I was thinking about how fast we can cool the test mass. No matter how we improve the emissivity of the test mass and the cryostat, there is a theoretical limitation. I wanted to calculate it as a reference to know how good the cooling is in an experiment.
We have a Si test mass of 300K in a blackbody cryostat with a 0K shield. How fast can we cool the test mass?
![m\,C_p(t)\,T'(t) = -\epsilon\,\sigma A\,[T(t)^4 - 0^4]](https://latex.codecogs.com/gif.latex?m%5C%2CC_p%28t%29%5C%2CT%27%28t%29%20%3D%20-%5Cepsilon%5C%2C%5Csigma%20A%5C%2C%5BT%28t%29%5E4%20-%200%5E4%5D)

Then assume the specific heat is linear as

The actual Cp follows a nonlinear function (cf Debye model), but this is not a too bad assumption down to ~100K.
Then the differential equation can be analytically solved:
,
where the characteristic time of t0 is
.
Here T_0 is the initial temperature, cp0 is the slope of the specific heat (Cp(T_0) = c_p0 T_0). epsilon is the emissivity of the test mass, sigma is Stefan Boltzmann constant, A is the radiating surface area, and m is the mass of the test mass.
Up to the characteristic time, the cooling is slow. Then the temperature falls sqrt(t) after that.
As the surface-volume ratio m/A becomes bigger for a larger mass, in general, the cooling of the bigger mass requires more time.
For the QIL 4" mass, Mariner 150mm mass, and the Voyager 450mm mass, t0 is 3.8hr, 5.6hr, and 33.7hr respectively.
- If the emissivity is not 1, just the cooling time is expanded by that factor. (i.e. The emissivity of 0.5 takes x2 more time to cool)
- And if the shields are not cooled fast or have a finite temperature in the end, of course, the cooling will require more time.
- 1.25 t0 and 8 t0 tell us how long it takes to reach 200K and 100K.
This is the fundamental limit for radiation cooling. Thus, we have to use conductive cooling if we want to accelerate the cooling further more than this curve. |