电机原理及拖动地全部问题详解

1、第一章直流电机原理PnUn14 103230二 60.87(A)P41 1 - 4 解:= 110 13=1430(W)Pn1430P = RPn =1430 1100=330(W)P42 1 - 21解:=1.53(A)Rf 150la = I N I f =69.6 1.53 =71.13(A)Ea=UN laRa =230 71.13 0.128 =239.1(V)2 2Pcu =IaRa =71.132 0.128 =647.6(W) Pm = Eal a =239.1 71.13 =17007(W)R1 N16 1030.855= 187135(W)P421- 29 解：la = I

2、N I 倶=40.60.683 = 39.92(A)Ea 二U N laRa =220 39.92 0.213 = 211.5(V)Pm =Eala =211.5 39.92 =8443.1(W)pcu = I ；Ra = 39.922 0.213 = 339.4(W)Pf =ItnUn =220 0.683 =150.26(W) R=UnIn=220 40.6 =8932(W)3血=75如=83.97%R 8932po 二Pm -Pn =8443.1 -7.5 103 =943.1(W)Tn= 9550= 9550nN8443.1/10003000= 26.88(N m)p7 5际二9550

3、 TN 9550 赢=23.88(" "ITo 二Tn -T2n =26.88 -23.88 =3(N m)或者 T。=9550 -PonN二 95503000m)P421-30 解:Pn27000110=245.5(A)I aN = I N - I fN = 245.5 5 = 250.5(A) Un * 丨 aN RanN110250.5 0.021150U N I aN Ra110 -250.5 0.020.1= 1050rpm或者：EaN 二Un GRa =11° 250.5 0.02 =115(V)Ea 二Un -laNRa =110 -250.5 0.

4、02 =105(V)nNEa nNE aN105 1150115=1050rpmP421- 31 解I：'EaN - Ce N nNn=nN- NI I aNEa 二 EaN 二Un -Ra =110 -13 1 = 97(V )U 二Ea -IzRa =97 -13 1 =84(V)P421-32 解:IaN =In -IfN = 255 -5 =250(A)uN - 1 aN Ra _ 440 -250 7.078Ce NU .OT 1nN500Ce> 丁CT N -30.Ce N 二 9.55 0.841 =8.032Ct3071(1)T2 N= 9550 弘=9550=

5、1833.6( N m)nN500(2) Tn = Ct la 8.032 250 =2008(N m)(3) T。=Tn -T2N =2008 -1833.6 =174.4( N m)noUn _ 440Ce N 0.841= 523rpmn。Ce NRl?t To = 523CeCT N0.078 174.40.841 8.032二 521rpm或者："Cv盟")n。Un - 1 ao RaCe N440 -21.71 0.0780.841=521rpmP421 33 解：P =UnIn =220 汇81.7 =17974(W)Pn 二 P N =17974 0.85

6、=15278(W)1 fNRf22088.8-2.48(A)laN “N -IfN =81.7 -2.48 = 79.22(A)Pcu =隰艮=79.2220.1 =627.6(W)Pf = l：Rf =2.482 88.8 =546(W)Z p = p PN = 17974 15278 = 2696(W)Pm PFe 八 P-Pcu - Pf = 2696 - 627.6 -546 = 1522(W)P. -Pn - pm - pFe =15278 1522 =16800(W)P421 34 解:p75(1) T2N =9550 = 9550955(N m)nN750(2) G “n 4 =

7、191-4=187(A)EaN =Un -laNRa =440 -187 0.082 = 424.7(V)Pm =EaNlaN =424.7 187 =79419(W)pM79 419TN =9550 -95501011.3(N m)nN750(3) Ce N =弘二坐47 =0.566nN 750un44L=777.4rpmCe N 0.566(4) T。二Tn -T2N =1011.3 -955 = 56.3(N m)aoToCT NT。9.55Ce N56.39.55 0.566= 10.42( A)n。UnaoRa 二 440 一10.42 0.082 = 776rpmCe N0.56

8、6P421-35 解:I aN = I N - I fN = 91 - 2.5 二 88.5(A)U n - 1 aN RanN220 -88.5 0.0741500= 0.142U N - 1 a RaCe 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:= 0.038(01 UnIn -R 1031 220 305-60 1032 22 IN23052U N - 1 aN RanN220 -305 0.0381000旦 2201055.7rpmCe N 0.2084Tn =9.55Ce NlaN =9.55 0.208

