POWER SYSTEM ANALYSIS AND DESIGN 7TH EDITION BY J. DUNCAN CLOVER, THOMAS J, OVERBYE, MULUKUTLA S. SA by QuentinAxel

AN INSTRUCTOR’S SOLUTIONS MANUAL TO ACCOMPANY

POWER SYSTEM ANALYSIS AND DESIGN SEVENTH EDITION J. DUNCAN GLOVER THOMAS J. OVERBYE MULUKUTLA S. SARMA Adam B. Birchfield

Contents Chapter 2............................................................................................................................. 1 Chapter 3........................................................................................................................... 24 Chapter 4........................................................................................................................... 63 Chapter 5........................................................................................................................... 85 Chapter 6......................................................................................................................... 125 Chapter 7......................................................................................................................... 160 Chapter 8......................................................................................................................... 166 Chapter 9.........................................................................................................................185 Chapter 10....................................................................................................................... 219 Chapter 11....................................................................................................................... 293 Chapter 12....................................................................................................................... 314 Chapter 13....................................................................................................................... 333 Chapter 14....................................................................................................................... 346 Chapter 15....................................................................................................................... 369

Chapter 2 Fundamentals

2.1

(a) 𝐴1̅ = 9∠30° = 9[𝑐𝑜𝑠 30° + 𝑗 𝑠𝑖𝑛 30°] = 7.8 + 𝑗4.5 5

(b) 𝐴̄2 = 4 + 𝑗5 = √16 + 25 ∠ 𝑡𝑎𝑛−1 4 = 6.40∠128.66° = 6.40𝑒 𝑗128.66° (c) 𝐴̅3 = (7.8 + 𝑗4.5) + (−4 + 𝑗5) = (11.8 + 𝑗9.5) = 15.15∠38.8° (d) ̅̅̅ 𝐴4 = (9∠30°)(6.4∠51.34°) = 57.6∠81.34° = 8.673 + 𝑗56.9 (e) ̅̅̅ 𝐴5 = 2.2

9∠30° = 1.41∠81.34° = 1.41𝑒 𝑗81.34° 6.4∠−52.34°

(a) 𝐼̄ = 500∠ − 30° = 433.01 − 𝑗250 (b) 𝑖(𝑡) = 4𝑠𝑖𝑛(𝜔𝑡 + 15°) = 4𝑐𝑜𝑠(𝜔𝑡 + 15° − 90°) = 4co s(𝜔𝑡 − 75°) 𝐼̅ =

4 ∠ − 75° = 2.83∠ − 75° = 0.73 − 𝑗2.73 √2

13 𝑉̅2 = 2 ∠(−(125 × 10−6 )(2𝜋60)) kV = 9.19∠ − 2.7° kV

2.4

(a) I1 = 100

√

− j6 6 − 90 = 10 = 7.5 − 90 A 8 + j6 − j6 8

I 2 = I − I1 = 100 − 7.3 − 90 = 10 + j 7.5 = 12.536.87 A V = I 2 ( − j 6 ) = (12.536.87 ) ( 6 − 90 ) = 75 − 53.13 V

(b)

2.5

(a)  (t ) = 277 2 cos ( t + 30) = 391.7cos ( t + 30 ) V ̅ 𝑉

(b) 𝐼 ̅ = 45 = 6.155∠30° A 𝑖(𝑡) = 8.7 cos (𝜔𝑡 + 30°) A (c) Z = j L = j ( 2 60 ) (10  10 −3 ) = 3.77190 

I = V Z = ( 277 30 ) ( 3.77190 ) = 73.46  − 60 A i(t ) = 73.46 2 cos (t − 60 ) =103.9cos (t − 60 ) A (d) Z = − j 25  I = V Z = ( 27730 ) ( 25 − 90 ) = 11.08120 A i(t ) = 11.08 2 cos ( t + 120 ) = 15.67cos ( t + 120 ) A

2.6

(a) 𝑉̄ = ( 2) ∠ − 15° = 53.03∠ − 15°;  does not appear in the answer. √

(b) 𝜐(𝑡) = 50√2 𝑐𝑜𝑠(𝜔𝑡 + 10°); with  = 377, 𝜐(𝑡) = 70.71 𝑐𝑜𝑠(377𝑡 + 10°) (c) 𝐴̄ = 𝐴∠𝛼; 𝐵̄ = 𝐵∠𝛽; 𝐶̄ = 𝐴̄ + 𝐵̄ c(t ) = a(t ) + b(t ) = 2 Re Ce jt 

