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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 energy balance equations for nodes 1 and 2 are: For node 1: \(\frac{kAD(T_2 - T_0)}{2\Delta x} + hP(T_1 - T_\infty) + \epsilon\sigma P(T_1^4 - T_{surr}^4) = 0\) For node 2: \(\frac{kAD(T_2 - T_1)}{2\Delta x} + hP(T_2 - T_\infty) + \epsilon\sigma P(T_2^4 - T_{surr}^4) = 0\)

Step by step solution

01

Setup energy balance formulation

Since we are using the energy balance approach, we need to set up the equations for the balance of heat conduction, convection, and radiation at each node. For steady-state conduction, the heat conducted into a node is equal to heat conducted out plus heat lost by convection and radiation. Starting with the general energy balance equation for a one-dimensional pin fin, we have: \(Q_{cond} + Q_{conv} + Q_{rad} = 0\) Here, \(Q_{cond}\) is the heat conduction, \(Q_{conv}\) is the heat convection, and \(Q_{rad}\) is the heat radiation.
02

Calculate the heat conduction term

The heat conduction term \(Q_{cond}\) can be calculated using Fourier's Law, given as: \(Q_{cond} = kA\frac{dT}{dx}\) where \(k\) is the thermal conductivity, \(A\) is the cross-sectional area of the fin, and \(\frac{dT}{dx}\) is the temperature gradient along the fin. Since we are using finite differences, we will approximate the temperature gradient using the central difference scheme: \(\frac{dT}{dx} \approx \frac{T_{i+1} - T_{i-1}}{2\Delta x}\) This approximation will be used to calculate the heat conduction term for each node.
03

Calculate the heat convection term

The heat convection term \(Q_{conv}\) can be calculated using Newton's Law of Cooling, given as: \(Q_{conv} = hP(T - T_\infty)\) where \(P\) is the perimeter of the fin and \(T_\infty\) is the ambient air temperature. Since we are only concerned about nodes 1 and 2, this term can be applied directly to each of the nodes.
04

Calculate the heat radiation term

The heat radiation term \(Q_{rad}\) can be calculated using Stefan-Boltzmann Law, given as: \(Q_{rad} = \epsilon\sigma P(T^4 - T_{surr}^4)\) where \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, and \(T_{surr}\) is the surrounding surfaces temperature. This term can also be applied directly to nodes 1 and 2.
05

Apply boundary conditions

Now that we have expressions for \(Q_{cond}\), \(Q_{conv}\), and \(Q_{rad}\), we need to apply the boundary conditions of specified temperature at the fin base (node 0) and negligible heat transfer at the fin tip (node 2). For node 0 (fin base): The temperature is specified, so we don't need an energy balance equation for this node. For node 1 (middle): The general energy balance equation holds true. We will need to combine the expressions for \(Q_{cond}\), \(Q_{conv}\), and \(Q_{rad}\) and solve for \(T_1\). For node 2 (fin tip): The heat transfer is negligible, so we will set \(Q_{cond}\), \(Q_{conv}\), and \(Q_{rad}\) to zero and solve for \(T_2\).
06

Solve for temperatures \(T_1\) and \(T_2\)

Now we have to solve the energy balance equations for both nodes 1 and 2. For node 1: \(\frac{kAD(T_2 - T_0)}{2\Delta x} + hP(T_1 - T_\infty) + \epsilon\sigma P(T_1^4 - T_{surr}^4) = 0\) For node 2: \(\frac{kAD(T_2 - T_1)}{2\Delta x} + hP(T_2 - T_\infty) + \epsilon\sigma P(T_2^4 - T_{surr}^4) = 0\) These are two non-linear equations with two unknowns \(T_1\) and \(T_2\). Depending on the specific problem, there might be different methods to solve the system of equations, such as numerical methods or using a computer program. In conclusion, we have derived the finite difference formulation for the given heat conduction problem in a pin fin system. By applying the boundary conditions and solving the energy balance equations for nodes 1 and 2, we can determine the temperatures \(T_1\) and \(T_2\).

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Most popular questions from this chapter

A stainless steel plate is connected to an insulation plate by square ASTM A479 904L stainless steel bars. Each square bar has a thickness of $1 \mathrm{~cm}\( and a length of \)5 \mathrm{~cm}$. The bars are exposed to convection heat transfer with a hot gas. The temperature of the hot gas is \(300^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of $25 \mathrm{~W} / \mathrm{m}^{2}$. . The thermal conductivity for ASTM A479 904L stainless steel is known to be \(12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The stainless steel plate maintains a uniform temperature of $100^{\circ} \mathrm{C}$. According to the ASME Code for Process Piping (ASME B31.3-2014, Table A-1M), the maximum use temperature for ASTM A479 904L stainless steel bar is \(260^{\circ} \mathrm{C}\). Using the finite difference method with a uniform nodal spacing of \(\Delta x=5 \mathrm{~mm}\) along the bar, determine the temperature at each node. Compare the numerical results with the analytical solution. Plot the temperature distribution along the bar. Would any part of the ASTM A479 904L bars be above the maximum use temperature of \(260^{\circ} \mathrm{C}\) ?

Two 3-m-long and \(0.4\)-cm-thick cast iron \((k=52 \mathrm{~W} /\) $\mathrm{m} \cdot \mathrm{K}, \varepsilon=0.8)\( steam pipes of outer diameter \)10 \mathrm{~cm}$ are connected to each other through two 1 -cm-thick flanges of outer diameter \(20 \mathrm{~cm}\). The steam flows inside the pipe at an average temperature of \(250^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The outer surface of the pipe is exposed to convection with ambient air at $12^{\circ} \mathrm{C}\( with a heat transfer coefficient of \)25 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}$ as well as radiation with the surrounding surfaces at an average temperature of \(T_{\text {surr }}=290 \mathrm{~K}\). Assuming steady onedimensional heat conduction along the flanges and taking the nodal spacing to be \(1 \mathrm{~cm}\) along the flange, \((a)\) obtain the finite difference formulation for all nodes, \((b)\) determine the temperature at the tip of the flange by solving those equations, and \((c)\) determine the rate of heat transfer from the exposed surfaces of the flange.

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 \(\alpha=0.39 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ). 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 \(h_{i}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) 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 min after the strip heaters are turned on. Also, determine the temperature distribution when steady conditions are reached.

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.

What is the cause of the discretization error? How does the global discretization error differ from the local discretization error?

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