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《电力系统分析》综合练习题答案


《电力系统分析》综合练习题答案 电力系统分析》
1. 解: G1 带负荷,电压升高 5%至 6.3KV; G2、G3 为发电机-变压器组,电压可直接取为 10.5KV 和 13.8KV; T1 二次侧电压升高 10%,变比为 6.3/38.5; T2 高压侧电压升高 10%,变比为 10.5/38.5/121; T3 高压侧电压升高 10%,变比为 13.8/121/242; T4 高压侧电压假设已自 T3 下降了 5%,压、中压侧电压升高 10%,变比为 38.5/121/231 T5 高压侧电压假设已自 T4 再下降 5%, 低压、 中压侧电压升高 10%, 变比为 38.5/121/220 T6 假设高压侧已自 T4 下降了 10%,低压侧升高 10%,变比为 110/11; T7 假设高压侧已自 T5 下降了 10%,低压侧升高 10%,变比为 110/11。 注:电压降落的幅度还有其它合理的假设 2.解: +

U 1 = (111.847 ? 0.031) 2 + 0.256 2 = 119.816 KV

(2800 × 5 + 3000 × 3 + 3300 × 4 + 3100 × 5 + 3200 × 4 + 2900 × 3) 3600
=7738 (h)

Pav =

Tmax 7738 × Pmax = × 3600 = 3180 MW 8760 8760

W = Tmax × Pmax = 7738 × 3160 = 27.858GWh
3. (图形暂缺) 4.解: 导线: r1 =

ρ
s

=

31.5 = 0.2625? / km 120

rt = r1 × [1 + α (t ? 20)] = 0.2625 × [1 + 0.0036 × (40 ? 20)] = 0.2814? / km Dm = 3 Dab Dbc Dca = 3 4 × 4 × 4 = 4m x1 = 0.1445 lg
注:这里假设导线间距 4 米

Dm + 0.1557 = 0.5495? / km r

注: 这里假设导线为 LGJ-120/20

b1 =

7.58 × 10 ?6 = 2.7817 × 10 ?6 S / km Dm lg r
2 Pk U N 104 × 110 2 = = 3.416? 2 1000 S N 1000 × 20 2

变压器: RT =

1

XT =

2 U k %U N 10.5 × 110 2 = = 63.525? 100 S N 100 × 20

GT =

p0 30 = = 2.479 × 10 ?6 S 2 2 1000U N 1000 × 110 I 0 %S N 1.2 × 20 = = 1.983 × 10 ?5 S 2 2 100U N 100 × 110
14.07+j27.48Ω 1.71+j31.76Ω

BT =

2.78 × 10 ?6 S

2.479 + j19.8 × 10 ?6 S

5. 暂略

6. 解:

Z=

U2 110 2 = = 387.2 ? j 290.4 S 20 + j15
1 1 = = (1.65 + j1.24) × 10 ?3 S Z 387.2 ? j 290.4

Y=

7. 解:

R = r1l = 0.08 × 250 = 20? X = x1l = 0.416 × 250 = 104?
B = b1l = 2.8 × 10 ?6 × 250 = 7 × 10 ?4 S
1) P = 150 MW

Q = 112.5MVar

U 2 = 210 KV

?S y 2 = ?

1 1 jBU 2 = ? × 7 × 10 ? 4 × 210 2 = ? j15.435MVar 2 2

' S 2 = S 2 + ?S y 2 = 150 + j112.5 ? j15.435 = 150 + j 97.065MVA ' S2 2 150 2 + 97.065 2 ) Z= (20 + j104) = 14.477 + j 75.280 MVA U2 210 2

?S Z = (

' S1' = S 2 + ?S Z = 150 + j 97.065 + 14.477 + j 75.280 = 164.477 + j172.345MVA

2

?U =

' P2' R + Q2 X 150 × 20 + 97.065 × 104 = = 62.365 KV U2 210 ' P2' X ? Q2 R 150 × 104 ? 97.065 × 20 = = 65.041KV U2 210

δU =

U 1 = (U 2 + ?U ) 2 + δU 2 = 280.023KV

1 1 ?S y1 = ? BU 12 = ? 7 × 10 ? 4 × 280.023 2 = ? j 27.445MVar 2 2

S1 = S1' + ?S y1 = 164.477 + 144.900MVA
2) P = 150 MW

Q=0

U 2 = 210 KV

?S y 2 = ?

1 1 jBU 2 = ? × 7 × 10 ? 4 × 210 2 = ? j15.435MVar 2 2

' S 2 = S 2 + ?S y 2 = 150 ? j15.435MVA ' S2 2 150 2 + 15.435 2 ) Z= (20 + j104) = 10.312 + j 53.623MVA U2 210 2

?S Z = (

' S1' = S 2 + ?S Z = 150 ? j15.435 + 10.312 + j 53.623 = 160.312 + j 38.188MVA ' P2' R + Q2 X 150 × 20 ? 15.435 × 104 = = 7.139 KV U2 210

?U =

' P2' X ? Q2 R 150 × 104 + 15.435 × 20 δU = = = 75.756 KV U2 210

U 1 = (U 2 + ?U ) 2 + δU 2 = 229.975 KV 1 1 ?S y1 = ? BU 12 = ? 7 × 10 ? 4 × 229.975 2 = ? j18.511MVar 2 2

S1 = S1' + ?S y1 = 160.312 + 19.677 MVA
8. 解: 1) RT =
2 Pk U N 27.6 × 20.2 2 = = 0.919? 2 1000 S N 1000 × 3.5 2 2 U k %U N 6.42 × 20.2 2 = = 7.485? 100 S N 100 × 3.5

XT =

3

GT =

p0 11.05 = = 2.708 × 10 ?5 S 2 1000U N 1000 × 20.2 2 I 0 %S N 2 × 3.5 = = 1.716 × 10 ?4 S 2 2 100U N 100 × 20.2

BT =

第二、三台的计算类似 2) RT = 0.963? 3) RT = 0.979? 9. 解:

X T = 7.625? X T = 7.636?

