Stupid question of the day....

[John Fields]

On Fri, 05 Aug 2005 04:43:00 GMT, TokaMundo
<TokaMundo@weedizgood.org> wrote:

On Thu, 04 Aug 2005 20:59:08 GMT, "daestrom"
daestrom@NO_SPAM_HEREtwcny.rr.com> Gave us:


No, you still don't see it.

No. You don't see "IT".

The heat in the center is conducted outwards
through an imaginary cylindrical surface separating the innermost core from
the next outer layer.

No. THAT is for cylinders where the heat source is at the center.
This heat source is throughout the medium.

---
If the surface of the conductor is cooler than the core, which it
_has_ to be by virtue of the fact that heat is radiating and being
convected away from it, then there will be a temperature gradient
from the conductor's center to its surface.

However, since the resistance of the cooler regions of the conductor
will be less than that of the hotter regions, the resistance of the
cooler regions will be lower, allowing more current to pass through
them. This will heat the cooler regions and more nearly make the
conductor isothermal.

So ya get a choice between a thermal gradient or a current gradient.
Or maybe some of both.



--
John Fields
Professional Circuit Designer

 

[Alexander]

"TokaMundo" <TokaMundo@weedizgood.org> schreef in bericht
news:i0r5f1hmahndt53joe66bufsditjor07se@4ax.com...

On 4 Aug 2005 05:36:24 -0700, "Autymn D. C." <lysdexia@sbcglobal.net
Gave us:

John Fields wrote:
And where are you, my dear?

there

Over there?
I was here, now i'm there.

You're not here so are yot there.
Wait a moment I will go there





Now I'm here which was there a moment ago and first here.
Strange a moment ago was there here and here there.

 

[Bob Myers]

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:gcr5f1lbi7u9hma8991a7ko1r8h484p9e1@4ax.com...

On Thu, 04 Aug 2005 19:24:11 GMT, "Bob Myers"
nospamplease@address.invalid> Gave us:

Is there no end of subjects in which you are willing to demonstrate
your ignorance? Io is NOT primarily heated by "the magnetic forces
of Jupiter," but rather by its gravity

I meant gravitational, not magnetic.

Of course you did. People are always mis-typing "magnetic"
when they mean "gravitational." The conclusion was still
wrong, either way.

Bob M.

 

[daestrom]

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in message
news:5425f1ph1v938jef1mahmlcrnbp56cbiph@4ax.com...

On Thu, 04 Aug 2005 20:59:08 GMT, "daestrom"
daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:


"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:ag83f199paf7od11jt0progsts34cc7nuj@4ax.com...
On Wed, 03 Aug 2005 15:54:26 -0500, John Fields
jfields@austininstruments.com> Gave us:

---
Sorry, Charlie, there'll still be a temperature gradient across the
diameter of the wire. There has to be, since the surface of the
wire will be radiating heat and being cooled by convection.
---

If it were modeled after the earth, with the heat source in the
center, I would agree. I feel, however, that it is more closely
modeled after Io, which is heated by the magnetic forces of Jupiter,
and more closely approximates an evenly heated body.

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.


No, you still don't see it. The heat in the center is conducted outwards
through an imaginary cylindrical surface separating the innermost core
from
the next outer layer. And the heat from the *that* layer, plus the heat
from the inner core is transmitted through the next concentric imaginary
cylindrical surface separating that layer from the area.

The surface area of each imaginary cylindrical surface increases
proportionally with the radius out from the center, but the heat that must
be conducted through each surface increases proportional to the radius
squared. So the temperature gradient across each imaginary cylindrical
surface gets stronger and stronger. Thus the *temperature* across the
cross-section is parabolic, even though the heat generation is flat/level.

This has *nothing* to do with the heat transfer from the outer most
surface
to it's surroundings. Heat removal from the outer surface by convection,
conduction, or radiation will *not* change the shape of the interior
temperature gradient.


That's very, very nearly true. But for any reasonably insulated wire,
the thermal conductivity of the insulation (including air) will be
minute compared to that of copper, so internal temp gradients will be
very low. To force a decent grad, you'd need to run a lot of current
through a wire directly in contact with water or something.

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.

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).

For good thermal conductors such as copper or Al, the gradient is tiny
unless you're working with extremely high powers. Poorer thermal conductors
like iron or ceramic-like allows, the gradient can be higher.

daestrom

 

[daestrom]

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:eir5f11sc016hfp6h3qn2vmj5cnn8ndvhe@4ax.com...

