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 \(0,1,2,3,4\), and 5 with a uniform nodal spacing of \(\Delta x\). Using the energy balance approach, obtain the finite difference formulation of the boundary nodes for the case of insulation at the left boundary (node 0 ) and radiation at the right boundary (node 5) with an emissivity of \(\varepsilon\) and surrounding temperature of \(T_{\text {carr }}\)

Short answer

Answer: The finite difference equation for node 0 is given by the insulation boundary condition, which states that the temperature is constant and there is no heat conduction out of the boundary. Therefore, the equation is: $$ T_{0} = T_{1} $$ For node 5, the finite difference equation is derived from the radiation boundary condition, where heat loss occurs due to radiation. The resulting equation is: $$ k \left(\frac{T_{5} - T_{4}}{\Delta x}\right) = \varepsilon \sigma (T_{5}^4 - T_{\text {carr}}^4) $$

Step-by-step solution

Heat conduction equation

The steady one-dimensional heat conduction in a plane wall can be modeled by the following equation: $$ k\frac{d^2T}{dx^2} - q_\text{gen} = 0$$ where \(k\) is the thermal conductivity, \(T\) is the temperature, \(x\) is the spatial coordinate, and \(q_\text{gen}\) is the heat generation per unit volume.

Finite difference formulation for boundary nodes

We can discretize the heat conduction equation using the central difference approximation for the second derivative: $$ \frac{d^2T}{dx^2} = \frac{T_{i-1} - 2T_{i} + T_{i+1}}{\Delta x^2}$$ This equation will be used to obtain the finite difference formulations for the boundary nodes 0 and 5.

Insulation boundary condition at node 0

The insulation boundary condition states that the heat flux at node 0 is zero, which means the temperature is constant and there is no heat conduction out of the boundary. The finite difference equation can be written as: $$ k\left(\frac{T_{1} - T_{0}}{\Delta x}\right) = 0$$ From this equation, we have: $$ T_{0} = T_{1} $$

Radiation boundary condition at node 5

The radiation boundary condition states that there is heat loss at node 5 due to radiation. It can be expressed as: $$ q_\text{rad} = \varepsilon \sigma (T_{5}^4 - T_{\text {carr}}^4) $$ The heat flux at the boundary can be approximated by: $$ q_\text{rad} = k\left(\frac{T_{5} - T_{4}}{\Delta x}\right) $$ By setting these equations equal, we have: $$ k \left(\frac{T_{5} - T_{4}}{\Delta x}\right) = \varepsilon \sigma (T_{5}^4 - T_{\text {carr}}^4) $$ These last two equations provide the finite difference formulation at nodes 0 and 5 required to solve the problem with the defined boundary conditions.