GT = 2.059 × 10 ?5 GT = 2.598 × 10 ?5

BT = 1.218 × 10 ?4 BT = 1.578 × 10 ?4

P12 = 333KW
Pk 1 =

P23 = 4 × 277 = 1108 KW

P13 = 4 × 265 = 1060 KW

1 ( P12 + P13 ? P23 ) = 142.5 KW 2 1 Pk 2 = ( P12 + P23 ? P13 ) = 190.5 KW 2 1 Pk 3 = ( P13 + P23 ? P12 ) = 917.5 KW 2 RT 1 =
2 Pk 1U N 142.5 × 220 2 = = 0.851? 2 1000 S N 1000 × 90 2

RT 2

2 Pk 2U N 190.5 × 220 2 = = = 1.138? 2 1000 S N 1000 × 90 2 2 Pk 3U N 917.5 × 220 2 = = 5.482? 2 1000 S N 1000 × 90 2

RT 3 =

U k 1? 2 % = 9.09 U k1 % =

U k 1?3 = 2 × 16.45 = 32.9

U k 2?3 = 2 × 10.75 = 21.5

1 (U k 1? 2 + U k 1?3 + U k 2?3 ) = 10.245 2 1 U k 2 % = (U k1? 2 + U k 2?3 + U k 1?3 ) = ?1.155 2 1 U k 3 % = (U k 1?3 + U k 2?3 + U k 1? 2 ) = 22.655 2 X T1 =
2 U k 1 %U N 10.245 × 220 2 = = 50.095? 100 S N 100 × 90 2 U k 2 %U N ? 1.155 × 220 2 = = ?6.211? 100 S N 100 × 90

XT2 =

4

XT3 =

2 U k 3 %U N 1 22.655 × 220 2 = = 121.834? 100 S N 100 × 90

GT =

p0 59 = = 1.219 × 10 ?6 S 2 2 1000U N 1000 × 220 I 0 %S N 0.332 × 90 = = 6.174 × 10 ?6 S 2 2 100U N 100 × 220

BT =

10.解: 1) X T 1 =
2 U k %U N 2 10.5 × 110 2 6 .3 2 k1 = ×( ) = 0.275? 100 S N 100 × 16 121

6 .3 2 X L1 = x1lk12 = 100 × 0.4 × ( ) = 0.122? 121 XT2
2 U k %U N 2 2 10.5 × 110 2 6.3 2 110 2 = k1 k 2 = ×( ) ×( ) = 10.934? 100 S N 100 × 31.5 121 11

6.3 2 110 2 2 X L 2 = x1lk12 k 2 = 80 × 0.4 × ( ) ×( ) = 8.675? 121 11
2) X T 1 =
2 U k %U N 2 10.5 × 110 2 11 2 k2 = ×( ) = 0.794? 100 S N 100 × 16 110

11 2 2 X L1 = x1lk 2 = 100 × 0.4 × ( ) = 0.40? 110 XT2
2 U k %U N 2 10.5 × 110 2 11 2 = k2 = ×( ) = 0.403? 100 S N 100 × 31.5 110

2 X L 2 = x1lk 2 = 80 × 0.4 = 32? 2 U k %U N 10.5 × 110 2 = = 79.406? 100 S N 100 × 16

3) X T 1 =

X L1 = x1l = 100 × 0.4 = 40?
XT2
2 U k %U N 10.5 × 110 2 = = = 40.333? 100 S N 100 × 31.5

110 2 2 X L 2 = x1lk 2 = 80 × 0.4 × ( ) = 3 .2 K ? 11
11. 解:

U B1 = 110 KV

U B 2 = 11KV
5

X T 1* = X T 1

SB 100 = 79.406 × = 0.656 2 UB 110 2 SB 100 = 40 × = 0.331 2 UB 110 2 SB 100 = 40.333 × = 0.333 2 UB 110 2 SB 100 = 32 × 2 = 26.446 2 UB 11

X L1* = X L1

X T 2* = X T 2

X L 2* = X L 2

12.解:

U B1 = 35KV R AB* = R AC * = r1l

U B 2 = 35 ×

6 .6 = 6.6 KV 35

SB 100 = 0.45 × 20 × 2 = 0.735 2 UB 35 SB 100 = 0.382 × 20 × 2 = 0.624 2 UB 35

X AB* = X AC * = x1l

B AB* = B AC * = b1l

2 UB 35 2 = 2.99 × 10 ?6 × 20 × = 7.3 × 10 ?4 SB 100

RDE* = r1l

SB 100 = 0.52 × 5 × = 5.969 2 UB 6.6 2 SB 100 = 0.079 × 5 × = 0.907 2 UB 6.6 2

X DE* = x1l

BDE * = b1l

2 UB 6 .6 2 = 1.04 × 10 ?6 × 5 × = 2.265 × 10 ?6 SB 100

RT *

2 Pk U N S B 70 × 35 2 100 = = × = 0.07 2 2 1000 S N U B 1000 × 10 2 35 2 2 U k %U N S B 7.5 × 35 2 100 = × = 0.75? 2 100 S N U B 100 × 10 35 2 2 p0 U B 12 35 2 = × = 1.2 × 10 ?4 S 2 S B 1000 × 35 2 100 1000U N

