On the Temperature Field of an Anisotropic Body with Internal Heat Generation
Abstract Generated abstract
This paper examines the steady temperature field in an anisotropic body with uniform internal heat generation, motivated by electrical engineering applications such as transformer cores and windings. The authors formulate the two dimensional heat conduction problem with unequal thermal conductivities along the coordinate axes, symmetry at the center, and convective heat transfer at the surfaces, then introduce a constant ratio assumption for the divergence components of the temperature gradient to obtain a parabolic temperature distribution. Using surface mean temperatures in the boundary conditions, they derive expressions for the central temperature and the parameter governing the partition of curvature between coordinate directions. The resulting formulas are reported to agree with temperatures calculated by an elementary balance method to within less than 1 percent.
Full Text
On the Temperature Field of an Anisotropic Body with Internal Heat Generation
G. P. Boikov, Yu. A. Korolenko
(Presented by Professor, Doctor N. D. Kutyavin)
In electrical engineering, the question very often arises of the temperature field of a body with internal heat generation \(w\ \dfrac{\mathrm{kcal}}{\mathrm{m}^3\ \mathrm{h}}\).
Often these bodies have unequal values of the thermal-conductivity coefficients \(\lambda\), \(\dfrac{\mathrm{kcal}}{\mathrm{m}\,\mathrm{h}\,^\circ\mathrm{C}}\), along the directions of the coordinate axes (cores and windings of transformers, etc.). The temperature distribution in such a body, with one infinite dimension and with the origin of coordinates at the center of the body, is described by the equations:
\[ \lambda_1 \frac{\partial^2 t(x;y)}{\partial x^2} + \lambda_2 \frac{\partial^2 t(x;y)}{\partial y^2} + w = 0; \tag{1} \]
\[ \frac{\partial t(0;y)}{\partial x}=0; \qquad \frac{\partial t(x;0)}{\partial y}=0; \tag{2} \]
\[ \lambda_1 \frac{\partial t(R_1;y)}{\partial x} = z\,[t(R_1;y)-t_f], \]
\[ -\lambda_2 \frac{\partial t(x;R_2)}{\partial y} = z\,[t(x;R_2)-t_f]. \tag{3} \]
Here \(2R_1 < 2R_2\) are the dimensions of the body, \(x\) and \(y\) are the current coordinates, \(t_f\) is the temperature of the surrounding medium, and \(z\) is the heat-transfer coefficient.
According to equation (1), one may write:
\[ \frac{\partial^2 t(x;y)}{\partial x^2} \left[ 1+ \frac{\lambda_2}{\lambda_1} \frac{ \dfrac{\partial^2 t(x;y)}{\partial y^2} }{ \dfrac{\partial^2 t(x;y)}{\partial x^2} } \right] = -\frac{w}{\lambda_1}, \]
\[ \frac{\partial^{2} t(x;y)}{\partial y^{2}} \left[ 1 : \frac{1}{ \frac{\lambda_{2}}{\lambda_{1}}\, \frac{\dfrac{\partial^{2} t(x;y)}{\partial y^{2}}} {\dfrac{\partial^{2} t(x;y)}{\partial x^{2}}} } \right] = -\frac{w}{\lambda_{2}} . \]
Considering, in accordance with [1], the ratio of the components of the divergence of the temperature gradient to be a constant quantity, we represent the last relations in the form:
\[ \frac{\partial^{2} t(x;y)}{\partial x^{2}}\,\xi' = -\frac{w}{\lambda_{1}}, \qquad \frac{\partial^{2} t(x;y)}{\partial y^{2}}\, \frac{\xi'}{\xi'-1} = -\frac{w}{\lambda_{2}} . \tag{4} \]
Then
\[ t(x;y)= -\frac{w x^{2}}{2\xi'\lambda_{1}} + f(y)x+\varphi(y), \]
\[ t(x;y)= -\frac{w y^{2}}{2\lambda_{2}}\,\frac{\xi'-1}{\xi'} + f(x)y+\varphi(x). \]
From the symmetry conditions (2), \(f(x)=f(y)=0\). Taking into account that for \(x=0\) or \(y=0\), we obtain \(t(0;y)=\varphi(y)\); \(t(x;0)=\varphi(x)\). Consequently,
\[ t(x;y)= -\frac{w x^{2}}{2\xi'\lambda_{1}} + t(0;y), \tag{5} \]
\[ t(x;y)= -\frac{\xi'-1}{\xi'}\,\frac{w y^{2}}{2\lambda_{2}} + t(x;0). \tag{6} \]
Putting respectively \(x=0\) and \(y=0\) in relations (4), we obtain the system of ordinary differential equations:
\[ \frac{d^{2}t(x;0)}{d x^{2}}\,\xi' = -\frac{w}{\lambda_{1}} \qquad \text{and} \qquad \frac{d^{2}t(0;y)}{d y^{2}}\, \frac{\xi'}{\xi'-1} = -\frac{w}{\lambda_{2}} . \]
Their solution:
\[ t(x;0)= -\frac{w x^{2}}{2\xi'\lambda_{1}}+D_{1}; \qquad t(0;y)= -\frac{\xi'-1}{\xi'}\,\frac{w y^{2}}{2\lambda_{2}}+D_{2}. \]
Since for \(x=0\) and \(y=0\), \(t(0;0)=t_{\mathrm{c}}\), then \(D_{1}=D_{2}=t_{\mathrm{c}}\), the temperature of the center of the body, and expressions (5) and (6) become identical:
\[ t(x;y)=t_{\mathrm{c}} -\frac{w x^{2}}{2\xi'\lambda_{1}} -\frac{\xi'-1}{\xi'}\,\frac{w y^{2}}{2\lambda_{2}} . \tag{7} \]
Equation (7) satisfies the differential equation of heat conduction (1) and the symmetry conditions (2).
