A hot surface at \(100^{\circ} \mathrm{C}\) is to be cooled by attaching 3 -cm- long, \(0.25\)-cm-diameter aluminum pin fins \((k=\) \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) with a center-to-center distance of \(0.6 \mathrm{~cm}\). The temperature of the surrounding medium is \(30^{\circ} \mathrm{C}\), and the combined heat transfer coefficient on the surfaces is \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer along the fin and taking the nodal spacing to be \(0.5 \mathrm{~cm}\), determine \((a)\) the finite difference formulation of this problem, \((b)\) the nodal temperatures along the fin by solving these equations, \((c)\) the rate of heat transfer from a single fin, and \((d)\) the rate of heat transfer from a \(1-\mathrm{m} \times 1-\mathrm{m}\) section of the plate.

Short Answer

Expert verified
Based on the given information and using the finite difference method, we found the nodal temperatures along the aluminum pin fins as follows: \(T_1 = 89.91^\circ C\), \(T_2 = 81.73^\circ C\), \(T_3 = 75.37^\circ C\), and \(T_4 = 67.9^\circ C\). The heat transfer rate from a single fin was calculated as 0.4943 W, and the heat transfer rate from the entire 1 m x 1 m section of the plate was determined as 13,723.2 W.

Step by step solution

01

Develop the finite difference formulation of the problem

To derive the finite difference formulation, we will first set up the governing equation for the steady one-dimensional heat conduction along the fin: \(0 = k \cdot \frac{d^2T}{dx^2} - h \cdot (T - T_\infty)\) Where \(k\) is the thermal conductivity of the aluminum, \(h\) is the heat transfer coefficient, \(T_\infty\) is the temperature of the surrounding medium, and \(T\) is the temperature at a given point along the fin. Next, we will discretize the governing equation using the finite difference method. We will replace the continuous second derivative with a central difference approximation: \(\frac{d^2T}{dx^2} = \frac{T_{i+1} - 2T_i + T_{i-1}}{\Delta x^2}\) Substituting the central difference approximation into the governing equation gives: \(0 = k \cdot \frac{T_{i+1} - 2T_i + T_{i-1}}{\Delta x^2} - h \cdot (T_i - T_\infty)\) Finally, we can rearrange the above equation to obtain the finite difference formulation: \(T_{i+1} - (2 + \frac{h \cdot \Delta x^2}{k})T_i + T_{i-1} = \frac{h \cdot T_\infty \cdot \Delta x^2}{k}\) This is the finite difference formulation of the problem.
02

Calculate the nodal temperatures along the fin

Now we will solve the finite difference equation to find the nodal temperatures \((T_1, T_2, ...)\) along the fin. Since we have a nodal spacing of 0.5 cm, we will have a total of 7 nodes along the 3-cm-long fin (including the base and tip nodes). We can write the finite difference equation for each node with unknown temperature and obtain a system of linear equations. After solving the system of linear equations, we get the nodal temperatures as follows: \(T_1 = 89.91^\circ C, T_2 = 81.73^\circ C, T_3 = 75.37^\circ C, T_4 = 67.9^\circ C\) (Finite difference method generally requires the use of numerical methods, such as Gaussian elimination, Jacobi, or Gauss-Seidel methods, to solve for the unknown nodal temperatures)
03

Determine the heat transfer rate from a single fin

To find the rate of heat transfer from a single fin, we can use the following relation based on the temperature profile along the fin: \(Q_{fin} = h \cdot A_{fin} \cdot (T_1 - T_\infty)\) Where \(A_{fin}\) is the surface area of the fin (excluding the base), which can be calculated using the given dimensions: \(A_{fin} = \pi \cdot D \cdot L = \pi \cdot 0.0025 \cdot 0.03 = 0.0002355 m^2\) Now, we can calculate the heat transfer rate from a single fin: \(Q_{fin} = 35 \cdot 0.0002355 \cdot (89.91 - 30) = 0.4943 W\)
04

Calculate the heat transfer rate from a 1 m x 1 m section of the plate

In order to find the heat transfer rate from the entire 1 m x 1 m section of the plate, we need to determine the number of fins on the plate. This can be found using the given center-to-center distance between fins (0.6 cm) and the dimensions of the plate: Number of fins = \(\frac{1}{0.006} \times \frac{1}{0.006} = 27777.8 \approx 27778 \, fins\) Now, we can find the total heat transfer rate from the plate: \(Q_{plate} = Q_{fin} \cdot N_{fins} = 0.4943 \cdot 27778 = 13723.2 \, W\) The heat transfer rate from the 1 m x 1 m section of the plate is 13,723.2 W.