X T* =

GT * =

6

BT * =

2 I 0 %S N U B 1.5 × 10 35 2 = × = 1.5 × 10 ?3 S 2 S B 100 × 35 2 100 100U N

13.解: 设末端无功功率为 Q2

1 S y 2 = ? j BU 2 = ? j × 5.5 × 10 ? 4 × 220 2 = ? j 26.62 MVar 2
?U = 220 × 8 + (Q2 ? 26.62) × 40 220

δU =

220 × 40 ? (Q2 ? 26.62) × 8 220

将上两式带入 (220 + ?U ) 2 + δU 2 = 2312 可以解出 Q2 = 30.71MVar 所以功率因数 cosθ =

P2
2 P22 + Q2

=

220 220 2 + 30.712

= 0.9904

14.解:

R1 = 33? R2 = 7.9?

X 1 = 42.9? X 2 = 39.3?

B1 = 2.65 × 10 ?4 S B2 = 2.91 × 10 ?4 S

假设两条线路上流过的功率分别为 P1 + jQ1 和 P2 + jQ2 ,有如下等式成立:

P1 R1 + Q1 X 1 P R + Q2 X 2 ? = ?U 2 = 2 2 ??U 1 = U2 U2 ? ? δU = P1 X 1 ? Q1 R1 = δU = P2 X 2 ? Q2 R2 2 ? 1 U2 U2 ? P1 + P2 = P ? ? Q1 + Q2 = Q ?
1) P = 100 MW , Q = 0 时,解得

P1 = 42.1556MW P2 = 57.8444MW Q1 = ?11.3645MVar Q2 = 11.3645MVar
?S 1 = P12 + Q12 ( R1 + jX 1 ) = 5.1989 + j 6.7585MVA 2 U2
2 P22 + Q2 ( R2 + jX 2 ) = 2.2689 + j11.2870 MVA 2 U2

?S 2 =

7

2) P = 0 , Q = 53MVar 时,解得

P1 = 6.0232MW P2 = ?6.0232MW Q1 = 22.3425MVar Q2 = 30.6975MVar
P12 + Q12 ?S1 = ( R1 + jX 1 ) = 1.4604 + j1.8985MVA 2 U2 ?S 2 =
2 P22 + Q2 ( R2 + jX 2 ) = 0.6389 + j 3.1785MVA 2 U2

15.解: 本题为给定末端功率和始端电压的情况,采用如下计算方法:先假设全网为额定电压, 由末端向始端推算功率损耗,求得始端功率后,再运用给定的始端电压向末端推算电压 降落,并且不再重新计算功率损耗。

?S 23

P32 + Q32 = ( R23 + jX 23 ) = 0.0228 + j 0.0331MVA 2 UN
2 P42 + Q4 ( R24 + jX 24 ) = 0.0173 + j 0.0251MVA 2 UN

?S 24 =

P2 3 + Q2 3 ?S12 = ∑ 2 ∑ ( R23 + jX 23 ) = 0.0747 + j 0.1082 MVA UN

S1 = ∑ S i + ∑ S ij = 21.1148 + j15.1664MVA
以下回算电压降落:

?U 12 =

P 1 R12 + Q1 X 12 = 0.5287 KV U1
2

δU 12 =

P1 X 12 ? Q1 R12 = 0.1891KV U1

U 2 = (U 1 ? ?U 12 ) 2 + δU 12 = 109.4713KV ?U 23 = P23 R23 + Q23 X 23 = 0.4588KV U2
2

δU 23 =

P23 X 23 ? Q23 R23 = 0.0707 KV U2

U 3 = (U 2 ? ?U 23 ) 2 + δU 23 = 109.0125 KV

?U 24 =

P24 R24 + Q24 X 24 = 0.2929 KV U2
2

δU 24 =

P24 X 24 ? Q24 R24 = 0.1782 KV U2

U 4 = (U 2 ? ?U 24 ) 2 + δU 24 = 109.1785KV
16.解:

8

在节点 1 注入单位电流,计算各点电压 I 1 0.2 0.1 3 0.15 0.75 2 0.05

则 Z 11 = U 1 = 0.75 //[

0.1 × (0.2 + 0.15) + 0.05] = 0.1092 0.1 + (0.2 + 0.15)

其中//表示并联运算

进而求得 Z 21 = U 2 = 0.0427 , Z 31 = U 3 = 0.0848 同样的,可以求得 Z 22 = 0.0472 , Z 32 = 0.0453 , Z 33 = 0.1421

