--
Tzortzakakis Dimitrios
major in electrical engineering, freelance electrician
FH von Iraklion-Kreta, freiberuflicher Elektriker
dimtzort AT otenet DOT gr
Ï "daestrom" Ýãñáøå óôï ìÞíõìá
...

"Dimitrios Tzortzakakis" wrote in message
...


--
Tzortzakakis Dimitrios
major in electrical engineering, freelance electrician
FH von Iraklion-Kreta, freiberuflicher Elektriker
dimtzort AT otenet DOT gr
Ï "Alexander" Ýãñáøå óôï ìÞíõìá
...

"TimPerry" schreef in bericht
...

"AllTel - Jim Hubbard" wrote in message
...
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.

"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.

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.
daestrom

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

On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
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.
You're right it needs also an electrlyt which is most of the times present.

On Tue, 02 Aug 2005 02:57:11 GMT, TokaMundo
wrote:



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.


Not nil. Do the math.

John

"John Fields" wrote in message
...
On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
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.

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

"John Fields" wrote in message
...
On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
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.
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.

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

On Tue, 02 Aug 2005 17:40:35 GMT, TokaMundo
wrote:

Corona is a function of voltage and the capacity for air to ionize.

Wow, and I thought it was a Mexican beer that tastes so bad you have
to add lime to it before consumption.

--
Owamanga!
http://www.pbase.com/owamanga

On Tue, 02 Aug 2005 17:24:34 GMT, TokaMundo
wrote:

On Mon, 01 Aug 2005 20:21:31 -0700, John Larkin
Gave us:

On Tue, 02 Aug 2005 03:00:06 GMT, TokaMundo
wrote:


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. At 60 Hz, copper skin depth is about 0.85 cm.

Did you even look at that number? That is 8.5 mm!


Not only did I look at it, but I calculated it, and typed it.

No? OK. So for all practical purposes that do not have 20 cm wire
involved (ie any normal residential application) there is NO skin
effect! Where in everyday life does a person use wire that has a
diameter greater than 8.5 mm that would present anything other than
100% current density in the conductor? The wave is just too slow for
anything other than full propagation. Hell, even a 25kW transformer
won't see any difference.


I just bought a building that has 3-phase, 800-amp service, and skin
effect certainly has affected the sizing of the main feeder wires. And
I work with people who build gigawatt 60 Hz power plants and jumbo-jet
400 Hz power systems. That's my "everyday life."

In big AC transmission lines, there's a complex optimization involving
wire weight, tensile strength, ohmic losses, skin effect, corona
losses, wire cost, and tower spacing/cost.

In ohms per foot DC or AC at 60Hz the value is the same for all wire
diameters that have a gauge number.

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.


What part of the word negligible, or not of any effect do you not
understand?

I'm an engineer, so I consider something to be "negligable" if I can
demonstrate, quantitatively, that it doesn't matter enough to affect a
system. 37% is therefore worth a second look.

To speak in your style, what part of "doing the math" do you not
understand?

John

"Alexander" wrote in message
...
Op [GMT+1=CET], hakte DBLEXPOSURE op ons in met:

"John Fields" wrote in message
...
On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
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.
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.

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


Your welcome, I thought you deserved a respectful reply...

"TokaMundo" wrote in message
...
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.

"TokaMundo" wrote in message
...
On Tue, 02 Aug 2005 10:54:26 -0700, John Larkin
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.

"TokaMundo" wrote in message
...

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.

On Tue, 2 Aug 2005 10:40:37 -0500, "DBLEXPOSURE"
wrote:


"John Fields" wrote in message
.. .
On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
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

"John Fields" wrote in message
...
On Tue, 2 Aug 2005 10:40:37 -0500, "DBLEXPOSURE"
wrote:


"John Fields" wrote in message
. ..
On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
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

On Tue, 2 Aug 2005 18:59:37 +0200, "Alexander"
wrote:

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

"John Fields" wrote in message
...
On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
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/p...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

On Tue, 02 Aug 2005 18:28:32 GMT, TokaMundo
wrote:

On Tue, 02 Aug 2005 10:54:26 -0700, John Larkin
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

On Tue, 02 Aug 2005 17:31:30 GMT, TokaMundo
wrote:

On Tue, 02 Aug 2005 04:41:38 -0500, John Fields
Gave us:

On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
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

On Tue, 2 Aug 2005 14:09:31 -0500, "DBLEXPOSURE"
wrote:


"John Fields" wrote in message
.. .
On Tue, 2 Aug 2005 10:40:37 -0500, "DBLEXPOSURE"
wrote:


"John Fields" wrote in message
...
On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
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

"TokaMundo" wrote in message
...
On Tue, 02 Aug 2005 00:45:43 GMT, "daestrom"
Gave us:


"Dimitrios Tzortzakakis" wrote in message
...


