An asymptotic approximation of the magnetic field and forces in electrical machines with rotor eccentricity

Abstract

In this contribution, a systematic asymptotic derivation of approximate solutions to the magnetic field problem in rotating electric machines with an eccentrically running rotor is proposed. It is consistent with the well-known permeance harmonic method and can be seen as an alternative view on the derivation of the magnetic flux density and resulting lateral forces in the air gap of the machine. The asymptotic expansion is based on a purely geometric small parameter and applies even for large rotor eccentricities. The advantage of this method lies in its extensibility towards higher approximation orders, and in the fact that it is possible to quantify the order of the involved approximation error. The discussion includes two examples and a numerical verification.

Appendix

Within this appendix, some peculiarities concerning the calculation of the electromagnetic force as described in Sects. 3 and 4 will be discussed and part of the magnetic potential of order $\mathcal O(\varepsilon ^2)$ will be shown.

To derive the Force, one can start from expression (28) for a first-order approach and insert the radial flux density (Eq. (26)). The force formula reads

$$\begin{aligned} \mathbf {F}_{\text {elm}, R} =&\frac{\mu _0 r_1 \ell }{2 \delta _\mathrm{m}^2} \int _0^{2 \pi } \underbrace{\frac{(V_2(\theta ,t) - V_1(\theta ,t) - V_0(t))^2}{(1-\tilde{e}\cos \theta )^2}}_{g(\theta )} \nonumber \\&(\cos \theta \mathbf {e}_\delta + \sin \theta \mathbf {e}_t) \text {d} \theta . \end{aligned}$$

(40)

Due to the general orthogonality of trigonometric functions and the instance that the term $g(\theta )$ is multiplied with $\cos \theta$ and $\sin \theta$, respectively, the involved task reduces to calculating the first harmonic of g. To do so, $h(\theta ) = \frac{1}{(1-\tilde{e}\cos \theta )^2}$ can be expanded into a Fourier series $\mathcal F(h(\theta ))$. After that, the product $g(\theta ) = \mathcal F((V_2(\theta ,t) - V_1(\theta ,t) - V_0(t))^2)\mathcal F(h(\theta ))$ is evaluated as a convolution of the Fourier coefficients.

The derivation of the Fourier series of h can be done using complex phasors. Its result

$$\begin{aligned} g(\theta ) = \frac{a_0}{2} + \sum _{n=0}^\infty a_n \cos (n \theta ), \end{aligned}$$

(41)

with

$$\begin{aligned} a_n = 2\frac{(1-\sqrt{(1-\tilde{e}^2)})^n(n \sqrt{1-\tilde{e}^2}+1)}{(1-\tilde{e}^2)^{\frac{3}{2}}\tilde{e}^n}, \end{aligned}$$

(42)

is similar to the Fourier series of the air gap permeance, as it is often used for the linear analysis in literature [13]. Furthermore, the similarity of the involved integrals and the Sommerfeld integrals in the theory of journal bearings [28] shall be mentioned. Their analogy is remarkable and also perspicuous, due to the similarity of the calculation procedure of the forces.

Evaluating as described above, formula (28) for the shown example can be evaluated in a straight forward manner, resulting in Eq. (34).

The second issue of this appendix is to provide the formula for the part of order $\mathcal O(\varepsilon ^2)$ of the magnetic potential. It reads

