Test sur les transferts thermiques

Question

Why is a metal a good heat conductor ?

Solution

Thermal conductivity goes hand in hand with electric conductivity.

Thus, metals, good electric conductors, are good thermal conductors thanks to their electrons that take part in the heat conduction phenomenon.

Question

  • Outline Fick's and Fourier's laws.

  • Give the dimensions of the thermal conductivity \(\lambda\) and \(D\), the coefficient of particle scattering.

Solution

  • Fourier's law :

    \(\vec j_{th}=-\lambda \vec {grad}T\)

  • Fick's law :

    \(\vec j_{diff}=-D \vec {grad}n^*\)

  • \(\lambda\) is in \(W.m^{-1}.K^{-1}\) and \(D\) in \(\)m^2.s{-1}.

Question

Give the volumetric power dissipated in an ohmic conductor (with \(\gamma\) its conductivity) carrying a volumetric current \(\vec j\).

Solution

The volumetric power is :

\(p=\vec j.\vec E=\gamma E^2=\frac{1}{\gamma}j^2\)

Question

  • Give the equation of heat diffusion, with and without source.

  • Give the equation of particle scattering, with and without source.

Solution

  • Equation of heat diffusion, with and without source :

    \(\frac{{{\partial ^2}T(x,t)}}{{\partial {x^2}}} + \frac{1}{\lambda }{p_s}(x,t) = \frac{{\rho \ c}}{\lambda }\ \frac{{\partial T(x,t)}}{{\partial t}}\)

    \(\frac{{{\partial ^2}T(x,t)}}{{\partial {x^2}}} = \frac{{\rho \ c}}{\lambda }\ \frac{{\partial T(x,t)}}{{\partial t}}\)

  • Equation of particle scattering, with and without source :

    \(D\frac{{{\partial ^2}n^*(x,t)}}{{\partial {x^2}}} + \sigma_s(x,t) = \frac{{\partial n^*(x,t)}}{{\partial t}}\)

    \(D\frac{{{\partial ^2}n^*(x,t)}}{{\partial {x^2}}} = \frac{{\partial n^*(x,t)}}{{\partial t}}\)

Question

What is it worth for a laterally isolated rod, with \(\lambda\) its conductivity?

Indice

The thermal resistance is defined by :

\(R_{th}=\frac {T_1-T_2}{\Phi_{th}}\)

Where \(\Phi_{th}\) is the thermal flow  :

\(\Phi_{th}=\iint_S\vec j_{th}.\vec n dS\)

For an isolated rod :

\(R_{th}=\frac{\ell}{\lambda S}\)

Where \(\ell\) is the length of the rod and \(S\) its surface.

Question

Outline Newton's law. \(h\) is the coefficient of convective exchange.

Solution

The thermal transfers between a nody and the external medium follow Newton's law is the density of the thermal flow (algebraically exiting through the surface of the material) is proportional to the temperature difference between the external medium and the surface of the material.

\({j_{conv}}=h({T_P}-{T_F})\)

\(h\) is called film coefficient of heat transfer.