X4?X5?Us10US10 ?SBSNSBSN== 10.75100?10060=0.179
X6?X7? ?= ?0.25100?10060??0.004
图3-3 110KV火电阻抗最简图
X8=X1//X2//X3=0.144
X9=(X4+X6)//(X5+X7)=0.0875
X10=X9+X8=0.232
即火电厂的阻抗为0.232。
2)又根据资料所得,可将变电所视为无限大电源所以取
\E?1 S变110?I*?SB
”I*?”S变110SBE?\650100?16.5?6.5
X变110?I*”?0.154
同理:因35KU变电所的短路容量为250MVA
所以 I”*?S变35SB?250100?2.5KA
X变35?EI\”*?12.5?0.4
火电厂到待设计的变电所距离12KM，阻抗为每千米0.4欧
X= Xl?SBU2?12?0.4?1001152?0.032
110KV变电所到到待设计的变电所距离9KM，阻抗为每千米0.4欧
X= Xl?SBU2 = Xl?SBU2?9?0.4?1001152?0.027
35KV变电所到到待设计的变电所距离7.5KM，阻抗为每千米0.4欧
X= Xl?SBU2?7.5?0.4?100372?0.219
待设计变电所中各绕组等值电抗
1US1% =（US(1-2)% + US(3-1)%?US(2-3)%）21 =（6.5?17?10.5)?6.52
1US2% =（US(1-2)% + US(2-3)%?US(3-1)%）21 =（6.5?10.5?17)?02
1US3% =（US(2-3)% + US(3-1)%?US(1-2)%）2 =12(10.5?17?6.5)?10.5SBSN10.510010020
XT1?Us10 ?= ??0.525
XT2?Us20 ?SBSN= 0100?10020?0
XT3?Us30 ?SBSN= 6.5100?10020?0.325
该变电所的两台型号规格一样所以另一个变压器的阻抗和
XT1，XT2，XT3相同。
根据主接线图可简化为以下图型
图3-4 主接线阻抗简化图
当K1点发生短路时将图四可转化为以下图行
图3-5 K1点短路阻抗图
X13?X1?X3?0.232+0.032=0.264
X14?X2?X4?0.154+0.027=0.181 X15?X5//X6?0.263 X16?X7//X8?0.163 X17?X9//X10?0
X18?X11?X12?=0.219+0.4=0.619
又因为E1是有限大电源(将0.263改为0.264)
3?25所以 Xjs?0.264?0.8?0.248 100\查短路电流周期分量运算曲线取T=0S ,可得I1*?4.324
I\2*?E2\X14?10.181?5.525
I\3*?E3\X15?X17?X18\\\?10.263?0?0.619?1.134
1003?115If?(I1*?I2*?I3*)?IB=(4.324+5.525+1.134)×\ =5.514KA
冲击系数取1.8
Iim?2?If?kim?\\\\2×5.514×1.8=14.034KA
S?(I1*?I2*?I3*)?SB=(4.324+5.525+1.134) ×100=1098.3MV.A
3.2.2 35KV侧短路计算
根据图四进行Y??变换
图3-6 星三角形转化图
图3-7 K2点短路阻抗图
X19?X13?X15?X13?X14?X14?X15X14
=
0.264?0.263?0.264?0.181?0.181?0.2630.181=0.910
X20??X13?X15?X13?X14?X14?X15X130.264X21??X13?X15?X13?X14?X14?X15X150.263 =0.627=0.625
0.264?0.263?0.264?0.181?0.181?0.263
0.264?0.263?0.264?0.181?0.181?0.263