Question

Starting with an energy balance on the volume element, obtain the steady three-dimensional finite difference equation for a general interior node in rectangular coordinates for \(T(x, y, z)\) for the case of constant thermal conductivity and uniform heat generation.

Short answer

Derive the steady three-dimensional finite difference equation for a general interior node in rectangular coordinates considering constant thermal conductivity and uniform heat generation. Solution: The steady three-dimensional finite difference equation for a general interior node in rectangular coordinates considering constant thermal conductivity and uniform heat generation is given by: \(T_{i,j,k} = \frac{1}{2k(\Delta x^2 + \Delta y^2 + \Delta z^2)}\Big((k\Delta y^2\Delta z^2)(T_{i+1,j,k} + T_{i-1,j,k}) + (k\Delta x^2\Delta z^2)(T_{i,j+1,k} + T_{i,j-1,k}) + (k\Delta x^2\Delta y^2)(T_{i,j,k+1} + T_{i,j,k-1}) - Q\Delta x^2\Delta y^2\Delta z^2\Big)\)

Step-by-step solution

Consider an arbitrary volume element

Consider an arbitrary volume element (\(\Delta x\), \(\Delta y\), \(\Delta z\)) within the 3-dimensional rectangular coordinate system. Our objective is to obtain the rate of heat transfer in and out of this volume element, taking into account the heat generation within the volume element.

Perform an energy balance on the volume element

Perform an energy balance on the volume element, considering heat transfer due to conduction along the x, y, z directions, and the heat generation within the volume element. The energy balance equation can be written as: \(\frac{\partial}{\partial x}(k_x\frac{\partial T}{\partial x}) + \frac{\partial}{\partial y}(k_y\frac{\partial T}{\partial y}) + \frac{\partial}{\partial z}(k_z\frac{\partial T}{\partial z}) + Q = 0 \) Since the thermal conductivity is constant, we can rewrite the energy balance equation as: \(k \Big(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2} \Big) + Q = 0\)

Discretize the spatial dimensions

We need to discretize the spatial dimensions (\(x, y, z\)) by dividing the domain into a grid with nodes representing the temperature \(T_i\). In this case, we can represent the location of each node by the general coordinates: \((x_i, y_j, z_k)\) where \(i\), \(j\), and \(k\) denote the spatial indices in each dimension.

Apply finite difference approximations

The second-order central difference scheme for the second derivatives in the spatial dimensions can be written as: \(\frac{\partial^2 T}{\partial x^2} = \frac{T_{i+1,j,k} - 2T_{i,j,k} + T_{i-1,j,k}}{\Delta x^2}\) \(\frac{\partial^2 T}{\partial y^2} = \frac{T_{i,j+1,k} - 2T_{i,j,k} + T_{i,j-1,k}}{\Delta y^2}\) \(\frac{\partial^2 T}{\partial z^2} = \frac{T_{i,j,k+1} - 2T_{i,j,k} + T_{i,j,k-1}}{\Delta z^2}\) Substituting these approximations into the energy balance equation, we obtain \(k\Big(\frac{T_{i+1,j,k} - 2T_{i,j,k} + T_{i-1,j,k}}{\Delta x^2} + \frac{T_{i,j+1,k} - 2T_{i,j,k} + T_{i,j-1,k}}{\Delta y^2} + \frac{T_{i,j,k+1} - 2T_{i,j,k} + T_{i,j,k-1}}{\Delta z^2}\Big) + Q = 0\)

Rearrange the finite difference equation for a general interior node

Rearrange the terms in the equation to obtain the final steady three-dimensional finite difference equation for a general interior node in rectangular coordinates considering constant thermal conductivity and uniform heat generation: \(T_{i,j,k} = \frac{1}{2k(\Delta x^2 + \Delta y^2 + \Delta z^2)}\Big((k\Delta y^2\Delta z^2)(T_{i+1,j,k} + T_{i-1,j,k}) + (k\Delta x^2\Delta z^2)(T_{i,j+1,k} + T_{i,j-1,k}) + (k\Delta x^2\Delta y^2)(T_{i,j,k+1} + T_{i,j,k-1}) - Q\Delta x^2\Delta y^2\Delta z^2\Big)\)