On the Temperature Field of Bodies with Internal Heat Generation
Abstract Generated abstract
This paper addresses the calculation of temperature fields in rectangular anisotropic bodies with uniform internal heat generation and different heat transfer coefficients on different surfaces, a problem relevant to electrical engineering components such as magnetic cores and capacitors. The authors formulate the steady heat conduction problem with symmetry and convective boundary conditions, then derive an approximate parabolic temperature distribution by assuming a constant ratio between components of the divergence of the temperature gradient. Expressions are obtained for the temperature at any cross sectional point, including parameters determined from averaged boundary conditions and the central temperature. A numerical comparison with the method of elementary balances shows good agreement for the tested case, with larger discrepancies mainly near the cooled surfaces.
Full Text
On the Temperature Field of Bodies with Internal Heat Generation
G. P. Boikov, Yu. A. Korochenko
(Presented by Prof., Doctor of Technical Sciences G. I. Fuks)
In electrical engineering the question of the temperature field of a body with internal heat generation very often arises. These bodies often have unequal values of the thermal-conductivity coefficients \(\lambda\) along the coordinate axes. In a number of cases (certain types of magnetic cores, capacitors, etc.) such anisotropic bodies have different heat-transfer coefficients \(\alpha\) on different portions of the surface of the bodies.
The temperature distribution in such a bar, with the origin of coordinates at the center of symmetry of the body, is described by the equations:
\[ \lambda_1 \frac{\partial^2 t(x;y)}{\partial x^2} + \lambda_2 \frac{\partial^2 t(x;y)}{\partial y^2} + w = 0; \tag{1} \]
\[ \frac{\partial t(0;y)}{\partial x}=0; \qquad \frac{\partial t(x;0)}{\partial y}=0; \tag{2} \]
\[ \lambda_1 \frac{\partial t(R_1;y)}{\partial x} = \alpha_1 \bigl[t(R_1;y)-t_{f_1}\bigr]; \]
\[ \lambda_2 \frac{\partial t(x;R_2)}{\partial y} = \alpha_2 \bigl[t(R_2;x)-t_{f_2}\bigr]. \tag{3} \]
Here \(2R_1 \le 2R_2\) are the dimensions of the body; \(x\) and \(y\) are the current coordinates of points of the body section; \(t_{f_1}\) and \(t_{f_2}\) are the temperatures of the surrounding medium; \(w\) is the internal heat generation, identical throughout the entire volume of the body.
Equation (1) can be rewritten in the form
\[ \frac{\partial^2 t(x;y)}{\partial x^2} \left[ 1+ \frac{\lambda_2}{\lambda_1} \frac{\dfrac{\partial^2 t(x;y)}{\partial y^2}} {\dfrac{\partial^2 t(x;y)}{\partial x^2}} \right] = -\frac{w}{\lambda_1}, \]
\[ \frac{\partial^2 t(x;y)}{\partial y^2} \left[ 1+ \frac{1}{ \dfrac{\lambda_2}{\lambda_1} \dfrac{\dfrac{\partial^2 t(x;y)}{\partial y^2}} {\dfrac{\partial^2 t(x;y)}{\partial x^2}} } \right] = -\frac{w}{\lambda_2}. \]
Assuming, in accordance with [1], that the ratio of the components of the divergence of the temperature gradient is a constant quantity, the last relations are written as follows:
\[ \frac{\partial^2 t(x;y)}{\partial x^2}\,\xi=-\frac{w}{\lambda_1}; \qquad \frac{\partial^2 t(x;y)}{\partial y^2}\,\frac{\xi'}{\xi'-1}=-\frac{w}{\lambda_2}. \tag{4} \]
The solution of equations (4) is
\[ t(x;y)=-\frac{wx^2}{2\xi\lambda_1}+f(y)x+z(y), \]
\[ t(x;y)=-\frac{\xi'-1}{\xi'}\,\frac{wy^2}{2\lambda_2}+f(x)y+z(x). \]
From the symmetry conditions (2) it follows that
\[ f(x)=f(y)=0. \]
Taking this into account, for \(x=0\) or \(y=0\) we obtain
\[ t(0;y)=\varphi(y); \qquad t(x;0)=\varphi(x). \]
Now
\[ t(x;y)=-\frac{wx^2}{2\xi\lambda_1}+t(0;y), \tag{5} \]
\[ t(x;y)=-\frac{\xi'-1}{\xi'}\,\frac{wy^2}{2\lambda_2}+t(x;0). \tag{6} \]
Putting respectively \(x=0\) and \(y=0\) in equations (4), we obtain the system of ordinary differential equations
\[ \frac{d^2 t(x;0)}{dx^2}\,\xi=-\frac{w}{\lambda_1}; \qquad \frac{d^2 t(0;y)}{dy^2}\,\frac{\xi'}{\xi'-1}=-\frac{w}{\lambda_2}. \]
Their solution is
\[ t(x;0)=-\frac{wx^2}{2\xi\lambda_1}+D_1; \qquad t(0;y)=-\frac{\xi'-1}{\xi'}\,\frac{wy^2}{2\lambda_2}+D_2. \]
Since at \(x=y=0\), \(t(0;0)=D\), it follows that \(D_1=D_2=D\), the temperature at the center of the body, and expressions (5) and (6) become identical:
\[ t(x;y)=D-\frac{wx^2}{2\xi\lambda_1} -\frac{wy^2}{2\lambda_2}\,\frac{\xi'-1}{\xi'}. \tag{7} \]
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 \(\xi'\) and \(D\) in the corresponding manner.
