Re: [ATM] undercorrecting?

[Mike Lockwood <melockwo@uiuc.edu>]

Arjan,

Sorry, I did not realize that the article
   http://home.hetnet.nl/~artm/atm/articles/cooling.html
was meant to analyze the effect of radiative cooling (I can't find 
where it says this) - I thought it was from convective cooling.

Below are a couple of messages from you.  The first one seems to 
indicate that the effects of radiative cooling are negligible.  I also 
heard someone else quote e = 0.01 rather than 0.04, so maybe that 
value should be verified.

In the second message you seem more convinced that there is an effect 
from the radiative cooling.  Is that correct?  If so, what made you 
change your mind?

I am curious what other factors could be in play with regard to the 
mirror mentioned on the Dutch mailinglist.

What seems to be overlooked here is how the front and back of the 
mirror are prepared.  The front is polished and coated with aluminum 
and silicon dioxide (what's the effective emissivity of the 
combination?), while the back is probably not.  What difference does 
that make for the emissivity of glass?  Is it larger than the effect 
of radiative cooling on the mirror?

Along that line of thinking, how about the insulating effect of the 
mirror cell's support points?

Seems many other factors could account for very small temperature 
gradients in the glass.  0.015K is not much.

This has been a great topic by itself, and it also started a great 
thread about how eyepieces affect the star test.  Lots of good stuff 
being investigated!

	Mike Lockwood

Arjan te Marvelde wrote on August 4th, 2006:
> Heat loss due to radiation is determined by the law of Stefan-Boltzmann:
> P = e*s*A*(T^4 - Tc^4)
> Where:
>   P is the effective radiated power [W/m^2]
>   e is the materials' emissivity
>   s is Stefans' constant (5.7 10^-8 W/m^2 K^4)
>   A is the objects' area [m^2]
>   T is the objects surface temperature [K]
>   Tc is the temperature of the surroundings.
> 
> The interesting parameters here are e and Tc. The emissivity of polished
> aluminium is about 0.04, the equivalent radiative temperature of the night
> sky is generally taken as 200K (not the 3K background radiation because of
> the atmosphere that's in between!).
> Assuming an ambient temperature (equal to the object temperature) of 293K,
> and 'grey body' radiation, the effective heat loss due to radiation is
> approximately:   E= 13 W/m^2
> When I plug this in my FlexPDE mirror cooling model (10"x3/4" plate), the
> front side indeed cools under ambient temperature, but only by 0.015K. The
> undercorrection expected to result from this is, in my opinion, negligible.

Arjan te Marvelde wrote on August 28th, 2006:
> In any case, thank for all the comments. Some of the issues I was 
> already planning to assess further, when time permits. Note that
> the paper was started because someone on a Dutch mailinglist could 
> not get good startest results, while bench testing was OK. He 
> suggested permanent deformation to compensate for the radiative
> cooling! Few people could believe that the radiation effect could
> possibly be so large that it would become visible. The quick
> FlexPDE model I whipped up suggests it can be...


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