Energy transfer by radiation

George Herold · 2014-02-24T18:33:13+00:00

Transfer of energy by thermal radiation. (thinking out loud at the SED.) So I've been calculating numbers for heat transfer. The radiation transfer between two surfaces at different temperatures is huge, (well a lot bigger than I expected.) Assuming emissivity equal to one, for a 1 K degree tem perature difference, the transfer rate at 300K is about 0.66 mW/cm^2. (6.6 W/m^2) I've been trying to understand how to think about this when I have a sample inside a radiation shield. (You can imagine concentric spheres, in practice it is closer to concentric cylinders.) The sample will have a much smaller area than the radiation shield surround ing it. (maybe a factor of ten or more.) So how important is the emissiv ity of the radiation shield? My gut says that that a low emissivity shie ld will only help (reduce the heat transfer) if the ( Area * emissivity) pr oduct of the shield is less than that of the sample. This equation for wik i seems to say that. (For a grey body). Now I'm not quite sure about the View factor, but I thi nk the view factor will be one for the inner of two concentric spheres. (a nd equal to the ratio of areas for the outer sphere.) OK no questions so far... I'm just hoping the physics types will correct me . But here's a fun question then. Say I have shield at 300 K, emissivity = 0.01 and a sample at 77 K, emiss ivity = 1.0. Now I cut a little hole in the radiation shield and measure the black body spectrum. What temperature will I see? (assume a factor o f ten area ratio.) George H.

Feb 24, 2014 22 Replies

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haiticare2011 12 years ago

George, This is not my understanding at all. If you create (in this thought experiment) surfaces at emissivity of zero, that means the surfaces don't emit. They emit nada. The only het transfer that's gonna happen off that (hypothetical) surface is from convection or conduction. Thus the "3" is totally wrong, since the emissivity controls everything. This is not just theory. Consider a thermos bottle or vacuum Dewar.

You can hold ln2 in these devices. Heat sneaks in via conduction, but the insulation suppresses nearly all transfer. Where did the 3 come from in that formula? s this a convection plus radiation formula? jb

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Martin Brown 12 years ago

I don't see where you get this equation from.

For equal areas the Wiki equation you reference simplifies to

Q_dot = S*Area*(T1^4 - T2^4)/(1/e1 + 1/e2 -1)

The generalisation from the black body equation that I gave is that the equilibrium shield temperature is independent of its emissivity, but the thermal flux across it decreases in direct proportion to its emissivity and so it is really worth aiming for the highest possible reflectivity you can get on a radiation shield.

Think of it this way: the hot side sees (1-e)T^4 reflected back at it + eS^4 the cold side sees (1-e)t^4 reflected back and e^S^4

It is not for nothing that they use a high mirror finish in cryo.

Regards, Martin Brown

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George Herold 12 years ago

ute zero

far

war,

t post,

et the following.

to zero, only reduces the heat flow by a factor of three. And if you fol low Martin's previous analysis with a shield, then if the shield has an emi ssivity of one you get a factor of 2 decrease in the heat flow, (compared t o no shield). As shown by Martin. And if you then assume a perfect shield , you (only) get another factor of two decrease.

be more.)

Oh, $h!t. OK ignore everything I said. (I somehow copied the wrong equation into my note book...)

OK I better try this again. T Thanks for the correction!

George H.

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