--
Tzortzakakis Dimitrios
major in electrical engineering, freelance electrician
FH von Iraklion-Kreta, freiberuflicher Elektriker
dimtzort AT otenet DOT gr
? "Alexander" ?????? ??? ??????
...

"TimPerry" schreef in bericht
...

"AllTel - Jim Hubbard" wrote in message
...
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

"Dimitrios Tzortzakakis" wrote in message
...

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

"TokaMundo" wrote in message
...
On Tue, 02 Aug 2005 00:45:43 GMT, "daestrom"
Gave us:


"John Fields" wrote in message
. ..
On Sat, 30 Jul 2005 10:50:24 -0700, John Larkin
wrote:

On Sat, 30 Jul 2005 09:39:58 -0700, John Larkin
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

"TokaMundo" wrote in message
...
On Tue, 02 Aug 2005 04:41:38 -0500, John Fields
Gave us:

On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
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

On Tue, 02 Aug 2005 21:46:40 GMT, TokaMundo
wrote:

On Tue, 02 Aug 2005 15:03:41 -0500, John Fields
Gave us:

On Tue, 02 Aug 2005 17:31:30 GMT, TokaMundo
wrote:

On Tue, 02 Aug 2005 04:41:38 -0500, John Fields
Gave us:

On Tue, 2 Aug 2005 06:42:29 +0200, "Alexander"
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.

---
Right. That's what the topic is about: galvanic corrosion, and
which is what all of us, except you, have been talking about.
---

The moniker I gave speaks about the process itself.

---
Crappola. You know nothing about the process, and when you butted
in with your ****, and with what you think the Navy thinks, you
thought that "galvanic action" was the right name for what it's
called. It's not, and now you're trying to cover your ass by doing
a little semantic "shuffle and smoke" routine. Typical for you, you
phony piece of ****. I suspect next you'll be off searching the web
for every possible thing you can find on galvanic corrosion just to
make it seem like, the next time you post, you knew it beforehand.
Hey, I'll even _give_ you a hand. Google "electrochemical series"
and suck on that for a while.
---

You're a ****ing jackass. Everybody speaks about that.

---
Make a list, mother****er.
---

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.

---
Because you knew nothing about it and couldn't manage to pull your
head out of your ass? Sounds to me more like reasons for folks to
plonk _your_ sorry ass out of existence.
---

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.

---
The reason I demand proofs from you is because you're a ****ing
liar.

I'm not, and when I say that something isn't true I can back it up
even if I don't choose to at the time I said it, for whatever
reason. The last one had to do with the poster's carelessness in
not declaring that an electrolyte was needed in order for galvanic
corrosion to proceed, and I figured that if I gave him a little prod
he'd figure it out for himself. I was a little surprised that he
was miffed at not having been given the answer on a silver platter,
but there ya go...

See, Tokey, one of the differences between you and I is that I've
got a solid technical background and can stand my ground without
having to resort to bull**** tactics, like you do, in order to try
to blow up my balloon.
---

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, dumb****.

---
LOL, you're ****ed because I didn't give the trick away early, so
that you could say that you knew it all along?

Tokamundo? More like Tokanada.

--
John Fields
Professional Circuit Designer

"TokaMundo" wrote in message
...

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,

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.


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 comment is particularly ironic, along with:

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.

given the following:

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

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

Further nonsense:

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.

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.

In article , TokaMundo wrote:
On Tue, 02 Aug 2005 10:54:26 -0700, John Larkin
Gave us:

What part of the word negligible, or not of any effect do you not
understand?

I'm an engineer, so I consider something to be "negligable" if I can
demonstrate, quantitatively, that it doesn't matter enough to affect a
system.

No ****.

37% is therefore worth a second look.

Your application of your "math" is what needs a second look.

To speak in your style, what part of "doing the math" do you not
understand?

What part of "you did the math wrong" do you not understand?

Before it would make a difference, the wire will have to be pretty
big (over 17mm diameter) , and before it will make a 37% difference,
it would have to be bigger still! Real simple math, there.

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

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 )

On Tue, 02 Aug 2005 21:48:11 GMT, TokaMundo
wrote:

On Tue, 02 Aug 2005 15:20:32 -0500, John Fields
Gave us:

Yup; wrong is wrong.

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

You're ****ed John. No way out.

---
Yeah, right. I need to listen to a scrawny, universally hated
misanthrope to tell me how to run my life so he'll feel more
comfortable in _his_ little 1000 kilocalorie world? Pedal on,
loser.

--
John Fields
Professional Circuit Designer

"TokaMundo" wrote in message
...
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.

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.

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

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

Bob M.

Some people say that there is no such thing as a stupid question. Obviously
there seems to be no shortage of stupid answers.

Bill

TokaMundo wrote:

It's called galvanic reaction.

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

The Navy also deals with lots of salt water. ;-)

On Wed, 03 Aug 2005 01:24:11 GMT, Repeating Rifle
wrote:

Some people say that there is no such thing as a stupid question. Obviously
there seems to be no shortage of stupid answers.