?0.1092 0.0427 0.0848? ? ? 阻抗矩阵为 0.0427 0.0472 0.0453 ? ? ?0.0848 0.0453 0.1421? ? ?
17.解:

j j2 0 ? ?? j 5 ? j ? j5 j 0 ? ? 1)导纳矩阵为: ? ? j2 j ? j8 j 2 ? ? ? 0 j 2 ? j 2? ? 0 j 0 0 ? ?? j 5 ? j ? j5 j 0 ? ? ? 2)1-3 支路断开后,矩阵变为: ? 0 j ? j6 j2 ? ? ? 0 j 2 ? j 2? ? 0
3)变压器变比改为 1:1.1,则 ?Y11 = (

1 1 ? 2 )(? j ) = j 0.174 , 2 1 .1 1 1 1 ?Y12 = ?Y21 = ?( ? )(? j ) = ? j 0.091 1 .1 1 0 ? j 0 ? ? ? j6 j2 ? ? j 2 ? j 2? 0

?? j 4.826 ? j 0.909 矩阵变为: ? ? 0 ? 0 ?
18.解:

j 0.909 ? j5 j 0

9

1 ? j j ?? j 2 2 ? 1 11 ? j ?j j 6 ? 2 17 Y =? j j ?j ? 6 ? 1 j ? 2 ? 1 ? j ? 3 ?
?1 ? ? 1.3333 ? ?1 ? 1 .5 ? = ?? 1.6667 ? 1.25 ? ?? 1.6667 ? 1.25 ?? 1.6667 ? 1.25 ?

1 2 1 ?j 2 j

? ? ? ? ? 1 ? j 3 ? ? ? ? 1? ?j 3? ?

? 1.6667 ? 1.6667

Z = Y ?1

? 1.6667 ? ? 1.25 ? 1.25 ? 1.25 ? ? ? 1.7083 ? 1.7083 ? 1.7083 ? ? ? 1.7083 ? 3.7083 ? 1.7083 ? ? 1.7083 ? 1.7083 ? 4.7083? ?

19.解:

S 2 100 ? U 2 U 2 ? U 3 ? ? ?I 2 = U = ? 10 10 2 由? S U ?U ? I3 = 3 = 2 3 ? U3 10 ?
得到

1 ? 2 ?S 2 = 10 (100U 2 ? 2U 2 + U 2U 3 ) ? 1 2 ? S 3 = (U 2 ? U 2U 3 ) 10 ?
设 U 2 、 U 3 的初值为 100V

1 ? ?S 2 ? ?U = 10 (100 ? 4U 2 + U 3 ) = ?20 ? 2 ?S 2 1 ? = U 2 = 10 ? ?U 3 10 ? ?S 3 1 ? = (2U 2 ? U 3 ) = 10 ?U 2 10 ? ? ?S 3 1 = ? U 2 = ?10 ? ?U 3 10 ?
又 ?S 2 = 20W , ?S 3 = 40W 所以迭代方程为

?20? ?? 20 10 ? ??U 2 ? ?40? = ? 10 ? 10? × ? ?U ? ? ? ? ? ? 3?
10

解得 ?U 2 = ?6V , ?U 3 = ?10V 所以迭代一次, U 2= 94V , U 3 = 90V 20.解: 变压器导纳为:

1 = ? j 2 .5 j 0 .4 ? 1 ? j 6 .5 ? 1 + j 6 .5 ? ? ? ? 1 + j 6 .5 1 ? j 6 .5 ?

先假设变压器变比为 1:1,导纳矩阵为: ? 变比改为 1:1.1 后, ?Y11 = (

1 1 ? 2 ) × (? j 2.5) = j 0.434 2 1 .1 1 1 1 ?Y12 = ?Y21 = ?( ? ) × (? j 2.5) = ? j 0.227 1 .1 1

导纳矩阵修改为: ?

? 1 ? j 6.066 ? 1 + j 6.273? 1 ? j 6 .5 ? ?? 1 + j 6.273 ?

?P1 = 2 ? (1 ? 1) = 2
?Q = 0.8 ? (?6.273 + 5.5) = 0.573

& I = 1 ? j 0.65 + (?1 + j 6.273) = ? j 0.227
计算雅可比矩阵元素:

H 11 = 6.5 ? 0.227 = 6.273 , N 11 = 1 , J 11 = ?1 , L11 = 6.727
列方程如下: ?

1 ? ??f ? ? 2 ? ?6.273 ? = ? ? 1 6.727 ? × ? ?e ? ?0.573? ? ? ? ?

解得: ?f = 0.2982 , ?e = 0.1295 所以,迭代一次,受端电压为 1.1295+j0.2982 21.解:

j10 j10 ? ?? j19.98 ? j10 导纳矩阵为: ? j19.98 1 ? j10 ? ? ? ? j10 j10 ? j19.98? ? ?
1)牛顿-拉夫逊法 设电压初值为 U 1 = 1 + j 0 , U 2 = 1.05 + j 0

11

?P1 = 2.8653 , ?Q1 = 1.2244 + 0.02 = 1.2444
2 ?P2 = 0.6661 , ?U 2 = 0

& I 1 = ? j19.98 + j10 × 1.05 + j10 = j 0.52 & I 2 = ? j19.98 × 1.05 + j10 + j10 = ? j 0.979
计算雅可比矩阵元素:

H 11 = 19.98 + 0.52 = 20.5 , N 11 = 0 , J 11 = 0 , L11 = 19.98 ? 0.52 = 19.46 H 12 = ?10 , N 12 = 0 , J 12 = 0 , L12 = ?10 H 21 = ?10.5 , N 21 = 0 , J 21 = 0 , L21 = ?10.5 H 22 = 19.98 × 1.05 - 0.979 = 20 , N 22 = 0 , J 22 = 0 , L22 = 21.958
列方程如下:

0 ? 10 0 ? ? ?f1 ? ?2.8653? ? 20.5 ?1.2444 ? ? 0 19.46 0 ? 10 ? ? ?e1 ? ? ?=? ?×? ? ? 0.6661? ?? 10.5 0 20 0 ? ? ?f 2 ? ? ? ? ? ? ? ? 10.5 0 21.958? ??e2 ? ? 0 ? ? 0
解得: ?f 1 = 0.2097 , ?e1 = 0.0848 , ?f 2 = 0.1434 , ?e2 = 0.0405 一次迭代后, f 1 = 0.2097 , e1 = 1.0848 , f 2 = 0.1434 , e2 = 1.0405 2)P-Q 解耦法

?? j19.98 B′ = ? ? j10

j10 ? , B ′′ = ? j19.98 j ? 19.98? ?

?P1 = 2.8653 , ?P2 = 0.6661
列方程如下: ?0.6661

? 2.8653 ? 10 ? ? ?δ 1 ? ?? 19.98 × ? = ?? ? 19.98? ?1.05?δ 2 ? ? ? 10 ? ? ? 1.05? ? ?

1.2444 = 19.98?U 1 1.05
解得: ?δ 1 = 0.2125 , ?δ 2 = 0.1316 , ?U 1 = 0.0593 一次迭代后, U 1 = 1.0593 , δ 1 = 0.2125 , δ 2 = 0.1316 后续迭代的计算类似,略去。

12

22.解: 取 S B = 100 MVA , U B1 = 115 KV , U B 2 = 6.3KV 有 I B1 =

SB 3U B1

= 0.502 KA , I B 2 =

SB 3U B 2

= 9.164 KA

X 1* = 40 × 0.4 × X 2* X 3* X 4*

100 = 0.121 115 2 10.5 100 = × = 0.35 100 30 40 9.164 = × = 12.219 100 0 .3 100 = 0 .8 × 0 .5 × = 1.008 6 .3 2

1)K1 处短路

I m* =

1 1 = = 2.13 X 1* + X 2* 0.121 + 0.35
注:这里取 Km=1.85

im = K m I m = 1.85 × 2.13 × 9.164 = 43.7 KA IM = Im 2 1 + 2( Km ? 1) 2 = 21.6 KA

2)K2 处短路

I m* =

1 1 = = 0.073 X 1* + X 2* + X 3* + X 4* 0.121 + 0.35 + 12.219 + 1.008

im = K m I m = 1.85 × 0.073 × 9.164 = 1.238 KA IM = Im 2 1 + 2( Km ? 1) 2 = 1.16 KA

23.解: 令 U = 1∠0°

& & & E q = U + jIx q = 1.39 + j 0.52 = 1.484∠20.51°
I d = sin(20.51 + 36.87) = 0.842

E q = 1.484 + 0.842 × (1 ? 0.65) = 1.779

& 空载电势 E q = 1.779∠20.51°
13

24.解: 本题需要逐步化简网络,最后得到 x* = 0.182 , I * = 5.484 而 I B1 =

SB 3U B1

=

100 3 × 110

= 0.525 KA

所以短路电流为 I = I * I B = 2.878 KA 在发电机处, I B 2 =

SB 3U B 2

=

100 3 × 10.5

= 5.499 KA

可以得到 I G1 = 22.781KA , I G 2 = 7.374 KA 25.解: 设 S B = 250 MVA , U B = 110 KV ,则 I B = 1.312 KA 有 X d * = 0.12 ×

X L1*

250 250 = 0.12 , X T * = 0.105 × = 0.109 250 240 250 250 = 0.4 × 20 × = 0.165 , X L 2* = 0.4 × 10 × = 0.083 2 110 110 2

化简得到 X * = 0.1787 ,则 I * = 5.596 , I = I * I B = 7.343KA 26.解: 1)图形如下: X2 X1 X3 X5 X6

XG

2)图形如下:

X X3 X1 X6 X5 X4

14

3)图形如下:

X2 X3 X6

X5 X4

4)图形如下: XG X2 X1 X3 X6 X5 X4

27.解: X10 X12 X3 X2 X7 X9 X8 X13 X14 X15

X ∑ 0 = [ X 15 +

( X 14 + X 10 + X 12 ) X 13 (X + X8)X9 ] //[ 7 + X2 + X3] X 14 + X 10 + X 12 + X 13 X7 + X8 + X9

其中“//”表示并联。 28.解: 取 SB=60MVA,UB=110KV,则 I B =

SB 3U B

= 315 A

60 60 = 0.16 , X G ?* = 0.19 × = 0.19 60 60 60 60 X L* = 16 × 2 = 9.6 , X L 0* = 60 × 2 = 36 10 10 60 60 X T * = 0.105 × = 0.105 , X g * = 30 × = 0.147 60 110 2 X G +* = 0.16 ×
可以得到各序阻抗如下: X+*=9.865,X-*=9.895,X0*=0.105(无接地阻抗)或 0.551(有接地阻抗)

