A cylindrical aluminum fin with adiabatic tip is attached to a wall with surface temperature of \(300^{\circ} \mathrm{C}\), and is exposed to ambient air condition of \(15^{\circ} \mathrm{C}\) with convection heat transfer coefficient of \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The fin has a uniform cross section with diameter of \(1 \mathrm{~cm}\), length of \(5 \mathrm{~cm}\), and thermal conductivity of \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assume steady one-dimensional heat transfer along the fin and the nodal spacing to be uniformly \(10 \mathrm{~mm},(a)\) obtain the finite difference equations for use with the Gauss-Seidel iterative method, and \((b)\) determine the nodal temperatures using the GaussSeidel iterative method and compare the results with the analytical solution.

Short Answer

Expert verified
Answer: The key steps in the solution are: 1) Calculate relevant fin parameters such as area and perimeter. 2) Develop the finite difference equation for the system considering steady-state heat transfer. 3) Implement the Gauss-Seidel iterative method for solving nodal temperatures. 4) Compare the obtained solution with the analytical solution by calculating the percentage error for each node, ensuring the error is small to confirm accuracy.

Step by step solution

01

Calculate relevant parameters

First, we will calculate the area and perimeter of the fin to be used for convection heat transfer. We are given the diameter \(d = 1 \mathrm{~cm}\), which is equal to \(0.01 \mathrm{~m}\). Area, \(A = \pi (\frac{d}{2})^2 = \pi (\frac{0.01}{2})^2\) Perimeter, \(P = \pi \times d = \pi \times 0.01\)
02

Develop the finite difference equation for the given system

We will now develop the finite difference equation for the given system assuming steady-state heat transfer along the fin: \(\frac{d^2T}{dx^2} - m^2(T-T_\infty) = 0\) Where \(m=\sqrt{\frac{Ph}{Ak}}\), \(P\) is the perimeter of the fin, \(h\) is the convection heat transfer coefficient, \(A\) is the cross-sectional area of the fin, \(k\) is the thermal conductivity of the fin, and \(T_\infty\) is the ambient temperature. Given nodal spacing = \(\Delta x = 10mm = 0.01m.\) The finite difference equation can be written as: \(T_i^{(k+1)} = \frac{1}{m^2\Delta x^2+2}\left[m^2\Delta x^2T_\infty+(T_{i-1}^{(k)}+T_{i+1}^{(k)})\right]\)(for i = 1, 2, 3, ...) With boundary conditions imposed as: - At \(x=0,\) \(T_0^{(k+1)} = T_s,\) where \(T_s=300^{\circ} \mathrm{C}\). - At \(x=L,\) \(T_L^{(k+1)}\) is adiabatic (\(q_{i+1}=0\)), so we replace \(T_{i+1}^{(k)}\) with \(T_{i-1}^{(k)}.\) Now, we can apply the Gauss-Seidel iterative method to solve for nodal temperatures.
03

Gauss-Seidel iterative method

To implement the Gauss-Seidel iterative method: 1. Set an initial guess for all temperatures. 2. Substituting the initial guess values or previous iteration values into the finite difference equation, update the temperatures. 3. Set a convergence criterion, e.g, an absolute difference between temperatures at successive iterations (\(|T_i^{(k+1)}-T_i^{(k)}|\)) less than a threshold value. 4. Repeat steps 2 and 3 until convergence is achieved.
04

Comparing the obtained solution with the analytical solution

The analytical solution for the temperature distribution of the fin can be given by: \(T(x) = T_s+\left(\frac{\cosh(m(L-x))}{\cosh(mL)}\right)(T_\infty-T_s)\) Compare the nodal temperature values obtained from the Gauss-Seidel method with those calculated using the analytical solution. This can be done by calculating the percentage error for each node, which should be small if the solution is accurate.

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Key Concepts

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

Understanding Finite Difference Equations
Finite difference equations are central to solving various types of differential equations that arise in engineering and physics, especially when dealing with heat transfer problems. In essence, these equations transform differential equations into a set of algebraic equations that approximate the original continuous problem into a discrete problem.

In the context of heat transfer, using finite difference equations enables us to predict how temperature changes over a solid object like a fin. By discretizing the fin into nodes and applying thermal balance equations to each node, we obtain our finite difference equations. Each equation relates the temperature at a point to its neighboring points, creating a system that can be solved iteratively.

