On Determining the Temperature on the Surfaces of an Infinite Plate during Radiation Heating
Abstract Generated abstract
This paper treats unsteady heating of an infinite plane plate by thermal radiation, where nonlinear radiative boundary conditions make a closed explicit solution difficult. The authors reduce the heat conduction problem to determining the time dependent temperatures of the two plate surfaces, approximate the temperature distribution by retaining the first term of the series solution, and derive integral and functional relations for symmetric and asymmetric heating in terms of the Fourier and Kirpichev criteria. The resulting formulas allow graphical or rapid calculation of surface temperatures and, subsequently, the full temperature field, with comparisons to numerical integration indicating the practical usefulness of the approximation.
Full Text
ON THE DETERMINATION OF TEMPERATURE ON THE SURFACES OF AN INFINITE PLATE UNDER RADIATIVE HEATING
Yu. V. Vidin, G. P. Boikov
(Submitted by Prof. Doctor of Technical Sciences G. I. Fuks)
Heating of a plane wall under the action of radiation for a whole range of practical conditions is described by the system of equations:
\[ \frac{\partial T(x,\tau)}{\partial \tau} = a \frac{\partial^2 T(x,\tau)}{\partial x^2}; \tag{1} \]
\[ \lambda \frac{\partial T(+R,\tau)}{\partial x} = \varepsilon_{n1} C_0 \left[ \left(\frac{T_{C1}}{100}\right)^4 - \left(\frac{T(+R,\tau)}{100}\right)^4 \right], \tag{2} \]
\[ -\lambda \frac{\partial T(-R,\tau)}{\partial x} = \varepsilon_{n2} C_0 \left[ \left(\frac{T_{C2}}{100}\right)^4 - \left(\frac{T(-R,\tau)}{100}\right)^4 \right]; \tag{3} \]
\[ T(x,0)=T_0=\mathrm{const}. \tag{4} \]
At present it is not possible to present the solution of this system in explicit form because of the complexity of the boundary conditions (2) and (3). Nevertheless, mathematically the problem is solved if, in some way, the laws of variation of the temperatures \(\bigl(T(+R,\tau)\) and \(T(-R,\tau)\bigr)\) on the surfaces of the plate are known.
Indeed, solving the system of equations:
\[ \frac{\partial T(x,\tau)}{\partial \tau} = a \frac{\partial^2 T(x,\tau)}{\partial x^2}; \tag{1′} \]
\[ T(+R,\tau)=T_1(\tau); \tag{2′} \]
\[ T(-R,\tau)=T_2(\tau); \tag{3′} \]
\[ T(x,0)=T_0=\mathrm{const}, \tag{4′} \]
one can obtain the temperature distribution over the cross section [1], [2]:
\[ T(x,\tau) = T_1(\tau) - \frac{R-x}{2R}\,[T_1(\tau)-T_2(\tau)] - \]
\[ -\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{1}{n} \exp\left[-\left(\frac{\pi n}{2}\right)^2\frac{a\tau}{R^2}\right] \sin\left(\pi n\frac{R-x}{2R}\right) \int_0^\tau \exp\left[\left(\frac{\pi n}{2}\right)^2\frac{a\vartheta}{R^2}\right] \times \]
\[ \times \left\{\frac{\partial [T_1(\vartheta)]}{\partial \vartheta} +(-1)^{n+1}\frac{\partial [T_2(\vartheta)]}{\partial \vartheta}\right\}\,d\vartheta . \tag{5} \]
Consequently, the problem of finding the integral of system (1)–(4) reduces to finding the temperature of the wall surfaces. The latter can be determined if we restrict ourselves to cases for which it is sufficient to use only one term of the infinite series. Under this assumption, relation (5) takes the form
