110kV变电站电气一次部分初步设计 下载本文

附录Ⅰ

短路电流的计算

根据说明书的一些要求选择基准容量:SB=100MVA 110KV级的UBI=UAL=115KV 35KV级的UBI=UAL=37KV 10KV级的UBI=UAL=10.5KV。

基准电抗:

UBI2XB1==132?

SBXB2= UB2=13.7?

SBXB3= UB3=1.1?

SB 各条线路的电抗:(各种线形的单位电抗统一选取0.4?)

XL1=L1?X1=80?0.4=32? XL2=L2?X2=10?0.4?4? XL3=L3?X3=60?0.4?24?

对应的标么值为:

XL1*= XL2*=

XL1=0.24 XB1XL2=0.03 XB1XL3=0.18 XB122XL3*=

变压器的电抗: US12%=10.5% US13%=17.5% U23%=6.5%

XT1*=

1S( US12%+ US13% -U23%) B 2SN1100=(10.5%+17.5%-6.5%) =0.43 225XT2*=

1S( US12%+ US23% -U13%)B 2SN1100=(10.5%+6.5%-17.5%)=-0.01 225

XT3*=

1S( US23%+ US13% -U12%)B 2SN1100=(6.5%+17.5%-10.5%)=0.27 225系统电抗:

100S Xs1*=XS1B=0.5×=0.05

1000SN2Xs2*=XS2

100SB=0.4×=0.07

600SN1

附图1-1 系统图

(1) 则d1点短路等效系统如图:

附图1-2 d1点短路

则网络变换得上图

转移电抗得计算Xa*:是S1对d1的转移电抗,Xb是S2对d2的转移电抗。

Xa**=X*11+X33+

**X11X33X22***=0.0.066+0.012+=0.012+0.166+

0.066?0.012=0.08

0.166Xb=X33+X22+计算电抗:

**X22*X33*X11*0.012?0.166=0.21

0.066Xjs1*=Xa*600Sn1=0.21×=1.26

100SB*Xjs2=Xb

*1000Sn2=0.52×=0.8

100SB查运算负荷曲线得:

1.当tk=0时,Is1* =0.8 Is2* =1.25

sI3uansII3uanI??=Is1*=0.8?+Is2*

6001000+1.26×=8.69KA

115?3115?32.当tk=1s时,Is1*=0.864 Is2*=1.243

Itk =Is1*sI3uan+Is2*sII3uan

=0.864×

6001000+1.243×=8.85KA

115?3115?33. 当tk/2=0.5s时,Is1*=0.78 Is2*=1.16

Itk/2=Is1*sI3uan+Is2*

sII3uan

=0.78×

6001000+1.16×=8.18KA

115?3115?34.当tk/2=∞时,Is1*=0.88 Is2*=1.47

I∞=Is1*sI3uan+Is2*sII3uan

=1.47×

6001000+0.88×=10KA

115?3115?3则冲击电流ish=2.55×8.69=22.16kA (2)则d2点短路等效系统图:

附图1-3 d2点短路

X55*=X11*?X44*?0.066?0.222X11*?X44*=0.066?0.222?=0.38

0.166X22* X66*=X22*?X44*?XJS1*=X55*XJS2*=X66*0.166?0.222X22*?X44*=0.166?0.222?=0.964

0.066X11*600s1N=0.964×=5.78 ?3.5

100sB1000s2N=0.38×=3.8 ?3.5

100sB11=0.173 Is2*= =0.263 5.783.8根据运算负荷曲线可以得到: 1.当tk=0时,Is1*=

I??=Is1*sI3uan+Is2*sII3uan

=0.173×

6001000+0.264×=5.72KA

37?337?3 因两个计算电抗都大于3.5,所以0s 3s 1.5s 及无穷大时曲线重合

则冲击电流ish=2.55×5.72=14.16Ka (3)则d3点短路等效系统图: