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 left surface boundary node exposed to constant heat flux and right surface boundary node exposed to 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

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Question: Derive the finite difference equation for a one-dimensional steady-state heat conduction problem with variable thermal conductivity and uniform internal heat generation. Answer: The finite difference equation for the interior nodes is: $$\frac{1}{\Delta x} \left[ k_{i+\frac{1}{2}}\left(\frac{T_{i+1}-T_{i}}{\Delta x}\right) - k_{i-\frac{1}{2}}\left(\frac{T_{i}-T_{i-1}}{\Delta x}\right) \right] = q_{gen}$$ where \(k_{i+\frac{1}{2}}\) and \(k_{i-\frac{1}{2}}\) are the average thermal conductivities between the nodes, which can be written as: $$ k_{i + \frac{1}{2}} = k_{o}(1 + \beta \frac{T_{i} + T_{i+1}}{2}) $$ $$ k_{i - \frac{1}{2}} = k_{o}(1 + \beta \frac{T_{i} + T_{i-1}}{2}) $$

Step by step solution

01

Formulate heat conduction equation with variable thermal conductivity and uniform heat generation

The one-dimensional heat conduction equation with uniform heat generation can be written as: $$\frac{d}{dx} \left(k \frac{dT}{dx}\right) = q_{gen}$$ For variable thermal conductivity, we have \(k(T) = k_{o}(1+\beta T)\). So, we can write the equation as: $$\frac{d}{dx} \left(k_{o}(1+\beta T) \frac{dT}{dx}\right) = q_{gen}$$
02

Express boundary conditions

The boundary conditions are: 1. Left surface (x=0): Constant heat flux \(q_{o}\). 2. Right surface (x=L): Exposed to convective environment with convective heat transfer coefficient \(h\) and ambient temperature \(T_{\infty}\). So, we have \(q_{conv} = h(T(L)-T_{\infty})\).
03

Approximate with finite difference method

To derive the finite difference approximation for the interior nodes, we can discretize the spatial domain using equally spaced grid points. Let \(T_{i}\) be the temperature at node \(i\), and \(\Delta x\) be the distance between the adjacent nodes. The finite difference approximation of the equation above can be expressed as: $$\frac{1}{\Delta x} \left[ k_{i+\frac{1}{2}}\left(\frac{T_{i+1}-T_{i}}{\Delta x}\right) - k_{i-\frac{1}{2}}\left(\frac{T_{i}-T_{i-1}}{\Delta x}\right) \right] = q_{gen}$$ where \(k_{i+\frac{1}{2}}\) and \(k_{i-\frac{1}{2}}\) are the average thermal conductivities between the nodes. Since \(k(T)=k_{o}(1+\beta T)\), we can write: $$ k_{i + \frac{1}{2}} = k_{o}(1 + \beta \frac{T_{i} + T_{i+1}}{2}) $$ $$ k_{i - \frac{1}{2}} = k_{o}(1 + \beta \frac{T_{i} + T_{i-1}}{2}) $$
04

Incorporate boundary conditions

With the boundary conditions, the finite difference approximations for the first node (i=1) and last node (i=n) become: 1. Left surface (i=1): \(\frac{k_{1} - k_{0}}{\Delta x} \frac{T_{2} - T_{1}}{\Delta x} = q_{o}\). 2. Right surface (i=n): \(\frac{k_{n} - k_{n-1}}{\Delta x} \frac{T_{n-1} - T_{n}}{\Delta x} = h (T_{n} - T_{\infty})\). By substituting the expressions for \(k_{i+\frac{1}{2}}\) and \(k_{i-\frac{1}{2}}\) into the equations above, we can obtain the finite difference equations for the first and last nodes incorporating the boundary conditions. These equations along with the finite difference approximation for the interior nodes can be solved to obtain the temperature distribution in the domain.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Variable Thermal Conductivity
Understanding variable thermal conductivity is crucial when analyzing heat transfer problems where the material's ability to conduct heat changes with temperature. In many materials, as they get hotter, their thermal conductivity increases. This relationship can often be described by the linear expression given in the exercise, which is \( k(T) = k_{o}(1 + \beta T) \), where \( k_{o} \) is the baseline thermal conductivity, \( \beta \) is a constant that provides how much the conductivity changes with temperature, and \( T \) represents the temperature.

This concept is significant because the change in thermal conductivity affects how heat flows through the material. As the conductivity varies linearly with temperature, it creates a non-uniform rate of heat transfer. In practice, engineers must consider this when designing systems that operate over a range of temperatures to ensure the safety and efficiency of the design. For instance, thermal barriers in engines or electronic equipment must account for variable conductivity to avoid overheating or material failure.