\[ \begin{aligned} T(x,\tau)=&\,T_1(\tau)-\frac{R-x}{2R}\,[T_1(\tau)-T_2(\tau)] \\ &-\frac{2}{\pi}\exp\left[-\frac{\pi^2}{4}\frac{a\tau}{R^2}\right] \sin\left(\pi\frac{R-x}{2R}\right) \int_0^\tau \exp\left[\frac{\pi^2}{4}\frac{a\vartheta}{R^2}\right]\times \\ &\times \left\{\frac{\partial [T_1(\vartheta)]}{\partial \vartheta} +\frac{\partial [T_2(\vartheta)]}{\partial \vartheta}\right\}\,d\vartheta . \end{aligned} \tag{5'} \]
Differentiating this expression with respect to the coordinate \(x\) and substituting the result into the boundary conditions (2) and (3), we have:
\[ \begin{aligned} &\frac{1}{2R}[T_1(\tau)-T_2(\tau)] +\frac{1}{R}\exp\left[-\frac{\pi^2}{4}\frac{a\tau}{R^2}\right] \int_0^\tau \exp\left[\frac{\pi^2}{4}\frac{a\vartheta}{R^2}\right]\times \\ &\times \left\{\frac{\partial [T_1(\vartheta)]}{\partial \vartheta} +\frac{\partial [T_2(\vartheta)]}{\partial \vartheta}\right\}\,d\vartheta = \frac{\varepsilon_{n1}\cdot C_0}{\lambda} \left[\left(\frac{T_{c1}}{100}\right)^4 -\left(\frac{T_1(\tau)}{100}\right)^4\right]; \end{aligned} \tag{6} \]
\[ \begin{aligned} &-\frac{1}{2R}[T_1(\tau)-T_2(\tau)] +\frac{1}{R}\exp\left[-\frac{\pi^2}{4}\frac{a\tau}{R^2}\right] \int_0^\tau \exp\left[\frac{\pi^2}{4}\frac{a\vartheta}{R^2}\right]\times \\ &\times \left\{\frac{\partial [T_1(\vartheta)]}{\partial \vartheta} +\frac{\partial [T_2(\vartheta)]}{\partial \vartheta}\right\}\,d\vartheta = \frac{\varepsilon_{n2}\cdot C_0}{\lambda} \left[\left(\frac{T_{c2}}{100}\right)^4 -\left(\frac{T_2(\tau)}{100}\right)^4\right]. \end{aligned} \tag{7} \]
According to (6) and (7),
\[ \begin{aligned} &\int_0^\tau \exp\left[\frac{\pi^2}{4}\frac{a\vartheta}{R^2}\right] \left\{\frac{\partial [T_1(\vartheta)]}{\partial \vartheta} +\frac{\partial [T_2(\vartheta)]}{\partial \vartheta}\right\}\,d\vartheta = \\ &=\frac{R}{2\lambda}\exp\left[\frac{\pi^2}{4}\frac{a\tau}{R^2}\right]\times \left\{ \varepsilon_{n1}\cdot C_0 \left[\left(\frac{T_{c1}}{100}\right)^4 -\left(\frac{T_1(\tau)}{100}\right)^4\right]\right. \\ &\left.\qquad\qquad\qquad +\varepsilon_{n2}\cdot C_0 \left[\left(\frac{T_{c2}}{100}\right)^4 -\left(\frac{T_2(\tau)}{100}\right)^4\right] \right\}. \end{aligned} \tag{8} \]
Substitution of this relation into (5′) gives
\[ \begin{aligned} T(x,\tau)=&\,T_1(\tau)-\frac{R-x}{2R}[T_1(\tau)-T_2(\tau)] \\ &-\frac{1}{\pi}\frac{R}{\lambda} \sin\left(\pi\frac{R-x}{2R}\right)\times \left\{ \varepsilon_{n1}\cdot C_0 \left[\left(\frac{T_{c1}}{100}\right)^4 -\left(\frac{T_1(\tau)}{100}\right)^4\right]\right. \\ &\left.\qquad\qquad\qquad +\varepsilon_{n2}\cdot C_0 \left[\left(\frac{T_{c2}}{100}\right)^4 -\left(\frac{T_2(\tau)}{100}\right)^4\right] \right\}. \end{aligned} \tag{9} \]
or
\[ \begin{aligned} T(x,\tau)=T_1(\tau)-\frac{R-x}{2R}\,[T_1(\tau)-T_2(\tau)]&\\ -\frac{1}{\pi}\sin\left(\pi\frac{R-x}{2R}\right) \Bigg\{\mathrm{Ki}_1(T_{c1}-T_0)\frac{T_{c1}^{4}-T_1^{4}(\tau)}{T_{c1}^{4}-T_0^{4}}&\\ +\mathrm{Ki}_2(T_{c2}-T_0)\frac{T_{c2}^{4}-T_2^{4}(\tau)}{T_{c2}^{4}-T_0^{4}}\Bigg\},& \end{aligned} \tag{9'} \]
where
\[ \mathrm{Ki}_1= \frac{\varepsilon_{n1}\cdot C_0\left[\left(\dfrac{T_{c1}}{100}\right)^4-\left(\dfrac{T_0}{100}\right)^4\right]\cdot R} {\lambda(T_{c1}-T_0)}, \]
\[ \mathrm{Ki}_2= \frac{\varepsilon_{n2}\cdot C_0\left[\left(\dfrac{T_{c2}}{100}\right)^4-\left(\dfrac{T_0}{100}\right)^4\right]\cdot R} {\lambda(T_{c2}-T_0)} \]
are the Kirpichev criteria [3], respectively for the left and right surfaces of the plate.
