POWER SYSTEM ANALYSIS SECOND EDITION PDF

Title	POWER SYSTEM ANALYSIS SECOND EDITION
Author	峻佑廖
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POWER SYSTEM ANALYSIS SECOND EDITION

ARTHUR R.BERGEN VIJAY VITTAL 考題整理

Name : Tel

:

Email :

Ch2-----------------------------1 Ch3 ---------------------------15 Ch4---------------------------33 Ch5-1 變壓器------------------64 Ch5-2 標么--------------------84 Ch9 ---------------------------101 Ch10---------------------------110 Ch11---------------------------142 Ch12---------------------------150 Ch14---------------------------199

第二章 -----------------------------------------------------------------------------------------------------------※One Port prob2.1 One Port [98 ] 1. In Figure 2.1, v(t) = 2 120 cos( t + 30) ,i(t) = 2 10 cos( t - 30) . a. Find p(t), P, and Q into the network. (12%) b. Find a simple ( two element ) series circuit consistent with the prescribed terminal behavior as described in this problem. (8%) i(t)  v(t) P(t) 

N

Figure 2.1 (sol) (a) p (t) =v(t)i(t)=2400 cos( t + 30) cos( t - 30)=1200 [cos(2 t )+ cos(60)]=600+1200cos(2 t )-----(4%) V=12030(V);I=10-30( A)

S  VI *  12030  1030  120060  600  j1039.23(VA) P  600(W )-------------------(4%) Q  1039.23(Var)-------------------(4%) (b) V 12030   1260  6  j10.39() I 10-30 R  6(); X  10.39()-------------(8%)[全對才給分 ]

Z=

6 10.39

----------------------

1

1

[96,98,97 ] i (t )  2 I sin t

Ex2.2 One Port

2.Consider a network with a driving-point impedance

Z  Z  Z

shown in Figure 2.

Assume i (t )  2 I sin t , find p(t) in terms of P and Q. (20%),

i (t )  2 I sin t

(Sol)： i (t )  2 I sin t Z  Z  Z

+ v(t )  2 Z I sin t   Z  -

Z P  Z I 2 cos  Z 2

Q  Z I sin  Z 2 p (t )  v(t )i (t )  2 Z I sin(t   Z ) sin t

2sin  sin   [cos(   )  cos(   )] 2sin(t   Z ) sin t    cos(2t   Z )  cos  Z  cos(   )  cos  cos   sin  sin  2 2 2 (a). S  VI *  ZII *  Z I  Re  Z I   jI m  Z I   P  jQ     2 2 2 2  P  Re  Z I   Z I cos  Z ; Q  I m  Z I   Z I sin  Z     (b).Assume i (t )  2 I sin t Then v(t )  2 Z I sin t   Z  2 2 p (t )  v(t )i (t )  2 Z I sin(t   Z ) sin t   Z I cos(2t   Z )  cos  Z  2   Z I (cos 2t cos  Z  sin 2t sin  Z )  cos  Z 

 Z I

2

 cos  Z (cos 2t  1)  sin 2t sin  Z 

  2 2    Z I cos  Z (cos 2t  1)  Z I sin  Z sin 2t      P Q      P (cos 2t  1)  Q sin 2t   P(1  cos 2t )  Q sin 2t p(t )  P(1  cos 2t )  Q sin 2t   P(cos 2t  1)  Q sin 2t (20%) 5.Consider

a

network

with

a

driving-point

impedance

Assume i (t )  2 I sin t , find p(t) in terms of P and Q. p(t)= (A) P 1  cos 2t   Q sin 2t (B) P 1  cos 2t   Q sin 2t (C) P (1  cos 2t )  Q sin 2 t (D) P (1  sin 2t )  Q cos 2t ----------------------

2

2

Z  Z  Z

shown

in

Fig.

2.

