电机原理及拖动的全部答案.docx

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1、第一章直流电机原理P411-3 解：In =艮=1403 =60.87(A)Un 230P 14P = =16.37( Kw)N 0.855P411 4 解：R =UnM =1103=1430(W)PNP1.1 1031430=76.92% p- P - N 1 430- 1 1 003/30()P42121 解:UnRf230150=1.53(A)精品资料Ia = I N If =69.6 1.53 =71.13(A)Ea=UN IaRa =230 71.13 0.128 =239.1(V).22Pcu =I2Ra =71.132 0.128 =647.6(W) Pm = Eal a =23

2、9.1 71.13 =17007(W)PN16 103P1 =-N187135/)N 0.855P421-29 解:I a = I N - I fN = 40.6 -0.683 = 39.92(A)Ea =Un -IaR =220 -39.92 0.213 = 211.5(V)PM =EaIa=211.5 39.92 =8443.1(W) pcu = I 2Ra = 39.922 0.213 = 339.4(W)Pf =1心=220 0.683 =150.26(W)P1二UnIn=220 40.6 -8 9 3(W)_ _，-3二压=75皿=83.97%P 8932po =PM -PN =84

3、43.1 -7.5 103 =943.1(W)TN =9550 =9550 8443.1/1000 = 26.88(N m) nN3000P7.5T2N =9550=9550-23.88(Nm)nN3000To =Tn -T2n =26.88 -23.88 =3(N m)或者 To = 9550 且=9550943.1/1000nN3000= 3(N m)P42P421-30 解:Pn27000UN110=245.5(A)IaN -IN I fN =245.5 5-250.5(A)UN IaN Ra110 250.5 0.02nN1150= 0.1U N I aN RaCeN110 -250.

4、5 0.020.1=1050rpm或者：EaN =UN IaNRa =110 250.5 0.02 =115(V)Ea =UN -IaNRa =110 -250.5 0.02 =105(V)EanN105 1150EaN115=1050 rpmP421-31解nnNEaNCe N nN n = nNEaN EaN I a - IaNEa = EaN =UN - INRa =110 -13 1=97(V)U =Ea -INRa =97 -13 1 =84(V)1-32 解:IaN -IN -I fN =255 -5 - 250(A)Ce N500U= 440 250 0.078 = 0.841n

5、NCeeJICT30. CT N =30 Ce N =9.55 0.841 =8.032(1)二 9550 口nN96= 95501833.6( N m)500(2)(3)TN =CT NIaN =8.032 250 =2008(N m)-TN -T2N =2008 -1833.6 =174.4( N m)n。UN440Ce N 0.841=523rpm(5) noUnRa CT - CeCT 始0.078 174.4 .To = 523 一 = 521rpm0.841 8.032或者:I aoTo174.4CT N 8.032=21.71(A)no 口-IaoRa440 -21.71 0.0

6、78Ce N0.841=521rpmP421 33 解:P =UnIn =220 x81.7 =1 7 97(4)PNN =17974 0.85 =1 52 7()叱Rf22088.8=2.48(A)IaN =In - IfN =81.7 -2.48 =79.22(A)Pcu -IaN Ra -79.222 0.1 -627.6(W)_2_2 pf =I；Rf =2.482 88.8 = 546(W)p p = R PN =1797415278 = 2696(W) pm pFe =、 P-Pcu -pf =2696-627.6-546 = 1522W)PM =PN pm PFe =15278

7、1522 -16800(W)P421-34解：(1) T2N=9550MPN = 9550M=955(N m)nN750(2) IaN =In IfN =1914 = 187(A)EaN =Un 一 I aN Ra =440 187 0.082 = 424.7(V)Pm =EaNIaN =424.7 187 =79419(W)P79.419TN =9550 =9550-1011.3(N m)nN750Ce%=ENnN424.7=0.566750no 工 W=777.4rpmCe N 0.566(4) T0=Tn -T2N =1011.3 955 =56.3(N m)aoToCT NTo9.55

8、Ce N56.39.55 0.566= 10.42( A)no =U =44082 =776rpmCe N0.566P421 35 解:IaN =In -I fN =91-2.5 = 88.5(A)Ce NUn - IaN Ra220 -88.5 0.0741500= 0.142nNnN-IaRaCe N220 -50 0.0740.142=1523rpm第三章直流电动机的电力拖动P105 3-28 解:(1)估算:D 1 UnIn -Pn 103Ra 二一2iN1 220 305-60 103二ZTT223052= 0.038()Un FRanN220 -305 0.0381000= 0.2

