电机原理及拖动的全部答案

1、第一章直流电机原理NRUnPn14 1060.87(A)230P411 - 4 解：R110 13 1430(W)PN 1430 1100 330(W)Pn 2将沧92%UnRf230伸厲Ea Un laRa 230 71.13 0.128239.1(V)2 2Pcu l；Ra 71.132 0.128 647.6(W) Pm Eala 239.1 71.13 17007(W)PnN16 1030.85518713.5(W)P421-29 解：la In I fN 40.6 0.683 39.92(A)Ea Un laRa220 39.92 0.213 211.5(V)PmEala 211.5

2、 39.92 8443.1(W)pcu lX 39.922 0.213 339.4(W)PfIfNUN 220 0.683 150.26(W) R UnIn 220 40.6 8932(W)駁 士383.97%7.5 103 943.1(W)P 89328443.1/100026.88(N m)3000Po PM PN 8443.1Tn 9550 乩 95507.53000 23.88(N m)nNPnT2N 95509550nNTo Tn T2n26.88 23.88 3(N m)PO或者 To 9550 nN9550943.1/100030003( N m)5I NCeU N I aN R

3、a110Pn27000U?110245.5(A)或者:EaUnEaNnNP42250.5(A)NnN250.5 0.0211500.1U N I aN RaCe NEaNU N1 aN Ra110Ean110250.5 0.021050rpm0.11 aN Ra110250.5 0.02Ea nNEaN250.5 0.02115(V)105(V)竺竺 1050rpm1151-31 解:N13 1I a I aN97(V)EaNCe N nN n nNEaEaN U NI N Ra110a Ra 9713 184(V)P421- 32 解:IaN In I fN 2555250(A)UN IaN

4、 Ra440 250 0.078Ce N0.841nN500CeCtN 30 Ce N 9.55 0.841 8.032CT30(1 )T2N95509695501833.6(N m)nN500(2)TnCTNIaN 8.032 2502008(N m)(3) ToTnT2N20081833.6174.4( N m)(4) n°Un440523rpmCen 0.841noUnCe N5230.078 174.40.841 8.032521rpm或者：G佥 器21.71(A)'U N I ao Ra n。Ce N440 21.71 0.0780.841521rpmP421-33

5、 解:R U NIN 220 81.717974(W)PnP n 17974 0.85 15278(W)I fNUnRf22088.82.48(A)IaN In IfN 81.7 2.4879.22(A)Pcul；NRa79.222 0.1 627.6(W)Pfl2NRf 2.482 88.8 546(W)p R &17974 15278 2696(W)PmPFePPcuPf2696 627.6 546 1522(W)PmPn Pm pFe 15278 1522 16800(W)P421-34 解:(1)T2n9550 Rl 9550nN75750955(N m)Tn9550nN955

6、079.4197501011.3(N m)Ce NE aN424.70.566nN750n。丛44 777.4rpmCe N 0.5661 aon。(4) T。Tn T2NToToCT N9.55Ce nUn I aoRa440Ce NP421 aNCe N1011.3 95556.39.55 0.56610.42 0.0820.5661-35 解:IN I fN 912.588.5(A)U N I aN RanN220 88.5 0.074150056.3(N m)10.42(A)776rpm0.142n Un 1尺Ce N220 50 0.0740.1421523rpm第三章直流电动机的电

7、力拖动P105 3-28 解:(1)估算. RU n丨N R 102 lN1 220 305 60 103230520.038()Ce NUn IaNRa220 305 0.038nN10000.2084noUnCe NTn9.55CeNIaN 9.55 0.2084 305607(N m)过(no = 1056rpm , T = 0),(n n = 1000rpmTn = 607 N.m )两点，可画出固有机械特性。n Un 1足Ce N(2) T = 0.75 Tn 时，Ia 0.75 :1014rpm220 0.75 305 0.0380.2084或者：nUnRa2T1055.7rpm0

8、.2084°.。380.75607pgrpmCe NCeCTN0.20849.550.2084UN Ce Nn 2200.2084 1100243.2(A)&0.038P1053-29 解:RnUn2203050.72(1 ) nRN n00.5Rn丁2 T N9.55(Ce n)21055.70.5 0.729.55 (0.2084)2607529rpm过（n, 0）, （529, 607）两点画直线。(2) nRN n02Rn-t-2 T N9.55(Ce n)21055.72 0.729.55 (0.2084)26071052 rpm过（n, 0）, （- 1052,