Ex2.2 One Port

i (t )  2 I cos t

ex2.2 (b).Assume i (t )  2 I cos t Then v (t )  2 Z I cos t  Z 

cos Z  cos(2t  Z ) 2  Z I  cos Z  cos 2t cos Z  sin 2t sin Z   P 1  cos 2t   Q sin 2t 2

p (t )  v(t )i (t )  2 Z I cos(t  Z ) cos t  Z I

2

---------------------One Port [91] In figure 1 v(t)= 2 Vcos(ωt+α)，i(t)= 2 Icos(ωt+β)， what is the instantaneous power network A to network B? What is the complex power S from network A to network B?

p(t) from

(B)9.In figure 1 v(t)= 2 Vcos(ωt+α)，i(t)= 2 Icos(ωt+β)， what is the instantaneous power p(t) from network A to network B? (A) VI[Cos(α-β)+Cos(2ωt-α-β)] (B) VI[Cos(α-β)+Cos(2ωt+α+β)] (C) VI[Cos(α+β)+Cos(2ωt-α-β)] (D) VI[Cos(α+β)+Cos(2ωt+α+β)] 。 (B)10.Repeat question 9，what is the Complex Power S from network A to network B? (A) VI∠(α+β) (B)VI∠(α-β) (C)2VI∠(α-β) (D) 2VI∠(α+β)。錯誤! 尚未定義書籤。 Figure 1 ---------------------One Port [100] 2. A single-phase load is supplied with a sinusoidal voltage v(t) = 200 cos(377t). The resulting instantaneous power is p(t) = 800 + 1000 cos(754t – 36.87。). (a) Find the complex power supplied to the load. (b) Find the instantaneous current i(t) supplied to the load. (c) Find the load impedance. Saadat 2.2 (18%) (sol) p(t) = 800 + 1000 cos(754t- 36.87) = 800 + 1000 (cos 36.87 cos 754t + sin 36.87sin 754t) = 800 + 800 cos 754t + 600 sin 754t = 800[1 + cos 2(377)t] + 600 sin 2(377)t = P[1 + cos 2(377)t] +Q sin 2(377)t thus S = P+jQ=800 + j600 = 1000∠ 36.87˚VA 1 2(100036.87)* (b) Using S  Vm I m*  I m   10  36.87A 2 200  i (t )  10 cos(377t  36.87)A V 2000  2036.87 (c) Z L   I 10  36.87

3

3

※平衡 Y 接負載 ---------------------Prob 2.6 平衡 Y 接負載 [92] The system shown in figure 2 is balanced (and positive sequence ). Z=100∠60°. What is Vbn=? Ic=? (B) 9.The system shown in figure 2 is balanced (and positive sequence ). Z=100∠60°. What is Vbn=? (A) 120  120 (B) 120  150 (C) 120120 (D) 10.Repeat question 11，Ic=? (A) 1.2  90 (B) 1.2  150

(C) 1.2  210

Vab=220∠0°, impedance

Vab=208∠0°, impedance (D) 120150 (D) 1.230

(D)11.The same as question 11 except the sequence is negative sequence. ∠0°. What is Vbn=? (A) 120  120 (B) 120  150 (C) 120120 (D) 120150 (B)12.Repeat question 13，Ic=? (A) 1.2  90 (B) 1.2  150

(C) 1.2  210 Ia Z

(D) 1.230

a Z n Z

c

Z=10060

b

Figure 2 ---------------------Prob 2.6 平衡 Y 接負載 [92,100 暑修,101,102] .m 1.The system shown in the Figure 1 is balanced (and positive sequence). Assume that Z  10  15 and Vca  208  120 . Find Vab, Vbc, Van, Vbn, Vcn, Ia, Ib, Ic, and S3 .(18%) a Z n Z

Ia Z

b

c

(sol)Vca    120

Figure 1

Vab  Vca   120  208  120  120  208  240  208120 Vbc  Vca 120  208  120120  2080 Van  Vab   30 / 3  208120  30 / 3  12090 Vbn  Van   120  12090  120  120  30 Vcn  Van 120  12090120  120210  120  150 V 12090 Ia  an   12105 Z 1015 I b  Ia   120  12105  120  12  15 Ic  Ia 120  12105120  12225  12  135 S3  3Van Ia *  3(12090)(12105)*  3(12090)(12  105)  4320(15)  4172.8  j1118.1(VA)

------------------------------------------------------4

4

1. Prob 2.12 平衡三相負載[99,100,104,105] 2. 圖 1 平衡系統中，負載電感器 ZL=j10Ω，負載電容器 ZC=-j10Ω，求(a)Ia。(b)Icap。(c)S3ψload。 2. The system shown in Figure 2. is balanced. Assume that load inductors, ZL=j10 load capacitors, ZC=-j10.  Find Ia, Icap, and S3Load (24%)

Ia

a

n

j1

 10  ICAP

j1

c

b

j1

Load Figure 2

(sol Problem 2.12) 1  0.2590 j1-j5 (4)Van  Ia  Zeq =(j0.25)(-j5)=1.25 (3)I a 

j1 a

E1  10

(5)Vab  3 Van 30=2.16530  1.25

(6)ICAP

(3)Ia

+

n

-

(2)Zeq =j10//(-j

+ (4)Van -

j10

10 )=-j5 3 a

(1)Z =-j n

V 2.16530  ab   0.2165120 ZC -j10

  3Van I a*  3  (1.25)(0.2590)*  0.9375  90 (7)S3Load

-------------------------------------------

5

5

10 3

EX 2.12 平衡三相負載[95] (A)1.The system shown in figure 2 is balanced (and positive sequence ). What is v1 t  =?

