ON ACCOUNTING FOR ELECTROMAGNETIC ENERGY LOSSES IN A CYLINDRICAL PRODUCT WHEN USING SLOT-TYPE EDDY-CURRENT SENSORS

D. V. MILYAEV, V. K. ZHUKOV

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

In article [1], included in the present collection, some questions concerning the theory of a slot-type sensor are considered. In the case of monitoring the dimensions or electrical conductivity of products, it is necessary to have a relation between the parameters of the sensor and the measured parameter. Most often, the dependence is found of the total impedance of the sensor on the parameters of the product and of the sensor, as well as on the frequency of the exciting field. This problem has been considered in detail by many authors for feed-through type sensors (a long solenoid). To solve the analogous problem for a slot-type sensor it is necessary to determine: a) the losses of electromagnetic energy in the cylindrical product being monitored; b) the energy losses in the air gap.

In doing so we assume:

1. The field in the gap is homogeneous.
2. The magnetic core at the gap has a rectangular cross section with its long side parallel to the axis of the product, \(b \gg a\).

To solve this problem we shall use the wave equations for the vector potentials in air and in the metal of the product

\[ \nabla^2 \overline{A}_2 = 0, \tag{1} \]

\[ \nabla^2 \overline{A}_1 + \kappa^2 \overline{A}_1 = 0, \tag{2} \]

where \(\overline{A}_2\) is the vector potential of the eddy currents in air,

\(\overline{A}_1\) is the vector potential of the eddy currents in the metal,

\[ \kappa = \sqrt{- i \omega \gamma_0 \mu_0}, \tag{3} \]

\(\gamma_0, \mu_0\) are the electrical conductivity and magnetic permeability of the product material.

The solutions of the wave equations are written most simply in a cylindrical coordinate system \((r,\varphi,z)\). Let the coordinate axes be chosen so that

\[ \overline{A}_1 = \overline{1}_z \dot{A}_1 \quad \text{and} \quad \overline{A}_2 = \overline{1}_z \dot{A}_2, \tag{4} \]

and, in addition, let \(b \gg a\). The wave equation for air is

\[ \frac{\partial^2 \dot{A}_2}{\partial r^2} + \frac{1}{r}\frac{\partial \dot{A}_2}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \dot{A}_2}{\partial \varphi^2} =0, \tag{5} \]

for the metal

\[ \frac{\partial^{2}\dot A_{1}}{\partial r^{2}}+\frac{1}{r}\frac{\partial \dot A_{1}}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}\dot A_{1}}{\partial \varphi^{2}}+\kappa^{2}\dot A_{1}=0. \tag{6} \]

The general solutions of the equations obtained have the form

\[ \dot A_{1}=(\dot N_{\varphi 1}\cos \lambda \varphi+\dot N'_{\varphi}\sin \lambda \varphi) \left[\dot C_{1}J_{n}(\kappa r)+\dot C'_{1}Y_{n}(\kappa r)\right], \tag{7} \]

\[ \dot A_{2}=(\dot N_{\varphi 2}\cos \lambda \varphi+\dot N'_{\varphi 2}\sin \lambda \varphi) \left[\dot C_{2}r^{n}+\dot C'_{2}r^{-n}\right]+\dot A_{0}, \tag{8} \]

where
\(\dot N_{\varphi 1};\ \dot N_{\varphi 2};\ \dot N'_{\varphi 1};\ \dot N'_{\varphi 2};\ \dot C_{1};\ \dot C_{2};\ \dot C'_{1};\ \dot C'_{2}\) are constants of integration,
\(\dot A_{0}\) is the vector potential of the exciting magnetic field, and \(J_{n}(\kappa r)\), \(Y_{n}(\kappa r)\) are Bessel functions.

Since the primary field does not contain \(\cos 2\varphi\), \(\cos 3\varphi,\ldots\), etc., the resultant field, i.e., the total field of the eddy currents and the exciting field, must contain only \(\cos \varphi\). In addition, \(\lambda\) is an integer, since \(\dot A_{1}\) and \(\dot A_{2}\) are even functions; consequently, \(\dot N'_{\varphi 1}=\dot N'_{\varphi 2}=0\). In a solid isotropic cylinder there is no reflected wave; one should set \(\dot C'_{1}=0\). The field of the eddy currents in air is equal to zero at infinity; it is necessary to take \(\dot C_{2}=0\). As a result, for the solution of the wave equations,

\[ \dot A_{1}=\dot N_{\varphi 1}C_{1}J_{1}(\kappa r)\cos \varphi, \tag{9} \]

\[ \dot A_{2}=\dot N_{\varphi 2}\dot C'_{2}r^{-1}\cos \varphi+\dot A_{0}. \tag{10} \]

