电机原理及拖动 (彭鸿才 著) 机械工业出版社 课后答案.pdf 31页

'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：I===60.87()ANU230NP14NP===16.37(Kw)1η0.855NP411－4解：P=UI=110131430()×=W1NN3P1.110×Nη===76.92%NP14301∑p=P1−PN=14301100−=330()WU230NP421－21解：I===.153(A)fR150fI=I+I=696.+.153=71.13(A)aNfE=U+IR=23071.130.128+×=239.1()VaNaa22p=IR=71.13×.0128=647(6.W)cuaaP=EI=2391.×71.13=17007(W)Maa3PN16×10P===18713(5.W)1η.0855NP421－29解：I=I−I=406.−.0683=39.92(A)aNfNE=U−IR=220−39.92×.0213=211(5.V)aNaaP=EI=2115.×39.92=8443(1.W)Maa22p=IR=39.92×.0213=339(4.W)cuaap=IU=220×.0683=150.26(W)ffNNP=UI=220×406.=8932(W)1NN3P5.7×10Nη===83.97%P893213p=P−P=84431.−5.7×10=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=26.8823.88−=3(Nm⋅)oN2NP943/1.1000o或者T=9550×=9550×=(3N⋅m)on3000NP27000NP421－30解：I===245(5.A)NU110NI=I+I=2455.+5=250(5.A)aNNfNU+IR110+2505.×.002NaNaCφ===1.0eNn1150NU−IR110−2505.×.002NaNan===1050rpmCφ1.0eN或者：E=U+IR=110+2505.×.002=115(V)aNNaNaE=U−IR=110250.50.02105()−×=VaNaNaEEEn105×1150aNaaNQ=∴n===1050rpmnnE115NaNP421－31解：QE=Cφnn=nφ=φI=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(A)aNNfNU−IR440−250×.0078NaNaCφ===.0841eNn500NCπ30eQ=∴Cφ=×Cφ=.955×.0841=.8032TNeNC30πTP96N（1）T=9550×=9550×=1833.6(Nm⋅)2Nn500N（2）T=CφI=.8032×250=2008(N⋅m)NTNaN2 彭鸿才《电机原理与拖动》习题解答（3）T=T−T=20081833.6174.4(−=Nm⋅)oN2NU440N（4）n===523rpmoCφ.0841eN"UNRa0.078174.4×(5)n=−×T=523−=521rpmo2oCφCCφ0.8418.032×eNeTNT1744.o或者：I===21.71(A)aoCφ.8032TN"UN−IaoRa440−21.71×.0078n===521rpmoCφ.0841eNP421－33解：P=UI=220×817.=17974(W)1NNP=Pη=17974×.085=15278(W)N1NU220NI===.248(A)fNR888.fI=I−I=817.−.248=79.22(A)aNNfN22p=IR=79.22×1.0=627(6.W)cuaNa22p=IR=.248×888.=546(W)ffNf∑p=P1−PN=17974−15278=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×.0082=424(7.V)aNNaNaP=EI=4247.×187=79419(W)MaNaNP79.419MT=9550×=9550×=1011(3.N⋅m)Nn750N3 彭鸿才《电机原理与拖动》习题解答E4247.aN（3）Cφ===.0566eNn750NU440Nn===777.4rpmoCφ0.566eN（4）T=T−T=1011.3955−=56.3(Nm⋅)oN2NTT563.ooI====10.42(A)aoCφ.955Cφ.955×.0566TNeN"UN−IaoRa440−10.42×.0082n===776rpmoCφ.0566eNP421－35解：I=I−I=91−5.2=88(5.A)aNNfNU−IR220−885.×.0074NaNaCφ===.0142eNn1500NU−IR220500.074−×Naan===1523rpmCφ0.142eNP1053－28解：331UI−P×1012203056010×−×NNN（1）估算：R=×==0.038()Ωa222I2305NU−IR220−305×.0038NaNaCφ===.02084eNn1000NU220Nn===10557.rpmoCφ.02084eNT=9.55CφI=9.550.2084305××=607(Nm⋅)NeNaN过（no=1056rpm，T=0），（nN=1000rpmTN=607N.m）两点，可画出固有机械特性。