ON THE SOLUTION OF TWO-DIMENSIONAL HEAT-CONDUCTION PROBLEMS

G. P. Boikov

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

The proposed method for solving two-dimensional problems can most conveniently be illustrated by the example of unsteady heat propagation under boundary conditions of the first and third kinds, since the final result is easily compared with the expression obtained on the basis of multiplying temperature criteria.

It is known that the process of heat propagation in an infinite bar of rectangular or square cross section is described by the heat-conduction differential equation of the form:

\[ a\left( \frac{\partial^2 t(x;y;\tau)}{\partial x^2} + \frac{\partial^2 t(x;y;\tau)}{\partial y^2} \right) = \frac{\partial t(x;y;\tau)}{\partial \tau}. \tag{1} \]

According to [2],

\[ a \frac{\partial^2 t(x;y;\tau)}{\partial x^2}\cdot \xi = \frac{\partial t(x;y;\tau)}{\partial \tau}; \]

\[ a \frac{\partial^2 t(x;y;\tau)}{\partial y^2}\cdot \frac{\xi}{\xi-1} = \frac{\partial t(x;y;\tau)}{\partial \tau}. \]

After partial integration, as a result of which the symmetry conditions are satisfied, we obtain

\[ v(x;y;\tau)=\varphi_1(y)\cos(K_n x)\cdot e^{-\xi aK_n^2\tau}; \]

\[ v(x;y;\tau)=\varphi_2(x)\cos(K_m y)\cdot e^{-\frac{\xi}{\xi-1}aK_m^2\tau}, \]

where \(v\) is the excess temperature of the body.

For \(x=0\) and, correspondingly, \(y=0\), we have

\[ v(0;y;\tau)=\varphi_1(y)\cdot e^{-\xi aK_n^2\tau}; \]

\[ v(x;0;\tau)=\varphi_2(x)\cdot e^{-\frac{\xi}{\xi-1}aK_m^2\tau}. \]

Consequently,

\[ v(x;\ y;\ \tau)=\varphi(0;\ y;\ \tau)\cdot \cos(K_n x); \]

\[ v(x;\ y;\ \tau)=\varphi(x;\ 0;\ \tau)\cdot \cos(K_m y). \]

The values \(v(0;\ y;\ \tau)\) and \(v(x;\ 0;\ \tau)\) can be found by solving the equations

\[ a\,\frac{\partial^2 v(x;\ 0;\ \tau)}{\partial x^2}\,\xi = \frac{\partial v(x;\ 0;\ \tau)}{\partial \tau}; \]

\[ a\,\frac{\partial^2 v(0;\ y;\ \tau)}{\partial y^2}\cdot \frac{\xi}{\xi-1} = \frac{\partial v(0;\ y;\ \tau)}{\partial \tau}. \]

From these equations it follows that

\[ v(x;\ 0;\ \tau)=B_1\cdot \cos(K_n x)\cdot e^{-\xi aK_n^2\tau}; \]

\[ v(0;\ y;\ \tau)=B_2\cdot \cos(K_m y)\cdot e^{-\frac{\xi}{\xi-1}aK_m^2\tau}. \]

Then

\[ v(x;\ y;\ \tau)=B_2\cdot \cos(K_m y)\cdot \cos(K_n x)\cdot e^{-\frac{\xi}{\xi-1}aK_m^2\tau}; \]

\[ v(x;\ y;\ \tau)=B_1\cdot \cos(K_n x)\cdot \cos(K_m y)\cdot e^{-\xi aK_n^2\tau}. \]

For \(x=y=0\) the temperature of the center \(v(0;\ 0;\ \tau)\) is obtained; therefore

\[ B_2\cdot e^{-\frac{\xi}{\xi-1}aK_m^2\tau} = B_1\cdot e^{-\xi aK_n^2\tau}. \]

Thus, the general solution is obtained in the form

\[ v(x;\ y;\ \tau)=B\cdot \cos(K_n x)\cdot \cos(K_m y)\cdot e^{-\xi aK_n^2\tau}. \tag{2} \]

