=Kx×Pe6=0.7×13.2=9.24 kw,Qjs6= Pjs6×tgψ=9.24×1.02=9.42 Kwar 6)总的计算负荷计算、干线同期系数取Kx=0.9 总的有用功率 Pjs=Kx×(Pjs1+Pjs2+``````+Pjsn) =0.9×(23.10+31.5+38.85+4.14+16.8+12.6+9.24)=122.6 kw 总的无用功率 Qjs=Kx×(Qjs1+Qjs2+``````+Qjsn) =0.9×(24.95+32.13+39.63+8.2+14.78+13.61+9.42)=128.5Kwar 总的视在功率Sjs=√Pjs2+Qjs2