Question

In many engineering applications variation in thermal properties is significant, especially when there are large temperature gradients or the material is not homogeneous. To account for these variations in thermal properties, develop a finite difference formulation for an internal node in the case of a three-dimensional, steady-state heat conduction equation with variable thermal conductivity.

Short answer

The final finite difference formulation for an internal node is given by: $$ \left(\frac{k_{E} + k_{W}}{\Delta x^2} + \frac{k_{N} + k_{S}}{\Delta y^2} + \frac{k_{U} + k_{D}}{\Delta z^2}\right)T_{i,j,k} = \frac{k_{E}T_{i+1,j,k} + k_{W}T_{i-1,j,k}}{\Delta x^2} + \frac{k_{N}T_{i,j+1,k} + k_{S}T_{i,j-1,k}}{\Delta y^2} + \frac{k_{U}T_{i,j,k+1} + k_{D}T_{i,j,k-1}}{\Delta z^2} $$ where \(k_E, k_W, k_N, k_S, k_U, k_D\) are the thermal conductivities at East, West, North, South, Up, and Down faces, respectively, and \((i, j, k)\) represents the indices of grid points in the x, y, and z directions.

Step-by-step solution

Write the general heat conduction equation for 3D, steady-state case

The general heat conduction equation for a three-dimensional, steady-state case is given as: $$ \nabla \cdot (k \nabla T) = 0 $$ Where \(k\) is the thermal conductivity, \(\nabla\) is the gradient operator and \(T\) is the temperature.

Expand the equation in Cartesian coordinates

The equation can be expanded in Cartesian coordinates (x, y, and z) as: $$ \frac{\partial}{\partial x}\left(k \frac{\partial T}{\partial x}\right) + \frac{\partial}{\partial y}\left(k \frac{\partial T}{\partial y}\right) + \frac{\partial}{\partial z}\left(k \frac{\partial T}{\partial z}\right) = 0 $$

Apply finite difference approximation to the derivatives

To convert this continuous equation into a discrete form, we need to apply finite difference approximation to the derivatives, using forward, backward, and central differences as appropriate. For instance, for the first term with respect to 'x', we can use central difference approximation as follows: $$ \frac{\partial}{\partial x}\left(k \frac{\partial T}{\partial x}\right) \approx \frac{k_{E}\left(\frac{T_{i+1,j,k} - T_{i,j,k}}{\Delta x}\right) - k_{W}\left(\frac{T_{i,j,k} - T_{i-1,j,k}}{\Delta x}\right)}{\Delta x} $$ Similarly, we can approximate the other two terms with respect to 'y' and 'z'.

Combine all terms into a single equation

Combine all the approximated derivatives in the finite difference form: $$ \frac{k_{E}\left(\frac{T_{i+1,j,k} - T_{i,j,k}}{\Delta x}\right) - k_{W}\left(\frac{T_{i,j,k} - T_{i-1,j,k}}{\Delta x}\right)}{\Delta x} + \frac{k_{N}\left(\frac{T_{i,j+1,k} - T_{i,j,k}}{\Delta y}\right) - k_{S}\left(\frac{T_{i,j,k} - T_{i,j-1,k}}{\Delta y}\right)}{\Delta y} + \frac{k_{U}\left(\frac{T_{i,j,k+1} - T_{i,j,k}}{\Delta z}\right) - k_{D}\left(\frac{T_{i,j,k} - T_{i,j,k-1}}{\Delta z}\right)}{\Delta z} = 0 $$ Where: - \(k_E, k_W, k_N, k_S, k_U, k_D\) are the thermal conductivities at East, West, North, South, Up and Down faces, respectively. - \((i, j, k)\) represents the indices of grid points in the x, y, and z directions.

Rearrange and simplify the equation to obtain the final finite difference formulation

Rearrange the equation and group terms associated with the unknown temperatures to obtain the final finite difference formulation: $$ \left(\frac{k_{E} + k_{W}}{\Delta x^2} + \frac{k_{N} + k_{S}}{\Delta y^2} + \frac{k_{U} + k_{D}}{\Delta z^2}\right)T_{i,j,k} = \frac{k_{E}T_{i+1,j,k} + k_{W}T_{i-1,j,k}}{\Delta x^2} + \frac{k_{N}T_{i,j+1,k} + k_{S}T_{i,j-1,k}}{\Delta y^2} + \frac{k_{U}T_{i,j,k+1} + k_{D}T_{i,j,k-1}}{\Delta z^2} $$ Now, we have derived the finite difference formulation for an internal node in the case of a three-dimensional, steady-state heat conduction equation with variable thermal conductivity.