9、607）两点画直线。或：nRNUN 2RnIn220 2 0.72 305Ce N0.20841052 rpm(3) n°n°0.5UnCe NRa0.5 n00.5 1055.7 528rpm2 Tn9.55(Ce n)5280.0389.55 (0.2084)607472 rpm过（n, 0） （472, 607）两点画直线。(4)Unn。0.5Cen。1055.72111.3rpmn°.50.5Ra0.038n n n°2 Tn2111.32 6071889 rpm9.55(0.5Ce,n)9.55 (0.5 0.2084)过（n0 , 0)(18

10、89,607)两点画直线。P1053 30 解:Un110I12 85.23%3 0.65 Ra- 0.129I1285.2,0.65()1.71.7Rst31.1In 93.72(A)符合要求。RST1Ra1.70.1290.219()RST2Rst!1.7 0.2190.372()rst1(1)Ra(1.7 1)0.1290.091()rst2rst11.70.0910.153()st3rst21.70.1530.26()P1053-31 解:Ia(1) Ce NUn 1 aN Ra“n22041 0.3761500降压瞬间，转速不变，有：nnNUCe N nNRa180 0.136 15

11、000.37663.83(A)9.55Ce N I a9.55 0.136 ( 63.83)82.9(N m)0.136丄空 180 41 严6 1210rpmCe NP1063 - 32 解:Q Ct 9.55Ce T Ct I a9.55Ce I aUn RaCeCeCTUn0.8Ce NRaUn0.8Ce NRa20.8 Ce NI aN29.55Ce N I aN9.55(0.8Ce N)2200.376 4121845rpm0.8 0.136 0.82 0.136或者:T Ct IaTNCT N I aNIa宀aNlaN0.8 n-4151.25(A)0.8Un I a Racn 1

12、845rpm220 525 °.3761845rpm0.8 0.136nN 1500rpm, I a 51.25 A I n 41A所以，电动机不能长期运行。n。P1063-33 解：(1)CU N 1 aN RaCe NnN220 68.7 0.22415000.1364nminUnCe N2201613rpm0.1364)1613 (10.3)1129rpmn maxnmin竺 1.331129RcUnCe Nn min1 NRa220 °1364 11290.2240.73768.7UnIn 220 68.715114(W)15.114kW忽略To,Tn = T2N,

13、T2n 9550 P 9550 盏 82.77(N m)因为额定负载不变，有:P2T2N nmin95509.785(kW)9550PcuRc 1 aN Rc68.720.7373478(W)P1063-34 解:(1) nN no nN 1613 1500 113rpmnNn 0 minmax113肓 376.7rpmnN376.7 113 263.7rpmP2(2)Dn max15005.69nmin263.7(3)U minCeN nomin376.7(4)P1U min 1n 51.468.7忽略To，则有：T2NTT2N nmin82.77263.70.136451.4(V)3531

14、.2(W)82.77N mn minn omin95509550P1063-35 解:CeUn gRa440 76 0.376。卅nN1000n。豈辯 1070'6rpmmax山 7060.28nomin 250(1)低速时采用降压调速:nNno nN 1070.6 100070.6rpmnmin(2)高速时采用弱磁调速，U=Un不变，允许最高转速时I a = InCe n。Ce Nn。maxc1070.6Ce15000.411 0.2935nominnN 250 70.6 179.4 rpm76213(N m)T Ct Ia 9.55Ce Ia 9.55 0.2935nmax 955

15、0 讯 9550291300rpmT213(3) D 皿空07.25nmin 179.4P106 3-36 解:Ce NU N 1 aN RanN110 35.2 0.357500.13EaNCen nNUnI aN Ra11035.20.3597.68(V)(1)Rcu1 a maxEaNRa97.68RcUEaN1 amaxRa能耗制动时:02 35.20.35 1.0375(110 97.682 35.20.352.6(1 anUCe N nRaRc0RaRc反接制动时:1 afCe NnRaRc110 02.9537.29(A)T CT N 1 af9.55 0.13 ( 37.29)

16、46.3( N m) T Tnla In 35.2ARcUCe N nIaRa0 0.13 ( 50O)0.3535.21.5()P1063-37 解:t他励机 Tl=Tn ,. I a=l n=68.7An=0, Ea=0Rc U Ea Ra 220 00.1953Ia68.7P1063-38解：采用电动势反接制动:QTl TnI aN I N 76 AUn I aN RaCe NnN440 76 0.37710000.411RcUnCe NnI aNRa440 0.411 ( 500)760.3778.12若采用能耗制动:能耗瞬间电流:0 0.411 ( 500)760.377 2.33U

