COMPARISON OF LOSSES INTRODUCED INTO A SOLENOID BY ELLIPTICAL AND CIRCULAR NONMAGNETIC CORES

V. K. ZHUKOV

(Presented by the scientific seminar of the Faculty of Automation and Computer Engineering)

One of the methods suitable for contactless measurement of the geometric dimensions of cylindrical products (such as wire, tubes) is the eddy-current method. The theory of this method has been developed rather well for the case of a circular cross section of the object being monitored. In practice, however, some noncircularity of the cross section is always present. It is therefore natural to pose the question of how the noncircularity of the product affects the parameters of the transducer and, consequently, the measurement result. It does not appear possible to solve this problem in general form. However, several special cases for which a solution exists can be considered, and the conclusions obtained can then be extended to the case of an arbitrary cross section. To this end, let us take elliptical and circular nonmagnetic cylinders with equal cross-sectional areas and compare the losses they introduce into an inductive transducer with a through coil. To simplify the derivations we shall regard the transducer as an infinitely long solenoid and shall consider the energy relations occurring in it per unit of its length.

Let us denote by $\dot W_1$ the power dissipated in the elliptical cylinder, and by $\dot W_2$ that dissipated in the circular cylinder, per unit of their length. The power in the gap between the transducer winding and the cylinder is equal: in the case of a circular cross section,

\[ \dot W_{\text{g.c.}} = \dot W_0 - \dot W_{20}, \]

and in the case of an elliptical one,

\[ \dot W_{\text{g.e.}} = \dot W_0 - \dot W_{10}, \]

where $\dot W_0$ is the power dissipated in the cavity of the transducer in the absence of the cylinder, $\dot W_{20}$ is the power in an air circular cylinder with an area equal to the area of the metallic cylinder, and $\dot W_{10}$ is the power in an air elliptical cylinder. The loss power of the solenoid with a circular core will be:

\[ \dot W_{\text{c}} = \dot W_2 + \dot W_{\text{g.c.}} = \dot W_0 - \dot W_{20} + \dot W_2, \tag{1} \]

and with an elliptical core:

\[ \dot W_{\text{e}} = \dot W_1 + \dot W_{\text{g.e.}} = \dot W_0 - \dot W_{10} + \dot W_1. \tag{2} \]

From relations (1) and (2) it is seen that the insertion of the cylinders causes

the change in the loss power in the solenoid by the amount \(\Delta \dot W_{\mathrm{k}}=\dot W_2-\dot W_{20}\) in one case and by \(\Delta \dot W_{\mathrm{e}}=\dot W_1-\dot W_{10}\) in the other case. The relative difference in power losses

\[ \frac{\Delta \dot W_{\mathrm{k}}-\Delta \dot W_{\mathrm{e}}}{\Delta \dot W_{\mathrm{k}}} = 1-\frac{\dot W_1-\dot W_{10}}{\dot W_2-\dot W_{20}}, \]

as is not difficult to show, will be equal to the relative difference of the introduced resistances

\[ \frac{\Delta \dot Z_{\mathrm{k}}-\Delta \dot Z_{\mathrm{e}}}{\Delta \dot Z_{\mathrm{k}}} = \frac{\Delta \Delta \dot Z}{\Delta \dot Z_{\mathrm{k}}}, \]

where \(\Delta \dot Z_{\mathrm{k}}\) and \(\Delta \dot Z_{\mathrm{e}}\) are the resistances introduced into the solenoid by the circular and elliptical cylinders, respectively. Thus, in order to compare the losses introduced by the cylinders into the solenoid, it is necessary to analyze the expression

\[ \frac{\Delta \Delta \dot Z}{\Delta \dot Z_{\mathrm{k}}} = 1-\frac{\dot W_1-\dot W_{10}}{\dot W_2-\dot W_{20}}, \]

for which it is necessary to find the dependences for \(\dot W_1,\dot W_2,\dot W_{10},\dot W_{20}\). Without significant error one may assume that \(\dot W_{10}=\dot W_{20}\); therefore

\[ \frac{\Delta \Delta \dot Z}{\Delta \dot Z_{\mathrm{k}}} = 1-\frac{\dot W_1-\dot W_{20}}{\dot W_2-\dot W_{20}} \tag{3} \]

and three quantities will have to be determined: \(\dot W_1,\dot W_2,\dot W_{20}\).

In determining the loss power in the elliptical cylinder we shall use the elliptic coordinate system \(\xi,\varphi,z\). The coordinate

Fig. 1

lines in this system (Fig. 1) are: \(\xi=\mathrm{const}\)—confocal ellipses, \(\varphi=\mathrm{const}\)—confocal hyperbolas. We arrange the coordinate system so that the surface of the specimen will be a

coordinate line \(\xi_0\). If the ellipse of the cross section of the specimen has a small eccentricity \(e\) and its area is considerably less than the area of the circle bounded by the solenoid winding, then it may be assumed that the winding coincides with the coordinate line \(\xi_n\).

