UDC 621.3.014.32.001.2

CALCULATION OF THE RESISTANCES OF COAXIAL TUBULAR CONDUCTORS TAKING THE SKIN EFFECT INTO ACCOUNT

V. S. SOKOLOV

(Presented by the scientific seminar of the Department of TOE)

The skin effect in cylindrical solid and tubular conductors has been considered in considerable detail in the literature on theoretical electrical engineering [1, 2, 3, etc.]. However, in the literature known to the author, final formulas suitable for direct engineering calculations are given only for the resistances of a solid single cylinder and of a coaxial cable with an opaque sheath.

At the same time, in practical calculations cases are possible in which the wall thickness of the tubes is comparable with the penetration depth, and complete attenuation of the wave in the thickness of the tube does not occur.

In the present article a calculation is given of the resistances of coaxial tubular conductors with allowance for the skin effect for the general case, when the wall thickness of the tubes is comparable with the penetration depth, that is, when the tubes are transparent or semitransparent to the electromagnetic wave.

Fig. 1

Let us consider a system of coaxial tubes (Fig. 1) serving as the outgoing and return conductors for a sinusoidal current \(I\) with angular frequency \(\omega\). In the inner tube the current flows “away from us,” and in the outer tube “toward us.” The radii of the inner tube are \(r_1\) and \(r_2\), and of the outer tube, \(r_3\) and \(r_4\).

The resistivity of the tube material is \(\gamma\), and the relative permeability is \(\mu\). In the calculation we assume that \(\mu\) is constant and does not depend on the magnetic-field strength, and that the phenomenon of hysteresis may be neglected. We also do not take into account displacement currents between the cylinders, that is, we assume that the current in the inner and outer tubes is the same in any cross section.

By virtue of the symmetry of the system, the vectors of the electric-field strength \(\vec E\) and current density \(\vec\delta\) in the tubes have only an axial component, while the vector of the magnetic-field strength \(\vec H\) is everywhere directed tangentially to circles concentric with the cross sections of the cylinder surfaces.

\[ \dot E = \dot E_z \quad \text{for } r_1 < r < r_2 \text{ and } r_3 < r < r_4,\quad \dot H = \dot H_\alpha. \]

The differential equation for the electric field in a conducting medium in complex form has the form

\[ \nabla^2 \dot{\vec E}=j\omega\mu\mu_0\gamma \dot{\vec E}. \tag{1} \]

Expanding the operator \(\nabla^2\) in the cylindrical coordinate system, taking into account that \(\dot{\vec E}\) depends only on the coordinate \(r\), we obtain:

\[ \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial \dot E}{\partial r}\right) = j\omega\mu\mu_0\gamma \dot E, \]

\[ \frac{d^2\dot E}{dr^2}+\frac{1}{r}\frac{d\dot E}{dr} - j\omega\mu\mu_0\gamma \dot E=0. \tag{2} \]

Replacing in equation (2) the variable \(r\) by the variable \(y=r\sqrt{-j\omega\mu\mu_0\gamma}\), we obtain the Bessel equation of zero order

\[ \frac{d^2\dot E}{dy^2}+\frac{1}{y}\frac{d\dot E}{dy}+\dot E=0, \tag{3} \]

whose solution can be written in the form

\[ \dot E=\dot A H_0^{(2)}(y)+\dot B J_0(y). \tag{4} \]

Here \(J_0(y)\) is the Bessel function of zero order of the first kind, \(H_0^{(2)}(y)\) is the Hankel function of zero order of the second kind (a Bessel function of the third kind), and \(\dot A\) and \(\dot B\) are constants of integration.

