Hi Bob:
I like the idea of a six inch on a side cube. An outer aluminum box,
styrofoam insulation, and a block of (copper, Aluminum, etc.) in the
center. A DS3231 or similar RTC and a PIC with a DB-9 connector on one face.
Have Fun,
Brooke Clarke
http://www.PRC68.com
Bob Camp wrote:
Hi
So getting back to the original idea.
Taking the other extreme - if you go for ~20 C / W on the insulation:
The mass of the block will be ~50X what it was before.
1.6" on a side goes to ~5" on a side.
Surface area goes up ~10X.
200 C/W with 2" foam goes to 20 C/W
Outer surface area goes to ~1.6X the inner surface area
Go to 4" of foam and you are pretty close.
Go to 8" of foam and you have some room left for things like wires and the like.
Resulting block is a bit under 2 feet on a side
Bob
On Mar 10, 2010, at 4:48 PM, Bruce Griffiths wrote:
Surely you meant 1 square inch of styrofoam with the specified thickness??
For a 100 hour (3.6E5 seconds) time constant and 1000 deg C/W thermal
resistance you need a thermal capacitance of 3.6E5/1E3 = 360 J/C
or 360/0.8371 = 430 gm of Aluminium which has a volume of 430/2.7 = 159 cc.
You still have the thermal radiation coupling problem through the infrared
transparent insulator.
Gold plating the aluminium will help.
Even if you solve the thermal radiation problem, the thermal conductance of any
leads bringing power or signals to or from the clock will be an major problem
even if you use constantan or austenetic stainless steel leads.
Bruce
Brooke Clarke wrote:
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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