Energy transfer by radiation

Yup. It's about the same as 4 mm or so of air at room temperature. (I had to worry about that when I was doing Footprints.) That's why you win with superinsulation--not only the high reflectance, but the radiation from the high temperature region drops by a lot as you work inwards.

If the shield is otherwise isolated, it'll have to come to radiative equilibrium, so eventually it'll be as though it wasn't there, regardless of what its characteristics are. The emissivity and stuff will determine its equilibrium temperature.

I doubt that you'd have a blackbody spectrum, and its detailed shape would be hard to predict unless you assume that the emissivities are perfectly flat.

Cheers

Phil Hobbs

It's weird (and maybe common), I knew the number (500W/m^2 at 300K), but I didn't *know* it. For low temperatures, 300K radiation is like having 1/2 your area exposed to full sunlight. That warms you up fast!

Oh sorry, the shield will be controlled. There's a question of how tightly controlled the temperature needs to be.

Yeah sure, assume flat emissivities. This is a "for fun" type question. But it seems to me that the radiation would be at an intermediate temperature. And closer to which every "thing" had the larger emissivity*Area.

So with really super mirrors on the walls and a *black* sample in the center, I see mostly the black sample temperature. Hey, say I then remove the black sample, how long till I get radiation to show the mirror temperature. (pretty darn fast I assume.)

George H.

That's not true; superinsulation works well because of the fact that two radiative couplings in series occur, and the relevant temperature difference for a double-radiation is 1/2 K, while that for a single-radiation is 1 K. The derivative of the radiation with respect to T (absolute temperature) is proportional to T**3, so a shield benefits both because you get the white-body effect twice, AND because you get up to a factor of (1/2)**3 = 0.125 from the diminution of temperature-offset from 1 K to 0.5 K.

Quite right, thanks. Half it is.

Cheers

Phil Hobbs

Hi whit3rd, K is the temperature difference? So the thermally isolated shield is at 1/2 the temperature difference? (well assuming all sorts of equal conditions.)

OK can you explain more about this point. (or direct me to a good book/link/) So if I hang a black conductive sheet, (copper painted black?) between two surfaces at differnt T's. The sheet comes to 1/2 the T difference. And the radiative heat flow is reduced by 1/8, 'cause it's gotta go from A->S->B rather than A->B

And then I would gain further by making the surfaces reflective. (Oh I'm assuming that everything's in vacuum, no other heat paths.)

Super insulation is a bit magical, (at least to me.) (I should get some aluminized mylar.) George H.

Does not sound right to me, regardless of the 'logic'. That would mean that you could make an essentially perfect insulator with a large number of sheets between the inner and outer surfaces.

Seems to me all you are doing is trading heat transfer by radiation for thermal conductivity.

Yup, many layers is exactly superinsulation, the mylar sheets are insulating on one side so that stops the thermal conductivity and the hanging layers don't touch too much. (Well at least in the one dewar I saw apart.) I've seen LN2 dewars with what looked to me like amazing hold times/ loss rates.

George H.

It does mean that. Do the arithmetic.

The thermal conductiity of 1nm of vacuum is the same as for 1km filling some of the distance with solid does no harm.

No.

It is in proper black body thermal equilibrium when the heat arriving from the hot side is equal to the heat leaving from the cold side.

So taking equal areas at temperatures hot, T, shield S, cold t

Hot Shield Cold

T S t | ~> > t then

S ~ 0.84*T(1 + (t/T)^4/4 )

No the other poster described it correctly the overall heat flow through the isolated shields position is reduced by a factor of two when it is interposed between the object at temperatures t and T.

If you can give it a mirror finish the temperature is the same but the emissivity is considerably less and the other surface mostly see their own reflected photons with a small contribution from the shield.

You can halve the flow again using the same trick but ultimately since you cannot have a perfect vacuum you fall foul of mean free path conductivity. Plenty of cryo systems use multiple heat shields.

To play about with it try the polystyrene foam backed aluminium or space blanket in sunshine vs black felt and you will quickly see.

Interestingly that isn't quite true. The EM field due to thermal excitations contains a lot of evanescent components, which can be sensed if you're much less than a wavelength from the surface. So putting them extremely close does increase the thermal conductance.

Cheers

Phil Hobbs

A bit academic, but would a low emissivity couple evanescent waves? Probably to the extent of it's absorptivity.

The issue of multiple layers as thermal insulators can be solved several ways. The most efficient is low e layers spaced from each other. At STP, the convective transfer goes up if they are too close, due to diffusive transfer.

Another solution is IR absorbing layers spaced as before. This works by requiring the IR to absorb and reradiate n times.

A final solution, at STP, is to use IR transparent layers (like polyethylene) with a low emissivity surface as heat source. This works by achieving a low e due to the T + R + A = 1 relation. Convective transfer will still occur, but the radiation relay component is suppressed.

At STP, a rule of thumb is that the radiative transfer is about equal to convective, for typical planar geometries like windows.

Another useful rule is Wien's Law, which is (wavelength microns) x (Kelvin Temperature) = 3000 or so. Thus STP heat rays have a 10 micron wl as the BB peak.

