电机原理及拖动的答案

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1、彭鸿才《电机原理与拖动》习题解答第一章直流电机原理3P1410NP411－3解：IA60.87()NU230NP14NP16.37(Kw)10.855NP411－4解：PUI110131430()W1NN3P1.110N76.92%NP14301pP1PN14301100330()WU230NP421－21解：I1.53(A)fR150fIII69.61.5371.13(A)aNfEUIR23071.130.128239.1()VaNaa22pIR71.130.12864

2、7.6(W)cuaaPEI239.171.1317007(W)Maa3P1610NP18713.5(W)10.855NP421－29解：III40.60.68339.92(A)aNfNEUIR22039.920.213211.5(V)aNaaPEI211.539.928443.1(W)Maa22pIR39.920.213339.4(W)cuaapIU2200.683150.26(W)ffNNPUI22040.68932(W)1NN3P7.510N83.97%P893213

3、pPP8443.17.510943.1(W)oMN1彭鸿才《电机原理与拖动》习题解答P8443.1/1000MT9550955026.88(Nm)Nn3000NP7.5NT9550955023.88(Nm)2Nn3000NTTT26.8823.883(Nm)oN2NP943.1/1000o或者T955095503(Nm)on3000NP27000NP421－30解：I245.5(A)NU110NIII245.55250.5(A)aNNfNUIR110250.50.02Na

4、NaC0.1eNn1150NUIR110250.50.02NaNan1050rpmC0.1eN或者：EUIR110250.50.02115(V)aNNaNaEUIR110250.50.02105()VaNaNaEEEn1051150aNaaNn1050rpmnnE115NaNP421－31解：ECnnnIIaNeNNNNaaNEEUIR11013197(V)aaNNNaUEIR9713184(V)aNaP421－32解：III2555250

5、(A)aNNfNUIR4402500.078NaNaC0.841eNn500NC30eCC9.550.8418.032TNeNCT30P96N（1）T955095501833.6(Nm)2Nn500N2彭鸿才《电机原理与拖动》习题解答（2）TCI8.0322502008(Nm)NTNaN（3）TTT20081833.6174.4(Nm)oN2NU440N（4）n523rpmoC0.841eN'URNa0.078174.4(5)nT523521rpmoo2C

6、CC0.8418.032eNeTNT174.4o或者：I21.71(A)aoC8.032TN'UNIaoRa44021.710.078n521rpmoC0.841eNP421－33解：PUI22081.717974(W)1NNPP179740.8515278(W)N1NU220NI2.48(A)fNR88.8fIII81.72.4879.22(A)aNNfN22pIR79.220.1627.6(W)cuaNa22pIR2.4888.8546(W)ffNfpP1PN179

7、74152782696(W)pmpFeppcupf2696627.65461522()WPPpp15278152216800()WMNmFeP421－34解：P75N（1）T95509550955(Nm)2Nn750N（2）III1914187(A)aNNfNEUIR4401870.082424.7(V)aNNaNaPEI424.718779419(W)MaNaN3彭鸿才《电机原理与拖动》习题解答P79.419MT955095501011.3(Nm)Nn750NE4

8、24.7aN（3）C0.566eNn750NU440Nn777.4rpmoC0