（2）T=0.75TN时，Ia=0.75IaN：U−IR220−.075×305×.0038Naan===1014rpmCφ.02084eNUR220.0038×.075×607Na或者：n=−×T=−=1014rpm22CφCCφ.02084.955×.02084eNeTNU−Cφn220−.02084×1100NeN（3）I===−243(2.A)R.0038a4 彭鸿才《电机原理与拖动》习题解答P1053－29解：U220NR===0.72ΩNI305N0.5R0.50.72×N（1）n=n−T=1055.7−×607=529rpmRN02N29.55(Cφ)9.55(0.2084)×eN过（n，0），（529，607）两点画直线。02R20.72×′=−N=−×=−（2）nnT1055.76071052rpmRN02N29.55(Cφ)9.55(0.2084)×eN过（n，0），（–1052，607）两点画直线。0U−2RI22020.72305−××′=NNN==−或：n1052rpmRNCφ0.2084eN0.5U′=N==×=（3）n0.5n0.51055.7528rpm00CφeNR0.038=′−a=−×=nnT528607472rpmVN02N29.55(Cφ)9.55(0.2084)×eN过（n′，0），（472，607）两点画直线。0Un1055.7′′=N=0==（4）n2111.3rpm00.5Cφ0.50.5eNR0.038=′′−a=−×=nnT2111.36071889rpmφN02N29.55(0.5Cφ)9.55(0.50.2084)××eN过（n′′，0），（1889，607）两点画直线。0P1053－30解：U110NR===0.65()Ωst3I285.2×1R0.65λ=3st3=3=1.7R0.129aI285.2×1I===100.2()1.1A>I=93.72()A2Nλ1.7符合要求。R=λR=1.70.129×=0.219()ΩST1aR=λR=1.70.219×=0.372()ΩST2ST15 彭鸿才《电机原理与拖动》习题解答r=(λ−1)R=(1.71)0.129−×=0.091()Ωst1ar=λr=1.70.0910.153()×=Ωst2st1r=λr=1.70.153×=0.26()Ωst3st2P1053－31解：U−IR220−41×.0376NaNa（1）Cφ===.0136eNn1500N降压瞬间，转速不变，有：n=nNU−CeφNnN180−.0136×1500I===−63.83(A)aR.0376aT=.955CφI=.955×.0136×(−63.83)=−82(9.N⋅m)eNaU−IaNRa180−41×.0376（2）n===1210rpmCφ.0136eNP1063－32解：QC=9.55CT=CφI=.955CφITeTaeaURURNaNan=−×T=−×.955CφI2N2eNaNCφCCφ8.0Cφ.9558.0(Cφ)eeTeNeNUR220.0376×41Na=−×I=−=1845rpm2aN28.0Cφ8.0Cφ8.0×.01368.0×.0136eNeN或者：QT=CφI=T=CφITaNTNaNφφ41NN∴I=I=I==51.25()AaaNaNφ0.8φ0.8NU−IR220−51.25×.0376Naan===1845rpmCφ8.0×.0136eQn=1845rpm>n=1500rpm,I=51.25A>I=41ANaN所以，电动机不能长期运行。U−IR220−687.×.0224NaNaP1063－33解：（1）Cφ===.01364eNn1500NU220Nn===1613rpmoCφ.01364eNn=n1(−δ)=1613×1(−)3.0=1129rpmmino6 彭鸿才《电机原理与拖动》习题解答n1500max（2）D===.133n1129minU−Cφn220−.01364×1129NeNmin（3）R=−R=−.0224=.0737ΩcaI687.N（4）P=UI=22068.715114()15.114×=W=kW1NN忽略To，TN=T2N，P13NT=9550×=9550×=82.77(Nm⋅)2Nn1500N因为额定负载不变，有：Tn82.771129×2NminP===9.785(kW)29550955022p=IR=687.×.0737=3478(W)cuRcaNcP1063－34解：（1）∆n=n−n=1613−1500=113rpmNoN∆n113Nn===3767.rpmominδ3.0maxn=n−∆n=3767.−113=2637.rpmminominNn1500max（2）D===.569n2637.min（3）U=Cφn=3767.×.01364=51(4.V)mineNomin（4）P=UI=514.×687.=3531(2.W)1minN忽略To，则有：T2N=TN=82.77Nm⋅Tn82.77263.7×2NminP===2.286(kW)295509550P1063－35解：U−IR440−76×.0376NaNaCφ===.0411eNn1000NU440Nn===10706.rpmoCφ.0411eN（1）低速时采用降压调速：∆n=n−n=10706.