On Determining the Irradiation Coefficient of an Elementary Cylindrical Ring on the Base of a Cylinder
Abstract Generated abstract
This paper addresses the calculation of the irradiation coefficient between an elementary cylindrical ring on the inner surface of a cylindrical cavity and the cylinder base, a geometric factor needed for radiant heat exchange models in elongated thermal apparatuses. Starting from the general double surface integral for view factors, the authors derive the relevant geometry, show that the exact subsequent integration is not expressible in elementary functions, and introduce an approximation based on the behavior of a coefficient that approaches unity for sufficiently large height to radius ratios. The resulting formula is checked against a derivation using ray algebra and the known irradiation coefficient between parallel circular surfaces. Both approaches lead to the same expression, with limiting behavior consistent with the physical interpretation of the coefficient.
Full Text
On the Problem of Determining the Irradiation Coefficient of an Elementary Cylindrical Ring on the Base of a Cylinder
Yu. M. Trifonov, Yu. A. Trifonova
(Presented by the Department of Machines and Apparatus of Chemical Production)
The study of the geometric characteristics of radiant heat exchange is necessary for an analytical description of the radiation field created by a given body, as well as for a mathematical description of the dynamic properties of thermal objects. Knowledge of the equations of dynamics, in turn, makes it possible to select automatic control systems at the design stage of the given object.
In a number of cases, for example, electric furnaces for continuous drying of wire insulation in the cable industry, heat-engineering units may be represented in the form of a conditionally closed cylindrical cavity whose length exceeds its hydraulic diameter by ten or more times.
Let us consider a cylindrical cavity with cross-sectional diameter \(D = 2R_0\), filled with a diathermic medium (Fig. 1). The inner surface of the cavity is assumed to be gray, isothermal, and optically homogeneous.
When the general formula for determining irradiation coefficients [1] is applied to the case of an elementary cylindrical ring and the base of a cylinder, we have
\[ \varphi_{1,2}=\frac{1}{F_1}\iint_{F_1 F_2} \frac{\cos\beta_N\cdot\cos\beta_M}{\pi\cdot s^2}\cdot dF_N\cdot dF_M, \tag{1} \]
where
\(dF_N\) and \(dF_M\) are elementary areas, respectively, on the cylindrical ring and on the base of the cylinder;
\(\beta_N\) and \(\beta_M\) are the angles formed by the direction of radiation with the normals to the areas \(dF_N\) and \(dF_M\);
\(s\) is the distance between the centers of the areas \(dF_N\) and \(dF_M\);
\(F_1\) and \(F_2\) are the areas, respectively, of the elementary cylindrical ring and of the base of the cylinder.
With the notation as in Fig. 1:
\[ \left. \begin{aligned} dF_N &= R_0\cdot dh\cdot d\alpha_N;\qquad dF_M = r\cdot dr\cdot d\alpha_M;\\[4pt] \cos\beta_N &= \frac{R_0-r\cdot\cos(\alpha_N-\alpha_M)}{s}; \qquad \cos\beta_M=\frac{h}{s}; \end{aligned} \right\} \tag{2} \]
From the geometric relations one may derive
\[ s=\sqrt{R_0^2+r^2+h^2-2R_0\cdot r\cdot\cos(\alpha_N-\alpha_M)}. \tag{3} \]
After substituting equations (2) and (3) into (1), we obtain:
\[ \varphi_{1,2}= \frac{1}{\pi F_1} \int_{0}^{R_0} \int_{0}^{2\pi} \int_{0}^{2\pi} \frac{\left|R_0-r\cdot\cos(\alpha_N-\alpha_M)\right|\cdot h\cdot R_0\cdot dh\cdot d\alpha_N\cdot r\cdot dr\cdot d\alpha_M} {\left[R_0^2+r^2+h^2-2R_0\cdot r\cdot\cos(\alpha_N-\alpha_M)\right]^2} \, . \tag{4} \]
Fig. 1. For calculating the irradiation coefficient
Having integrated with respect to \(\alpha_N\) and denoting the result of the first integration by \(I_1\), we have
\[ I_1= \frac{2a\,(R_0\cdot b+r\cdot c)} {\sqrt{(b^2-c^2)^3}} \cdot \arctan \left( \sqrt{\frac{b-c}{b+c}}\cdot \tan\frac{\alpha}{2} \right) \bigg|_{-\alpha_M}^{2\pi-\alpha_M}, \tag{5} \]
where
\[ a=\frac{h\cdot R_0\cdot dh\cdot r\cdot dr\cdot d\alpha_M}{\pi\cdot F_1}; \qquad b=h^2+R_0^2+r^2; \]
\[ c=-2R_0\cdot r; \qquad \alpha=\alpha_N-\alpha_M . \]
It can be shown that subsequent integration of expression (5) does not make it possible to resolve it in elementary functions [2].
