Temperature Field in a Rectangular Bar with Internal Heat Generation
Abstract Generated abstract
This paper derives an exact steady state solution for the temperature field in an infinitely long rectangular bar with anisotropic thermal conductivity and uniform internal heat generation, with convective heat transfer from all faces. By transforming the anisotropic heat equation to an equivalent isotropic form and applying separation of variables with symmetry and Newton cooling boundary conditions, the author obtains a series expression for the temperature at any cross-sectional point and a transcendental equation for the eigenvalues. The analysis indicates that, for many practical cases, the first term of the series is sufficient, and recommends orienting the coordinate axis to improve convergence. A numerical example for a magnetic core shows the method’s calculation of the maximum temperature and illustrates differences from an earlier approximate approach.
Full Text
TEMPERATURE FIELD IN A RECTANGULAR BAR WITH INTERNAL HEAT GENERATION
L. G. Fuks
(Submitted by the Department of Boiler Construction and Boiler Installations)
In a number of electrical-engineering structures, materials are widely used whose thermal conductivity is not the same in different directions. Such structures include, for example, transformer cores, betatron magnetic cores, etc., operating under conditions of internal heat generation. The thermal conductivity of steel sheets along the stack is higher than in the transverse direction, so that, as a whole, the core may be regarded as an anisotropic body.
The object of the calculation is to find the temperature field over the cross section of the magnetic core and, in particular, to determine the maximum overheating temperature.
Approximate solutions of the problem are available, for example, in [1]. These solutions are based on the assumption of a large difference in the thermal conductivities of the anisotropic material or on the use of approximate methods of numerical integration. Below an exact solution is given for a bar of infinite length.
If the axes \(x\) and \(y\) are directed along the axes of symmetry of the transverse section of the bar (Fig. 1), then the steady-state temperature field is described by the equation
\[ \lambda_x \frac{\partial^2 t}{\partial x^2}+\lambda_y \frac{\partial^2 t}{\partial y^2}+q_v=0. \tag{1} \]
The boundary conditions for heat removal according to Newton’s law may be expressed as
\[ \alpha_x (t-t_0)=-\lambda_x \left(\frac{dt}{dx}\right)\quad \text{for } x=\pm a, \tag{2} \]
\[ \alpha_y (t-t_0)=-\lambda_y \left(\frac{dt}{dy}\right)\quad \text{for } y=\pm b. \tag{3} \]
Here the notation is as follows:
\(t\) — temperature at an arbitrary point of the bar,
\(t_0\) — temperature of the surrounding medium,
\(2a, 2b\) — linear dimensions of the cross section of the bar;
\(\lambda_x, \lambda_y\) are the thermal conductivities in the \(x\) and \(y\) directions,
\(q_v\) is the internal heat generation,
\(\alpha_x, \alpha_y\) are the heat-transfer coefficients from the surface of the bar in the \(x\) and \(y\) directions.
Introduce into equation (1) and the boundary conditions (2), (3) a new function and independent variables
\[ \vartheta = t - t_0, \qquad x = u\sqrt{\lambda_x}, \qquad y = v\sqrt{\lambda_y}. \]
Fig. 1. Computational scheme of the problem.
