Thermal resistance

Fondamental : Definition of the thermal resistance

Definition of the thermal resistance of an insulated cylindrical rod :

We want to determine, in stationnary mode, the temperature in a homogeneous cylindrical rod, with \(L\) its length, \(S\) its section and its edges are at temperatures \(T_1\) and \(T_2<T_1\).

Let's suppose that the lateral surface is insulated.

The heat equation is simply

\(\frac {d^2T(x)}{dx^2}= 0\)

So :

\(T(x) = \frac{{{T_2} - {T_1}}}{L}x + {T_1}\)

The thermal flow through the rod is :

\(\Phi=j_{th}S=-\lambda S\frac {dT(x)}{dx}=-\lambda S\frac {T_2-T_1}{L}\)

We can defined a thermal resistance \(R_{th}\) such as :

\({T_1} - {T_2} = {R_{th}}\Phi\)

That is to say :

\({R_{th}} = \frac{1}{\lambda }\frac{L}{S}\)

We can also define thermal conductance :

\({G_{th}} = 1/{R_{th}}\)

Attention : Definition of the thermal resistance

\({T_1} - {T_2} = {R_{th}}\Phi\)

Exemple : Insulation of the walls of a house

Blockworks, polystyrene, plasterboard and wallpapers : serial thermal resistances
Likewise with a window added : shunt thermal resistances.

Let's take the case of a double-glazing with a window (surface \(S\), \(e/3\) its thickness and \(\lambda\) its conductivity), a gas layer with \(\lambda'\) its conductivity, and second window similar to the first one.

It's the serial association of three heat resistances, hence the total resistance :

\({R_{th}} = \frac{{e/3}}{{\lambda S}} + \frac{{e/3}}{{\lambda 'S}} + \frac{{e/3}}{{\lambda S}} = \frac{e}{{\lambda S}}\left( {\frac{2}{3} + \frac{1}{3}\frac{\lambda }{{\lambda '}}} \right)\)

Where \(e/\lambda S\) is the resistance of a window of thickness \(e\).

With \(\lambda'=\lambda/100\), it appears that double-glazing allows us to multiply the thermal resistance by \(102/3=34\) and thus to divide by \(34\) the losses through the windows.

Complément : Analogy between Fourier's and Ohm's phenomenological laws

The classical Drude model affords to interpret local Ohm's law in the metals :

\(\vec j = \sigma \vec E\)

Let's remember that the expression of the heat resistance of a metal wire, with \(L\) its length, \(S\) its section and \(\sigma\) its conductivity (\(\rho\) its resistivity) :

\(R = \rho \frac{L}{S} = \frac{1}{\sigma }\frac{L}{S}\)

Obviously, we can make the analogy with the thermal resistance of a one-dimensional straight-lined bar.