北航网络教育-《电力系统分析》开卷考试考前试题与答案(二)

北航网络教育-《电力系统分析》开卷考试考前试题与答案(二)
北航网络教育-《电力系统分析》开卷考试考前试题与答案(二)

北航网络教育-《电力系统分析》开卷考试考前试题与答案(二)

一、计算题

1.有一台110/11,容量为20000KVA 的三相双绕组变压器,名牌给出的实验数据为:△P K =135KW, P 0=22KW ,U k %=10.5,I 0%=0.8,试计算折算到一次侧的Γ型变压器参数,画出等值电路。

2. 无穷大电源供电系统如图1所示,在输电线中点发生三相短路,求短路冲击电流最大有效值和短路功率,已知:Ug=110kV,降压变压器参数为Z T =0.5003+j7.7818Ω,Z L =1.1+j402Ω。

图1

3.电力线路长80km ,线路采用LGJ-120导线,其参数为Z l =21.6+j33Ω,B l /2=1.1*10-4

S ,线路额定电压为110KV ,末端接有一台容量为20MVA 、变比为110/38.5KV 的降压变压器,折算到变压器高压侧的变压器参数为: Z T =4.93+j63.5Ω,Y T =(4.95-j49.5)*10-6

S ,变压器低压侧负荷为15+j11.25MVA ,正常运行时要求线路始端电压U 1为117.26KV 。试画出网络等值电路,计算该输电网的潮流、电压分布(有名值、标幺值均可)。

4.计算图3网络导纳矩阵(图中数据是阻抗值)

图 3

5.单相对地短路故障如图4所示,故障边界条件为:.

.

.

0,0fb fa fc U I I ===, 其中:Z 0

=j0.25, Z 1

=j0.14, Z 2

=j0.15, U F

=1.05∠0

(a )画出序网络的连接关系图,(b)写出正、负、零序故障电流、电压计算式。

T 1

L

23

3

S

36KV

图4

电力线路长100km ,线路采用LGJ-120导线,如图3所示,其参数为Z l =27+j41.2Ω,B l /2=j1.38×10-4

S ,线路额定电压为110kV ,末端接有一台容量为20MVA 、变比为110/38.5kV 的降压变压器,折算到变压器高压侧的变压器参数为:Z T =4.93+j63.5Ω,Y T =(4.95-j49.5)×10-6

S ,变压器低压侧负荷为10+j5MVA ,正常运行时要求线路始端电压U 1为118kV 。(1)试画出网络参数为有名值时的等值电路;(2)计算各元件标幺值并绘制标幺值等值电路;(3)计算该输电网的潮流、电压分布(有名值、标幺值均可)。

图3

二、综合题

1. 某一网络进行潮流计算后的数据如图5所示,说明该系统中发电机、变压器及负载个数,利用图中数据说明发电机总的发电功率,负荷总的用电功率,计算网络损耗和效率,网络中各节点电压是否满足稳态运行要求,若要增大节点电压可采取哪种措施?

图5

2、电力线路长80km ,线路采用LGJ-120导线,如图3所示,其参数为Z l =21.6+j33Ω,B l /2=j1.1×10-4

S ,线路额定电压为110kV ,末端接有一台容量为20MVA 、变比为110/38.5kV 的降压变压器,折算到变压器高压侧的变压器参数为: Z T =4.93+j63.5Ω,Y T =(4.95-j49.5)×10-6

S ,变压器低压侧负荷为10+j6MVA ,正常运行时要求线路始端电压U 1为118 kV 。(1)试画出网络参数为有名值时的等值电路;(2)计算各元件标幺值并绘制标幺值等值电路;(3)计算该输电网的潮流、电压分布(有名值、标幺值均可)。

T 1

L

23

3

S 36KV

图3

3.电力网络如图2所示,各元件参数用纯电抗表示,试绘制该电力网正序、负序及零序等值电路,写出网络f -1(n )

点发生单相接地

故障时的正序、负序及零序等效电抗表达式,绘制该点单相接地短路时的三序网的连接关系图。

图2

4、如图3所示的某三相系统中k 点发生单相接地故障,已知

1j a1Σ=E ,

Z 1∑=j0.4, Z 2∑=j0.5,Z 0∑=j0.25,Z k =j0.35。求a 相经过阻抗Z k 接地短路时短路点的各相电流、电压。(20分)

图3

5、额定电压110kV 的辐射形电网各段阻抗及负荷如图4所示。已知电源A 的电压为121kV ,求功率分布和各母线电压。(注:考虑

功率损耗,可以不计电压降落的横分量U δ)。 (15分)