Thus, the electric-field intensity for the inner tube is equal to

\[ \dot E_1=\dot A_1 H_0^{(2)}(y)+\dot B_1 J_0(y) \tag{5} \]

and, correspondingly, for the outer tube,

\[ \dot E_2=\dot A_2 H_0^{(2)}(y)+\dot B_2 J_0(y). \tag{6} \]

The electric and magnetic field intensities are related to each other by Maxwell’s second equation:

\[ \operatorname{rot}\dot{\vec E}=-j\omega\mu\mu_0\dot{\vec H}. \]

Representing the curl in cylindrical coordinates and taking into account that in the present case \(\dot{\vec E}\) and \(\dot{\vec H}\) depend only on the coordinate \(r\) (the variable \(y\)), we obtain

\[ \dot H=\frac{1}{jz_c}\frac{d\dot E}{dy}, \tag{7} \]

where \(z_c=\sqrt{\dfrac{j\omega\mu\mu_0}{\gamma}}\) is the wave impedance of the medium.

Performing the differentiation in equation (7), we obtain for the inner tube:

\[ \dot H_1=\frac{1}{jz_c}\left[\dot A_1 H_1^{(2)}(y)+\dot B_1 J_1(y)\right] \tag{8} \]

and for the outer tube:

\[ \dot H_2=\frac{1}{jz_c}\left[\dot A_2 H_1^{(2)}(y)+\dot B_2 J_1(y)\right], \tag{9} \]

where \(H_1^{(2)}(y)\) is the Hankel function of the second kind of first order, and \(J_1(y)\) is the Bessel function of the first kind of first order.

The reduction formulas used were

\[ \frac{d}{dy} J_0(y)=-J_1(y) \quad \text{and} \quad \frac{d}{dy}H_0^{(2)}(y)\simeq -H_1^{(2)}(y). \]

We determine the integration constants \(\dot A_1,\dot B_1,\dot A_2,\dot B_2\) from equations (8) and (9), using the boundary conditions:

\[ \begin{aligned} &1)\quad \dot H=0 \quad \text{for } r=r_1,\\ &2)\quad \dot H_1=\dot I/2\pi r_2 \quad \text{for } r=r_2,\\ &3)\quad \dot H_2=\dot I/2\pi r_3 \quad \text{for } r=r_3,\\ &4)\quad \dot H_2=0 \quad \text{for } r=r_4. \end{aligned} \]

As a result of solving the system of equations (10), we obtain

\[ \dot A_1=-\frac{j z_c \dot I J_1(y_1)}{2\pi r_2\cdot D_1}, \qquad \dot B_1=\frac{j z_c \dot I H_1^{(2)}(y_1)}{2\pi r_2 D_1}, \tag{11} \]

\[ \dot A_2=-\frac{j z_c \dot I J_1(y_4)}{2\pi r_3 D_2}, \qquad \dot B_2=\frac{j z_c \dot I H_1^{(2)}(y_4)}{2\pi r_3 D_2}, \tag{12} \]

where

\[ D_1= \begin{vmatrix} J_1(y_1) & H_1^{(2)}(y_1)\\ J_1(y_2) & H_1^{(2)}(y_2) \end{vmatrix} \quad \text{and} \quad D_2= \begin{vmatrix} J_1(y_4) & H_1^{(2)}(y_4)\\ J_1(y_3) & H_1^{(2)}(y_3) \end{vmatrix}. \tag{13} \]

To determine the total resistance of the conductor, it is necessary to determine the flux of the Poynting vector penetrating through its surface over a length of 1 m and divide it by the square of the current passing through the conductor,

\[ Z=R+jx= \frac{-\displaystyle\oint \left[\dot{\vec E}_{\mathrm{surf}}\cdot \dot{\vec H}_{\mathrm{surf}}^{\,*}\right]d\vec s}{I^2}. \tag{14} \]

Energy penetrates into the inner conductor only from the outer surface \((r=r_2)\), and into the outer conductor only from the inner surface \((r=r_3)\), since \(\dot H=0\) at \(r=r_1\) and \(r=r_4\).