To obtain an approximate solution, let us investigate the behavior of the coefficient preceding \(\operatorname{tg}\dfrac{\alpha}{2}\) as the ratio \(h_i/R_0\) varies. Denoting this coefficient by \(\kappa\), we substitute into it the expressions for \(b\) and \(c\), and after transformations obtain
\[ \kappa = \sqrt{ \frac{ \left(\dfrac{h}{R_0}\right)^2+\left(\dfrac{r}{R_0}+1\right)^2 }{ \left(\dfrac{h}{R_0}\right)^2+\left(\dfrac{r}{R_0}-1\right)^2 } }. \tag{6} \]
Obviously, as the ratio \(\dfrac{h}{R_0}\) varies over the interval \([0,\infty]\), the values of the coefficient \(k\) lie in the interval \([+\infty, 1]\).
Table 1 gives the data of the calculation of the coefficient \(k\) by formula (6) for various values of the ratio \(h/R_0\).
Comparison of the results shows:
Table 1
| \(h_i/R_0 \backslash r/R_0\) | 0 | 0.2 | 0.4 | 0.6 | 0.8 | 1.0 |
|---|---|---|---|---|---|---|
| 0 | 1.00 | 1.50 | 2.34 | 4.00 | 9.00 | [[unclear: blank cell]] |
| 0.5 | 1.00 | 1.38 | 1.90 | 2.82 | 3.46 | 4.12 |
| 1 | 1.00 | 1.22 | 1.48 | 1.75 | 2.02 | 2.24 |
| 2 | 1.00 | 1.08 | 1.17 | 1.26 | 1.34 | 1.41 |
| 3 | 1.00 | 1.04 | 1.08 | 1.12 | 1.16 | 1.20 |
| 4 | 1.00 | 1.025 | 1.05 | 1.07 | 1.10 | 1.12 |
| 5 | 1.00 | 1.015 | 1.034 | 1.049 | 1.063 | 1.077 |
| 6 | 1.00 | 1.01 | 1.02 | 1.034 | 1.039 | 1.049 |
| 10 | 1.00 | 1.035 | 1.007 | 1.01 | 1.015 | 1.020 |
1) if the ratio \(h/R_0=\mathrm{idem}\), then the maximum value of \(\kappa\) is attained at \(r/R_0=1\), and the minimum at \(\dfrac{r}{R_0}=0\);
2) for \(\dfrac{h}{R_0}>6\), the coefficient \(\kappa\) differs from unity by no more than \(5\%\) and assumes values within the limits
\[ 1 \leqslant \kappa \leqslant 1.049. \]
Denoting \(\operatorname{arctg}\left(\kappa \operatorname{tg}\dfrac{\alpha}{2}\right)\) by \(y\), we calculate the absolute \(\Delta y\) and relative \(\delta\) errors that arise when replacing \(\kappa=1.05\) by \(\kappa=1\). The results of the calculation are given in Table 2.
Taking \(\kappa=1\), expression (5) can be written as
\[ I_1=\frac{2\pi a\left(R_0 b+rc\right)}{\sqrt{(b^2-c^2)^3}}. \tag{7} \]
Integration of relation (4) over the limits from \(\alpha_M=0\) to \(\alpha_M=2\pi\), from \(r=0\) to \(r=R_0\), gives
\[ \varphi_{1,2}=\frac{\pi h\cdot dh}{F_1}\cdot \left[ \frac{h^2+2R_0^2}{\sqrt{h^4+4h^2R_0^2}}-1 \right], \tag{8} \]
where \(F_1=2\pi R_0\cdot dh\).
After transformation, formula (8) takes the form:
\[ \varphi_{1,2}= \frac{\left(\dfrac{h}{D}\right)^2+0.5} {\sqrt{\left(\dfrac{h}{D}\right)^2+1}} -\frac{h}{D}. \tag{9} \]
For \(h/D\to\infty\), the coefficient \(\varphi_{1,2}\to 0\), while for \(h/D\to 0\), the coefficient \(\varphi_{1,2}\to 0.5\), which corresponds to the physical meaning of the irradiation coefficient.
Table 2
| \(\dfrac{\alpha}{2}\) | \(0^\circ\) | \(1^\circ\) | \(6^\circ\) | \(12^\circ\) | \(15^\circ\) |
|---|---|---|---|---|---|
| \(\Delta y\) | 0 | \(3'\) | \(18'\) | \(37'\) | \(45'\) |
| \(\delta\) [%] | 0 | 4,75 | 4,76 | 4,9 | 4,75 |
| \(\dfrac{\alpha}{2}\) | \(30^\circ\) | \(45^\circ\) | \(60^\circ\) | \(75^\circ\) | \(90^\circ\) |
| \(\Delta y\) | \(1^\circ13'\) | \(1^\circ24'\) | \(1^\circ13'\) | \(0^\circ42'\) | \(0^\circ\) |
| \(\delta\) [%] | 3,92 | 3,02 | 1,99 | 0,93 | 0 |
To verify the result obtained, let us solve the stated problem by the method of ray algebra [4].
