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Chapter 5: Problem 70

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.

Short Answer

Expert verified
Answer: The three steps involved are: 1. Apply the finite difference method and obtain the difference equations for the problem. 2. Solve the system of linear equations to determine the nodal temperatures. 3. Calculate the rate of heat transfer from the block to the iced water.

Step by step solution

01

Finite Difference Formulation

To obtain the finite difference formulation, we will use the 2D steady-state heat conduction equation without any heat generation: \(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0\) Considering the symmetry of the problem, we can use a quarter of the domain with a grid spacing of \(\Delta x = \Delta y = 0.1\:m\). We will use central difference scheme for both the space derivatives: \(\frac{T_{i+1,j} - 2T_{i,j} + T_{i-1, j}}{\Delta x^2} + \frac{T_{i, j+1} - 2T_{i,j} + T_{i, j-1}}{\Delta y^2} = 0\) Since \(\Delta x = \Delta y\), we can simplify the equation as: \(T_{i+1, j} + T_{i-1, j} + T_{i, j+1} + T_{i, j-1} - 4T_{i, j} = 0\) Now, we can create separate difference equations for the interior nodes and the boundary nodes.
02

Solving for Nodal Temperatures

Set up a system of linear equations using the difference equations obtained in step 1 for each interior and boundary node. Solve the system of linear equations to find the nodal temperature values.
03

Calculating the Heat Transfer Rate

To determine the rate of heat transfer from the block to the iced water, we can use the heat transfer coefficient, the temperature of the surface nodes (\(T_{i, j}\)), and the temperature of the iced water (\(T_{\infty} = 0^{\circ}C\)). Using Newton's law of cooling, we can calculate the heat transfer rate as: \(q_{conv} = h \cdot A \cdot (T_{i, j} - T_{\infty})\) Sum up the heat transfer rate for all the surface nodes in contact with iced water to determine the total rate of heat transfer from the block to the iced water.

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Most popular questions from this chapter

Consider a house whose windows are made of \(0.375\)-in-thick glass \(\left(k=0.48 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\alpha=\) \(4.2 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\) ). Initially, the entire house, including the walls and the windows, is at the outdoor temperature of \(T_{o}=35^{\circ} \mathrm{F}\). It is observed that the windows are fogged because the indoor temperature is below the dew-point temperature of \(54^{\circ} \mathrm{F}\). Now the heater is turned on and the air temperature in the house is raised to \(T_{i}=72^{\circ} \mathrm{F}\) at a rate of \(2^{\circ} \mathrm{F}\) rise per minute. The heat transfer coefficients at the inner and outer surfaces of the wall can be taken to be \(h_{i}=1.2\) and \(h_{o}=2.6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), respectively, and the outdoor temperature can be assumed to remain constant. Using the explicit finite difference method with a mesh size of \(\Delta x=0.125\) in, determine how long it will take for the fog on the windows to clear up (i.e., for the inner surface temperature of the window glass to reach \(54^{\circ} \mathrm{F}\) ).

The wall of a heat exchanger separates hot water at \(T_{A}=90^{\circ} \mathrm{C}\) from cold water at \(T_{B}=10^{\circ} \mathrm{C}\). To extend the heat transfer area, two-dimensional ridges are machined on the cold side of the wall, as shown in Fig. P5-76. This geometry causes non-uniform thermal stresses, which may become critical for crack initiation along the lines between two ridges. To predict thermal stresses, the temperature field inside the wall must be determined. Convection coefficients are high enough so that the surface temperature is equal to that of the water on each side of the wall. (a) Identify the smallest section of the wall that can be analyzed in order to find the temperature field in the whole wall. (b) For the domain found in part \((a)\), construct a twodimensional grid with \(\Delta x=\Delta y=5 \mathrm{~mm}\) and write the matrix equation \(A T=C\) (elements of matrices \(A\) and \(C\) must be numbers). Do not solve for \(T\). (c) A thermocouple mounted at point \(M\) reads \(46.9^{\circ} \mathrm{C}\). Determine the other unknown temperatures in the grid defined in part (b).

A 1-m-long and 0.1-m-thick steel plate of thermal conductivity \(35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is well insulated on its both sides, while the top surface is exposed to a uniform heat flux of \(5500 \mathrm{~W} / \mathrm{m}^{2}\). The bottom surface is convectively cooled by a fluid at \(10^{\circ} \mathrm{C}\) having a convective heat transfer coefficient of \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming one dimensional heat conduction in the lateral direction, find the temperature at the midpoint of the plate. Discretize the plate thickness into four equal parts.

Write a two-page essay on the finite element method, and explain why it is used in most commercial engineering software packages. Also explain how it compares to the finite difference method.

How can a node on an insulated boundary be treated as an interior node in the finite difference formulation of a plane wall? Explain.
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