On Thu, 04 Aug 2005 20:59:08 GMT, "daestrom"
daestrom@NO_SPAM_HEREtwcny.rr.com> Gave us:


No, you still don't see it.

No. You don't see "IT".

The heat in the center is conducted outwards
through an imaginary cylindrical surface separating the innermost core
from
the next outer layer.

No. THAT is for cylinders where the heat source is at the center.
This heat source is throughout the medium.

No, it is *not* for cylinders with only a central source. You jump to the
wrong conclusion and stop reading without applying 'the little grey cells'.

It is for when the heat source is throughout the medium, just like a copper
wire. Heat is generated in each unit cross-section of the wire. The heat
in the central area must be conducted through the surrounding material. The
surrounding material also generates heat which must also be conducted away.
The further out from the center you get, the more and more heat must be
conducted outward through the surrounding material. Hence a larger
temperature gradient as you move from the center.

Maybe if you read through the whole explanation instead of just snipping it
off, you'd learn something new. Maybe you should open up your mind a bit.
Maybe you should study heat-transfer in a heat producing medium for a few
years.

daestrom

 

[daestrom]

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:eir5f11sc016hfp6h3qn2vmj5cnn8ndvhe@4ax.com...

On Thu, 04 Aug 2005 20:59:08 GMT, "daestrom"
daestrom@NO_SPAM_HEREtwcny.rr.com> Gave us:


No, you still don't see it.

No. You don't see "IT".

The heat in the center is conducted outwards
through an imaginary cylindrical surface separating the innermost core
from
the next outer layer.

No. THAT is for cylinders where the heat source is at the center.
This heat source is throughout the medium.

That makes the rest of your supposition incorrect so... SNIP.

To put it another way, if there is *any* heat generated in the central core
of the wire, the only way it can be conducted to the surrounding material is
if it is at a higher temperature than the surrounding material.

If the central core is *exactly* the same temperature as the surrounding
material, then *ZERO* heat would be transferred by conduction away from the
central core. And if heat is generated 'throughout the medium' (which it
is), then the heat in the central core will raise the temperature of the
central core. Now its temperature is higher than the surroundings and heat
begins to conduct away from the central core.

Often called the 'zeroth law of thermodynamics', no net heat is transferred
without a temperature difference.

daestrom

 

[John Larkin]

On Fri, 05 Aug 2005 21:17:47 GMT, "daestrom"
<daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in message
news:5425f1ph1v938jef1mahmlcrnbp56cbiph@4ax.com...
On Thu, 04 Aug 2005 20:59:08 GMT, "daestrom"
daestrom@NO_SPAM_HEREtwcny.rr.com> wrote:


"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:ag83f199paf7od11jt0progsts34cc7nuj@4ax.com...
On Wed, 03 Aug 2005 15:54:26 -0500, John Fields
jfields@austininstruments.com> Gave us:

---
Sorry, Charlie, there'll still be a temperature gradient across the
diameter of the wire. There has to be, since the surface of the
wire will be radiating heat and being cooled by convection.
---

If it were modeled after the earth, with the heat source in the
center, I would agree. I feel, however, that it is more closely
modeled after Io, which is heated by the magnetic forces of Jupiter,
and more closely approximates an evenly heated body.

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.


No, you still don't see it. The heat in the center is conducted outwards
through an imaginary cylindrical surface separating the innermost core
from
the next outer layer. And the heat from the *that* layer, plus the heat
from the inner core is transmitted through the next concentric imaginary
cylindrical surface separating that layer from the area.

The surface area of each imaginary cylindrical surface increases
proportionally with the radius out from the center, but the heat that must
be conducted through each surface increases proportional to the radius
squared. So the temperature gradient across each imaginary cylindrical
surface gets stronger and stronger. Thus the *temperature* across the
cross-section is parabolic, even though the heat generation is flat/level.

This has *nothing* to do with the heat transfer from the outer most
surface
to it's surroundings. Heat removal from the outer surface by convection,
conduction, or radiation will *not* change the shape of the interior
temperature gradient.


That's very, very nearly true. But for any reasonably insulated wire,
the thermal conductivity of the insulation (including air) will be
minute compared to that of copper, so internal temp gradients will be
very low. To force a decent grad, you'd need to run a lot of current
through a wire directly in contact with water or something.


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.

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.

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.