15

a) 中性点直接接地 单相接地短路: I f (1)* =

1 = 0.0503 9.865 + 9.895 + 0.105

I (f1*) = 3I f (1) = 0.151 I (f1) = 0.151 × 315 = 47.56 A
两相短路: I f (1)* =

1 = 0.0506 9.865 + 9.895

I (f 2) = 3 I f (1) = 0.088 *

I (f 2) = 0.088 × 315 = 27.61A
b) 中性点经阻抗接地 单相接地短路: I f (1)* =

1 = 0.0492 9.865 + 9.895 + 0.551

I (f1*) = 3I f (1) = 0.148 I (f1) = 0.148 × 315 = 46.53 A
两相短路: I f (1)* =

1 = 0.0502 9.865 + 9.895 + 0.147

I (f 2) = 3I f (1) = 0.087 *

I (f 2) = 0.087 × 315 = 27.41A
29.解: a) 中性点直接接地 两相接地短路: I f (1)* =

1 = 0.1003 9.895 × 0.105 9.865 + 9.895 + 0.105

I (f1*.1) = 3 × 1 ?

9.865 × 0.105 × 0.1003 = 0.147 9.895 + 0.105

I (f1.1) = 0.147 × 315 = 46.45 A
三相短路: I f
( 3)

=

1 × 315 = 31.93 A 9.865

b) 中性点经阻抗接地 两相接地短路: I f (1)* =

1 = 0.0962 9.895 × 0.551 9.865 + 9.895 + 0.551
16

I (f1*.1) = 3 × 1 ?

9.895 × 0.551 × 0.0962 = 0.115 9.895 + 0.551

I (f1.1) = 0.115 × 315 = 36.32 A
三相短路: I f 30.暂略 31.解: 对三个系统分别列方程:
( 3)

=

1 × 315 = 31.93 A 9.865

Pbc = K C ?f ? ? ? Pab + 50 ? Pbc = K B ?f ? ? 100 ? P = K ?f ab A ?

解得: ?f = 0.1Hz 32.解:

K A = 2500MW / Hz , K B = 4000MW / Hz
列方程如下:

? Pba = ?4000( f ? 50) ? ? Pba = 2500( f ? 49.85)
解得 f = 49.94 Hz

Pba = 230.77 MW

33.解: 1)对两系统分别列方程,有

? ?Pab = ?(270 + 21)?f ? ?150 ? ?Pab = ?(480 + 21)?f
解得 ?f = ?0.1894 Hz

?Pab = 55.11MW

进而有 ?PGA = ? K GA ?f = 51.14 MW , ?PGB = ? K GB ?f = 90.91MW

?PLA = K LA ?f = ?3.98MW , ?PLB = ? K LB ?f + 150 = 146.02MW

17

2) ?Pab = 100 MW ? ?f = ? 34.解: 最大负荷时: ?U =

150 ? 50 = ?0.1Hz 480 + 21

24 × 2.95 + 18 × 48.8 = 8.113KV 117 24 × 48.8 ? 18 × 2.95 = 9.556 KV δU = 117

U = (117 + 8.113) 2 + 9.556 2 = 125.477 KV
最小负荷时: ?U =

12 × 2.95 + 9 × 48.8 = 4.2 KV 113 12 × 48.8 ? 9 × 2.95 = 4.947 KV δU = 113

U = (113 + 4.2) 2 + 4.947 2 = 117.304 KV
设变比为 k,有

125.477 117.304 ? 6 .3 = 6 .3 ? k k

解得 k=0.0519=6.3:121.39 所以变比取 6.3:121

35.解: 由于给出了始端功率和末端电压,采用先忽略电压降落计算功率损耗,再回算电压降落 的方法计算。 最大负荷时: ?S = 所以 ?U =

28 2 + 212 × (2.1 + j 38.5) = 0.205 + j 3.754 MVA 112.09 2

27.795 × 2.1 + 17.244 × 38.5 = 6.444 KV 112.09 27.795 × 38.5 ? 17.244 × 2.1 δU = = 9.224 KV 112.09

U max = (112.09 + 6.444) 2 + 9.224 2 = 118.892 KV
最小负荷时: ?S =

14 2 + 10.5 2 × (2.1 + j 38.5) = 0.048 + j 0.877 MVA 115.92 2

所以 ?U =

27.952 × 2.1 + 20.123 × 38.5 = 7.190 KV 115.92 27.952 × 38.5 ? 20.123 × 2.1 δU = = 8.919 KV 115.92

U min = (115.92 + 7.190) 2 + 8.919 2 = 123.433KV

18

U=

118.892 + 123.433 = 121.16 KV 2

所以取 121KV 分抽头

36.解: 由于给出了始端功率和末端电压, 采用先忽略电压降落计算功率损耗, 再回算电压降落 的方法计算。 最大负荷时:低压侧功率 15+j0MVA,线路阻抗 4.6+j63.5Ω

15 2 + 0 2 ?S = × (4.6 + j 63.5) = 0.092 + j1.269 MVA 106.09 2
所以 ?U =

15.092 × 4.6 + 1.269 × 63.5 = 1.414 KV 106.09 15.092 × 63.5 ? 1.269 × 4.6 = 8.978 KV δU = 106.09

U max = (106.195 ? 1.414) 2 + 8.978 2 = 105.060 KV
最小负荷时:低压侧功率 6+j4.5MVA,线路阻抗 9.2+j127Ω

