上海理工大学学报  2022, Vol. 44 Issue (5): 464-472 PDF

Calculation of eddy current loss of permanent magnet synchronous machine based on analytical model
ZHU Tuo, LI Zheng, LI Zi
School of Mechanical Engineering, University of Shanghai for Science and Technology, Shanghai 200093, China
Abstract: A rapid analytical model of rotor eddy current loss for surface mounted permanent magnet synchronous machine (PMSM) with retaining sleeve was proposed, accounting for the stator space-time harmonics, eddy current reaction and circumferential segmentation of permanent magnet. When considering the circumferential segmentation of permanent magnet, the secondary harmonics and their coupling effects were ignored to simplify the calculation process. The analytical model was applied to a six phase 24 slot 14 pole PMSM. Firstly, the convergence of the results was analyzed to reduce the truncation error and improve the calculation efficiency. Then the accuracy was verified by time-stepping finite element method. The average eddy current loss obtained is in good agreement with the analytical solution of the model. Finally, the eddy current density map in the permanent magnet layer was drawn by using the analytical model. This analytical model can quickly obtain the response surface of eddy current loss, and provide theoretical basis for motor design and optimization iteration.
Key words: surface mounted permanent magnet synchronous machine     eddy current loss     analytical model     permanent magnet segmentation     eddy current reaction     stator space-time harmonics

1 定子磁动势涡流损耗解析模型

 图 1 6相24槽14极永磁同步电机截面图 Fig. 1 Cross-sectional schematic of a 6-phase 24-slot 14-pole PMSM

1.1 简化假设

a. 将气隙、保护套和永磁体区域看作半无限平面；

b. 忽略端部效应，感应涡流只有轴向分量；

c. 忽略定子槽开口引起的气隙磁导变化；

d. 定子绕组电流由分布在定子槽开口的等效电流片模型表示；

e. 各个媒质是均匀且各向同性的，其电导率、磁导率是不变的常数；

f. 定子和转子铁心磁导率无穷大，电导率为0，忽略定子和转子铁心中产生的涡流；

 图 2 考虑分段的涡流扩散模型 Fig. 2 Eddy current diffusion model considering segmentation

1.2 等效电流片模型

 $\begin{split}{J_{\rm A}} =& \frac{{2N{i_{\rm A}}}}{{\text{π} {R_{\rm s}}}}\sum\limits_{\nu = 1}^\infty {{k_{{\rm o}}}} {k_{\rm p }}{k_{\rm d}}\sin (\nu \alpha ) =\\ &\frac{{2N{i_{\rm A}}}}{{\text{π} {R_{\rm s}}}}\sum\limits_{\nu = 1}^\infty {{k_{{\rm o} }}} {k_{\rm w }}\sin (\nu \alpha ) \end{split}$ (1)

6相绕组总的合成等效电流片为

 $\begin{split} J =& {J_{\rm A}} + {J_{\rm B}} + {J_{\rm C}} + {J_{\rm U}} + {J_{\rm V}} + {J_{\rm W}} = \\ & \left\{ {\begin{array}{*{20}{c}} - \displaystyle \sum\limits_{k = 1}^\infty {\displaystyle \sum\limits_{\nu = 1}^\infty {{J_{k\nu }}\sin (kp{\omega _{\rm r}}t - \nu \alpha )} }\qquad {k \ne 3n}, {\nu = 6n + k} \\ \displaystyle \sum\limits_{k = 1}^\infty {\displaystyle \sum\limits_{\nu = 1}^\infty {{J_{k\nu }}\sin (kp{\omega _{\rm r}}t + \nu \alpha )} } \qquad{k \ne 3n},{\nu = 6n - k} \\ \qquad0\qquad\qquad\qquad\qquad\nu \ne 6n \pm k \\ \begin{gathered} \sum\limits_{k = 1}^\infty {\sum\limits_{\nu = 1}^\infty {{J_{k\nu }}[\sin (kp{\omega _{\rm r}}t + \nu \alpha ) - \sin (kp{\omega _{\rm r}}t - \nu \alpha )]} } \;\;\; \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;k = 3n,\nu = 6n \pm k \\ \end{gathered} \end{array}} \right. \\ \end{split}$ (2)

