电气自动化毕业论文(2)

X4?X5?X6?X7?10.75100Us1%SB?== =0.179 ?10060100SNUS1%SB?0.25100 ?= ???0.004 100SN10060

图5.3 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

S变110650I???6.5

SB100”*X变110E\1?”??0.154 I*6.5同理:因35KU变电所的短路容量为250MVA

S变35250??2.5KA 所以 I?SB100”*X变35E\1?”??0.4 I*2.5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?SB100?12?0.4??0.032 22U115110KV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?SBSB100 = X??9?0.4??0.027 l222UU11535KV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?SB100?7.5?0.4??0.219 22U37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)?6.521US2% =（US(1-2)% + US(2-3)%?US(3-1)%）2

1 =（6.5?10.5?17)?021US3% =（US(2-3)% + US(3-1)%?US(1-2)%）2

1 =(10.5?17?6.5)?10.52XT1?Us1%SB10.5100= ??0.525 ?10020100SNXT2?Us2%SB0100 ?= ??0 100SN10020

XT3?Us3%SB6.5100 ?= ??0.325 100SN10020 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5.4 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5.5 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)

25所以 Xjs?0.264?0.8?0.248

1003?查短路电流周期分量运算曲线取T=0S ,可得I1\*?4.324

\E1\I2*?2??5.525

X140.181\E31I???1.134

X15?X17?X180.263?0?0.619\3*\\I\f?(I1\*?I2*?I3*)?IB=(4.324+5.525+1.134)×100 =5.514KA

3?115冲击系数取1.8

Iim?2?I\f?kim?2×5.514×1.8=14.034KA

\\S?(I1\*?I2*?I3*)?SB=(4.324+5.525+1.134) ×100=1098.3MV.A

5.2.2 35KV侧短路计算

根据图四进行Y??变换

图5.6 星三角形转化图

图5.7 K2点短路阻抗图

X19?X13?X15?X13?X14?X14?X15

X14=

0.264?0.263?0.264?0.181?0.181?0.263=0.910

0.181X20??X13?X15?X13?X14?X14?X15X130.264?0.263?0.264?0.181?0.181?0.263=0.6250.264X13?X15?X13?X14?X14?X15X21?X15?

0.264?0.263?0.264?0.181?0.181?0.263 =0.6270.263