The current hold time is 25 hrs, for 9.5 L of He.
relevant constants:
L He density: 124.96 kg/m^3
He heat of vaporization: 20.3 kJ/kg
Heat load on the dewar now:
9.5 L * (124.96kg /m^3 )(m^3/1000 L)*(20.3 kJ)= 24.1 kJ per 25 hr --> 0.27 W
The copper wire that will be installed is a 2m long 36 AWG (0.005 inches in diameter).
Lakeshore* has values for the thermal conductivity of Cu at cryogenic temperatures:
4K 300 W/m/K
10K 700 W/m/K
20K 1100 W/m/K
80K 800 W/m/K
The wire connects the 4K He dewar to the 77 N dewar.
As a worst case, I adopt the 1100 W/m/K.
The heat load from the wire is:
1100 W/mK *(77-4 K) A / L, where A is the cross-sectional area of the wire and L is the length, 2 m.
A= (0.005/2 inches)^2 *pi * (2.54 cm/inch * 1 m / 100 cm)^2 = 1.27e-8 m^2
1100 W/mK * 73 K * 1.27e-8 m^2 /2 m = 0.00051 W, compared to 0.27 W currently in the dewar.
This will reduce the hold time by: 0.05 hrs or 3 min.
* http://www.lakeshore.com/temp/acc/am_wirets.html
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1 comment:
Thanks for doing this. I would like to say that the relevant length is between the LHe and LN2 surfaces. I doubt those are 2m apart, although I'd be happy if I'm wrong. If we want to do a worst case, I'd guess 25 cm (10"). That means the 3 min is now 24 min.
Is there only one of these Cu wires? I thought we were talking about a ground for each cable bundle, although that doesn't make sense. We only really need one, right?
So, our 25 hours may be shorter, but probably remembering to turn off the detector heater at the end of the day is more important. Certainly pumping on the LN2 is more important.
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