Since, in the case under consideration, the magnetic-field intensity on the surface of the inner conductor is \(\dot H=\dot I/2\pi r_2\), and the surface element is \(ds=2\pi r_2 l\), then

\[ Z_{\mathrm{in}}=R+jx= \frac{\dot E_{\mathrm{surf}}\cdot \dot I^{\,*}\cdot 2\pi r_2}{\dot I\dot I^{\,*}\cdot 2\pi r_2} = \frac{\dot E_{\mathrm{surf}}}{\dot I}. \tag{15} \]

Analogously, \(Z_{\mathrm{out}}\) is also determined.

Using (4), as well as (11) and (12), from formula (15) we determine the resistances of the tubes: of the inner one,

\[ Z_{\mathrm{in}}= \frac{jz_c}{2\pi r_2} \left[ \frac{ J_1(y_1)H_0^{(2)}(y_2)-H_1^{(2)}(y_1)J_0(y_2) }{ J_1(y_2)H_1^{(2)}(y_1)-H_1^{(2)}(y_2)J_1(y_1) } \right] \tag{16} \]

and of the outer one,

\[ Z_{\mathrm{out}}= \frac{jz_c}{2\pi r_3} \left[ \frac{ J_1(y_4)H_0^{(2)}(y_3)-H_1^{(2)}(y_4)J_0(y_3) }{ J_1(y_3)H_1^{(2)}(y_4)-H_1^{(2)}(y_3)J_1(y_4) } \right]. \tag{17} \]

From formulas (16) and (17) it is easy to obtain formulas for the resistances of the conductor and the opaque sheath of a coaxial cable [1]. The resistan-

we obtain for a solid conductor if in formula (15) we set \(r=0\). Since \(J_1(0)=0\), then

\[ Z_{\mathrm{in}}=-\frac{j z_c}{2\pi r_2}\cdot \frac{J_0(y_2)}{J_1(y_2)} . \tag{18} \]

For the case of an opaque sheath of a coaxial cable, in formula (17) one should take \(r_4 \to \infty\) \((y_4 \to \infty)\), since if the wave is completely attenuated in the thickness of the tube, then the distribution of current density in it does not depend on the outer radius and, in particular, must remain the same also for \(r \to \infty\).

Since \(H_1^{(2)}(y)=0\) as \(y \to \infty\), we obtain

\[ Z_{\mathrm{in}}=-\frac{j z_c}{2\pi r_3}\, \frac{H_0^{(2)}(y_3)}{H_1^{(2)}(y_3)} . \tag{19} \]

Formulas (18) and (19) are completely identical, apart from somewhat different notation, to formulas (6–90б) and (6–97) [1].

Formula (16) for the inner tube can be used directly to calculate the resistance of a single tube, since the magnetic field of the current of the outer tube does not penetrate into the inner tube.

Formula (17) takes into account the proximity effect, which is manifested in the fact that, under the influence of the magnetic field of the current of the inner tube, the current density in the outer tube decreases with increasing radius, rather than increasing as in the case of a single tube.

Tables of Bessel and Hankel functions of a complex argument, as well as formulas for their approximate determination, are given in [1, 2, 5, 6].

REFERENCES

1. A. V. Netushil, K. M. Polivanov. Fundamentals of Electrical Engineering. Part 3, Gosenergoizdat, 1956.

2. K. M. Polivanov. Theoretical Foundations of Electrical Engineering. Part 3, “Energiya,” 1969.

3. L. A. Bessonov. Theoretical Foundations of Electrical Engineering. “Vysshaya shkola,” 1967.

4. L. R. Neiman and K. S. Demirchyan. Theoretical Foundations of Electrical Engineering. Vol. 2, “Energiya,” 1967.

5. E. Jahnke, F. Emde. Tables of Functions with Formulas and Curves. Fizmatgiz, 1959.

6. G. B. Dwight. Tables of Integrals and Other Mathematical Formulas. “Nauka,” 1964.