9、4 30 607( N m)n = 607 N.m )过(no = 1056rpm ，T = 0)， ( n n = 1000rpm T两点，可画出固有机械特性。(2)T = 0.75 T N 时，la =0.75iaN :U N 一 I a RaCe'N220 -0.75 305 0.0380.2084= 1014rpm或者:RaCeCT N 0.20840.038 0.75 60729.55 0.20841014rpmUN- Ce N nRa220 0.2084 11000.038二工43.2(A)P105 3 - 29 解:Rn =UnI N220 =0.72 门305(1)nR

10、N°.5rn丁9.55(Ce n )2 N= 1055.7 -0.5 °72 26079.55 (0.2084)2529 rpm过（n°,0),(529, 607)两点画直线。¥nRN2Rn9.55(Ce N )2 Tn= 1055.72 0.722 6079.55x(0.2084)21052 rpm过（n°,0),（1052, 607）两点画直线。或:FnRNU N - 2RN 1 NCe N220 -20.72305 052rpm0.2084n°-0.5n0 -0.5 1055.7 =528rpmCe'NIVN 二 n0

11、-9.55(Ce n)Ra2 Tn 5280.0382 607 二 472rpm9.55 (0.2084)过(n。, 0) (472,607）两点画直线。Un n0.5Ce Nn°0.51055.72111.3rpm0.5木 2 TN9.55(0.5Ce N)Ra0.038= 2111.3 -29.55汇(0.5 汉 0.2084)6071889 rpm过(n；, 0), (1889,607)两点画直线。P105 3 - 30 解:-0.65C1)-1.7I2 _ 二100.2(A)1.1IN1.7= 93.72( A)符合要求。Rst ' Ra =1.7 0.129 = 0

12、.219()Rst2 二Rsti -1.7 0.219 = 0.372()rst1 =( 1)Ra =(1.7-1) 0.129 =0.091(门) rst2 = ' rst1 =1.7 0.091 = 0.153(11) rst3 =，rst2 =1.7 0.153 = 0.26( |)P105 3 - 31 解:(1) CeNU N 1 aN RanN220 -41 0.3761500降压瞬间，转速不变，有：n二nNIaU Gn18° 心50063.83(A)Ra0.376T =9.55Ce NIa =9.55 0.136 (-63.83) - -82.9(N m)0.1

13、36(2) n 上 沁80-41 °.376 =1210rpmP1063 - 32 解:V Cy = 9.55CeT = CT I a = 9.55Ce I an UnRaUnRax Tz = 疋 9 55C © z I zCJCeCT 2 N 0.8Ce N 9.55(0.8C； n)2 9.55Ce N aNUnRa0.8Ce N 0.82Ce NI aN2200.376 4121845rpm0.8 0.1360.82 0.136或者：；T=CJIa=TNCt nI aN© I IaN0&N4 =51.25( A)0.8Ce0.8 0.136n =18

14、45rpm nN =1500rpm, la = 51.25A I n = 41A所以，电动机不能长期运行。P106 3 - 33 解：(1)c总UN-laNRa 220 -68.7 0.2240.1364n。Un2201613rpmCe N 0.1364nN1500nmin =n o(1 -、.)=1613 (1-0.3) =1129rpmD =归=1500 =1.33nmin1129RcU N _ Ce N nminRa220 "1364 1129 - 0.224 = 0.73768.7P 二UnIn =220 68.7 =15114(W) =15.114kW忽略 To, Tn =

15、 T 2N,p13际=9550小9550阪nN= 82.77( N m)因为额定负载不变，有:P2 =T2Nlin =82.7L_1129 =9.785(kW)95509550PcuRc =I；nRo =68.720.737 =3478(W)P106 3 - 34 解:(1) nN =no -n N = 1613 -1500 = 113rpmno min沁二空=376.7rpm'max 0.3n min n。min - nN =376.7 -113 =263.7rpmD二500 =5.69nmin263.7Umin =Ce N nomin =376.7 0.1364 = 51.4(V)

16、Pi = U min In =51.4 68.7 =3531.2(W)忽略To,则有：T2n 二 Tn =82.77N m卩2Nnmin 8277263.795502.286(kW)9550P106 3 - 35 解:= 0.411Un-laNRa440 -76 0.376nN1000Un4401070.6rpmCe N 0.411(1)低速时采用降压调速：=nN = n° - nN = 1070.6 -1000 = 70.6rpmmaxn omin70.6250= 0.28nmin =n°min f nN =250 70.6 =179.4rpm(2)高速时采用弱磁调速，U

