On the Analytical Solution of the Plane Heat Conduction Problem for Porous Bodies
Abstract Generated abstract
This paper develops an analytical approximation for plane heat conduction in a thin porous body whose pores are assumed to have infinite thermal resistance, so that heat is conducted only through the solid skeleton. Using complex potentials and a superposition argument, the authors model the disturbance of heat flow around a pore and neighboring pores in an elementary cell, then relate the resulting temperature gradient to porosity. The derivation yields an expression for the effective thermal conductivity relative to the matrix conductivity, with a simplified approximate form \(\lambda_{\mathrm{eff}}=\lambda_m(1-\sqrt{m})(1+0.35m)\). The calculated dependence on porosity is reported to agree with results obtained from an electrical analogue model.
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ON THE ANALYTICAL SOLUTION OF THE PLANE HEAT-CONDUCTION PROBLEM FOR POROUS BODIES
G. M. Serykh, O. A. Ponomarev
Presented by Professor-Doctor G. I. Fuks
As was indicated in [5], the propagation of heat in a porous body is a very complicated process. A large part of the heat is transferred through the skeleton of the porous body.
In connection with this, the analytical investigation of the problem of heat propagation in a porous body acquires special importance under the condition that the pores have thermal resistance \(R = \infty\).
To find the dependence \(\lambda = \lambda(m)\), we shall apply the method of superposition. The essence of the method is as follows: we assume that in each selected element of space there exists a certain function independent of external actions. The influence from other elements of space (external action) is effected by the propagation of the functions of these elements to the selected element. The resulting composite function is then equal to the sum of the functions of the selected element and of the disturbances.
FORMULATION OF THE PROBLEM
We have a porous body in the form of an unbounded plate of very small thickness. The pores conduct no heat at all. Heat propagation through the thickness of the plate is absent.
Let us first cut out only one pore on the plate. Place the coordinate axes at the center of the pore (Fig. 1). For a certain temperature gradient, the complex potential of the flow around the pore by the heat flux has the form [1]:
\[ w = v\left(z + \frac{R^2}{z}\right), \tag{1} \]
where: \(v\) is a certain constant, \(R\) is the radius of the pore.
Let us fill the plane with pores so that a certain porosity \(m\) is created. In the porous field, select an elementary cell (Fig. 2) with boundary \(\Gamma\). Consider the complex potential in zone \(A\) in the immediate vicinity of the pore.
According to the superposition principle adopted above:
\[ w = w_n + w_b, \tag{2} \]
where: \(w_n\) is the potential of the pore, \(w_b\) is the potential of the disturbances.
\[ w_b = w_1 + w_2 - w_3 - w_4 - w_5 + w_6 + w_7 \tag{3} \]
Fig. 1.
Fig. 2.
We have:
\[ w_n=v\left(z+\frac{R^2}{z}\right); \tag{4} \]
