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.

Short Answer

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Short answer question: After applying the explicit finite difference method to the given heat conduction equation, calculate the centerline temperature of the long solid bar with a square cross-section at (a) 20 minutes and (b) steady-state conditions.

Step by step solution

01

Understand the Explicit Finite Difference Method

The explicit finite difference method (EFDM) is a numerical method used to approximate the solution of partial differential equations (PDEs) such as the heat conduction equation. In this method, we will discretize the time and space into intervals and solve the temperature for each node using the neighboring nodes' values. We will be using the heat conduction equation for this purpose.
02

Discretize the heat conduction equation using the explicit finite difference method

The heat conduction equation for a 2-D Cartesian coordinate system is given by: \(\frac{\partial T}{\partial t} = \alpha(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2}) + \frac{\dot{e}}{k}\) The explicit finite difference equation for this PDE is: \( T_{i,j}^{n+1} = T_{i,j}^n + \Delta t \left[ \alpha(\frac{T_{i-1,j}^n - 2 T_{i,j}^n + T_{i+1,j}^n}{\Delta x^2} + \frac{T_{i,j-1}^n - 2 T_{i,j}^n + T_{i,j+1}^n}{\Delta y^2}) + \frac{\dot{e}}{k} \right] \) We will use this equation to find the centerline temperature of the bar.
03

Set up the initial conditions and mesh

Given that the bar has an initial uniform temperature of 32°C, we can assign this as the initial temperature at each node in our mesh. The given mesh size is \(\Delta x = \Delta y = 10 \mathrm{~cm} = 0.1 \mathrm{~m}\).
04

Calculate the temperature after 20 minutes

For part (a), we need to find the temperature after 20 minutes (i.e., 1200 seconds). We first need to calculate Δt from the mesh size and given thermal diffusivity α: \( \Delta t = \frac{\Delta x^{2}}{4\alpha} \) \( \Delta t = \frac{(0.1)^2}{4 \times 12 \times 10^{-6}} \) \( \Delta t = 2.08 \) seconds Now, we will use the EFDM equation from Step 2 to find the temperature at each node after 20 minutes (1200 seconds). We will update the temperature values at each iteration until 1200 seconds.
05

Compute steady-state temperature

For part (b), we need to find the steady-state temperature. For steady-state conditions, we require the change in temperature at each node to be minimal: \(\frac{\partial T}{\partial t} = 0\) Using the equation from Step 2 for these conditions, we calculate the centerline temperature after the steady-state is achieved.
06

Calculate the centerline temperature

At the end of Steps 4 and 5, we get temperature values for each node in the mesh. To calculate the centerline temperature of the bar, we take the average of the temperatures at the intersection of the two centerlines after 20 minutes and steady-state conditions. Following the steps mentioned above and using the given parameters, you can find the centerline temperature of the bar using the explicit finite difference method for both cases: (a) after 20 minutes and (b) after steady conditions are established.

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Most popular questions from this chapter

Consider steady one-dimensional heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes \(0,1,2,3\), and 4 with a uniform nodal spacing of \(\Delta x\). Using the finite difference form of the first derivative (not the energy balance approach), obtain the finite difference formulation of the boundary nodes for the case of uniform heat flux \(q_{0}\) at the left boundary (node 0 ) and convection at the right boundary (node 4) with a convection coefficient of \(h\) and an ambient temperature of \(T_{\infty}\).

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

Consider steady two-dimensional heat transfer in a long solid bar \((k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of square cross section \((3 \mathrm{~cm} \times 3 \mathrm{~cm})\) with the prescribed temperatures at the top, right, bottom, and left surfaces to be \(100^{\circ} \mathrm{C}, 200^{\circ} \mathrm{C}, 300^{\circ} \mathrm{C}\), and \(500^{\circ} \mathrm{C}\), respectively. Heat is generated in the bar uniformly at a rate of \(\dot{e}=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Using a uniform mesh size \(\Delta x=\Delta y=1 \mathrm{~cm}\) determine \((a)\) the finite difference equations and \((b)\) the nodal temperatures with the Gauss-Seidel iterative method.

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 a 5 -m-long constantan block \((k=23 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) \(30 \mathrm{~cm}\) high and \(50 \mathrm{~cm}\) wide (Fig. P5-70). The block is completely submerged in iced water at \(0^{\circ} \mathrm{C}\) that is well stirred, and the heat transfer coefficient is so high that the temperatures on both sides of the block can be taken to be \(0^{\circ} \mathrm{C}\). The bottom surface of the bar is covered with a low-conductivity material so that heat transfer through the bottom surface is negligible. The top surface of the block is heated uniformly by a 8-kW resistance heater. Using the finite difference method with a mesh size of \(\Delta x=\Delta y=10 \mathrm{~cm}\) and taking advantage of symmetry, (a) obtain the finite difference formulation of this problem for steady twodimensional heat transfer, \((b)\) determine the unknown nodal temperatures by solving those equations, and \((c)\) determine the rate of heat transfer from the block to the iced water.

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