Let's we have two equal sized conductor and an insulator materials, which are perfectly joined together. We have two senerios:
1. Insulator is heated to 300 C and temperature is measured at conductor end (let's say Point A) after 180 seconds
2. Conductor is heated to 300 C and temperature is measured at insulator end (let's say Point B) after 180 seconds
Question is: Will the temperature at point A and B will be same (as total length covered is same) or there will be different temperatures at points A and B due to difference of dT in both cases?
Hello,
I understand you have 2 sheets, one is a conductor and one is an insulator.
So your simplified problem is 1D conduction (perpendicular to sheet) and we suppose conduction independent of temperature.
In that case, in steady state, the global thermal conductivity is the same.
But you are in transcient conduction and I don't know the answer. You should write the differential equations. It is simple to solve with excel or other table : x in lines t in columns for exemple.
Start with large steps then try again with smaller steps untill convergence.
Hi,
Assuming that your composite material is made of two layers initially at room temperature T_0, a temperature T_A is suddenly applied to the external surface of layer 1 (point A) whereas the external surface of layer 2 is insulated (point B). The transient evolution of the temperature at point B is proportional to the temperature difference (T_A-T_0) and depends non-linearly on time and on 3 parameters: the square root of diffusion time (SRDT) in layer 1 , (i.e. thickness divided by the thermal diffusivity of layer 1, denoted by xi1), the SRDT in layer 2 (denoted by xi2), and the ratio of the thermal effusivities of layer 1 (i.e. b1) and layer 2 (i.e. b2). By applying the quadrupole method we can find that the normalised temperature in the Laplace space is expressed by 1/(ch1*ch2+b2/b1*sh1*sh2) where, for example, ch1 denotes cosh(xi1*sqrt(p)) and sh2 denotes sinh(xi2*sqrt(p)), where p is the Laplace variable. The Laplace inversion can be performed easily with the Gaver-Stehfest method.
I can send you a short matlab script to calculate the temperature evolution, given the 3 parameters.
As an example, if the diffusion time is the same in layer 1 and 2 (i.e. xi1=xi2), the temperature rise at point B reaches the following values at the particular time corresponding to (xi1+xi2)^2 :
- 29% of the temperature difference if b2=10*b1
- 99.7% of the temperature difference if b2=0.1*b1
In summary, point B will come faster close to the imposed temperature if the layer with the highest effusivity is at position 1. The main reason for this is that more power is required to impose T_A on the material with high effusivity. This is valid at least at the beginning of the process; this is however enough to make the difference. As a consequence, the asymptotic equilibrium temperature T_A is then reached faster on the opposite surface B.
This result remains valid for other values of the ratio xi2/xi1.
BR.
Jean-Claude
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