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Thermal properties - Experimental Design

  1. Jan 5, 2014 #1
    Apologies for the break from format, but I'm not sure how to make this fit!

    I'm designing an experiment to compare the insulation properties of several materials.

    G1JsDU6.jpg

    I would like to find the Thermal Conductivity (K) or the U (or R) values of these materials.

    However essentially all I am after is a recognised way to compare the insulation properties of the different panels I am investigating. As one of the materials is a bio-composite, there is no listed K value, so I can't seem to calculate U.

    I am coming from a built environment background and by no means is Thermodynamics my strong suit. If we need something insulated we just select the material with a suitable U value!

    I'm trying to use
    U = QA / deltaT,
    but I could be on totally the wrong track. Also tried
    Q/t = k A (T1 - T2) / L
    which I thought might work if I read off the temperatures at the same point in time? But then I still don't have a k value I can use.

    If anyone knows the equation I could use with this experimental setup, or modifications to the setup to make it viable, It would be hugely appreciated. Can't seem to rearrange them to suit.

    Thank you in advance,
     
    Last edited: Jan 5, 2014
  2. jcsd
  3. Jan 5, 2014 #2

    mfb

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    The transmitted thermal power in equilibrium is Q/t = k A deltaT / d with the surface area A and the thickness d (assuming a uniform plate of your material, uniform temperatures on both sides and so on).
     
  4. Jan 5, 2014 #3
    I follow you, but I don't have any value for thermal conductivity (k). Any way around this?

    And I take it if I had a thermostat to regulate the temperature inside the box, and took outside the box to be room temperature and constant, that would work?

    But then where does time fit in? Does it even have any place in this experiment?
     
  5. Jan 5, 2014 #4

    mfb

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    I thought you plan to measure that?

    Sure.

    Assuming your box has a perfect insulation, Q/t is given by your heating element. Otherwise, you can consider adding a second layer with known conductivity to measure the temperature difference across this (to get Q/t).
     
  6. Jan 5, 2014 #5
    Ah thank you, I see where my confusion was I think. I should be using:

    k = (Q/t) * (d / A deltaT). I didn't realise I had Q/t.

    Thank you very much mfb. I'll see how I get on from here.
     
  7. Jan 5, 2014 #6
    Time comes in because you have to wait until the system comes to steady state. After that, both thermocouples should level off at final values and no longer change with time. Your equation only applies at steady state. You can also use the transient temperature variations of the thermocouples to calculate the thermal conductivity of the insulation, but to do that you need to know other properties (such as the heat capacity and density), and you need to solve some more complicated equations.
     
  8. Jan 6, 2014 #7
    Thank you Chestermiller.

    Do you agree with mfb that Q/t will be the output of my heating element? When I have used the equation before, Q/t in Watts has been the heat flow through the material itself, where k has been known.
     
  9. Jan 6, 2014 #8
    Yes, I agree, as long as none of the heat from the heater element flows through the other walls of the container.

    Chet
     
  10. Mar 6, 2014 #9
    Progress... and stuck again

    So the experiment was modified somewhat.

    The 6 inch by 6 inch material panel forms a small part (approximately 1 in 224) of a 5 sided box (the left hand box). The excel file shows the dimensions of the panels, along with 4 temperature readings.

    It was my hope to show the K (thermal conductivity) of the bio-composite in comparison to polystyrene with:

    Q/t = k A (dT) / L

    However both K and the Q/t is unknown.
    Q/t is unknown because although the heater is 1200W, it is controlled by an automated switch to keep the internal temperature within a set range, so it is only on some of the time, not all of it. Further to that, something like metal (which I used as a base line) will have a higher conductivity, and therefore be transmitting more power than an insulating material. I'm not sure I have enough data here to give a meaningful comparison.

    Is there anything useful at all I can do with this data to show that the bio-composite is in the same league if not as good as polystyrene as an insulator?
     

    Attached Files:

    Last edited: Mar 6, 2014
  11. Mar 6, 2014 #10

    mfb

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    Due to thermal losses elsewhere and if the setup is very similar in both cases, a smaller temperature difference should correspond to a better heat conductivity, but it is hard to quantify differences based on that setup.
     
  12. Mar 7, 2014 #11

    BvU

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    According to your excel file, steel is a better insulator than polystyrene. I do find that a little hard to believe.
    You have a pretty fancy setup if I look at the pictures. So if you need Q/t, you'd better measure the time the heater is on; that can't be too complicated.

    Then: do you check your assumptions? And the assumptions of your helpers? I am not all that convinced the box with the heater is perfectly insulated. Understatement. (that is the left box, right? So what is all that black stuff on the box on the right ? -- Otherwise, provide us with a new diagram of your setup -- what is the function of the right box ?):
    If you want to let 99% of the heat from the heater to go through the 0.0225 m2 of the test material, the remaining 4 or 5 m2 should isolate some 20000 times better. (4/0.02 /1%) No way.

    Measuring the outside temperature of the box(es) is also sensible.

    But I am still searching for the rationale behind this setup. Good for measuring a ΔT with 0.01 C precision, but no idea how much heat goes through ?

    Show me wrong!
     
  13. Mar 7, 2014 #12

    mfb

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    I think it's the other way round - the box is not a good insulator, so the heating power and the temperature inside does not depend significantly on the test material. That allows to see the temperature difference as function of the conductivity - a smaller difference indicates a better conductivity (as more heat is emitted at the outside <-> more heat is conducted, at a smaller temperature difference.
     
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