To solve problems involving variable thermal conductivity using the finite difference method, we approximate the temperature-dependent conductivity terms around the point of interest to discretize the heat equation. This allows us to turn a complex differential equation into a set of algebraic equations that can be solved numerically, providing a practical approach to deal with the changing conductivity in the heat conduction analysis.
Uniform Internal Heat Generation
Uniform internal heat generation refers to the scenario where heat is being produced at a constant rate throughout a material. This might occur due to chemical reactions, electrical resistance, or radioactive decay, for instance. It is an essential factor in thermal analyses because it directly impacts the temperature distribution within a material.

In the context of our problem, the internal heat generation is represented by the term \( q_{gen} \) and is assumed to be constant. This simplification is common in engineering to make the mathematical modelling of heat conduction manageable. It means that every unit volume of the material generates the same amount of heat irrespective of its position. When combined with variable thermal conductivity, this represents a more complex, yet realistic scenario where the material not only generates heat internally but its capacity to transfer heat varies with temperature.

The finite difference method, used to approximate the behavior of the system, needs to handle this internal heat generation by integrating it into the set of algebraic equations derived from the heat conduction equation. Not accounting for internal heat generation or assuming it's non-uniform when it is, in fact, uniform, could lead to inaccurate models and potential design failures.
Boundary Conditions
Boundary conditions in heat conduction problems are the constraints that define the behavior of a system at its boundaries. They are critical to solving heat conduction problems because they provide information about the thermal interaction between the system and its environment. There are different types of boundary conditions, like specified temperature, specified heat flux, or convective boundary conditions.

In our exercise, the left boundary at \( x=0 \) has a constant heat flux, denoted as \( q_{o} \), representing, for example, a constant rate of heat being added to or taken from the material surface. On the right boundary at \( x=L \), we have a convective environment, which suggests that heat is lost to the surroundings at a rate proportional to the difference between the material's surface temperature and the ambient temperature, using Newton's law of cooling characterized by the convective heat transfer coefficient \( h \).

When applying the finite difference method, the boundary conditions must be incorporated into the formulation explicitly to ensure that the numerical solution adheres to the physical constraints of the real-world scenario being modeled. Failing to properly implement these boundary conditions could result in a flawed temperature distribution, which might not reflect the true performance of the material under study.
Temperature Distribution
Temperature distribution within a material indicates how temperature varies across the material. In heat conduction problems, finding the temperature distribution is often the primary goal as it impacts material properties, structural integrity, and system performance. In the exercise, we aim to discover how the temperature changes from one side of the material to the other, accounting for variable thermal conductivity and uniform internal heat generation.

The finite difference method is a tool that allows us to approximate the temperature at discrete points within the domain. By breaking down the continuous material into a finite number of nodes, we convert the differential heat conduction equation into a set of linear algebraic equations. These equations can then be solved to find the temperature at each node, which in turn helps us understand the temperature distribution across the entire material.

To obtain accurate results, we employ the calculated expressions for thermal conductivity at each node and apply the boundary conditions at the edges. The resulting temperature profile helps engineers and scientists predict how a material will behave in various thermal scenarios and thus, design more effective heating or cooling strategies, leading to improved material performance systems.

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Most popular questions from this chapter

Suggest some practical ways of reducing the roundoff error.

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.

Write a two-page essay on the finite element method, and explain why it is used in most commercial engineering software packages. Also explain how it compares to the finite difference method.

A common annoyance in cars in winter months is the formation of fog on the glass surfaces that blocks the view. A practical way of solving this problem is to blow hot air or to attach electric resistance heaters to the inner surfaces. Consider the rear window of a car that consists of a \(0.4\)-cm-thick glass \(\left(k=0.84 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=0.39 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\). Strip heater wires of negligible thickness are attached to the inner surface of the glass, \(4 \mathrm{~cm}\) apart. Each wire generates heat at a rate of \(25 \mathrm{~W} / \mathrm{m}\) length. Initially the entire car, including its windows, is at the outdoor temperature of \(T_{o}=-3^{\circ} \mathrm{C}\). The heat transfer coefficients at the inner and outer surfaces of the glass can be taken to be \(h_{i}=6\) and \(h_{o}=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Using the explicit finite difference method with a mesh size of \(\Delta x=\) \(0.2 \mathrm{~cm}\) along the thickness and \(\Delta y=1 \mathrm{~cm}\) in the direction normal to the heater wires, determine the temperature distribution throughout the glass \(15 \mathrm{~min}\) after the strip heaters are turned on. Also, determine the temperature distribution when steady conditions are reached.

Consider steady heat conduction in a plane wall whose left surface (node 0 ) is maintained at \(30^{\circ} \mathrm{C}\) while the right surface (node 8 ) is subjected to a heat flux of \(1200 \mathrm{~W} / \mathrm{m}^{2}\). Express the finite difference formulation of the boundary nodes 0 and 8 for the case of no heat generation. Also obtain the finite difference formulation for the rate of heat transfer at the left boundary.

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