Calculating thermal conductivity with measured data

Question

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 |

user115350 · Answer

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.

Chet Miller · Answer

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.

Answered by Chet Miller on November 28, 2020

Calculating thermal conductivity with measured data

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