有一台发电机中性点为小电流接地系统,发电机额定电压为15.75kV,电压互感器变比为(15000/1.732)/(100/1.732)/(100/3),若在发电机运行电压为15.75kV时在定子线圈某处发生单相直接接地时,若定子100%接地保护继电器动作,且测得的电压为30V,求这时接地点距发电机中性点为多少? (α为几)
正确答案:
动作电压 U=36/3*(15000/1.732)/(100/3)=5400/1.732V
α=(5400/1.732)/(15000/1.732)*100%=36%
α=(5400/1.732)/(15000/1.732)*100%=36%
答案解析:有
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