Steady-State Heat Conduction under Boundary Conditions Expressed by the Stefan–Boltzmann Law and Internal Heat Generation

G. P. Boikov, Yu. A. Korolenko

(Presented by Professor, Doctor I. L. Kutyavin)

As an example, consider a body of infinite length and rectangular cross section, for which

\[ \frac{\partial^2 t(x;y)}{\partial x^2} + \frac{\partial^2 t(x;y)}{\partial y^2} + \frac{w}{\lambda} =0 \qquad \left( \begin{array}{c} -R_1 \leq x \leq R_1\\ -R_2 \leq y \leq R_2 \end{array} \right); \tag{1} \]

\[ \frac{\partial t(0;y)}{\partial x}=0; \qquad \frac{\partial t(x;0)}{\partial y}=0; \tag{2} \]

\[ -\lambda \frac{\partial t(R_1;y)}{\partial x} = \varepsilon_n C_0 \left[ \left(\frac{T(R_1;y)}{100}\right)^4 - \left(\frac{T_f}{100}\right)^4 \right], \]

\[ -\lambda \frac{\partial t(x;R_2)}{\partial y} = \varepsilon_n C_0 \left[ \left(\frac{T(R_2;x)}{100}\right)^4 - \left(\frac{T_f}{100}\right)^4 \right]. \tag{3} \]

Here \(w\), \(\dfrac{\text{kcal}}{m^3\text{hr}}\), is the internal heat generation, the same throughout the entire volume of the body,

\(\lambda\), \(\dfrac{\text{kcal}}{m\cdot\text{hr}\cdot\text{deg}}\), is the coefficient of thermal conductivity,

\(\varepsilon_n\) is the reduced degree of blackness of the system,

\(T_f\) is the temperature of the surrounding medium, \(R_1 \leq R_2\) are the dimensions of the body, and \(x\) and \(y\) are current coordinates.

Equation (1) can be given the form

\[ \frac{\partial^2 t(x;y)}{\partial x^2} \left[ 1+ \frac{ \dfrac{\partial^2 t(x;y)}{\partial y^2} }{ \dfrac{\partial^2 t(x;y)}{\partial x^2} } \right] = -\frac{w}{\lambda}, \]

\[ \frac{\partial^2 t(x;y)}{\partial y^2} \left[ 1+\frac{1}{\dfrac{\dfrac{\partial^2 t(x;y)}{\partial y^2}}{\dfrac{\partial^2 t(x;y)}{\partial x^2}}} \right] = -\frac{w}{\lambda}. \]

Assuming, according to [1], that the ratio of the components of the divergence of the temperature gradient is a constant quantity, we write the last relations in the form:

\[ \frac{\partial^2 t(x;y)}{\partial x^2}\,\xi=-\frac{w}{\lambda}; \qquad \frac{\partial^2 t(x;y)}{\partial y^2}\,\frac{\xi}{\xi-1}=-\frac{w}{\lambda}. \tag{4} \]

Hence

\[ t(x;y)=-\frac{w x^2}{2\xi\lambda}+f(y)x+\varphi(y), \]

\[ t(x;y)=-\frac{\xi-1}{\xi}\,\frac{w y^2}{2\lambda}+f(x)y+\varphi(x). \]

On the basis of the symmetry conditions (2),

\[ f(x)=f(y)=0 \]

and

\[ t(x;y)=-\frac{w x^2}{2\xi\lambda}+\varphi(y), \]

\[ t(x;y)=-\frac{\xi-1}{\xi}\,\frac{w y^2}{2\lambda}+\varphi(x). \]

For the central conditions (when \(x=0\) or \(y=0\)) we obtain, respectively,

\[ t(0;y)=\varphi(y);\qquad t(x;0)=\varphi(x) \]

and

\[ t(x;y)=-\frac{w x^2}{2\xi\lambda}+t(0;y), \tag{5} \]

\[ t(x;y)=-\frac{\xi-1}{\xi}\,\frac{w y^2}{2\lambda}+t(x;0). \tag{6} \]

