电力系统短路电流计算

C相短路为例,此时复合序网是正序和负序的并联РIa1=-Ia2=EX1+X2=0.94j(0.2622+0.2622)=-j1.793РUa1=Ua2=X1Ia1=j0.2622*-j1.793=0.47Р故障相的短路电流为РIb=-j3Ia1=-j3*-j1.793=-3.106РIc=j3Ia1=j3*-j1.793=3.106Р其有名值为РIb=Ic=3.106*1003*115=1.559(KA)Р故障点各相对地电压为РUa=Ua1+Ua2=0.47+0.47=0.94РUb=Uc=(a2+a)Ua1=-0.47Р有名值为Ua=0.94*115=108.1(KV) Ub=Uc=0.47*115=54.05(KV)Р Р3当节点5发生两相接地短路Р假设是B、C相接地计算,此时复合序网是正负零序网络的并联,因此РIa1=EX1+X2*X0X2+X0=0.94j(0.2622+0.2622*0.320.2622+0.32)=-j2.313РIa2=X0X2+X0Ia1=0.320.2622+0.32*j2.313=j1.271РIa0=X2X2+X0Ia1=0.26220.2622+0.32*j2.313=j1.042РUa1=Ua2=Ua0=X2*X0X2+X0Ia1=0.333Р因此,故障相的短路电流为РIb=a2Ia1+aIa2+Ua0=2.313∠150+1.271∠210+j1.042= -3.1038 + 1.563i=3.475∠153.27РIc=aIa1+a2Ia2+Ua0=2.313∠30+1.271∠330+j1.042= 3.1038 + 1.563i=3.475∠26.72Р其有名值是РIb=Ic=3.475*1003*115=1.745(KA)Р非故障相对地电压为РUa=3Ua1=0.333*3=1Р化为有名值是Ua=1*115=115(KV)