k≠3n时，将式(2)变形如下：

 \begin{aligned} J =& \sum\limits_{k = 1}^\infty {\sum\limits_{\nu = - \infty }^\infty {{J_{k\nu }}\sin (kp{\omega _{\rm r}}t + \nu \alpha )} } {\text{ = }} \\ &\sum\limits_{k = 1}^\infty {\sum\limits_{\nu = - \infty }^\infty {{\rm {Im}} \left\{ {{J_{k\nu }}\exp ({\rm j} kp{\omega _{\rm r}}t)\exp \left({\rm j}\dfrac{{\nu \text{π} }}{\tau }x\right)} \right\}} } \end{aligned} (3)

 $J = \sum\limits_{k = 1}^\infty {\sum\limits_{\nu = - \infty }^\infty {{\rm {Im}} \left\{ {{J_{k\nu }}\exp [{\rm j}(kp + \nu ){\omega _{\rm r}}t]\exp \left({\rm j}\frac{{\nu \text{π} }}{\tau }{x_{\rm r}}\right)} \right\}} }$ (4)

 ${\dot {J}(x)}\text={J}_{k\nu }{\rm e}^{{\rm j}\beta {x}_{\rm r}}$ (5)

1.3 各计算域中的控制方程及通解

a. 气隙域

 ${\Delta}{A}_{\rm g}=0$ (6)

 $\Delta {A}_{\rm g}\text=\frac{{\partial }^{2}{A}_{\rm g}}{\partial {x}^{2}}+\frac{{\partial }^{2}{A}_{\rm g}}{\partial {y}^{2}}$ (7)

 ${A}_{\rm g}={\displaystyle \sum _{k=1}^{\infty }{\displaystyle \sum _{\nu =-\infty }^{\infty }[A{\rm e}^{\beta y}+B{\rm e}^{-\beta y}]}}{\rm e}^{{\rm j}\beta x}$ (8)

 ${B_x} = \frac{{\partial {A_z}}}{{\partial y}}$ (9)

 ${H}_{x{\rm g}}=\frac{1}{{\mu }_{0}}{B}_{x{\rm g}}\text={\displaystyle \sum _{k=1}^{\infty }{\displaystyle \sum _{\nu =-\infty }^{\infty }\frac{\beta }{{\mu }_{0}}[A{\rm e}^{\beta y}-B{\rm e}^{-\beta y}]}}{\rm e}^{{\rm j}\beta x}$ (10)

b. 保护套域

 $\Delta {A}_{\rm s}={\mu }_{0}{\mu }_{\rm{s}}{\sigma }_{\rm{s}}\frac{\partial {A}_{\rm s}}{\partial t}={\rm j}(kp+\nu ){\omega }_{\rm r}{\mu }_{0}{\mu }_{\rm{s}}{\sigma }_{\rm{s}}{A}_{\rm s}$ (11)

 ${A}_{\rm s}={\displaystyle \sum _{k=1}^{\infty }{\displaystyle \sum _{\nu =-\infty }^{\infty }[C{\rm e}^{{\lambda }_{\rm s}y}+D{\rm e}^{-{\lambda }_{\rm s}y}]}}{\rm e}^{{\rm j}\beta x}$ (12)

