TEMPERATURE DISTRIBUTION IN A BODY OF ELLIPTICAL CROSS-SECTION WITH INTERNAL HEAT SOURCES

V. V. Ivanov

(Presented by Prof. G. I. Fuks, Doctor of Technical Sciences)

The article calculates the temperature field in a body of elliptical cross-section with internal heat generation. The boundary conditions are specified by the heat-transfer coefficient \(h\) and by the temperature of the surrounding medium, taken to be zero. The solution is given in elliptical coordinates.

It is known that, for better cooling of heat-generating elements (electrical conductors, rods of nuclear reactors, etc.), it is necessary to have a large heat-transfer surface. An increase in surface area can be achieved either by ribbing or by replacing rods of circular cross-section, which have the minimum heat-transfer surface, with rods of other cross-sections, for example oval or elliptical ones.

The problems of calculating temperature fields in bodies of elliptical cross-section in the presence of internal heat sources have been treated in works [1, 2]. However, the results obtained cannot be considered final because of a number of assumptions adopted to simplify the investigation.

Fig. 1.

We shall seek the temperature distribution in an infinitely long body whose cross-section is an ellipse with semiaxes \(a\) and \(b\) (Fig. 1). The body under consideration is in a medium with zero tempera-

tube. Inside the body there acts a heat source \(q_v\) of the same magnitude (per each \(m^3\)); the thermal-conductivity coefficient \((\lambda)\) and the heat-transfer coefficient \((h)\) are constant quantities. To find the temperature distribution, it is necessary to solve Poisson’s equation

\[ \Delta T+\frac{q_v}{\lambda}=0 \]

with the boundary condition \(-\lambda \operatorname{grad} T=hT\) on the surface of the body.

To obtain a formula describing the temperature field, we shall use the system of elliptic coordinates \(\alpha,\beta\) [3,4], \(0\le \alpha<\infty\), \(-\pi<\beta<\pi\). If \(\alpha=\alpha_0\) is the equation of the surface of the body, then

\[ a=c\,\operatorname{ch}\alpha_0,\quad b=c\,\operatorname{sh}\alpha_0,\quad c=\sqrt{a^2-b^2}. \]

Poisson’s equation in elliptic coordinates has the form

\[ \frac{1}{c^2(\operatorname{ch}^2\alpha-\cos^2\beta)} \left( \frac{\partial^2 T}{\partial \alpha^2} + \frac{\partial^2 T}{\partial \beta^2} \right) +\frac{q_v}{\lambda}=0, \tag{1} \]

and the boundary condition is given by the relation: for \(\alpha=\alpha_0\)

\[ -\lambda\, \frac{1}{c\sqrt{\operatorname{ch}^2\alpha_0-\cos^2\beta}} \cdot \frac{\partial T}{\partial \alpha} =hT. \tag{2} \]

The substitution

\[ U(\alpha,\beta)=T(\alpha,\beta)+\frac{q_v c^2}{4\lambda}\left(\operatorname{sh}^2\alpha+\cos^2\beta\right) \]

transforms equation (1) into Laplace’s equation

\[ \frac{\partial^2 U}{\partial \alpha^2} + \frac{\partial^2 U}{\partial \beta^2} =0, \tag{3} \]

and the boundary condition (2) takes the form, for \(\alpha=\alpha_0\),

\[ \frac{1}{\sqrt{\operatorname{ch}^2\alpha_0-\cos^2\beta}} \left( \frac{\partial U}{\partial \alpha} - \frac{q_v}{4\lambda}c^2\operatorname{sh}2\alpha_0 \right) = Bi\left[ \frac{q_v}{4\lambda}c^2 \times \right. \]

\[ \left. \times\left(\operatorname{sh}^2\alpha_0+\cos^2\beta\right)-U \right], \quad Bi=\frac{hc}{\lambda}. \tag{4} \]

The solution of equation (3) is given by the relation [3]

\[ U=\sum_{n=0}^{\infty} A_n \operatorname{ch} n\alpha \cos n\beta + B_n \operatorname{sh} n\alpha \sin n\beta, \]

where in our case, from considerations of symmetry, one should set \(B_n=0\). The constants \(A_n\) are found from the boundary condition (4)

\[ \frac{1}{\sqrt{\operatorname{ch}^2\alpha_0-\cos^2\beta}} \left( \sum_{n=0}^{\infty} nA_n\operatorname{sh} n\alpha_0\cos n\beta - \frac{q_v}{4\lambda}c^2\operatorname{sh}2\alpha_0 \right) = \]

\[ = Bi\left[ \frac{q_v}{4\lambda}c^2 \left(\operatorname{sh}^2\alpha_0+\cos^2\beta\right) - \sum_{n=0}^{\infty} A_n\operatorname{ch} n\alpha_0\cos n\beta \right]. \tag{5} \]

