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# Thermal conductivity of a "vacuum"

## Thermal conductivity of a "vacuum"

(OP)
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

### RE: Thermal conductivity of a "vacuum"

There is a 1/15 of the number of air molecules but they will travel further between collisions.

### RE: Thermal conductivity of a "vacuum"

The thermal conductivity of gases doesn't change much with pressure, but they do with temperature, which affects molecular motion. It varies (and so does sound velocity) with the square root of the absolute temperature.

I suggest you pay a visit to the NIST chemistry webbook looking for nitrogen, for example, and may find that the thermal conductivity at, say, 25 deg C doesn't change much between 1 atm abs and a high vacuum.

### RE: Thermal conductivity of a "vacuum"

I think you may actually be interested in coefficient of convection vs. pressure more than "k".

### RE: Thermal conductivity of a "vacuum"

(OP)
Well my question now is: why does anybody bother to make Thermos's...what's the point?

I'm pretty sure a Thermos worke quite well, so well in fact that I've burned my lips with them a few times not thinking coffee would still be hot 8 hours later...

Plastics Industry

### RE: Thermal conductivity of a "vacuum"

Three modes of heat transfer. Conduction is just one.

### RE: Thermal conductivity of a "vacuum"

I'm not sure I follow your conclusion.  Thermos(TM) is true vacuum path, hence the complications with glass or metal vecuum flasks.

The comments about independence of pressure only applies for a limited range.  Obviously, there can be no conduction if there's no gas, so it can't be completely independent over the entire span of pressures.

http://www.thermco.com/ss119.asp and
http://www.rdmag.com/ShowPR.aspx?PUBCODE=014&ACCT=1400000100&ISSUE=0410&RELTYPE=PR&ORIGRELTYPE=Pho&PRODCODE=00000000&amp;PRODLETT=U&CommonCount=0

At sufficiently low pressure, thermal conductivity is directly affected by pressure, otherwise, there are a lot of vacuum pressure meters that shouldn't be working.  The pressure stated in the OP is in-between, so there might be some amount of reduction in thermal conductivity.

### RE: Thermal conductivity of a "vacuum"

(OP)
Here's another question (sorry for all these "layman-like" questions)...if the "k" of air is .024, and the "k" of fiberglass insulation is .04, why do we insulate our houses with fiberglass?  Why not just use empty stud bays?  Obviously there's an easy answer to this...

I guess the only thing the fiberglass does is disrupt convection?

Plastics Industry

### RE: Thermal conductivity of a "vacuum"

That's part of it.  The other is that houses are actually quite leaky, so not only do you prevent convection, but also intrusion of external air.

### RE: Thermal conductivity of a "vacuum"

(OP)
But on an industrial vessel, where there is no intrusion of outside air, and even if there was it's not super cold, why use fiberglass...why not just use an air gap?

Plastics Industry

### RE: Thermal conductivity of a "vacuum"

You still need to prevent internal convection.  Any gap larger than about 1/2 in will experience natural convection.

That's why foamed insulators are so effective, the cells keep the air from circulating.

### RE: Thermal conductivity of a "vacuum"

Other gas thermophysical properties besides thermal conductivity, k, not much affected by pressure are: absolute viscosity μ, specific heat capacity cp and, as a result also the Prandtl number which is cp.μ/k.

### RE: Thermal conductivity of a "vacuum"

(OP)
So there really is no "plug and chug" equation for calculating the thermal conductivity of a gas (air) at a given pressure...bummer!

Plastics Industry

### RE: Thermal conductivity of a "vacuum"

(OP)
Nice links.  Bummer they don't go down into my pressure range, which is .007 MPa...

Thanks.

-Plasmech

Mechanical Engineer, Plastics Industry

### RE: Thermal conductivity of a "vacuum"

At the that pressure, you should be able to get some information from the hot-wire pressure gauge community, since that is the range that the hot-wire gauges can operate at.

In fact, at that pressure, you should already have a thermocouple gauge that's doing the calculation of pressure from the thermal conductivity of whatever's left in the chamber.

### RE: Thermal conductivity of a "vacuum"

OK, that's too high a pressure for that to work then.

NIST has a calculator for nitrogen:
http://webbook.nist.gov/cgi/fluid.cgi?Action=Data&Wide=on&;ID=C7727379&Type=IsoBar&amp;P=.0047&THigh=303&TLow=273&amp;TInc=1&amp;RefState=DEF&TUnit=K&amp;PUnit=MPa&DUnit=mol%2Fl&amp;HUnit=kJ%2Fmol&amp;WUnit=m%2Fs&VisUnit=uPa*s&STUnit=N%2Fm

Hopefully, that comes out correct.  The second to the last column is thermal conductivity, which is pretty close to what you had originally proposed.

If the link doesn't work, the values were:

273    0.023989
274    0.024059
275    0.024128
276    0.024197
277    0.024266
278    0.024335
279    0.024404
280    0.024473
281    0.024542
282    0.02461
283    0.024679
284    0.024747
285    0.024815
286    0.024883
287    0.024952
288    0.02502
289    0.025088
290    0.025155
291    0.025223
292    0.025291
293    0.025358
294    0.025426
295    0.025493
296    0.025561
297    0.025628
298    0.025695
299    0.025762
300    0.025829
301    0.025896
302    0.025963
303    0.026029

The pressure wound up at 0.0047 MPa

### RE: Thermal conductivity of a "vacuum"

(OP)
OK, so based on this data, having a vacuum jacket at my pressure (.006-.007 MPa) is NOT going to do anything for me in terms of CONDUCTION.  But...air still conducts less than fiberglass.  And if I I do pull a vacuum in that air gap, there will be no air movement to speak of (there will be mollecular movement, yes, but no "mass flow" of air whatsoever), so my CONVECTION heat transfer coefficient is going to be very small, probably less than 10 W / (m^2*C).

I think.

-Plasmech

Mechanical Engineer, Plastics Industry

### RE: Thermal conductivity of a "vacuum"

It'll be well less than that.

A 20-mm gap will will just get it up to 1 W/m^2-ºC, based on the data above.

### RE: Thermal conductivity of a "vacuum"

(OP)
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

### RE: Thermal conductivity of a "vacuum"

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.

### RE: Thermal conductivity of a "vacuum"

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

-Plasmech

Mechanical Engineer, Plastics Industry

### RE: Thermal conductivity of a "vacuum"

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

### RE: Thermal conductivity of a "vacuum"

(OP)
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

### RE: Thermal conductivity of a "vacuum"

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.

### RE: Thermal conductivity of a "vacuum"

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

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

and that lambda = 1/n?

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

-Plasmech

Mechanical Engineer, Plastics Industry

### RE: Thermal conductivity of a "vacuum"

Plasmech,

k = (somefactor)(n)(v)(lambda)

and that lambda = (someotherfactor)/n

so k = (andnow4somethingcompletelydifferent)v

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

### RE: Thermal conductivity of a "vacuum"

(OP)
um...should we mark that last post as spam?  ;)

-Plasmech

Mechanical Engineer, Plastics Industry

### RE: Thermal conductivity of a "vacuum"

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

### RE: Thermal conductivity of a "vacuum"

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

The same kinetic theory of ideal gases tells us that vaverage = factor * (T0.5) 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 * vaverage

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.

### RE: Thermal conductivity of a "vacuum"

(OP)
Thanks 25362! (what does your handle mean).

-Plasmech

Mechanical Engineer, Plastics Industry

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