Developing Finite Difference Equations

To develop finite difference equations, we make use of Taylor's expansion to approximate the temperature at points near our point of interest. For example, the temperature at a node, often labeled as \(T_i\), will be determined by its neighboring nodes' temperatures such as \(T_{i-1}\) and \(T_{i+1}\). The relationship between these temperatures will include terms that account for thermal properties of the fin, such as its thermal conductivity and the heat transfer coefficient.
Convection Heat Transfer Coefficient Explained
The convection heat transfer coefficient, symbolized as \(h\), is a parameter that quantifies the convective heat transfer between a solid surface and a fluid in contact with it. It is a measure of how effectively heat is transferred from the solid surface to the fluid or vice versa.

In layman's terms, imagine you're holding a warm cup of coffee. The heat that you feel on your hands is being transferred from the cup to your hands through a process called convection, and the effectiveness of this heat transfer depends on a variety of factors including the fluid's velocity, its properties, and the temperature difference.

Significance in Heat Transfer Problems

The value of \(h\) is critical when calculating the rate of heat loss from an object. A higher heat transfer coefficient means that the object will lose heat more quickly to its surroundings. This is particularly important when designing systems that requirer thermal regulation, such as the fins mentioned in the exercise, since the goal is often to maximize or control the rate at which heat is dissipated.
Adiabatic Tip in Heat Transfer
An adiabatic tip in heat transfer refers to the end condition of an object like a fin where the tip is perfectly insulated. This means that no heat is transferred at that point to the surrounding environment.

To visualize this, think of an adiabatic tip as the end of a rod that is wrapped in an insulation layer so thick that none of the heat from the rod can escape through that end. In practical terms, it's a theoretical construct because perfect insulation is impossible, but it serves as an ideal scenario to simplify calculations.

Implications for Temperature Distribution

In the fin example, an adiabatic tip means that the heat transfer through the tip is zero, which affects the temperature distribution along the fin. Mathematically, this translates to a boundary condition that impacts the finite difference equations. For example, the absence of heat transfer at the tip means that the temperature gradient at the last node is zero, which affects the temperature calculations of the nodes close to it.

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Most popular questions from this chapter

Consider a large plane wall of thickness \(L=0.3 \mathrm{~m}\), thermal conductivity \(k=2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=24 \mathrm{~m}^{2}\). The left side of the wall is subjected to a heat flux of \(\dot{q}_{0}=350 \mathrm{~W} / \mathrm{m}^{2}\) while the temperature at that surface is measured to be \(T_{0}=60^{\circ} \mathrm{C}\). Assuming steady one-dimensional heat transfer and taking the nodal spacing to be \(6 \mathrm{~cm},(a)\) obtain the finite difference formulation for the six nodes and (b) determine the temperature of the other surface of the wall by solving those equations.

Express the general stability criterion for the explicit method of solution of transient heat conduction problems.

Consider steady one-dimensional heat conduction in a pin fin of constant diameter \(D\) with constant thermal conductivity. The fin is losing heat by convection to the ambient air at \(T_{\infty}\) with a convection coefficient of \(h\), and by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}\). The nodal network of the fin consists of nodes 0 (at the base), 1 (in the middle), and 2 (at the fin tip) with a uniform nodal spacing of \(\Delta x\). Using the energy balance approach, obtain the finite difference formulation of this problem to determine \(T_{1}\) and \(T_{2}\) for the case of specified temperature at the fin base and negligible heat transfer at the fin tip. All temperatures are in \({ }^{\circ} \mathrm{C}\).

Consider a medium in which the finite difference formulation of a general interior node is given in its simplest form as $$ T_{\text {node }}=\left(T_{\text {left }}+T_{\text {top }}+T_{\text {right }}+T_{\text {bottom }}\right) / 4 $$ (a) Is heat transfer in this medium steady or transient? (b) Is heat transfer one-, two-, or three-dimensional? (c) Is there heat generation in the medium? (d) Is the nodal spacing constant or variable? (e) Is the thermal conductivity of the medium constant or variable?

Starting with an energy balance on a disk volume element, derive the one- dimensional transient implicit finite difference equation for a general interior node for \(T(z, t)\) in a cylinder whose side surface is subjected to convection with a convection coefficient of \(h\) and an ambient temperature of \(T_{\infty}\) for the case of constant thermal conductivity with uniform heat generation.

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