The resultant has the same frequency . 2.7

(a) The circuit diagram is shown below:

(b) 𝑍̄ ̶ = 3 + 𝑗8 − 𝑗4 = 3 + 𝑗4 = 5∠53.1° 𝛺 (c) I = (1000 ) ( 553.1 ) = 20 − 53.1 A The current lags the source voltage by 53.1

Power Factor = cos53.1 = 0.6 Lagging

2.8

Z LT = j ( 377 ) ( 30.6  10 −6 ) = j11.536 m  Z LL = j ( 377 ) ( 5  10 −3 ) = j1.885  ZC = − j V=

1 = − j 2.88  ( 377 ) ( 921  10−6 )

120 2 2

 − 30 V

The circuit transformed to phasor domain is shown below:

2.9

KVL : 1200 = ( 600 )( 0.1 + j 0.5 ) + VLOAD  VLOAD = 1200 − ( 600 )( 0.1 + j 0.5 ) = 114.1 − j 30.0 = 117.9 − 14.7 V 

2.10 (a) p(t ) =  (t )i (t ) =  400cos ( t + 30 )  100cos ( t − 80 )

1 ( 400 )(100 ) cos110 + cos ( 2 t − 50 ) 2 = −6840.4 + 2  104 cos ( 2 t − 50 ) W =

(b) P = VI cos ( −  ) = ( 282.84 )( 70.71) cos ( 30 + 80 ) = −6840 W Absorbed = +6840 W Delivered

(d) The phasor current (−𝐼̄) = 70.71∠ − 80° + 180° = 70.71 ∠100° A leaves the positive terminal of the generator. The generator power factor is then 𝑐𝑜𝑠(30° − 100°) = 0.3420 leading 2.11 (a) p(t ) =  (t )i(t ) = 391.7  19.58cos 2 ( t + 30 ) 1 = 0.7669  10 4   1 + cos ( 2 t + 60 )  2 = 3.834  103 + 3.834  10 3 cos ( 2 t + 60 ) W P = VI cos ( −  ) = 277  13.85cos0 = 3.836 kW Q = VI sin ( −  ) = 0 VAR Source Power Factor = cos ( −  ) = cos ( 30 − 30 ) = 1.0

(b) p(t ) =  (t )i(t ) = 391.7  103.9cos ( t + 30 ) cos ( t − 60 ) 1 = 4.07  10 4   cos90 + cos ( 2 t − 30 )  2 = 2.035  10 4 cos ( 2 t − 30 ) W P = VI cos ( −  ) = 277  73.46 cos ( 30 + 60 ) = 0 W Q = VI sin ( −  ) = 277  73.46sin 90 = 20.35 kVAR pf = cos ( −  ) = 0 Lagging

(c) p(t ) =  (t )i(t ) = 391.7  15.67cos ( t + 30 ) cos ( t + 120 ) 1 = 6.138  103   cos ( −90 ) + cos ( 2 t + 150 )  = 3.069  10 3 cos ( 2 t + 150 ) W 2 P = VI cos ( −  ) = 277  11.08cos ( 30 − 120 ) = 0 W Q = VI sin ( −  ) = 277  11.08sin ( −90 ) = −3.069 kVAR Absorbed = +3.069 kVAR Delivered pf = cos ( −  ) = cos ( −90 ) = 0 Leading

2.12 (a) pR (t ) = ( 359.3cos  t )( 35.93cos  t )

= 6455 + 6455cos2 t W (b) px (t ) = ( 359.3cos  t ) 14.37cos ( t + 90 )  = 2582 cos ( 2 cot + 90 ) = −2582sin 2 t W

(

) 10 = 6455W Absorbed

(

) 25 = 2582 VARS Delivered

(e) (  −  ) = tan −1 ( Q / P ) = tan −1 ( 2582 6455 ) = 21.8 Power factor = cos ( −  ) = cos ( 21.8 ) = 0.9285 Leading

2.13

Z = R − jxc = 10 − j 25 = 26.93  − 68.2 

i(t ) = ( 359.3 / 26.93 ) cos ( t + 68.2 ) = 13.34 cos ( t + 68.2 ) A

(a) pR (t ) = 13.34 cos ( t + 68.2 )  133.4cos ( t + 68.2 ) = 889.8 + 889.8cos 2 ( t + 68.2 )  W

