40m QIL Cryo_Lab CTN SUS_Lab TCS_Lab OMC_Lab CRIME_Lab FEA ENG_Labs OptContFac Mariner WBEEShop
  PSL  Not logged in ELOG logo
Entry  Sun Feb 24 01:16:26 2013, tara, DailyProgress, RefCav, refcav and mount assembly photo(9).JPGphoto(8).JPG
    Reply  Mon Feb 25 01:02:58 2013, tara, DailyProgress, Vacuum, Heater for refcav heater.pngphoto(10).JPG
       Reply  Wed Feb 27 09:57:33 2013, tara, DailyProgress, TempCtrl, Heater for refcav 
          Reply  Thu Feb 28 09:12:22 2013, not tara, DailyProgress, TempCtrl, Heater for refcav 
             Reply  Wed Mar 6 15:44:26 2013, Zach, DailyProgress, TempCtrl, Heater for refcav 
Message ID: 1104     Entry time: Mon Feb 25 01:02:58 2013     In reply to: 1102     Reply to this: 1105
Author: tara 
Type: DailyProgress 
Category: Vacuum 
Subject: Heater for refcav 

I did the calculation to estimate the required power for heating up the cavity by 20 K above room temperature. More details are coming.


Since we plan to tune the beat frequency by tuning the cavity length with thermal expansion, we need to know how much power is needed to heat one cavity up by 20K above room temp (see LIGO-1200057 for more detail).


To simplify the calculation, I use 2-D model for the cavity and the shield. Assuming the system is in equilibrium. In the calculation, I considered the effect from reflected radiation from both the shield, and the cavity itself.



  •   At room temperature of 29 C, depending on the emissivity of the shield, the tube has to be from 20 C - 100 C above room temperature inorder to keep the cavity at 20 C above room temp. Since the emissivity of the copper can vary from, .03 (highly polished)-0.77 (oxide),  see ref. Our cavity is polished a bit, so the reasonable value should be ~ 0.1-0.2.  Thus, the shield temperature should be able to reach at least 50 [C]. 




  •  Next is to estimate how long the wire should be for bringing the shield up to 50 degree C. If all the heating power from the wire goes to radiation, the radiated power from 50[C] shield, at room temp is sigma*epsilon*area*(T^4 - T0^4) which is ~ 0.6 - 1 Watt Note: I checked the calculation and Frank's note in PSL:765, the calculated number is reasonable.
ELOG V3.1.3-