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this is my first post here. I hope I do and format everything correctly.

I was given a spreadsheet which was automatically created and filled by a Cone Calorimeter. The sheet contains the following columns:

  • time [s], every five seconds the following values were noted
  • heat flux [W/m²]
  • surface temperature [°C]
  • back side temperature [°C]
  • some other values which I think are not relevant, including the temperature of the heater, etc.

I put a short excerpt at the end of my question to make it clear.

The tested sample

  • constant cross section: 0,157 m x 0,164 m
  • Area: A = 0,025748 m²
  • 0,0065 m thickness
  • homogeneous material
  • density: 1004,768107 kg/m³

was heated from one side (surface). The edges are isolated so that no heat can escape, thermal conduction only in direction of the thickness. You could imagine a slice of bread which is roasted from one side.

Is it possible to determine the thermal conductivity of that sample with only the described data? And if so, how?

I tried my best juggling equations but just can't finde a viable solution.


spreadsheet:

google spreadsheet

|time [s] | heat flux [W/m²] | surface [°C] | back side [°C] |
| 0 | 40000 | 22 | 22 |
| 5 | 40000 | 28 | 23 |
| 10 | 40000 | 70 | 25 |
...
| 75 | 40000 | 283,647036| 94,8174358 |
...
| 375 | 40000 | 480 | 456 |

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  • $\begingroup$ What you need is to have steady state conditions where over a period of time you note that temperatures at each end of you sample stay sensibly the same. You can then reason that none of the heat which you are supplying to the sample is changing its temperature rather it is just being transported through the sample, ie heat flux in is equal to heat flux out. Having satisfied yourself about this then there is sufficient data. $\endgroup$ – Farcher Aug 28 '17 at 13:25
  • $\begingroup$ @Farcher: steady state sounds plausible. But our tests were done because of different reasons than thermal conductivity. Trying to calculate it now is just a nice-to-have-gimmick I was told to find. I assume there's no way to do that with inconstant temperatures on both sides during the whole testing time? $\endgroup$ – hiwifgf Aug 28 '17 at 13:51
  • $\begingroup$ This is a transient heat conduction problem for a slab, involving a heat flux at one of its boundaries. The problem is pretty straightforward, but it is definitely not a steady state heat transfer situation (as exemplified by your data). There is another unknown besides the thermal conductivity, and that is the heat transfer coefficient at the far boundary. Both these parameters should be obtainable by calibrating the heat transfer model to the experimental data. Other parameters that would be useful to know are the density and heat capacity. Any chance of providing all the temp. data? $\endgroup$ – Chet Miller Aug 28 '17 at 18:48
  • $\begingroup$ @ChesterMiller: I added the density and a spreadsheet with all the temperatures. $\endgroup$ – hiwifgf Aug 29 '17 at 8:24
  • $\begingroup$ I plotted the data up and, beyond about 200 min, the behavior looked very suspicious. At long times, the far-side temperature even exceeded the near-side temperature. Any comments on what might have been happening? I think we can work with the data at <200 min. to get the thermal conductivity. $\endgroup$ – Chet Miller Aug 29 '17 at 12:31
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This is only a crude estimate of the thermal conductivity of the plaster (derived in view of the very crude quality of the experimental data). Assuming that the heat loss rate at the far boundary is negligible (after water at the boundary has been been evaporated), and, based on the detailed heat transfer analysis presented in the following thread https://www.physicsforums.com/threads/time-taken-for-heat-transfer.921537/#post-5814526 (post # 14), the temperature difference between the two boundaries (during the main part of the heating) is approximated by:$$\Delta T=\frac{qH}{2k}$$where q is the heat flux, H is the thickness of the sheet, and k is the thermal conductivity. From the data on the plot, the temperature difference over the main part of the heating curve is approximately 200 C. Substituting the data values into this equation gives a rough value of 0.65 W/(m-C) for the thermal conductivity.

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    $\begingroup$ The outcome of your equation looks good. I edited the spreadsheet in my question: I added two plots for the thermal conductivity, one for the original data (1) and the other one for a second set of data (2), both with the same heat flux but different temperatures and slightly different thickness. Just have to ignore the extreme values at the beginning and the end. $\endgroup$ – hiwifgf Sep 4 '17 at 10:14
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The thermal conductivity is calculated using, by definition, the following equation. $$\frac{W}{\triangle T \cdot L}$$

The heat transfer rate is the product of heat flux and area. You didn't provide area in your post.

Thus at time=0s, $\lambda = 0002583*A/(0*0.0065) = \infty$. This might be because the system is far from reaching equilibrium.

At time=5s, $\lambda = 0003149*A/(5*0.0065) = 0.097 \cdot A$

At time=10s, $\lambda = 0002889*A/(45*0.0065) = 0.010 \cdot A$

At time=375s, $\lambda = 0002889*A/(24*0.0065) = 0.018 \cdot A$

So basically you can draw a curve for the thermal conductivity over time and calculate mean of that for the material's property.

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  • $\begingroup$ With your equation and the now correct heat flux (I edited the question..), lambda is getting way to big. For example at time=75s: 40000*0,025748/188,83*0,0065 = 839 W/(K m). I would be happy with anything around 1 W/(K m). $\endgroup$ – hiwifgf Aug 29 '17 at 8:14
  • $\begingroup$ @ChesterMiller answer suggested that the area calculation that you made is questionable. It is not cross section area but the cone surface area. $\endgroup$ – user115350 Aug 29 '17 at 23:06

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