We take
\[ -\lambda_1 \frac{\partial t(R_1;y)}{\partial x} = \alpha_1\left[t(R_1;y)_{\mathrm{av}}-t_{f_1}\right], \]
\[ -\lambda_2 \frac{\partial t(x;R_2)}{\partial x} = \alpha_2\left[t(x;R_2)_{\mathrm{av}}-t_{f_2}\right], \tag{8} \]
where
\[ t(R_1;y)_{\mathrm{av}} = \frac{1}{R_2}\int_{0}^{R_2} t(R_1;y)\,dy = D-\frac{wR_1^2}{2\xi'\lambda_1} -\frac{\xi'-1}{\xi'}\frac{wR_2^2}{6\lambda_2}, \]
\[ t(x;R_2)_{\mathrm{av}} = \frac{1}{R_1}\int_{0}^{R_1} t(x;R_2)\,dx = D-\frac{wR_1^2}{6\xi'\lambda_1} -\frac{\xi'-1}{\xi'}\frac{wR_2^2}{2\lambda_2}. \]
Substituting the values \(t(R_1;y)_{\mathrm{av}}\) and \(t(x;R_2)_{\mathrm{av}}\) into (8), after transformations we obtain
\[ \xi' = \frac{ \dfrac{R_1}{4\alpha_1} + \dfrac{R_2}{4\alpha_2} + \dfrac{R_1^2}{3\lambda_1} + \dfrac{R_2^2}{3\lambda_2} }{ \dfrac{R_2}{4\alpha_2} + \dfrac{R_2^2}{3\lambda_2} - \dfrac{t_{f_1}-t_{f_2}}{W} }. \tag{9} \]
and the temperature of the center of the body will be
\[ D = \frac{wR_1}{4\xi'\alpha_1} + \frac{wR_1^2}{2\xi'\lambda_1} + \frac{\xi'-1}{\xi'}\frac{wR_2^2}{6\lambda_2} + t_{f_1}. \tag{10} \]
The temperature values found at various points of the cross section of the bar by formulas (7), (9), and (10) agree well with the data obtained by the method of elementary balances.
The comparison results are given in Table 1.
In the calculations it was assumed that:
\[ 2R_1=0.5\,\text{m}, \qquad 2R_2=1.0\,\text{m}, \qquad \lambda_1=1\,\frac{\text{kcal}}{\text{m}\,\text{h}\,^\circ\mathrm{C}}, \]
\[ \lambda_2=50\,\frac{\text{kcal}}{\text{m}\,\text{h}\,^\circ\mathrm{C}}, \qquad \alpha_1=1000\,\frac{\text{kcal}}{\text{m}^2\,\text{h}\,^\circ\mathrm{C}}, \]
\[ \alpha_2=10\,\frac{\text{kcal}}{\text{m}^2\,\text{h}\,^\circ\mathrm{C}}, \qquad t_{f_1}=20^\circ\mathrm{C}, \qquad t_{f_2}=40^\circ\mathrm{C}, \]
\[ w=500\,\frac{\text{kcal}}{\text{m}^3\,\text{h}}. \]
Table 1
| No. | Coordinates of points (m) | Coordinates of points (m) | Temperatures of points, °C | Temperatures of points, °C | Discrepancy, % |
|---|---|---|---|---|---|
| $x$ | $y$ | by formulas (7), (9), and (10) | by the method of elementary balances | ||
| 1 | 0 | 0 | 101.3 | 101.3 | 0 |
| 2 | 0.1 | 0 | 88.34 | 88.34 | 0 |
| 3 | 0.2 | 0 | 49.46 | 49.65 | 0.38 |
| 4 | 0 | 0.2 | 100.34 | 100.34 | 0 |
| 5 | 0.1 | 0.2 | 87.38 | 87.38 | 0 |
| 6 | 0.2 | 0.2 | 48.50 | 48.79 | 0.6 |
| 7 | 0 | 0.4 | 97.44 | 99.70 | 2.2 |
| 8 | 0.1 | 0.4 | 84.48 | 86.89 | 2.8 |
| 9 | 0.2 | 0.4 | 45.60 | 47.40 | 3.8 |
Conclusion
The approximate method presented for calculating the temperature field of an anisotropic bar under complex cooling conditions is distinguished by its simplicity and sufficient accuracy, except for points lying on the surface of the bar, i.e., at $x=R_1$ and $y=R_2$, and close to them.
References
- Boikov G. P. Heating of bodies of finite dimensions under the action of radiant heat. Izvestiya TPI, vol. 101, Tomsk, 1958.