John

 

[TokaMundo]

On Fri, 05 Aug 2005 18:26:03 GMT, "Bob Myers"
<nospamplease@address.invalid> Gave us:

"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:gcr5f1lbi7u9hma8991a7ko1r8h484p9e1@4ax.com...
On Thu, 04 Aug 2005 19:24:11 GMT, "Bob Myers"
nospamplease@address.invalid> Gave us:

Is there no end of subjects in which you are willing to demonstrate
your ignorance? Io is NOT primarily heated by "the magnetic forces
of Jupiter," but rather by its gravity

I meant gravitational, not magnetic.

Of course you did. People are always mis-typing "magnetic"
when they mean "gravitational." The conclusion was still
wrong, either way.

Even after I admit I was wrong, you are still an asshole.
Fuck you.... Bob. Like I said, I likely knew about Io long before
you did. ALL about it.

 

[NunYa Bidness]

On Fri, 05 Aug 2005 06:31:42 -0500, John Fields
<jfields@austininstruments.com> Gave us:

On Fri, 05 Aug 2005 04:36:15 GMT, TokaMundo
TokaMundo@weedizgood.org> wrote:


John Fields
Professional Self Aggrandizer

Your lame ass also accuses people of "self aggrandizement".
I have yet to see one post from you where you don't do the same.
Funny, since you're no more than a fat tub of lard.

---
LOL, if I were to write the single line: "Tokamundo is a good guy."
and post it, you'd critcise it in your boring, predictable way.

Probably by calling me a liar.

You didn't even get it. You jump on others when your sig is
exactly what you claim others to be. Self aggrandizing bullshit.

 

[NunYa Bidness]

On Fri, 5 Aug 2005 17:03:29 +0200, "Alexander"
<electricdummy@hotmail.com> Gave us:

"TokaMundo" <TokaMundo@weedizgood.org> schreef in bericht
news:i0r5f1hmahndt53joe66bufsditjor07se@4ax.com...
On 4 Aug 2005 05:36:24 -0700, "Autymn D. C." <lysdexia@sbcglobal.net
Gave us:

John Fields wrote:
And where are you, my dear?

there

Over there?
I was here, now i'm there.
You're not here so are yot there.
Wait a moment I will go there





Now I'm here which was there a moment ago and first here.
Strange a moment ago was there here and here there.

If today was yesterday...

 

[NunYa Bidness]

On Fri, 05 Aug 2005 21:24:21 GMT, "daestrom"
<daestrom@NO_SPAM_HEREtwcny.rr.com> Gave us:

Often called the 'zeroth law of thermodynamics', no net heat is transferred
without a temperature difference.

It is in motion. Always. Whether in local thermal equilibrium or
not.

 

[John Fields]

On Sat, 06 Aug 2005 10:03:17 GMT, NunYa Bidness
<nunyabidness@nunyabidness.org> wrote:

On Fri, 05 Aug 2005 06:31:42 -0500, John Fields
jfields@austininstruments.com> Gave us:

On Fri, 05 Aug 2005 04:36:15 GMT, TokaMundo
TokaMundo@weedizgood.org> wrote:


John Fields
Professional Self Aggrandizer

Your lame ass also accuses people of "self aggrandizement".
I have yet to see one post from you where you don't do the same.
Funny, since you're no more than a fat tub of lard.

---
LOL, if I were to write the single line: "Tokamundo is a good guy."
and post it, you'd critcise it in your boring, predictable way.

Probably by calling me a liar.


You didn't even get it. You jump on others when your sig is
exactly what you claim others to be. Self aggrandizing bullshit.

---
The reason I didn't 'get' it is because my sig, you insignificant
pile of shit, merely points out how I make a living. What rankles
your ass and the asses of your ilk is the fact that I choose to call
myself a professional, in public, while you can't and not expect to
be laughed out of Dodge if you did.

Well, I'll tell ya what, buddy-boy, my sig isn't going to change
anytime soon and if galls you to have to read it because you can't
stand to plonk me, then that makes my day!

Thanks, asshole!
--
John Fields
Extra Special Super-Duper Extremely Professional Circuit Designer

 

[John Fields]

On Sat, 06 Aug 2005 10:05:23 GMT, NunYa Bidness
<nunyabidness@nunyabidness.org> wrote:

On Fri, 5 Aug 2005 17:03:29 +0200, "Alexander"
electricdummy@hotmail.com> Gave us:


"TokaMundo" <TokaMundo@weedizgood.org> schreef in bericht
news:i0r5f1hmahndt53joe66bufsditjor07se@4ax.com...
On 4 Aug 2005 05:36:24 -0700, "Autymn D. C." <lysdexia@sbcglobal.net
Gave us:

John Fields wrote:
And where are you, my dear?

there

Over there?
I was here, now i'm there.
You're not here so are yot there.
Wait a moment I will go there





Now I'm here which was there a moment ago and first here.
Strange a moment ago was there here and here there.