Putting in (4), respectively, \(x=0\) and \(y=0\), we obtain a system of ordinary differential equations

\[ \frac{d^2 t(x;0)}{dx^2}\,\xi=-\frac{w}{\lambda}; \qquad \frac{d^2 t(0;y)}{dy^2}\,\frac{\xi}{\xi-1}=-\frac{w}{\lambda}. \]

Their solution is

\[ t(x;0)=-\frac{w x^2}{2\xi\lambda}+D_1; \qquad t(0;y)=-\frac{\xi-1}{\xi}\,\frac{w y^2}{2\lambda}+D_2. \]

Since for \(x=0\) and \(y=0\) we obtain \(t(0;0)=D\), it follows that \(D_1=D_2=D\), and expressions (5) and (6) become identical:

\[ t(x;y)=D-\frac{wx^2}{2\xi\lambda}-\frac{\xi-1}{\xi}\,\frac{wy^2}{2\lambda}. \tag{7} \]

Solution (7) satisfies the differential equation of heat conduction (1) and the symmetry conditions (2).

We determine the unknown constants \(\xi\) and \(D\) from equations (3), substituting in them

\[ T(R_1;y)=T_{\mathrm{cp}}(R_1;y)=\frac{1}{R_2}\int_0^{R_2}T(R_1;y)\,dy =D-\frac{wR_1^2}{2\xi\lambda}-\frac{\xi-1}{\xi}\,\frac{wR_2^2}{6\lambda}, \]

\[ T(x;R_2)=T_{\mathrm{cp}}(x;R_2)=\frac{1}{R_1}\int_0^{R_1}T(x;R_2)\,dx =D-\frac{\xi-1}{\xi}\,\frac{wR_2^2}{2\lambda}-\frac{wR_1^2}{6\xi\lambda}. \]

Substituting the corresponding temperature values into (3), we obtain:

\[ \frac{wR_1}{\xi\varepsilon_n C_0} +\left(\frac{T_f}{100}\right)^4 = \left( \frac{ D-\dfrac{wR_1^2}{2\xi\lambda} -\dfrac{\xi-1}{\xi}\,\dfrac{wR_2^2}{6\lambda} }{100} \right)^4, \]

\[ \frac{\xi-1}{\xi}\,\frac{wR_2}{\varepsilon_n C_0} +\left(\frac{T_f}{100}\right)^4 = \left( \frac{ D-\dfrac{\xi-1}{\xi}\,\dfrac{wR_2^2}{2\lambda} -\dfrac{wR_1^2}{6\xi\lambda} }{100} \right)^4. \tag{8} \]

Solving the system of equations (8) with respect to \(D\) and \(\xi\), we obtain:

\[ D=100\sqrt[4]{\frac{wR_1}{\xi\varepsilon_n C_0} +\left(\frac{T_f}{100}\right)^4} +\frac{wR_1^2}{2\xi\lambda} +\frac{\xi-1}{\xi}\,\frac{wR_2^2}{6\lambda}. \tag{9} \]

\[ \sqrt[4]{\frac{wR_1}{\xi\varepsilon_n C_0} +\left(\frac{T_f}{100}\right)^4} - \sqrt[4]{\frac{wR_2}{\varepsilon_n C_0}\,\frac{\xi-1}{\xi} +\left(\frac{T_f}{100}\right)^4} = -\frac{w}{300\lambda}\left(\frac{\xi-1}{\xi}R_2^2-\frac{R_1^2}{\xi}\right). \tag{10} \]

Equation (10) is used to determine \(\xi\), since all the other quantities entering into it are known from the initial conditions.

Then, from equation (9), \(D\)—the temperature at the center of the body—is calculated, and from equation (7) the entire temperature field of the body is calculated.

The temperature values found from formulas (7), (9), and (10) at various points of the cross section of the metal bar coincide with the data obtained by the method of elementary balances.

REFERENCES

1. G. P. Boikov. Heating of bodies of finite dimensions under the action of radiant heat. Izvestiya TPI, vol. 101, Tomsk, 1958.