9、084no =-UN- =20- =1055.7rpmCe N 0.2084Tn =9.55Ce nLn =9.55 0.2084 305 = 607( N m)Tn = 607 N.m )过(no = 1056rpm , T = 0 ), (n n = 1000rpm两点，可画出固有机械特性。(5) T = 0.75 T N 时，Ia=0.75laN：=1014rpmUn -IaRa _220 -0.75 305 0.038Ce N0.2084或者:nRa丁 220-T T =CeCT N 0.20840.038 0.75 607_29.55 0.2084二1014rpmUN -Ce Nn

10、220 -0.2084 1100Ra0.038- -243.2(A)P1053-29 解:Rn =Un220=0.72-J305(1)nRN-n00.5Rn9.55(Ce n)2Tn =1055.7 -0.5 0.729.55 (0.2084)2607 =529rpm0),（529, 607）两点画直线。(2)FnRN二n02Rn9.55(Ce n)2TN =1055.72 0.729.55 (0.2084)2 607 = -1052rpm0),（-1052 , 607）两点画直线。或:FnRNUn -2RNIN 220 -2 0.72 305CeN0.2084-1052 rpm(3)n00.

11、5UnCe n= 0.5n0 =0.5 1055.7 =528rpmnVN =n0-Ra0.038过(n09.55(Ce n)2TN =528 2 607 =472rpm9.55 (0.2084)20), (472,607）两点画直线。Unn。1055.70.5Ce n0.50.5=2111.3rpmRan n =n0 -9.55(0.5Ce n)= 2111.30.0389.55 (0.5 0.2084)5 607 -1889rpmn”，0), (1889 ,607）两点画直线。P1053-30 解:Rst3Un 110I12 85.2=0.65(j)，岳斯7I12 85.2=100.2(A

12、) 1.1IN = 93.72(A)1.7符合要求。RST1 =，Ra=1.7 0.129 =0.219)RST2 = RST1 =1.7 0.219 =0.372(-J)rst1 =( -1)Ra -(1.7 -1) 0.129 =0.091)rst2 =二%1 二1.7 0.091 = 0.153(c) rst3 = j =1.7 0.153 =0.26(J)P1053-31 解:(1) Ce 罪U N - IaNRa220 -41 0.376= 0.136nN1500降压瞬间，转速不变，有：n =nNU -Ce NnN . 180 一 0.136 1500 3.83(A) Ra0.376

13、T =9.55Ce NIa =9.55 0.136 (-63.83) - -82.9(N m)(2)U aNRaCe n180 -41 0.3760.136=1210rpmP1063-32 解:?CT =9.55CeT -CT I a =9.55Ce IaaNUnRaT _ UnCeCeCT 2 N _0.8Ce Na9.55Ce nI9.55(0.8Ce N)Un0.8Ce NRa220一 2 八 .- I aN 二0.82Ce N 0.8 0.1360.376 410.82 0.136二1845rpm或者： T =CTIa =TN =CT NIaNI aNNI0.8 NIaN41=51.2

14、5(A)0.8n 二”=磔工36mCe0.8 0.136n =1845rpm nN =1500rpm, Ia = 51.25A I n = 41A所以，电动机不能长期运行。P1063-33 解：(1) Ce*U n IN 二 nNaNRa220 -68.7 0.22415000.1364noUn220Ce N 0.1364=1613rpmnmin =no(1 -、.)=1613 (1 - 0.3) =1129rpm(2)n 1500D =-max1.33nmin 1129(3)U N - Ce N nminRc = j- Ra =I N220 -0.1364 1129 0.224 : 0.73

15、7.168.7R =UnIn =220 68.7 =15114(W) =15.114kW忽略To, Tn = T2N,T2N =9550 PNnN13二 9550 = 82.77( N m)1500因为额定负载不变，有:P2 二T2Nnmin82.77 112995509550= 9.785(kW)22Pcurc =I：nR =68.72 0.737 =3478(W)P106334 解:(1) AnN =no -nN =1613 -1500 =113rpmn e.=o min、maxnN 1130.3=376.7rpmnmin = nomin - En = 376.7 一 113 = 263.

