Conductivity of Heat Through Copper?

depends on ambient temperature and grade of copper

About 25 C, I guess, and whatever they make standard plumbing pipe out of.

Ballparkish figures are ok.

It's copper pipe which I've cut lengthwise and flattened. It's air on both sides. I'm using it as a fin to conduct heat from the hotspot of a parabolic trough to another, thinner, copper tube containing ethanol.

I had a look at Fourier's law, but I don't know the hot or less hot temperature, I was hoping to use the wattage to calculate them.

Speaking of which; how do I calculate the temperature of one end of a plate, given a wattage gradient across it? What do I consider the wattage per volume to be?

I don't know. I'm as dumb as a cat, and I hate converting everything into units I can understand. The only metric unit I can relate to is 9mm.

Using table from lynlynch...

The thermal resistance of your piece of copper as described is

0.03[m] / (400 [W/m-degC] x 0.15[m] x 0.001[m]) = 0.5 [degC/W]

If you have a source of power delivering 50 [W] to one edge, 50 [W] will flow to the other edge (ignoring side losses) and the temperature RISE across the copper piece should be

50 [W] x 0.5 [degC/W] = 25 [degC]

So the hot side should rise to a temperature 25 [degC] above the cold side (containing the ethanol) when 50 [watts] of heat is flowing through the copper.

Suggest reviewing literature describing thermal calculations for electronics and heat sinks.
Something like this might be useful... http://www.altera.com/literature/an/an185.pdf

You gotta have a temperature difference. And watts is power, not energy, so that's not the way to approach it. But, temperature will drive the process.

Assuming no losses you will need a temperature difference of about 27ºC to transfer 50 w over 30 mm of a 150 mm x 1 mm conductor. The radiation loss for the average temperature rise will be about 1.4 W.

Ball park figure will be about 48.6W delivered to ethanol.

Bioramani

Oh, ok, cool. So the losses are pretty small then?

Out of interest, how would that look for aluminium? (250 W/m/K)

Quote:how would that look for aluminium? (250 W/m/K) Unquote

Since aluminum conductivity is about half that of copper the temperature rise would be nearly double, say about 50ºC, for which the radiation loss will also be nearly double, say 2.8 to 3 W.

Bioramani

Gotcha, in order to transfer 50 watts (minus side radiation) through the metal I need to establish a hot end temp of 103 C (78 + 25).
(I'm guessing a greater hot temp will not increase the rate of flow, since there's a maximum of 50 watts going in?)

So final question: how do I calculate the rate of temperature increase on the hot end? I know what 50 watts will do to 150 X 30 X 1 mm copper (4.5 ml, 40 g), but how do I figure in the gradient?

The difference needed to transfer 50W is 25 degC. If the cold end is at a fixed temperature the gradient is automatically fixed by the wattage. Either the cold end or the hot end should be constant if the wattage is fixed. Since the cold end is ethanol, probably the cold end will react its boiling point. Gradient is directly controlled by the wattage. If this is a fixed 50W the hot end is automatically fixed. Curious to know how a fixed wattage is input into the hot end. Bioramani

It depends upon how little heat is lost to the environment between the two ends.