17、 = U不变，允许最高转速时I a = I NnoCe nno max二!0706 0.411=0.2935 1500T 二Ct la =9.55Ce la =9.55 0.2935 76 = 213(N m) nmax =9550 Pn =9550 公 1300rpmT213呜 7.25nmin 179.4P106 3 - 36 解:Ce =Un laNRa11035.2 江 0.35=013750nNEaN 二Ce NnN 二 U n 一 I a"Ra = 110 - 35.2 0.35 = 97.68(V)(1)Rc _U -EaNI amax0 一 97.68 一 0.35

18、=1.0375()-2 35.2R _U - EaN R RcRa11097.68I amax-2 35.20.35 =2.6(")能耗制动时:I anU -Ce n n0RaRcRaRcT 二Ct (an =0反接制动时:1 afU GnJzII37.29(A)Ra2.95T =CT Nlaf =9.55 0.13 (£7.29) - -46.3(N m)(4) T =Tn. 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 二m Tl2 =0.8Tn T.2Tn =-0.4Tn a - -0.41 N - -0.4 76 - -30.4(

19、A)此时电机工作在正向回馈制动状态，不串电枢电阻时:= I = 35.2ARc=U-CeNn_Ra=0-0.13_500)_o.35=1.5(Ia35.2P106 3 37 解：'他励机 Tl=Tn,. I a=I n=68.7An=0, Ea=0IaRa220 -068.7 0.195 =3"P106 3 38 解：采用电动势反接制动:t TL =TNI a - I - 76 ACe NUn laNRa _440 76 汉 0.377 _04111000nNRc N -Ce NRaI aN440 -°.411 (一500)_0.377 ®276若采用能耗

20、制动:Rc =U CM-RaIa0-0.411 (一500)_0.377 = 2.3376能耗瞬间电流:IaRaRc0-0.411 10000.3772.33二-151.83(A)Iamax =8In =1.8 76 =136.8(A)Ia Alamax 二不能采用能耗制动，只能采用电势反接制动。p1063 39 解：0.5Ce N440 -（-3。4）。劝二阴如0.411_ l a (RaRc)Ce N440 -(-30.4)(°.377°.5)“35rpm0.411P106 3-40 解:440 -60。劝借阳如） Cn0.411T =CT NIa =9.55Ce Nl

21、a =9.55 0.411 60 =235.5(N m)R 二 Ula 二-440 60 = -26400(W)(2)R _ U N Ce Nn R _ 440 - 0.411( -850)Ia60- 0.377 =12.78(")R=UNla=440 60 =26400(W)Pcuc= ljRc=602 12.78 =46008(W)(3)Rc=UN-CeNn_Ra=0_0.4!(_300)_0.377.678(l a60Pcuc=1；艮=602 1.678 =6040.8(W)p107 3-41 解：（1）降压瞬间，n来不及改变，曰不变Ea =Un -laRa =440 -0.8

22、In 0.377 =417(V).'U -Ea*Ra400-4170.377-45( A)= 9.55Ce Nl '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 -IaRaCe%40°0加了护0.377 /0.411（1）EaN 二UP107 3-42 解:N -|aNRa =220 -22.3 0.91 =199.7(V),'U - EaNl aRa&込摯42.35(A)0.91 9EaN199.71000= 0.1997t' =Ct Nl a =9.55Ce Nl

23、 a =9.55 0.1997 (-42.35) =80.77( N m)(2) n = 0 时，Ea = 0，有：U -EaRaRc-220 - 00.91922.2(A)T''二CT NIa =9.55Ce Nl "a =9.55 0.1997 (-22.2) =-42.34(N m)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。U1 :- 1 o(Zml'Zmll即有:1 o ZmlJu31En=I

24、o ZmlU13E1ii- 1 o Zmll3U 21E1I147k211U 2 IIHIo(ZmI - 2ZmJ =3l°ZmI440=147(V)440小23-293(V)_ E1II _ 293k由此可见，空载电压不相等。2 MV)第四章变压器P1414-12 解：Z .U1NU 1NZmI Zmll I oI1 ollV IoI= 2I°ii二 ZmIU 1NU 1 NZmllZmll = 2Z lol2Ioii2顺极性串联时，I。一样，忽略漏阻抗压降，有:P141 4-13 解:Uu = 4.44 fN1。= 220V2204.44 fN1kN2220 =2110

