旋转磁场的同步转速 n s = 60 f 1 p n_s = \frac{60f_1}{p} ns​=p60f1​​

f 1 f_1 f1​：电源频率 p p p：极对数

f 2 = s f 1 f_2 = sf_1 f2​=sf1​

Ω = 2 π n 60 \Omega = 2\pi\frac{n}{60} Ω=2π60n​

s m = ± R 2 ′ R 1 2 + ( X 1 + X 2 ′ ) 2 s_m =\pm \frac{R_2'}{\sqrt{R_1^2 + (X_1 +X_2')^2}} sm​=±R12​+(X1​+X2′​)2 ​R2′​​

s m = s N ( K T + K T 2 − 1 ) s_m = s_N(K_T + \sqrt{K_T^2 - 1}) sm​=sN​(KT​+KT2​−1 ​)

s N = n s − n N n S s_N = \frac{n_s - n_N}{n_S} sN​=nS​ns​−nN​​

T = P Ω T= \frac{P}{\Omega} T=ΩP​

T e = p 2 π f 1 m 1 E 2 ′ I 2 ′ c o s φ 2 ′ T_e = \frac{p}{2\pi f_1}m_1E'_2I'_2cos\varphi'_2 Te​=2πf1​p​m1​E2′​I2′​cosφ2′​ 代 入 E 2 ′ = 2 π f 1 N 1 k w 1 Φ m ， 代入E'_2 = \sqrt{2}\pi f_1N_1k_{w1}\Phi_{m}， 代入E2′​=2 ​πf1​N1​kw1​Φm​， T e = C T 1 Φ m I 2 ′ c o s φ 2 ′ T_e = C_{T1}\Phi_mI'_2cos\varphi_2' Te​=CT1​Φm​I2′​cosφ2′​

其中， C T 1 = p m 1 N 1 k w 1 2 C_{T1} = \frac{pm_1N_1k_{w1}}{\sqrt2} CT1​=2 ​pm1​N1​kw1​​

P m e c h Ω = P e Ω s = T e \frac{P_{mech}}{\Omega} = \frac{P_e}{\Omega_s} = T_e ΩPmech​​=Ωs​Pe​​=Te​

Ω \Omega Ω：转子的机械角速度。 Ω \Omega Ω：旋转磁场的机械角速度。

T 0 = p m e c h + p Δ Ω T_0=\frac{p_{mech}+p_\Delta}{\Omega} T0​=Ωpmech​+pΔ​​

T o = P 2 Ω T_o = \frac{P_2}{\Omega} To​=ΩP2​​

T m a x = ± m 1 Ω s U ϕ 2 2 [ ± R 1 2 + R 1 2 + ( X 1 + X 2 ′ ) 2 ] T_{max} = \pm \frac{m_1}{\Omega_s} \frac{U_\phi^2}{2[\pm R_1^2 + \sqrt{R_1^2 + (X_1 + X_2')^2}]} Tmax​=±Ωs​m1​​2[±R12​+R12​+(X1​+X2′​)2 ​]Uϕ2​​

T m a x = K T T N T_{max} = K_TT_N Tmax​=KT​TN​

T N = P N Ω = 9550 P N n N T_N = \frac{P_N}{\Omega} = 9550\frac{P_N}{n_N} TN​=ΩPN​​=9550nN​PN​​

P N P_N PN​单位： K W KW KW n N n_N nN​单位： r / m i n r/min r/min

（1）主电动势 E ˙ 1 = − j 4.44 f 1 N 1 k w 1 Φ ˙ m \dot{E}_1 = -j4.44f_1N_1k_{w1}\dot{\Phi}_m E˙1​=−j4.44f1​N1​kw1​Φ˙m​

E ˙ 1 = − I ˙ m Z m = − I ˙ m ( R m + j X m ) \dot{E}_1 = -\dot{I}_mZ_m = -\dot{I}_m(R_m + jX_m) E˙1​=−I˙m​Zm​=−I˙m​(Rm​+jXm​)

（2）漏电动势 公式一： E ˙ 1 σ = − j 4.44 f 1 N 1 k w 1 Φ ˙ 1 σ \dot{E}_{1\sigma} = -j4.44f_1N_1k_{w1}\dot{\Phi}_{1\sigma} E˙1σ​=−j4.44f1​N1​kw1​Φ˙1σ​ 公式二 E ˙ 1 σ = − j I ˙ 1 X 1 \dot{E}_{1\sigma} = -j\dot{I}_1X_1 E˙1σ​=−jI˙1​X1​

（1）主电动势【 E ˙ 2 s \dot{E}_{2s} E˙2s​】 E ˙ 2 s = − j 4.44 f 2 N 2 k w 2 Φ ˙ m = − j 4.44 f 1 N 2 k w 2 Φ ˙ m s \dot{E}_{2s} =-j4.44 f_2N_2k_{w2}\dot{\Phi}_m=-j4.44 f_1N_2k_{w2}\dot{\Phi}_ms E˙2s​=−j4.44f2​N2​kw2​Φ˙m​=−j4.44f1​N2​kw2​Φ˙m​s （2）漏电动势【 E ˙ 2 σ s \dot{E}_{2\sigma s} E˙2σs​】 E ˙ 2 σ s = − j 4.44 f 2 N 2 k w 2 Φ ˙ 2 σ s \dot{E}_{2\sigma s} =-j4.44 f_2N_2k_{w2}\dot{\Phi}_{2\sigma s} E˙2σs​=−j4.44f2​N2​kw2​Φ˙2σs​ 由 于 f 2 = s f 1 带 入 得 由于f_2 = sf_1带入得 由于f2​=sf1​带入得，

E ˙ 2 σ s − j 4.44 s f 1 N 2 k w 2 Φ ˙ 2 σ s \dot{E}_{2\sigma s} -j4.44 sf_1N_2k_{w2}\dot{\Phi}_{2\sigma s} E˙2σs​−j4.44sf1​N2​kw2​Φ˙2σs​

P 1 = m 1 U 1 I 1 c o s φ 1 P_1 = m_1U_1I_1cos \varphi_1 P1​=m1​U1​I1​cosφ1​

p C u 1 = m 1 I 1 2 R 1 p_{Cu1} = m_1I_1^2R_1 pCu1​=m1​I12​R1​

p F e = m 1 I m 2 R m p_{Fe} = m_1I_m^2R_m pFe​=m1​Im2​Rm​

P e = m 1 E 2 ′ I 2 ′ c o s φ 2 ′ = m 1 I 2 ′ 2 R 2 ′ s P_{e} = m_1E'_2I'_2cos\varphi'_2 = m_1I_2'^2\frac{R'_2}{s} Pe​=m1​E2′​I2′​cosφ2′​=m1​I2′2​sR2′​​

p C u 2 = m 1 I 2 ′ 2 R 2 ′ p_{Cu2} = m_1I_2'^2R_2' pCu2​=m1​I2′2​R2′​ p C u 2 = s P e p_{Cu2} = sP_e pCu2​=sPe​

P m e c h = ( 1 − s ) P e P_{mech} = (1-s)P_e Pmech​=(1−s)Pe​