图4

6、电力线路长100km ,线路采用LGJ-120导线,如图3所示,其参数为Z l =27+j41.2Ω,B l /2=j1.38×10-4

S ,线路额定电压为110kV ,末端接有一台容量为20MVA 、变比为110/38.5kV 的降压变压器,折算到变压器高压侧的变压器参数为:Z T =4.93+j63.5Ω,Y T =(4.95-j49.5)×10-6

S ,变压器低压侧负荷为10+j5MVA ,正常运行时要求线路始端电压U 1为118kV 。(1)试画出网络参数为有名值时的等值电路;(2)计算各元件标幺值并绘制标幺值等值电路;(3)计算该输电网的潮流、电压分布(有名值、标幺值均可)。

T 1

L

23

3

S 36KV

T 1

L

23

3

S 36KV

图3

7、在图4所示的电力网络中,当降压变电所kV 5.10母线上发生了三相短路时,可将系统视为无限大容量电源,试求此时短路点的冲击电流im i ,短路电流的最大有效值im I 和短路容量t S 。

图4

电力系统分析复习题答案

一、计算题

1.解:计算变压器阻抗 1)串联电阻(归算到121kV 高电压侧) 2)串联电抗

3)励磁回路(并联)导纳

参数等效在高压侧

2.解:短路电流周期分量有效值

22

22

(/2)(/2)38.5/3

(0.50030.55)(7.7818 2.01)2.2571(7.7818 2.01)/314.160.02970.50030.55

g

T L T L U I R R X X kA

L T s

R =+++=

+++=+===+

()

()

22

2

2110135 4.0810********k N T N P U R S ===Ω()()

22110%0.10563.5310020k N

T N

U U X S =?=?=Ω

高()

()

3602222

*10 1.8210110T N P G S U --===?高()()

6

022

%200.00813.210100110N T N

I S B S U -=?=?=?高(4.0863.53)T T T Z R jX j =+=+Ω

6(1.8213.2)10T T T Y G jB j S

-=-=-?

冲击系数: 0.01/1 1.714T ch

k e -=+=

短路冲击电流、最大有效值及短路功率

221.7142 2.2571 5.470312(1) 2.257112(1.7141) 3.20763337 2.2571144.648ch ch m ch ch t N i k I kA

I I k kA S U I MVA

==??==+-=?+-===??= 3.解:由已知条件绘制网络等值电路为

先设全网各节点电压为线路额定电压 ,求潮流分布(由末端向始端推算功率损耗及功率分布)。 1)从末端向始端推算各元件的功率损耗和全网功率分布而不计算各节点电压。 变压器功率损耗:

()()2

2222

33322

1511.25 4.9363.50.14 1.85110ZT T T N N ZT ZT

S P Q S Z R jX j j U U P j Q ??++?==+=+=+ ???

=?+? ()()*

2

26110(4.9549.5)100.060.6

yT N T N T T N

S U Y U G jB U j j -?==+=+?=+

线路末端导纳支路功率损耗:

()*

2

2421110( 1.1)10 1.3322

y N N N Y S U U G jB U j j -???==-=?-?=- ???

线路阻抗支路末端功率:

'232

(150.140.06)(11.25 1.850.6 1.33)15.212.37

ZT yT y S S S S S j j =+?+?+?=+++++-=+ 线路阻抗损耗:

()()2

222222

22

2

15.212.3721.6330.69 1.056

110Z N N S P Q S Z R jX U U j j ??'''+?==+ ???+=+=+ 线路首端导纳支路功率损耗:

()*

2

2411110( 1.1)10 1.3322

y N N N Y S U U G jB U j j -???==-=?-?=- ???

首端功率:

1S '2S 2S 12341.3810j S -?2741.2j +Ω 4.9363.5j +Ω6(4.9549.5)10j S --?1511.25j MVA +

()()12

1 15.212.37(0.69 1.056) 1.33 15.8912.1

z y S S S S j j j j '=+?+?=++++-=+ 2)待求得首端功率后,由给定的首端电压,根据网络功率分布从首端至末端推算各元件的电压损耗和各节点电压,不再重新计算

全网的功率分布。

()()22

211

2115.8921.612.133

6.33,

117.26

0117.26 6.43110.93

U U U U U U U

tg U U

δδδδ-?+??==≈=

-?+≈-=-=≈-?

()()

22

321

3215.14 4.9313.163.5

8.17,0

110.93

110.938.17102.760U U U U U U U tg U U

δδδδ-?+?''?==≈''=

-?+=-='

-=≈'-?