We seek the temperature of the center of the body \(t_{\mathrm{c}}\) and \(\xi'\) from conditions (3), considering the temperatures on the surface of the body \(t(R_{1};y)\) and \(t(x;R_{2})\) to be constant,
equal to the mean integral temperatures of the corresponding surfaces:
\[ -\lambda_1 \frac{\partial t(R_1;y)}{\partial x} = \alpha \left[t_{\mathrm{cp}}(R_1;y)-t_f\right]; \qquad -\lambda_2 \frac{\partial t(x;R_2)}{\partial y} = \alpha \left[t_{\mathrm{cp}}(x;R_2)-t_f\right], \tag{8} \]
where
\[ t_{\mathrm{cp}}(R_1;y) = \frac{1}{R_2}\int_{0}^{R_2} t(R_1;y)\,dy = t_c-\frac{\bar w R_1^2}{2\xi'\lambda_1} - \frac{\xi'-1}{\xi'}\frac{\bar w R_2^2}{6\lambda_2} = t_c-K_*, \]
\[ t_{\mathrm{cp}}(x;R_2) = \frac{1}{R_1}\int_{0}^{R_1} t(x;R_2)\,dx = t_c-\frac{\bar w R_1^2}{6\xi'\lambda_1} - \frac{\xi'-1}{\xi'}\frac{\bar w R_2^2}{2\lambda_2} = t_c-K_{**}. \]
Knowing the value of \(t_{\mathrm{cp}}\) and taking (7) into account, we write equation (8) in the form:
\[ \frac{\bar w R_1}{\xi' \alpha}=t_c-K_*-t_f \qquad\text{and}\qquad \frac{\bar w R_2}{\alpha}\frac{\xi'-1}{\xi'}=t_c-K_{**}-t_f. \]
Subtracting the second expression from the first, we obtain:
\[ \frac{\bar w R_1}{\xi'\alpha} - \frac{\xi'-1}{\xi'}\frac{\bar w R_2}{\alpha} = \frac{\bar w R_1^2}{6\xi'\lambda_1} + \frac{\xi'-1}{\xi'}\frac{\bar w R_2^2}{2\lambda_2} - \frac{\bar w R_1^2}{2\xi'\lambda_1} - \frac{\bar w R_2^2}{6\lambda_2}\frac{\xi'-1}{\xi'}, \]
whence
\[ \xi' = \frac{ \dfrac{R_1+R_2}{\alpha} - \dfrac{\dfrac{\lambda_2}{\lambda_1}R_1^2+R^2}{3\lambda_2} }{ \dfrac{R_2}{\alpha} - \dfrac{R_2^2}{3\lambda_2} }. \tag{9} \]
The temperature at the center of the body is
\[ t_c = t_f+\bar w\left( \frac{R_1}{\xi'\alpha} + \frac{R_1^2}{2\xi'\lambda_1} + \frac{\xi'-1}{\xi'}\frac{R_2^2}{6\lambda_2} \right). \tag{10} \]
The temperature values found from formulas (7), (9), and (10) differ from the temperatures obtained by the method of elementary balances by less than \(1\%\).
REFERENCES
- G. P. Boikov. Heating of finite bodies under the action of radiant heat, Tomsk. Izvestiya TPI, vol. 101, 1958.