The function \(f(\beta)=\sqrt{\operatorname{ch}^2\alpha_0-\cos^2\beta}\) is even with respect to \(\beta\); we expand it in a cosine series

\[ \sqrt{\operatorname{ch}^{2}\alpha_{0}-\cos^{2}\beta} =\frac{a_{0}}{2}+\sum_{n=1}^{\infty}a_{n}\cos n\beta, \]

where

\[ a_{n}=\frac{2}{\pi}\int_{0}^{\pi} \sqrt{\operatorname{ch}^{2}\alpha_{0}-\cos^{2}\beta}\,\cos n\beta\,d\beta. \]

Since
\[ \cos n\beta=\cos^{n}\beta-C_{n}^{2}\cos^{\,n-2}\beta\sin^{2}\beta +C_{n}^{4}\cos^{\,n-4}\beta\sin^{4}\beta-\ldots, \]
it is not difficult to show that all \(a_n\) with odd \(n\) vanish. Therefore

\[ \sqrt{\operatorname{ch}^{2}\alpha_{0}-\cos^{2}\beta} =\frac{a_{0}}{2}+\sum_{n=1}^{\infty}a_{2n}\cos 2n\beta. \]

Rewriting (5) in the form

\[ \sum_{n=0}^{\infty} nA_{n}\operatorname{sh} n\alpha_{0}\cos n\beta -\frac{q_{v}}{4\lambda}c^{2}\operatorname{sh}2\alpha_{0} = Bi\,\frac{q_{v}}{4\lambda}c^{2} \left(\operatorname{sh}^{2}\alpha_{0} \sqrt{\operatorname{ch}^{2}\alpha_{0}-\cos^{2}\beta} +\right. \]

\[ \left. +\sqrt{\operatorname{ch}^{2}\alpha_{0}-\cos^{2}\beta}\,\cos^{2}\beta \right) -Bi\,\frac{a_{0}}{2}\sum_{n=0}^{\infty} A_{n}\operatorname{ch} n\alpha_{0}\cos n\beta \]

\[ -Bi\sum_{n=1}^{\infty}a_{2n}\cos 2n\beta \sum_{n=0}^{\infty}A_{n}\operatorname{ch} n\alpha_{0}\cos n\beta \]

and integrating with respect to \(\beta\) from \(0\) to \(\pi\), we obtain

\[ -\frac{q_{v}}{4\lambda}c^{2}\operatorname{sh}2\alpha_{0}\,\pi = Bi\,\frac{q_{v}}{4\lambda}c^{2} \left( \operatorname{sh}^{2}\alpha_{0} \int_{0}^{\pi}\sqrt{\operatorname{ch}^{2}\alpha_{0}-\cos^{2}\beta}\,d\beta +\right. \]

\[ \left. +\int_{0}^{\pi}\sqrt{\operatorname{ch}^{2}\alpha_{0}-\cos^{2}\beta}\, \cos^{2}\beta\,d\beta \right) -Bi\,\frac{\pi}{2} \left(a_{0}A_{0}+\operatorname{ch}2n\alpha_{0}\,a_{2n}A_{2n}\right). \]

Whence

\[ A_{2n}= \left[ \frac{q_{v}}{2\lambda}c^{2}\operatorname{sh}2\alpha_{0}\,\pi + Bi\,\frac{q_{v}}{2\lambda}c^{2} \left( \operatorname{sh}^{2}\alpha_{0} \int_{0}^{\pi}\sqrt{\operatorname{ch}^{2}\alpha_{0}-\cos^{2}\beta}\,d\beta +\right. \right. \]

\[ \left. \left. +\int_{0}^{\pi}\sqrt{\operatorname{ch}^{2}\alpha_{0}-\cos^{2}\beta}\, \cos^{2}\beta\,d\beta \right) \right] :Bi\,\pi\,\operatorname{ch}2n\alpha_{0}\,a_{2n}, \qquad n=0,1,2\ldots \]

To compute the integrals in the last formula we use the substitution \(x=\pi/2-\beta\). Then

\[ \int_{0}^{\pi}\sqrt{\operatorname{ch}^{2}\alpha_{0}-\cos^{2}\beta}\,d\beta = \int_{\pi/2}^{-\pi/2} \operatorname{ch}\alpha_{0} \sqrt{1-\frac{\sin^{2}x}{\operatorname{ch}^{2}\alpha_{0}}}\,d(-x) = \]

\[ =2\operatorname{ch}\alpha_{0} E\left(\frac{1}{\operatorname{ch}\alpha_{0}},\,\frac{\pi}{2}\right), \]

where \(E\) is the complete elliptic integral of the second kind.

