Question

Consider steady one-dimensional heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes \(0,1,2,3\), and 4 with a uniform nodal spacing of \(\Delta x\). Using the finite difference form of the first derivative (not the energy balance approach), obtain the finite difference formulation of the boundary nodes for the case of uniform heat flux \(\dot{q}_{0}\) at the left boundary (node 0 ) and convection at the right boundary (node 4) with a convection coefficient of \(h\) and an ambient temperature of \(T_{\infty}\) -

Short answer

Question: Write down the finite difference formulations for the boundary nodes of a plane wall with variable heat generation and constant thermal conductivity, given uniform heat flux at the left boundary (Node 0) and convection at the right boundary (Node 4). Answer: For Node 0: \(k \frac{0 - 2T_0 + T_1}{(\Delta x)^2} = -q_{gen}\) For Node 4: \(k \frac{T_3 - 2T_4 + 0}{(\Delta x)^2} = -q_{gen}\)

Step-by-step solution

Write down the general heat conduction equation

The general one-dimensional heat conduction equation with heat generation is given by: \(\frac{d}{dx}\left( k \frac{dT}{dx} \right) = -q_{gen}\) Where \(k\) is the constant thermal conductivity, \(T\) is the temperature and \(q_{gen}\) is the heat generation. Since the thermal conductivity is constant, we can rewrite the above equation as: \(k \frac{d^2 T}{dx^2} = -q_{gen}\)

Apply Finite Difference Formulation on the conduction equation

Now, apply the central finite difference approximation for the second derivative, \(\frac{d^2 T}{dx^2}\), as follows: \(\frac{d^2 T}{dx^2} \approx \frac{T_{i-1} - 2T_i + T_{i+1}}{(\Delta x)^2}\) Where \(T_i\) is the temperature at node \(i\), and \(Δx\) is the uniform nodal spacing. Substitute the finite difference formula for the second derivative in the general heat conduction equation: \(k\frac{T_{i-1} - 2T_i + T_{i+1}}{(\Delta x)^2} = -q_{gen}\)

Apply boundary conditions

We need to apply the given boundary conditions at node 0 and node 4. Node 0: uniform heat flux \(\dot{q}_{0}\) \(\dot{q}_{0} = -k \frac{dT}{dx}\Big|_{x=0}\), use forward finite difference approximation of first derivative, \(\dot{q}_{0} \approx -k \frac{T_1 - T_0}{\Delta x}\) Node 4: convection with convection coefficient \(h\) and ambient temperature \(T_{\infty}\) Due to convection, \(\dot{q}_{conv} = h(T_{\infty} - T_4)\) At node 4, the outward heat flow due to conduction \(\dot{q}_{cond}\), is equal to the heat removed due to convection \(\dot{q}_{conv}\): \(-k \frac{dT}{dx}\Big|_{x= x_4} = h(T_{\infty} - T_4)\) Using a backward finite difference approximation of the first derivative, \(-k \frac{T_4 - T_3}{\Delta x} = h(T_{\infty}- T_4)\)

Finite Difference Formulation for boundary nodes

Using the above equations, we can now write the finite difference formulations for the boundary nodes. Node 0: \(k\frac{T_{i-1} - 2T_i + T_{i+1}}{(\Delta x)^2} = -q_{gen}\) Substituting the known values, we get: \(k \frac{0 - 2T_0 + T_1}{(\Delta x)^2} = -q_{gen}\) Node 4: \(k\frac{T_{i-1} - 2T_i + T_{i+1}}{(\Delta x)^2} = -q_{gen}\) Substituting the known values, we get: \(k \frac{T_3 - 2T_4 + 0}{(\Delta x)^2} = -q_{gen}\) Finally, we have obtained the finite difference formulation for the boundary nodes of the plane wall with given boundary conditions.