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Entry  Thu Aug 5 11:53:26 2021, Radhika, DailyProgress, Cryo vacuum chamber, Cooldown model fitting for MS model_fit_v_data.pdf
    Reply  Wed Aug 11 14:58:47 2021, Radhika, DailyProgress, Cryo vacuum chamber, Cooldown model fitting for MS rad_model_fit_v_data.pdfrad_cond_model_fit_v_data.pdf
       Reply  Wed Aug 11 18:00:19 2021, Koji, DailyProgress, Cryo vacuum chamber, Cooldown model fitting for MS 
          Reply  Fri Aug 13 15:14:14 2021, Radhika, DailyProgress, Cryo vacuum chamber, Cooldown model fitting for MS rad_295_model_fit_v_data.pdf
             Reply  Fri Aug 13 21:01:42 2021, Radhika, DailyProgress, Cryo vacuum chamber, Cooldown model fitting for MS 
                Reply  Thu Aug 19 14:34:10 2021, Radhika, DailyProgress, Cryo vacuum chamber, Cooldown model fitting for MS model_fit_tm_painted.pdfmodels_painted.pdf
                   Reply  Fri Aug 20 13:44:58 2021, rana, DailyProgress, Cryo vacuum chamber, Cooldown model fitting for MS 
                      Reply  Fri Aug 20 14:05:45 2021, Radhika, DailyProgress, Cryo vacuum chamber, Cooldown model fitting for MS 
Message ID: 2641     Entry time: Wed Aug 11 14:58:47 2021     In reply to: 2636     Reply to this: 2642
Author: Radhika 
Type: DailyProgress 
Category: Cryo vacuum chamber 
Subject: Cooldown model fitting for MS 

I've made simplifications to the testmass cooling model. Assuming 2 possible cooling mechanisms, radiative and conductive, the ODE must be a function of only (T_{coldplate}^4 - T_{testmass}^4) and (T_{coldplate} - T_{testmass}). If conductive cooling/heating of the testmass is treated as negligible, as we previously assumed, then:

\ddot{Q}_{testmass} \approx {\color{Blue} a} \sigma (T_{coldplate}^4 - T_{testmass}^4), where a is the fit parameter. I include \sigma in the equation because it would appear as a constant in any radiative transfer model. I use the measured coldplate/inner shield temperature data for T_{coldplate}. Note that \frac{dT}{dt} = \frac{\ddot{Q}}{Cp*m}, and here and throughout I use a temperature-dependent Cp_{Si}(T).

The best fit parameter is a = 0.014, and the result of this fit can be seen in Attachment 1. The disagreement between the best-fit model and data suggest that cooling is not only dependent on (T_{coldplate}^4 - T_{testmass}^4), i.e. it cannot be radiative alone. I added back a conductive heating/cooling component:

\ddot{Q}_{testmass} = {\color{Red} a} \sigma (T_{coldplate}^4 - T_{testmass}^4) + {\color{Red} b}(T_{coldplate} - T_{testmass}), where a and b are the parameters of the fit. 

The result of the fit can be seen in Attachment 2. The best fit parameters [a, b] = [0.022, -0.0042]. This model matches the data much better than the purely radiative model, but the negative sign on b is non-physical (I think) since (T_{coldplate} - T_{testmass}) should always be < 0, so the sign of the conductive term should also be < 0. I'm not sure how to interpret this result, but it almost seems like there is conductive heating to the test mass even though physically this shouldn't be the case. 

Attachment 1: rad_model_fit_v_data.pdf  31 kB  Uploaded Wed Aug 11 17:33:12 2021  | Hide | Hide all
rad_model_fit_v_data.pdf
Attachment 2: rad_cond_model_fit_v_data.pdf  37 kB  Uploaded Wed Aug 11 17:33:20 2021  | Hide | Hide all
rad_cond_model_fit_v_data.pdf
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