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 \(q_{0}\) 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

Answer: The finite difference formulations for the boundary nodes are: $$ T_{0} = T_{1} - \frac{q_{0} \Delta x}{\kappa} $$ $$ T_{4} = \frac{h \Delta x T_\infty + \kappa T_{3}}{h \Delta x + \kappa} $$

Step-by-step solution

Write the general heat conduction equation

Consider the steady one-dimensional heat conduction with variable heat generation, the general equation can be given as $$-\kappa \frac{d^{2} T}{d x^{2}}=q_{g}(x)$$ where \(\kappa\) is the constant thermal conductivity and \(q_{g}(x)\) is the heat generation term.

Write the finite difference form of the first derivative

Using the central difference approximation for the second derivative, the finite difference form can be written as $$ \frac{d^2 T}{d x^2} \approx \frac{T_{i+1} - 2T_{i} + T_{i-1}}{(\Delta x)^2} $$ Now, substituting the finite difference form back to the general equation, we get $$ -\kappa \frac{T_{i+1} - 2T_{i} + T_{i-1}}{(\Delta x)^2} = q_{g}(x) $$

Apply the boundary conditions and find the finite difference formulations for the boundary nodes

The given boundary conditions are: 1. Uniform heat flux \(q_{0}\) at the left boundary (node 0) 2. Convection at the right boundary (node 4) with a convection coefficient \(h\) and ambient temperature \(T_{\infty}\) To incorporate Boundary condition 1, we apply the heat flux relation: $$ q_{0} = -\kappa \frac{dT}{dx} \approx -\kappa \frac{T_{1}-T_{0}}{\Delta x} $$ We want to express \(T_{0}\) in terms of the other variables, so we rearrange the equation as: $$ T_{0} = T_{1} - \frac{q_{0} \Delta x}{\kappa} $$ Now, for boundary condition 2, we apply the convection condition at node 4: $$ -\kappa \frac{dT}{dx} = h (T_{4}-T_{\infty}) $$ Approximate the first derivative at node 4 as $$ -\kappa \frac{dT}{dx} \approx -\kappa \frac{T_{4}-T_{3}}{\Delta x} $$ Therefore, $$ -\kappa \frac{T_{4}-T_{3}}{\Delta x} = h (T_{4}-T_{\infty}) $$ Rearrange the equation to express \(T_{4}\) in terms of other variables, we get $$ T_{4} = \frac{h \Delta x T_\infty + \kappa T_{3}}{h \Delta x + \kappa} $$ The finite difference formulations for the boundary nodes are: $$ T_{0} = T_{1} - \frac{q_{0} \Delta x}{\kappa} $$ $$ T_{4} = \frac{h \Delta x T_\infty + \kappa T_{3}}{h \Delta x + \kappa} $$

Key concepts

Finite Difference Method

The Finite Difference Method is a numerical technique used to approximate solutions to differential equations. In heat conduction problems, it helps us discretize the continuous domain into a set of discrete points or nodes. This approach allows us to turn differential equations into a set of algebraic equations that are easier to solve.

In the context of the given exercise, we're dealing with a one-dimensional system. The finite difference method involves approximating derivatives by differences between neighboring nodes. For instance, we use the central difference approximation for the second derivative:

• \( \frac{d^2 T}{d x^2} \approx \frac{T_{i+1} - 2T_{i} + T_{i-1}}{(\Delta x)^2} \)

This equation helps us replace the continuous differential terms with discrete counterparts, making the problem solvable with respect to each node.

Boundary Conditions

Boundary conditions are essential in solving heat conduction problems as they describe how the system exchanges heat with its surroundings. In this exercise, two specific boundary conditions are considered:

• **At the left boundary (node 0)**: There is a uniform heat flux \(q_{0}\). This condition is related to the rate at which heat is flowing into or out of the material. It is mathematically expressed as:
  • \( q_{0} = -\kappa \frac{T_{1}-T_{0}}{\Delta x} \)
• **At the right boundary (node 4)**: Convection occurs with the surroundings, characterized by a convection coefficient \(h\) and an ambient temperature \(T_{\infty}\). This is given by:
  • \( -\kappa \frac{T_{4}-T_{3}}{\Delta x} = h (T_{4}-T_{\infty}) \)

These conditions are crucial for determining the temperature distribution across the nodes and ensure that the physical constraints of the problem are respected.

Heat Generation

Heat generation refers to the internal production of heat within a material. It can vary spatially (as \(q_{g}(x)\) in our equation) and influences the temperature distribution.

In the differential heat conduction equation:

• \( -\kappa \frac{d^{2} T}{d x^{2}} = q_{g}(x) \)

The function \(q_{g}(x)\) represents this generated heat. The presence of heat generation adds complexity to the problem as it introduces another source affecting the temperature. The finite difference method allows approximating the effect of this term directly on the nodes, helping calculate how much each point is heated from within.

Thermal Conductivity

Thermal conductivity, denoted by \(\kappa\), is a material property that measures a material's ability to conduct heat. It is critical in defining how heat moves through a solid.

In the equation:

• \( -\kappa \frac{d^2 T}{d x^2} = q_{g}(x) \)

\(\kappa\) helps determine the rate of heat flow per unit area per temperature gradient. In this exercise, it is constant, simplifying the calculations.

A higher \(\kappa\) indicates that the material conducts heat better, affecting both internal heat distribution and how it interacts with boundary conditions. The finite difference equations use \(\kappa\) to calculate temperature differences between nodes, providing insight into flow behavior across the material.