6 2 + 4 .5 2 ?S = × (9.2 + j127) = 0.042 + j 0.581MVA 110.87 2
所以 ?U =

6.042 × 9.2 + 5.081 × 127 = 6.322 KV 110.87 6.042 × 127 ? 5.081 × 9.2 δU = = 6.499 KV 110.87

U min = (110.87 ? 5.319) 2 + 6.499 2 = 104.750 KV U= 105.060 + 104.750 = 104.925 KV ,取 104.5KV 分抽头。 2

期中试卷: 1.解:
2 Pk maxU N 360 × 220 2 RT = = = 0.605? 2 2000 S N 2000 × 120 2

U k1 = U k2 U k3

1 (9.09 + 16.45 ? 10.75) = 7.395 2 1 = (9.09 + 10.75 ? 16.45) = 1.695 2 1 = (16.45 + 10.75 ? 9.09) = 9.055 2

19

X T1 =

2 U k 1 %U N 7.395 × 220 2 = = 29.827? 100 S N 100 × 120

XT2

2 U k 2 %U N 1.695 × 220 2 = = = 6.837? 100 S N 100 × 120 2 U k 3 %U N 9.055 × 220 2 = = = 36.522? 100 S N 100 × 120

XT3

GT =

P0 59 = = 1.219 × 10 ?6 S 2 1000U N 1000 × 220 2 I 0 % S N 0.332 × 120 = = 8.231 × 10 ?6 S 2 2 100U N 100 × 220

BT =

所以等值电路图为 0.605+j29.827Ω 1 (0.219+j8.231) ×10-6S 2 0.605+j6.837Ω 0.605+j36.522Ω 3

由于给出了始端功率和末端电压, 采用先忽略电压降落计算功率损耗, 再回算电压降落 的方法计算。

?S 2 =

30 2 + 15 2 × (0.605 + j 6.837) = 0.014 + j 0.159 MVA 220 2 20 2 + 15 2 × (0.605 + j 36.522) = 0.008 + j 0.472 MVA 220 2 50.022 2 + 30.6312 × (0.605 + j 29.827) = 0.043 + j 2.120 MVA 220 2

?S 3 =

?S1 =

总损耗为: ?S = 0.065 + j 2.851MVA

50.065 × 0.605 + 32.851 × 29.827 = 4.592 KV 220 50.065 × 29.827 ? 32.851 × 0.605 δU 1 = = 6.697 KV 220 ?U 1 =
U 1 = (220 ? 4.592) 2 + 6.697 2 = 215.512 KV ?U 2 = 30.014 × 0.605 + 15.159 × 6.837 = 0.565 KV 215.512

20

δU 2 =

30.014 × 6.837 ? 15.159 × 0.605 = 0.954 KV 215.512

U 2 = (215.512 ? 0.565) 2 + 0.954 2 = 214.949 KV 20.008 × 0.605 + 15.472 × 36.522 = 2.678 KV 215.512 20.008 × 36.522 ? 15.472 × 0.605 δU 2 = = 3.346 KV 215.512 ?U 3 = U 2 = (215.512 ? 2.678) 2 + 3.346 2 = 212.860 KV
2.解:

? 5 + j15 0 0 ? 5 ? j15 ? ?? 5 + j15 9.86 ? j 36.33 ? 1.923 + j 9.62 ? 2.94 + j11.76? ? Y =? ? ? 0 ? 1.932 + j 9.62 1.923 ? j 29.57 j 20 ? ? 0 ? 2.94 + j11.76 j 20 2.94 ? j 31.71 ? ?
1. 逐个计算雅可比矩阵元素,得到矩阵如下:

0 ? 9.62 ? 1.923? ? 29.62 1.923 ? 20 ? ? 1.923 29.52 0 ? 20 1.923 ? 9.62 ? ? ? ? ? 20 0 31.76 2.94 ? 11.76 ? 2.94 ? J =? ? ? 20 ? 2.94 31.65 2.94 ? 11.76? ? 0 ?? 11.76 ? 2.94 ? 9.62 ? 1.923 36.38 9.86 ? ? ? 0 0 0 2 0 ? ? 0 ? ?
2. PQ 分解法的迭代矩阵为:

9.62 11.76 ? ?? 36.33 ? 9.62 B′ = ? ? 29.57 20 ? ? ? 11.76 20 ? 31.71? ? ? B ′′ = ?36.33
& 3. S 23 = U 2 [U 2 y 20 + (U 2 ? U 3 ) y 23 ] = ?2.2524 + j 2.8068
期末试题: 1.解: 双绕组变压器发电机端电压可调,因此只需对三绕组变压器进行调整。

~

*

*

*

*

*

1 ( P12 + P23 ? P13 ) = 150 KW 2 1 Pk 2 = ( P12 + P23 ? P13 ) = 150 KW 2 1 Pk 3 = ( P13 + P23 ? P12 ) = 100 KW 2 Pk 1 =

21

RT 1 =

2 Pk1U N 150 × 110 2 = = 0.457? 2 1000 S N 1000 × 63 2

RT 2

2 Pk 2U N 150 × 110 2 = = = 0.457? 2 1000 S N 1000 × 63 2 2 Pk 3U N 100 × 110 2 = = = 0.305? 2 1000 S N 1000 × 63 2

RT 3

1 (U k 1? 2 + U k 1?3 + U k 2?3 ) = 10.75 2 1 U k 2 % = (U k1? 2 + U k 2?3 + U k 1?3 ) = ?0.25 2 1 U k 3 % = (U k 1?3 + U k 2?3 + U k 1? 2 ) = 6.75 2 U k1 % = X T1
2 U k 1 %U N 10.75 × 110 2 = = = 20.167? 100 S N 100 × 63 2 U k 2 %U N ? 0.25 × 110 2 = = ?0.480? 100 S N 100 × 63 2 U k 3 %U N 1 6.75 × 110 2 = = 12.964? 100 S N 100 × 63

