Question

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

Short answer

#Short Answer# The finite difference equation for a general interior node (i, j) in the case of variable thermal conductivity and uniform heat generation is: $$ k_{i+\frac{1}{2},j}(T_{i+1,j} - T_{i,j}) - k_{i-\frac{1}{2},j}(T_{i,j} - T_{i-1,j}) + k_{i,j+\frac{1}{2}}(T_{i,j+1} - T_{i,j}) - k_{i,j-\frac{1}{2}}(T_{i,j} - T_{i,j-1}) = Gh^2 $$

Step-by-step solution

Write the energy balance equation for the given conditions

We consider a general interior node \((x, y)\) of a volume element. The energy balance equation states that the heat generation (Q) is equal to the conduction heat loss (C), under steady-state conditions: $$ Q = C $$ Considering variable thermal conductivity (\(k(x, y)\)) and uniform heat generation (\(G\)), the equation can be written as: $$ G = \nabla \cdot (k(x, y) \nabla T(x, y)) $$

Expand the divergence term in rectangular coordinates

Let's now expand the divergence term in rectangular coordinates: $$ \nabla \cdot (k(x, y) \nabla T(x, y)) = \frac{\partial}{\partial x}(k(x, y) \frac{\partial T}{\partial x}) + \frac{\partial}{\partial y}(k(x, y) \frac{\partial T}{\partial y}) $$

Replace the derivatives with finite difference approximations

We will replace the partial derivatives in the previous equation with finite difference approximations for a general interior node \((i, j)\) in a grid with mesh spacing \(h\) in both \(x\) and \(y\) direction. Let's consider the finite-difference approximations for the required derivatives: $$ \frac{\partial}{\partial x}(k(x, y) \frac{\partial T}{\partial x}) \approx \frac{(k_{i+\frac{1}{2},j} (T_{i+1,j} - T_{i,j}) - k_{i-\frac{1}{2},j} (T_{i,j} - T_{i-1,j}))}{h^2} $$ $$ \frac{\partial}{\partial y}(k(x, y) \frac{\partial T}{\partial y}) \approx \frac{(k_{i,j+\frac{1}{2}} (T_{i,j+1} - T_{i,j}) - k_{i,j-\frac{1}{2}} (T_{i,j} - T_{i,j-1}))}{h^2} $$ Now, substitute these approximations in the energy balance equation and rearrange terms:

Obtain the finite difference equation

Substituting the approximations in the energy balance equation, we get: $$ G \approx \frac{(k_{i+\frac{1}{2},j} (T_{i+1,j} - T_{i,j}) - k_{i-\frac{1}{2},j} (T_{i,j} - T_{i-1,j}))}{h^2} + \frac{(k_{i,j+\frac{1}{2}} (T_{i,j+1} - T_{i,j}) - k_{i,j-\frac{1}{2}} (T_{i,j} - T_{i,j-1}))}{h^2} $$ Rearranging terms, the finite difference equation for a general interior node \((i, j)\) in the case of variable thermal conductivity and uniform heat generation becomes: $$ k_{i+\frac{1}{2},j}(T_{i+1,j} - T_{i,j}) - k_{i-\frac{1}{2},j}(T_{i,j} - T_{i-1,j}) + k_{i,j+\frac{1}{2}}(T_{i,j+1} - T_{i,j}) - k_{i,j-\frac{1}{2}}(T_{i,j} - T_{i,j-1}) = Gh^2 $$