(A) 368cos  ωt+45°  (B) 368 2cos  ωt+45°  (C) 386cos  ωt+45°  (D) 386 2cos  ωt+45° 

(B)2.Repeat question 2， What is i2 t  =?

(A) 319cos  ωt+75°  (B) 319cos  ωt+165°  (C) 319 2cos  ωt+75°  (D) 319 2cos  ωt+165°  j 0.1

E1 

350 2

45

j1.0 V 1

1  j 0.01

n

a

a

a

 j2

n

i2 b

c

c

c

b

b

Fig 1

 2   j2 〈Sol〉： Zeq   j1 //  j     j2  3 3  2 2 (1). Z    j 3 (2). Consider the per phase ckt (for phase a) in figure 2.12(b) Z eq 368 45 Va ' n '  V1  Ea  1.05Ea  j 0.1  Z eq 2 638 75 2 V 319  a 'b '  165  ,  j2 2

Va 'b '  3Va ' n '30  i2  I a 'b ' 

Va 'b ' Z

i2 (t )  319 cos   t  165  , v1 (t )  368cos   t  45  j 0.1

a

+ E1 

n

-

350 2

a Z  

a

+ j1.0 V 1 n

45 1  j 0.01

 j2

Z 2 j 3 3

Z n

i2 b

c

c

b

2.12(a)

Fig

a

n

j 0.1

Zeq 350 E1  45 j1.0 V1 2 Fig

a

Z   j n

2 3

2.12(b)

-------------------------------------------

6

6

c

b

EX 2.12 平衡三相負載[96] Ean  100 (D) 1. Given the unbalanced three-phase system shown in Fig. 1. The voltage VazL is (A) 0 (B) -52.6316 +j52.6316 (C) -52.6316 -j52.6316 (D) 105.2632 (V) (A) 2.Repeat question 1, The current I nn is (A) 0 (B) -26.3158 -j26.3158 (C) j 52.6316 (D)  j105.2632 (A) a

a

a

Ia

Z line  j 0.1

Ean  100 n

I azc ZL  j

I nn  ? 1  j 0.01

Zc   j

VazL  ?

2 3

n

n

Ebn  50  j 50 c

Ecn  50  j 50

b

Z line

c

ZL

b

ZL

c

Zc

Zc

Z line

b

Fig. 1

(sol) Ean 100 Z an   j1.9 j 0.1 Z eq   j 2   j 52.6316(A) a Z an  j1.9 Ia Vzaline I aZL VazL  I a Z eq  ( j 52.6316)(-j2)=105.2632(V) j1.0 VazL  ? Ean  100 VazL   j105.2632 I aZL  n j1 V I azc  azL  j157.89  j2 / 3

 Ean  100; I a 

 Ebn  50  j 50; I b 

a I azc 2 Zc   j 3 n

Ebn 100  =-26.3158 -j26.3158 Z bn  j1.9

VbzL  I b Z eq  (-26.3158 -j26.3158)(-j2) =-52.6316 +j52.6316(V) VbzL  52.6316 +j52.6316 j1 V I bzc  bzL  -78.9474 -j78.9474  j2 / 3 Vzbline =I b Zline =2.6316 -j2.6316

Z bn   j1.9 j 0.1 Z eq   j 2

I bZL 

 Ecn  50  j 50; I c 

b

Ib

Vzbline

Ebn  50  j 50 n

I bZL j1.0

VbzL

b I bzc 2 Zc   j 3 n

Ecn 50  j 50 =26.3158 -j26.3158  Z cn  j1.9

VczL  I c Z eq  (26.3158 -j26.3158)(-j2) =-52.6316 -j52.6316(V) VczL  -52.6316 +j52.6316 j1 V I czc  czL  78.9474 -j78.9474  j2 / 3 Vzcline =I c Zline =2.6316 +j 2.6316 I cZL 

c

Z cn   j1.9 j 0.1 Z eq   j 2 Ic

Vzcline

Ecn  50  j 50

I cZL j1.0

VczL

n

I nn  ( I a  I b  I c )  ( j 52.6316-26.3158 -j26.3158+26.3158 -j26.3158)=0 ------------------------------------------7

