Question

For a one-dimensional, steady-state, variable thermal conductivity heat conduction with uniform internal heat generation, develop a generalized finite difference formulation for the interior nodes, with the left surface boundary node exposed to constant heat flux and the right surface boundary node exposed to a convective environment. The variable conductivity is modeled such that the thermal conductivity varies linearly with the temperature as \(k(T)=k_{o}(1+\beta T)\) where \(T\) is the average temperature between the two nodes.

Short answer

Question: Develop a finite difference formulation for a one-dimensional, steady-state heat conduction problem with variable thermal conductivity and uniform internal heat generation. Consider linear temperature-dependence of thermal conductivity. The left surface boundary node is exposed to a constant heat flux, and the right surface boundary is exposed to a convective environment. Answer: The finite difference formulation for interior nodes (2 to N-1) is given by: \(k_{o}(1+\beta T_{i})\frac{T_{i+1}-2T_{i}+T_{i-1}}{(\Delta x)^2} + q_{gen} = 0\) For the boundary nodes, we have: 1. Left boundary node (i=1): \(k_{o}(1+\beta T_{1})\frac{T_{2}-T_{1}}{\Delta x} = q_{f}\) 2. Right boundary node (i=N): \(k_{o}(1+\beta T_{N})\frac{T_{N}-T_{N-1}}{\Delta x} = h(T_{N}-T_{\infty})\) These equations, combined with the given boundary conditions, can be used to solve for the temperature distribution across the spatial domain.

Step-by-step solution

Understand the problem and given data

We are given a one-dimensional, steady-state heat conduction problem with variable thermal conductivity and uniform internal heat generation. The conductivity varies linearly with the temperature as \(k(T)=k_{o}(1+\beta T)\). The left surface boundary node is exposed to a constant heat flux, and the right surface boundary is exposed to a convective environment.

Write the general heat conduction equation

For one-dimensional, steady-state heat conduction with variable thermal conductivity and uniform internal heat generation, the general heat conduction equation is: \(\frac{d}{dx}\left[k(T)\frac{dT}{dx}\right] + q_{gen} = 0\)

Apply the finite difference method

We can discretize the spatial domain into N nodes, and the equation becomes: \(\frac{d}{dx}\left[k(T_{i})\frac{T_{i+1}-T_{i-1}}{2\Delta x}\right] + q_{gen} = 0\)

Rearrange the equation for interior nodes and incorporate the variable conductivity

We will now rewrite the equation for the interior nodes by substituting the given conductivity expression and rearrange to obtain an expression for node i: \(k_{o}(1+\beta T_{i})\frac{T_{i+1}-2T_{i}+T_{i-1}}{(\Delta x)^2} + q_{gen} = 0\)

Implement boundary conditions

We need to apply the boundary conditions at the left and right surface nodes. Suppose that at the left boundary, we have a constant heat flux \(q_{f}\) and we can use a forward difference approximation: \(k_{o}(1+\beta T_{1})\frac{T_{2}-T_{1}}{\Delta x} = q_{f}\) For the right boundary, assume that it is exposed to a convective environment with convective heat transfer coefficient \(h\) and ambient temperature \(T_{\infty}\). Then, we apply the Newton's law of cooling to the right boundary node, N: \(k_{o}(1+\beta T_{N})\frac{T_{N}-T_{N-1}}{\Delta x} = h(T_{N}-T_{\infty})\)

Develop the finite difference formulation

With all the derived expressions and boundary conditions, we can construct a linear system of N equations for N unknown temperatures. The finite difference formulation for interior nodes (2 to N-1) is given by: \(k_{o}(1+\beta T_{i})\frac{T_{i+1}-2T_{i}+T_{i-1}}{(\Delta x)^2} + q_{gen} = 0\) For the boundary nodes, we have: 1. Left boundary node (i=1): \(k_{o}(1+\beta T_{1})\frac{T_{2}-T_{1}}{\Delta x} = q_{f}\) 2. Right boundary node (i=N): \(k_{o}(1+\beta T_{N})\frac{T_{N}-T_{N-1}}{\Delta x} = h(T_{N}-T_{\infty})\) We can solve this system of linear equations to obtain the temperature distribution across the spatial domain under the specified conditions.