Calculating thermal conductivity

In summary, it is not possible to calculate the thermal conductivity of a material given only the delta-T at steady state and the material thickness. The heat flux through the material must also be known, which can be determined by measuring the temperature difference in a reference piece of material with a known thermal conductivity. This information can then be used to calculate the thermal conductivity of the material in question. However, care must be taken to ensure that the heat flux is equal for both materials. In some cases, such as with metals, the calculated thermal conductivity may seem unrealistic and could be due to a mistake in the problem or calculation.
  • #1
marchithermal
3
0
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.
 
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  • #2
From a simplified version of Fourier's law in 1 dimension,

[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
Yeti08 said:
From a simplified version of Fourier's law in 1 dimension,

[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?
 
  • #4
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
Excellent, thank you for that clarification. I will experiement with a 1/2 inch piece of copper.
 
  • #6
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
 

Attachments

  • Exam 2008.pdf
    184.1 KB · Views: 470
  • #7
rdt24 said:
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

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
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
 

Attachments

  • Exam 2008.pdf
    273.1 KB · Views: 402
  • #9
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
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
 

Related to Calculating thermal conductivity

1. What is thermal conductivity?

Thermal conductivity is the measure of a material's ability to conduct heat. It is the rate at which heat energy flows through a material per unit time and per unit area when there is a temperature gradient present.

2. How is thermal conductivity calculated?

Thermal conductivity is calculated by dividing the amount of heat energy transferred through a material by the product of the cross-sectional area and the temperature gradient. The resulting value is expressed in units of watts per meter per Kelvin (W/mK).

3. What factors affect thermal conductivity?

Several factors can affect the thermal conductivity of a material, including its composition, density, temperature, and the presence of impurities. Generally, materials with higher densities and lower thermal resistivity have higher thermal conductivity.

4. What is the unit of measurement for thermal conductivity?

The unit of measurement for thermal conductivity is watts per meter per Kelvin (W/mK). However, other units, such as BTU per hour per square foot per degree Fahrenheit (BTU/h·ft²·F), are also commonly used.

5. Why is thermal conductivity important?

Thermal conductivity is an essential property in many fields, including engineering, physics, and materials science. It helps in the design and optimization of heating and cooling systems, as well as in predicting the behavior of materials under different thermal conditions. It is also crucial in understanding heat transfer and energy efficiency.

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