From: Hans Jostlein (email@example.com)
Date: Fri Jun 24 2005 - 13:41:40 CDT
I have done a quick and dirty estimate of the thermal time constant of a
spherical Braidwood detector,
e.g. for moves in the middle of winter or summer.
The spreadsheet is below:
Sphere diameter 7 m
Oil heat capacity 0.45 BTU/ (F #) 1.9 J/g K
Insulation R factor 6 (sf F hr) / BTU
Oil Mass 151871 kg 334117 #
Steel Mass 7269 kg 15992 #
Oil total heat capacity 150353 BTU/F
Temperature difference 30 F
Sphere surface area 154 m^2 1657 sf
Heat flow 8285 BTU/hr
Time constant 544 hr
Window glass has an R value of 0.9, and it has two air interfaces.
The outer one is exposed to air motion and has a lower R value.
Steel, by the sam reckoning, will have an R value of 0.3 or so for an
This gives a thermal time constant of 27 hours. I ignored the contribution
from the steel.
With modest insulation (R=6) we would get 544 hours time constant, or over
We may not use any insulation, because the natural cooling counteracts the
heat produced in the PMT bases.
Even without insulation, the temperature of the oil will vary quite slowly
during the moves.
I expect we will heat trace the tank with fluid tubing,
which makes it possible to keep the temperature constant quite easily and
Some summary of the above may be useful for NUSAG question #9, perhaps.
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