2
$\begingroup$

Let's say you have two cylinders one flatter and one longer, both having equal surface area and both having equal volume. Which will arrive at the surrounding temperature first? Assume that the method of cooling doesn't benefit either shape.

The dimensions are r = 1.20214, h = 5.944 and r = 2.13946, h = 1.877

This provides two cylinders with same volumes and surface areas. Quite interesting. Also the ratio of surface area to volume is 54/27 or exactly 2. Just for funzies.

For any volume and surface area there are always two cylinders that will fit that volume-surface area combo.

Reason why: I simply just wanted to figure out how long I should leave my soda in the freezer and how I would calculate that. Then I got sidetracked thinking about this question.

$\endgroup$
  • $\begingroup$ Wait a minute. Same volume and same surface area and both cylinders but different dimensions? $\endgroup$ – C. Towne Springer May 10 '17 at 6:49
  • $\begingroup$ Ah I know I was quite confused for a second as well. I will post the dimensions. $\endgroup$ – A A Ron May 10 '17 at 6:51
  • $\begingroup$ For the case of minimum surface area, the two solutions must be the same? $\endgroup$ – C. Towne Springer May 10 '17 at 7:01
  • $\begingroup$ Oh, so they're not "equal". Only "approximately" equal. This has gone from a physics problem to an engineering problem. BTW, care to post the mathematical proof of the last statement? $\endgroup$ – Oscar Bravo May 10 '17 at 7:17
  • $\begingroup$ @OscarBravo If $V/S=Q$ then $1/2Q=1/R+1/H$ , so there should be plenty of $R,H$ combinations satisfying a given $Q$, is it not? $\endgroup$ – mikuszefski May 10 '17 at 7:21
2
$\begingroup$

If we can assume that there is no significant loss of energy due to radiation (which I'm not sure we can):

I think that aside from surface area you need to take into account the rate at which heat diffuses to the surface of the object.

The heat flux given by Fourier's law of thermal conduction: $$q=-k\frac{dT}{dx}$$

where q is the heat flux density and k is the conductivity of the material. The derivative on the RHS is a measure of the local difference in temperature, i.e. the difference between the surface of the object and the outside air (assuming that there is air or some other material to 'accept' the heat energy).

My point is that the surface temperature is not necessarily equal for both objects, since heat must be conducted from the internals of the object to the surface, and therefore the heat flux density is not equal.

You must solve the 3D heat equation for each object with the appropriate boundary conditions.

$\endgroup$
  • $\begingroup$ So I think I get what you're saying. Even though they have the same volume and therefore quanity of heat, and even though the area exposed is identical, the way the heat radiates to the surface of the shape determines the loss rate. $\endgroup$ – A A Ron May 10 '17 at 7:04
  • $\begingroup$ Yes, but just to clarify: heat conduction towards the surface is modeled by the heat equation (not sure if 'radiates' is the proper wording) $\endgroup$ – YoA May 10 '17 at 7:06
  • $\begingroup$ I think this sheds some light into the issue here. It depends on what you assume the heat source to be. Often times in (at least engineering) heat transfer we would be given a surface temperature for the cylinders; so this wouldn't change it a whole lot. If you were doing a thorough analysis of say a fluid in the cylinder; it would make the difference. $\endgroup$ – JMac May 10 '17 at 11:02
2
$\begingroup$

In the squat lumpy piece, heat has to move further to reach a surface and be conducted away. In the long skinny one you can think of the the material all being closer to the surface, no more than 0.6 away from a surface, and it will cool faster.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.