| | #1 (permalink) | |
| Thinking | Hi clever peeps! I conducted an experiment whereby a tea light candle was placed on an aluminium block with a temperature probe in it, and the candle lit and the temperature monitored for 1hr. I have to calculate the heat transfer through the aluminium block and am just having a bit of trouble getting to the bottom of which equation to use. I have found this one: qx'' = -k (T2 - T1)/L which seems to make sense, but a colleague told me it was qx'' = kA (T2 - T1)/L where A is the area of the block. Can anyone advise me on this? I don't have to account for heat loss or any other variable at this stage. Thanks muchly! joho ---------------- I don't agree with animal testing: they get all nervous and give the wrong answers! Stephen Fry | |
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| | #2 (permalink) | |
| Creating | Re: Heat transfer of aluminium I think the second equation is better since the total heat transfer will be dependent on the size of the block. The A and L give the volume. If we were looking at the rate of heat transfer, then the A gets factored out, since it will become expressed per CM2. This is the first equation. If you want to get fancier, the candle flame is not the same distance from every point on the surface. The edges are farther away. As a better approximation you can divide the surface into CM2, use the first equation, and then add or integrate them all. The A of the second equation becomes a summation of each CM2 of first equation. | |
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| | #3 (permalink) | |
| Creating | Re: Heat transfer of aluminium The equation for heat conduction is where k is a constant of conductivity, A is area in units squared, is the hotter thing, and is the colder thing, and L is the length of conduction.![]() //edited to remove rambling on about specific heat Here is a good source for what you're doing: Heat Transfer: Conduction, Convection, Radiation or Heat Transfer -modest ---------------- Last edited by modest; 04-29-2008 at 11:14 AM. Reason: shown | |
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