(b) px (t ) = 13.34 cos ( t + 68.2 )  333.5cos ( t + 68.2 − 90 ) = 2224sin 2 ( t + 68.2 )  W

(

2 10 = 889.8 W

(

2 25 = 2224 VAR S

)

(e) pf = cos  tan −1 ( Q / P )  = cos  tan −1 (2224 / 889.8)  = 0.3714 Leading

2.14 (a) 𝐼̄ = 2∠0°𝑘𝐴 V = Z I = ( 3 − 45 )( 20 ) = 6 − 45 kV

 (t ) = 6 2 cos ( t − 45 ) kV p (t ) =  (t )i (t ) = 6 2 cos ( t − 45 )   2 2 cos  t  1 = 24   cos ( −45 ) + cos ( 2 t − 45 )  2 = 8.49 + 12cos ( 2 t − 45 ) MW

(b) 𝑃 = 𝑉𝐼 𝑐𝑜𝑠(𝛿 − 𝛽) = 6 × 2 𝑐𝑜𝑠(−45° − 0°) = 8.49 𝑀𝑊 Delivered (c) Q = VI sin ( −  ) = 6  2sin ( −45 − 0 )

= −8.49MVAR Delivered = + 8.49MVAR Absorbed (d) pf = cos ( −  ) = cos ( −45 − 0 ) = 0.707 Leading

(

2.15 (a) I =  4 

)

2 60 ( 230 ) = 2 30 A 

i(t ) = 2 cos ( t + 30 ) A with  = 377 rad/s p(t ) =  (t )i(t ) = 4 cos30 + cos ( 2 t + 90 )  = 3.46 + 4 cos ( 2 t + 90 ) W

(b) (t), i(t), and p(t) are plotted below. (c) The instantaneous power has an average value of 3.46 W, and the frequency is twice that of the voltage or current.

2.16 (a) 𝑍̄ ̶ = 10 + 𝑗120𝜋 × 0.04 = 10 + 𝑗15.1 = 18.1∠56.4° 𝛺

pf = cos56.4 = 0.553 Lagging (b) V = 120 0 V The current supplied by the source is I = (120 0 ) (18.156.4 ) = 6.63 − 56.4 A

The real power absorbed by the load is given by

P = 120  6.63  cos56.4 = 440 W which can be checked by I 2 R = ( 6.63) 10 = 440 W 2

The reactive power absorbed by the load is Q = 120 x 6.63 x sin56.4° = 663 VAR (c) Peak Magnetic Energy = W = LI 2 = 0.04 ( 6.63 ) = 1.76 J 2

Q = W = 377  1.76 = 663VAR is satisfied. ̶ ̄|2 = 𝑗𝜔𝐿𝐼 2 2.17 (a) 𝑆̄ = 𝑉̄ 𝐼̄∗ = 𝑍̄𝐼̶ ̄𝐼̄∗ = 𝑍̄|𝐼

Q = Im[ S ] =  LI 2  (b)  (t ) = L

di = − 2 L I sin ( t +  ) dt

p(t ) =  (t )  i(t ) = −2 L I 2 sin (t +  ) cos (t +  ) = − L I 2 sin 2 (t +  )  = − Q sin 2 (t +  ) 

Average real power P supplied to the inductor = 0 

Instantaneous power supplied (to sustain the changing energy in the magnetic field) has a maximum value of Q.  2.18 (a) S = V I * = Z I I * = Re  Z I 2  + j Im  Z I 2 

= P + jQ P = Z I cos Z ; Q = Z I 2 sin Z  2

(b) Choosing i(t ) = 2 I cos  t , Then  (t ) = 2 Z I cos (t + Z )  p(t ) =  (t )  i(t ) = Z I 2 cos ( t + Z )  cos t = Z I 2  cos Z + cos ( 2 t + Z ) 

= Z I 2  cos Z + cos2 t cos Z − sin 2 t sin Z  = P (1 + cos2 t ) − Q sin 2 t  1

where 𝑄𝐿 = 𝜔𝐿𝐼 2 and 𝑄𝐶 = − 𝜔𝐶 𝐼 2 which are the reactive powers into L and C, respectively. Thus p(t ) = P (1 + cos2 t ) − QL sin 2 t − QC sin 2 t  If  2 LC = 1, Then