\[ \begin{aligned} w_1&=v\left(re^{i135^\circ}+z+\frac{R^2}{re^{i135^\circ}+z}\right);\\ w_2&=v\left(-r+z+\frac{R^2}{-r+z}\right);\\ w_3&=v\left(re^{i225^\circ}+z+\frac{R^2}{re^{i225^\circ}+z}\right);\\ w_4&=v\left(re^{i270^\circ}+z+\frac{R^2}{re^{i270^\circ}+z}\right);\\ w_5&=v\left(re^{-i45^\circ}+z+\frac{R^2}{re^{-i45^\circ}+z}\right);\\ w_6&=v\left(r+z+\frac{R^2}{r+z}\right);\\ w_7&=v\left(re^{i45^\circ}+z+\frac{R^2}{re^{i45^\circ}+z}\right). \end{aligned} \tag{5} \]
Substituting (4) and (5) into (2) and reducing like terms, we obtain:
\[ \begin{aligned} w=v\Bigg(&2z+\frac{R^2}{z}+\frac{4r}{\sqrt{2}}\,i +\frac{R^2}{-\dfrac{r}{\sqrt{2}}+\dfrac{r}{\sqrt{2}}\,i+z} -\frac{R^2}{-r+z}\\ &-\frac{R^2}{-\dfrac{r}{\sqrt{2}}-\dfrac{r}{\sqrt{2}}\,i+z} -\frac{R^2}{-ri+z} -\frac{R^2}{\dfrac{r}{\sqrt{2}}-\dfrac{r}{\sqrt{2}}\,i+z}\\ &+\frac{R^2}{r+z} +\frac{R^2}{\dfrac{r}{\sqrt{2}}+\dfrac{r}{\sqrt{2}}\,i+z} \Bigg). \end{aligned} \tag{7} \]
Separate the real and imaginary parts:
\[ \begin{aligned} \varphi = v \Bigg\{&2x+\frac{R^2x}{x^2+y^2} +\frac{R^2\left(x-\frac{r}{\sqrt{2}}\right)} {\left(x-\frac{r}{\sqrt{2}}\right)^2+\left(\frac{r}{\sqrt{2}}+y\right)^2} +\frac{R^2(x-r)}{(x-r)^2+y^2} \\ &+\frac{R^2\left(\frac{r}{\sqrt{2}}-x\right)} {\left(\frac{r}{\sqrt{2}}-x\right)^2+\left(\frac{r}{\sqrt{2}}-y\right)^2} -\frac{R^2x}{x^2+(y-r)^2} -\frac{R^2\left(\frac{r}{\sqrt{2}}+x\right)} {\left(\frac{r}{\sqrt{2}}+x\right)^2+\left(y-\frac{r}{\sqrt{2}}\right)^2} \\ &+\frac{R^2(r+x)}{(r+x)^2+y^2} +\frac{R^2\left(\frac{r}{\sqrt{2}}+x\right)} {\left(\frac{r}{\sqrt{2}}+x\right)^2+\left(\frac{r}{\sqrt{2}}+y\right)^2} \Bigg\}. \end{aligned} \tag{8} \]
The function \(\varphi\) is identified with the temperature curve \(\tau \equiv \varphi\).
The temperature gradient in the section \(AA\) (Fig. 2) is expressed by the formula:
\[ \begin{aligned} \left(\frac{\partial \tau}{\partial x}\right)_{x=0} = v \Bigg\{& 2+\frac{R^2}{y^2} +\frac{2R^2} {\left(\frac{r}{\sqrt{2}}\right)^2+\left(\frac{r}{\sqrt{2}}+y\right)^2} +\frac{2R^2}{r^2+y^2} \\ &-\frac{2R^2r^2} {\left[\frac{r^2}{2}+\left(\frac{r}{\sqrt{2}}+y\right)^2\right]^2} -\frac{2\cdot 2 R^2r^2}{(r^2+y^2)^3} +\frac{-R^2} {\frac{r^2}{2}+\left(\frac{r}{\sqrt{2}}-y\right)^2} \\ &+\frac{2R^2r^2} {\left[\frac{r^2}{2}+\left(\frac{r}{\sqrt{2}}-y\right)^2\right]^2} -\frac{R^2}{(x-r)^2} -\frac{R^2} {\frac{r^2}{2}+\left(y-\frac{r}{\sqrt{2}}\right)^2} \Bigg\}. \end{aligned} \tag{9} \]
With a sufficient degree of accuracy one may take:
\[ \begin{aligned} \left(\frac{\partial \tau}{\partial x}\right)_{AA} \simeq \left(\frac{\partial \tau}{\partial x}\right)_{\substack{x=0\\ y=R}} = v \Bigg(& 3+\frac{2} {\frac{r^2}{R^2}+2\frac{r}{R}\frac{1}{\sqrt{2}}+1} \\ &-\frac{2R^2r^2} {\left(r^2+\frac{2}{\sqrt{2}}rR+R^2\right)^2} +\frac{2R^2}{r^2+R^2} -\frac{4R^2r^2}{r^4+2r^2R^2+R^4} \\ &-\frac{2R^2} {r^2-\frac{2r}{\sqrt{2}}R+R^2} +\frac{2R^2r^2} {\left(r^2-\frac{2}{\sqrt{2}}rR+R^2\right)^2} -\frac{R^2}{R^2-2Rr+r^2} \Bigg) \end{aligned} \tag{10} \]
Let us determine the quantity \(\dfrac{r^2}{R^2}\) through the porosity.
