Magma Sanctum

複素関数論による連続体力学

誘導電動機トルクの数理的導出

 

Δ0（'26/02/16）書きかけの記事です．数式だけ羅列していますが，後日体裁を整えていきます．

 

固定子 r = 1 ，回転子 r = q < 1 とする系 z = re^{i\theta} について考える．

\frac{\mathrm{d}F}{\mathrm{d}{z}} = C_0 + \displaystyle\sum_{n = -\infty}^{\infty} C_n z^n

\mathrm{Re} \left( \left. \frac{\mathrm{d}F}{\mathrm{d}{z}} \right|_{r=1} \right) = H_{1\theta}(\theta)

\mathrm{Re} \left( \left. \frac{\mathrm{d}F}{\mathrm{d}{z}} \right|_{r=q} \right) = H_{2\theta}(\theta)

とすると，これを満たす係数 C_n は，

C_0 = \frac{1}{2\pi} \displaystyle\int_0^{2\pi}H_{1\theta}\,\mathrm{d}\theta + i \mathrm{Im}(C_0) = \frac{1}{2\pi} \displaystyle\int_0^{2\pi}H_{2\theta}\,\mathrm{d}\theta + i \mathrm{Im}(C_0)

C_n = \frac{1}{\pi(1-q^{2n})} \left\{ \displaystyle\int_0^{2\pi}H_{1\theta}e^{-in\theta}\,\mathrm{d}\theta - q^n \displaystyle\int_0^{2\pi}H_{2\theta}e^{-in\theta}\,\mathrm{d}\theta \right\}

\frac{\mathrm{d}F}{\mathrm{d}{z}} = C_0 + \displaystyle\sum_{n = -\infty}^{\infty}\frac{1}{\pi(1-q^{2n})} \left\{ \displaystyle\int_0^{2\pi}H_{1\theta}e^{-in\theta}\,\mathrm{d}\theta - q^n \displaystyle\int_0^{2\pi}H_{2\theta}e^{-in\theta}\,\mathrm{d}\theta \right\} z^n

H_\theta(r,\theta) = \mathrm{Re} \left( \frac{\mathrm{d}F}{\mathrm{d}z} \right)

H_r(r,\theta) = \mathrm{Im} \left( \frac{\mathrm{d}F}{\mathrm{d}z} \right) = \mathrm{Im}(C_0) + \frac{1}{\pi} \displaystyle\sum_{n = -\infty}^{\infty} \frac{r^n}{1-q^{2n}} \left\{ \left( \displaystyle\int_0^{2\pi} H_{1\theta}\cos{n\theta}\,\mathrm{d}\theta \right) \sin{n\theta} - \left( \displaystyle\int_0^{2\pi} H_{1\theta}\sin{n\theta}\,\mathrm{d}\theta \right) \cos{n\theta} \right\} - \frac{q^nr^n}{1-q^{2n}} \left\{ \left( \displaystyle\int_0^{2\pi} H_{2\theta}\cos{n\theta}\,\mathrm{d}\theta \right) \sin{n\theta} - \left( \displaystyle\int_0^{2\pi} H_{2\theta}\sin{n\theta}\,\mathrm{d}\theta \right) \cos{n\theta} \right\}

以下，回転子 r = q について考える．

H_{2\theta}(\theta) = H_{2\theta,0} + \frac{1}{\pi} \displaystyle\sum_{n = 1}^{\infty} \left\{ \left( \displaystyle\int_0^{2\pi} H_{2\theta}\cos{n\theta}\,\mathrm{d}\theta \right) \cos{n\theta} + \left( \displaystyle\int_0^{2\pi} H_{2\theta}\sin{n\theta}\,\mathrm{d}\theta \right) \sin{n\theta} \right\}

H_{2r}(\theta) = \mathrm{Im}(C_0) + \frac{1}{\pi} \displaystyle\sum_{n = 1}^{\infty} \frac{2q^n}{1-q^{2n}} \left\{ \left( \displaystyle\int_0^{2\pi} H_{1\theta}\cos{n\theta}\,\mathrm{d}\theta \right) \sin{n\theta} - \left( \displaystyle\int_0^{2\pi} H_{1\theta}\sin{n\theta}\,\mathrm{d}\theta \right) \cos{n\theta} \right\} - \frac{1+q^{2n}}{1-q^{2n}} \left\{ \left( \displaystyle\int_0^{2\pi} H_{2\theta}\cos{n\theta}\,\mathrm{d}\theta \right) \sin{n\theta} - \left( \displaystyle\int_0^{2\pi} H_{2\theta}\sin{n\theta}\,\mathrm{d}\theta \right) \cos{n\theta} \right\}

