电力拖动自动控制系统 第四版 课后答案 下载本文

?is??3?cos(?t)?

?Im??i??2?sin(?t)??s??两相旋转坐标系中的电流

?isd??cos?sin???is??3?cos?sin???cos(?t)???Im??i????i?????sin?cos?2??s????sin?cos???sin(?t)? ?sq??3?cos?cos(?t)?sin?sin(?t)?3?cos(?t??)??Im??Im??2?cos?sin(?t)?sin?cos(?t)?2??sin(?t??)?当

d???1时,???1t,两相旋转坐标系中的电流 dt?3??isd?3?cos(?t??)??Im?

Im??i?????2?sin(?t??)2sq??????0??电流矢量幅值

is?isd?

3Im 26.4笼型异步电动机铭牌数据为:额定功率

PN?3kW,额定电压

UN?380V,额定电流

IN?6.9A,额定转速

,转子电阻

nN?1400r/min,额定频率

fN?50Hz,定子绕组Y联接。由实验测得定子电阻

Rs?1.85?Rr?2.658?,定子自感Ls?0.294H已折合到定子侧,系统的转动惯量

,转子自感

Lr?0.2898H,定、转子互感

Lm?0.2838H,转子参数

J?0.1284kg?m2,电机稳定运行在额定工作状态,试求:转子磁链?r和按转子磁链定向的定子

电流两个分量sm、st。

解:由异步电动机稳态模型得额定转差率

iisN?额定转差

n1?nN1500?14001??

n1150015?sN?sN?1?sN2?fN?电流矢量幅值

100?rad/s

1522is?ism?ist?3Im?3?6.9A 2由按转子磁链定向的动态模型得

Ldyr1=-yr+mismdtTrTrws=LmistTryr

稳定运行时,

dyr=0,故yr=Lmism, dtist=wsTryr100p0.2898=wsTrism=?ismLm152.6582.2835ism

22is?ism?ist?1?2.28352ism?2.493ism?3?6.9

解得

ism?3?6.9?4.79A

2.493ist=2.2835ism=2.2835?4.79=10.937A

转子磁链

yr=Lmism=0.2838?4.79=1.359Wb