We shall now find the constant quantity \(\xi\) so that condition (1) is satisfied. Substitution of the general solution (2) into the differential heat-conduction equation (1) gives

\[ \xi=1+\frac{K_m^2}{K_n^2}; \]

\[ v(x;\ y;\ \tau)=B\cdot \cos(K_n x)\cdot \cos(K_m y)\cdot e^{-a(K_n^2+K_m^2)\tau}. \tag{3} \]

According to (3) and the boundary conditions of the third kind, we find

\[ \operatorname{ctg}\mu_n=\frac{\psi_n}{Bi_n}; \qquad \operatorname{ctg}\mu_m=\frac{\psi_m}{Bi_m}, \]

where

\[ \mu_n=K_n R_1;\qquad \mu_m=K_m R_2. \]

This means that the solution (3) must be represented by the infinite series

\[ \vartheta(x;\,y;\,\tau)= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} B\cos\left(\mu_n\frac{x}{R_1}\right)\cdot \cos\left(\mu_m\frac{y}{R_2}\right)\, e^{-\mu_n^2\frac{a\tau}{R_1^2}}\cdot e^{-\mu_m^2\frac{a\tau}{R_2^2}} . \]

To determine the constant \(B\), we use the initial conditions

\[ \vartheta(x;\,y;\,0)= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} B\cdot \cos\left(\mu_n\frac{x}{R_1}\right)\cdot \cos\left(\mu_m\frac{y}{R_2}\right)=\vartheta_0, \]

having first integrated them with respect to \(y\) and with respect to \(x\):

\[ \int_{-R_1}^{+R_1}\int_{-R_2}^{+R_2} \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} B\cos\left(\mu_n\frac{x}{R_1}\right)\cdot \cos\left(\mu_m\frac{y}{R_2}\right)\cdot \cos\left(\mu_{n^*}\frac{x}{R_1}\right) \]

\[ {}\times \cos\left(\mu_{m^*}\frac{y}{R_2}\right)\,dy\,dx = \vartheta_0 \int_{-R_1}^{+R_1}\int_{-R_2}^{+R_2} \cos\left(\mu_{n^*}\frac{x}{R_1}\right)\cdot \cos\left(\mu_{m^*}\frac{y}{R_2}\right)\,dy\,dx . \]

According to the orthogonality conditions, all terms of the series, except the case \(\mu_n=\mu_{n^*}\) and \(\mu_m=\mu_{m^*}\), are equal to zero, and therefore

\[ \int_{-R_1}^{+R_1}\int_{-R_2}^{+R_2} B\cdot \cos^2\left(\mu_n\frac{x}{R_1}\right)\cdot \cos^2\left(\mu_m\frac{y}{R_2}\right)\,dy\,dx = \]

\[ \vartheta_0 \int_{-R_1}^{+R_1}\int_{-R_2}^{+R_2} \cos\left(\mu_n\frac{x}{R_1}\right)\cdot \cos\left(\mu_m\frac{y}{R_2}\right)\,dy\,dx . \]

Hence

\[ B=\vartheta_0 \frac{2\sin\mu_n}{\mu_n+\sin\mu_n\cdot \cos\mu_n} \cdot \frac{2\sin\mu_m}{\mu_m+\sin\mu_m\cdot \cos\mu_m} = \vartheta_0 A_n\cdot A_m . \]

The final solution of the problem takes the form

\[ \vartheta(x;\,y;\,\tau)= \vartheta_0\sum_{n=1}^{\infty} A_n\cdot \cos\left(\mu_n\frac{x}{R_1}\right)\cdot e^{-\mu_n^2\frac{a\tau}{R_1^2}} \times \]

\[ {}\times \sum_{m=1}^{\infty} A_m\cdot \cos\left(\mu_m\frac{y}{R_2}\right)\cdot e^{-\mu_m^2\frac{a\tau}{R_2^2}} . \]

Exactly the same expression is obtained when multiplying the temperature criteria of two infinite plates of thickness \(2R_1\) and \(2R_2\) [1]. Verification has shown that the method presented can be satisfactorily used for solving problems of steady-state heat conduction with internal heat generation.

References

1. Lykov A. V. Theory of Heat Conduction. GITL, Moscow, 1952.
2. Boikov G. P. Heating of bodies of finite dimensions under the action of radiant heat. “Izv. GPI,” vol. 101, 1958.