In this case (1), (2), and (3) will take the form
\[ \frac{\partial^2 \vartheta}{\partial u^2} + \frac{\partial^2 \vartheta}{\partial v^2} + q_v = 0, \tag{4} \]
\[ \vartheta = -\gamma_x \left(\frac{\partial \vartheta}{\partial u}\right)_{u=a_1}, \qquad \vartheta = -\gamma_y \left(\frac{\partial \vartheta}{\partial v}\right)_{v=b_1}, \tag{5},(6) \]
where
\[ \gamma_x = \frac{\sqrt{\lambda_x}}{\alpha_x}, \qquad \gamma_y = \frac{\sqrt{\lambda_y}}{\alpha_y}, \qquad a_1 = \frac{a}{\sqrt{\lambda_x}}, \]
\[ b_1 = \frac{b}{\sqrt{\lambda_y}}. \]
We shall seek the solution in the form
\[ \vartheta = \varphi(u) + U(u)\cdot V(v). \tag{7} \]
Substitute function (7) into equation (4):
\[ UV'' + VU'' + \varphi'' + q_v = 0. \tag{8} \]
Choose \(\varphi\) so that
\[ \varphi'' + q_v = 0, \]
and determine the constants of integration from the boundary conditions. We find the following form of the function:
\[ \varphi(u) = \frac{q_v a_1^2}{2} \left[ 1 - \left(\frac{u}{a_1}\right)^2 + \frac{2\gamma_x}{a_1} \right]. \tag{9} \]
With this choice of \(\varphi\), from (8), after separation of variables, we obtain
\[ V''/V=-U''/U=\beta^2 . \tag{10} \]
To find a particular solution, we shall use the symmetry conditions, which in the case under consideration are written as
\[ \left(\frac{\partial \vartheta}{\partial u}\right)_{u=0}=0, \tag{11} \]
\[ \left(\frac{\partial \vartheta}{\partial v}\right)_{v=0}=0. \tag{12} \]
The complete solution of the differential equation (4) under the boundary conditions (5) and (6) will be
\[ \vartheta(u,v)=\frac{q_v a_1^2}{2} \left[ 1-\left(\frac{u}{a_1}\right)^2+\frac{2\gamma_x}{a_1} \right] +\sum_{n=1}^{\infty} A_n \cos \beta_n u \cdot \operatorname{ch}\beta_n v . \tag{13} \]
The constants \(A_n\) and the eigenvalues \(\beta_n\) are found from the boundary conditions in the usual way. The final solution of the problem after passing to the original arguments has the form
\[ \begin{aligned} \vartheta(x,y)=\frac{q_v a^2}{2\lambda_x} \Bigg[ &1-\left(\frac{x}{a}\right)^2+\frac{2\lambda_x}{a\alpha_x} \\ &-4\sum_{n=1}^{\infty} \frac{ \sin\mu_n\cdot \cos\mu_n\frac{x}{a}\operatorname{ch}\mu_n\frac{y}{a} \sqrt{\frac{\lambda_x}{\lambda_y}} }{ \mu_n^3 \left( \mu_n\frac{\lambda_y}{a\alpha_y} \sqrt{\frac{\lambda_x}{\lambda_y}}\operatorname{sh}\mu_n\frac{b}{a} \sqrt{\frac{\lambda_x}{\lambda_y}} +\operatorname{ch}\mu_n\frac{b}{a} \sqrt{\frac{\lambda_x}{\lambda_y}} \right) } \\ &\times \frac{1}{\left(1+\frac{\sin 2\mu_n}{2\mu_n}\right)} \Bigg], \end{aligned} \tag{14} \]
where the eigenvalues \(\mu_n\) are the roots of the transcendental equation
\[ \frac{\alpha_x a}{\lambda_x}\operatorname{ctg}\mu=\mu . \tag{15} \]
The values of the first two roots of equation (15) are given in Fig. 2. More detailed tables are available, for example, in [2].
Analysis of equation (14) shows that in the majority of practical calculation cases it is sufficient to restrict oneself to only the first term of the infinite sum. At least, if the quantity
\[ \frac{\alpha_x a}{\lambda_x}<0.5, \]
then, by discarding the second term, we make an error of less than \(0.2\%\) of the entire sum.
When working with the exact solution, the \(x\)-axis should be placed in the direction in which the quantity
\[ \frac{\alpha_x a}{\lambda_x} \]
will be the smallest, since in this case the convergence of the series will be better.
As an example, we present a comparison of the proposed calculation with the method adopted in [1], from which the data given below have been taken.
