STATIONARY TEMPERATURE FIELD IN A SEGMENT

V. V. Ivanov

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

In some applications it is necessary to know the stationary distribution of temperature in an infinite bar whose cross-section is a segment with central angle \(2\beta\) and chord length \(2R_0\). The surfaces corresponding to the arc and the chord of the segment are maintained respectively at temperatures \(T\) and zero. Between the surfaces there is thermal insulation: the points \(iR_0\), \(-iR_0\) (Fig. 1a).

To solve this problem we shall use conformal mappings.

The fractional-linear mapping

\[ z_1 = R_0 \frac{iR_0 - Z}{iR_0 + Z} \tag{1} \]

maps the region bounded by the segment onto the interior of the angle \(\beta\) (Fig. 1b). In this case the zero and \(T\) isotherms of the \(Z\)-plane pass in the \(z_1\)-plane respectively into the positive real half-axis and the ray \(\varphi_1=\beta\).

With the aid of the function

\[ z_2 = z_1^{\pi/2\beta} \tag{2} \]

we transform the angle \(\beta\) into a straight angle (Fig. 1c) and use the mapping

\[ z = iR_0^{\pi/2\beta}\, \frac{z_2 + R_0^{\pi/2\beta}}{z_2 - R_0^{\pi/2\beta}}, \tag{3} \]

which maps the straight angle into a semicircle of radius \(R_0^{\pi/2\beta}\) (Fig. 1g). Replacing, for convenience of notation, \(R_0^{\pi/2\beta}\) by \(R\), we find that the function

\[ W = \frac{1}{i}\ln \frac{z}{R}, \quad \text{where } W = u + iv, \]

maps the semicircle onto a half-strip (Fig. 1d). The insulating points \(iR\) and \(-iR\), the isotherms \(T\) and zero of the \(z\)-plane are mapped in the \(W\)-plane respectively into the points \(\pi/2\), \(-\pi/2\), into the segment \((-\pi/2,\ \pi/2)\) and the straight lines \(u=-\pi/2\) and \(u=\pi/2\) \((\operatorname{Im} W>0)\).

Fig. 1.

(a) Domain \(\mathfrak{Z}\): axes \(X\), \(Y\); parameters \(\beta\), \(\theta\), \(\rho\), \(h_0\), \(-h_0\), \(\kappa\).

(b) Domain \(\mathfrak{Z}\): axes \(x\), \(y\); parameter \(\gamma\); endpoints \(i h_0^{\pi/\beta}\), \(-i h_0^{\pi/\beta}\).

(c) Domains \(\mathfrak{Z}_1\) and \(\mathfrak{Z}_2\): axes \(x_1,y_1\) and \(x_2,y_2\); angles \(\delta\), \(\delta\); rays labeled \(s_1\), \(s_2\); angle \(\beta\).

(d) Domain \(M\): coordinate \(u\); marked points \(-\pi/2\), \(0\), \(\pi/2\).

The temperature field in this half-strip is given by the equation [1]

\[ t(W)=\frac{2T}{\pi}\operatorname{arctg}\frac{\cos u}{\operatorname{sh}v}. \tag{4} \]

Since \(z=Re^{iw}\), then, putting \(z=re^{i\varphi}\), we obtain \(u=\varphi\), and \(v=-\ln r/R\). Substituting the values of \(u\) and \(v\) in (4), we find the temperature distribution in a semicircle of radius \(R\):

\[ t(z)=t(r,\varphi)=\frac{2T}{\pi}\operatorname{arctg}\frac{2Rr\cos\varphi}{R^{2}-r^{2}}. \tag{5} \]

Solving (3) with respect to \(z_2\), for \(z=r(\cos\varphi+i\sin\varphi)\), we obtain that

\[ \varphi_2=\operatorname{arctg}\frac{y_2}{x_2} =\operatorname{arctg}\frac{2Rr\cos\varphi}{R^{2}-r^{2}}, \]

and

\[ t(z_2)=\frac{2T}{\pi}\varphi_2. \tag{6} \]

Combining (2) and (6), we determine the temperature field inside the angle \(\beta\):

\[ t(z_1)=\frac{T}{\beta}\varphi_1. \]

Substituting \(Z=r(\cos\Theta+i\sin\Theta)\) in (1), we have that

\[ \varphi_1=\operatorname{arctg}\frac{y_1}{x_1} =\operatorname{arctg}\frac{2R_0r\cos\Theta}{R_0^{2}-r^{2}}. \]

Hence the desired temperature distribution in the segment has the form

\[ t(Z)=t(r,\Theta)=\frac{T}{\beta}\operatorname{arctg} \frac{2R_0r\cos\Theta}{R_0^{2}-r^{2}}. \]

The heat through the segment

\[ Q_{\beta}=-\lambda\int_S \operatorname{grad}t\,ds \left[\frac{\text{kcal}}{\text{m}\cdot\text{hour}}\right] \]

can be determined on the basis of equivalence with respect to the heat flux of the given segment and a semicircle \(R=R_0^{\pi/2\beta}\), the temperature field in which is given by equation (5),

\[ \operatorname{grad}t=\left(\frac{\partial t}{\partial r}\right)_{r=R} =\frac{2T}{\pi R\cos\varphi},\qquad ds=R\,d\varphi. \]

The angle \(\varphi\) must vary from zero to \(\pi/2-\left(\dfrac{\Delta\delta}{R_0}\right)^{\pi/2\beta}\), where \(\Delta\delta\) is the insulating segment \((iR_0,K)\), which passes on the plane \(z\) into the segment \((\Delta\delta)^{\pi/2\beta}\). Consequently,

\[ Q_{\beta} = 2\lambda \int_{0}^{\frac{\pi}{2}-\left(\frac{\Delta\delta}{R_0}\right)^{\pi/2\beta}} \frac{2TR\,d\Theta}{\pi R\cos\Theta} = \frac{4\lambda T}{\pi}\ln\operatorname{tg} \left[ \frac{\pi}{2} - \frac{1}{2} \left(\frac{\Delta\delta}{R_0}\right)^{\pi/2\beta} \right]. \tag{7} \]

We checked formula (7) by the method of electrothermal analogies on a plane electrical model of the segment, made of electrically conductive paper. Along the contours of the model, 2 copper buses were fastened, imitating the isothermal surfaces \(T\) and zero [2].

According to Ohm’s law, the current flowing in the model is

\[ I=\frac{h}{\rho}\,\Delta V\cdot \Phi . \tag{8} \]

Here \(\Delta V\) is the voltage on the buses, \(\rho\) is the specific electrical resistance of the paper of thickness \(h\), and \(\Phi\) is the dimensionless shape factor; on the other hand,

\[ Q_t=\lambda T\Phi . \]

Therefore

\[ \Phi=\frac{4}{\pi}\ln \operatorname{tg}\left[ \frac{\pi}{2}-\frac{1}{2}\left(\frac{\Delta \delta}{R_0}\right)^{\pi/2\beta} \right]. \tag{9} \]

The check was carried out by comparing the values of \(\Phi\) obtained by calculation from (9) and from experiment by (8). The experiments were performed for segments with angles \(\beta\) of \(68^\circ\) and \(110^\circ\). The results are summarized in the table.

LITERATURE

1. Lavrent’ev M. A., Shabat B. V. Methods of the Theory of Functions of a Complex Variable. Gostekhizdat, 1958.

2. Petukhov B. S. Experimental Study of Heat-Transfer Processes. Gosenergoizdat, 1952.