Rate of thermal conduction in various copper structures

steveh721

A cylinder is constructed from a super insulator. The dimensions of the cylinder are Length L: 1 M; inside diameter i.d.: 12 mm. The entire Length of the cylinder contains copper (cases described below).

Heat (T > 100C) is continuously applied to the copper on one end of the cylinder. There is a heat sink attached to the copper at the opposite end of the cylinder--it can dissipate the heat at or above the rate of heat absorption.

What will be the difference in rate of conduction, if any, for:

a) the cylinder contains solid copper (12mm dia. X L)

b) the cylinder is packed with copper powder (say, 30 microns)

c) the cylinder is packed with small diameter copper rods (say, 1mm dia. X L)

d) the cylinder is packed with small diameter copper tubes (say, 5mm o.d., 4m i.d. X L)

e) the cylinder contains a uniform copper lattice (honeycomb X L) structure surrounded by air

If there's a difference, which case results in the highest conduction rate and why? Lowest?

CWatters

The thermal conductivity will be proportional to cross sectional area and inversely proportional to length. So cases a) c) and d) should be easy enough.

Case b) and e) is tricky because instead of having a uniform cross section you have lots of voids in three dimensions. I'm not sure how you go about calculating the thermal conductivity but it will obviously be lower than for case a) which is a solid rod.

In a material with small pores the mean free path of air molecules (eg Nitrogen) might be less than the pore size. So they are more likely to collide with the walls of the pore than each other. So the thermal energy will be transferred to the conducting copper pore walls rather than travel through the air in the pores. I think this means b) is likely to have increased conductivity compared to e). Edit: and perhaps quite close to that of a).

steveh721

Thank you for your comments.

I was anticipating responses would be more toward d) and/or e) because the infrared spectrum is short wavelength/high frequency electro-magnetic energy. And, at least in the RF spectrum, high frequency energy propagates on the surface--and increased surface area has reduced impedance. Consequently, I imagined that more surface area would also result in reduced propagation time for infrared, too.

Could there be any validity to this idea?

CWatters

Pretty sure copper is opaque to infrared :-)

Conduction usually beats radaition unless the temperature difference is very high.