Sin(2x 所以kn=Sm=x90 =0933 2Si Sim(2×3×0) k, 3×30 Smn(2×5× ks=Sm×5×90 =0.067 2 Sin 5×30 Sm(2×7×) kn=Sm=×7×90 0.067 7×30 (4)E=4449yk En=444×72×50×89×10×0.933=133y E3=444×72×50×89×10×03×(-0.5)=-213V E5=444×72×50×89×10×02×0067=19V En=444×72×50×89×10×0.15×0067=143V 所以E,=√E+E2+E+E2=1347V E=3E+E+E 中7 2304 14补充:解:每极每相槽数=2Pm2x3=5槽 槽距角a4=P×360=12 12 n=Bm15×90=09511 k, k =0.9567 ql 5×Si所以 0.933 2 2 ) 2 (2 6 5 30 30 90 0 0 0 1 = = Sin Sin k Sin w 0.5 2 2 ) 2 (2 3 3 6 5 3 30 30 90 0 0 0 3 = − = Sin Sin k Sin w 0.067 2 2 ) 2 (2 5 5 6 5 5 30 30 90 0 0 0 5 = = Sin Sin k Sin w 0.067 2 2 ) 2 (2 7 7 6 5 7 30 30 90 0 0 0 7 = = Sin Sin k Sin w (4) E V k wr wf = 4.44 4.44 72 50 8.9 10 0.933 133 3 1 = = − E V 4.44 72 50 8.9 10 0.3 ( 0.5) 21.3 3 3 = − = − − E V 4.44 72 50 8.9 10 0.2 0.067 1.9 3 5 = = − E V 4.44 72 50 8.9 10 0.15 0.067 1.43 3 7 = = − E V 所以 134.7 2 7 2 5 2 3 2 1 E = E +E +E +E = V 3 230.4 2 7 2 5 2 1 E = E +E +E = V 14 补充:解:每极每相槽数 q= 5 2 3 30 2 = = Pm Z 槽 槽距角 12 360 0 0 1 = = Z P 0.9511 15 12 900 1 k = Sin = y 0.9567 2 5 2 5 12 12 0 0 1 = = Sin Sin k q