10KV的450KW (4级)电机,额定电流是31A，怎么算出来的啊？ 哈理工电气工程及其自动化各方向排名与解释（高压电技术、电机电...
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P=√3UIcosφ
式中U表示为线电压、I表示为线电流，cosφ为电机的功率因数关系。
代入所给数据
450KW=√3x10KVxIcosφ
I=450KW/(√3x10KWxcosφ)=25.98/cosφA
电机的功率因数一般在0.8-0.9之间，如取0.84，线电流就是31A.
功率是45*1000W，额定电压是10*1000V，峰值电压,是10*1000*1.732V（根号3），一般这种电网的功率因数在0.82-0.85之间，所以电流=（450*1000）/【（10*1000*1.732）*0.84】=30.93A,这样解释，希望对你有帮助。
I=P/(1.732*U*cosΦ)=450/10/1.732/0.85=30.56
450/10/1.732/0.85=30.56
约31A
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