# Calculating thermal conductivity

1. Jan 12, 2010

### marchithermal

Is it possible to calculate thermal conductivity of a material given just the delta-T at steady state and the material thickness? I have material that is 1/2 inches thick and has a 145 degree-C diffential @ 220C. Can I calculate its k-factor from that data alone?

Thank you.

2. Jan 12, 2010

### Yeti08

From a simplified version of Fourier's law in 1 dimension,

$$q^{''}=\frac{k \Delta T}{L}$$

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. Jan 12, 2010

### marchithermal

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?

4. Jan 12, 2010

### Yeti08

By knowing the change in temperature and the conductivity of the reference material (copper is a good one because you can typically assume uniform temperature perpendicular to the heat flow) you can determine the heat flux. This will be the same heat flux that will go through the other material, thus allowing the calculation of the thermal conductivity. You would need pieces with equal cross sections for the heat flux (W/m2) statement to be valid. Otherwise you would calculate the heat (W) and assume that to be equal, but then you'd also have to take into accound 2-D conduction.

5. Jan 12, 2010

### marchithermal

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

6. Jan 15, 2010

### rdt24

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 = -kc x (Temp gradient across copper bar) = -kA x (Temp gradient across metal sample)

So in this case,

-396 x 4.2 = -kA x 0.55
kA = 3024 W/mK

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

#### Attached Files:

• ###### Exam 2008.pdf
File size:
184.1 KB
Views:
235
7. Jan 15, 2010

### Yeti08

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. Jan 15, 2010

### rdt24

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

#### Attached Files:

• ###### Exam 2008.pdf
File size:
273.1 KB
Views:
183
9. Jan 15, 2010

### Yeti08

From how the problem is stated, you are correct even though it may not be a realistic value for a metal (though a single crystal at cryogenic temperatures can have very high conductivity). The temperature gradient values might have been reversed when the problem was written which, if that were the case, would give a conductivity of 51.9 W/m-K for the unknown sample - a much more believable number.

10. Jan 15, 2010

### rdt24

Yeah I was playing around with it because I didn't believe my original answer and I came up with 51.9 W/mK too. I'll have to ask my lecturer what happened there.

Thanks for the confirmation
Rhys