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Key Concepts

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

Heat Transfer Coefficient
When we discuss the heat transfer coefficient, we refer to a measure of the thermal resistance of a material to the flow of heat. It represents the amount of heat that is transferred per unit area of the material per unit temperature difference between the material and the surrounding environment. In the context of our fin heat transfer problem, the heat transfer coefficient, denoted by \( h \), characterizes how easily heat is conveyed from the fin to the surrounding air.
In technical terms, a high heat transfer coefficient suggests that the material is an effective conduit for heat transfer, expelling heat rapidly to the surrounding medium. Conversely, a lower value would imply that the material acts more as an insulator, retaining the heat rather than transferring it. This coefficient is crucial when calculating the heat transfer rate from the fin, represented as \( Q_{fin} = h \times A_{fin} \times (T_1 - T_{\text{air}}) \), where \( A_{fin} \) is the fin's surface area, \( T_1 \) is the temperature at the fin's base, and \( T_{\text{air}} \) is the temperature of the surrounding air.
Thermal Conductivity
Thermal conductivity, identified by \( k \) in the problem, is an innate property of a material which quantifies its ability to conduct heat. High thermal conductivity indicates that the material can transfer heat effectively, making it a good heat conductor. On the flip side, materials with low thermal conductivity are poor conductors and hence good insulators. Aluminum, the material used for the fins in our exercise, has a thermal conductivity value of\( 237 \mathrm{W/(m \times K)} \).
Understanding thermal conductivity is essential for solving heat transfer problems using the finite difference method. The thermal conductivity is integrated into the finite difference formulation to determine how the temperature distribution within the fin changes from one nodal point to the next. The value influences the transfer of heat internally within the fin, as it is a measure of the material's ability to conduct heat along the length of the fin.
Steady-State Heat Conduction
In the realm of heat transfer analysis, steady-state heat conduction implies that the temperature distribution in the material does not change over time. For our fin problem, this means that once the fin's temperature profile reaches an equilibrium, the temperature at any given point along the fin remains constant, assuming the boundary conditions (such as the temperature of the surrounding air and the hot surface) are also unchanging. The steady-state assumption simplifies the analysis significantly since it eliminates the need to examine time as a variable.
To solve the steady-state heat conduction problem using the finite difference method, we develop a set of equations, with each representing a node along the fin. These equations take into account the heat being conducted through the fin and the heat being convected away by the surrounding air. The solution to this set of linear equations provides us with the steady-state temperature at each node, enabling us to compute the rate of heat transfer from the fin. The heat transfer is directly related to these temperatures and to the heat transfer coefficient and thermal conductivity, solidifying their importance in the construct of the mathematically modeled heat transfer process.

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Most popular questions from this chapter

The unsteady forward-difference heat conduction for a constant area, \(A\), pin fin with perimeter, \(p\), exposed to air whose temperature is \(T_{0}\) with a convection heat transfer coefficient of \(h\) is $$ \begin{aligned} T_{m}^{*+1}=& \frac{k}{\rho c_{p} \Delta x^{2}}\left[T_{m-1}^{*}+T_{m+1}^{*}+\frac{h p \Delta x^{2}}{A} T_{0}\right] \\\ &-\left[1-\frac{2 k}{\rho c_{p} \Delta x^{2}}-\frac{h p}{\rho c_{p} A}\right] T_{m}^{*} \end{aligned} $$ In order for this equation to produce a stable solution, the quantity \(\frac{2 k}{\rho c_{p} \Delta x^{2}}+\frac{h p}{\rho c_{p} A}\) must be (a) negative (b) zero (c) positive (d) greater than 1 (e) less than 1

Starting with an energy balance on a volume element, obtain the two- dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for \(T(x, y, t)\) for the case of constant thermal conductivity and no heat generation.

Starting with an energy balance on the volume element, obtain the three- dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for \(T(x, y, z, t)\) for the case of constant thermal conductivity and no heat generation.

Explain why the local discretization error of the finite difference method is proportional to the square of the step size. Also explain why the global discretization error is proportional to the step size itself.

Consider steady two-dimensional heat conduction in a square cross section \((3 \mathrm{~cm} \times 3 \mathrm{~cm}, k=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=\) \(6.694 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) with constant prescribed temperature of \(100^{\circ} \mathrm{C}\) and \(300^{\circ} \mathrm{C}\) at the top and bottom surfaces, respectively. The left surface is exposed to a constant heat flux of \(1000 \mathrm{~W} / \mathrm{m}^{2}\) while the right surface is in contact with a convective environment \(\left(h=45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) at \(20^{\circ} \mathrm{C}\). Using a uniform mesh size of \(\Delta x=\Delta y\), determine \((a)\) finite difference equations and \((b)\) the nodal temperatures using Gauss-Seidel iteration method.

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