## Heat Conduction in a Large Plane Wall

**Heat Conduction in a Large Plane Wall**

Consider the plane wall of thickness 2L, in which there is **uniform and constant heat generation** per unit volume, **q _{V }[W/m^{3}]**. The centre plane is taken as the origin for x and the slab extends to + L on the right and – L on the left. For constant thermal conductivity k, the appropriate form of the heat equation, is:

The general solution of this equation is:

where C_{1} and C_{2} are the constants of integration.

1)

Calculate the temperature distribution, T(x), through this thick plane wall, if:

- the temperatures at both surfaces are 15.0°C
- the thickness this wall is 2L = 10 mm.
- the materials conductivity is k = 2.8 W/m.K (corresponds to uranium dioxide at 1000°C)
- the volumetric heat rate is q
_{V}= 10^{6}W/m^{3}

In this case, the surfaces are maintained at given temperatures T_{s,1} and T_{s,2}. This corresponds to the Dirichlet boundary condition. Moreover, this problem is thermally symmetric and therefore we may use also thermal symmetry boundary condition. The constants may be evaluated using substitution into the general solution and are of the form:

The resulting temperature distribution and the centerline (x = 0) temperature (maximum) in this plane wall at these specific boundary conditions will be:

The heat flux at any point, q_{x} [W.m^{-2}], in the wall may, of course, be determined by using the temperature distribution and with the **Fourier’s law**. Note that, with heat generation the heat flux is no longer independent of x, the therefore:

**Example of Heat Equation – Problem with Solution**