Thermal conductivity of a "vacuum" 1

[Plasmech]

How would I calculate the thermal conductivity, k, of a "vacuum", meaning obviously a "thickness" of air that is not at atmospheric pressure? Would I simply proportion down the "k" of air? I other words, if my "vacuum" was 1/15th of an atmosphere, would I say that k=.025 (w/m*k) / 15 = .00166 (w/m*k)?

Plastics Industry

 

[Plasmech]

What would a 4 inch gap give me in terms of convective coefficient? Can't imagine it would be much different...

-Plasmech

Mechanical Engineer, Plastics Industry

 

[IRstuff]

Should be down to about 0.25 W/m^2-ºC. Sounds small, but time will still be a killer if you're trying to maintain temperature passively.

TTFN

FAQ731-376

 

[Plasmech]

Thanks yet again IR. How did you come up with the .25 may I ask?

-Plasmech

Mechanical Engineer, Plastics Industry

 

[IRstuff]

There's a Mathcad e-book that does similar calculations, cranking the Nusselt, Raylegh, and Prandtl calculations.



TTFN

FAQ731-376

 

[Plasmech]

IR, (bringng this topic up again),

do you know why thermal conducticity of a gas does not decrease proportionally with pressure as one (like me) would intuitively think it would?

-Plasmech

Mechanical Engineer, Plastics Industry

 

[25362]

Grosso modo explanation: from the kinetic theory of gases the thermal conductivity k of a gas at a given temperature is proportional to n.v.[λ]

where n = number of molecules per unit volume, indeed directly related to pressure
v = mean velocity of molecules
[λ] = mean free path between collisions proportional to 1/n

Thus, the "density" factor vanishes and k is independent of the pressure of the gas.

 

[Plasmech]

So if I assume the temperature remains constant, can I assume the "v" remains constant?

Are you stating that k=**(lambda) ?

and that lambda = 1/n?

wouldn't the "n's" cancel out?



-Plasmech

Mechanical Engineer, Plastics Industry

 

[btrueblood]

Plasmech,

k = (somefactor)(v)(lambda)

and that lambda = (someotherfactor)/n

so k = (andnow4somethingcompletelydifferent)v

..and I'd better quit writing equations now.

 

[Plasmech]

um...should we mark that last post as spam?

-Plasmech

Mechanical Engineer, Plastics Industry

 

[btrueblood]

Yes, and I'll leave now, and take the Vikings with me.

 

[25362]

"Come, friend Watson, the curtain rings up for the last act" Sherlock Holmes.

The same kinetic theory of ideal gases tells us that v_average = factor * (T^0.5) for any particular (ideal) gas with T being absolute temperature.

n, in fact a "density" depending on pressure, cancels out as btrueblood pointed out leaving

k = new factor * v_average​

independent of pressure.

It has been claimed that with pressures down to 0.1 torr random collisions indeed follow the above statements.
At higher vacuum levels when the mean free path length becomes of the same order of magnitude as the distance between the walls, intermolecular collisions are less important and ballistic trajectories become preeminent, with pressures again affecting the conductivity.

Nothing clears up a case so much as stating it to another person Sherlock Holmes.

 

[Plasmech]

Thanks 25362! (what does your handle mean).

-Plasmech

Mechanical Engineer, Plastics Industry