17、Ce N nNRaRc0 0.411 10000.3772.33151.83(A)Iamax 1.8In 8 76136.8(A)I a I amax不能采用能耗制动，只能采用电势反接制动。P1063 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 l Tl1 Tl2 0.8T n 1.2T n0.4TnIa 0.4In 0.4 7630.4(A)此时电机工作在正向回馈制动状态，不串电枢电阻时:nUnlaRaCe N440 ( 30.4) 0.3770.4111098rpm串0.5 Q电阻时:UnI a (RaRc)Ce N440 ( 30.4) (&

18、#176;.3770.5)10.411P1063-40 解:(1) nUlaRaCe N440 60 0.3770.4111125.6(r/mi n)CT N I a9.55Ce NIa9.55 0.411 60235.5(N m)UIa440 6026400(W)(2)U n Ce N ncRa 440 °411(850)600.37712.78()UnJ440 6026400(W)pcucIaRc602 12.7846008(W)(3)Un Ce Nn0 0.411 ( 300)%RaI a600.3771.678()Pcuc1；艮 602 1.6786040.8(W)P1073

19、-41 解：(1)降压瞬间，n来不及改变，Ea不变EaUn IaRa 4400.8In 0.377417(V)IauEaRa400 417茴45(A)CT N I a9.55Ce NI a 9.55 0.411 ( 45)176.63(N m)电动机从电动t正向回馈制动t正向电动。(2)n Un laRaCe N400 O'876 O.377917.5rpm0.411P1073-42 解:(1)EaNUI aN Ra22022.3 0.91 199.7(V)IaUEaN220 199.7RaRc0.91 942.35(A)CeEaNnN199 7一-0.19971000CT NI a9

20、.55Ce nI a9.55 0.1997 ( 42.35)80.77( N m)(2)n = 0 时,Ea = 0，有:IaU Ea220 0RaRc0.91 922.2(A)Ct N 1 a9.55Ce nI a9.55 0.1997 ( 22.2)42.34(N m)电动机电磁转矩均为拖动转矩，在|T”+Tl作用下,电机反向起动,第四章T “ <0,位能转矩 Tl>0，二者方向一致，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。变压器P1414-12 解:ZmIU1NZmIIU 1NI ollZ mlU 1NlolU 1N2IoiiZmll2顺极性串

21、联时,I O 一样，忽略漏阻抗压降，U1o(ZmIZmll )1 o (ZmI2Z ml )即有：lo ZmlU131440EnI oZ ml-U313147(V)E1ii1 o Zmll3U144032293(V)E1l147U 21k273(V)E1II293U 211147(V)k2由此可见，空载电压不相等。p1414-13解：Uu 4.44fN1o220V有:ZmllIoI2Ioii3 1 oZmI2ZmI2204.44 fN1N12N2N1 N2顺串时:UU1U2U1U2J%4.44fN1UlU2o 4.44 fN3'-4.44 fN1 o 330V4.44 fN1因为磁通不

22、变，所以磁势不变,IoN1 Io(N1 N2)N1I0Io自。逆联时：UU1u2Uuu2 UU1U24.44 fN1 o 4.44 fN?14.44fN110V2204.44 fN 1因为磁通不变，所以磁势不变,IoN1 Io(N1 N2)N1I0Io2IoP1414 -18 解:rmU 2N0.38U 2N380I 2o39.5Po1100109.62(U 1NZm39.52l2o26.30.705(rmk2*26.32 0.705487.6()Xmzm2rm66542487.626636(Uk450Zk22.5()I k20Pk4100rk.210.25()Ik202Xk、zi2 rk、2

23、2.52210.2520()6654()234.575).0C)rk75'Zmk2Zm 26.32 9.6210.2512.2(234.5 25Zk75° C2X2k75°Xk12.2220223.4(P1414 -20 解:(1)U1 NU 2N230115rkr1k2r20.3 22 0.050.5(xkX1x2x-ik2x220.8 22 0.1 1.2(Zk0.52 1.221.3()2 2(2) rk ri r2n/k a 0.3/20.05 0.125()xk x-i x2 x-i /k2 x20.8/ 220.10.3()Zk、k2 X0.1252 0

24、.32 0.325()(3)IiN3 10323013.044(A)1 2NU2N3 10311526.09(A)乙NU1NI1N230-17.63(1)U 2NZ2NI 2N1154.41()13.04,26.09*rk0.5*Xk1.2rk0.028Xk0.068Z1N17.63Z1N17.63*Zk1.3Zk0.074Z1N17.63*rk0.125*Xk0.3*rkk0.028 rkXk0.068XkZ2N4.41Z2N4.41*Zk0.325*Zk0.074ZkZ2N4.41(4) U kNI1NZk 13.044 1.316.957(V)U UkNkN16.9570.074 ZkU

