gave us:


"Dimitrios Tzortzakakis" <dimtzort@otenet.gr> wrote in message
news:dcl731$u0$1@usenet.otenet.gr...


--
Tzortzakakis Dimitrios
major in electrical engineering, freelance electrician
FH von Iraklion-Kreta, freiberuflicher Elektriker
dimtzort AT otenet DOT gr
? "Alexander" <electricdummy@hotmail.com> ?????? ??? ??????
news:dcfqoa$dbo$1@news1.zwoll1.ov.home.nl...

"TimPerry" <timperry@noaspamadelphia.net> schreef in bericht
news:WrqdndPloplTh3bfRVn-3w@adelphia.com...

"AllTel - Jim Hubbard" <reply@newsgroups.com> wrote in message
news:232f7$42eaf0d2$97d59ba4$23531@ALLTEL.NET...
I am curious about what would happen to an electrical current in 2
situations.....

Assume that you have 2 wires that, when joined, complete a closed
electrical
DC circuit with electrons flowing thusly.....

------------ ============
eeeeeeeeee eeeeeeeeeeeeeee
------------ ============


If you flattened out the end of each wire where they connect , would
the
resulting electron paths be more like figure A or Figure B?


neither ... research "skin effect"

Most of the times this just aplies to AC (high frequency) circuits
Or of line-to-line voltage equal or above 220 kV.Therefore transmission
lines of 400 kV are always designed with a double conductor, thus to
reduce
the corona discharge due to skin effect.

Oh boy, you have a 'couple of crossed wires' there.

"Skin effect" is the phenomenon where electric current flow is forced out
from the center of a conductor due to the self-inductance in the conductor
when carrying AC current. The higher the frequency, the more pronounced
the
current shift to the exterior. It's mostly a problem with high current
situations, even if the voltages are so low that corona discharge is not a
problem.

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.

"Corona discharge" is *NOT* caused by AC or skin effect. Corona discharge
is caused by a high voltage gradient in the space around a conductor.
This
is a combination of the voltage applied to the conductor and the effective
radius of the conductor. A high voltage, or very small effective radius
can
increase the gradient to the point where the air is ionized. Simple proof
is that corona discharge is a problem with high DC voltage systems as well
as AC.

Sometimes hollow tubes are used for high frequency power conductors. This
reduces the weight and cost by eliminating the central part of the
conductor, where 'skin effect' has rendered the impedence high anyway. So
little admittance is lost for a great savings in material/weight.

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]

On Tue, 02 Aug 2005 13:48:47 -0500, John Fields
<jfields@austininstruments.com> Gave us:

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]

On Tue, 02 Aug 2005 15:03:41 -0500, John Fields
<jfields@austininstruments.com> Gave us:

On Tue, 02 Aug 2005 17:31:30 GMT, TokaMundo
TokaMundo@weedizgood.org> wrote:

On Tue, 02 Aug 2005 04:41:38 -0500, John Fields
jfields@austininstruments.com> Gave us:

On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
electricdummy@hotmail.com> wrote:


If you connect Au to Cu and put a Current through it, for best results AC,
the Cu starts corroding at the transistion from Cu to Au. This is always the
case when putting to metals together, the greater the difference between the
metals the faster the corroding will be.

---
That's not true.

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...

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]

On Tue, 02 Aug 2005 15:20:32 -0500, John Fields
<jfields@austininstruments.com> Gave us:

Yup; wrong is wrong.

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

You're fucked John. No way out.

 

[daestrom]

"Dimitrios Tzortzakakis" <dimtzort@otenet.gr> wrote in message
news:dcnpft$3pu$1@usenet.otenet.gr...

snip

And for high voltage systems, multiple parallel conductors are used to
give
a larger 'effective radius', thereby reducing the corona losses.

But the two phenomenon are not related, and the two techniques used are
not
really related.

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]

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:atnte1h4fuokfj67a2eu9ck8mtmnoj2osj@4ax.com...

On Tue, 02 Aug 2005 00:45:43 GMT, "daestrom"
daestrom@NO_SPAM_HEREtwcny.rr.com> Gave us:


"John Fields" <jfields@austininstruments.com> wrote in message
news:tbune11gjuu0qst4fv4446d89ejojeu1rq@4ax.com...
On Sat, 30 Jul 2005 10:50:24 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 30 Jul 2005 09:39:58 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


At higher frequency AC, current in a wire tends to avoid the center
and crowd near the surface, "skin effect."


Hmmm...

Copper does have a weak Hall effect. And the current through a round
wire does make a circular/transverse magnetic field. So, at very high
DC currents, is the current density a bit non-uniform?

---
I would think that simple thermal effects would cause charge to flow
closer to the surface just because that part of the conductor would
be cooler, ergo lower resistance than the hotter interior.


An interesting point. *IF* the current density is uniform across the
conductor, then the heat generated would be uniform in each unit
cross-section. And a uniform heat generation in a cylindrical rod leads
to
a parabolic temperature profile, the highest exactly at the centerline,
dropping of as you move outward along any radial line.

Of course, in an AC line, the current density isn't uniform, so neither is
the heat generation. So when it comes to skin effect, it tends to lower
the
peak, centerline temperature.

Now, given that both copper and aluminum are excellent heat conductors, it
might be interesting to calculate how big a temperature profile could be
expected, and from this calculate the variation in resistivity.

I suspect the work has been done before, and that the difference is rather
modest for all but the largest cylindrical conductors.

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]

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:mhbve1p34rfo2cpma8te55gjbgclte2a0j@4ax.com...

On Tue, 02 Aug 2005 04:41:38 -0500, John Fields
jfields@austininstruments.com> Gave us:

On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
electricdummy@hotmail.com> wrote:


If you connect Au to Cu and put a Current through it, for best results
AC,
the Cu starts corroding at the transistion from Cu to Au. This is always
the
case when putting to metals together, the greater the difference between
the
metals the faster the corroding will be.

---
That's not true.

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