QL + QC = Q = 0

    p(t ) = P (1 + cos2t )   ∗

150 5 2.19 (a) 𝑆̄ = 𝑉̄ 𝐼̄∗ = ( ∠10°) ( ∠ − 50°) = 375∠60° √2

√2

= 187.5 + 𝑗324.8 P = Re S = 187.5 W Absorbed Q = Im S = 324.8 VAR SAbsorbed

(b) pf = cos ( 60 ) = 0.5 Lagging (c) QS = P tan QS = 187.5tan cos−1 0.9 = 90.81VAR S

QC = QL − QS = 324.8 − 90.81 = 234 VAR S

2.20

Y1 =

1 1 = = 0.05 − 30 = ( 0.0433 − j 0.025 ) S = G1 − jB1 Z1 2030

Y2 =

1 1 = = 0.04 − 60 = ( 0.02 − j 0.03464 ) S = G2 + jB2 Z 2 2560

P1 = V 2 G1 = (100 ) 0.0433 = 433 W Absorbed 2

Q1 = V 2 B1 = (100 ) 0.025 = 250 VAR S Absorbed 2

P2 = V 2 G2 = (100 ) 0.02 = 200 W Absorbed 2

Q2 = V 2 B2 = (100 ) 0.03464 = 346.4 VAR SAbsorbed 2

2.21 (a)

L = cos−1 0.6 = 53.13 QL = P tan L = 500 tan 53.13 = 666.7 kVAR S = cos−1 0.9 = 25.84 QS = P tan S = 500 tan 25.84 = 242.2 kVAR QC = QL − QS = 666.7 − 242.2 = 424.5 kVAR SC = QC = 424.5 kVA

(b) The Synchronous motor absorbs Pm =

( 500 ) 0.746 = 414.4 kW and Q = 0 kVAR 0.9

Source PF = cos tan −1 ( 666.7 914.4 )  = 0.808 Lagging

1 𝑍1

2.22 (a) 𝑌̄1 = ̄ ̶ = (4+𝑗5) =

1 = 0.16∠ − 51.34° 6.4∠51.34°

= (0.1 − 𝑗0.12)𝑆 1 1 𝑌̄2 = 𝑍̄ ̶ = 10 = 0.1𝑆 2

𝑃 1000 𝑃 = 𝑉 2 (𝐺1 + 𝐺2 ) ⇒ 𝑉 = √ =√ = 70.71 V (0.1 + 0.1) 𝐺1 + 𝐺2 𝑃1 = 𝑉 2 𝐺1 = (70.71)2 0.1 = 500𝑊 𝑃2 = 𝑉 2 𝐺2 = (70.71)2 0.1 = 500𝑊 (b) 𝑌̄𝑒𝑞 = 𝑌̄1 + 𝑌̄2 = (0.1 − 𝑗0.12) + 0.1 = 0.2 − 𝑗0.12 = 0.233∠ − 30.96° 𝑆 𝐼𝑆 = 𝑉 𝑌𝑒𝑞 = 70.71(0.233) = 16.48𝐴 2.23

𝑆̅ = 𝑉̅ 𝐼 ∗̅ = (120)(12∠ − 30°) = 1440∠ − 30° = 1247 − 𝑗720 𝑃 = 𝑅𝑒(𝑆̅) = 1247 𝑊 Delivered 𝑄 = 𝐼𝑚𝑎𝑔(𝑆̅) = 720 var Absorbed

2.24 (a) Apparent power = S = (120 V)(8 A) = (960 VA). Real power = P = S * pf = 960 * 0.96 = 825.6 W. Reactive power = Q = √𝑆 2 − 𝑃2 = √9602 − 825.62 = −490 var. (b) R = 𝑉 2 /𝑃 = (120 V)2 /825.6W = 17.442 Ω, 𝑉2

(120 V)2

𝑋𝑐 = 𝑄 = 490 var = 29.39 Ω. 𝐶= 2.25

1 1 = = 90.2 𝜇F 𝜔𝑍𝑐 2𝜋60⋅29.39 Ω 𝑉 −𝑉2 100−100∠𝜃 = = 17.68 sin 𝜃 + 𝑗(−17.68 + 1.77 cos 𝜃). 𝑍 𝑗(2⋅𝑝𝑖⋅60)(0.015)

𝐼= 1

𝑆 = 𝑉𝐼 ∗ = 1768 sin 𝜃 + 𝑗 (1768 − 177 cos 𝜃). 𝑀𝑎𝑥(𝑅𝑒𝑎𝑙(𝑆)) = 1768 W. 2.26 (a) The problem is modeled as shown in figure below: PL = 120 kW pfL = 0.85Lagging