−1000=706.rpmNoN7 彭鸿才《电机原理与拖动》习题解答∆n706.Nδ===.028maxn250ominn=n−∆n=250−706.=1794.rpmminominN（2）高速时采用弱磁调速，U=UN不变，允许最高转速时Ia=INCφn10706.eoQ=∴Cφ=×.0411=.02935eCφn1500eNomaxT=CφI=9.55CIφ=9.550.293576××=213(Nm⋅)TaeaP29Nn=9550×=9550×=1300rpmmaxT213n1300max（3）D===.725n1794.minP1063－36解：U−IR110−352.×.035NaNaCφ===.013eNn750NE=Cφn=U−IR=11035.20.35−×=97.68()VaNeNNNaNaU−E097.68−aN（1）R=−R=−0.351.0375()=ΩcaI−×235.2amaxU−E−11097.68−aN（2）R=−R=−0.35=2.6()ΩcaI−×235.2amaxU−Cφn0eN（3）能耗制动时：I===0anR+RR+RacacT=CφI=0TNanU−Cφn−1100−eN反接制动时：I===−37.29()AafR+R2.95acT=CφI=.955×.013×(−37.29)=−46(3.N⋅m)TNaf（4）QT=T∴I=I=352.ANaNU−Cφn00.13(500)−×−eNR=−R=−0.351.5()=ΩcaI35.2aP1063－37解：∵他励机TL=TN，∴Ia=IN=68.7An=0，Ea=08 彭鸿才《电机原理与拖动》习题解答U−E2200−aR=−R=−0.195=Ω3caI68.7aP1063－38解：采用电动势反接制动：QT=T∴I=I=76ALNaNNU−IR440−76×.0377NaNaCφ===.0411eNn1000NU−Cφn4400.411(500)−×−NeNR=−R=−0.377=8.12ΩcaI76aN若采用能耗制动：U−CeφNn0−.0411×(−500)R=−R=−.0377=.233ΩcaI76a"U−CeφNnN0−.0411×1000能耗瞬间电流：I===−151.83(A)aR+R.0377+.233acI=8.1I=8.1×76=136(8.A)amaxN"QI>I∴不能采用能耗制动，只能采用电势反接制动。aamaxP1063－39解：因为下坡时TL2与n正方向相同，故为负；TL1与n正方向相反，故为正，所以：T=T+T=8.0T−2.1T=−4.0TLL1L2NNNI=−4.0I=−4.0×76=−30(4.A)aN此时电机工作在正向回馈制动状态，不串电枢电阻时：U−IR440−(−30)4.×.0377Naan===1098rpmCφ.0411eN串0.5Ω电阻时：U−I(R+R)440−(−30)4.×.0(377+)5.0Naacn===1135rpmCφ.0411eNP1063-40解：U−IR−440600.377−×aa（1）n===−1125.6(/min)rCφ0.411eNT=CφI=.955CφI=.955×.0411×60=235(5.N⋅m)TNaeNaP=UI=−440×60=−26400(W)1aU−Cφn4400.411(850)−×−NeN（2）R=−R=−0.37712.78()=ΩcaI60a9 彭鸿才《电机原理与拖动》习题解答P=UI=440×60=26400(W)1Na22p=IR=60×12.78=46008(W)cucacU−Cφn00.411(300)−×−NeN（3）R=−R=−0.3771.678()=ΩcaI60a22p=IR=60×.1678=6040(8.W)cucacP1073-41解：（1）降压瞬间，n来不及改变，Ea不变E=U−IR=440−8.0I×.0377=417(V)aNaaN"U−Ea400417−I===−45()AaR0.377a""T=CTφNIa=.955CeφNIa=.955×.0411×(−45)=−176.63(N⋅m)电动机从电动→正向回馈制动→正向电动。U−IR400−8.0×76×.0377Naa（2）n===9175.rpmCφ.0411eNP1073-42解：（1）E=U−IR=22022.30.91199.7()−×=VaNNaNa"U−EaN−220199.7−I===−42.35()AaR+R0.919+acE1997.aNCφ===.01997eNn1000N"""T=CφI=9.55CφI=9.550.1997(42.35)××−=−80.77(Nm⋅)TNaeNa（2）n=0时，Ea=0，有：""U−Ea−2200−I===−22.2()AaR+R0.919+ac""""""T=CφI=9.55CφI=9.550.1997(22.2)××−=−42.34(Nm⋅)TNaeNa电动机电磁转矩T“<0，位能转矩TL>0，二者方向一致，均为拖动转矩，在|T”|+TL作用下，电机反向起动，进入反向电动状态，最后稳态运行在反向回馈制动状态，以很高的速度下放重物。P1414-12解：UU1N1NZ=Z=mImIIIIoIoIIQI=2IoIoII10 彭鸿才《电机原理与拖动》习题解答UUZ1N1NmII∴Z===⇒Z=2ZmImIImII2I2oIoII顺极性串联时，IO一样，忽略漏阻抗压降，有：U=I(Z+Z)=I(Z+2Z)=3IZ1omImIIomImIomI1即有：IZ=UomI131440E=IZ=U==147(V)1IomI1332440E=IZ=U=×2=293(V)1IIomII133E1471IU===73(V)2Ik2E2931IIU===147(V)2IIk2由此可见，空载电压不相等。P1414-13解：220U=.444fNφ=220V⇒φ=UU1oo12.444fN1N22031k===2N=2NN+N=N12121N11022""3"顺串时：U=U+U=.444fNφ+.444fNφ=×.444fNφ=330VUuU1U2u1u21o2o1o122"220φ==φoo.444fN1"3""2因为磁通不变，所以磁势不变，IN=I(N+N)=NI⇒I=Io1o121ooo23""""1""逆联时：U=U−U=.444fNφ−.444fNφ=×.444fNφ=110VUuU1U2u1u21o2o1o122""220φ==φoo.444fN1""1""""因为磁通不变，所以磁势不变，IN=I(N−N)=NI⇒I=2Io1o121ooo2P1414–18解：U101Nk===26.3U0.382N"U2N380Z===9.62()ΩmI39.52o"po1100r===0.705()Ωm22I39.52o11 彭鸿才《电机原理与拖动》习题解答2"2Z=kZ=26.3×9.62=6654()Ωmm2"2r=kr=26.3×0.705=487.6()Ωmm2222x=Z−r=6654−487.6=6636()ΩmmmU450kZ===22.5()ΩkI20kp4100kr===10.25()Ωk22I20k2222X=Z−r=22.5−10.25=20()Ωkkk234.575+ro=×10.2512.2()=Ωk75C234.525+2222Zk75oC=rk750+Xk=12.2+20=23.4()ΩP1414–20解：U2301N（1）k===2U1152N22r=r+r′=r+kr=0.32+×0.05=0.5()Ωk121222x=x+x′=x+kx=0.82+×0.11.2()=Ωk12122222Z=r+X=0.5+1.2=1.3()Ωkkk22（2）r′=r′′+r=rk/+r=0.3/2+0.05=0.125()Ωk121222x′=x′′+x=xk/+x=0.8/2+0.10.3()=Ωk12122222Z′=r′+X′=0.125+0.3=0.325()Ωkkk33S310×S310×NN（3）I===13.044()AI===26.09()A1N2NU230U1151N2NU230U1151N2NZ===17.63()ΩZ===4.41()Ω1N2NI13.044I26.091N2N*rk0.5r===0.028kZ17.631N*Xk1.2X===0.068kZ17.631N12 彭鸿才《电机原理与拖动》习题解答*Zk1.3Z===0.074kZ17.631N*rk′0.125*r′===0.028=rkkZ4.412N*Xk′0.3*X′===0.068=XkkZ4.412N*Zk′0.325*Z′===0.074=ZkkZ4.412N(4)U=IZ=13.0441.316.957()×=VkN1Nk∗UkN16.957∗U===0.074=ZkNkU2301N（5）满载时β=1cosϕ=1时，sinϕ=0：22**∆U%=β(rcosϕ+xsinϕ)1(0.02810=××+=0.028=2.8%k2k2cosϕ=0.8落后时，sinϕ=0.6：22**∆U%=β(rcosϕ+xsinϕ)1(0.0280.80.0680.6=××+×=0.0632=6.32%k2k2cosϕ=0.8超前时，sinϕ=−0.6：22**∆U%=β(rcosϕ+xsinϕ)1(0.0280.80.0680.6=××−×=−0.0184=−1.84%k2k2P1414–21解：U1Nϕ10/3（1）Y,y结：k===25noU2Nϕ/4.03"U2N/3400/3空载试验：Z===3.849()ΩmI602o"po3800r===3.519()Ωm223I360×2o2"2"Z=kZ=2405.63()Ωr=kr=219.91()Ωmmmm22x=Z−r=2395.55()Ωmmm13 