10KV系统短路电流整定计算表格

1、10KV电网短路电流计算,假定电网短路容量为300MVA,基准电压Uj=10KV,基准电流Ij=9.16KA,冲击系数Kch.s取1.8. 各元件有效电阻较小,短路电路中总电阻不大于总电抗的1/3,不予考虑.,10KV|-2.0Km 3X300电缆-10KV变配电所SS2 |F1-2.5Km 3X185电缆-F4|SS1 10/0.4KV变压器|F5 b |-SS2 10/0.4KV变压器|F2 |- 0.8Km 3X70电缆-10KV电机|F6 b |-1.5Km 3X70电缆-10KV电机|F3 b |-1.5Km 3X185电缆-F7|SS3 10/0.4KV变压器|F8 b |-1.0

2、Km 3X70电缆-10KV电机|F9,序号,电路元件,变压器/电网容量SrT/Sks MVA,电缆长度Km,变压器Uk%,短路电路总电抗Xjs,超瞬变短路电流KA,短路容量MVA,短路全电流KA,折算到高压侧的短路电流(KA),阻抗/Km,计算公式,短路元件电抗(标值),Is=I0.2=Ik= Ij/Xjs*,Ss= Sj/Xjs*,ich.s= 1.414*Kch.s*I,I1= I2/(1.732*nl)1,电网,300,10,X1*=Sj/Sks,0.333 ,2,进线电缆,YJV-3X300,2,0.079,X2*=L*X(Sj/Uj2),0.143 ,短路点F1,SS2 10KV母

3、线,Xjs1*=X1*+X2*,0.477 ,19.218 ,209.80 ,48.91 ,3,SS2变压器(基准电压10KV),0.63,4.50%,X3*=(Uk%/100)*(Sj/SrT)*(0.42/102),0.0114 , ,短路点F2,SS2变压器低压侧,Xjs2*=X1*+X2*+X3*,0.345 ,26.569 ,290.06 ,67.62 ,0.614 4,SS2-10KV电机,YJV-3X70,0.8,0.071,X4*=L*X(Sj/Uj2),0.052 ,短路点F3,Xjs3*=X1*+X2*+X4*,0.528 ,17.343 ,189.34 ,44.14 ,

4、5,SS2-SS1电缆,YJV-3X185,2.5,0.063,X5*=L*X(Sj/Uj2),0.143 ,短路点F4,SS1 10KV母线,Xjs4*=X1*+X2*+X5*,0.620 ,14.786 ,161.42 ,37.63 ,6,SS1变压器(基准电压10KV),0.25,4.50%,X6*=(Uk%/100)*(Sj/SrT)*(0.42/102),0.0288 ,短路点F5,SS1变压器低压侧,Xjs5*=X1*+X2*+X5*+X6*,0.648 ,14.129 ,154.25 ,35.96 ,0.326 7,SS1-10KV电机,YJV-3X70,0.8,0.071,X7

5、*=L*X(Sj/Uj2),0.052 ,短路点F6,Xjs3*=X1*+X2*+X5*+X7*,0.671 ,13.651 ,149.03 ,34.74 ,8,SS2-SS3电缆,YJV-3X185,1.5,0.063,X8*=L*X(Sj/Uj2),0.086 ,短路点F7,SS3 10KV母线,Xjs3*=X1*+X2*+X8*,0.562 ,16.289 ,177.82 ,41.46 ,9,SS3变压器(基准电压10KV),0.5,4.50%,X9*=(Uk%/100)*(Sj/SrT)*(0.42/102),0.0144 ,短路点F8,SS3变压器低压侧,xjs4*=X1*+X2*+X8*+X9*,0.577 ,15.882 ,173.38 ,40.42 ,0.367 10,SS3-10KV电机,YJV-3X70,0.8,0.071,X10*=L*X(Sj/Uj2),0.052 ,短路点F9,Xjs3*=X1