16、7rpm(2)nmaxnmin1500263.7= 5.69(6) Umin =Ce*Nnmin =376.7 乂 0.1364 =51.4(V)(7) n =UminIN =51.4X68.7 =3531.2(W)忽略 To ,则有：T2N =Tn =82.77N mP2 =T2Nnmin955082.77 263.79550= 2.286(kW)P106335 解:Ce n=4401376 =0.411 nN1000no = UN = 440 =1070.6rpmCe N 0.411(1)低速时采用降压调速：En = no - nN =1070.6 -1000 =70.6rpmmax.Wn

17、no min70.6250= 0.28T =5 3 =0精品资料nminnominn = 250 - 70.6 = 179.4rpm(2)高速时采用弱磁调速，U=Un不变，允许最高转速时I a = I.CeCenono max_ .1070.6.Ce = 0.411 = 0.29351500T =CT 1a =9.55Ce 1a =9.55 0.2935 76 = 213(N m)nmax= 9550 史=9550 -29- =1300rpm T213(3)nmaxD 二nmin1300179.4= 7.25P106 3-36 解:URaJ10-35.2 .35=0.13 nN750EaN =

18、Ce NnN = U n - I aNRa = 110 - 35.2 0.35 = 97.68(V )U -EaN0-97.68(1 ) Rc =a-Ra =-0.35=1.03759)I2 35 2a max乙 uu.4(2)Rc 二U aN-110-97.68I amax-2 35.20.35 u 2.6()(3)能耗制动时:I anu -Ce NnRaRc反接制动时：IafU -Ce Nn -110-0RaRc2.95-37.29(A)T =CT NIaf =9.55 0.13 (-37.29) =-46.3(N m)(4) T=TnIa = In =35.2A0 -0.13 (-500

19、)35.2-0.35 =1.5精品资料P1063-37 解:.他励机Tl=Tn,%=卜=68.7人n=0, Ea=0&:匕且一一0.195:3Ia68.7P1063-38解：采用电动势反接制动:t TL -TNIaN = IN =76ACe N 二 UI nNaN Ra440 - 76 0.377二二 0.4111000Rc =UNCe Nn-RI aN440 -0.411 ( -500)-0.377 =8.12口76若采用能耗制动:Rc =U -Ce nIa-Ra0-0.411 (-500)0.377 : 2.33 176能耗瞬间电流:IaU -Ce nN0-0.411 1000RaRc0.

20、377 2.33=-151.83(A)I amax = 1.8In= 1.8 76 =136.8(A)丁 IaI amax二不能采用能耗制动，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:Tl F Tl2 =0.8Tn -1.2Tn =-0.4Tn1a - -0.4IN - -0.4 76 - 一30.4(A)此时电机工作在正向回馈制动状态，不串电枢电阻时:Un -IaRa 440 -(-30.4) 0.377CeN0.411=1098rpm串0.5 电阻时:Un f(Ra Rc)Ce n440 -(-30.4) (0

21、.377 0.5)0.411=1135rpmP1063-40 解:(1)U -IaRa-440 -60 0.3770.411=-1125.6(r/min)T =Ct NIa =9.55Ce NIa =9.55 0.411 60 =235.5(N m)R -UIa - -440 60 - -26400(W)(2)Rc =Un -Ce NnIa440 -0.411 (-850)60-0.377 =12.78(J)R =UnL =440 60 =26400(W)Pcuc =I；E =602 12.78 = 46008(W)(3)R=UN：CeNn-Ra =I a0 -0.411 (-300)60-0

22、.377 = 1.678()Pcuc =I；Rc =602 1.678 =6040.8(W)P1073-41 解:(1)降压瞬间，n来不及改变，Ea不变Ea =Un - IaRa =440 -0.8In 0.377 =417(V)u -EaRa400 -4170.377=-45( A)T =CT NIa =9.55Ce NIa =9.55 0.411 (-45) -176.63(N m)电动机从电动一正向回馈制动一正向电动。(2) n =Un -IaRa 400 -0.8 76 0.377Ce n0.411=917.5rpmP1073-42 解:(1) EaN =Un -IaNRa =220

23、22.3x0.91 =199.7(V)IaU - EaN-220 -199.7RaRc0.91 9= 42.35( A)199.7=0.1997nN1000T = Ct n1 a= 9.55Ce NI a =9.55 0.1997 (-42.35) =-80.77( N m)(2) n = 0 时，Ea = 0，有:IaU - Ea-220 - 0Ra Rc0.91 9=-22.2(A)， ，T =CT nI第四章变压器P1414-12 解:a -9.55Ce N I a -9.55 0.1997 (-22.2) -42.34(N m)电动机电磁转矩 T0,二者方向一致，均为拖动转矩，在|T”