The vector \(\dot{\overline A}_{0}\) is related to the magnetic induction vector \(\dot{\overline B}_{0}\) of the exciting field by the relation

\[ \operatorname{rot}\dot{\overline A}_{0}=\dot{\overline B}_{0}. \tag{11} \]

In a cylindrical coordinate system, \(\operatorname{rot}\dot A\)

\[ \overline 1_{r}\frac{1}{r}\frac{\partial \dot A_{0}}{\partial \varphi} -\overline 1_{\varphi}\frac{\partial \dot A_{0}}{\partial r} =\dot{\overline B}_{0}, \tag{12} \]

On the other hand,

\[ \dot{\overline B}_{0}=\overline 1_{r}\dot B_{0}\sin \varphi-\overline 1_{\varphi}\dot B_{0}\cos \varphi. \tag{13} \]

From (12) and (13)

\[ \dot A_{0}=-\dot B_{0}r\cos \varphi. \tag{14} \]

To determine the constants of integration it is necessary to use the equalities

\[ \dot{\overline A}_{1}=\dot{\overline A}_{2},\qquad \frac{\partial \dot A_{1}}{\partial r}=\frac{\partial A_{2}}{\partial r}, \tag{15} \]

which are valid for \(r=r_{0}\), i.e., at the interface between the two media. After solving the system of equations, the constants are determined:

\[ \dot N_{2}=\dot B_{0}r_{0}^{2} \frac{\kappa r_{0}J'_{1}(\kappa r_{0})-J_{1}(\kappa r_{0})} {\kappa r_{0}J'_{1}(\kappa r_{0})+J_{1}(\kappa r_{0})}, \tag{16} \]

\[ \dot N_{1}=\frac{-2\dot B_{0}r_{0}} {\kappa r_{0}J'_{1}(\kappa r_{0})+J_{1}(\kappa r_{0})}. \tag{17} \]

The vector potentials in air and in the metal, with allowance for (16) and (17), are rewritten as follows:

\[ \bar{\dot A}_1=\bar{\mathbf 1}_z\, \frac{-2\dot B_0 r_0 J_1(kr_0)\cos\varphi} {kr_0 J_1'(kr_0)+J_1(kr_0)}, \tag{18} \]

\[ \bar{\dot A}_2=\bar{\mathbf 1}_z\, \frac{\dot B_0 r_0^2[kr_0J_1'(kr_0)-J_1(kr_0)]\,r^{-1}} {[kr_0J_1'(kr_0)+J_1(kr_0)]}\cos\varphi+\bar{\dot A}_0 . \tag{19} \]

From the vector potentials it is necessary to pass to the components of the electromagnetic field: the magnetic- and electric-field intensities. As is known,

\[ \dot{\bar H}=\frac{1}{\mu_0}\operatorname{rot}\bar{\dot A}, \tag{20} \]

\[ \bar{\dot E}=-i\omega\bar{\dot A}. \tag{21} \]

Substituting the expressions for the vector potentials into (20) and (21) and carrying out the necessary operations, one can obtain the distribution of \(\bar E\) and \(\bar H\) in air and in the metal:

\[ \bar{\dot E}_1=\bar{\mathbf 1}_z\, \frac{2i\omega\dot B_0 r_0 J_1(kr_0)} {J_1(kr_0)+kr_0J_1'(kr_0)}\cos\varphi, \tag{22} \]

\[ \bar{\dot E}_2=\bar{\mathbf 1}_z \left\{ \frac{-i\omega\dot B_0[kr_0J_1'(kr_0)-J_1(kr_0)]\,r^{-1}r_0^2} {kr_0J_1'(kr_0)+J_1(kr_0)}\cos\varphi -i\omega\dot A_0 \right\}, \tag{23} \]

\[ \bar{\dot H}_1=\bar{\mathbf 1}_r \frac{2\dot B_0 r_0J_1(kr_0)\sin\varphi} {\mu_0 r\,[J_1(kr_0)+kr_0J_1'(kr_0)]} +\bar{\mathbf 1}_{\varphi} \frac{2\dot B_0 r_0 kJ_1'(kr_0)\cos\varphi} {\mu_0[J_1(kr_0)+kr_0J_1'(kr)]}, \tag{24} \]

\[ \bar{\dot H}_2=\bar{\mathbf 1}_r \left\{ \frac{\dot B_0}{\mu_0}\sin\varphi -\frac{\dot B_0 r_0^2[kr_0J_1(kr_0)-J_1(kr_0)]} {\mu_0 r^2[kr_0J_1'(kr_0)+J_1(kr_0)]}\sin\varphi \right\} +\rightarrow \]

\[ \rightarrow+\bar{\mathbf 1}_{\varphi} \left\{ \frac{\dot B_0}{\mu_0}\cos\varphi +\frac{\dot B_0 r_0^2[kr_0J_1'(kr_0)-J_1(kr_0)]} {\mu_0 r^2[kr_0J_1'(kr_0)+J_1(kr_0)]}\cos\varphi \right\}. \tag{25} \]

