电力系统分析基础部分习题答案

1 电力系统分析部分习题答案(参考) 稳态部分 第一章 电力系统的基本概念 1-2 (1)各个元件的额定电压 G：10.5kV T：10.5/121kV，10.5kV/399V, 110/38.5/11(10.5)kV,35/6.6(6.3)kV (2) a)发电机和变压器的额定电压 G:10.5kV T1:10.5/121kV T2:10.5/38.5kV T3:35/11(10.5)kV b)变压器的实际变比 T-1：121(1+2.5%)/10.5 T-2:110/38.5 T3:35(1-5%)/11(10.5) (3) G:10.5kV T1:10.5/242kV T2:220/121/38.5kV T3:35/6.3kV T4:220/38.5kV T5:10.5/3.15kV D1:3kV D2:6kV 第二章 电力网络元件的参数和等值电路 2-2-1 解：RT=Pk*UN2/(1000SN2)=100*1102/(1000*31.52)=2.4389(Ω) XT=Uk%*UN2/(100SN)=10.5*1102/(100*31.5)=40.33(Ω) GT=P0/(1000UN2)=86/(1000*1102)=7.1074*10-6(s) BT=I0%*SN/(100UN2)=2.7*31.5/(100*1102)=7.0289*10-5(s) 等值电路为： 2-2-3 解：RT(100)=Pkmax*UN2/(2000SN2)=163*1102/(2000*202)=2.4654(Ω) RT(50)= 2RT(100) =2*2.4654=4.9307(Ω) UK1%=1/2*( UK1-2%+ UK1-3%- UK2-3%)=1/2*(10.5+17-6)=10.75 UK2%=1/2*( UK1-2%+ UK2-3%- UK1-3%)=1/2*(10.5+6-17)=-0.25 UK3%=1/2*( UK1-3%+ UK2-3%- UK1-2%)=1/2*(17+6-10.5)=6.25 XT1=Uk1%*UN2/(100SN)=10.75*1102/(100*20)=65.0375(Ω) XT2=Uk2%*UN2/(100SN)=-0.25*1102/(100*20)=-1.5125(Ω) XT3=Uk3%*UN2/(100SN)=6.25*1102/(100*20)=37.8135(Ω) GT=P0/(1000UN2)=75/(1000*1102)=6.1983*10-6(s) BT=I0%*SN/(100UN2)=3.3*20/(100*1102)=5.4545*10-5(s) 2.4389+j40.33Ω (7.1074-j70.289)*10-6S 2 2-2-5 解：Pk13=4P’k13=4*52=208kW Pk23=4P’k23=4*47=188kW Pk1= (Pk12+Pk13- Pk23)/2=(152.8+208-188)/2=86.4kW Pk2= (Pk12+Pk23- Pk13)/2=(152.8+188-208)/2=66.4kW Pk3= (Pk23+Pk13- Pk12)/2=(188+208-152.8)/2=121.6kW RT1=Pk1*UN2/(1000SN2)=86.4*1212/(1000*202)=3.16(Ω) RT2=Pk2*UN2/(1000SN2)=66.4*1212/(1000*202)=2.43(Ω) RT3=Pk3*UN2/(1000SN2)=121.6*1212/(1000*202)=4.45(Ω) UK1%=1/2*( UK1-2%+ UK1-3%- UK2-3%)=1/2*(10.5+18-6.5)=11 UK2%=1/2*( UK1-2%+ UK2-3%- UK1-3%)=1/2*(10.5+6.5-18)=-0.5 UK3%=1/2*( UK1-3%+ UK2-3%- UK1-2%)=1/2*(18+6.5-10.5)=7 XT1=Uk1%*UN2/(100SN)=11*1212/(100*20)=80.53(Ω) XT2=Uk2%*UN2/(100SN)=-0.5*1212/(100*20)=-3.66(Ω) XT3=Uk3%*U...