課堂作業(yè)及答案-第四章作業(yè)

1:ZT1:ZT1ZZ1:KKK*:ZZZK*:4-144-4解：題中四種情況的形等效電路分別為Z1k*k*k*12

ykk*Z

kk*kyTy*k*ykk*2k*2Z1k**k y*k*k*ZZ kZ1Z1 k*k*1k

* 4-6如圖4-16所示，已知 ZT=j0.3，求電路的π形等值圖 ZZ1:K4-164-6 k*1

1.1 1j0.30k k*

1k*

1 1

k k*

yT1

kk*

1ZT1ZT1:1.2

2 1S1j19.98U2j10U1U2j10U1U1 S2j10U2U1j19.98U22j10U2U 3S3j10U3U1j10U3U2j19.98U3試用-法求U

、U

，由U

U310U1 U1

31

14-184-9

S1j0.32S31j0.5S322

0 10jU110jU3

1 10jU110jU2 發(fā)電機(jī)輸出功率SG0.8j0.6，負(fù)荷功率SL1j0.8，輸電線路導(dǎo)納y=1-j4。試用潮流計(jì)算 法寫出第一次迭代時(shí)的直角坐標(biāo)修正方程式（電壓迭代初值11y12L4-224-15 2SSS0.8j0.61j0.80.2 YB 11j0 11j0U21

22

11j0e P0 e0 0 f0f0Gf0Be0n ije Bij ij ijjQ0 f0 0 f0e0Gf0Be0n i

ije

Bij ij ij 21 21 21 21P0e0Ge0 f0f0 21 21 21 21 22 22 22 22e0Ge0 f0f0 f 22 22 22 22111401111011 21 21 21 21Q0f0Ge0Bf0e0G 21 21 21 21 eG Bf0 eG B 22 22 22 22110411104144P0PP00.20 Q0QQ00.20 I0 a0i i I0

0a02 2HPiB

G

bH0

ii

ii

B

22

41100NPiG

Bf

ii N0Ge0 f0a011400 22 22 JQiG

Bf

ii

ii J0

i0 f0a011400G 22 22 LQiB

GfBBL0

ii

41100 22 22 H N0 2 J02

J

J01

122

1 1 1

17 1 1 10.2 2

J Q0

17

2

2

0

2 ee222 e0 ee222

0.94122i P Ni Q LePi

J

N22f

2

22 2SL220j10

0,PG2=15U2=1G11XG11XB2B2S4-234-16

jB2

j0.1 1jB

j0.1

1

Y 12l Y 12l Y 22 率方程 PPneG B G

j

ij i i U2U2 f

PeG21e1Bf 21f1B21e121 22

22

B22f2

22f

B22e22其中e1 f10將其代入2 2 22 22 22 22PPe fBeGe 2 22 22 22 22U2U2e2f2