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This is a work-related question. A warm steel torus of a given diameter & thickness is left in a room held at a controlled temperature, how long does it take to reach equilibrium? Assume the air is circulating so that the torus doesn't create a warm spot. Of course it never quite gets there but this is a practical problem so almost is fine.

For example, the torus is 250mm diameter and 50mm thick (so that it has a 150mm hole). When taken from processing it is 22°C and the room is 20°C, how long to reach 20.01°C? It is resting on a measuring instrument supported by 2 steel dowels so assume it is just in mid-air.

(Such small differences in temperature are important here because we are taking very close measurements. For something this size each 1°C changes the diameter of the hole about 0.0007 mm which to us is a big deal.

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This question is not answerable. The convective cooling largely depends on how fast the air is flowing. –  Bernhard Apr 5 '13 at 17:41
    
Depending on the diameter of those steel support dowels, they may be acting as a heat sink. Because steel conducts heat well, this effect might not be negligible compared to convective cooling. –  OSE Apr 5 '13 at 18:21
    
@Bernhard if you think this question is unanswerable, I'd be interested to hear what you think the practicality of physics is. –  zhermes Apr 5 '13 at 20:10
    
@zhermes It is unanswerable because there is too little information. We can make estimates, but he clearly stated in the question that that is not good enough. –  Bernhard Apr 5 '13 at 20:20
    
@Bernhard the question states the air is 'circulating' effectively---if that's accurate, it means the Rayleigh number is small and the explicit air velocity isn't important (especially once you're getting to a 0.01C temperature difference). In any case, you can't know for sure unless you solved the differential equations for this particular geometry. –  zhermes Apr 5 '13 at 20:28
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1 Answer 1

I don't think there is a simple way to calculate this from scratch.

The dominant cooling mechanism for your torus will be convective cooling by air currents. This is a complex process, but in most cases it can be approximated by Newton's law of cooling i.e. the rate of change of temperature is proportional to the difference in temperature between the object and it's surroundings. This gives you an exponential decay of temperature:

$$ T - T_{room} = (T_{initial} - T_{room}) \space e^{-t/\tau} $$

The problem is determining the constant $\tau$, which is sort of a cooling half life. There isn't a simple way to calculate it precisely, and in any case at very small temperature differences the exponential dependance probably breaks down.

Really, the only reliable way forward is to measure the temperature vs time curve for your system and use that to establish how long the cooling takes.

Later:

It's just occurred to me that maybe you have toruses (tori?) of different sizes and you want an equation to calculate the cooling time from the torus size. You probably could do something along these lines by making experimental measurements for a few sample objects to establish the required constants of proportionality, and using dimensional analysis or some other method to find n quation linking cooling to torus size.

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Thank you for your answer John Rennie. I had that feeling that it had no easy theoretical answer. We will be getting into this so after some samples I'll post what we get. One complicating thing is that they are not really toruses that is just the simplest shape approximation. They are ring gage blanks in size #s 13-18 as shown here: link One thing that experiments have shown is that it doesn't take take nearly as long as everyone around here thought before the size of the gage starts fluctuating in lock-step… –  plh Apr 9 '13 at 20:04
    
…with the ambient temperature. At that point it's a thermometer! For example a 3.5" ring warmed to 73F took about 2-3 hours to start following the room (which was 67.5F). When I use cooling plates to force them close to the ambient temperature they start following right away. –  plh Apr 9 '13 at 20:04
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