Bill


There are no stupid questions, only stupid people.

John

In article , TokaMundo wrote:
On Tue, 2 Aug 2005 22:58:33 +0000 (UTC), (Don
Klipstein) Gave us:

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.


The table I saw shows the AC and DC resistance as being exactly the
same for both.

Maybe your table rounds? Please cite source as well as I did.

Your flaw is where you failed to note the topic given in the CRC
handbook.

I did note this, as you quoted above.

60 Hz is NOT high frequency... at all.

Try some calculations at 100 kHz and you'll see that those
frequencies down near zero (ie 60Hz) yield very nearly nil difference.

The formulas are functions of wire diameter, wire resistivity and
frequency, and do not lose validity merely because a thick wire has AC
resistance greater than DC resistance at a frequency that is easy to label
"NOT high".

And as you asked... Ratio of AC resistance to DC resistance of 17 mm
diameter copper wire at 100 KHz is about 21.5. This does not invalidate
the calculation for 60 Hz.

You would have been better off claiming that resistance 2% higher at 60
Hz than at DC is a negligible increase.

- Don Klipstein )

John Fields wrote:
I see. Instead of reason, you prefer insult.

I will neither read your "proof" nor will I shut up, and if you
don't like it, you miserable son of a bitch, you can go ****
yourself.

I'm not a son, you illiterate and uncouth obscurantist troll. I
already used reason in the proof, which you wilfully ignore. It uses
maths too, which puts a limiting case on the Lorentz corrections,
remembering that GR and QM are incompatible. It looks like my other
replier didn't read it too or he would be forced to agree with me. The
people here are retards.

-Aut

John Larkin wrote:
There are no stupid questions, only stupid people.

There are stupid questions, those that could be easily found on one's
own.

Stop swearing or I'll beat your head in, the part that causes swearing.

Jasen Betts wrote:
I looked at it, and you're right, the posting at that URL is wrong.
here's another time wasting URL http://www.geocities.com/jasen_betts/Autymn.txt

You can't prove anything. You're wrong in everything.

On Wed, 03 Aug 2005 01:20:29 GMT, TokaMundo
wrote:

---
Nothing of consequence, or eloquence, and ended it with:
---

You're also retarded. Seek help, asshole.


--
John Fields
Professional Circuit Designer

On 2 Aug 2005 21:58:47 -0700, "Autymn D. C."
wrote:

John Fields wrote:
I see. Instead of reason, you prefer insult.

I will neither read your "proof" nor will I shut up, and if you
don't like it, you miserable son of a bitch, you can go ****
yourself.

I'm not a son, you illiterate and uncouth obscurantist troll. I
already used reason in the proof, which you wilfully ignore. It uses
maths too, which puts a limiting case on the Lorentz corrections,
remembering that GR and QM are incompatible. It looks like my other
replier didn't read it too or he would be forced to agree with me. The
people here are retards.

---
And where are you, my dear?

--
John Fields
Professional Circuit Designer

On 2 Aug 2005 22:09:22 -0700, "Autymn D. C."
wrote:

John Larkin wrote:
There are no stupid questions, only stupid people.

There are stupid questions, those that could be easily found on one's
own.

---
Example???

--
John Fields
Professional Circuit Designer

On 2 Aug 2005 22:27:12 -0700, "Autymn D. C."
wrote:

Stop swearing or I'll beat your head in, the part that causes swearing.

---
Violent little bitch, eh?

Why don't you leave a snippet of the post you're replying to intact
so that those of us who aren't mind-readers don't have to go
searching to find out whom you're threatening?

BTW, don't be surprised if the law comes knocking at your door...
Threats of physical violence are looked on very seriously these
days.

--
John Fields
Professional Circuit Designer

On Wed, 03 Aug 2005 07:19:12 GMT, TokaMundo
wrote:


Sometimes I find it hard to believe that we actually call ourselves
sentient beings with the way some of you act.

---
I don't think anyone has ever called you sentient, much to their
credit.

--
John Fields
Professional Circuit Designer

On Wed, 03 Aug 2005 07:24:21 GMT, TokaMundo
wrote:

On 2 Aug 2005 21:58:47 -0700, "Autymn D. C."
Gave us:

John Fields wrote:
I see. Instead of reason, you prefer insult.

I will neither read your "proof" nor will I shut up, and if you
don't like it, you miserable son of a bitch, you can go ****
yourself.

I'm not a son, you illiterate and uncouth obscurantist troll. I
already used reason in the proof, which you wilfully ignore. It uses
maths too, which puts a limiting case on the Lorentz corrections,
remembering that GR and QM are incompatible. It looks like my other
replier didn't read it too or he would be forced to agree with me. The
people here are retards.


Hey, you lysdexic dufus! I was on your side!

---
Autymn and Tokey up a tree,
k-i-s-s-i-n-g

Now _there's_ a match made in heaven, lol...




--
John Fields
Professional Circuit Designer