彭鸿才《电机原理与拖动》习题解答S750NI===43(3.A)1N3U3×101NU/3440/31NZ===.5867(Ω)kI433.1kp10900kr===1.938()Ωk223I343.3×1k2222X=Z−r=5.867−1.938=5.538()Ωkkk234.575+ro=×1.938=2.357()Ωk75C234.520+2222Zk75oC=rk750+Xk=2.357+5.538=6.02()Ω22PkN=3Ir1Nk750=×343.3×2.35713257()13.26(=w=kw)"11r1=r2=rk750=×2.3571.179=Ω22"1x=x=×5.538=2.769()Ω1σ2σ2U31Nϕ1010/3×（2）Z===133.34()Ω1NI43.31Nϕ*rk7502.357r0===0.0177k75Z133.341N*Xk5.538X===0.0415kZ133.341N*Zk7506.02Z0===0.045k75Z133.341Na）cosϕ=8.0滞后时，sinϕ=0.6：22**∆U=β(rk750cosϕ2+xksinϕ2)1(0.01770.80.04150.6)=××+×=0.03913.91%=βScosϕ17500.8××N2η===97.24%22βScosϕ+βp+p17500.81××+×13.263.8+N2kNob）cosϕ=1时，sinϕ=0：22*∆U=βrk750cosϕ2=×10.017710.01771.77%×==βScosϕ17501××N2η===97.78%22βScosϕ+βp+p175011××+×13.263.8+N2kNo14 彭鸿才《电机原理与拖动》习题解答c）cosϕ=8.0超前时，sinϕ=−0.6：22**∆U=β(rk750cosϕ2+xksinϕ2)1(0.01770.80.04150.6)=××−×=−0.01074=−1.074%效率与cosϕ=8.0滞后时相同：2βScosϕ17500.8××N2η===97.24%22βScosϕ+βp+p17500.81××+×13.263.8+N2kNoP1424–22解：22p+βp6.61+×21.20kNη=−1=−1=98.1%22βScosϕ+p+βp118000.86.61××++×21.2N2okNp6.60β===0.558mp21.2kN2p26.6×0η=−1=−1=98.5%maxβScosϕ+2p0.55818000.826.6××+×mN2oP1424–24解：15 彭鸿才《电机原理与拖动》习题解答16 彭鸿才《电机原理与拖动》习题解答P1334–33解：U220111k===U18092I2002I===163.64()A1k11/9I=I−I=36.36()A1221S=IU=36.36×180=6544(8.W)电磁122S=IU=163.64×180=29455(W)传导12第五章三相异步电动机原理P1895–5解：00000槽距角α=20−0=40−20=20；每极每相槽数q=3oqα320×sinsin22∴k===0.9598p1oα20qsin3sin22∵元件电势Ey=40V∴元件组电势E=qEk=×3400.9598115.2()×=Vqyp1P1895–6解：Z=362p=4m=31ooZ36p⋅3602×360oq===3α===202pm4×3Z3617 彭鸿才《电机原理与拖动》习题解答oqα320×sinsiny1o22k=k⋅k=sin90×=×1=0.96w1y1p1oτα20qsin3sin22pqN2×3×10每相匝数：N===60(匝)1a1E=.444fNkφ=.444×50×60×.096×.0172=219.94(V)111w1P1895–7解：y070K=sin×90=sin×90=0.9397y1τ900p×3602360×0α===20Z361Z361q===3(槽)2mp232××0qα320×sinsin22k===0.9598p10α20qsin3sin22k=k×k=0.93970.9598×=0.902w1y1p1P1905–26解：60f160×50取整数p===.308⎯⎯→⎯3n=1000rpm1n9701n−n1000−9751Ns===.0025Nn100013P75×1075000Nη====942.%NP13UNINcosϕN3×380×139×.087P1905–27解：3P=P−p−p=8.610×−425210−=7965()WM1cu1Fep=sP=0.057965×=398.25()Wcu2MP=P−p=7965398.25−=7566.75()WmMcu2P1905–29解：（1）Qn=950rpm∴n=1000rpmN1n−n1000−9501NS===.005Nn10001（2）P=P+p+p=281.10++=29.1(kW)mNms18 彭鸿才《电机原理与拖动》习题解答P29.1mP===30.63(kW)⇒p=sP=0.0530.631.532(×=kW)Mcu2M1−S10.05−N（3）P=P+p+p=30.632.2+=32.83(kW)1Mcu1FeP282η===853.%NP32.8313P32.83×101（4）I===56.68(A)N3Ucosϕ3×380×.088NNP1905–30解：3P=P−p−p=10.710×−45020010050()−=WM1cu1Fep=sP=0.02910050×=291.45()Wcu2MP=P−p=10050291.45−=9758.55()WmMcu2∵四极电机，∴n1=1500r/minP10050MT=9.55=9.55×=63.985(Nm⋅)n15001P1905–31解：（1）Qn=1455rpm∴n=1500rpmN1n−n1500−14551NS===.003Nn15001（2）P=P+p+p=100.20510.205(+=kW)mNms1−SSP.003×10.205mP=p⇒p===.0316(kW)mcu2cu2S1−S1−.003P10N（3）T=9550=9550×=65.64(N⋅m)Nn1455Np+p0.205msT=9550=9550×=1.346(Nm⋅)on1455NP10.205mT=9550=9550×=67(Nm⋅)n1455NP1905–33解：U/3380/30Z===60.3()Ω0I3.640p−p26411−0mr===6.4()Ω0223I33.64×019 