$$\begin{aligned}&\mathcal V^{(2)} = \bigg [F_1(\psi )\left( \frac{{\tilde{s}}}{{\tilde{\delta }}^{(0)}}\right) ^2 + F_2(\psi )\frac{{\tilde{s}}}{{\tilde{\delta }}^{(0)}} + F_3(\psi )\bigg ] \nonumber \\&\quad \frac{{\tilde{s}}}{{\tilde{\delta }}^{(0)}}\big (V_1 + V_0 - V_2^{(0)}\big ) + \bigg [ F_4(\psi )\left( \frac{{\tilde{s}}}{{\tilde{\delta }}^{(0)}}\right) ^2 + F_5(\psi ) \nonumber \\&\quad \frac{{\tilde{s}}}{{\tilde{\delta }}^{(0)}} + F_6(\psi ) \bigg ]\frac{{\tilde{s}}}{{\tilde{\delta }}^{(0)}}\frac{\partial }{\partial \psi }\big (V_1 + V_0 - V_2^{(0)}\big ) + \frac{({\tilde{\delta }}^{(0)})^2}{6} \nonumber \\&\quad \bigg [ \left( \frac{{\tilde{s}}}{{\tilde{\delta }}^{(0)}}\right) ^2 - 1\bigg ] \frac{{\tilde{s}}}{{\tilde{\delta }}^{(0)}}\frac{\partial ^2}{\partial \psi ^2}\big (V_1 + V_0 - V_2^{(0)}\big ) + F_7(\psi )\nonumber \\&\quad \left( \frac{{\tilde{s}}}{{\tilde{\delta }}^{(0)}}\right) ^2 + F_8(\psi ) \frac{{\tilde{s}}}{{\tilde{\delta }}^{(0)}}, \end{aligned}$$

(43)

where

$$\begin{aligned} F_1(\psi ) =&\, \frac{1}{12}\bigg ( 4 ({\tilde{\delta }}^{(0)})^2 + 4 \bigg (\frac{\partial \delta ^{(0)}}{\partial \psi }\bigg )^2 - 2\frac{\partial ^2 {\tilde{\delta }}^{(0)}}{\partial \psi ^2} {\tilde{\delta }}^{(0)} \bigg ), \\ F_2(\psi ) =&\, \frac{1}{4}\bigg ( 4 {\tilde{\delta }}^{(0)}{\tilde{e}} \cos \psi + 2 {\tilde{\delta }}^{(1)} - ({\tilde{\delta }}^{(0)})^2 - 4 \frac{\partial {\tilde{\delta }}^{(0)}}{\partial \psi } {\tilde{e}} \sin \psi \bigg ), \\ F_3(\psi ) =&\, \frac{1}{12}\bigg ( 2 {\tilde{\delta }}^{(0)} \frac{\partial ^2 {\tilde{\delta }}^{(0)}}{\partial \psi ^2} + 12 \frac{\partial {\tilde{\delta }}^{(0)}}{\partial \psi }{\tilde{e}} \sin \psi - ({\tilde{\delta }}^{(0)})^2 \\&- 12 {\tilde{\delta }}^{(0)}{\tilde{e}} \cos \psi - 4\bigg (\frac{\partial {\tilde{\delta }}^{(0)}}{\partial \psi }\bigg )^2 + 12 \bigg (\frac{{\tilde{\delta }}^{(1)}}{{\tilde{\delta }}^{(0)}}\bigg )^2\bigg ), \\ F_4(\psi ) =&\, -\frac{1}{3} {\tilde{\delta }}^{(0)} \frac{\partial {\tilde{\delta }}^{(0)}}{\partial \psi }, \quad F_5(\psi ) =\, {\tilde{\delta }}^{(0)}{\tilde{e}} \sin \psi , \\ F_6(\psi ) =&\, \frac{1}{3}\bigg ({\tilde{\delta }}^{(0)} \frac{\partial {\tilde{\delta }}^{(0)}}{\partial \psi } - 3 {\tilde{\delta }}^{(0)} {\tilde{e}} \sin \psi \bigg ), \\ F_7(\psi ) =&\, \frac{1}{2}\bigg ({\tilde{\delta }}^{(0)}V_2^{(1)} + ({\tilde{\delta }}^{(0)})^2 \frac{\partial ^2 V_2^{(0)}}{\partial \psi ^2}\bigg ), \\ F_8(\psi ) =&\, \frac{1}{2}\bigg ( 2 \frac{{\tilde{\delta }}^{(1)}}{{\tilde{\delta }}^{(0)}}V_2^{(1)} - {\tilde{\delta }}^{(0)}V_2^{(1)} - ({\tilde{\delta }}^{(0)})^2 \frac{\partial ^2 V_2^{(0)}}{\partial \psi ^2}\bigg ). \end{aligned}$$

As one can see, the solution becomes very unhandy.