 ${H}_{x{\rm s}} = \frac{1}{{\mu }_{0}{\mu }_{\rm{s}}}{B}_{x{\rm s}} = {\displaystyle \sum _{k=1}^{\infty }{\displaystyle \sum _{\nu =-\infty }^{\infty }\frac{{\lambda }_{{\rm s}}}{{\mu }_{0}{\mu }_{\rm{s}}}[C{\rm e}^{{\lambda }_{{\rm s}}y} - D{\rm e}^{-{\lambda }_{{\rm s}}y}]}}{\rm e}^{{\rm j}\beta x}$ (13)
 ${J}_{\rm s} = -{\sigma }_{\rm {s}}\frac{\partial {A}_{\rm s}}{\partial t} = -{\displaystyle \sum _{k=1}^{\infty }{\displaystyle \sum _{\nu =-\infty }^{\infty }{\rm j}{\sigma }_{\rm {s}}(kp + \nu ){\omega }_{{\rm{r}}}[C{\rm e}^{{\lambda }_{\rm s}y} + D{\rm e}^{-{\lambda }_{\rm s}y}]}}{\rm e}^{{\rm j}\beta x}$ (14)

c. 永磁体域

 ${x_{{\rm l}\lambda \rho }} = (\rho - 1){\tau _{\rm m}} - \frac{{{b_{\rm m}}}}{2} + (\lambda - 1)\frac{{{b_{\rm m}}}}{s} = {x_{{\rm r}\lambda \rho }} - \frac{{{b_{\rm m}}}}{s}$ (15)

 $\Delta {A}_{\rm m}={\mu }_{0}{\mu }_{\rm m}{\sigma }_{\rm m}(x)\frac{\partial {A}_{\rm m}}{\partial t}-{\mu }_{0}{\mu }_{\rm m}{J}_{\rm a}(x)$ (16)

 ${J}_{\rm a}(x) = \left\{\begin{array}{*{20}{c}}\displaystyle \sum _{k=1}^{\infty }\displaystyle \sum _{\nu =-\infty }^{\infty }\frac{{\sigma }_{\rm m}s}{{b}_{\rm m}m}\displaystyle {\int }_{{x}_{\rm l\lambda \rho }}^{{x}_{\rm r\lambda \rho }}\displaystyle {\int }_{0}^{m}\frac{\partial {A}_{\rm m}}{\partial t}{\rm d}y{\rm d}x & x\in ({x}_{\rm l\lambda \rho },{x}_{\rm r\lambda \rho })\\ 0 & x\notin ({x}_{\rm l\lambda \rho },{x}_{\rm r\lambda \rho })\end{array}\right.$ (17)
 ${\sigma _{\rm m}}(x) = \left\{ {\begin{array}{*{20}{c}} {\sigma _{\rm m}}&x \in ({x_{\rm l\lambda \rho }},{x_{\rm r\lambda \rho }}) \\ 0&x \notin ({x_{\rm l\lambda \rho }},{x_{\rm r\lambda \rho }}) \end{array}} \right.$ (18)

 ${J_{\rm a}}(x) = \sum\limits_{k = 1}^\infty {\sum\limits_{\nu = - \infty }^\infty {{J_{\rm a}}{{\rm e}^{{\rm j}\beta x}}} }$ (19)
 ${\sigma _{\rm m}}(x) = {a_{\rm p}}{\sigma _{\rm m}}$ (20)

 ${A}_{\rm m}={\displaystyle \sum _{k=1}^{\infty }{\displaystyle \sum _{\nu =-\infty }^{\infty }\left[E{\rm e}^{{\lambda }_{\rm m}y}+F{\rm e}^{-{\lambda }_{\rm m}y}+\frac{{\mu }_{0}{\mu }_{ \rm m}}{{\lambda }_{\rm m}^{2}}{J}_{\rm a }\right]}}{\rm e}^{{\rm j}\beta x}$ (21)

 ${H}_{x{\rm m}}\text={\displaystyle \sum _{k=1}^{\infty }{\displaystyle \sum _{\nu =-\infty }^{\infty }\frac{{\lambda }_{\rm m}}{{\mu }_{0}{\mu }_{ m}}[E{\mathrm{e}}^{{\lambda }_{\rm m}y}-F{\mathrm{e}}^{-{\lambda }_{\rm m}y}]}}{\mathrm{e}}^{{\rm j}\beta x}$ (22)