The losses of electromagnetic energy in the metal, as before, are determined through the specific energy flux

\[ \bar{\dot \Pi}_{\mathrm{iz}}=\frac{1}{2}\,[\bar{\dot E}_1\bar{\dot H}_1^{*}]. \tag{26} \]

The power absorbed by the inspected product is found as in the previous case, i.e., first the energy flux is determined at \(r=r_0\), and then this flux is integrated over the lateral surface of the cylinder, since only through this surface does the electromagnetic field penetrate into the cylinder.

\[ P_{\mathrm{iz}}=\int_s \bar{\dot \Pi}_{\mathrm{iz}}\,\bar{ds}. \tag{27} \]

But, on the other hand,

\[ P_{\mathrm{iz}}=I^2 z_{\mathrm{iz}}, \tag{28} \]

where \(z_{\mathrm{iz}}\) is the electrical resistance of the cylinder, characterizing the energy losses in it.

Integrating expression (28) and equating it to (29), one can obtain the electrical resistance.

\[ z_{\mathrm{i}3}=\pi \omega \mu_0 w^2 i_0 r_0^2 \left\{ \left[ \frac{2\operatorname{ber}_2(\kappa r_0)\operatorname{bei}(\kappa r_0)-2\operatorname{bei}_2(\kappa r_0)\operatorname{ber}(\kappa r_0)} {\operatorname{ber}^2 \kappa r_0+\operatorname{bei}^2 \kappa r_0} \right] +\right. \]
\[ \left. {}+i\left[ \frac{\operatorname{ber}^2(\kappa r_0)+\operatorname{bei}^2(\kappa r_0)-\operatorname{ber}_2^2(\kappa r_0)-\operatorname{bei}_2^2(\kappa r_0)} {\operatorname{ber}^2(\kappa r_0)+\operatorname{bei}^2(\kappa r_0)} \right] \right\}. \tag{29} \]

In the last expression the so-called Kelvin functions have been introduced.

The adopted assumption that the vector potential in the air gap has only the \(z\)-component is valid only when the poles of the ferromagnetic core have a rectangular cross section and one of the dimensions is larger than the other. Consequently, two plane waves propagate in the gap. The field components \(E_z\) and \(H_\varphi\), found earlier, must be rewritten for a rectangular coordinate system chosen so that the \(z\)-axes of both systems coincide, and the \(x\)-axis is directed along the line \(\varphi=0\).

Using the coordinate-transformation formulas, the distribution of the field intensities in air can be rewritten in the rectangular system.

In addition, let us determine the losses in the air gap when the inspected product is absent from the gap. To do this, in the expressions for the field intensities we put \(r=0\), after which we obtain

\[ \bar{H}_2=\bar{I}\frac{B_0}{\mu_0}, \tag{30} \]

\[ \bar{E}_2=-\kappa i\omega B_0 x. \tag{31} \]

Next it is necessary to find the specific energy flux as

\[ \Pi_{\mathrm{a.g}}=\frac{1}{2}\left[E_2 H_2^*\right]. \tag{32} \]

Substituting (30) and (31) into (32) and taking into account that the product of conjugate complex quantities is equal to the square of the modulus of the complex number, we obtain the specific flux

\[ \Pi_{\mathrm{a.g}}=-\frac{1}{2}i\omega \frac{B_0^2}{\mu_0}x, \]

which must be determined at \(x=\dfrac{b}{2}\).

The power absorbed by the gap is

\[ P_{\mathrm{a.g}}=2\int_0^a \Pi_{\mathrm{a.g}}\cdot dz. \]

On the other hand,

\[ P_{\mathrm{a.g}}=I^2 z_{\mathrm{a.g}}, \]

where \(z_{\mathrm{a.g}}\) is the total resistance of the air gap; consequently,

\[ z_{\mathrm{a.g}}=\frac{i\omega B_0^2 b\cdot a}{2\mu_0 I^2}. \]

By the total-current law,

\[ B_0=I w_{10}\mu_{\mathrm{e}}, \]

then the expression for the total resistance of the air gap will have the form

\[ z_{\mathrm{a.g}}=i\omega\mu_0 b\cdot a w_{10}^2. \]

The total losses in a slot-type eddy-current sensor in the absence of the product being inspected are characterized by the resistance

\[ z_0 = z_{\mathrm{v.z}} + z_{\mathrm{f.s}} + z_{\mathrm{ob}}. \]

The relative change in the total resistance of the sensor in the presence of the product is then determined as the ratio \(z_{\mathrm{iz}}/z_0\).

REFERENCES

1. D. V. Milyaev, V. K. Zhukov. Accounting for electromagnetic energy losses in a slot-type eddy-current sensor (the article is included in the present collection).