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Chapter 5: Problem 10

In many engineering applications variation in thermal properties is significant especially when there are large temperature gradients or the material is not homogeneous. To account for these variations in thermal properties, develop a finite difference formulation for an internal node in case of a three dimensional steady state heat conduction equation with variable thermal conductivity.

Short Answer

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Question: Develop a finite difference formulation for an internal node in a three-dimensional steady-state heat conduction equation with variable thermal conductivity. Answer: The finite difference formulation for an internal node (i,j,k) in a three-dimensional steady-state heat conduction problem with variable thermal conductivity can be expressed as: 0 ≈ k(i+1/2,j,k) * (∇^2T)_x + k(i,j+1/2,k) * (∇^2T)_y + k(i,j,k+1/2) * (∇^2T)_z where k(i+1/2,j,k) = (k(i+1,j,k) + k(i,j,k)) / 2, and the second derivatives of temperature (∇^2T)_x, (∇^2T)_y, and (∇^2T)_z are obtained using the central difference method. This equation, along with similar equations for all internal nodes, forms a system of linear equations that can be solved using numerical methods such as Gauss-Seidel or Conjugate Gradient.

Step by step solution

01

Discretize the spatial domain

Divide the three-dimensional domain into small elements or finite differences, using uniform grid-spacing in all three directions: x, y, and z. Denote the increments with Δx, Δy, and Δz, respectively. The Cartesian coordinates of an internal node will be (i*Δx, j*Δy, k*Δz), where i, j, and k are integer indices for the nodes.
02

Approximate the derivatives with central differences

First, calculate the partial derivatives of the variable thermal conductivity with respect to x, y, and z. Then, approximate them with central differences. For example, the partial derivative with respect to x is: (∇k)_x = (k(i+1,j,k) - k(i-1,j,k)) / (2*Δx) Do the same for y and z directions. Next, approximate the second derivatives of the temperature with respect to x, y, and z using the central difference method. For example, the second derivative with respect to x is: (∇^2T)_x = (T(i+1,j,k) - 2*T(i,j,k) + T(i-1,j,k)) / (Δx^2) Repeat this process for the y and z directions as well.
03

Combine the derivatives into the finite difference equation

Substitute the approximated derivatives (Step 2) into the 3D steady-state heat conduction equation: (∇(k(x,y,z)∇T(x,y,z))) ≈ ( (∇k)_x * (∇T)_x + (∇k)_y * (∇T)_y + (∇k)_z * (∇T)_z
04

Incorporate the variable thermal conductivity

Multiply the thermal conductivity with the second derivatives of the temperature in each direction, and sum these terms: 0 ≈ k(i+1/2,j,k) * (∇^2T)_x + k(i,j+1/2,k) * (∇^2T)_y + k(i,j,k+1/2) * (∇^2T)_z Here, we use interpolated values of the thermal conductivity at the cell faces, e.g., k(i+1/2,j,k) = (k(i+1,j,k) + k(i,j,k)) / 2.
05

Simplify and solve for the internal temperature node

Rearrange the equation from Step 4 to express T(i, j, k) - the internal node temperature - as a function of its neighboring nodes' temperatures and thermal conductivities. This equation will become a part of the system of linear equations that can be solved with methods such as Gauss-Seidel or Conjugate Gradient.

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Most popular questions from this chapter

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.

Consider steady heat conduction in a plane wall whose left surface (node 0 ) is maintained at \(30^{\circ} \mathrm{C}\) while the right surface (node 8 ) is subjected to a heat flux of \(1200 \mathrm{~W} / \mathrm{m}^{2}\). Express the finite difference formulation of the boundary nodes 0 and 8 for the case of no heat generation. Also obtain the finite difference formulation for the rate of heat transfer at the left boundary.

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 \(T_{\infty}\).

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 \(0,1,2\), and 3 with a uniform nodal spacing of \(\Delta x\). The temperature at the left boundary (node 0 ) is specified. Using the energy balance approach, obtain the finite difference formulation of boundary node 3 at the right boundary for the case of combined convection and radiation with an emissivity of \(\varepsilon\), convection coefficient of \(h\), ambient temperature of \(T_{\circ}\), and surrounding temperature of \(T_{\text {surr }}\). Also, obtain the finite difference formulation for the rate of heat transfer at the left boundary.

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 \(54^{\circ} \mathrm{F}\) ).
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