(4)相电势E=44k.f,中 所以E=444kfy =444108×09452×50×1015×10 23002V p, =4.440ko3 3f o =444×108×(-0.5714)×3×50×0.66×102 44k4.5f 40.236V Eo =4.440 kor o =9.163V 相电势EE+E2E2,+E2=02y 线电势中E内相互抵消了,所以 E=√3×E2+E2+E2=4048 14-7 解:发电机极距=Z÷2P=27 每极每相槽数q=Z÷2Pm=54÷(2×1×3)=9 槽角度a=360=3560=667 Sin9× 667 所以k=Sm×90x =0.91533 9×Sin 6.67 2 每极串联的匝数为W=Z×2/(2m)=(2×54)/(2×3)=18 6300 空载时每相感应的电势E =3637.3V 根据E9=44k得每极基波磁通量 中 36373 0.9944Wb 1444k,f1444×48×09153x0(4)相电势 r V V E k f V = 4.44 所以 11 1 4.44 1 E = k f 10 2 4.44 108 0.9452 50 1.015 − = = 230.02 V 3 4.44 33 3 E = k f =4.44×108×(-0.5774)×3×50×0.66×102 =-274.1 V 5 4.44 55 5 E = k f =40.236 V 7 4.44 7 7 7 E = k f =9.163V 相电势 E E E E E 2 2 2 2 1 3 5 7 = + + + =360.2 V 线电势中 E 3 相互抵消了,所以 E= 3 2 2 2 404.8 1 3 7 + + = E E E V 14-7 解:发电机极距 =Z÷2P=27 每极每相槽数 q=Z÷2Pm=54÷(2×1×3)=9 槽角度 6.67 360 360 0 0 0 1 54 = = = Z P 所以 0.91533 2 9 2 9 27 22 6.67 6.67 90 0 0 0 1 = = Sin Sin k Sin w 每极串联的匝数为 W=Z×2/(2m)=(2×54)/(2×3)=18 空载时每相感应的电势 3637.3 3 6300 3 0 1 = U = = E V 根据 11 1 4.44 1 E = k f 得每极基波磁通量 0.9944 4.44 18 0.91533 50 3637.3 4.44 1 1 1 1 = = = k f E Wb