24、|+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。ZmIU inIoIUin _ZmII2Ioii 2ZmII =2ZmIU1NU U 1NmI = ZmII =I oII oII顺极性串联时，I O一样，忽略漏阻抗压降，有:Ui =Io(ZmI ZmII) =Io(ZmI 2ZmI) =3内_1即有：IoZmI =-Ui 31440Eii = IoZmI =-Ui - =147(V)332440Eiii =IoZmii =_Ui =- 2 =293(V) 33U21 =号=手=73(V)k 2U 2 IIEm293= 147(V)2200

25、二4.44 fN1k=NN2220=2N1 =2N2110N1 N2N1顺串时:Uuu2 =Uu U 叫= 4.44fNi 0 4.44fN2 03,. =_ 4.44 fN1 0 = 330V2o =.2204.44 fN1因为磁通不变，所以磁势不变,3I0N1 =Io(N1 N2) =-N1lo,2 .=,10 = - 103逆联时：UU1u2= UUU2 -UU1U2-4.44 fN10-4.44 fN21.4.44 fN1 0 =110V2112200 一 4.44 fN1二，o因为磁通不变，所以磁势不变,1= I。=21。由此可见，空载电压不相等。1414-13 解:UMJ| =4.

26、44fN1 0 =220VUU 21414 -18解:k=U =1=26.3U2N 0.38ZmU2N 380I2039.5= 9.62()rmp01100I22039.52=0.705(c)Zm =k2Zm =26.32 9.62 =6654(ij) rm -k2rm -26.32 0.705 =487.6()xm= Z； - rm =、66542 - 487.62 = 6636( 1)ZkUk 450Ik 20=22.5()rkpk 4100, 22I k 20= 10.25)Xk，. Z2 - rk2 = 22.52 -10.252 = 20()rk75oC234.5 75 10.25

27、=12.2)234.5 25Zk7 = ；75。X2 = ,12.22 202 =23.4)P1414 - 20解:(1) k=UNU 2N230 115rk=rr2 =r k2r2 =0.3 22 0.05 =0.5)Xk三x1x2 Mxik2x2 =0.8 22 0.1 =1.2(J)Zk=.rk2 X；= 0.52 1.22 : 1.3(c)(2) r； = r；H+r2 =r1/k2 +r2 =0.3/ 22 +0.05 = 0.125(0)xk=X x2=x1/k2 x2 =0.8/22 0.1 =0.3)Zk =Xk2 = 0.1252 0.32 =0.325()(3) I 1NS

28、N3 103U1N 230=13.044(A)I 2NSn3 103Z1NU1N 230I1N 13.044-17.63(1.Z2NU 2N115=26.09(A)U2N 115I2N 26.09= 4.41（）*rk0.50.028 XkZ1N17.63XkZ1N1.2=0.06817.63*ZkZkZ1N黑3=.074rk0.125Z2N4.41*=0.028 = rkXk*Xk 0.3 八、, -0.068 - XZ2N4.41ZkZk0.325Z2N4.41_ _ _*=0.074 = Zk(4) UkN =I1NZk =13.044 1.3 =16.957(V)UkNU kN16.9

29、57U1n 230=0.074 = Zk(5)满载时3 =1cos % =1 时，sin% =0 : _ _c *_._ _U % = ：(rk cos 2 xksin 2) =1 (0.028 1 0 =0.028 =2.8%cos52 =0.8 落后时，sin52=0.6：一 一G *.*._ .U % 二.-(rk cos :2 xksin 2)=1 (0.028 0.8 0.068 0.6 =0.0632 =6.32%cos 平2 =0.8 超前时，sin 中2 =-0.6： _ _c *_ _ U% = ：(rk cos 2 xksin 2) =1 (0.028 0.8 -0.068

30、 0.6 = -0.0184 =-1.84%P1414 - 21解:U1N :10/*J3 Y, yno结：k =千=25U 2N 0.4/, 3 , 仝载试验：Zm =U2N/ 3 400/ 一 3I 2o60-3.849（，）Po38002- =2 =3.519(,)3I2o 3 60Zm22 =k Zm =2405.63(，)rm =k rm =219.91(J)Xm=、z V =2395.55()SN.3U1N75043.3(A).3 10ZkU1N/.3 440/311k43.3= 5.867。Pk3I1：109002 = 1.938(-:)3 43.3Xk= Z -r： = ,5.