25、N2N23N1N2N12顺串时:DU"=UUUiU=4.44fN14.44fN2 0 =3 4.44 fN； 0 = 330V2U1U204.44 fN1逆联时：'' '' 1 ''Ud1u2 =Uuu2 _Uu1u4.44fN1 o -4.44 fN/o4.44。= 110V因为磁通不变，所以磁势不变,loNi =l°(Ni N2) =3Nil。= lO -2Io23IIo220-4.44 fN,'' 1 ' ''因为磁通不变，所以磁势不变,ION1 = IO(N1_N2匕 N1l&#

26、176; = lo=21。P141 4- 18 解rm1026.3U 2N'0.38 -U 2N380l 2o_39.5Po1100=9.62(门)Zmk U1NZm =k2Zm =26.32 9.62 =6654(门)rm =k2rm =26.32 0.705 =487.6(门)Xm=Z； _r；66542 -487.62 6636(")Zk5450 .22.5)lk 20rk= 10.25(门)Xk=22.52 -10.252 =20(门)rk75',o234 5 +7510.25 =12.2(")C 234.5 25Zk75o Cr爲-X：二 12.2

27、2202 二 23.4(门)P141 420 解:U 2N叫2115rkk2r2 =0.3 22 0.05 =0.5(")xkxx1k2x2 =0.8 22 0.1 =1.2(")Zk rk2 X：二 0.52 1.22 =1.3,)2 2(2) r=ri a 二 ri/kr 0.3/ 20.05 = 0.125(xk =论 x2 =论 /k2 x2 = 0.8/ 22 0.1 = 0.3(.)ZkrJXk = 0.1250.3 =0.325()(3)1 1NS 3 103U = 230-13.044(A)1 2NSn _ 3 103= 115=26.09(A)Un230乙

28、N17.63(")I1N13.044*rk0.5rk =0.028X；Z1N17.63*Zk1.3Zk0.074Z1N17.63*rk r；0.125*0.028 = r；Z2N4.41*zk0.325*z;0.074 - ZkZ2N4.41；x哙疇U 2N115Z2N -4.41(")I 2N26.09X；1.20.068Z1N 17.63Y,yno结:空载试验：ZmU2n/'、3l2o400/-360= 3.849,)ZmPo3l；o3800k3®9)XmN-r； =2395.55()SN750Zk5n/ 3440/ 3 =5.867(11k43.3X

29、k.z" -rk25.8672 -1.9382 5.538()3U1nPk3li；r"k75oC3 10二 43.3(A)10900EW)234 5+752 ' 2 '=k Zm =2405.63(")咕二 k =219.91(门)k75°Crk：50X： =、2.3572 5.5382 =6.02(")PkN =31；匚750 =3 43.32 2.357 = 13257(w) =13.26(kw)H =D =1心50 冷H.357 =1.179°Xv - x2；-=? 5.538 = 2.769(11)(2)Z1N

30、Um 10103/133.34)1 1N !43.3rk750rk750遁=0.0177乙N 133.34XkXkZ1N進 70415133.34Zk750Zk750Z1N1骼 0.045a) cos ;2 =0.8 滞后时， sin 2=06n*：U = .：(rk750 cos 2 XkSin » =1 (0.0177 0.8 0.0415 0.6) = 0.0391 = 3.91%PS N cos2：S N cos ；2 ：2pkN Po1 汉 750 汉 0.821 750 0.8 113.26 3.8= 97.24%b) cos 2 =1 时，sin 2 =0=1 0.01

31、77 1 =0.0177=1.77%-97.78%卩 S N cos1 S N cos 2： 2 PkN Po1750" 2 750 1 113.26 3.8c) cos 2 =0.8 超前时，sin 2 二-0.6 :U = ' (r*750 cos 2 xksin 2)=1 (0.0177 0.80.0415 0.6) =0.01074 =1.074% 效率与cos 2 =0.8滞后时相同:= 97.24%PS N cos护2-S N cos ；2 2 PkN Po仆750汉0.821 750 0.8 113.26 3.8P02PkNP1424- 22解二12BSn CO

32、S®2 +PoPkN=1竺工半 98.1%1 1800 0.8 6.6 12 21.2J P 'I5582P02 6.6max 工101fSnCOS 2 2po-98.5%0.558 1800 0.8 2 6.642 424解:PooXYz33解：pi33 4k斗丝U21809IiI2 _ 200k 11/9= 163.64( A)l12 = I2 _ h = 36.36(A)S 电t =I12U2 =36.36 180 =6544.8(W)S传导二 IJ2 =163.64 180 = 29455(W)第五章三相异步电动机原理P189 5_5解槽距角：_ 20° _