The issue of a large area low e chamber surrounding a high e small area absorber is counter-intuitive, at least to me. I believe the answer is the chamber e is effectively high for calculations, but could be wrong. Olbers paradox is related.

In the past, as a student, I made a 5 layer solar collector from Silicone sandwiched between FEP clear teflon. This collector achieved 425 F. Then I developed "heat mirror" coatings, which are light-transparent, but IR reflecting. This is achieved by creating an interference filter sandwiched around a conductive metal film. The IF matches though the metal in the vis, but goes away in the IR. Nifty metallurgical furnaces are made with concentric Vycor tube and a gold-based version of this coating, allowing visual inspection close up. JCC Fan did the theoretical and lab work originally, has anyone run into him?

Oh, Duh (head slap) 1/2 it is then.

OK thanks for the above Martin, (I'm still a bit confused on exactly how it changes when the shield is reflective.. but I'll think about it.)

George H.

Niote: I believe you should have used t**4, not the derivative. So that factor becomes .064, if one believes the analysis. If so, I thin k you would multiply the .064 by 2, and get the .125 overall.

Say, a trick question to ask physicists: (innocently) Say, I heard heat rises, but I have seen forest fires race DOWN hills. How does that happen?

jb

Er, no, I was thinking that 1 K was the nominal temperature difference between the inner and outer layer; then adding an intermediate layer, which will equilibrate at 1/2 degrees Kelvin from each of the two original layers.

That step depends on the layer-to-layer temperature drop being modest compared to the absolute temperature of the layers.

So, the net heat flow is dependent on the deriative with respect to temperature of the total heat flow, times the temperature difference (in this case, one-half Kelvin).

OK, here's where we diverge. I wasn't assuming either side near absolute zero but rather was concerning myself with doubling-the-layers in a superinsulation blanket, thus halving the interlayer temperature drop. My calculation was intended to apply to small temperature variations, far from zero Kelvin temperatures. My model was for tepid coffee in a Dewar, not for liquid He boiloff.

Radiative and conductive heat transfer may be roughly equal at STP, but for est fires are a lot hotter than STP - radiative heat transfer is what burns you if you get too close to a forest fire (unless a burning tree falls on you, as happens all too frequently).

There's also fairly vigorous convective circulation in the vicinity of a fo rest fire - the burning embers initially fly upwards, but as the surroundin g air rises it also cools and the still-burning embers come down all around the fire front - behind and in front of it when the fire-front is a more o r less linear.

On the downhill side the burning embers may have further to fall, but not e nough to make much of a difference.

orest fires are a lot hotter than STP - radiative heat transfer is what bur ns you if you get too close to a forest fire (unless a burning tree falls o n you, as happens all too frequently).

--

Yes, exactly. The radiative transfer dominates due to t**4 transfer, and radiation happy to go down hill. Incidentally, you can make a superior fir e place that way. Just arrange your logs to form an open box facing into the room. Then when the interior of the cavity gets hot, the rad xfer into room much better than the usual convection. Kelvin missed that one.

zero

Hi guys, I just wanted to cross the tee's of this discussion. So taking the radiative transfer equation form the wiki link in my first po st, And assuming surfaces with equal areas, and a viewing factor of 1. I get t he following.

Q_dot = S*Area*(T1^4 - T2^4)/(3-e1-e2)

Where S is the Stefan-Boltzmann constant and e1, e2 are the emissivity's.

So this is interesting. Going from all e's equal to one, to all equal to z ero, only reduces the heat flow by a factor of three. And if you follow Martin's previous analysis with a shield, then if the shield has an emissiv ity of one you get a factor of 2 decrease in the heat flow, (compared to no shield). As shown by Martin. And if you then assume a perfect shield, yo u (only) get another factor of two decrease. (Well not that factors of two are to be ignored, but I thought it would be more.)

George H.

te zero

ar

ar,

post,

the following.

zero, only reduces the heat flow by a factor of three. And if you follo w Martin's previous analysis with a shield, then if the shield has an emiss ivity of one you get a factor of 2 decrease in the heat flow, (compared to no shield). As shown by Martin. And if you then assume a perfect shield, you (only) get another factor of two decrease.

e more.)

I guess I'm not following this analysis - If you are talking only about radiative transfer, and if you put a perfect reflector in the way (R = 1) , then the rad transfer will be 0.

In STP practice, the convective transfer and radiative are about equal, so for a double pane window, you can only hope to get 1/2. Commercial home insulation has been marketed as Al-ized mylar layers.

I am sure I've missed the train of thought here... JB

Here's the equation again, Q_dot = S*Area*(T1^4 - T2^4)/(3-e1-e2) (Q_dot = dQ/dt)

So if both e's are one I get, S*Area*(T1^4 - T2^4)/1 and if both e's are zero. I get, S*Area*(T1^4 - T2^4)/3.

OK emmissivity equal to zero is not possible. So you can put in 0.01 or some other small number.

I didn't post the algebra for the shield case. (I figured anyone interested could follow Martin's analysis and do their own algebra. :^)

George H.