The wave equation for the magnetic-field intensity vector in the metal

\[ \nabla^2 \overline{H}+k^2\overline{H}=0 \tag{4} \]

in the elliptic coordinate system is transformed into two equations with variable coefficients, whose solutions are special functions—the Mathieu functions. The general solution of the wave equation (4) will have the form

\[ \overline{H}=\sum_{n=0}^{\infty}\overline{C}_{2n}\cdot Ce_{2n}(\xi,q)\cdot ce_{2n}(\varphi,q), \]

where \(Ce_{2n}(\xi,q)\) and \(ce_{2n}(\varphi,q)\) are Mathieu functions, \(q=-j\dfrac{1}{4}\omega\cdot\mu_0\cdot\gamma\cdot h^2\) is the argument of the functions, and \(h\) is the distance between the foci of the cross-section ellipse. The integration constant \(C_{2n}\) is determined from the boundary condition: \(\overline{H}=\overline{H}_0\) at \(\xi=\xi_0\). We take the magnetic-field intensity inside the coil to be constant and equal to \(\overline{H}_0\). After determining the integration constant, we shall have:

\[ \overline{H} = 2\pi\overline{H}_0 \sum_{n=0}^{\infty} \frac{ A_0^{(2n)}\cdot Ce_{2n}(\xi,q)\cdot ce_{2n}(\varphi,q) }{ Ce_{2n}(\xi_0,q)L_{2n} }. \tag{5} \]

Here

\[ L_{2n}=\int_{0}^{2\pi} ce_{2n}^{2}(\varphi,q)\cdot d\varphi, \]

and \(A_0^{(2n)}\) are the coefficients entering into \(Ce_{2n}(\xi,q)\) and \(ce_{2n}(\varphi,q)\).

To determine the energy losses in an elliptic cylinder, it is necessary to know the electric-field intensity \(\overline{E}\). We find the vector \(\overline{E}\) from the expression \(\operatorname{rot}\overline{H}=\gamma\overline{E}\), into which, after substituting the value of \(\overline{H}\) from (5), we obtain:

\[ \overline{E} = \overline{\xi}\, \frac{2\pi\overline{H}_0}{\gamma\cdot h_{\varphi}} \sum_{n=0}^{\infty} \frac{ A_0^{(2n)}\cdot Ce_{2n}(\xi,q)\cdot ce'_{2n}(\varphi,q) }{ Ce_{2n}(\xi_0,q)L_{2n} } \]

\[ - \overline{\varphi}\, \frac{2\pi\overline{H}_0}{\gamma\cdot h_{\xi}} \sum_{n=0}^{\infty} \frac{ A_0^{(2n)}\cdot Ce'_{2n}(\xi,q)\cdot ce_{2n}(\varphi,q) }{ Ce_{2n}(\xi_0,q)\cdot L_{2n} }, \tag{6} \]

where \(\overline{\xi}\) and \(\overline{\varphi}\) are unit vectors, and \(h_{\varphi}\) and \(h_{\xi}\) are the Lamé metric coefficients. The energy flux through the lateral surface of the cylinder will be determined by the Umov–Poynting vector,

\[ \overline{P}=\frac{1}{2}\left[\overline{E}\cdot\overline{H}^{*}\right], \]

where \(\overline{H}^{*}\) is the complex conjugate quantity with respect to \(\overline{H}\). In determining the vector \(\overline{P}\), the values of \(\overline{H}\) and \(\overline{E}\) on the surface of the cylinder are substituted, i.e., at \(\xi=\xi_0\). The power \(W_1\) released

absorbed in the elliptical cylinder will be determined by integration over the entire surface of the unit cylinder:

\[ \dot W_1=\int_S \bar P\cdot d\bar S . \]

After carrying out all operations and transformations, we find:

\[ \dot W_1=\frac{2\pi\cdot \dot H_0^{\,2}}{\gamma} \sum_{n=0}^{\infty} \frac{Ce'_{2n}(\xi_0,q)} {Ce_{2n}(\xi_0,q)\,L_{2n}} . \tag{7} \]

To determine the power absorbed by a circular cylinder, let us find the electric-field intensity vector \(\bar E\) and the magnetic-field intensity vector \(\bar H\) in the cylinder. Then we determine the energy-flux vector \(\bar P\) through the surface of the cylinder and, by integration over the entire surface, find \(\dot W_2\).

The vector \(\bar H\) is obtained as a result of solving the wave equation (1), written in a cylindrical coordinate system. The solution will have the form:

\[ \bar H=\bar H_0 \frac{I_0\!\left(\sqrt j\,m_k\right)} {I_0\!\left(\sqrt j\,m_{k.0}\right)} , \tag{8} \]

where \(m_{k.0}=r_0\sqrt{\omega\cdot\gamma\cdot\mu_0}\), \(r_0\) is the radius of the circular cylinder, and \(m_k=r\sqrt{\omega\cdot\gamma\cdot\mu_0}\).