33、00 =40° _200 =20° ；每极每相槽数q=3sin兰.a qsi n 2T元件电势Ey=40V.3 20o sin2 = 0.959820o3si n2二元件组电势Eq-qEy kp1=3 40 0.9598= 115.2(V)P189 56解7Zj =362p =4362 pm43 =3p 360Ot =2360 = 20 o36kw1 = k y1kp1 =si n/90osin兰aqsin2.3 20o sin二 0.9620o3si n2每相匝数：NpqN23 102 3 iu =60(匝)Ei 二 4.44 £ N1kw1= 4.44 50

34、60 0.96 0.172 = 219.94(V)P189 57解7y 0Ky1 =sin 90二 sin? 90° =0.93979p 3600 2 輕=20。36362mp 2 3 2二 3（槽）kp1aqsi n 2.3 200 sin二 0.95982003sin -2g =ky1 kp1 =0.9397 0.9598 =0.902'190 526解：n1取整数 >3n1970二 1000rpm1000-975n11000=0.02575 103_ PN_ =N R . 3U n I n cos n75000= 94.2%3 380 139 0.87P190 5

35、27解：3Pm = R Pcu1 PFe =8.6 汉 10 425210 = 7965(W)Pcu2 =sPm =0.05 7965 =398.25(W)Pm 二 R - Pcu2 二 7965 -398.25 二 7566.75(W)P190 529解- n” 950rpm.m TOOOrpmSN -n1 f10归空=0.051000ni Pm -Pn Pm Ps =28 1.1 0 =29.1(kW)29 130.63(kW)1 -SN 1-0.05PmPcu2 二 sPM = 0.05 30.63 =1.532(kW) R =PMpcu1 - pFe =30.63 2.2 =32.83

36、(kW)N = 285.3%-32.83Pi32.83 10356.68(A) 3UnC0S n 3 380 0.88190 5- 30 解：PM = P pcu1 - PFe =10.7 勺03 450 200 =10050(W) pcu2 二sPM =0.029 10050 =291.45(W)Pm =PM - pcu2 =10050 -291.45 =9758.55(W):四极电机，n=1500r7minPM10050T =9.55 旦=9.5563.985( N m)n11500P190 5解-1) nN =1455rpmn _! =1500rpmn11500 -14551500= 0

37、.03 Pm =PN Pm Ps =1° 0.205 =10.205(kW)1 -Spcu2pcu 2SFm _ 0.03 10.2051 -S 1-0.03= 0.316(kW)P103)Tn =9550=955065.64(N m)nN1455"95507=9550 鬻 十46小m)nNnNZ。u。/、3I0380/=60.3(03.64°= P°-Pm264 -11 =6.4)23 3.64X。=60.32 -6.42 =60(")ZkUk/、3Ik葺乞 8.25（）Pk4703Ik23 72= 3.2(XkZ： -r：.8.252 -3

38、.22= 7.6(")P10 205T =9550=955067(N m)1455PI90 5- 33 解:r2 二 rk -匚=3.2-2 =1.2(") x x2 =1 x 1 7.6 =3.8(")2 2rm 二 r° -n = 6.4-2 = 4.4(') xm = x0 - x1 = 60 - 3.8 =56.2(')第六章 三相异步电动机的电力拖动p251 6-33解：Un =380/、3= 219.4(V) tmax2'1u22'211(Xy X2；J2219.42 2 二 502.082.082 (3.12

39、 4.25)2= 70.84( N m)1000 一957 “043ni1000 SmaxUi* 2s2 2 ' 2 (ri-) - (x、；一 X2；_)s''219.421.530.043(2.081.53 )20.043)(3.12 4.25)2maxTn= 2.2151.53=33.335( N m)2 ' 2 1(X!；X2；J0.1998.2.082(3.124.25)2(1)(2)(3)Sn=370 胡.04750P251 634解7(4)Sm =Sn( m -1) =0.04(2.4.2.4Tmax = 'mTN =2.4 994.8 =

40、2387.52( N.m) -1) =0.1832Tmax2 2387.52Sm .旦SSm0.183. s (Nm)s 0.183(5)Sn E2N3 1 2N0 04 H 2130.04 213 =0.0224()3 220p252 6- 35 解： Sn =3n1込叫0.044750SN E2N0.044 223.5 31 2N3 47= 0.12(门)m T) =0.044(3.153.152 -1) =0.27:sm =2 +RSmSm 二Sm4 0.27=2.520.12!/'mTNSA = Sm'mTn、2一 TlTl百=0.27輕一0.7Tn0.7Tn=0.03

