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 ATF eLog Not logged in Message ID: 2617     Entry time: Sun Jul 25 21:45:46 2021     In reply to: 2614
 Author: Koji Type: Summary Category: Cryo vacuum chamber Subject: About the radiation heat transfer model

The following radiation cooling model well explained the cooling curve of the test mass (until ~150K)

where dQ/dt is the heat removed from the test mass, A is the surface area of the test mass, is the Stefan-Boltzmann constant, T_SH and T_TM are the temperatures of the surrounding shield and the test mass.

Can we extract any information from this "0.15"?

I borrowed "Cryogenic Heat Transfer (2nd Ed)" by Randall F. Barron and Gregory F. Nellis (2016) from the library.
P.442 Section 8.5 Radiant Exchange between Two Gray Surfaces can be expressed by Eq 8.44

where T_i is the temperature of objects 1 and 2. For us, OBJ1 is the test mass and OBJ2 is the shield. A1 is the surface area of A1. F_1,2 is the view factor and is unity if all the heat from the OBJ1 hits OBJ2. (It is the case for us.)

is an emissivity factor.

The book explains some simple cases in P 443:

Case (a): If OBJ2 is much larger than OBJ1, where the e_i is the emissivity of OBJi. This means that the radiated heat from OBJ1 is absorbed or reflected by OBJ2. But this reflected heat does not come back to OBJ1. Therefore the radiative heat transfer does not depend on the emissivity of OBJ2.

Case (b): If OBJ1 and OBJ2 has the same area, . The situation is symmetric and the emissivity factor is influenced by the worse emissivity between e1 and e2. (Understandable)

Case (c): For general surface are ratio,  . OBJ2 receives the heat from OBJ1 and reradiates it. But only a part of the heat comes back to OBJ1. So the effect of e2 is diluted.

For our case, OBJ1 is the Si mass with DxH = 4in x 4in, while the shield is DxH = 444mm x 192mm. A1/A2 = 0.12.
We can solve this formula to be Fe=0.15. e1 = (0.147 e1)/(e2-0.0178).

Our inner shield has a matte aluminum surface and is expected to have an emissivity of ~0.07. This yields the emissivity of the Si test mass to be e1~0.2

How about the sensitivity of e1 on e2? d(e1)/ d(e2) = -0.95 (@e2=0.07).

Depending on the source, the emissivity of Aquadag varies from 0.5 to 1.
e.g. https://www.infrared-thermography.com/material-1.htm / https://www.mdpi.com/1996-1944/12/5/696/htm

• Assuming Aquadag's emissivity is ~1
• If only the test mass is painted, F_e increases from 0.15 to 0.39 (x2.6)
• If the inner shield is also painted, F_e increases to 1, of course. (pure black body heat transfer)
• If shield panels are placed near the test mass with the inner surface painted, again F_e is 1.
• Assuming Aquadag's emissivity is ~0.5
• If only the test mass is painted, F_e increases from 0.15 to 0.278
• If the inner shield is also painted, F_e increases to 0.47.
• If shield panels are placed near the test mass with the inner surface painted, F_e is 0.33 assuming the area ratio between the test mass and the shield panels to be unity.

It seems that painting Aquadag to the test mass is a fast, cheap, and good try.

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