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Starting with an energy balance on the volume element, obtain the three- dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for \(T(x, y, z, t)\) for the case of constant thermal conductivity and no heat generation.

Short Answer

Expert verified
The explicit finite difference equation for a general interior node in rectangular coordinates for temperature \(T(x, y, z, t)\) for the case of constant thermal conductivity and no heat generation can be written as: \(T(i, j, k, n+1) = T(i, j, k, n) + \Delta t \alpha(\frac{T(i+1, j, k, n) - 2T(i, j, k, n) + T(i-1, j, k, n)}{(\Delta x)^2} + \frac{T(i, j+1, k, n) - 2T(i, j, k, n) + T(i, j-1, k, n)}{(\Delta y)^2} + \frac{T(i, j, k+1, n) - 2T(i, j, k, n) + T(i, j, k-1, n)}{(\Delta z)^2})\) Here, \(\Delta t\) is the time step, \(\alpha\) is the thermal diffusivity, and \(i\), \(j\), and \(k\) are indices of grid points in the x, y, and z directions, respectively. The equation describes how the temperature at an interior node \((i, j, k)\) changes from time step \(n\) to \(n+1\).

Step by step solution

01

Write down the transient heat conduction equation

The transient heat conduction equation for constant thermal conductivity (\(k\)) and no heat generation is given as: \(\frac{\partial T}{\partial t} = \alpha(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2})\) where \(\alpha = \frac{k}{\rho c_p}\) is the thermal diffusivity, \(\rho\) is the density, and \(c_p\) is the specific heat capacity at constant pressure.
02

Discretize the domain using grid points

We discretize the domain into a grid with uniform spacing \(\Delta x\), \(\Delta y\), and \(\Delta z\) along x, y, and z directions respectively. The grid points can be represented as \((i, j, k)\), corresponding to the spatial coordinates \((x_i, y_j, z_k)\) where \(i=1, 2, ..., N_x\), \(j=1, 2, ..., N_y\), and \(k=1, 2, ..., N_z\). The time variable is also discretized into time steps \(\Delta t\).
03

Apply central finite differences to the spatial second derivatives

We apply central finite differences to approximate the second derivatives of the temperature with respect to x, y, and z. For example, the second derivative with respect to x can be approximated as follows: \(\frac{\partial^2 T}{\partial x^2} \approx \frac{T(i+1, j, k, n) - 2T(i, j, k, n) + T(i-1, j, k, n)}{(\Delta x)^2}\) Similarly, the second derivatives with respect to y and z can be approximated as: \(\frac{\partial^2 T}{\partial y^2} \approx \frac{T(i, j+1, k, n) - 2T(i, j, k, n) + T(i, j-1, k, n)}{(\Delta y)^2}\) \(\frac{\partial^2 T}{\partial z^2} \approx \frac{T(i, j, k+1, n) - 2T(i, j, k, n) + T(i, j, k-1, n)}{(\Delta z)^2}\)
04

Apply forward finite difference to the time derivative

We apply forward finite difference to approximate the time derivative of the temperature: \(\frac{\partial T}{\partial t} \approx \frac{T(i, j, k, n+1) - T(i, j, k, n)}{\Delta t}\)
05

Substitute the finite difference approximations into the transient heat conduction equation

Substitute the finite difference approximations obtained in Steps 3 and 4 into the transient heat conduction equation: \(\frac{T(i, j, k, n+1) - T(i, j, k, n)}{\Delta t} = \alpha(\frac{T(i+1, j, k, n) - 2T(i, j, k, n) + T(i-1, j, k, n)}{(\Delta x)^2} + \frac{T(i, j+1, k, n) - 2T(i, j, k, n) + T(i, j-1, k, n)}{(\Delta y)^2} + \frac{T(i, j, k+1, n) - 2T(i, j, k, n) + T(i, j, k-1, n)}{(\Delta z)^2})\)
06

Rearrange the equation to find \(T(i, j, k, n+1)\)

We can rearrange the above equation to obtain the temperature at the next time step for the interior node \((i, j, k)\): \(T(i, j, k, n+1) = T(i, j, k, n) + \Delta t \alpha(\frac{T(i+1, j, k, n) - 2T(i, j, k, n) + T(i-1, j, k, n)}{(\Delta x)^2} + \frac{T(i, j+1, k, n) - 2T(i, j, k, n) + T(i, j-1, k, n)}{(\Delta y)^2} + \frac{T(i, j, k+1, n) - 2T(i, j, k, n) + T(i, j, k-1, n)}{(\Delta z)^2})\) This is the three-dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for \(T(x, y, z, t)\) for the case of constant thermal conductivity and no heat generation.

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Most popular questions from this chapter

Consider a refrigerator whose outer dimensions are $1.80 \mathrm{~m} \times 0.8 \mathrm{~m} \times 0.7 \mathrm{~m}$. The walls of the refrigerator are constructed of \(3-\mathrm{cm}\)-thick urethane insulation $(k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( and \)\alpha=0.36 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}$ ) sandwiched between two layers of sheet metal with negligible thickness. The refrigerated space is maintained at \(3^{\circ} \mathrm{C}\), and the average heat transfer coefficients at the inner and outer surfaces of the wall are \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and $9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. Heat transfer through the bottom surface of the refrigerator is negligible. The kitchen temperature remains constant at about \(25^{\circ} \mathrm{C}\). Initially, the refrigerator contains \(15 \mathrm{~kg}\) of food items at an average specific heat of $3.6 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$. Now a malfunction occurs, and the refrigerator stops running for \(6 \mathrm{~h}\) as a result. Assuming the temperature of the contents of the refrigerator, including the air inside, rises uniformly during this period, predict the temperature inside the refrigerator after \(6 \mathrm{~h}\) when the repairman arrives. Use the explicit finite difference method with a time step of $\Delta t=1 \mathrm{~min}\( and a mesh size of \)\Delta x=1 \mathrm{~cm}$, and disregard corner effects (i.e., assume one-dimensional heat transfer in the walls).

Using appropriate software, solve these systems of algebraic equations. (a) \(\quad 3 x_{1}-x_{2}+3 x_{3}=0\) $$ \begin{array}{r} -x_{1}+2 x_{2}+x_{3}=3 \\ 2 x_{1}-x_{2}-x_{3}=2 \end{array} $$ (b) $$ \begin{aligned} 4 x_{1}-2 x_{2}^{2}+0.5 x_{3} &=-2 \\ x_{1}^{3}-x_{2}+x_{3} &=11.964 \\ x_{1}+x_{2}+x_{3} &=3 \end{aligned} $$ Answers: \((a) x_{1}=2, x_{2}=3, x_{3}=-1,(b) x_{1}=2.33, x_{2}=2.29\), \(x_{3}=-1.62\)

Is there any limitation on the size of the time step \(\Delta t\) in the solution of transient heat conduction problems using \((a)\) the explicit method and \((b)\) the implicit method?

Consider transient 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 $\Delta x$. The wall is initially at a specified temperature. Using the energy balance approach, obtain the explicit finite difference formulation of the boundary nodes for the case of uniform heat flux \(\dot{q}_{0}\) 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 \(T_{\infty}\). Do not simplify.

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 \(T_{\infty}\), and surrounding temperature of $T_{\text {sarr }}$. Also, obtain the finite difference formulation for the rate of heat transfer at the left boundary.

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