電機學作業(yè)答案胡虔生版

1、電機作業(yè)題第九章解：效率為二9.1Pn7.5 燈 03,3U n I n cos3 380 16.4 0.789-2、設(shè)有一50Hz,4極三相異步電機，請?zhí)顫M表9-1的空格n(r/min)15401470-600s10f2(Hz)工作狀態(tài)解:n(r/min)1540147001500-600s-0.0270.02101.4f2(Hz)1.35150070工作狀態(tài)發(fā)電機電動機起動或堵轉(zhuǎn)理想空載電磁制動9.4解：(1)已知額定轉(zhuǎn)速為1457r/min,因為額定轉(zhuǎn)速略低于同步轉(zhuǎn)速極對數(shù)為60 fn 1500額定時轉(zhuǎn)差率為nAn 1500 -1457-0.287m1500故該機的同步轉(zhuǎn)速為n=1500

2、(r/min)(2)額定輸入功率為日90100.559(kW)0.895額定輸入電流為I1NF23U 1NCOSA N90 (60100.559 10 3=22.503(A)、3 3000 0.863-590.167( N_m)額定輸出轉(zhuǎn)矩T2:(3)求歸算變比先計算每相有效匝數(shù)2門2二145760定子每相梢數(shù)Zi48定子梢距角:-12Pmi23x3p360023600i548a定子繞組因數(shù)Kd=sinq4sin7.sin3O766qs三=i22pkp=cos0.966kzi=kpkd=0.925定子繞組每相匝數(shù)Ni=pqNc=2440=320電動勢變比KA2NiKNA23200.925=59

3、2電流變比Kj/pgNiKNi223320o92559.2Z260阻抗變比KeKj=35046.4(4)額定時電動勢為Ei=0.95=0.93000i558.846(V)Ei每極磁通：？：m=4.44fiNhi-0.0237氣隙磁密最大值Bm=0.752(T)2l(5)轉(zhuǎn)子每相梢數(shù)q=乙=52Pm223600n=i20Z2ot2繞組因數(shù)=0.9567.aqsin2kp2-ikNi=kpkd=0.9567轉(zhuǎn)子繞組每相匝數(shù)N2=60=203每相轉(zhuǎn)子感應(yīng)電動勢E2二4.44sfiN2KNaa=2.89轉(zhuǎn)子電動勢的頻率f2二sfi=0.028750=1.4335(Hz)N 1 kz10.93 I =4

4、495.5P TOC o 1-5 h z 33(6)定子基波旋轉(zhuǎn)磁動勢的振幅Fm221000-975-0.025解：額定轉(zhuǎn)差率SN1000n19.5設(shè)以u:為參考軸，則U1=1000乙300(1)應(yīng)用T形等效電路計算定子電流：h =乙轉(zhuǎn)子電流：-|2U1Z2Z=1004 - 30.39Z ZI1Z1 =88.45 49.00(2)應(yīng)用較準確近似等效電路計算修正系數(shù):X1G =11 =1.04轉(zhuǎn)子電流:=88.50 /-(A q叵)j (為歟2 )/112.03c勵磁電流:33.99-84.27定子電流：h=im2=100.8.30.76ACi(3)應(yīng)用簡化等效電路計算轉(zhuǎn)子電流:勵磁電流:?Ui

5、-121=91.88/-12.255j(X2)Im533.99-84.27(H+rm)+j(N+Xm)定子電流：h=1m-I2=1074-29.78A9-6解:(1)電磁內(nèi)功率:P二Pm%-Pfe=5811.5-237.5=5574w輸出功率:F2=PPAc_Pad=55744529=5500w電機效率:旦-3=87.03%R6.3210Pr-5.32kW展二舍0.041同步轉(zhuǎn)速：由Sn=Pm5811.5m二型-1500r/minn,一n可得空4=0.041x代一FSeufPft-L67.5W曠1500額定轉(zhuǎn)速為n=1439 r/min(3)電磁轉(zhuǎn)矩：T =Pi =55746037.00NLm

6、2兀勺439P29-7機械轉(zhuǎn)矩：T236.52NLm-1n用較準確的近視等效電路計算轉(zhuǎn)子電流:勵磁電流:定子電流:?UiI2iF?LriImi(rirm)j(xixm)J=lm-12=i7.23.-29.54Ci=i5.42/i4.224.89/-82.80、Ui2輸入功率：R=mUilicosA=i7089w電磁功率:定子銅耗：定子鐵耗：轉(zhuǎn)子銅耗：Pmm1122r2=529w巳二呻222=i5879ws2=miliA=66iw2Pfe=mJmrm=646w最大轉(zhuǎn)矩:mip2抽2(x,張)2193.79N*m額定轉(zhuǎn)矩:TnmiPU12SN=101.05N?(門-嚴)2(xie1X2)2SN9-

7、8過載能力：km臨界轉(zhuǎn)差率：Sk由Sk=1得:m=i.9i8tn0.i27(xiex?)2rj(xiqx2)2C=5.9362=r2-r2=5.936-0.7425.i94解：(i)nA60f=750r/minSN=0.037由簡化轉(zhuǎn)矩公式Sk=SN(KKm-0.148Tn二一3439N?mTm=KTN=7325.1Nm2602S由T-Tm可得SkS=0.01時T=989.89NmS=0.02時T=1979.78NmS=0.03時T=2969.67Nm9-10C1)PM=p2FfnecRdRu2=155.9kWPcu2額定轉(zhuǎn)差率：Sn空=0.014PM同步轉(zhuǎn)速：m=60f1500r/minP額

8、定轉(zhuǎn)速：nN=(1-Sn)m=1479r/min(2)p=pM-pc2u=153.7kwp電磁轉(zhuǎn)矩：TL=992.38NmQUr38。一(J20%5.V5最大轉(zhuǎn)矩：T扁=卓=2137.9N m隅2?+Jrj+(治+X2)2F臨界轉(zhuǎn)差率：Sk0.096?(X1X2)2負載轉(zhuǎn)矩：“牛觀49968.49220.096可得2 0.096TmSk2Sk2137.9S0.096Sk0.096S解得S=0.020一Sk電機能夠正常運行轉(zhuǎn)速n=Ri(1-s)=1470r/min=651.3Nm電機學第十章10.3解：(1)當起動轉(zhuǎn)矩最大時，有Sk-1修正系數(shù)&=1+Xl=1.04XmI則由Q:Gr2,=1+(

9、治+GX2)2解得r2=0.3984則需串電阻為r;二r2-r2=0.32612此時起動電流為St52-Ulri+rm+j(X!+Xm)GA+GD+j(GXi+0X(2)當需要限制起動電流不超過額定電流的2倍時即Is：21nUirirmj(X1Xm)st十U%jgC2X2)s此時的起動轉(zhuǎn)矩為T：-mpU12(ri+&2)2+j(Xi+GX2)起動轉(zhuǎn)矩倍數(shù)210.6Ui+S22rirm?jXiXm)Xy2廨上叫加據(jù)麗為Snj一“1500147015000.02r21.047門ni15001300廠598.9A2)j1X1CX2)0.1333niFCU2匯Pn=0.612kW1-SNp2=0.0769531則r,=0.4361(2)調(diào)速電阻上功率損耗為P;=312r；=3.745kW1000 -980 =0.021000SN210.7Sni-nN解：額定轉(zhuǎn)差率