Hi Bruce:
If a square meter of Styrofoam is 1.5 deg C/W then a cubic inch would be
39.37 * 39.37 * 1.5 or 2,325 deg C/W
The DS3231 dissipates about 1 mw when running. I'm not sure how to come
up with an allowable temp increase, but suspect it's based on not
exceeding the max allowed operating temp. But for this part the Temp
vs. freq curve is flattest at room temp so best to minimize the temp rise.
So: (1 deg C) / (0.001 W) = 1000 deg C/W, the same number you had.
That requires 185 g of Aluminum or about 69 cc volume or 4.25 cubic
inches. That's a block about 1.62" on a side.
How does that look?
Have Fun,
Brooke Clarke
http://www.PRC68.com
Bruce Griffiths wrote:
Thermal resistance is measured in degrees (C or K or ..) per watt.
Its inversely proportional to area and proportional to thickness.
I think the clueless clown who created that table means that for a 1
square meter panel of the specified thickness the thermal resistance
is the tabulated value measured in Kelvin/Watt.
Bruce
Brooke Clarke wrote:
Hi Bruce:
What does m2K/W mean? See:
http://building.dow.com/europe/uk/proddata/styrofoam/thermal.htm
50 mm it's about 1.5 and for 100 mm it's about 3.
Have Fun,
Brooke Clarke
http://www.PRC68.com
Bruce Griffiths wrote:
Bruce Griffiths wrote:
Brooke Clarke wrote:
Hi:
For some time I've considered surrounding a free running 32678 Hz
oscillator (like a Dallas 32khz, or one of the newer Maxim units)
with thermal mass and insulation in order to get the time constant
into the range of some days. To get a feel for it a simple
experiment shows that a half inch diameter brass rod 3.75" long
(102 grams) has a thermal time constant of about 6 min 35 seconds
when wrapped lightly in a towel.
Is there a way to calculate the amount of aluminum and Styrofoam
needed to get a time constant of say 100 hours?
This came up in relation to WWVB clocks that free run for most of
the time. When you compare WWVB clocks it's not uncommon to see
tens of seconds difference between them.
http://www.prc68.com/I/Loop.shtml#TC
Start with the maximum thermal resistance the application can
withstand (determined by internal dissipation and acceptable
temperature rise above ambient).
If for example the dissipation is 10mW and acceptable temperature
rise 10C then thermal resistance will be about 1000C/W.
The thermal capacity required can then be calculated from the time
constant:
C= 3.6E5/1E3 = 360 J/C requires about 7.2 kg of aluminium.
The required thickness of styrofoam can then be calculated from the
surface area of the aluminium block.
Achieving a thermal resistance of 1000C/W may be a little difficult
without using radiation shields.
Bruce
Oops, the specific heat of Alum9inium is about 0.2Cal/gm/C or 0.8371
J/gm/C so the mass of Aluminium required would be 430gm if the
thermal resistance to ambient were 1000K/W. With a thermal
resistance of 100C/W you need 4.3Kg of aluminium ....
Bruce
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