 $\begin{split} {J}_{{\rm m},\lambda \rho }=&-{\sigma }_{\rm m}\frac{\partial {A}_{\rm m}}{\partial t}+\displaystyle \sum _{k=1}^{\infty }\displaystyle \sum _{\nu =-\infty }^{\infty }{J}_{\rm a,\lambda \rho }(x) =\\ &\displaystyle \sum _{k=1}^{\infty }\displaystyle \sum _{\nu =-\infty }^{\infty }-{\rm j}{\sigma }_{\rm m}(kp+\nu ){\omega }_{\rm r}\Biggr(E{\rm e}^{{\lambda }_{\rm m}y}+\\ &F{\rm e}^{-{\lambda }_{\rm m}y}+\dfrac{{\mu }_{0}{\mu }_{\rm m}{J}_{\rm a}}{{\lambda }_{\rm m}^{2}}\Biggr){\rm e}^{{\rm j}\beta x}+{J}_{\rm a,\lambda \rho }(x)\end{split}$ (23)
1.4 各计算域间的边界条件

 $当y= g+l+m\;时，{H}_{x{\rm g}}=J$ (24)
 $当y=l+m\;时，{A}_{\rm g}={A}_{\rm s}\;，\;{H}_{x{\rm g}}={H}_{x{\rm s}}$ (25)
 $当y=m\;时，{A}_{\rm s}={A}_{\rm m}\; ，\; {H}_{\rm s}={H}_{x{\rm m}}$ (26)
 $当y=0\;时，{H}_{x{\rm m}}=0$ (27)

1.5 源项系数求解

 ${J}_{\rm a,\lambda \rho }(x)=\frac{{\sigma }_{\rm m}s}{{b}_{\rm m}m}{\displaystyle {\int }_{{x}_{\rm l\lambda \rho }}^{{x}_{\rm r\lambda \rho }}{\displaystyle {\int }_{0}^{m}\frac{\partial {A}_{\rm m}}{\partial t}{\rm d}y{\rm d}x}}$ (28)

 ${F_{\rm {cr}}} = \frac{1}{\tau }\sum\limits_{\rho = 1}^{2p} {\sum\limits_{\lambda = 1}^s {\int_{{x_{\rm l\lambda \rho }}}^{{x_{\rm r\lambda \rho }}} {{{\rm {Re}}} \left\{ {{J_{\rm a,\lambda \rho }}(x)} \right\}\cos (\beta x){\rm d} x} } }$ (29)
 ${F_{\rm{sr}}} = \frac{1}{\tau }\sum\limits_{\rho = 1}^{2p} {\sum\limits_{\lambda = 1}^s {\int_{{x_{\rm l\lambda \rho }}}^{{x_{\rm r\lambda \rho }}} {{{\rm{Re}}} \left\{ {{J_{\rm a,\lambda \rho }}(x)} \right\}\sin (\beta x){\rm d}x} } }$ (30)
 ${F_{\rm {ci}}} = \frac{1}{\tau }\sum\limits_{\rho = 1}^{2p} {\sum\limits_{\lambda = 1}^s {\int_{{x_{\rm l\lambda \rho }}}^{{x_{\rm r\lambda \rho }}} {{{\rm{Im}}} \left\{ {{J_{\rm a,\lambda \rho }}(x)} \right\}\cos (\beta x){\rm d}x} } }$ (31)
 ${F_{\rm {si}}} = \frac{1}{\tau }\sum\limits_{\rho = 1}^{2p} {\sum\limits_{\lambda = 1}^s {\int_{{x_{\rm l\lambda \rho }}}^{{x_{\rm r\lambda \rho }}} {{{\rm{Im}}} \left\{ {{J_{\rm a,\lambda \rho }}(x)} \right\}\sin (\beta x){\rm d}x} } }$ (32)