\[ \int_{0}^{\pi}\sqrt{\operatorname{ch}^{2}\alpha_{0}-\cos^{2}\beta}\,\cos^{2}\beta\,d\beta = \int_{\pi/2}^{\pi/2}\operatorname{ch}\alpha_{0} \sqrt{1-\frac{\sin^{2}x}{\operatorname{ch}^{2}\alpha_{0}}}\,\sin^{2}x\,dx = \]

\[ = 2\operatorname{ch}\alpha_{0}\left[ \frac{\operatorname{sh}^{2}\alpha_{0}}{3} F\left(\frac{1}{\operatorname{ch}\alpha_{0}},\frac{\pi}{2}\right) + \left(\frac{2-\operatorname{ch}^{2}\alpha_{0}}{3}\right) E\left(\frac{1}{\operatorname{ch}\alpha_{0}},\frac{\pi}{2}\right) \right], \]

where \(F\) is the complete elliptic integral of the first kind [5].

Let us now determine \(a_{2n}\)

\[ a_{2n} = \frac{2}{\pi}\int_{0}^{\pi} \sqrt{\operatorname{ch}^{2}\alpha_{0}-\cos^{2}\beta} \left(\cos^{2n}\beta-C_{2n}^{2}\cos^{2n-2}\beta\sin^{2}\beta+\ldots\right)d\beta = \frac{4\operatorname{ch}\alpha_{0}}{\pi}\times \]

\[ \times \int_{0}^{\pi/2} \sqrt{1-\frac{\sin^{2}x}{\operatorname{ch}^{2}\alpha_{0}}} \left(\sin^{2n}x-C_{2n}^{2}\sin^{2n-2}x\cos^{2}x+\ldots\right)dx = \]

\[ = \frac{2\operatorname{ch}\alpha_{0}}{\pi} \left[ B\left(\frac{2n+1}{2},\frac{1}{2}\right)\times \right. \]

\[ \left. \times F_{1}\left(\frac{2n+1}{2},-\frac{1}{2},n+1;\frac{1}{\operatorname{ch}^{2}\alpha_{0}}\right) - \right. \]

\[ \left. - C_{2n}^{2} B\left(\frac{2n-1}{2},\frac{3}{2}\right) F_{1}\left(\frac{2n-1}{2},-\frac{1}{2},n+1;\frac{1}{\operatorname{ch}^{2}\alpha_{0}}\right) +\ldots \right], \]

where \(B(y,z)\) is the beta function, \(F_{1}(y,z,m;k)\) is the hypergeometric function [5].

Finally, the desired temperature distribution is described by the equation

\[ T(\alpha,\beta) = \sum_{n=0}^{\infty} A_{2n}\operatorname{ch}2n\alpha\cos2n\beta - \frac{q_{v}}{4\lambda}c^{2} \left(\operatorname{sh}^{2}\alpha+\cos^{2}\beta\right), \]

in which \(A_{2n}\) is found from the relation

\[ A_{2n} = \left\{ \frac{q_{v}}{2\lambda}c^{2}\operatorname{sh}2\alpha_{0}\,\pi + Bi\,\frac{q_{v}}{\lambda}c^{2}\operatorname{ch}\alpha_{0} \left[ \operatorname{sh}^{2}\alpha_{0} E\left(\frac{1}{\operatorname{ch}\alpha_{0}},\frac{\pi}{2}\right) + \right.\right. \]

\[ \left.\left. + \frac{\operatorname{sh}^{2}\alpha_{0}}{3} F\left(\frac{1}{\operatorname{ch}\alpha_{0}},\frac{\pi}{2}\right) + \left(\frac{2-\operatorname{ch}^{2}\alpha_{0}}{3}\right) E\left(\frac{1}{\operatorname{ch}\alpha_{0}},\frac{\pi}{2}\right) \right] \right\} : \]

\[ : Bi\,\pi\,\operatorname{ch}2n\alpha_{0}\,a_{2n}. \]

For \(h\to\infty\) (this corresponds to the temperature distribution in a rod of elliptical cross-section when its surface is maintained at zero temperature), we obtain the solution given in [3].

REFERENCES

1. Boikov G. P. Temperature fields in bodies of finite dimensions with internal heat generation. IFZh, No. 5, 1959.

2. Boikov G. P., Korolenko Yu. A. Temperature field in a bar of elliptical cross-section with internal heat generation. IFZh, No. 12, 1960.

3. Lebedev N. N., Skalskaya I. P., Uflyand Ya. S. Collection of Problems in Mathematical Physics. Gostekhizdat, 1955.

4. Morse Ph. M. and Feshbach H. Methods of Theoretical Physics, Vol. II, GIIL, 1960.

5. Gradshteyn I. S. and Ryzhik I. M. Tables of Integrals, Sums, Series, and Products. Gostekhizdat, 1962.