华北电力大学电力系统分析14年真题及答案汇总

'Y33?y23?y34?b33?b33?(0.32?j3.97)?(0.44?j4.64)?j0.055?j0.11?0.76?j10.45Y34??y34??0.44?j6.64

Y44?y34?b44?(0.44?j6.64)?j0.055?0.44?j6.59 由以上参数得系统节点导纳矩阵得:

00?1.4?j18.66?1.4?j18.66???1.4?j18.661.72?j22.52?0.32?j3.97?0? Ybus???0?0.32?j3.970.76?j10.45?0.44?j6.64???00?0.44?j6.640.44?j6.59??(2)

U11I1k:1U’1I1’ZI2U22

推导此类理想变压器的Π型等值电路:

<1> 列写电压电流方程:

11?1'?I?U?U'??U1?kU1,I1?I1?1k2Z1kZ2?用U1、U2表示I1、I2??(1)k??U'?I'Z?U,I'??I?I??kI??1U?1U2122112?11?kZZ?

<2> 推导Π型等值电路标准形式,如下:

U11I1I10y10I12y12I21I20y20I2U22

根据标准形式的电压电流方程,如下:

?I1?I10?I12?U1?y10?(U1?U2)?y12?I1?(y10?y12)?U1?y12?U2??(2) ?I?I?I?U?y?(U?U)?yI??y?U?(y?y)?U220211212120122?22021?2式(2)与式(1)对应,可得:

?111?k1?1.04?y10?k2Z?kZ?k2Z?1.042?(0.004?j0.0533)??0.05?j0.69??11k?11.04?1y?????0.05?j0.72 ?20ZkZkZ(0.004?j0.0533)??11??1.35?j17.94?y12?kZ1.04?(0.004?j0.0533)?修改节点导纳矩阵:

Y11??y12?y10?(1.35?j17.94)?(?0.05?j0.69)?1.3?j17.25Y12??y12??1.35?j17.94Y22?y12?y20?b22?(1.35?j17.94)?(0.05?j0.72)?j0.11?1.72?j22.52

其余矩阵元素不改变,则新生成的节点导纳矩阵Y如下: ?1.35?j17.9400?1.3?j17.25???1.35?j17.941.72?j22.52?0.32?j3.97?0? Y???0?0.32?j3.970.76?j10.45?0.44?j6.64???00?0.44?j6.640.44?j6.59??

29. (10分)计算系统的三相短路电流的非周期分量及冲击电路。已知断路器CB的断

路容量为25000MVA。

GGTd(3)LCBSS30MWcosφ=0.8xd’’=1.0答:

40MW10.5/121Uk(%)=1180kmx1=0.4Ω/km

将发电机,变压器,线路和系统S的参数标幺化: 取SB=40MVA,UB=Uav

'则xG?x'd?PG/cos?4040?1.0??1.067,xT?0.11??0.11

SB30/0.840xL?x'L?SB40?0.4?80??0.097 22Uav115当CB断路器右侧短路时,xS?111???0.0016 IkSCB/3Uav2500040S/3UBav画出系统等值电路图,如下:

j1.067Gj0.113)I(dkj0.097j0.0016S

3)? 由上图可计算出:I(dk1?11.01

(j1.067?j0.11)//(j0.097?0.0016)3)3)?I(dk? 将标幺值转化为实际值:I(DKSB3Uav?11.01?403?115?2.21(kA)

3)?1.8?2.21?3.98(kA) 取冲击系数KM=1.8,则冲击电流为:iM?KM?I(DK

四、 计算题(80分)

?;如30. (20分)长度为200km的单回输电线路,已知线路末端负荷S2和末端电压U2图所示,线路参数为:r1=0.15 Ω/km,x1=0.45 Ω/km,b1=2.5?10?6S/km。试求

~?和始端功率S。 线路始端电压U11~?U1??210kVU2~S1

答:本题为已知末端电压与末端功率,求始端电压与始端功率,采用从后向前推的方法,计算过程如下:

(1) 画出线路Π型等值电路图:

~S1?120?20MVAU1S1S’1ΔSZR+jXS’2U2S21jB2ΔS1ΔU+δUBj2ΔS22

R?0.15?200?30(?),jX?j0.45?200?j90(?),j(2) 求功率损耗:

B?j0.00025(S)2

~S2?120?j20(MVA)

??U??0?U22

~B?*?2?S2?(j?U11.025(MVA)2)?U2??j0.00025?210??j2

~'~~S2?S2??S2?120?j20?j11.025?120?j8.975(MVA)

~P2'2?Q'221202?8.9752?SZ??(R?jX)?(30?j90)?9.85?j29.55(MVA)2U22102

~'~'~S1?S2??SZ?(120?j8.975)?(9.85?j29.55)?129.85?j38.525(MVA)

此时需要先计算电压降落,并计算出U1,由此来计算出节点1侧导纳的损耗。

(3) 求电压降落:

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