41、04SB=Sa = 0.0304TnP16=955宀9550 亍213(Nm)nNTb2 3.15 213sm. SbSbSm16.19(N.m)2.520.0304+0.03042.52mTN= 2.52 泄-0.7Tn(3阿)2_10.7Tn二 0.284nc =(1-sc)m =(1 -0.284) (750) =537(r/mi n)PM =Tcc =Tl=0.7 213 2 -537 =8384.6(W)6060P2 : Pm =(1 -sJF =(1 -0.284) 8384.6 =6003.4(W)Pcu2 二 PM -Pm =8384.6 -6003.4 =2381.2(W)R

42、1P代齐 P"寸 2381."2126(W)(3)转子串电阻前后 Tl =0.7Tn 没有变,据 T =CT mI 2 cos :2Ct没有变、U不变故m不变、COS2基本不变,二转子电流基本不变。p252 6- 36 解：Sn 二0.027 3995 1161500 -14601500= 0.027= 0.0536(")Sm =SN(m . m T) =0.027(2.82.82 -10.146sJ500 -500=0.671500I FTnIT+ (込)2_1= 4.59(擁一1) °.053".63(徴：S = Sm一 Tl('m

43、Tn )2 _1Tl= 0.0213= 0.146込一严丐0.8Tn V 0.8TnS0.67R, =(1)r2 =(1) 0.0536 =1.63()S0.0213结果相同】92526- 38 解:I N20 /3 = 11 .547(A)1)R =V3|NUN COS®N =7320x380x0.87=1145232(0)p2“9550汁備 屠"6(N m)直接起动时：Tst二KtTn二1.4Tn降压起动时，设：U1 二u“/k ，则有:Tst(UN/k)212 2TstUn k2Tst1 .4Tn'TstTn-1.4k =1.183801.18= 322(V)

44、111(3)y/ a起动：| st Ist k| I N 7 20 =46.6(A)333111TstTst二KtTn 二-1.4 Tn =0.46 ： 0.5Tn333所以，不可以半载起动。'1(4)TstTst3tst =0.5TnTst =1.4Tnk2 =2.8k =1.67Ist Ist kilN 7 20 = 50(A)k 2.82.8U N _ 380k 1.67= 227.5(V)252 643解一ni750 一 726 "032750SN E2N0.032 285 =0.081() .3 65-1) =0.032(2.8 一 2.82 -1) =0.173I

45、 '-mTNSa = Sm ILTl5Tn )2 /|=0.173 |Tl2.8Tn0.8Tn（环= 0.025Sb = 2 - SA = 2 - 0.025 =1.975Sm二亘迟Sm丿081212 0.173 =4.70.081TN =9550 PnN30=9550394.6(N m)7262 ' mTN Tb :為+ SBSBSm2 2.8 394.64.71.975 =789.2(N.m)1.9754.7mTNTl4,箫= 0.68nc = (1 - Sc)m = (1 - 0.68) (-750) = -240( r / mi n)253 6- 44 解：750 sN

46、 二 n1 - nN =750 _ 726 =0.032转子电阻：2E2N SN 31 2N285_°.032 =0.081 、3 65Sa =Sm固有特性上的临界转差率:Sm 二SN('m.叮 T) =0.032 (2.82.82 -1) =0.1733固有特性上 T = Tn时的转差率:)2 -1 =0.1733 (2.8- . 2.82 -1) =0.032Rc =(S J"二3S -S,PS3 1S 3=(1) 0.032 =1.2567r20.081n =(1 S )n j =(1 1.2567) 750 = _192.5rpm(2)旦二 550更 -P2

47、 = Tn (-192.5) 30 = _ 7.9545( kw)PnTn nNTn nNTn 7269550总机械功率： Pmec = P2 + Pmec + Ps = 7.9545 +1.05 =.9045( kW)Pmec-SSPcu2pcu 2SPnec1 -S1.2567 (-6.9045)1 -1.2567= 33.8(kW)外串转子电阻的损耗=33.8 - 1.5 = 32.3 kW3)P2 - -7.9545(kWP253 6-45解-n17522=0.04750E2NSN Sm七学Tst(tTN)2“證和-121? 0.04 =0.0224(")3I2N .3 220S = Sn ( ' m + g -1) =0.04x(2.4+J2.4 -1 =2.85 Rst =(Sm -1)r2 =(-