Temperature Field in a Short Cylinder with Internal Heat Generation: First Report
Abstract Generated abstract
This paper develops an approximate engineering method for calculating the steady temperature field in a short cylindrical coil with uniform internal heat generation and convective heat exchange at its outer surfaces. Using the method of parallel integration, the heat conduction equation is reduced by assuming a constant ratio between components of the divergence of the temperature gradient, yielding simple formulae for the temperature distribution, the central temperature, and an auxiliary parameter determined from averaged boundary conditions. A numerical example for a coil is compared with results from the method of elementary balances, showing generally close agreement, with larger discrepancies near the surface. The authors conclude that the approach is simple and sufficiently accurate, particularly for small Biot numbers.
Full Text
Temperature Field in a Short Cylinder with Internal Heat Generation
Yu. A. Korotenko, G. P. Boikov
(Presented by Professor, Doctor G. I. Fuks)
Coils with internal heat generation, made in one form or another, are an important part of electrical devices. A number of works ([2] and others) are devoted to the problem of calculating the temperature field of such coils.
The computational relations given in these works, as a rule, are very complicated in form and inconvenient for engineering practice.
Let us apply the method of parallel integration [3] to the study of the temperature field of a cylindrical coil. The temperature field of the coil is described by the equations:
\[ \frac{\partial^{2} t(r;y)}{\partial r^{2}} + \frac{1}{r}\frac{\partial t(r;y)}{\partial r} + \frac{\partial^{2} t(r;y)}{\partial y^{2}} + \frac{w}{\lambda} =0, \tag{1} \]
\[ \frac{\partial t(0;y)}{\partial r}=0; \qquad \frac{\partial t(r;0)}{\partial y}=0; \tag{2} \]
\[ \lambda \frac{\partial t(R_{1};y)}{\partial r} = \alpha \left[t(R_{1};y)-t_f\right]; \]
\[ -\lambda \frac{\partial t(r;R_{2})}{\partial y} = \alpha \left[t(r;R_{2})-t_f\right]. \tag{3} \]
Here \(2R_1\) and \(2R_2\) are the dimensions of the body, \(r\) and \(y\) are the current coordinates of points of the body cross section, \(t_f\) is the temperature of the surrounding medium, and \(w\) is the internal heat generation, the same throughout the entire volume of the coil.
Equation (1) can be rewritten in the form:
\[ \left[ \frac{\partial^{2}t(r;y)}{\partial r^{2}} + \frac{1}{r}\frac{\partial t(r;y)}{\partial r} \right] \left[ 1+ \frac{ \frac{\partial^{2}t(r;y)}{\partial y^{2}} }{ \frac{\partial^{2}t(r;y)}{\partial r^{2}} + \frac{1}{r}\frac{\partial t(r;y)}{\partial r} } \right] = -\frac{w}{\lambda}, \]
\[ \frac{\partial^{2}t(r;y)}{\partial y^{2}} \left[ 1+ \frac{1}{ \frac{ \frac{\partial^{2}t(r;y)}{\partial y^{2}} }{ \frac{\partial^{2}t(r;y)}{\partial r^{2}} + \frac{1}{r}\frac{\partial t(r;y)}{\partial r} } } \right] = -\frac{w}{\lambda}. \]
Taking, according to [1], the ratio of the components of the divergence of the temperature gradient to be a constant quantity, we write the last formulas as follows:
\[ \frac{\partial^2 t(r;\,y)}{\partial r^2}+\frac{1}{r}\frac{\partial t(r;\,y)}{\partial r} =-\frac{w}{\lambda \xi}, \]
\[ \frac{\partial^2 t(r;\,y)}{\partial y^2} =-\frac{\xi-1}{\xi}\frac{w}{\lambda}. \]
The solution of equations (4)
\[ t(r;\,y)=-\frac{w}{4\lambda \xi}r^2+\varphi_1(y)\ln r+\varphi_2(y), \]
\[ t(r;\,y)=-\frac{\xi-1}{\xi}\frac{w}{2\lambda}y^2+\varphi_1(r)y+\varphi_2(r). \]
From the symmetry conditions (2) it follows that
\[ \varphi_1(y)=\varphi_1(r)=0. \]
Taking into account the central conditions for \(r=0\), and then \(y=0\),
\[ t(0;\,y)=\varphi_2(y), \qquad t(r;\,0)=\varphi_2(r). \]
Then
\[ t(r;\,y)=-\frac{w}{4\lambda \xi}r^2+t(0;\,y); \tag{5} \]
\[ t(r;\,y)=-\frac{\xi-1}{\xi}\frac{w}{2\lambda}y^2+t(r;\,0). \tag{6} \]
Putting respectively \(r=0\) and \(y=0\) in equations (4), we obtain a system of ordinary differential equations:
\[ \frac{d^2 t(r;\,0)}{dr^2}+\frac{1}{r}\frac{dt(r;\,0)}{dr} =-\frac{w}{\lambda \xi}, \]
\[ \frac{d^2 t(0;\,y)}{dy^2} =-\frac{\xi-1}{\xi}\frac{w}{\lambda}. \]
Their solution:
\[ t(r;\,0)=-\frac{w}{4\lambda \xi}r^2+D_1, \]
\[ t(0;\,y)=-\frac{\xi-1}{\xi}\frac{wy^2}{2\lambda}+D_2. \]
Since for \(r=y=0\), \(t(0;\,0)=D\) (the temperature of the center), it follows that \(D_1=D_2=D\), and equations (5) and (6) become identical:
\[ t(r;\,y)=D-\frac{w}{4\lambda \xi}r^2-\frac{\xi-1}{\xi}\frac{wy^2}{2\lambda}. \tag{7} \]
The solution (7) satisfies the differential equation of heat conduction (1) and the symmetry condition (2).