4.解 根据节点导纳矩阵的定义,求导纳矩阵各元素:

11101213211212133113222012232332333013111

0.935 4.262300.080.400.120.5

0.481 2.4040.454 1.891

111

1.069 4.742300.080.400.10.40

0.588 2.353Y y y y j j j j Y Y y j Y Y y j Y y y y j j j j Y Y j Y y y y =++=

++=--++==-=-+==-=-+=++=++=--++==-+=++2334

344334443411

0.454 1.8910.588 2.353 1.0697.54329303.3333.333

y j j j j j Y Y y j Y y j +=

-++-+=--==-===- Y 14=Y 24=Y 41=Y 42=0

111213

1421

2223243132333441

4243

44Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y ??????=??????

5.

解:根据故障类型有

.

120120...3.

..0

b

fb fb fb fb fb fb I I I I U U U ?

?===?

=>??++=??

1

2

120.

1.050.

.

.

2.1 2.1(90)()0.20.150.15F

f f f j j j j j U I I I Z Z Z

∠=====-=∠-++++

201

12022

12020

1200.150.15.. 1.050.630.20.150.150.15.. 1.050.3150.20.150.150.15.. 1.050.315

0.20.150.15f F

f F

f F

j j j j j j j j j j j j j Z Z U

U Z Z Z Z U

U Z Z Z Z U U Z Z Z ∑

?++==?=?

++++??

?=-=-?=-?

++++??

?=-=-?=-?++++? 一、 综合题

1、解:发电机总的发电功率,S G =377+j136+520+j354=897+j490MV A 负荷总的用电功率,S L =60+j30+800+j300=860+j330MV A

网络损耗功率为△S= S G - S L =(897+j490)-(860+j330)=37+j160MVA

节点1、3、4、5电压分别为1、相角00;1.05、相角-0.0340;1.017、相角-2.3730;0.970、相角-4.3360,满足节点电压在额定值±5%的运行要求。

节点2电压低于0.816、相角-22.5540,低于0.95,不满足节点电压在额定值±5%的运行要求,可以安装无功功率补偿设备提高该点电压。

2、 电力线路长80km ,线路采用LGJ-120导线,如图3所示,其参数为Z l =21.6+j33Ω,B l /2=j1.1

×10-4

S ,线路额定电压为110kV ,末端接有一台容量为20MV A 、变比为110/38.5kV 的降压变压器,折算到变压器高压侧的变压器参数为: Z T =4.93+j63.5Ω,Y T =(4.95-j49.5)×10-6S ,变压器低压侧负荷为S D =15+j11.25MV A ,正常运行时要求线路始端电压U 1为117.26kV 。(1)试画出网络参数为有名值时的等值电路;(2)计算各元件标幺值并绘制标幺值等值电路;(3)计算该输电网的潮流、电压分布(有名值、标幺值均可)。 解:(1)由已知条件绘制网络参数为有名值时的等值电路为

T 1

L

23

3S

36KV

(2)计算各元件标幺值

设基准容量为S B =100MVA ,基准电压为U B =110kV , 线路参数标幺值如下: 变压器参数标幺值如下:

负荷标幺值为:

标幺值等值电路为:

(3)计算该输电网的潮流、电压分布

用有名值计算,先设全网各节点电压为线路额定电压110kV ,求潮流分布(由末端向始端推算功率损耗及功率分布)。 1)从末端向始端推算各元件的功率损耗和全网功率分布而不计算各节点电压。 变压器阻抗中的功率损耗为:

()()2

2222

33322

1511.25 4.9363.50.14 1.85110ZT T T N N ZT ZT

S P Q S Z R jX j j U U P j Q ??++?==+=+=+ ???

=?+? 变压器导纳中的功率损耗为:

()()*

2

26110(4.9549.5)100.060.6

yT N T N T T N

S U Y U G jB U j j -?==+=+?=+

变压器首端功率:

22100(21.633)0.17850.2727110

B l l B S Z Z j j U *

=?=+?=+22

4110(1.110)0.0133100B l l

B U Y Y j j S -*=?=??=22100(4.9363.5)0.0410.525110B T T B S Z Z j j U *=?=+?=+22

6

110(4.9549.510)0.00060.006100B T T

B

U Y Y j j S -*=?=-??=+1511.250.150.1125100D D B S j S j S *

+===+

23(150.140.06)(11.25 1.850.6)15.213.7

ZT yT

S S S S j j =+?+?=+++++=+ 线路末端导纳支路功率损耗:

()*

2

2421110( 1.1)10 1.3322

y N N N Y S U U G jB U j j -???==-=?-?=- ???

