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Thermal conduction
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Heat transport equation
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The heat transport equation describes how a temperature difference dT over a distance dz in a body of section S leads to the displacement of a dQ heat in a dt time and the thermal conduction constant

$\displaystyle\frac{ dQ }{ dt }= S \lambda \displaystyle\frac{ dT }{ dz }$

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Thermal conduction as energy exchange
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If the temperature at a point z is T (z) and at a neighboring point z + dz it is T (z + dz) it will be taken that the particles at a distance of a free path l can redistribute the energy:

$\displaystyle\frac{f}{2}k\Delta T=\displaystyle\frac{f}{2}k(T(z+dz)-T(z))$



The number of particles participating in this process is equal to those found in a volume of section S and height equal to the free path l :

$S,l,c_n$



Therefore, the heat flowing dQ will be equal to the change in energy over time, that is

$ dQ =-\displaystyle\frac{ f }{2} S l c_n m k_B dT $

where the negative sign is due to the heat flowing from the area with the highest to the lowest temperature.

ID:(3947, 0)


Microscopic thermal conduction
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If you consider the dQ heat flux created by mixing particles of different energy

$ dQ =-\displaystyle\frac{ f }{2} S l c_n m k_B dT $



where S in the section, l the free path, c_n the concentration, m the mass and f the free degrees of the particle and dT . This expression can be re-formulated dividing it by time and entering the distance dz traveled by the heat

$\displaystyle\frac{dQ}{dt}=-\displaystyle\frac{f}{2}S,l,c_nm\displaystyle\frac{dt}{dz}\displaystyle\frac{dz}{dt}$



The derivative of the z position with respect to time can be modeled using

$\displaystyle\frac{dz}{dt}=v_z=\displaystyle\frac{1}{3}\sqrt{\langle v^2\rangle}$



In this way the force created by the mixture of moments like

$\displaystyle\frac{dQ}{dt}=-\displaystyle\frac{1}{3}S,l,c_nm\sqrt{\langle v^2\rangle}\displaystyle\frac{dT}{dz}$



If you compare this expression with the viscous force

$\displaystyle\frac{dQ}{dt}=-S\lambda\displaystyle\frac{dT}{dz}$



it is concluded that thermal conduction has to be

$\lambda=\displaystyle\frac{f}{6}k\,c_nl\sqrt{\langle v^2\rangle}$

ID:(3949, 0)


Thermal conduction depending on the temperature
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If thermal conduction is

$\lambda=\displaystyle\frac{f}{6}k\,c_nl\sqrt{\langle v^2\rangle}$



with l the free path, c_n the concentration, k_B the Boltzmann constant and \ langle v ^ 2 \ rangle the expected value of the square of the velocity. With the expression for the free way

$l=\displaystyle\frac{1}{\sqrt{2}\pi d^2c_n}$



the viscosity as a function of the temperature will be:

$\lambda=\displaystyle\frac{1}{6\pi d^2}\sqrt{\displaystyle\frac{f^3k^3T}{2m}}$

where the negative sign is because the force is opposite to the direction of flow.

ID:(3950, 0)