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Thank you.

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- Thread starter marchithermal
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- #1

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Thank you.

- #2

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[tex]q^{''}=\frac{k \Delta T}{L}[/tex]

you can see that you need to know the heat flux through the material as well. This can be done by knowing the temperature difference in a reference piece of some material where the thermal conductivity is well known, and is thermally in series with the material in question.

- #3

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[tex]q^{''}=\frac{k \Delta T}{L}[/tex]

you can see that you need to know the heat flux through the material as well. This can be done by knowing the temperature difference in a reference piece of some material where the thermal conductivity is well known, and is thermally in series with the material in question.

Interesting, thank you for the information. As far as using a known material in the same thermal conditions, it is simple to measure the delta-T across, say, a piece of copper or aluminum in the same environment. Forgive my ignorance but how would I use that information to determine the q'' of the insulation under test?

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- #5

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Excellent, thank you for that clarification. I will experiement with a 1/2 inch piece of copper.

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I have attached a problem I am having trouble with in PDF format. Could anyone take a look and tell me what they think please? It's Question 1a part (iii).

As far as I can tell, the heat flux through the network should be constant, so Fourier's law for heat conduction gives:

q = -k_{c} x (Temp gradient across copper bar) = -k_{A} x (Temp gradient across metal sample)

So in this case,

-396 x 4.2 = -k_{A} x 0.55

k_{A} = 3024 W/mK

Clearly, this value for conductivity is far too high. As it's a metal sample so I'm guessing k_{A} should be something between 100 and 400 W/mK.

I just can't see why this gives such a ridiculous answer! Can anyone help?

Thanks

Rhys

As far as I can tell, the heat flux through the network should be constant, so Fourier's law for heat conduction gives:

q = -k

So in this case,

-396 x 4.2 = -k

k

Clearly, this value for conductivity is far too high. As it's a metal sample so I'm guessing k

I just can't see why this gives such a ridiculous answer! Can anyone help?

Thanks

Rhys

- #7

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I have attached a problem I am having trouble with in PDF format. Could anyone take a look and tell me what they think please? It's Question 1a part (iii).

As far as I can tell, the heat flux through the network should be constant, so Fourier's law for heat conduction gives:

q = -k_{c}x (Temp gradient across copper bar) = -k_{A}x (Temp gradient across metal sample)

So in this case,

-396 x 4.2 = -k_{A}x 0.55

k_{A}= 3024 W/mK

Clearly, this value for conductivity is far too high. As it's a metal sample so I'm guessing k_{A}should be something between 100 and 400 W/mK.

I just can't see why this gives such a ridiculous answer! Can anyone help?

Thanks

Rhys

I don't see anything but control problems in the attached pdf. For your conduction problem - I don't see the length (i.e. thickness) in your equations - is that a typo or calculation mistake?. Your answer, if it is a metal, should be less than about 400 W/m-K (silver is about 429 though), and could be less than 100 W/m-K such as the case for steels.

- #8

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Did I attach the wrong file? Here's the right one.

Sorry about that. The equation I quoted takes the thickness into account.

Temperature gradient = (T1 - T2)/x

If you substitute that back into the last equation I gave, it becomes Fourier's law again. The metal sample is an unknown metal, and its thermal conductivity must be calculated. I was just guessing between 100 and 400, but i know 3024 is DEFINITELY wrong!

Thanks

Rhys

Sorry about that. The equation I quoted takes the thickness into account.

Temperature gradient = (T1 - T2)/x

If you substitute that back into the last equation I gave, it becomes Fourier's law again. The metal sample is an unknown metal, and its thermal conductivity must be calculated. I was just guessing between 100 and 400, but i know 3024 is DEFINITELY wrong!

Thanks

Rhys

- #9

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- #10

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Thanks for the confirmation

Rhys

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