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
专业：2002级专升本电力系统及其自动化
姓名：
学校：东北电力学院
设计计算书
1、 计算电路图和等值电路图
系统阻抗标幺值：设：SJ=100MVA
X1=X2=X3=0.2
X4=X5=X6=(Ud／100 )*(Sj/Se)=(14.1/100)*(100/240)=0.59
X7=X8=X9=X10=Xd*”*(Sj/Se)=0.167*(100/300/0.85)=0.0473
X7=X8=X9=X10= ( Ud% / 100 )*(Sj/Se)=(14.6/100)*(100/360)
=0.0406
X15=X16=X* Sj / Up² = 0.4*150*( 100 / 230² ) = 0.1134
X17=X18=X* Sj / Up² = 0.4*100*( 100 / 230² ) = 0.0756
根据主变的选择SFPSLO-240000型变压器，可查出：
UdI-II% =14.6、UdI-III% =6.2、UdII-III% =9.84
X19=X22=1/200*( UdI-II%+ UdI-III%- UdII-III%)*(Sj/Se)
=1/200*(14.6+6.2-9.84)*(100/240)=0.0228
X20=X23=1/200*( UdI-II%+ UdII-III%- UdI-III%)*(Sj/Se)
=1/200*(14.6+9.84-6.2)*(100/240)=0.0379
X20=X23深圳挤出机用螺杆组合=1/200*( UdI-III%+ UdII-III%- UdI-II%)*(Sj/Se)
=1/200*(6.2+9.84-14.6)*(100/240)=0.003
(1)、d1点短路电流的计算：
X25=（X1+X4）/3=0.0863
X26=（X7+X11）/4=0.02198
X27=X15/2=0.0567
X28=X17/2=0.0378
X29=X25+ X27=0.143
X30=X26+ X28=0.05978
用个别法求短路电流
1 水电厂 S–1:
Xjss–1= X29*( SN∑1/ Sj )=0.143 * ( 3*200/0.875/100 ) = 0.98极压高温润滑脂
2 水电厂 H–1:
XjsH–1= X30*( SN∑1/ Sj )=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
IS-1”= ( I*S-1” * SNS-1)／(√3 * Uj )
=1.061*( 3*200/0.875)／(√3 * 230 )=1.826KA
IchS-1= IS-1”*√[1+2*(Kch-1) ²]
=1.826*√[1+2*(1.85-1) ²]=2.855KA
IH-1”= (I*H-1”* SNH-1)/(√3 * Uj )
=1.242*(4*300/0.85)/(√3 * 230 )=4.402KA
IchH-1= IH-1”*√[1+2 * (Kch-1) ²]
=4.402*√[1+2 * (1.85-1) ²]=6.883KA
I”= IS-1”+ IH-1”=1.826+4.402=6.288KA
Ich1= IchS-1+ IchH-1=2.855+6.833=9.738KA
t=2”时
I*t=2s-1”=1.225
I*t=2H-1”=1.36
It=2s-1”= (I*t=2s-1”* SNS-1)/ (√3 * Uj )
=1.225*(3*200/0.875)/ (√3 * 230 )=2.109KA
It=2H-1”=(I*t=2H-1”*SNH-1)/(√3 * Uj )
=1.36*(4*300/0.85)/( √3 * 230 )=4.8198KA
It=2”= It=2s-1”+ It=2H-1”=2.109+4.8198=6.928KA
T=4”时
I*t=4s-1”=1.225
I*t=4H-1”=1.375
It=4s-1”= (I*t=4s-1”* SNS-1)/ (√3 * Uj )
=1.225*(3*200/0.875)/ (√3 * 230 )=2.109KA
It=4H-1”=(I*t=4H-1”*SNH-1)/(√3 * Uj )
=1.375*(4*300/0.85)/( √3 * 230)=4.873KA
It=4”= It=4s-1”+ It=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
①水电厂 S–1:
Xjss–1= X32*( SN∑1/ Sj )=0.246 *( 3*200/0.875/100 ) = 1.687
②水电厂 H–1:
