"TokaMundo" wrote in message
...
On Sun, 07 Aug 2005 12:05:21 -0700, John Larkin
Gave us:

On Sun, 07 Aug 2005 16:10:37 GMT, "daestrom"
wrote:


"John Larkin" wrote in
message
...
On Fri, 05 Aug 2005 21:17:47 GMT, "daestrom"
wrote:

snip


Well I *almost* agree with you. To get a severe gradient, you do need
to
run a lot of current. But it still does *NOT* matter what is in
contact
with the outside surface.

Sure it does. If you run a copper wire in air, and dump in enough
current to produce a decent radial gradient, it will vaporize. You'd
have to water cool it (boiling water is ideal) to sustain the power
levels necessary for a non-trivial gradient.

Copper conducts heat about 12,000 times as well as air, and there's a
lot more air available than copper in most situations. So, very
roughly, a 1mm copper wire surrounded by a 10 mm air gap, with enough
current flowing to create a 1 deg C internal gradient, will have a
surface temp of 120,000 C.



Nah... The thermal conductivity of a film coefficient for air is not the
same as the thermal conductivity of air. The thermal conductivity is
only
relavent in a very thin layer against the surface (much less than 10 mm).
Moving outward, the viscosity and velocity of the air become dominant.
Given the film coefficient of air against a vertical surface of about 25
W/(m^2-K), I make it out to only be about 8,000 C. ;-) Having forced air
convection (or a good 'stiff' wind) can improve the film coefficient to
almost 200 W/(m^2-K) (down to 1,000 C ;-). Water cooling can be as high
as
5000 to 10000 W/(m^2-K) (as low as 20 C).



I did hedge my number with "very roughly", figuring I could be 2
orders of magnitude off and still make the point.


But larger wires, and those of Al can develop such a gradient more
easily.
And true, boiling heat transfer can be several orders of magnitude
better,
but one then has to worry about exceeding the critical heat flux (also
known
as 'departure from nucleat boiling', 'boiling transition', or 'dryout').
Whether the water is circulating or not, and how far the bulk water
temperature is from saturation also become important (i.e. becomes a real
engineering nightmare).

A spiral of #10 bare copper wire in a plastic garbage can full of
water makes an impressive dummy load, up until the water gets hot
enough to melt the plastic can. Then the hot water gets loose. Keep a
good chair handy.


The industry has a long history of success using pressurized hydrogen.
Most
large generators and their connections to step-up transformers are cooled
this way. Much better cooling than plain air, allowing much higher
current
densities. And with the same material properties, stronger temperature
gradients.

Except all of the H2-cooled gen-xfmr leads that I've seen use hollow
conductors with H2 forced through the center as well as surrounding the
outside. Similarly, the water-cooled conductors that I've seen are those
found in generators and the water flows down the center of the hollow
conductor. Not much of a temperature profile when the cross-section is
mostly cooling water ;-)

The internal gradient is a function of the heat
generated per unit mass and the thermal conductivity of the material.
Period. Nothing else.


Not once it's gaseous.

True, but one usually designs to avoid melting, much less boiling.

Fact is, in 60hz applications, the usual design restrictions regarding
skin-effect overshadow any problems with centerline temperature concerns.
Perhaps engineers working with high-current DC applications are more
concerned with the temperature gradient issues. But I suspect it is
still
small for good thermal conductors like copper.

I jumped into this fray when 'TokaMundo' said, "In a wire,....would show
the
wire at the same temp from center to outer surface". I think we agree
this
is wrong. And I agree that the temperature gradient is not severe for
conductors made of Cu or Al under normal circumstance such as air
cooling.
But *some* gradient *must* exist, otherwise the centerline temperature
must
increase (due to heat generated and not conducted away) until a gradient
begins to conduct heat away as fast as it's created by the electric
current.

Wonder how bad it is for graphite rods used in electric furnaces? Of
course
graphite has a much higher melting temperature so it can withstand a
strong
gradient. But graphite, with its lower thermal conductivity and higher
resistivity, probably develops a very strong gradient. Coupled with the
temperature coefficient of resistivity, it might make for an interesting
current distribution. Even for DC applications.


The external medium will determine the exact temperature of the outer
surface, and by virtue of the gradient for the specific material/power,
the
centerline temperature. But the shape and relative height of the
gradient
is irrespective of the external surface (as long as the thermal
conductivity
and heat produced are assumed constant).

Thermal conductivity is itself a function of temperature, so the
gradient does depend mildly on the absolute temperature of the whole
rig. Especially after the copper melts.


True. But below the melting point, it isn't hard to approximate the
variance with a low-order polynomial using temperature alone as the
independent variable. I would think this would make it relatively easy
to
incorporate into the integration. Haven't tried it though, so who
knows???

daestrom



My conclusion from this thread is that skin effect can be important at
60 Hz in entirely practical situations, and thermal gradients in
copper or aluminum conductors are inconsequential unless the current
is high and the cooling novel. We're doing some thermocouple stuff
just now (a simulator module and a complementary measurement gadget,
for jet engine testing) so thermal stuff is on my mind.

I've done a little superconductive/cryo work, where things are very
different. Here, the thermal conductivity of metals changes radically
as a function of temperature, so the net heat flow of, say, a
stainless or manganin leadwire from 4K up to to room temp is
determined by a complex integral (the bottom line of which,
fortunately, you can just look up.)

Yeah, the Toka guy is weird. He insists on crudely insulting anyone
who disagrees with him, and he's usually wrong. Some people seek and
need public humiliation: Usenet pain sluts, as it were.


That's funny since your position supports what I said about the
gradient being negligible, not the full on slope that daystruck
equates.

Here, does this look familiar???
8/4/2005 1:24 AM TokaMundo wrote....
In the wire, since the heat is generated throughout the medium via
current flow, even from low currents on up to my cherry red scenario
would show the wire at the same temp from center to outer surface.

You claim that because the 'heat is generated throughout the medium...' the
wire would be at the *same* temperature from center to outer surface??
That's just plain wrong, and you're the one that said it.

Further down in the same post you also said...
The "thermal skin" of the wire that would be slightly cooler due to
surface convection is very thin and beneath it the medium has even
temperature, not a gradient to the center.

The temperature gradient is *not* limited to some imaginary 'very thin'
"thermal skin" as you tried to say here. Again you're just plain wrong.

And as far as what *I* said, on 8/2/2005 6:01PM
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.

And later to John Larkin I replied...
Well I *almost* agree with you. To get a severe gradient, you do need to
run a lot of current. But it still does *NOT* matter what is in contact
with the outside surface. The internal gradient is a function of the heat
generated per unit mass and the thermal conductivity of the material.
Period. Nothing else.

I haven't said any where that there is a 'full on slope' (whatever that's
supposed to mean).

Having a little trouble admitting your mistakes are you??

daestrom