25、1N230(5)满载时B =1cos 21 时，sin 20 :U %(r；cos 2*xk sin2) 1 (0.0281 0 0.0282.8%cos 20.8落后时，sin 20.6 :U %*(rk cos 2*Xk sin2)1(0.0280.8 0.0680.60.06326.32%cos 20.8超前时，sin 20.6 :(rk(0.028 0.80.0680.60.01841.84%cos 2 xk sin 2) 1U%P1414 -21解:Y,yno结:U 1NU 2N10/ .30.4/ . 3空载试验：ZmU2n/'、3400/-3603.849()ZmZkXk

26、Po37!单 3.519()3 6022 'k Zm 2405.63()2rmXmU1N/ .311k2oCZk753I1Nrk750X"、2.3572 5.53826.02('1r1r22rk750(2)Z1NU1NI 1Nrk750Zk750rm k2 'rm219.91()2395.55(440/. 343.3SN3U1n75043.3(A).3 105.867(Pk31：3 43.31090役 1.938().5.8672 1.93825.538()r'k75oC43.3" 2.357 13257(w)2.3571.17910 103

27、 / . 3"2.357 乙 N 133.34Zk7506.02Z1Na) cos43.30.0177133.34 0.04520.8 滞后时，sin 20.6(rk750 cos 2 XkSin 2)S n cos 2S n cos 22pkN Pob) cos 21 时，sin13.26(kw)X1X2133.34()1 (0.0177XkUrk750 cos 21 0.0177S n cos22S N cos 2pkN po234.5 75 1.9382.357()20234.5XkZ1N0.8 0.04151 750 0.82750 0.8 113.260.0177 1.77

28、%1 750 15.5382.769()進 0.0415133.340.6) 0.0391 3.91%97.24%3.81 750 1 12 13.26 3.897.78%c) cos 20.8 超前时， sin 20.6 :U(rk750 cos 2 XkSin 2) 1 (0.0177 0.8 0.0415 0.6)0.010741.074%cos0.8滞后时相同:S n cos 2S n cos 22PkNPo1 750 0.81 750 0.8 12 13.26 3.897.24%P1424-22解2pkN2Po0.558maxpkN6.6 12 21.21 1800 0.8 6.6

29、12 21.298.1%2p° mS n CoS 2 2p°2 6.698.5%0.558 1800 0.8 2 6.6P1424-24 解：ZY/y-SbSTP1334 -33解：kU122011I1 且型 163.64(A)U21809k 11/9S电磁112U 236.36 1806544.8(W)I12 I2 I136.36(A)S传导I1U2163.6418029455(W)第五章三相异步电动机原理P1895 -5 解：槽距角0 0 0 0 0200402020每极每相槽数q=3si nJ二 kp1qsi n2T元件电势Ey=40V.3 20osin220o 3s

30、i n20.9598二元件组电势EqqEykpl40 0.9598115.2(V)5-6解7362p 4z2 pm364 3p 360 o2 360 o 20 o36每相匝数:E1ky1N1kp1 sin 生 90osinq2qsi n2pqN4.44刑水肋P18957解-y 0Ky1 sin 90q 2mp36kw1 ky1P1905 -6解-n12 3 1060(匝)4.447 sin 9509002 3 23（槽）.3 20o sin0.9620o3sin260 0.960.9397kp1kp1 0.9397 0.95980.9023.08970取整数 30.172219.94(V)p

31、3600sin 2qsi n 23620022003sin 2sin31000rpm1000 975n110000.025PNP175 10375000P1905 -7解：- 3U N I N cos N3 380 139 0.8794.2%3R Pcu1 PFe 8.6 10425 2107965(W)-2000.9598PmPm Pcu2 7965 398.25 7566.75(W)PT905-t29解：1) n”950 rpm lOOOrpmn1 nN 1000 Sn1000空 0.05PmPnpm ps 28 1.1 0 29.1(kW)(3) R(4) InPm1SN29.11Pcu

32、l両 3°.63(kW)Pcu2sFM0.05 30.63pFe 30.63 2.232.83(kW)1.532(kW)2832.8385.3%Pl32.83 103. 3U N COS56.68(A)380 0.88P1905 -30解-PMR Pcu1 pFe10.7103 450 20010050(W)Pcu2 sPm 0.02910050291.45(W)Pm PMPcu2 10050 291.45 9758.55(W)t四极电机，二 n=1500r/minT 9.55 也9.5510050150063.985( N m)P1905 -31 解：1) nN 1455 rpmq