Since the zone of action of a pore includes 8 semicircles and the pore itself, then:
\[ m=5\frac{\pi R^2}{\pi r^2}=a\frac{R^2}{r^2}; \tag{11} \]
\[ \frac{r^2}{R^2}=\frac{a}{m}; \qquad \frac{r}{R}=\sqrt{\frac{a}{m}} . \tag{12} \]
Substituting the values (11) and (12) into (10), we have
\[ \left(\frac{\partial \tau}{\partial x}\right)_{AA} = v\left( 3+ \frac{2}{\dfrac{a}{m}+\sqrt{2}\sqrt{\dfrac{a}{m}}+1} - \frac{2}{\dfrac{m}{a}+4+\dfrac{x}{m}+4\sqrt{\dfrac{m}{2a}}+4\sqrt{\dfrac{2a}{m}}} + \frac{2}{\dfrac{a}{m}-1} - \frac{4}{\dfrac{x}{m}+2+\dfrac{m}{a}} \right. \]
\[ \left. - \frac{2}{\dfrac{a}{m}-\sqrt{\dfrac{2a}{m}}+1} + \frac{2}{\dfrac{a}{m}+4+\dfrac{m}{a}-4\sqrt{\dfrac{m}{2a}}-4\sqrt{\dfrac{2a}{m}}} - \frac{1}{\dfrac{a}{m}-2\sqrt{\dfrac{a}{m}}+1} \right) = v\bigl(3+m\cdot f(m)\bigr). \tag{13} \]
where \(f(m)\ne\infty\) for \(m=0\).
Let us determine the amount of heat transferred at a unit gradient for the fictitious body:
\[ q=S\lambda_{m}\left(\frac{\partial \tau}{\partial x}\right)_{AA} = \lambda_{\mathrm{eff}}\cdot S_{\mathrm{total}}\cdot 1 \tag{15} \]
The quantity \(v\) is determined from the condition:
\[ \frac{\lambda_{\mathrm{eff}}}{\lambda_{m}}=1 \qquad \text{for } m=0. \tag{16} \]
Hence
\[ v=\frac{1}{3}. \tag{17} \]
Then
\[ \frac{\lambda_{\mathrm{eff}}}{\lambda_{m}} = (1-\sqrt{m})\left[1+\frac{1}{3}\,m\cdot f(m)\right]. \tag{18} \]
Let us evaluate the function for \(a\approx 5\), \(m\approx 0.8\), since one pore is then screened from the influence of the main pore, i.e., is located in the shadow zone (Fig. 2):
\[ \frac{1}{3}f(m) = \frac{1}{3}\left( \frac{2}{a+m\sqrt{\dfrac{2a}{m}}+m} - \frac{2}{\dfrac{m^2}{a}+4m+a+4\sqrt{\dfrac{m}{2a}}\,m+4\sqrt{\dfrac{a}{2m}}\,m} \right. \]
\[ \left. + \frac{2}{a+m} - \frac{4}{a+2m+\dfrac{m^2}{a}} - \frac{2}{a-2m\sqrt{\dfrac{a}{2m}}+m} \right. \]
\[ \left. + \frac{2}{a+4m+\dfrac{m^2}{a}-4m\sqrt{\dfrac{m}{2a}}-4m\sqrt{\dfrac{a}{2m}}} - \frac{1}{a-2m\sqrt{\dfrac{a}{m}}+m} \right). \tag{19} \]
\[ \frac{1}{3} f(m)\cdot m=\frac{m}{3}(0.4-0.2+0.4-0.5-0.6+2-0.5)= \]
\[ \frac{1}{3} f(m)\cdot m \approx 0.35\,m . \tag{20} \]
Then expression (18), for approximate calculation, takes the form:
\[ \lambda_{\mathrm{eff}}=\lambda_m(1-\sqrt{m})(1+0.35m). \tag{21} \]
In Fig. 3 the dependence \(\dfrac{\lambda_{\mathrm{eff}}}{\lambda_m}=f(m)\) is given. Curve 2 was obtained experimentally on an electrical model. Curve 1 was obtained from formula (21). As can be seen from Fig. 3, the approximate formula gives coincident results.
REFERENCES
- N. E. Kochin, I. A. Kibel, N. V. Roze, Theoretical Hydromechanics, Parts I–II, GITTL, 1955.
- V. I. Smirnov, A Course of Higher Mathematics, Vol. III, Part II, GITTL, 1953.
- B. A. Fuks and B. V. Shabat, Functions of a Complex Variable, GITTL, 1949.
- M. A. Mikheev, Fundamentals of Heat Transfer, GEI, 1949.
- G. M. Serykh, “On the question of the thermal conductivity of porous materials,” Izvestiya TPI, Vol. 101.