Example: An infinite bar of rectangular cross-section \(0.48 \times 0.16 \, \mathrm{m}^2\), whose thermal-conductivity coefficients are respectively equal to
Fig. 2. Roots of the transcendental equation
\[ \frac{\alpha_x a}{\lambda_x}\operatorname{ctg}\mu=\mu . \]
\(45.4 \, \dfrac{\mathrm{W}}{\mathrm{m}\,^{\circ}\mathrm{C}}\) and \(1.16 \, \dfrac{\mathrm{W}}{\mathrm{m}\,^{\circ}\mathrm{C}}\), is heated by internal heat generation of intensity \(30200 \, \dfrac{\mathrm{W}}{\mathrm{m}^3}\). The bar is cooled with the same heat-transfer coefficient \(62.8 \, \dfrac{\mathrm{W}}{\mathrm{m}^2\,^{\circ}\mathrm{C}}\) on all faces.
The ambient air temperature is \(35^\circ\mathrm{C}\). Find the temperature of any point of the bar and determine the maximum temperature under conditions of steady-state heating.
Let us determine two quantities
\[ \frac{54\cdot 0.24}{39}=0.332,\qquad \frac{54\cdot 0.08}{1}=4.32. \]
Thus, it is more convenient to direct the \(x\)-axis parallel to the side \(0.48\ \mathrm{m}\). In this case \(\psi_1=0.9547\) and \(\psi_2=3.24\).
The data of the problem will be as follows:
\[ a=0.24\ \mathrm{m},\qquad b=0.08\ \mathrm{m},\qquad \lambda_x=45.4\ \mathrm{W}/\mathrm{m}^{\circ}\mathrm{C}, \]
\[ \lambda_y=1.16\ \mathrm{W}/\mathrm{m}^{\circ}\mathrm{C},\qquad \alpha_x=\alpha_y=\alpha=62.8\ \mathrm{W}/\mathrm{m}^2{}^{\circ}\mathrm{C}, \]
\[ q_v=30200\ \mathrm{W}/\mathrm{m}^3,\qquad t_0=35^\circ\mathrm{C}. \]
We successively compute the calculation complexes
\[ \frac{q_v a^2}{2\lambda_x}=19.2^\circ\mathrm{C};\qquad \sqrt{\frac{\lambda_x}{\lambda_y}}=6.25; \]
\[ \psi_1\frac{b}{a}\sqrt{\frac{\lambda_x}{\lambda_y}}=1.138;\qquad \frac{2\lambda_x}{\alpha_x\cdot a}=6.02; \]
\[ \psi_1\frac{\lambda_y}{\alpha_y\cdot a}\sqrt{\frac{\lambda_x}{\lambda_y}}=0.263 \]
and, substituting these quantities into (14), taking into account only the first term of the sum, after transformations we obtain
\[ t(x,y)=169.8-333x^2-64.7\cos 2.28x\cdot \operatorname{ch}14.25y. \]
To determine the temperature, the values of \(x\) and \(y\) must be taken in meters.
The maximum temperature will be at the point \(x=y=0\),
\[ t_{\max}=105.1^\circ\mathrm{C}. \]
(according to the calculation given in work [1], \(t_{\max}=98^\circ\mathrm{C}\)).
Calculation of the correction associated with neglecting the second term of the series gave the small value \(0.2\cdot 10^{-3}\ ^\circ\mathrm{C}\).
Conclusions
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When calculating the temperatures of an anisotropic rectangular bar with internal heat generation, one should use the proposed exact solution, which gives the temperature value at any point of the bar.
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The use of the exact solution is especially recommended for close values of the Biot criterion in two mutually perpendicular directions of the bar cross-section, since in this case approximate methods may give a significant error.
References
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A. A. Gurchenok. Calculation of the heating temperature of concrete magnetic cores. Izvestiya vysshikh uchebnykh zavedenii, Elektromekhanika, No. 3, 1959.
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A. V. Lykov. Theory of Heat Conduction. GITTL, Moscow, 1952.