彭鸿才《电机原理与拖动》习题解答2222x=Z−r=60.3−6.4=60()Ω000U/3100/3KZ===8.25()ΩkI7kp470kr===3.2()Ωk223I37×K2222x=Z−r=8.25−3.2=7.6()Ωkkk"r=r−r=3.221.2()−=Ω2k1"11x=x=x=×7.6=3.8()Ω12k22r=r−r=6.42−=4.4()Ωm01x=x−x=603.856.2()−=Ωm01P2516–33解：U=380/3=219(4.V)Nϕn−n1000−9571N（1）S===.0043Nn10001m1p21（2）T=⋅U⋅max12ωr+r2+(x+x")21111σ2σ23×32194.=×=70.84(N⋅m)2×2π×50.208+.2082+.3(12+.425)2"r2m1p2s（3）T=⋅U⋅N1"ω1r22"2(r+)+(x+x)11σ2σs2.1532194.×3×3.0043=×=33.335(N⋅m)2π×50.15322.2(08+)+.3(12+.425).0043Tmaxλ==.2215mTN"r.1532（4）S===.01998max2"222r+(x+x).208+.3(12+.425)11σ2σP2516–34解：20 彭鸿才《电机原理与拖动》习题解答P75N（1）T=9550=9550×=994.8(Nm⋅)Nn720N（2）T=λT=2.4994.8×=2387.52(.)NmmaxmNn−n750720−1N（3）S===0.04Nn750122（4）s=s(λ+λ−1)=0.04(2.4+2.4−1)=0.183mNmm2T22387.52×maxT==(.)Nmss0.183sm++sss0.183msE0.04213×N2N（5）r===0.0224()Ω23I3220×2NP2526–35解：n−n750717−1N(1)S===0.044Nn7501sE0.044223.5×N2Nr===0.12()Ω23I347×2N22s=s(λ+λ−1)=0.044(3.15+3.15−1)=0.27mNmms′r+Rm2∵=srm2r+R0.121+′=2=×=∴ss0.272.52mmr0.122⎡λmTNλMTN2⎤⎡3.15TN3.15TN⎤s=s⎢−()−1⎥=0.27⎢−()−1⎥=0.0304Am2TT0.7T0.7T⎣LL⎦⎣NN⎦s=s=0.0304BAP16NT=9550=9550×=213(Nm⋅)Nn717N2λT23.15213××mNT===16.19(.)NmB′ss2.520.0304mB++ss′0.03042.52Bm⎡λmTNλMTN2⎤⎡3.15TN3.15TN2⎤(2)s=s′⎢−()−1⎥=2.52⎢−()−1⎥=0.284cmTT0.7T0.7T⎣LL⎦⎣NN⎦21 彭鸿才《电机原理与拖动》习题解答n=(1−sn)=(10.284)(750)−×=537(/min)rcc12πn2π×537CP=TΩ=T×=0.7213××=8384.6()WMCCL6060P≈P=(1−sP)=(10.284)8384.6−×=6003.4()W2mcMp=P−P=8384.66003.4−=2381.2()Wcu2MmR1p=p=×2381.2=2126()WcuRcu2r+R0.121+2（3）转子串电阻前后T=0.7T没有变，LN据T=CTφmI2cosϕ2，CT没有变、U1不变故Φm不变、cosΦ2基本不变，∴转子电流基本不变。P2526–36解：n−n15001460−1NS===0.027Nn15001sE0.027399×N2Nr===0.0536()Ω23I3116×2N22s=s(λ+λ−1)=0.027(2.8+2.8−1)=0.146mNmmn−n′1500500−′=1==s0.67n15001⎡λmTNλMTN2⎤⎡2.8TN2.8TN⎤s′=s′⎢+()−1⎥=0.67⎢+()−1⎥=4.59m2TT0.8T0.8T⎣LL⎦⎣NN⎦"S4.59mR=(−1)r=(−×1)0.05361.63()=Ωc2S0.146m⎡λmTNλMTN2⎤【或：s=s⎢−()−1⎥mTT⎣LL⎦⎡2.8T2.8T⎤NN=0.146⎢−()−1⎥=0.021320.8T0.8T⎣NN⎦S′0.67R=(−1)r=(−×1)0.05361.63()=Ωc2S0.0213结果相同】P2526–38解：I=20/3=11.547(A)Nϕ（1）P=3IUcosϕ=3×20×380×.087=11452.32(W)1NNN22 彭鸿才《电机原理与拖动》习题解答P=Pη=11452.32×.0875=10020(8.W)21P10.0208NT=9550=9550×=66(N⋅m)Nn1450N（2）直接起动时：T=KT=4.1TstTNN"降压起动时，设：U=U/k，则有：1N"2Tst(UN/k)12Tst4.1TN==⇒k===4.1⇒k=.11822"TUkTTstNstN"UN380U===322(V)k.118"111（3）Y/Δ起动：I=I=kI=×7×20=46(6.A)ststIN333"111T=T=KT=×4.1×T=.046<5.0TststTNNN333所以，不可以半载起动。"