If today was yesterday...

---
You'd be a little less retarded than you are today?

Just a little snack for ya, troll boy!


--
John Fields
Professional Circuit Designer

 

[John Fields]

On Sat, 06 Aug 2005 03:57:13 GMT, TokaMundo
<TokaMundo@weedizgood.org> wrote:

On Fri, 05 Aug 2005 18:26:03 GMT, "Bob Myers"
nospamplease@address.invalid> Gave us:


"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:gcr5f1lbi7u9hma8991a7ko1r8h484p9e1@4ax.com...
On Thu, 04 Aug 2005 19:24:11 GMT, "Bob Myers"
nospamplease@address.invalid> Gave us:

Is there no end of subjects in which you are willing to demonstrate
your ignorance? Io is NOT primarily heated by "the magnetic forces
of Jupiter," but rather by its gravity

I meant gravitational, not magnetic.

Of course you did. People are always mis-typing "magnetic"
when they mean "gravitational." The conclusion was still
wrong, either way.


Even after I admit I was wrong, you are still an asshole.
Fuck you.... Bob. Like I said, I likely knew about Io long before
you did. ALL about it.

---
You're a goddam liar.

You never said you were wrong, all you said was that you transposed
two words. And, you don't know _all_ about Io, no one does.

What happened was that Bob caught you with your hand in the cookie
jar and you tried to slime out of it in your typical fashion.

What a fuggin' maroon!!!

--
John Fields
Professional Circuit Designer

 

[TokaMundo]

On Sat, 06 Aug 2005 06:42:34 -0500, John Fields
<jfields@austininstruments.com> Gave us:

On Sat, 06 Aug 2005 03:57:13 GMT, TokaMundo
TokaMundo@weedizgood.org> wrote:

On Fri, 05 Aug 2005 18:26:03 GMT, "Bob Myers"
nospamplease@address.invalid> Gave us:


"TokaMundo" <TokaMundo@weedizgood.org> wrote in message
news:gcr5f1lbi7u9hma8991a7ko1r8h484p9e1@4ax.com...
On Thu, 04 Aug 2005 19:24:11 GMT, "Bob Myers"
nospamplease@address.invalid> Gave us:

Is there no end of subjects in which you are willing to demonstrate
your ignorance? Io is NOT primarily heated by "the magnetic forces
of Jupiter," but rather by its gravity

I meant gravitational, not magnetic.

Of course you did. People are always mis-typing "magnetic"
when they mean "gravitational." The conclusion was still
wrong, either way.


Even after I admit I was wrong, you are still an asshole.
Fuck you.... Bob. Like I said, I likely knew about Io long before
you did. ALL about it.

---
You're a goddam liar.

You are, fuckhead.

You never said you were wrong, all you said was that you transposed
two words.

There was only one word used, so I didn't say anything about
transposition. What I said was that I indeed meant gravitational,
even though I said magnetic. Your fucked in the head assessment of it
means nothing.

And, you don't know _all_ about Io, no one does.

I know all there is in current understanding of it. That is usually
what one means when one says such a thing. I wouldn't expect a
retarded bastard like you to understand, however. In fact, I knew
your retarded troll ass would chime in on that very phrase the moment
I wrote it, asswipe.

What happened was that Bob caught you with your hand in the cookie
jar and you tried to slime out of it in your typical fashion.

You're an idiot, in typical lard ass, Johnny boy fashion.

What a fuggin' maroon!!!

You insult Bugs Bunny every time you use that word, troll boy,
and you're a fucking wussy when it comes to cussin' on usenet.

 

[John Larkin]

On Fri, 05 Aug 2005 17:33:59 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

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.

John

Which leads into some interesting stuff about thermal conductivity
integrals and Richard Garwin.

www.lakeshore.com/pdf_files/Appendices/LSTC_appendixI_l.pdf

RG sounds like he was quite a guy, involved in all sorts of stuff
including development of the FFT. He's worth googling.