XjsH–1= X33*( SN∑1/ Sj )= 0.103*( 4*300/0.85/100 ) = 1.454
查运算曲线 ：
t=0”时
I*S-1”=0.616
I*H-1”=0.71
IS-1”= ( I*S-1” * SNS-1)／(√3 * Uj )
=0.616*( 3*200/0.875)／(√3 * 230 )=1.06KA
IchS-1= IS-1”*√[1+2*(Kch-1) ²]
=1.06*√[1+2*(1.85-1) ²]=1.657KA
IH-1”= (I*H-1”* SNH-1)/(√3 * Uj )
=0.71*(4*300/0.85)/(√3 * 230 )=2.516KA
IchH-1= IH-1”*√[1+2 * (Kch-1) ²]
=2.516*√[1+2 * (1.85-1) ²]=3.934KA
I”= IS-1”+ IH-1”=1.06+2.516=3.576KA
Ich1= IchS-1+ IchH-1=1.657+3.934=5.591KA
t=2”时
I*t=2s-1”=0.649
I*t=2H-1”=0.74
It=2s-1”= (I*t=2s-1”* SNS-1)/ (√3 * Uj )
=0.649*(3*200/0.875)/ (√3 * 230 )=1.117KA
It=2H-1”=(I*t=2H-1”*SNH-1)/(√3 * Uj )
=0.74*(4*300/0.85)/( √3 * 230 )=2.623KA
It=2”= It=2s-1”+ It=2H-1”=1.117+2.623=3.74KA
T=4”时
I*t=4s-1”=0.649
I*t=4H-1”=0.74
It=4s-1”= (I*t=4s-1”* SNS-1)/ (√3 * Uj )
=0.649*(3*200/0.875)/ (√3 * 230 )=1.117KA
vagooIt=4H-1”=(I*t=4H-1”*S乙基氯化物NH-1)/(√3 * Uj )
=0.74*(4*300/0.85)/( √3 * 230)=2.623KA
It=4”= I阻燃双面胶t=4s-1”+ It=4H-1”=1.117+2.623=3.74KA
⑶、d3点短路电流的计算：
X34=（X19+X21）/2=0.0129
X35=X29+X34+ X29*X34/ X30
=0.143+0.0129+0.143*0.0129/0.0598=0.187
X36=X30+X34+ X30*X34/ X29
=0.0598+0.0129+0.0598*0.0129/0.143=0.078
用个别法求短路电流
①水电厂 S–1:
Xjss–1= X35*( SN∑1/ Sj )=0.187 *( 3*200/0.875/100 ) = 1.282
②水电厂 H–1:
XjsH–1= X36*( SN∑1/ Sj )= 0.078*( 4*300/0.85/100 ) = 1.101
查运算曲线 ：
t=0”时
I*S-1”=0.810
I*H-1”=0.94
IS-1”= ( I*S-1” * SNS-1)／(√3 * Uj )
=0.810*( 3*200/0.875)／(√3 * 230 )=1.394KA
IchS-1= IS-1”视频门禁系统*√[1+2*(Kch-1) ²]
=1.394*√[1+2*(1.85-1) ²]=12.18KA
IH-1”= (I*H-1”* SNH-1)/(√3 * Uj )
=0.94*(4*300/0.85)/(√3 * 230 )=3.331KA
IchH-1= IH-1”*√[1+2 * (Kch-1) ²]
=3.331*√[1+2 * (1.85-1) ²]=5.21KA
I”= IS-1”+ IH-1”=1.394+3.331=4.725KA
Ich1= IchS-1+ IchH-1=2.81+5.21=7.39KA
t=2”时
I*t=2s-1”=0.888
I*t=2H-1”=1.011
It=2s-1”= (I*t=2s-1”* SNS-1)/ (√3 * Uj )
=0.888*(3*200/0.875)/ (√3 * 230 )=1.529KA
It=2H-1”=(I*t=2H-1”*SNH-1)/(√3 * Uj )
=1.011*(4*300/0.85)/( √3 * 230 )=3.583KA
It=2”= It=2s-1”+ It=2H-1”=1.529+3.583=5.112KA
T=4”时
I*t=4s-1”=0.888
I*t=4H-1”=1.011
It=4s-1”= (I*t=4s-1”* SNS-1)/ (√3 * Uj )
=0.888*(3*200/0.875)/ (√3 * 230 )=1.529KA
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