To satisfy the boundary conditions (3), it is necessary to determine \(D\) and \(\xi\) accordingly.
We use conditions (3), rewriting them approximately in the form
\[ -\lambda \frac{\partial t(R_1;y)}{\partial r} = \alpha \left[t(R_1;y)_{\mathrm{av}}-t_f\right], \]
\[ \lambda \frac{\partial t(r;R_2)}{\partial y} = \alpha \left[t(r;R_2)_{\mathrm{av}}-t_f\right], \tag{8} \]
where
\[ t(R_1;y)_{\mathrm{av}} = \frac{1}{R_2}\int_0^{R_2} t(R_1;y)\,dy = D-\frac{wR_1^2}{6\lambda} -\frac{\xi-1}{\xi}\frac{wR_2^2}{4\lambda}, \]
\[ t(r;R_2)_{\mathrm{av}} = \frac{2}{R_1^2}\int_0^{R_1} t(r;R_2)\,r\,dr = D-\frac{wR_1^2}{8\lambda \xi} -\frac{\xi-1}{\xi}\frac{wR_2^2}{2\lambda}. \]
Substituting the values of \(t(R_1;y)_{\mathrm{av}}\) and \(t(r;R_2)_{\mathrm{av}}\) into (8), after transformations we obtain
\[ \xi = 1+ \frac{\dfrac{R_1}{2\alpha}+\dfrac{R_1^2}{8\lambda}} {\dfrac{R_2}{\alpha}+\dfrac{R_2^2}{3\lambda}}. \tag{9} \]
The temperature at the center of the body is
\[ D = \frac{wR_1}{2\alpha \xi} + \frac{\xi-1}{\xi}\frac{wR_2^2}{6\lambda} + \frac{wR_2^2}{4\lambda \xi}. \tag{10} \]
| Coil Nos. | Coordinates of points (mm), \(r\) | Coordinates of points (mm), \(y\) | Temperature of points, by 7, 9, 10 | Temperature of points, by the method of elementary balances | Discrepancy, % |
|---|---|---|---|---|---|
| 1 | 0 | 0 | 61.98 | 62.0 | 0 |
| 2 | 5 | 0 | 60.32 | 60.4 | 0.14 |
| 3 | 10 | 0 | 55.33 | 55.3 | 0.05 |
| 4 | 13 | 0 | 50.75 | 50.4 | 0.7 |
| 5 | 0 | 10 | 58.63 | 58.7 | 0.12 |
| 6 | 5 | 10 | 56.97 | 57.05 | 0.14 |
| 7 | 10 | 10 | 51.98 | 52.2 | 0.42 |
| 8 | 13 | 10 | 47.40 | 47.9 | 1.05 |
| 9 | 0 | 20 | 48.55 | 48.6 | 0.4 |
| 10 | 5 | 20 | 46.92 | 46.3 | 1.34 |
| 11 | 10 | 20 | 41.93 | 42.5 | 1.0 |
| 12 | 13 | 20 | 37.35 | 39.1 | 4.5 |
The value of the temperatures found from formulas (7), (9), and (10) for various points of the coil agrees well with the data obtained by the method of elementary balances.
The results of the comparison are given in Table 1.
In the calculations it was assumed:
outer diameter of the coil \(D = 2R_1 = 30\ \mathrm{mm}\);
length of the coil \(L = 2R_2 = 50\ \mathrm{mm}\);
coefficient of thermal conductivity of the coil winding
\[ \lambda = 0.3\ \frac{\mathrm{kcal}}{\mathrm{m}\cdot \mathrm{h}\cdot {}^\circ\mathrm{C}}; \]
coefficient of heat transfer of the coil \(\alpha = 12\ \mathrm{kcal}/\mathrm{m}^2\cdot\mathrm{h}\cdot{}^\circ\mathrm{C}\); volumetric heat generation in the coil \(w = 100000\ \mathrm{kcal}/\mathrm{m}^3\cdot\mathrm{h}\); temperature of the surrounding medium \(t_f = 0^\circ\mathrm{C}\).
Conclusion
The approximate method presented for calculating the temperature field of a cylindrical coil with internal heat generation is distinguished by its simplicity and sufficient accuracy (except for points lying on the surface of the coil and close to it). The accuracy of calculations by formulas 7, 9, and 10 is especially high for small values of the Biot criterion (Bi).
LITERATURE
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G. P. Boikov. Heating of bodies of finite dimensions under the action of a heat source. Izvestiya TPI, vol. 101, 1958.
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Schneider. Engineering Problems of Heat Conduction. Foreign Literature Publishing House, Moscow, 1960.
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G. P. Boikov, Yu. A. Korolenko. Temperature field in bodies of elliptical cross-section with internal heat sources. Engineering-Physics Journal, No. 12, 1960.