彭鸿才《电机原理与拖动》习题解答

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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P411－3P14´10333彭鸿才《电机原理与拖动》习题解答解：I=N==60.87(A)33彭鸿才《电机原理与拖动》习题解答UNN230hP=PN=14=16.37(Kw)N10.855P411－4解：P1=UNIN=110´13=1430(W)33彭鸿才《电机原理与拖动》习题解答NNh=P=1.1´103=33彭鸿才《电机原理与拖动》习题解答76.92%P11430åp=P1-PN=1430-1100=330(W)33彭鸿才《电机原理与拖动》习题

2、解答P421－21解：If=UNRf=230=1.53(A)15033彭鸿才《电机原理与拖动》习题解答Ia=IN+If=69.6+1.53=71.13(A)33彭鸿才《电机原理与拖动》习题解答Ea=UN+IaRa=230+71.13´0.128=239.1(V)p=I2R=71.132´0.128=647.6(W)cuaaPM=EaIa=239.1´71.13=17007(W)33彭鸿才《电机原理与拖动》习题解答h1P=PNN=16´1030.855=18713.5(W)33彭鸿才《电机原理与拖

3、动》习题解答P421－29解：Ia=IN-IfN=40.6-0.683=39.92(A)Ea=UN-IaRa=220-39.92´0.213=211.5(V)PM=EaIa=211.5´39.92=8443.1(W)p=I2R=39.922´0.213=339.4(W)cuaapf=IfNUN=220´0.683=150.26(W)P1=UNIN=220´40.6=8932(W)33彭鸿才《电机原理与拖动》习题解答h=PN=P17.5´1038932=83.97%33彭鸿才《电机原理与拖动》习题解答p

4、o=PM-PN=8443.1-7.5´103=943.1(W)33彭鸿才《电机原理与拖动》习题解答33彭鸿才《电机原理与拖动》习题解答NT=9550´PMnN=9550´8443.1/1000=26.88(N×m)300033彭鸿才《电机原理与拖动》习题解答33彭鸿才《电机原理与拖动》习题解答T2N=9550´PNnN=9550´7.53000=23.88(N×m)33彭鸿才《电机原理与拖动》习题解答To=TN-T2N=26.88-23.88=3(N×m)33彭鸿才《电机原理与拖动》习题解

5、答或者To=9550´PonN=9550´943.1/1000=3(N×m)300033彭鸿才《电机原理与拖动》习题解答33彭鸿才《电机原理与拖动》习题解答PP421－30解：IN=NUN=27000=245.5(A)11033彭鸿才《电机原理与拖动》习题解答IaN=IN+IfN=245.5+5=250.5(A)33彭鸿才《电机原理与拖动》习题解答CefN=UN+IaNRanN=110+250.5´0.02=0.1115033彭鸿才《电机原理与拖动》习题解答33彭鸿才《电机原理与拖动》习题解答n

6、=UN-IaNRaCefN=110-250.5´0.02=1050rpm0.133彭鸿才《电机原理与拖动》习题解答或者：EaN=UN+IaNRa=110+250.5´0.02=115(V)Ea=UN-IaNRa=110-250.5´0.02=105(V)33彭鸿才《电机原理与拖动》习题解答QEaN=Ea=EanN=105´1150=1050rpm33彭鸿才《电机原理与拖动》习题解答nNnEaN11533彭鸿才《电机原理与拖动》习题解答P421－31解：33彭鸿才《电机原理与拖动》习题解答QE

7、aN=CefNnNn=nNf=fNIa=IaN33彭鸿才《电机原理与拖动》习题解答Ea=EaN=UN-INRa=110-13´1=97(V)U=Ea-INRa=97-13´1=84(V)P421－32解：IaN=IN-IfN=255-5=250(A)33彭鸿才《电机原理与拖动》习题解答CefN=UN-IaNRanN=440-250´0.078=0.84150033彭鸿才《电机原理与拖动》习题解答33彭鸿才《电机原理与拖动》习题解答QCe=pCf=30´Cf=9.55´0.841=8.03

8、233彭鸿才《电机原理与拖动》习题解答CT30TNpeN33彭鸿才《电机原理与拖动》习题解答（1）T2N=9550´PNnN=9550´96500=1833.6(N×m)33彭鸿才《电机原理与拖动》习题解答（2）TN=CTfNIaN=8.032´250=2008(N×m)（3）To=TN-T2N=2008-1833.6=174.4(N×m)33彭鸿才《电机原理与拖动》习题解答（4）no=UNCf=440=523r