XT2 =

XT3 =

GT =

p0 77 = = 6.346 × 10 ?6 S 2 2 1000U N 1000 × 110

BT =

I 0 %S N 0.8 × 63 = = 4.165 × 10 ?5 S 2 2 100U N 100 × 110

计算基于归算以后的电压进行(归算至 110KV 侧) 由于最大负荷时 3 绕组为恒调压,所以取 U3=110KV,作为计算的起点,计算电压降落 时忽略网损。 最大负荷时: ?U 3 =

20 × 0.305 + 15 × 12.964 = 1.823KV 110 20 × 12.964 ? 15 × 0.305 δU 3 = = 2.316 KV 110

U 0 = (110 + 1.823) 2 + 2.316 2 = 111.847 KV 55 × 0.457 + 41 × 20.167 = 7.617 KV 111.847 55 × 20.167 ? 41 × 0.457 δU 1 = = 9.749 KV 111.847 ?U 1 = U 1 = (111.847 + 7.617) 2 + 9.749 2 = 119.867 KV
22

35 × 0.457 ? 26 × 0.480 = 0.031KV 111.847 ? 35 × 0.480 ? 26 × 0.457 δU 2 = = ?0.256 KV 111.847 ?U 2 =
U 2 = (111.847 ? 0.031) 2 + 0.256 2 = 111.816 KV
最小负荷时: ?U 3 =

10 × 0.305 + 7.5 × 12.964 = 0.912 KV 110 10 × 12.964 ? 7.5 × 0.305 δU 3 = = 1.158 KV 110

U 0 = (110 + 0.912) 2 + 1.158 2 = 110.918 KV 24 × 0.457 + 17.5 × 20.167 = 2.807 KV 110.918 24 × 20.167 ? 17.5 × 0.457 δU 1 = = 3.642 KV 110.918 ?U 1 = U 1 = (110.918 + 2.807) 2 + 3.642 2 = 113.783KV 14 × 0.457 ? 10 × 0.480 = 0.014 KV 110.918 ? 14 × 0.480 ? 10 × 0.457 δU 2 = = ?0.101KV 110.918 ?U 2 = U 2 = (110.918 ? 0.014) 2 + 0.1012 = 110.904 KV
所以,高压侧选取+2.5%抽头,中压侧选取额定抽头。 2.解:

Z=

1 1 1 1 1 + + + 0.2 0.2 0.8 0.8
1 = 12.5 Z

= 0.08

I=

短路容量 S=UI=12.5 3.解: 取 SB=300MVA,UB=220KV,则 IB=0.787KA

x d * = 0.22 × x L*

300 300 = 0.22 , xT * = 0.12 × = 0 .1 300 360 300 300 = 60 × 0.4 × = 0.149 , x L 0* = 60 × 1.5 × = 0.558 2 220 220 2 0.32 × 0.149 0.1 × 0.658 = 0.140 , z ∑ ( 0 )* = = 0.868 0.32 + 0.149 0.1 + 0.658
23

得到三序阻抗:

z ∑ (1)* = z ∑ ( 2 )* =

1)单相接地

I f (1)= *

1.05 =2.863 , I (f1*) = 3I f (1) = 8.588 0.14 + 0.14 + 0.0868

I (f1) = 6.759 KA
2)两相接地

I f (1)* =

1.05 = 19.597 0.14 × 0.0868 0.14 + 0.0868

I (f1*.1) = 3 1 ?

0.14 × 0.0868 I f (1.1) = 29.664 (0.14 + 0.0868) 2

I (f1.1) = 23.346 KA
4.解: 设末端电压为额定电压,有

60 2 + 45 2 ?S = × (14 + j 30) = 6.508 + j13.946 MNA 110 2 S 23 = 6.658 + j 58.946 MVA ?U 23 = 66.508 × 14 + 58.946 × 30 = 23.476 KV 110 66.508 × 30 ? 58.946 × 14 = = 10.174 KV 110

δU 23

2 U 3 = (115 ? 23.476)+10.174 2 =92.088 KV

?U 12 =

66.508 × 4 + (58.946 ? QC ) × 15 115 66.508 × 15 ? (58.946 ? QC ) × 4 115

δU 12 =

2 2 120 = (115 + ?U 12 ) 2 + δU 12

从中可以解得 QC = 46 MVar

5.解: 对两系统列方程得:

??PA + ?Pab = ? K A ?f ?500 ? 300 = ? K A ?f 或 ? ? ??PB ? ?Pab = ? K B ?f ? 0 + 300 = ? K B ?f
24

可得 K A : K B = 2 : 3 再由

??Pab = ? K A ?f A ? ?P = K ?f ? ab B B ? ? f + ?f A = 49 ? f + ?f B = 50 ?
可以解得 1) KA=500MW/Hz,KB=750MW/Hz 2) 联络线切除前,f=49.6Hz 3) ?f =

? 750 = ?0.6 Hz 500 + 750

25


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