线路阻抗支路末端功率:

'

232

(150.140.06)(11.25 1.850.6 1.33)15.212.37

ZT yT y S S S S S j j =+?+?+?=+++++-=+ 线路阻抗损耗:

()()2

222222

222

15.212.3721.6330.69 1.056

110Z N N S P Q S Z R jX U U j j ??'''+?==+ ???+=+=+

线路首端导纳支路功率损耗:

()*

2

2411110( 1.1)10 1.3322

y N N N Y S U U G jB U j j -???==-=?-?=- ???

首端功率:

()()12

1 15.212.37(0.69 1.056) 1.33 15.8912.1

z y S S S S j j j j '=+?+?=++++-=+ 2)待求得首端功率后,由给定的首端电压,根据网络功率分布从首端至末端推算各元件的电压损耗和各节点电压,不再重新计算

全网的功率分布。 线路电压降落:

()()

22

2112115.8921.612.133

6.33,

117.26

0117.26 6.43110.83

U U U U U U U

tg U U

δδδδ-?+??==≈=

-?+≈-=-=≈-?

变压器电压降落:

()()22

321

3215.2 4.9313.763.5

8.52,0

110.93

110.938.52102.410

U U U U U U U tg U U δδδδ-?+?''?==≈''=

-?+=-='

-=≈'

-?

变压器低压侧电压为:

338.5

102.4135.84110

U =?

= 3、电力网络如图6所示,各元件参数用纯电抗表示,试绘制该电力网正序、负序及零序等值电路,写出网络的f -1(n )

点发生单相不

对称短路时正序、负序及零序等效电抗表达式,绘制三序网的连接关系图。(10分)

2

正序等效

零序等效

4、如图3所示的某三相系统中k 点发生单相接地故障,已知1j a1Σ=E ,

Z 1∑=j0.4, Z 2∑=j0.5,Z 0∑=j0.25,Z k =j0.35。求a 相经过阻抗Z k 接地短路时短路点的各相电流、电压。(20分)

图3

解:单相接地时复合网为串联型,于是可知

5.4264.00.353j0.25j0.5j0.41

j 3k

021a1Σ

a1

a1a1=?+++=

+++===∑∑∑

Z Z Z Z E I I I

)0.353j0.25j0.5(5.4264.0)3(k

02a1a1?++=++=∑∑ Z Z Z I U

7883.05.3529.15.4264.0=?=

5.471

6.09025.05.4264.05.4732.0905.05.4264.00a0a02a2a2-=?-=-=-=?-=-=∑

Z I U Z I U

求各相电流和电压为

1561144.0j 993.05.7416.02405.7432.01207883.05.249.037.0j 82.05.7416.01205.7432.02407883.05.4267.035.05.4292.105.4292.15.4264.033a0

a22a1c a0

a2a12b k a a c

b

a1a0a2a1a1-=-=-++-++=++=-=-=-++-++=++==?=====?==++=U U a U a U U U a U a U Z I U I I

I I I I I

5、额定电压110kV 的辐射形电网各段阻抗及负荷如图4所示。已知电源A 的电压为121kV ,求功率分布和各母线电压。(注:考虑功率损耗,可以不计电压降落的横分量U δ)。 (15分)

图4

解 设U

C =U N = 110 kV 22BC

2108(20j30)0.271j0.407110S +?=+=+ 407.22j 207.30953.7j 793.930j 40593.7j 793.9407.0j 271.0)8j 10(B B B

BC C B +=--+=''?+='--=++--=?+-=''S S S S S S 083.27j 545.32j4.676338.2407.22j 207.30676.4j 338.2)40j 20(110407.22207.30AB

B A 2

2

2

AB

+=+++=?+'=+=++=?S S S S 已知U A =121kV

)

kV (64.110972.3668.106)

kV (972.3668

.10630

)593.7(20)793.9()kV (668.106332.14121)

kV (332.14121

40

083.2720545.32BC B C BC AB A B AB =+=?-=-=?-+?-=?=-=?-==?+?=

?U U U U U U U U

6、电力线路长100km ,线路采用LGJ-120导线,如图3所示,其参数为Z l =27+j41.2Ω,B l /2=j1.38×10-4

S ,线路额定电压为110kV ,末端接有一台容量为20MVA 、变比为110/38.5kV 的降压变压器,折算到变压器高压侧的变压器参数为:Z T =4.93+j63.5Ω,