33、 1500rpmSn1 nN 1500SN145515000.03 PmPnPmPs 100.205 10.205(kW)P 1 SPms Pcu 2Pcu2盏°.°3 仁05 0.316(kW)1 S 10.033) Tn 9550nN95501065.64(N m)To 9550Pm 冬nN9550 僭14551.346(N m)T 9550PmnN955010.20567(N1455m)Z。ZkI03.64P0Pm264 11引023 3.642Zor。2.60.32Uk/、3100/i3Uo/、36.4()roX)7Ik6.42 60(T905 -33 解：380/

34、60.3()8.25()XkrmXmPk31k2rkroXo47头 3.2(8.252 3.227.6(ri3.2 21.2()1F 7.6 3(ri6.4 24.4()x160 3.8 56.2(第六章 三相异步电动机的电力拖动P2516 -33 解：Un 380/、3219.4(V)100095710000.043 TmaxU1211一(X1X2 )22219.42 2502.08.2.082(3.12 4.25)270.84(N m)me 2U11' 2(X1X2 )350219.421.530.043(4) STmax 2.215TnrgmaX .A2(X1X2 )2P2516

35、34解：（i）Tn9550 乩nN759550 -720(2)TmaxmTN(3)SNni(4)SN ( m2Tmaxsm ss sm(5)SN E2N3I 2NP2526 -35解：ninNSnn1SN E2N3I 2NSN ( m% a Rsmr233.335( N(3.12 4.25)21.53,2.082(3.12 4.25)2994.8(N m)2.4 994.82387.52( N.m)750 7200.04750, m 1) 0.04(2.4 、2.42 1) 0.1832 2387.520.1830.04m)0.1998s (N.m)0.183213.3 220750 7177

36、500.044 223.50.0224()0.0440.12(2m 1) 0.044(3.15、3.152 1) 0.27SBTnTbncP2Pcu2PcuRRsm晋J 0.270.122.52mTNTlmTn )21Tl0.27 吐Tn0.7Tn(30加 10.0304Sa 0.03049550 PnN2 mT NSmSbSBsmmTNTLTCC Tl9550 卫 213(N m)7172 3.15 2132.52 0.03040.03042.5216.19(N.m)MTn、2Tl(1 0.284) (750)60Pm(1 Sc)FMi(1Pm Pm 8384.6R箱 Pcu22.52(3.

37、15Tn)20.7TnY 0.7Tn537(r / min)25370.7 2138384.6(W)600.284) 8384.66003.4(W)6003.42381.2(W)寸 2382 2126(W)(3)转子串电阻前后 Tl0.7Tn 没有变,据TCt mI 2 COS 2，Ct没有变、U1不变故m不变、cos2基本不变，二转子电流基本不变。P2526 -36解:SN空心°0.0271500smSN E2N' 3I 2NSN ( mnin10.2840.°27 3990.0536()”3 116m 1) 0.027(2.82.82 1) 0.146吨趣0.6

38、7 15000.67 28TnSmRc4.594 59(诙1) 00536 1.63()【或:mTNSSm-TlM T- 2TlRc0.1462.8Tn0.8Tn(0.8Tn)2 10.0213S(S 1)r20.67(聞 1)°05361.63(Tn 9550RnnN9550 屠 66(N m)结果相同】32526 38 解：I -20/311.547 (A) R V3lNUNcos n 品 20 380 0.87 11452.32(W)P2 P11452.32 0.87510020.8(W)直接起动时：TstKTTn1 .4TnTslTst(U-/k)2U-1k2k2Tst1.4

39、T-1.4TstT-1U N380U322(V)k1.181(3) Y/ 起动：| stlist1klki 1 N1 72046.6(A)3331111Tst-Tst-KtTn- 1.4Tn0.460.5T-333所以，不可以半载起动。降压起动时，设：U1 U-/k ，则有:k 1.18'1 '(4) TstTst0.5T-1 st 1 .41 Nk22.8k1.671st1 |k2lst1 ki I n2.8172.82050(A)U-k型 227.5(V)1.67SnEnNn1750 7260.032750smSBTnTbncSN E2N°.°32 285 0.081(.3 65SN ( mmTNTL2Sa2r2 R29550 PnN2 mT NSmSbSm(1SBmTNTlSc)mP2536 -44 解：转子电阻：2m 1) 0.032(2.8MTN、2Tl0.0250.0812.821)2 1 °.173 誥1.9750.081竺 0.1734.7309550394.6(N m)7262 2.8 394.64.71.9751.9754.7789.2(N.m)0.173(mTn )2(1 0.68)(750726750E2N SN 31 2N固有特性上的临界转