计算答案.docx

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1、电机及拖动(第二版)计算A卷答案一、计算分析题(每题1分)1.Rj57.5/=R=23x0:AINUN2301.=n+,=(100+4)=104APQJH=ii042=1081.6w(1)转速nEa=Un+凡1.=(230+0.1X104)=240,4VC=-=0.1603JnN150()电动机运行：及二UN-R=(220F1x104)=209.6VC=C=-01603=0.15331.AV2301.e230Ea209.6.n=J=1367rminCt01533(2)电磁功率P=FI=209.6104=21.798W.ema/a.a(3)电磁转矩2.解：(1)原来高压绕组的匝数UJUinp=N

2、JN1因为630,卢=M,所以n=630匝。4003402v,(2)新的高压绕组匝数因为100003_TVrl43-40所以N=0匝_U,v-R/n_220-0.1x158.511N1000=0.204V(rmin)n_URJn_U108RJnlCCv220-0.8x0.1x158.5=1016r/m.n0.2042.UScO8(R+Rs儿=22O-O.8x(O.I+O.3)x158.5=83(WminCv0.2043 .降压瞬间n不突变，Ea不突变，电流突变为:188-0.204x1016=_193A稳态后电流Ia恢复到原来值(0.8In),稳态后转速为:0.204，一R/J88-。.8x0

3、.1x158.5=859“tnin4。根据r,=C丁N1.=G1。得，=/=-o.8/=/=158.5A1 a,1.a0.8,vu-E“u-c/RaNIN=1251rminUbR/.220-01x158.5Cl,0.80.2044 .解：（1）折算到高压侧的变压器“T”形等效电路参数并画出等效电路图（设RI=R2,X=X2,）由空载试验数据求励磁参数R1.曾吃。%CXn=Z2tn-R2,n=3.852-0.352=383Q折算到高压侧有：9-77一）?Zm=kZm=253.85=240625,Rn,=kRlt,=25035=218.75xJ=2X.=2523.83=2393,75由短路试验数据

4、求短路参数=UOM=WOl逐S-八.43.3=587C,RS=岭噢“94CIs43.3?R=R产；RS=gxl94=097则Xs=JzMK=5/5.87=1.94=5.540x1=x=Xs=554=277折算到高压侧的变压器T”形等效电路如图JS0307所示：（2）额定负载且cos62=0.8（滞后）时的二次端电压Z=JOoOo/G=i33.3,R*=旦=1=0.0146乙I.43.3KSZJI33.3密封线*=X1.=也I=OO4I6Q,ASZm133.3AU=X*scos*ssin9Ji00%=1(0.01460.8+0.04160.6)100%=3.664%U、=(1-M)U,N=(1-

5、003664)x400=385.3V5.起动电流比为：=工二J=1.7J1R1V285.20.13各级起动电阻为：Hm=(万一DR”=(1.7-I)XO.13=0.09ICHm=/RM=I7xOO91=O.155CR,3二/Rn=1.7X0.155=0.263C6.解：（1）额定电流j=PNJXIo1NUN230=73.91A(2)电枢电动势F=CZ7v=-11M=l31021500=241V口UJ6()4N60lIU7.在固有特性曲线上:=0.41lV(rmin)UbRJN_440-0.38x76一1500Mo=UN.440C.v。411=1070rmin=%=1070-1000=70r/m

6、in在降压人为特性曲线上：11n,n=。一=250-70=180rminn7()静差率为：%=l.23Xs*=0.039Rs*=0.0225(C)(2)u=h*(Rk*cos+Xk*sin)100%=4.14%U2=(1-u)U2N=383.44V15.解：(1)p2=pN=GkWIr力=竺4Rf177=1.3AP=A=177x1.3W=299WIn6000230A=26.IA乙=八-/=(26.1+1.3)A=27.4APQta=RE=057X27.42W=427.9W所以：Pel=P2Pcua+P/=(6000427,9299)W=6726.9WP67269T=2_-=_777=M3N77

7、71 em2145060(2)P.=P,+P陪+PFe=(6726.9+234+61)W=7021.9W=100%=-100%=85.45%P17021.916.解：Z；=0.063=渭厂但X；=依一辰2=。056Z/=1/0.045=22.222R11=Po*Io*2=(1000/180000)0.0452=2.743Xm*=Jzf-2=22.05(2) AU=R；Cosq+X；sin/)=2220.8+0.0560.6=5.136%密封线(3)=1Q-SV,100%(=1-I-SNCoS化+Po+PsN,X100%1000+05240000.5X180X1030.8+1OOO+0.5240

8、0()/=97.3%17.解：(1)输出功率：(小心+&+吃+匕。=6.32-(0.0290.045+0.2375+0.1675+0.341)=5.5%W效率：7=oo%=-100%=87.03%Pi6.32(2)电磁功率：Pem=P2+Pad+Pmec+Pcu2=5.5+0.029+0.045+0.2375=5.8115&W转差率：SN=当30.041Pem5811.5(3)转速：n=(1-s)n.=(1-5)=(1-0.041)x50=1438.5r/minP2(4)空载损耗：E)=Ptnec+Pad=0.045+0.029=0.074kWP74空载转矩：=9550=9550=0.49Nmn1438.5P55(5)输出转矩：7；=9550-=9550:=36.5INmn1