电力系统及其自动化毕业论文
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
d2
d2
①水电厂 S–1:
X jss–1= X32*( S N∑1/ S j )=0.246 *( 3*200/0.875/100 ) = 1.687 ②水电厂 H–1:
X js H–1= X33*( S N∑1/ S j )= 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
I S-1”= ( I*S-1” * S NS-1)／(√3 * U j )
=0.616*( 3*200/0.875)／(√3 * 230)=1.06KA
I ch S-1= I S-1”*√[1+2*(K ch-1)2]
=1.06*√[1+2*(1.85-1)2]=1.657KA
I H-1”= (I*H-1”* S NH-1)/(√3 * U j )
=0.71*(4*300/0.85)/(√3 * 230 )=2.516KA
I ch H-1= I H-1”*√[1+2 * (K ch-1)2]
=2.516*√[1+2 * (1.85-1)2]=3.934KA
I”= I S-1”+ I H-1”=1.06+2.516=3.576KA
I ch1= I ch S-1+ I ch H-1=1.657+3.934=5.591KA
t=2”时
I*t=2s-1”=0.649
I*t=2H-1”=0.74
I t=2s-1”= (I*t=2s-1”* S NS-1)/ (√3 * U j )
=0.649*(3*200/0.875)/ (√3 * 230)=1.117KA I t=2H-1”=(I*t=2H-1”*S NH-1)/(√3 * U j )
=0.74*(4*300/0.85)/( √3 * 230)=2.623KA
I t=2”= I t=2s-1”+ I t=2H-1”=1.117+2.623=3.74KA
T=4”时
I*t=4s-1”=0.649
I*t=4H-1”=0.74
I t=4s-1”= (I*t=4s-1”* S NS-1)/ (√3 * U j )
=0.649*(3*200/0.875)/ (√3 * 230)=1.117KA I t=4H-1”=(I*t=4H-1”*S NH-1)/(√3 * U j )
=0.74*(4*300/0.85)/( √3 * 230)=2.623KA
I t=4”= I t=4s-1”+ I t=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:
X jss–1= X35*( S N∑1/ S j )=0.187 *( 3*200/0.875/100 ) = 1.282 ②水电厂 H–1:
X js H–1= X36*( S N∑1/ S j )= 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
I S-1”= ( I*S-1” * S NS-1)／(√3 * U j )
=0.810*( 3*200/0.875)／(√3 * 230)=1.394KA
I ch S-1= I S-1”*√[1+2*(K ch-1)2]
=1.394*√[1+2*(1.85-1)2]=12.18KA
I H-1”= (I*H-1”* S NH-1)/(√3 * U j )
=0.94*(4*300/0.85)/(√3 * 230 )=3.331KA
I ch H-1= I H-1”*√[1+2 * (K ch-1)2]
=3.331*√[1+2 * (1.85-1)2]=5.21KA
I”= I S-1”+ I H-1”=1.394+3.331=4.725KA
I ch1= I ch S-1+ I ch H-1=2.81+5.21=7.39KA
t=2”时
I*t=2s-1”=0.888
I*t=2H-1”=1.011
I t=2s-1”= (I*t=2s-1”* S NS-1)/ (√3 * U j )
=0.888*(3*200/0.875)/ (√3 * 230)=1.529KA I t=2H-1”=(I*t=2H-1”*S NH-1)/(√3 * U j )
=1.011*(4*300/0.85)/( √3 * 230)=3.583KA I t=2”= I t=2s-1”+ I t=2H-1”=1.529+3.583=5.112KA
T=4”时
I*t=4s-1”=0.888
I*t=4H-1”=1.011
I t=4s-1”= (I*t=4s-1”* S NS-1)/ (√3 * U j )
=0.888*(3*200/0.875)/ (√3 * 230)=1.529KA I t=4H-1”=(I*t=4H-1”*S NH-1)/(√3 * U j )
=1.011*(4*300/0.85)/(√3 * 230)=3.583KA
I t=4”= I t=4s-1”+ I t=4H-1”=1.529+3.583=5.112KA
电气设备的选择与校验
一、断路器的选择与校验，隔离开关的选择与校验
1、220KV电压等级
断路器
S n=240MVA
最大工作电流：I max =1.05* S n/(√3 * U n )
=1.05*240/(1.732*220)=661A
选SW2-220型断路器
假定主保护动作时间为0.05”，后备保护3.9”。则：断路计算时间：t k=0.05+0.39+0.05=4”
短路点选主变压器与断路器之间，流经该断路器的短路电流为最大。
1）热稳定校验;
I”=6.228KA 、I2.0=6.928KA 、I4.0=6.56KA
Q P=(t K/12)*(I”2+10I tk/22I t k 2)
=(4/12)*(6.2282+10*6.9282+6.562)
=187.26[(KA)2.S]
由于t K＞1秒故不计Q NP
所以：Q K= Q P=187.26[(KA)2.S]
I t2*t=212*4=1764[(KA)2.S]
Q K＜I t2*t热稳定满足。
2）动稳定校验：
i sh=K sh√2 * I”=2.55* I”=2.55*6.228=15.8814KA
i es=59KA
因为i es＞ i sh ，所以动稳定满足。
3） I”=6.228KA＜I Nbr=15.8KA
i sh=15.8814KA＜I Ncl=59KA