John

 

[John Fields]

On Sat, 06 Aug 2005 19:08:29 GMT, TokaMundo
<TokaMundo@weedizgood.org> wrote:

On Sat, 06 Aug 2005 06:42:34 -0500, John Fields
jfields@austininstruments.com> Gave us:

You're a goddam liar.

You are, fuckhead.

---
No you are, you little parrotting pipsqueak.
---

You never said you were wrong, all you said was that you transposed
two words.

There was only one word used, so I didn't say anything about
transposition. What I said was that I indeed meant gravitational,
even though I said magnetic. Your fucked in the head assessment of it
means nothing.

---
I was using the two words that you tried to use to make it seem like
you knew what you were talking about all along when you didn't. You
always do that, wait until somebody gives you the answer and then
you pretend you knew it all along, ya drooling knuckle-dragger.
---

And, you don't know _all_ about Io, no one does.

I know all there is in current understanding of it. That is usually
what one means when one says such a thing.

---
Maybe, but that's sure as hell not you! You want everybody to think
you're right up there with all the facts at your fingertips, so when
someone explains something that you _didn't_ understand or know
about, you say, "Oh yeah, I knew about that", and if you get caught
you yell and scream and stamp your feet after you try the "Oh,
wait, that's not what I meant..." dance and you get busted on that
one too, ya slobbering, lightweight goon.
---

I wouldn't expect a
retarded bastard like you to understand, however. In fact, I knew
your retarded troll ass would chime in on that very phrase the moment
I wrote it, asswipe.

---
And yet, you wrote it anyway? Awwww... How sweet! You must be
thinking of me constantly and trying to figure out how to beguile me
into responding to that tripe that you try to pass off as
"writing", ya queer faggot cocksucker. (Not that there's anything
wrong with that...)
---

What happened was that Bob caught you with your hand in the cookie
jar and you tried to slime out of it in your typical fashion.

You're an idiot, in typical lard ass, Johnny boy fashion.

---
Hey, phonyphuck, I'm not the one backpedalling my ass off trying to
lie my way out some shit I got myself into by trying to blame
everyone else for it, so I think it's pretty clear who's wearing
idiot shoes around here. Oh, yeah... just in case you can't figure
out who that is, bonehead, its _you_, ya pathetic clown.
---

What a fuggin' maroon!!!

You insult Bugs Bunny every time you use that word, troll boy,
and you're a fucking wussy when it comes to cussin' on usenet.

---
LOL, and you make Ol' Wiley look like a genius after pulling some of
that sorry-ass shit _you_ think is gonna work.

"Wussy"? Are you so afraid to be labeled a mysoginist that you
can't even write "pussy", ya bony-ass fuckin' pussy coward?

--
John Fields
Professional Circuit Designer

 

[TokaMundo]

On Sat, 06 Aug 2005 15:23:30 -0500, John Fields
<jfields@austininstruments.com> Gave us:

Maybe, but that's sure as hell not you! You want everybody to think
you're right up there with all the facts at your fingertips, so when
someone explains something that you _didn't_ understand or know
about, you say, "Oh yeah, I knew about that",

You're full of shit, boy. If you had read the thread, you would
have noted where I said that I have had Nasa's Laser Disc on the
subject for over 15 years.

Did your lame ass even know what decent video was back then? Did
you get into laser discs? I doubt it. You're a twit. Nothing more.

and if you get caught
you yell and scream and stamp your feet after you try the "Oh,
wait, that's not what I meant..."

You're an idiot.

dance and you get busted on that
one too, ya slobbering, lightweight goon.

You still like mouthing off when you're more than an arm's length
away. You're still also full of shit, Stinky.

>---

 

[daestrom]

"NunYa Bidness" <nunyabidness@nunyabidness.org> wrote in message
news:0539f1934hof6069l96hma7s749gvjo6v9@4ax.com...

On Fri, 05 Aug 2005 21:24:21 GMT, "daestrom"
daestrom@NO_SPAM_HEREtwcny.rr.com> Gave us:

Often called the 'zeroth law of thermodynamics', no net heat is
transferred
without a temperature difference.

It is in motion. Always. Whether in local thermal equilibrium or
not.

That's why I said *net* heat transfer.

daestrom

 

[daestrom]

"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in message
news:6d08f1d39hgu458cpg4cjchqgr91ln1l12@4ax.com...

On Fri, 05 Aug 2005 21:17:47 GMT, "daestrom"
daestrom@NO_SPAM_HEREtwcny.rr.com> 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).

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).

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