1"（4）T=TT=5.0TT=4.1TstststNstN32k=8.2⇒k=.167"111I=I=kI=×7×20=50(A)st2stINk8.28.2"UN380U===227(5.V)k.167P2526–43解：n−n750726−1N(1)s===0.032Nn7501sE0.032285×N2Nr===0.081()Ω23I365×2N22s=s(λ+λ−1)=0.032(2.8+2.8−1)=0.173mNmm⎡λmTNλMTN2⎤⎡2.8TN2.8TN⎤s=s⎢−()−1⎥=0.173⎢−()−1⎥=0.025Am2TT0.8T0.8T⎣LL⎦⎣NN⎦s=−2s=−20.0251.975=BAs′r+Rm2∵=srm2r+R0.0812.12+′=2=×=∴ss0.1734.7mmr0.081223 彭鸿才《电机原理与拖动》习题解答P30NT=9550=9550=394.6(Nm⋅)Nn726N2λT22.8394.6××mNT===789.2(.)NmB′ss4.71.975mB++ss′1.9754.7Bm⎡λmTNλMTN2⎤⎡2.8TN2.8TN⎤(2)s=s′⎢−()−1⎥=4.7⎢−()−1⎥=0.68cm2TT0.8T0.8T⎣LL⎦⎣NN⎦n=(1−sn)=(10.68)(750)−×−=−240(/min)rcc1P2536–44解：n−n750−7261N（1）S===.0032Nn7501ES2850.032×2NN转子电阻：r===0.081()Ω23I365×2N固有特性上的临界转差率：22S=S(λ+λ−)1=.0032×8.2(+8.2−)1=.01733mNmm固有特性上TL=TN时的转差率：λmTNλmTN22S=S[−()−]1=.01733×8.2(−8.2−)1=.0032AmTTLLSS−SRS1.31c1QR=(−)1r=r∴S=+S=(+)1×.0032=.12567c221SSr.0081112n=1(−S)n=1(−.12567)×750=−1925.rpm1""Tn2"""P29550T2n"TN×(−192)5.（2）==⇒P=×30=−.79545(kW)2PTnTnT×726NNNNNN9550"总机械功率：P=P+p+p=−.79545+.105=−.69045(kW)mec2mecs1−SSP.12567×(−.69045)mecP=p⇒p===33(8.kW)meccu2cu2S1−S1−.12567外串转子电阻的损耗＝33.8-1.5=32.3kW"（3）P=−.79545(kW2P2536–45解：n−n750720−1N（1）S===0.04Nn750124 彭鸿才《电机原理与拖动》习题解答ES2130.04×2NNr===0.0224()Ω23I3220×2N22S=S(λ+λ−1)=0.04(2.4×+2.4−1)=0.183mNmm起动时：s=1T=1.5TststN"λmNTλmNT2λmNTλmNT2S=S[+()−1][=+()−1]mstTT1.5T1.5TststNN2.42.42=[+()−1]=2.851.51.5"S2.85mR=(−1)r=(−×1)0.0224=0.326()Ωst2S0.183m(2)s=−2s=−2s=−20.041.96=BANs=s时，T=T=2TBBN′′=λmNT+λmNT2−=×2.4+2.42−=SS[()1]1.96[()1]3.65mB2T2T22NNS′′3.65mR=(−1)r=(−×1)0.0224=0.424()Ω2S0.183mP2536–46解：n−n1000−9601N（1）S===.004Nn10001ES2200.04×2NN转子电阻：r===0.0261()Ω23I3195×2N固有特性上的临界转差率：22S=S(λ+λ−)1=.004×5.2(+5.2−)1=.0192mNmmP60NT=9550=9550×=596.9(Nm⋅)Nn960N（1）T=530Nm时的转差率：λmNTλmNT22.5596.9×2.5596.9×2S=S[−()−1]0.192[=×−()−1]0.0352=AmTT530530LLn=(1−Sn)=(10.0352)1000−×=965(/min)rAA1（2）电动机的负载转矩：T=530N⋅mL25 彭鸿才《电机原理与拖动》习题解答实际位能负载转矩：T=T⋅η=530×.087=461(1.N⋅m)L1L摩擦转矩：∆T=T−T=530−4611.=68(9.N⋅m)LL1"下降时电动机轴上的等效负载转矩：T=T−∆T=4611.−689.