We find the electric-field intensity vector \(\bar E\) from Maxwell’s first equation

\[ \operatorname{rot}\bar H=\gamma\cdot\bar E . \]

After substituting into it the value of \(\bar H\) from (8) and performing transformations, we obtain:

\[ \bar E=\bar\varphi\, \frac{(1+j)\dot H_0} {\gamma\sqrt{2\omega\cdot\gamma\cdot\mu_0}}\, \frac{I_1\!\left(\sqrt j\,m_k\right)} {I_0\!\left(\sqrt j\,m_{k.0}\right)} . \tag{9} \]

\(I_1\!\left(\sqrt j\,m_k\right)\) and \(I_0\!\left(\sqrt j\,m_k\right)\) in expressions (8) and (9) denote Bessel functions of a complex argument.

Knowing \(\bar H\) and \(\bar E\), it is not difficult to find the power \(\dot W_2\) absorbed by the circular cylinder:

\[ \dot W_2=\frac{\pi\cdot \dot H_0^{\,2}}{\gamma}\cdot m_k\cdot \frac{I_1\!\left(\sqrt j\,m_{k.0}\right)} {I_0\!\left(\sqrt j\,m_{k.0}\right)} . \tag{10} \]

It remains to find the power losses \(\dot W_{20}\) in the air circular cylinder. The method remains the same as in the case of the metallic cylinder. Carrying out all the transformations, we find:

\[ \dot W_{20}=j\,\frac{\pi\cdot \dot H_0^{\,2}}{\gamma} \sqrt{\frac{\mu_0}{\varepsilon_0}}\, r_0\cdot\gamma\cdot \frac{I_1(m_{\mathrm v})}{I_0(m_{\mathrm v})}, \tag{11} \]

where \(m_{\mathrm v}=r_0\cdot\omega\sqrt{\varepsilon_0\cdot\mu_0}\).

Let us express \(q\) and \(m_{\mathrm v}\) through \(m_k\), for which we find \(\omega\) from the expression for \(m_k\) and substitute it into the other two. After substitution we find

\[ q=-j\,\frac{1}{4}\left(\frac{a}{b}-\frac{b}{a}\right)\cdot m_k^2 \tag{12} \]

and

\[ m_{\mathrm v}=\frac{1}{r_0\cdot\gamma} \sqrt{\frac{\varepsilon_0}{\mu_0}}\cdot m_k^2 . \tag{13} \]

Fig. 2.
Vertical axis: \(\Delta X / X\), \(\%\).
Horizontal axis: \(m_{r}a\).
Curves: 1, 2, 3.

Fig. 3.
Vertical axis: \(\Delta R / R\), \(\%\).
Horizontal axis: \(m_{r}k\).
Curves: 1, 2, 3.

Here \(b\) and \(a\) denote the minor and major semiaxes of the ellipse of the cross section.

Since \(m_{\mathrm{v}} \ll 1\), \(J_1(m_{\mathrm{v}})=\dfrac{m_{\mathrm{v}}}{2}\), and \(I_0(m_{\mathrm{v}})=1\); therefore

\[ \dot W_{20}=j\,\frac{\pi H_0^2}{2\gamma}\,m_{\mathrm{k}}^2 . \tag{14} \]

The dependences obtained still do not answer the question posed; therefore we shall plot graphically the dependence

\[ \frac{\Delta \dot Z}{\Delta \dot Z_{\mathrm{k}}}=f(m_{\mathrm{k}}) \]

for three values of the ratio \(b/a\), using formulas (7, 10, 14). Since Mathieu functions have not been tabulated and their calculation is complicated, especially in the case of a complex argument, it was necessary to use the Minsk-1 electronic computer. The calculation results are presented in the graphs (Figs. 2 and 3). Each curve corresponds to a definite value of \(b/a\): 1 — \(b/a=0.9\); 2 — \(b/a=0.8\); 3 — \(b/a=0.7\). The length of the curves is limited, since the calculation was carried out for values of \(q\) not exceeding 0.6. Nevertheless, even from such incomplete graphs the following conclusions can be drawn:

1. For values of the argument \(m_{\mathrm{k}}=2.5\), the circular and elliptical cylinders introduce the same active resistance into the sensor. Consequently, at this value of \(m_{\mathrm{k}}\), the active component of the sensor resistance responds to the cross-sectional area of the article and does not depend on the shape of the section.

2. At large values of the argument \(m\), the reactive component of the sensor resistance ceases to sense the shape of the article’s cross section and responds only to the area.

REFERENCES

1. McLachlan. Theory and Application of Mathieu Functions, Foreign Literature Publishing House, 1953.
2. A. M. Vainberg. Induction Melting Furnaces, Gosenergoizdat, 1960.