From the integral equations (6) and (7) one can derive the functional relations:
\[ T_1(\tau)-T_2(\tau)= \mathrm{Ki}_1(T_{c1}-T_0)\frac{T_{c1}^{4}-T_1^{4}(\tau)}{T_{c1}^{4}-T_0^{4}} -\mathrm{Ki}_2(T_{c2}-T_0)\frac{T_{c2}^{4}-T_2^{4}(\tau)}{T_{c2}^{4}-T_0^{4}}; \tag{10} \]
\[ \begin{aligned} &-\ln \frac{ \mathrm{Ki}_1(T_{c1}-T_0)\dfrac{T_{c1}^{4}-T_1^{4}(\tau)}{T_{c1}^{4}-T_0^{4}} +\mathrm{Ki}_2(T_{c2}-T_0)\dfrac{T_{c2}^{4}-T_2^{4}(\tau)}{T_{c2}^{4}-T_0^{4}} }{ \mathrm{Ki}_1(T_{c1}-T_0)+\mathrm{Ki}_2(T_{c2}-T_0) } \\ &\quad -4\int_{T_0}^{T_1} \frac{ 1+4\mathrm{Ki}_1(T_{c1}-T_0)\dfrac{[T_1^*(\tau)]^3}{T_{c1}^{4}-T_0^{4}} }{ f(T_1^*)-T_1^*+2\mathrm{Ki}_1(T_{c1}-T_0)\dfrac{T_{c1}^{4}-[T_1^*(\tau)]^4}{T_{c1}^{4}-T_0^{4}} }\,dT_1^* = -\frac{\pi^2}{4}\mathrm{Fo}_0 . \end{aligned} \tag{11} \]
These equations make it possible to find the surface temperatures
\[ T(+R,\tau)=T_1(\tau)\quad\text{and}\quad T(-R,\tau)=T_2(\tau)=f(T_1). \]
It should be noted that the graphical determination of the surface temperature from expression (11) does not create any considerable difficulties, since the Fourier criterion entering the right-hand side of the equation is not connected by any functional dependence.
In the case when the temperatures on the surfaces of the plate vary according to identical laws \(T_1(\tau)=T_2(\tau)=T(\tau)\) (i.e., when symmetric heating conditions hold), relations (9′) and (11) take the form
\[ T(x,\tau)=T(\tau)-\frac{2}{\pi}\sin\left(\pi\frac{R-x}{2R}\right)\mathrm{Ki}(T_c-T_0)\times \]
\[ \times\frac{T_c^4-T^4(\tau)}{T_c^4-T_0^4}; \tag{9''} \]
\[ \ln \left(\frac{T_c^4 - T^4(\tau)}{T_c^4 - T_0^4}\right) - \frac{T_c^4 - T_0^4}{\operatorname{Ki}(T_c - T_0)\cdot T_c^3} \times \]
\[ \times \left[ \ln \frac{[T_c + T(\tau)]\cdot [T_c - T_0]} {[T_c - T(\tau)]\cdot [T_c + T_0]} + 2\left( \arctg \frac{T(\tau)}{T_c} - \arctg \frac{T_0}{T_c} \right) \right] = -\frac{\pi^2}{4}Fo_0 . \tag{11′} \]
Figure 1 gives the curves of variation of the relative temperature of the surfaces, as well as the temperature difference across the thickness of the plate (dashed lines), constructed according to (9″) and (11′). For comparison, curves obtained by calculation under the same conditions by the method of numerical integration [4] are also plotted here (solid lines).
Fig. 1. Variation of the relative temperature of the surface and of the temperature difference over the cross section under symmetric heating by radiation.
Figure 2 shows the temperature-field pattern in an infinite plate for \(\operatorname{Ki}_1 = 1.0\); \(\operatorname{Ki}_2 = 0.1\); \(T_{c1} = 1700^\circ K\); \(T_{c2} = 1190^\circ K\); \(T_0 = 300^\circ K\), obtained according to (9′), (10), and (11) (dashed lines). The solid lines show the temperature distribution found by the method of numerical integration for the same initial data.
Thus, owing to the structural form of the regularities (11) and (11′), it becomes possible to carry out rapid calculations of the surface temperature under symmetric and asymmetric heating. This, in turn, makes it possible to determine the entire temperature field.
Fig. 2. Picture of the temperature field in a flat wall under asymmetric radiative heating.
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