=392(2.N⋅m)LL1λmNTλmNT22.5596.9×2.5596.9×2S=−S[−()−1]=−0.192[×−()−1]=−0.0257Bm""TT392.2392.2LLn=(1−Sn)=(1(0.0257))(1000)−−×−=−1025.7(/min)rBB1n−n1000−(−280)1C（3）S===.128Cn10001"λmNTλmNT22.5596.9×2.5596.9×2S=S[+()−1]1.28[=×+()−1]9.57=mcC""TT392.2392.2LL"S9.57mcR=(−1)r=(−×1)0.02611.275()=Ωc2S0.192m"（4）T=4611.N⋅mS=1LD""λmNTλmNT22.5596.9×2.5596.9×2S=S[+()−1]1[=×+()−1]6.31=mDD""""TT461.1461.1LL""S6.31mDR=(−1)r=(−×1)0.02610.832()=ΩD2S0.192m26 彭鸿才《电机原理与拖动》习题解答P2747–12解：11（1）λ===20sinθsin30NEUoN1（2）负载转矩不变，即：T=T=msinθ不变。NNΩx1cE0maxU10msin90TΩxEEmax1c0max0maxλ=====4TNmEUoN1sinθEoNsinθN0.5EoNNΩx1cIEfmax0max==×40.5=2IEfNoN励磁电流应增加到原值的2倍。Tmax【或：∵T=T不变，∴λ=∝T∝E∝INmax0maxfmaxTN∵当I=I时，λ=2，∴λ′=2λ=4时，I′=I=2I】ffNffmaxfNP2747–13解：P1300N（1）P===1420.77(Kw)1Nη0.9153P1420.7710×1Ncosϕ===0.899NSN36000152××0ϕ=26sinϕ=0.438NN（2）Q=3UIsinϕ=360001520.438×××=691.9(var)kNNNN27 彭鸿才《电机原理与拖动》习题解答U6000N（3）U===3464.1()VNϕ3322E=(Ucosϕ)+(Usinϕ+Ix)oNNϕNNϕNNc22=(3464.10.899)×+(3464.10.43815222.8)×+×=5876()VIxcosϕ15222.80.899××Ncsinθ===0.53NE5876oN当输出功率P2变为1000Kw时，P10002P===1093(kW)1η0.915忽略电枢绕组电阻，PM≈P1。因为励磁电流If不变，EoN不变。EUoNϕmsinθNPPxP10931NMNcM≈=⇒sinθ=sinθ=×0.53=0.4077NP1PMEUoNϕPMN1420.77msinθxc−10θ=sin0.4077=24.122220IX=E+U−2EUcosθ=5876+3464.1−×258763464.1cos24.1××=3060.4()V1c0NNϕ0NNϕIX3060.41cI===134.2()A1X22.8c3P109310×1cosϕ===0.7843UI36000134.2××N1−1sinϕ=sin(cos0.784)=0.62Q=3UIsinϕ=36000134.20.62×××=864.7(var)kNNNNP2747–14解：−1o（1）U=10000/3=5773(5.V)ϕ=cos9.0=25.842NϕP=3IUcosϕ=3×417×10000×/9.01000=6500(kW)1NNNP6300Nη===96.923%NP65001o（2）Q=Ptgϕ=6500×tg25.842=3148(1.kVar)1122（3）E=(Ucosϕ)+(Usinϕ+Ix)oNNNNc22=(57735.×)9.0+(57735.×.04359+417×14)=9838(7.V)28 彭鸿才《电机原理与拖动》习题解答Ixcosϕ417×14×9.0Ncsinθ===.0534NE98387.oP6300NP===6500(kW)1η.096923P65001I===375(3.A)3Ucosϕ3×10×12222E=U+(Ix)=57735.+(3753.×14)=7806(V)ocE7806×404oI=I==320.54(A)ffNE98387.oN（4）忽略电枢绕组电阻，PM=P1。因为励磁电流If不变，Eo不变。P4000.0/96923Msinθ=sinθ=×.0534=.0339NP6500MN−10δ=sin0.33919.82=22IX=U+E−2UEcosδsNφ0NNφN220=5773.5+9838.7−×25773.59838.7cos19.82××=4822()VIx4822sI===344.43()Ax14sS=3UI=310000344.435965.7(××=KVA)N2222Q=S−P=5965.7−4127=4307.8(Kvar)114000EUo（5）P==msinθ=3UIcosϕ⇒M0.96923xc4000ocosϕ==0.5714⇒ϕ=55.1530.9692310417×××4000oQ=Ptgϕ=×tg55.15=5927(kVar)11.09692322E=(Ucosϕ)+(Usinϕ+Ix)oNc2o2=(57735.×.05714)+(57735.×sin55.15+417×14)=11078(6.V)E110786.×404oI=I==454.91(A)ffNE98387.oN29 彭鸿才《电机原理与拖动》习题解答30 彭鸿才《电机原理与拖动》习题解答31'