 ${\rm {Re}} \left\{ {{J_{\rm a}}} \right\} = \frac{1}{2}({F_{\rm {cr}}} + {F_{\rm{ci}}} + {F_{\rm {sr}}} + {F_{\rm{si}}})$ (33)
 ${\rm{Im}} \left\{ {{J_{\rm a}}} \right\} = \frac{1}{2}({F_{\rm{cr}}} + {F_{\rm{ci}}} - {F_{\rm{sr}}} - {F_{\rm{si}}})$ (34)

 ${J_{\rm a}}{\text{ = }}{\rm Re} \left\{ {{J_{\rm a}}} \right\} + {\rm Im} \left\{ {{J_{\rm a}}} \right\}{\rm j}$ (35)

 ${A}_{\rm m}\text=f({J}_{\rm a}\text{)=}f(\mathrm{Re}\left\{{J}_{\rm a}\right\}+\mathrm{Im}\left\{{J}_{\rm a}\right\}\rm j)$ (36)

 $\left\{\begin{array}{c}\mathrm{Re}\left\{{J}_{\rm a}\right\}={{g}}_{1}({A}_{\rm m}\text{)=}{{g}}_{1}(f(\mathrm{Re}\left\{{J}_{\rm a}\right\}+\mathrm{Im}\left\{{J}_{\rm a}\right\}{\rm j})\text{)}\\ \mathrm{Im}\left\{{J}_{\rm a}\right\}={{g}}_{2}({A}_{\rm m}\text{)=}{{g}}_{2}(f(\mathrm{Re}\left\{{J}_{\rm a}\right\}+\mathrm{Im}\left\{{J}_{\rm a}\right\}{\rm j})\text{)}\end{array}\right.$ (37)

1.6 涡流损耗求解

 ${P}_{{\rm m},\lambda \rho }={\displaystyle \sum _{k=1}^{\infty }{\displaystyle \sum _{\nu =-\infty }^{\infty }\frac{{L}_{\rm a}}{2{\sigma }_{\rm m}}}}{\displaystyle {{\int }^{m}_{0}}{\displaystyle{{\int }^{{x}_{\rm r\lambda \rho }}_{{x}_{\rm l\lambda \rho }}}{J}_{{\rm m},\lambda \rho }{J}_{{\rm m},\lambda \rho }^{\ast }}}{\rm d}x{\rm d}y$ (38)

 ${P}_{\rm m}\text=2ps{P}_{{\rm m},\lambda \rho }$ (39)

 ${P}_{\rm s}={\displaystyle \sum _{k=1}^{\infty }{\displaystyle \sum _{\nu =-\infty }^{\infty }\frac{{L}_{\rm a}p{\tau }_{\rm s}}{{\sigma }_{\rm s}}}}{\displaystyle {{\int }^{m+l}_{m}}{J}_{\rm s}}{J}_{\rm s}^{\ast }{\rm d}y$ (40)

2 计算结果及分析 2.1 电机参数

2.2 收敛性分析

 图 3 转子涡流损耗收敛曲线 Fig. 3 Convergence curve of rotor eddy current loss

2.3 精度分析

 图 4 转子涡流损耗对比 Fig. 4 Comparison of rotor eddy current loss

 图 5 不同分段数下的永磁体涡流损耗 Fig. 5 Eddy current loss of permanent magnet under different segment numbers

 图 6 永磁体层中的涡流电密 Fig. 6 Eddy current density in permanent magnet layer
2.4 影响因素分析

 图 7 绕组节距和极弧系数示意图 Fig. 7 Schematic diagram of winding pitch and pole arc coefficient

 图 8 永磁体周向分段示意图 Fig. 8 Circumferential segmentation diagram of permament magnet

 图 9 不同永磁体分段数下涡流损耗响应面 Fig. 9 Response surface of eddy current loss under different segments of permanent magnet
3 结　论

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