Y T =(4.95-j49.5)×10-6

S ,变压器低压侧负荷为15+j11.25MVA ,正常运行时要求线路始端电压U 1为118kV 。(1)试画出网络参数为有名值时的等值电路;(2)计算各元件标幺值并绘制标幺值等值电路;(3)计算该输电网的潮流、电压分布(有名值、标幺值均可)。

图3

解:(1)由已知条件绘制网络参数为有名值时的等值电路为

T 1

L

23

3S

36KV

(2)计算各元件标幺值并绘制标幺值等值电路 设基准容量为S B =100MVA ,基准电压为U B =110kV , 线路参数标幺值如下:

变压器参数标幺值如下:

负荷标幺值为:

标幺值等值电路为:

(3)计算该输电网的潮流、电压分布

用有名值计算,先设全网各节点电压为线路额定电压110kV ,求潮流分布(由末端向始端推算功率损耗及功率分布)。 1)从末端向始端推算各元件的功率损耗和全网功率分布而不计算各节点电压。 变压器阻抗中的功率损耗为:

()()2

2222

33322

1511.25 4.9363.50.14 1.85110ZT T T N N ZT ZT

S P Q S Z R jX j j U U P j Q ??++?==+=+=+ ???

=?+? 变压器导纳中的功率损耗为:

()()*

2

26110(4.9549.5)100.060.6

yT N T N T T N

S U Y U G jB U j j -?==+=+?=+

变压器首端功率为:

22100

(2741.2)0.22310.3405110B l l B S Z Z j j U *=?=+?=+224110(1.3810)0.0167100

B l l B U Y Y j j S -*

=?=??=22100(4.9363.5)0.0410.525110

B T T B S Z Z j j U *

=?=+?=+22

6110(4.9549.510)0.00060.006100B T T

B U Y Y j j S -*=?=-??=+1511.250.150.1125100D D B S j S j S *

+===+

23(150.140.06)(11.25 1.850.6)15.213.7

ZT yT

S S S S j j =+?+?=+++++=+ 线路末端导纳支路功率损耗:

()*

2

2421110( 1.1)10 1.3322

y N N N Y S U U G jB U j j -???==-=?-?=- ???

线路阻抗支路末端功率:

'

232

(150.140.06)(11.25 1.850.6 1.33)15.212.37

ZT yT y S S S S S j j =+?+?+?=+++++-=+ 线路阻抗损耗:

()()2

222222

222

15.212.372741.20.8625 1.3184

110Z N N S P Q S Z R jX U U j j ??'''+?==+ ???+=+=+

线路首端导纳支路功率损耗:

()*

2

2411110( 1.38)10 1.6722

y N N N Y S U U G jB U j j -???==-=?-?=- ???

首端功率:

()()12

1 15.212.37(0.8625 1.3184) 1.67 16.0612.02

z y S S S S j j j j '=+?+?=++++-=+ 2)待求得首端功率后,由给定的首端电压,根据网络功率分布从首端至末端推算各元件的电压损耗和各节点电压,不再重新计算

全网的功率分布。 线路电压降落:

()()

22

2112115.062712.0241.2

8.16,

118

01188.16119.84

U U U U U U U

tg U U

δδδδ-?+??==≈=

-?+≈-=-=≈-?

变压器电压降落:

()()22

321

3215.2 4.9313.763.5

8.61,0

109.84

119.848.61101.230

U U U U U U U tg U U δδδδ-?+?''?==≈''=

-?+=-='

-=≈'

-?

变压器低压侧电压为:

338.5

101.2335.43110

U =?

= 7、在图4所示的电力网络中,当降压变电所kV 5.10母线上发生了三相短路时,可将系统视为无限大容量电源,试求此时短路点的冲击电流im i ,短路电流的最大有效值im I 和短路容量t S 。

图4

解 取MVA S B

100=、n aV B U U .=,等值电路如下:

首先计算各元件参数的标幺值电抗

525.020

100

1005.10100%1=?==

N B k S S U X

292.037100

104.02

2.12=??==n aV B U S l

x X 19.22

.3100

1007100%43=?=?==N B k S S U X X

取1=E ,作等值网络如图(b)所示。 短路回路的等值电抗为

短路电流周期分量的有效值为

kA

系数8.1=im

K ,则冲击电流为

若取冲击

34.788.255.228.1=?=?=ωI i im kA

短路电流的最大有效值为

38.488.252.152.1=?==ωI I im kA

短路功率为

3.52100523.0=?==*B t S I S ω MVA

11

0.5231.912

I X ω*∑

===100

0.523 2.88

310.5

B I I I ωω*==?=?1

0.5250.292 2.19 1.9122

X ∑=++

?=

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