N。和。。。pp是。。、、調(diào)和。UU。UU=NT和emZ。矩。ss在0mT/T。mN，F(xiàn)a√）m√）(4)(3）安mmm(3)(2)(51安(3)13）2。）SN、P=U=kkNNIN）NN??UφU）INNastNNNafmecFePW,adNNfIIIAAaNfIRWWaaIRWWCufffPP+P706.9)W22002.9Wem2CuaCufCuaCufmecFead12P2NNNNNa）輸入功率P和效率;1N電樞銅損耗P、電磁功率P和機械損耗P;C201NNP246200N(2)電樞銅損耗電磁功率P=IR=2100.075W=3307.5W22CaaCNPP(3)電磁轉(zhuǎn)矩NNPP22NN02NNNNn。解（1a1NNN2I2a2N0.099NNnENNTNENN0.099ENR0.1721.84aCCENNn=n-T=(1111-1.84T)r/0emem0TemNTNNPT9.55LNNnNT58.652.56.1NmLN0n=nT(11111.846.1)r/r/000NNNN21時32UIP2NNN3I3a2NNNnENNUNCENRaC)22ENnTT0ememem0emNENNr/N(2)R時nr/S0=aSC)22ENnTememN1200em當T=Tnr/r/emNP10kW,UV,I53,n1000r/R，NNNNaNaNaNEa2IBaNECanENNaaENR得BBNNNa工作在回饋制動狀態(tài)，勻速下放重物，電樞回路不串電阻，求電動機轉(zhuǎn)速。解：提升重物時電機運行于正向電動狀態(tài)，下放重物時電動機運行于反向回饋制動狀態(tài)。因為磁通不變，故=0.4NaNnENNUR根據(jù)反向回饋制動機械特性可求得轉(zhuǎn)速nnI=-Na0aCENENNNNaN0.136NaanENN(1)Ian=NaS0.136EN(2)與（）相同，I=IaNn=a0.136EN）因為TCICIemTNNTa1N0.9aNNCENNNNN1N負載時的轉(zhuǎn)差率s;(4)該電機的機座中心高是多少？N60f121pA1.5ANUcosNNNNnn轉(zhuǎn)差率s=0.07311500NNN頻率f50Hz,額定轉(zhuǎn)速n962r/min,額定負載時功率因數(shù)cos，定子銅損耗N1Cu1N22120em.解：（）額定轉(zhuǎn)差率sN1p30.038110002轉(zhuǎn)子銅損耗P227625N2N11Cu1P1UcosN120em.P75009629.552nNPPNm1.24Nmn962NTTT(74.451.24)Nm75.69Nm20NNT2N2NLN1n150011N13mT大于（1.11.2）T1N2.222NL0.1080.2412R2R22R3R3P32RRR