一台三相异步电动机,额定功率为10KW,额定电压为380V,额定转速980r/min,额定工作效率η=95%? 一台三相异步电动机的额定功率Pn=10kw,额定电压Un=3...

\u6709\u4e00\u53f0\u4e09\u76f8\u5f02\u6b65\u7535\u52a8\u673a,\u989d\u5b9a\u529f\u738710kW\uff0c\u8f6c\u901f1440r/min,\u989d\u5b9a\u7535\u538b380V\uff0c\u989d\u5b9a\u9891\u738750Hz\uff0c\u989d\u5b9a\u6548\u73870.90\uff0c

\uff081\uff09\u7535\u52a8\u673a\u7684\u78c1\u6781\u5bf9\u6570P\u4e3a\u591a\u5c11\uff1f---------------------------------\u78c1\u6781\u5bf9\u6570P=2\uff1b
\uff082\uff09\u7535\u52a8\u673a\u63a5\u4e8e380V\u7535\u6e90\u65f6\uff0c\u5b9a\u5b50\u7ed5\u7ec4\u5e94\u91c7\u7528\u4ec0\u4e48\u63a5\u6cd5\uff1f---\u5b9a\u5b50\u7ed5\u7ec4\u63a5\u6cd5\u5e94\u770b\u94ed\u724c\uff0c\u5047\u5982\u4e0a\u9762\u6807\u6709\uff1a380/220V Y/\u0394 \u5219\u5e94\u91c7\u7528Y\u63a5\u6cd5\uff1b
\uff083\uff09\u7535\u52a8\u673a\u989d\u5b9a\u8fd0\u884c\u65f6\u8f93\u5165\u529f\u7387\u4e3a\u591a\u5c11\uff1f-----------------------\u8f93\u5165\u529f\u7387Pr=Pc/\u03b7=10/0.9=11.11KW\u3002

\u4f60\u597d\uff1a
\u2014\u2014\u26051\u3001\u4e09\u76f8\u7535\u52a8\u673a\u989d\u5b9a\u7535\u6d41\u7684\u8ba1\u7b97\u516c\u5f0f\u4e3a\uff1a\u2160\uff1d P \u00f7 \uff081.732 \u00d7 380 \u00d7 Cos \u03c6 \u00d7 \u03b7\uff09\u3002

\u2014\u2014\u26052\u3001\u8fd9\u53f0\u7535\u52a8\u673a\u7684\u989d\u5b9a\u7535\u6d41\u4e3a\uff1a\u2160\uff1d 10 KW \u00f7 \uff081.732 \u00d7 380 \u00d7 0.75 \u00d7 0.86\uff09\uff1d 23.6 A \u3002

首先,我们来定义一些变量:
P_n = 额定功率 = 28kW
U_n = 额定电压 = 380V
f_n = 额定频率 = 50Hz
n_n = 额定转速 = 950r/min
P_Cu + P_Fe = 铜耗及铁耗 = 2.2kW
P_m = 机械损耗 = 1.1kW
P_add = 附加损耗 = 186W

计算结果为:转差率 s = 0,效率 η = -5.760714285714286,定子电流 I_s = 0 A。



电机转矩计算公式:M=9550P/N,M转矩,P电机额定功率,N电机额定转速。根据公式此电机额定转矩M=9550*10/980=97.4N.M。

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