电力系统潮流短路 计算课程设计(5)

S1 1 =

′SLD14

+

SLD14′2UNB22

R2+X2 jUN

= 4.217+j1.197 + 0.0333 j1.6305 MVA= 4.2503 j0.433 MVA

⑥ 1.3间120Km线路损耗 SLD13

1010

5.651+j5.154 = 2.569+j2.343 MVA =S3=

B12

jUN= 2.569+j0.3102 MVA

SLD13′2UNB12

R1+X1 jUN

SLD13′=SLD13S1 2 =

′SLD13

+

= 2.596+j0.3102 + 0.0139+j0.0266 j2.0328MVA= 2.61 j1.696 MVA

⑦ 2.4间70Km线路损耗 SLD24

1010

10.2405+j7.02 = 6.0238+j4.1294 MVA =S4=

B42

jUN= 6.0238+j2.9436 MVA

SLD24′2UNB42

R4+X4 jUN

SLD24′=SLD24

S2 1 =SLD24′+

= 6.0238+j2.9436 + 0.0546+j0.104 j1.1856MVA= 6.0784+j1.862 MVA

⑧ 2.3间100Km线路损耗 SLD24

1212

5.651+j5.154 = 3.082+j2.811 MVA =S=

4B32

jUN= 3.082+j1.1172 MVA

SLD23′=SLD23