《电机学》(华中科大出版社,辜承林,第二版)课后答案 下载本文

(4)?P?PCU?PCU2?PFe?Pmec?Pad=474+316+213+45+37.5=1103.5W

?=P2P27500???87.17% P1P2??P7500?1103.5(5)PU1I1cos?1?3?380I1?0.824 1?37500?1103.5?I1??15.86A

3?380?0.8245.36 一台4极中型异步电动机,PN?200kW,UN?380V,定子Δ联接,定

子额定电流IN?385A,频率50Hz,定子铜耗pCu2?5.12kW,转子铜耗

pCu2?2.85kW,铁耗pFe?3.8kW,机械损耗pmec?0.98kW,附加损耗pad?3kW,R1?0.0345?,Xm?5.9?。正常运行时X1??0.202?,

???0.195?;起动时,由于磁路饱和与趋肤效应的影响,??0.022?,X2R2???0.11?。试求: ??0.0715?,X2X1??0.1375?,R2 (1)额定负载下的转速、电磁转矩和效率;

(2)最大转矩倍数(即过载能力)和起动转矩倍数。 解:(1)Pmec?p2?pmec?pad=200+0.98+3=203.98kw PCU2?SPem Pmec?(1?S)Pem

pS2.85S?cu2? 即 ?pmec1?S203.981?SS=0.01378

n?(1?S)n1?(1?0.01378)?1500?1479rmin

p2.85Pem?cu2??206.82kw

S0.01378p206.82?103?60Tem?em??1317N.M

?12??1500PP2200?=2???92.7%

P1P2??P200?5.12?2.85?3.8?0.98?3(2)Tmaxm1u121?

2?12(R1?R12?(X1??X2??))603?3802???3186N.m

222??15002?(0.0345?0.0345?(0.202?0.195))PN200?103?60TN???1291.97N.m

?2??1479T3186km?max??2.466

TN1292Tst??m1PU12R222???(X1??X2??)?2?f1??R1?R2??1?0.04?3??0.34?20.52?10287.72W0.04

?3?2?3802?0.07152??50???0.0345?0.0715???0.1375?0.11????22?2721.5N.m

kst?Tst2721.5??2.11 TN1292

PN?200kW,UN?380V,f?50Hz,5.37 一台三相8极异步电动机的数据为:

nN?722r/min,过载能力kM?2.13。试求: (1)产生最大电磁转矩时的转差率;

(2)s=0.02时的电磁转矩。 (1)

Sm?SN(km?k2m?1 推倒如下:

TN21(即Tem?TN时) ??TmaxSN?SmkmSmSN Sm2?2kmSNSm??N2?0 求一元二次方程即可

n?n750?722??0.03733 SN?1n1750Sm?0.03733?(2.13?2.132?1)?0.1497

(2)

pN200?103?60??2646.6N.m TN??2??722Tmax?kmTN?2.13?2646.6?5637.2N.m Tem2 ?TmaxSm?SSSmTem22???0.2625 0.14970.025637.27.485?0.1336?0.020.1497Tem?5637.2?0.2625?1480N.m

5.38 一台三相4 极异步电动机额定功率为28kW,UN?380V,?N?90%,

cos??0.88,定子为三角形联接。在额定电压下直接起动时,起动电流为额定电流的6倍,试求用Y-Δ起动时,起动电流是多少? 解:

PN28?103 IN???53.72A

3UNcos?N?3?380?0.85?0.9直接起动时的起动电流:

Ist?6IN?6?53.72?322.3A

用Y-△起动时: I Ist??st?107.4A

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