Heat Flow

The problems consiste in determining the temperature variation along a bar of length $l$ at any instant of time, given the initial gradient of temperature.

Figure 13: A bar with the ends in contact with heat reservoirs.

The problem is described by the ``Heat Equation'' that can be derived as follows:

At any instant of time, the heat flow through the bar equals the variatioon of energy inside the bar:

$$\mathrm{Heat Flow=Variation of internal energy}$$

or

$$\nabla \cdot Q = -\frac{dE}{dt}.$$ (60)

The variation of the internal energy is given by the body's ability to store heat by raising its temperature:

$$\frac{dE}{dt} = \rho c \frac{dT}{dt},$$ (61)

where $\rho$ is the density, and $c$ is the specific heat of the material. Fourier's Law of heat conduction states that

$$Q=-K\nabla T,$$ (62)

where $K$ is the thermal conductivity. Hence, we obtain

$$\nabla \cdot \left( K \nabla T\right) = \rho c \frac{dT}{dt},$$ (63)

or

$$\frac{dT}{dt}=\alpha \nabla^2T,$$ (64)

with $\alpha = K/(c\rho)$. In 1d this equation is written

$$\frac{\partial T(x,t)}{\partial t} = \alpha \frac{\partial ^2T}{\partial x^2}.$$ (65)

We must solve this equation given the initial condition

$$T(x,t=0) = f(x) \mathrm{(initial condition)}$$ (66)

and the boundary condition

$$T(x=0,t) = T_1,$$ (67)

$$T(x=l,t) = T_2.$$ (68)