31、8672 -1.9382 =5.538(，)kLfyMC)k75oCrk2750 X； = . 2.3572 5.5382 =6.02()_2_2 、PkN =3I1Nrk750 =3 43.3 2.357 = 13257(w) = 13.26(kw)1X1.-X25.538 =2.769()-2(2)J 也上& 133.34I1N :43.3C1)*rk750rk750 乙N需=0.0177*XkXk153534 =0.0415*Zk7506.02133.34=0.045a) cos% = 0.8滞后时， sin中2 =0.6_ _c_*_.一_ _ _:S N cos 2- S N CoS

32、 ；：2 2 PkN PoU = ：(rk750 cos 2 xk sin 2) =1 (0.0177 0.8 0.0415 0.6) = 0.0391 = 3.91%= 97.24%1 750 0.8 22 一 一 1 750 0.8 113.26 3.8b) cos % =1 时 sin% =0_ _n *._一_一_ _ U =工 750 cos 2 =1 0.0177 J0.0177： 750 0.8 12 13.26 3.8P1424 一 22 解:.77%:S N cos 2- S N cos 2- 2 PkN Po1 750 11 750 1 12 13.26 3.8= 97.7

33、8%_ : S N cos ：2:S N cos ： 2： 2 PkN Po=1Po“PkN：S N cos 2Po： 2 PkN26.6 12 21.21 1800 0.8 6.6 12 21.2-98.1%max2Po-mSNcos 22 Pon2 6.60.558 1800 0.8 2 6.6=98.5%c) cos 2 =0.8超前时，sin 中 2 = -0.6U = P(rk750 cos%+xksin%) =1父(0.0177M0.80.0415M0.6) = 0.01074=1.074% 效率与cos邛2 =0.8滞后时相同:)1424 24 解:)1334 33 解:U U1

34、220k =U2180119Ii20011/9-163.64(A)I12 = I2 - I 1=36.36( A)S 电磁=I12U2 =36.36 180 =6544.8(W)S 传导=I1U2 =163.64 180 = 29455(W)第五章三相异步电动机原理)1895 5 解:00000槽距角 a =20 -0 =40 -20 =20每极每相槽数q=3kp1一 q：sin 一2_.3 20o sin 2a qsin23sin变 2=0.9598.元件电势Ey=40V . 元件组电势 Eq =qEykp1 =3 40 0.9598 = 115.2(V)1895 - 6 解:Z1 =363

35、6p 360 o2360 o2 pm36=20 okw1 =ky1kD1 =sin 丛 90o psinsin3 20oaqsin 20o3sin 2= 0.96每相匝数:NipqNa2 3 10= 60(匝)E1 = 4.44 flMk. =4.44 50 60 0.96 0.172 = 219.94(V)91895 一 7 解:Ky1 = sin 90=s” 90=0.9397 9p 3600 2 36000二二 2036362mp 2 3 2=3(槽)kp1_ q： sin 一2,3 2C0 sin2. qsin 万3sin 迎 2=0.9598kw1 =ky1 kp1 =0.9397

36、0.9598 =0.902,1905 -26 解:60 fln160 x50 取整数 八=3.083970n1 = 1000rpm1000-9751000-0.025750003 380 139 0.87= 94.2%_ Pn _75 103P13U N I N COS N,1905 -27 解：_3PM =p -pcu1-pFe =8.6父10 -425-210 = 7965(W)pcu2 usPM -0.05 7965 =398.25(W)Pm = Pl _ pcu2 = 7965 -398.25 = 7566.75(W)P1905 -29 解:(1); nN =950rpmn1 =100

37、0 rpmSNn 一nNn11000 -9501000= 0.05 Pm=PN +pm+ps =28+1.1 +0=29.1(kW)P 29.1PM =-m= =30.63(kW)=pcu2 =sPl =0.05 30.63 -1.532(kW)1 -Sn1 -0.05 R = PM + pcu1 +pFe =30.63 +2.2 =32.83(kW)n = = = 85.3%R 32.83P,3U n cos n32.83 1033 380 0.88=56.68(A)1905 30 解：_3PM = P pcu1 pFe =10.7父10 450200 =10050(W)Pcu2 =sPm

38、=0.029 10050 =291.45(W)Pm = PM - pcu2 =10050 -291.45 =9758.55(W)四四极电机，；n1=1500r/minPM10050T =9.55 =9.5563.985( N m)n11500P1905 31 解:(1); nN =1455rpm.n1 =1500 rpmn1 -nN 1500 -1455 -一N 1 N0.03n11500 Pm =PN + pm +ps =10 +0.205=10.205(kW)0.316(kW)1 -SSPn0.03 10.205Pm = Pcu2= Pcu2 =S1 -S 1-0.03Pn10 Tn =95501=9550M= 65.64(N m)nN1455pm Ps0.205To -9550- =9550 =1.346(N m)