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 \({ }^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Answer: The temperatures T1 and T2 of the fin can be determined using the finite difference method by following these steps: 1. Formulate the energy balance equation for each node on the fin (node 1 and node 2), considering conduction, convection, and radiation. 2. Write the heat fluxes for conduction, convection, and radiation in terms of the nodal temperatures (T0, T1, T2) and other known parameters. 3. Replace the continuous values with discrete approximations to obtain the finite difference equations for node 1 and node 2. 4. Solve the system of equations using a suitable numerical method (e.g., Newton-Raphson or relaxation method) to find the values of T1 and T2.

Step by step solution

01

Formulate the energy balance equation

First, we need to balance the incoming and outgoing heat energy in the fin. This includes conduction inside the fin, convection to the ambient air, and radiation to the surrounding surfaces. The energy balance equation for the first node (node 1) is: \(q_{cond,01} - q_{cond,12} - q_{conv,1} - q_{rad,1} = 0\) And for the second node (node 2) it's: \(q_{cond,12} - q_{conv,2} - q_{rad,2} = 0\) We need to express these heat fluxes by substituting the respective formulae and temperatures of the corresponding nodes.
02

Write heat fluxes

For conduction heat flux between nodes 0 and 1, and nodes 1 and 2, we can write: \(q_{cond,01} = k \frac{T_{0} - T_{1}}{\Delta x}\) \(q_{cond,12} = k \frac{T_{1} - T_{2}}{\Delta x}\) For convection heat flux in nodes 1 and 2: \(q_{conv,1} = h (T_{1} - T_{\infty})\) \(q_{conv,2} = h (T_{2} - T_{\infty})\) For radiation heat flux in nodes 1 and 2: \(q_{rad,1} = \sigma \cdot A \cdot (T_{1}^4 - T_{surr}^4)\) \(q_{rad,2} = \sigma \cdot A \cdot (T_{2}^4 - T_{surr}^4)\) Here, \(A\) is the fin's surface area and \(\sigma\) is the Stefan-Boltzmann constant.
03

Apply the finite difference method

Next, we will be replacing all the continuous values with discrete approximations, to obtain the finite difference equations: Node 1: \(k \frac{T_{0} - T_{1}\left(1\right)}{\Delta x} - k \frac{T_{1}\left(1\right) - T_{2}\left(1\right)}{\Delta x} - h (T_{1}\left(1\right) - T_{\infty}) - \sigma \cdot A \cdot (T_{1}\left(1\right)^4 - T_{surr}^4) = 0\) Node 2: \(k \frac{T_{1}\left(1\right) - T_{2}\left(1\right)}{\Delta x} - h (T_{2}\left(1\right) - T_{\infty}) - \sigma \cdot A \cdot (T_{2}\left(1\right)^4 - T_{surr}^4) = 0\)
04

Solve the system of equations for \(T_{1}\) and \(T_{2}\)

To solve the system of equations above, we can use a suitable numerical method (e.g., Newton-Raphson or relaxation method). This will give us the values of \(T_{1}\) and \(T_{2}\) at the middle and the fin tip, respectively. This concludes our analysis. The temperature profiles of the fin at the two nodes can be determined using the finite difference method which considers heat conduction, convection to the ambient air, and radiation to the surrounding surfaces.

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Most popular questions from this chapter

Consider a long solid bar \((k=28 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=\) \(12 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) of square cross section that is initially at a uniform temperature of \(32^{\circ} \mathrm{C}\). The cross section of the bar is \(20 \mathrm{~cm} \times 20 \mathrm{~cm}\) in size, and heat is generated in it uniformly at a rate of \(\dot{e}=8 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). All four sides of the bar are subjected to convection to the ambient air at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the explicit finite difference method with a mesh size of \(\Delta x=\Delta y=10 \mathrm{~cm}\), determine the centerline temperature of the bar \((a)\) after \(20 \mathrm{~min}\) and \((b)\) after steady conditions are established.

A hot brass plate is having its upper surface cooled by impinging jet of air at temperature of \(15^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(220 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The \(10-\mathrm{cm}-\) thick brass plate \(\left(\rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=\right.\) \(110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=33.9 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) had a uniform initial temperature of \(650^{\circ} \mathrm{C}\), and the lower surface of the plate is insulated. Using a uniform nodal spacing of \(\Delta x=\) \(2.5 \mathrm{~cm}\) determine \((a)\) the explicit finite difference equations, (b) the maximum allowable value of the time step, \((c)\) the temperature at the center plane of the brass plate after 1 minute of cooling, and \((d)\) compare the result in \((c)\) with the approximate analytical solution from Chapter 4 .

Consider one-dimensional transient heat conduction in a composite plane wall that consists of two layers \(A\) and \(B\) with perfect contact at the interface. The wall involves no heat generation and initially is at a specified temperature. The nodal network of the medium consists of nodes 0,1 (at the interface), and 2 with a uniform nodal spacing of \(\Delta x\). Using the energy balance approach, obtain the explicit finite difference formulation of this problem for the case of insulation at the left boundary (node 0 ) and radiation at the right boundary (node 2 ) with an emissivity of \(\varepsilon\) and surrounding temperature of \(T_{\text {surr }}\).

Can the global (accumulated) discretization error be less than the local error during a step? Explain.

Consider a long concrete dam \((k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), (es \(\alpha_{s}=0.7\) ) of triangular cross section whose exposed surface is subjected to solar heat flux of \(\dot{q}_{s}=\) \(800 \mathrm{~W} / \mathrm{m}^{2}\) and to convection and radiation to the environment at \(25^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The \(2-\mathrm{m}\)-high vertical section of the dam is subjected to convection by water at \(15^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer through the 2-m-long base is considered to be negligible. Using the finite difference method with a mesh size of \(\Delta x=\Delta y=1 \mathrm{~m}\) and assuming steady two-dimensional heat transfer, determine the temperature of the top, middle, and bottom of the exposed surface of the dam.

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