电机与拖动基础课后习题答案(部分) 1 第一章作业解答参考 1—8.解:)(630230101453AUPINNN )(1619.0145KWPPNN 1—20.解:(1)UVnapNEa)(186150001.0160372260 是电动机状态。 (2))(46.163208.0186220AREUIaaa )(63.19315002604.30)(4.3046.163186mNPTKWIEPMaaM (3))(6.5208.046.16322KWRIpaacua )(79.292403624.30)(366.54.3021KWppPPKWpPPFaMcuaM %823679.2912 PP 1—23.解:(1))(112.54300014.326010172603mNnPPTNNNNN %2.82942201017)(812.557.1112.54)(7.1344014.32608.2)316.08.2220(26030'000000NNNNNaaMIUPmNTTTmNnIEPT (2)NeeCC0忽略电枢反应影响,恒定。 00'0eaaNCRIUn, 0636,03440316.08.22200eC )min(34590636.022000rCUneN 电机与拖动基础课后习题答案(部分) 2 (3) eMCC55.9 )min(27860636.0)15.0316.0(89.91220)()(89.910636.055.9812.55rCRRIUnITTACTIeaaNaZMa不变不变, 第二章 习题解答参考 2—6.解:(1)TTTnCCRRCUnNmeaNeN64.1115819.055.94.006.01158202 (2)TTTnCCRCUnNmeaNe21.057921.019.011002 (3)TTTnCCRCUnmeaeN35.0146717.006.015.022002 19.0NeC N8.0 15.019.08.0eC 2NmeCC=255.9NeC =219.055.9=0.28 2meCC=17.028.08.02 2—16.解:(1) VRIUEaNNa20425.064220 ARREUIZaaNa84.67625.0204220max 29.0700204 NaNNNenRIUC 76.229.055.955.9NeNmCC mNICTanaxNm23.187)84.67(76.2max 停机时 n=0 0nCENea ARRUIZaNa2.35625.0220 mNICTaNm15.97)2.35(76.2 此时反抗性负载 mNICTNNmZ64.1766476.2 由于 TTZ 故系统不会反向起动。 (2)ZNmeZaTCCRRnn20 电机与拖动基础课后习题答案(部分) 3 ...