7

c I czc 2 Zc   j 3 n

EX 2.13 平衡三相負載[97] [102,103].m 2. A three-phase line has an impedance of 0.6+j3.0 Ω/ψ as shown in Figure 2. The line feeds three balanced three loads that are connected in parallel. The first load is absorbing a total of 156kW and 117kVAR magnetizing voltamperes. The second load is △ connected and has an impedance of 144-j42 Ω/ψ.The third load is 115KVA at 0.6 PF leading. The line to neutral voltage at the load end of the line is 2600V. Find (a) I1 , I 2 , I3 and Iline .(20%) (b) the magnitude of the line voltage at the source end of the line. (5%) EX.2.13. Iline

Zline =0.6+j3(Ω/phase) I2

I1

Vline =?

I3

Zline Zline

P3  156kW Q3  117kVAR VP  2600V

Load2

Load1

Load3

115kVA PF=0.6 leading

Z   144  j 42( / phase)

Figure 2 (Sol)：

156  j117 *  52  j 39( KVA)  6536.87( KVA)  V  I1 3 2600 0  52  j39(KVA) I1   20  j15(A)  25-36.87(A)(5%) 26000 Z Load2： Z   Z Y 【For per phase】 ZY    48  j14  50-16.26() 3   V 26000 26000 I2     49.92+j14.56(A)  5216.26(A)(5%) ZY 48  j14 50-16.26

Load1： S1 

Load3： S3  115  cos 1 0.6(kVA)  115  53.13  (69 -j92)(kVA)=3  2600 I 3*  115 53.13( KVA)  I3  3  8.846  j11.795( A)  14.753.13( A)(5%) 2600 Source: I l  I1  I 2  I 3  78.766  j11.355=79.58048.2032( A)(5%) Van  26000  I l Z line  26000  (79.58048.2032)(0.6  j 3) =2613.195+j243.111  2624.55.315(V ) Vab  3Van 30 =3709.25+j2627.76  4545.735.315(V )(5%)

8

8

※不平衡 ---------------------Prob 2.15 不平衡 Y 接負載 [101 暑修]

2. In Figure 2, assume that E a = 245, E b =1  90 , E c =1180and I a =1  10. The load is symmetrical and Z is not given. The source are not balanced, but we note that  E a + E b + E c =0. Find Z and S3load . (22%) Ia

a

+ Ea

Z

-

Ec

-

n

-

Eb

+

Z

c

Z

n

+ b

Load

Figure 2 (sol2.15)  E a + E b + E c =0.

 neutrals are at the same potential. E n =E n connect: node n and n . a

E a = 245

Z

I a =1  10

a

Va 245   255 I a 1  10

Z

E c =1180 n

E b =1  90 c

c

Z 

n

Z

b

b

Va 245   255 (11%) I a 1  10 2

V V S=VI  V( )*  * Z Z *

Va

Sa 



*

Z

Sb  Sc 

2

Vb

2



*

Z Vc

*

2



245

2

( 255)

*

1  90

2

( 255)* 1180

 

2

2  55 1 2  55 1

 255  

1 2 1

2  55 2 1 1  =Sa  Sb  Sc  ( 2  )55  S3load  2 2 Z

( 255)*



2

Z

55 55

 2 255 (11%) 9

9

-----------------------Prob 2.8 不平衡 Y 接負載 [104,103] 4. In the system shown in Figure 4, Za=Zb=j1.0, Zc=j0.9. Find Ia, Ib, and Ic . . (21%) a

+

Ia

j0.1

10 1120 c

+

j0.1

n

1  120

+

Za

b

n

Zc

Ib

j0.1

Ic

j0.1

j1 Zb

j0.9 j1

Figure 4 (sol2.8) (a) use loop analysis node n:I n  Ia  I b  Ic loop1:Van =1=(j0.1+ Za +j0.1)Ia +(j0.1)I n =(j1.2)I a +(j0.1) I n =(j1.2)Ia +(j0.1)I b +(j0.1)Ic   Ia  I b  Ic

j1

loop2:Vbn =1-120=(j0.1+ Zb +j0.1)I b +(j0.1)I n =(j1.2)I b + I n =(j0.1)I a +(j1.2)I b +(j0.1)Ic   Ia  I b  Ic

j1

loop3:Vcn =1120=(j0.1+ Zb +j0.1)Ic +(j0.1)I n =(j1.1)Ic +(j0.1) I n =(j0.1)Ia +(j0.1)I b +(j1.1)Ic   Ia  I b  Ic

j0.9

 j1.2 j0.1 j0.1  Ia   1   Ia   0.9123-90.3507            j0.1 j1.2 j0.1  I b   1-120   I b   0.9123150.3507   j0.1 j0.1 j1.1  Ic   1120   Ic   0.992930 

