电力系统及其自动化毕业论文
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东北电力学院毕业设计论文
220kV变电所电气部分一次系统设计
设计计算书
专业:电力系统及其自动化
姓名:
学校:东北电力学院
设计计算书
短路电流计算
1、计算电路图和等值电路图
TS900/296-32QFS300-2
SSP-360/220 SSPSL-240/220
100KM
150KM
I II III III
I
II
230KV
115KV
KV
KV
d1
d2
d3
X1 X4X2X3X7X8X9X10 X5X6X11X12X13X14
X15
X19
X20X16X17X18
X22
X23
d1
d2
d3
230KV
10.5KV
115KV X21X24
系统阻抗标幺值:设:SJ=100MVA
X1=X2=X3=0.2
X4=X5=X6=(Ud/100 )*(S j/S e)=(14.1/100)*(100/240)=0.59
X7=X8=X9=X10=X d*”*(S j/S e)=0.167*(100/300/0.85)=0.0473
X7=X8=X9=X10= ( Ud% / 100 )*(S j/S e)=(14.6/100)*(100/360) =0.0406
X15=X16=X* S j / U p²= 0.4*150*( 100 / 230²) = 0.1134
X17=X18=X* S j / U p²= 0.4*100*( 100 / 230²) = 0.0756
根据主变的选择SFPSLO-240000型变压器,可查出: U dI-II % =14.6、U dI-III % =6.2、U dII-III % =9.84 X 19=X 22=1/200*( U dI-II %+ U dI-III %- U dII-III %)*(S j /S e )
=1/200*(14.6+6.2-9.84)*(100/240)=0.0228
X 20=X 23=1/200*( U dI-II %+ U dII-III %- U dI-III %)*(S j /S e )
=1/200*(14.6+9.84-6.2)*(100/240)=0.0379
X 20=X 23=1/200*( U dI-III %+ U dII-III %- U dI-II %)*(S j /S e )
=1/200*(6.2+9.84-14.6)*(100/240)=0.003
(1)、d 1点短路电流的计算:
d1
X28
X26X27
X25X29
X30
d1
230KV
230KV
X 25=(X 1+X 4)/3=0.0863 X 26=(X 7+X 11)/4=0.02198 X 27=X 15/2=0.0567 X 28=X 17/2=0.0378 X 29=X 25+ X 27=0.143 X 30=X 26+ X 28=0.05978 用个别法求短路电流 ① 水电厂 S –1:
X jss –1= X 29*( S N ∑1/ S j )=0.143 * ( 3*200/0.875/100 ) = 0.98
②水电厂 H–1:
X js H–1= X30*( S N∑1/ S j )=0.0598 *( 4*300/0.85/100 ) = 0.844 查运算曲线:
t=0”时
I*S-1”=1.061
I*H-1”=1.242
I S-1”= ( I*S-1” * S NS-1)/(√3 * U j )
=1.061*( 3*200/0.875)/(√3 * 230)=1.826KA
I ch S-1= I S-1”*√[1+2*(K ch-1)²]
=1.826*√[1+2*(1.85-1)²]=2.855KA
I H-1”= (I*H-1”* S NH-1)/(√3 * U j )
=1.242*(4*300/0.85)/(√3 * 230 )=4.402KA
I ch H-1= I H-1”*√[1+2 * (K ch-1)²]
=4.402*√[1+2 * (1.85-1)²]=6.883KA
I”= I S-1”+ I H-1”=1.826+4.402=6.288KA
I ch1= I ch S-1+ I ch H-1=2.855+6.833=9.738KA
t=2”时
I*t=2s-1”=1.225
I*t=2H-1”=1.36
I t=2s-1”= (I*t=2s-1”* S NS-1)/ (√3 * U j )
=1.225*(3*200/0.875)/ (√3 * 230)=2.109KA
I t=2H-1”=(I*t=2H-1”*S NH-1)/(√3 * U j )
=1.36*(4*300/0.85)/( √3 * 230)=4.8198KA I t=2”= I t=2s-1”+ I t=2H-1”=2.109+4.8198=6.928KA
T=4”时
I*t=4s-1”=1.225
I*t=4H-1”=1.375
I t=4s-1”= (I*t=4s-1”* S NS-1)/ (√3 * U j )
=1.225*(3*200/0.875)/ (√3 * 230)=2.109KA I t=4H-1”=(I*t=4H-1”*S NH-1)/(√3 * U j )
=1.375*(4*300/0.85)/( √3 * 230)=4.873KA
I t=4”= I t=4s-1”+ I t=4H-1”=2.109+4.873=6.982KA
⑵、d2点短路电流的计算:
X31=(X19+X20)/2=0.03035
X32=X29+X31+ X29*X31/ X30
=0.143+0.03035+0.143*0.03035/0.0598=0.246
X33=X30+X31+ X30*X31/ X29
=0.0598+0.03035+0.0598*0.03035/0.143=0.103
用个别法求短路电流