ここで，直交関数列の性質を利用して，

\displaystyle\int_0^{2\pi} \cos{m\theta}\cos{n\theta}\,\mathrm{d}\theta = \displaystyle\int_0^{2\pi} \sin{m\theta}\sin{n\theta}\,\mathrm{d}\theta = \pi\delta_{m,n}

\displaystyle\int_0^{2\pi} \cos{m\theta}\sin{n\theta}\,\mathrm{d}\theta = 0

より，

\displaystyle\int_0^{2\pi} H_{2\theta}(\theta) \cdot H_{2r}(\theta)\,\mathrm{d}\theta = 2\pi H_{2\theta,0}\mathrm{Im}(C_0) + \frac{1}{\pi} \displaystyle\sum_{n = 1}^{\infty} \frac{2q^n}{1-q^{2n}} \left\{ \left( \displaystyle\int_0^{2\pi} H_{1\theta}\cos{n\theta}\,\mathrm{d}\theta \right) \left( \displaystyle\int_0^{2\pi} H_{2\theta}\sin{n\theta}\,\mathrm{d}\theta \right) - \left( \displaystyle\int_0^{2\pi} H_{2\theta}\cos{n\theta}\,\mathrm{d}\theta \right) \left( \displaystyle\int_0^{2\pi} H_{1\theta}\sin{n\theta}\,\mathrm{d}\theta \right) \right\}

よって，マクスウェル応力に基づく回転子トルクは

T = R^2 \mu_0 \displaystyle\int_0^{2\pi} H_{2\theta}(\theta) \cdot H_{2r}(\theta)\,\mathrm{d}\theta

となる．

 

誘導電動機について考える．

0 = - R_b J_{r,k-1} + 2 R_b J_{r,k} - R_b J_{r,k+1} + \frac{\mathrm{d}\Phi_{r,k}}{\mathrm{d}{t}} =R_b (i_{b,k}-i_{b,k-1}) + \frac{\mathrm{d}\Phi_{r,k}}{\mathrm{d}{t}}

0 = R_b \dfrac{i_{b,k}-i_{b,k-1}}{\Delta\theta} + qR\dfrac{1}{\Delta\theta} \displaystyle\int_{\Delta\theta} \frac{\mathrm{d}}{\mathrm{d}{t}}b_{2r}(\theta)\,\mathrm{d}\theta

i_2 = \frac{i_b}{\Delta\theta} より，

0 = R_b \Delta\theta\dfrac{i_2(\theta_k)-i_2(\theta_{k-1})}{\Delta\theta} + qR\dfrac{1}{\Delta\theta} \displaystyle\int_{\Delta\theta} \frac{\mathrm{d}}{\mathrm{d}{t}}b_{2r}(\theta)\,\mathrm{d}\theta

R_2 = \dfrac{3}{N_2}R_b より， R_b\Delta\theta = R_b\dfrac{2\pi}{N_2} = \dfrac{2}{3} \pi R_2 なので，

両辺の極限 \Delta\theta \rightarrow 0 をとることで，

0 = \dfrac{2}{3} \pi R_2 \dfrac{\mathrm{d}i_2}{\mathrm{d}\theta} + qR \dfrac{\mathrm{d}b_{2r}}{\mathrm{d}t}

i_1 = i_{1m} \sin{(p\theta + \Delta\theta_1)}

i_2 = i_{2m} \sin{(p\theta + \Delta\theta_2)}

\Delta\theta = ps \omega_1 t + \mathrm{const.}

とすると，

H_{1\theta} = - \dfrac{1}{R} \dfrac{\mathrm{\partial}F_1}{\mathrm{\partial}\theta} = - \dfrac{i_1}{R}

H_{2\theta} = - \dfrac{1}{qR} \dfrac{\mathrm{\partial}F_2}{\mathrm{\partial}\theta} = - \dfrac{i_2}{qR}

より，（ただし， p ：極対数， \omega_1 = \dfrac{2 \pi f}{p} ：同期角速度）

b_{2r}(\theta) = \dfrac{2q^p}{1-q^{2p}} \dfrac{\mu_0}{R} i_{1m} \cos (p\theta + \Delta\theta_1) + \dfrac{1 + q^{2p}}{1 - q^{2p}} \dfrac{\mu_0}{qR} i_{2m} \cos (p\theta + \Delta\theta_2) \sim \dfrac{2}{1 - q^{2p}} \dfrac{\mu_0}{R} \left\{ i_{1m} \cos (p\theta + \Delta\theta_1) + i_{2m} \cos (p\theta + \Delta\theta_2) \right\}

合成関数の微分より，

\dfrac{\mathrm{\partial}b_{2r}}{\mathrm{\partial}t} = - \dfrac{2}{1 - q^{2p}} \dfrac{\mu_0}{R} \left\{ i_{1m} \sin (p\theta + \Delta\theta_1) + i_{2m} \sin (p\theta + \Delta\theta_2) \right\} \cdot p s \omega_1

0 = \dfrac{2}{3} \pi R_2 \dfrac{\mathrm{d}i_2}{\mathrm{d}\theta} + q R \dfrac{\mathrm{\partial}b_{2r}}{\mathrm{\partial}t}

より，

\dfrac{2}{3} \pi R_2 \dfrac{\mathrm{d}i_2}{\mathrm{d}\theta} + q R \dfrac{\mathrm{\partial}b_{2r}}{\mathrm{\partial}t} = \dfrac{2 \pi}{3} p R_2 i_{2m} \cos (p \theta + \Delta \theta_2) - \dfrac{2}{1 - q^{2 p}} \mu_0 p s \omega_1 \left\{ i_{1 m} \sin (p \theta + \Delta \theta_1) + i_{2 m} \sin (p \theta + \Delta \theta_2) \right\} = 0

\left\{ \dfrac{2 \pi}{3} R_2 i_{2 m} \cos \Delta \theta_2 - \dfrac{2}{1 - q^{2 p}} \mu_0 s \omega_1 (i_{1 m} \sin \Delta \theta_1 + i_{2 m} \sin \Delta \theta_2) \right\} \cos p \theta - \left\{ \dfrac{2 \pi}{3} R_2 i_{2 m} \sin \Delta \theta_2 + \dfrac{2}{1 - q^{2 p}} \mu_0 s \omega_1 (i_{1 m} \cos \Delta \theta_1 + i_{2 m} \cos \Delta \theta_2) \right\} \sin p \theta= 0

上式が \theta に関係なく成り立つため，左辺第一項係数 = 0 より，

\dfrac{2}{1 - q^{2 p}} \mu_0 s \omega_1 i_{1 e} \sin \Delta \theta_1 = \left( \dfrac{2 \pi}{3} R_2 \cos \Delta \theta_2 - \dfrac{2}{1 - q^{2 p}} \mu_0 s \omega_1 \sin \Delta \theta_2 \right) i_{2 e} = \left[ \left( \dfrac{2 \pi}{3} R_2 \right)^2 + \left( \dfrac{2}{1 - q^{2 p}} \mu_0 s \omega_1 \right)^2 \right]^{1/2} i_{2 e}

よって，

\sin (\Delta \theta_1 - \Delta \theta_2) = \sin \alpha = \dfrac{2 \pi}{3} R_2 \left[ \left( \dfrac{2 \pi}{3} R_2 \right)^2 + \left( \dfrac{2}{1 - q^{2 p}} \mu_0 s \omega_1 \right)^2 \right]^{- 1/2}  

\dfrac{2}{1 - q^{2 p}} \mu_0 s \omega_1 i_{1 e} = \left[ \left( \dfrac{2 \pi}{3} R_2 \right)^2 + \left( \dfrac{2}{1 - q^{2 p}} \mu_0 s \omega_1 \right)^2 \right]^{1/2} i_{2 e}

トルクは，

T = \pi R^2 \mu_0 \dfrac{2 q^p}{1 - q^{2 p}} H_{1 \theta m} H_{2 \theta m} \sin (\Delta \theta_1 - \Delta \theta_2)

より，

T = - 4 \pi \mu_0 \dfrac{q^{p-1}}{1 - q^{2 p}} i_{1 e} i_{2 e} \sin (\Delta \theta_1 - \Delta \theta_2) \sim - \dfrac{4 \pi^2}{3} \dfrac{R_2}{s \omega_1} i_{2 e}^2 \sim - \dfrac{\left( \frac{2}{1 - q^{2 p}} \mu_0 s \omega_1 \right)^2}{\left( \frac{2 \pi}{3} R_2 \right)^2 + \left( \frac{2}{1 - q^{2 p}} \mu_0 s \omega_1 \right)^2} \dfrac{4 \pi^2}{3} \dfrac{R_2}{s \omega_1} i_{1 e}^2

T = - \dfrac{1}{3} \dfrac{R_2}{s \omega_1} N_2^2 i_{b e}^2 \sim - \dfrac{\left( \frac{2}{1 - q^{2 p}} \mu_0 s \omega_1 \right)^2}{\left( \frac{2 \pi}{3} R_2 \right)^2 + \left( \frac{2}{1 - q^{2 p}} \mu_0 s \omega_1 \right)^2} \dfrac{1}{3} \dfrac{R_2}{s \omega_1} N_1^2 i_{1 e}^2

ただし， N_1 は一次全巻数， N_2 は二次スロット数

 

 

#EOF