10

10

method 2 Zan =j0.1+Za =j0.1+j1=j1.1;Zbn  =j0.1+Zb =j0.1+j1=j1.1;Zcn =j0.1+Zc =j0.1+j0.9=j1;Znn  =j0.1; (

1 1 1 1 1 1 1    )Vn   Van  Vbn  Vcn  0 Zan  Zbn  Zcn  Znn  Zan  Zbn Zcn 

1 1 1 1 1 1 1    )Vn   Van  Vbn  Vcn  0 j1.1 j1.1 j1 j0.1 j1.1 j1.1 j1 -j0.1 Vn   Vcn  0.00355  j0.00615  0.0071120 -j14.1 V  Vn  1  0.0071120 1.0036-0.351 Ia  an    0.9123-90.351(7%)  0.0056  j0.912 Zan j1.1 1.190

(

Ib  Ic 

Vbn  Vn 1  120  0.0071120 1.0036 -119.649  0.9123150.351(7%)  0.793  j0.45   1.190 Zbn j1.1 Vcn  Vn  1120  0.0071120 0.9929 120    0.9929 30(7%)  0.86  j0.496 Zcn j1 190

I n  Ia  I b  Ic  0.0709 30 In =

Vn  0.0071120 = =0.071 30 j0.1 j0.1 j0.1

Ia

a

+

loop1

10

1120 c

n

j0.1

In

+ 1  120

Za

+

b

loop2

Ib IC

n

loop3 j0.1

Zb

Zc

j0.1

Figure P2.8

node n: I n  Ia  I b  Ic

=================

11

11

(20%) 不平衡Y接負載 [105新增] 3. The system shown in Figure 3 is unbalanced. (a) Find the currents Ia, Ib, and Ic . (b) Find the short circuit current Ign when terminals g nare short-circuited. (20%) Ia

a

j

1

a

+

11

120 _ 0

120120

_

_

g

+

_

Vng

120-120

-j13

+

c

b

11

n

b

Ib

1

j

Ic

1

j

(sol)Node analysis 1 1 1 1 1 1 a. [ + + ]Vng  Vag  Vbg  Vcg =0 12 12-j12 12+j12 Zan Zbn Zcn

-j 11

+

c

g

_

Vag -Vng

120-81.960 =3.170(5%) 12 12 Vbg -Vng 120-120-81.960 = =10.37-98.79(5%) I bb = 12-j12 12-j12 Vcg -Vng 120120-81.960 Icc = =  10.3798.79(5%) 12+j12 12+j12 Iaa  =

=

1

j a  11

b Ib

1

j

b 11

-j13

1

j c 11

j11



120  120

_

c Ic



120120

_ Vng 

Zth

b.將ng短路時Ign =? E th =Vng,OC =81.960  ng兩端的戴維寧等效電路    Z th  [12 //(12  j12) //(12  j12)]  6  81.960 ng短路時Ign =-I ng ==13.66180(A)(5%) 6 另解(a) method2 use loop analysis loop1:Vag -Vbg =1200  120  120=(1+j+11-j1)Ia -(1+j+11-j13)I b

 6 E th 81.96 _

Vag -Vbg =120 330=12Ia -(12-j12)I b loop2:Vbg -Vcg =120  120  120120=(1+j+11-j13)I b -(1+j+11+j11)Ic Vbg -Vcg =120 3  90=(12-j12)I b -(12+j12)Ic 0   Ia   3.170  1 1 1   Ia    12 -(12-j12)   I   120 330    I   10.37-98.79 0   b    b     0 (12-j12) -(12+j12)   Ic  120 3  90  Ic   10.3798.79   

12

12

-j

a Ia

 1200

1 1 1 120 120-120 120120 0 + + ]Vng  12 12-j12 12+j12 12 12-j12 12+j12 _ Vng =81.96...