For a closed system of four bodies \(F_1, F_2, F_3, F_4\) (Fig. 2), we may write the determining equations:
\[ \left\{ \begin{aligned} \overline{\varphi}_{1,1}+\overline{\varphi}_{1,2}+\overline{\varphi}_{1,3}+\overline{\varphi}_{1,4}&=1\\ \overline{\varphi}_{2,1}+\overline{\varphi}_{2,2}+\overline{\varphi}_{2,3}+\overline{\varphi}_{2,4}&=1\\ \overline{\varphi}_{3,1}+\overline{\varphi}_{3,2}+\overline{\varphi}_{3,3}+\overline{\varphi}_{3,4}&=1\\ \overline{\varphi}_{4,1}+\overline{\varphi}_{4,2}+\overline{\varphi}_{4,3}+\overline{\varphi}_{4,4}&=1 \end{aligned} \right. \tag{10} \]
To solve the system of equations (10) with respect to \(\overline{\varphi}_{1,2}\), we introduce a partition \(F_D\) at the boundary of the surfaces \(F_1\) and \(F_3\), and use the expression for the irradiation coefficient of a circle onto a circle [3]:
\[ \varphi=\left[\frac{h}{D}-\sqrt{1+\left(\frac{h}{D}\right)^2}\right]^2 . \tag{11} \]
As a result we obtain:
\[ \overline{\varphi}_{1,2}=\left(\overline{\varphi}_{D,2}-\overline{\varphi}_{4,2}\right)\frac{D}{4\cdot \Delta h}, \tag{12} \]
where
\[ \begin{aligned} \overline{\varphi}_{1,2} &\text{— irradiation coefficient of an infinitely small cylindrical ring of height } \Delta h \text{ onto the base } F_2;\\ \overline{\varphi}_{D,2} &\text{— irradiation coefficient between the partition } F_D \text{ and the base } F_2;\\ \overline{\varphi}_{4,2} &\text{— irradiation coefficient between the bases } F_4 \text{ and } F_2. \end{aligned} \]
The values \(\overline{\varphi}_{D,2}\) and \(\overline{\varphi}_{4,2}\) are determined from (11); then the irradiation coefficient in finite increments can be written as follows:
Fig. 2. Closed system of four surfaces
\[ \overline{\varphi}_{1,2} = \left\{ \left[x-\sqrt{1+x^2}\right]^2 - \left[(x+\Delta x)-\sqrt{1+(x+\Delta x)^2}\right]^2 \right\} \frac{1}{4\Delta x}, \tag{13} \]
where
\[ x=\frac{h}{D}\text{ — dimensionless height,} \]
\[ \Delta x=\frac{\Delta h}{D}\text{ — increment of the dimensionless height.} \]
As \(\Delta x \to 0\), \(\overline{\varphi}_{1,2}\) is written in the following form:
\[ \varphi_{1,2} = \lim_{\Delta x\to 0}\overline{\varphi}_{1,2} = \frac{1}{4}\lim \left[ \frac{\varphi(x)-\varphi(x+\Delta x)}{\Delta x} \right], \tag{14} \]
where
\[ \varphi(x)=\left[x-\sqrt{1+x^2}\right]^2, \]
\[ \varphi(x+\Delta x)=\left[(x+\Delta x)-\sqrt{1+(x+\Delta x)^2}\right]^2. \]
In differential form:
\[ \varphi_{1,2}=-\frac{1}{4}\cdot \frac{d\varphi}{dx} \]
or
\[ \varphi_{1,2} = -\frac{1}{4}\cdot \frac{d\varphi}{d\left(\dfrac{h}{D}\right)} . \tag{15} \]
Ultimately, taking (11) into account, we have
\[ \varphi_{1,2}=\frac{\left(\dfrac{h}{D}\right)^2+0.5}{\sqrt{1+\left(\dfrac{h}{D}\right)^2}}-\frac{h}{D}. \tag{16} \]
Conclusion
The last equality (16) coincides with equality (9). This shows that determining the irradiation coefficient of an elementary cylindrical ring on the base of a cylinder by approximate integration and by the method of ray algebra, when using the generally accepted relation (11), leads to the same result.
References
- T. Hobler. Heat Transfer and Heat Exchangers. GKhI, 1961.
- I. S. Gradshteyn and I. M. Ryzhik. Tables of Integrals, Sums, Series, and Products. Fizmatgiz, 1962.
- S. N. Shorin. Heat Transfer. “Vysshaya Shkola” Publishing House, 1964.
- A. S. Nevsky. Heat Exchange by Radiation in Metallurgical Furnaces and Boiler Fireboxes. Metallurgizdat, 1958.