 L = cos−1 0.85 = 31.79 Power triangle for the load:

SL = PL + jQL = 141.1831.79 kVA

QL = PL tan ( 31.79 )

I = SL / V = 141,180 / 480 = 294.13A

= 74.364 kVAR

Real power loss in the line is zero. Reactive power loss in the line is 𝑄𝐿𝐼𝑁𝐸 = 𝐼 2 𝑋𝐿𝐼𝑁𝐸 = (294.13)2 1 = 86.512 kVAR

 SS = PS + jQS = 120 + j ( 74.364 + 86.512 ) = 200.753.28 kVA

The input voltage is given by VS = SS / I = 682.4 V(rms) The power factor at the input is cos53.28 = 0.6 Lagging (b) Applying KVL, 𝑉̄𝑆 = 480∠0° + 𝑗1.0(294.13∠ − 31.79°) = 635 + j 250 = 682.421.5 V (rms) ( pf )S = cos ( 21.5 + 31.79 ) = 0.6 Lagging

2.27 The circuit diagram is shown below:

Pold = 50 kW; cos−1 0.8 = 36.87; OLD = 36.87; Qold = Pold tan (old ) = 37.5kVAR ∴ 𝑆̄𝑜𝑙𝑑 = 50,000 + 𝑗37,500

 new = cos−1 0.95 = 18.19; Snew = 50,000 + j 50,000 tan (18.19 ) = 50,000 + j16,430 Hence Scap = Snew − Sold = − j 21,070 VA

C = 2.28

21,070

( 377)( 220 )

= 1155 F 

S1 = 15 + j 6.667 S 2 = 3 ( 0.96 ) − j 3 sin ( cos −1 0.96 )  = 2.88 − j 0.84 S3 = 15 + j 0 STOTAL = S1 + S 2 + S3 = ( 32.88 + j 5.827 ) kVA

(i) Let 𝑍̄ ̶ be the impedance of a series combination of R and X

𝑉̄ ∗

𝑉2

Since 𝑆̄ = 𝑉̄ 𝐼̄∗ = 𝑉̄ (𝑍̄ )̶ = 𝑍̄ ∗̶ , it follows that 𝑍̄ ∗̶ =

(240)2 𝑉2 = = (1.698 − 𝑗0.301) 𝛺 (32.88 + 𝑗5.827)103 𝑆̄

∴ 𝑍̄ ̶ = (1.698 + 𝑗0.301)𝛺

←

(ii) Let 𝑍̄ ̶ be the impedance of a parallel combination of R and X

( 240 ) = 1.7518  R= ( 32.88)103 2 240 ) ( X= = 9.885  ( 5.827 )103 2

Then

 Z = (1.7518 j 9.885 )  

2.29 Since complex powers satisfy KCL at each bus, it follows that 𝑆̄13 = (1 + 𝑗1) − (1 − 𝑗1) − (0.4 + 𝑗0.2) = −0.4 + 𝑗1.8 ∗ 𝑆̄31 = −𝑆̄13 = 0.4 + 𝑗1.8

←

Similarly, 𝑆̄23 = (0.5 + 𝑗0.5) − (1 + 𝑗1) − (−0.4 + 𝑗0.2) = −0.1 − 𝑗0.7 ∗ 𝑆̄32 = −𝑆̄23 = 0.1 − 𝑗0.7

←

At Bus 3, 𝑆̄𝐺3 = 𝑆̄31 + 𝑆̄32 = (0.4 + 𝑗1.8) + (0.1 − 𝑗0.7) = 0.5 + 𝑗1.1

←

2.30 (a) For load 1: 1 = cos−1 (0.28) = 73.74 Lagging

S1 = 12573.74 = 35 + j120 S2 = 10 − j 40 S3 = 15 + j 0 STOTAL = S1 + S2 + S3 = 60 + j80 = 10053.13 kVA = P + jQ  PTOTAL = 60 kW; QTOTAL = 80 kVAR; kVA TOTAL = STOTAL = 100 kVA.  Supply pf = cos ( 53.13 ) = 0.6 Lagging 

(b) ITOTAL =

S * 100  103  − 53.13 = = 100 − 53.13 A V* 10000

At the new pf of 0.8 lagging, PTOTAL of 60kW results in the new reactive power 𝑄 ′ , such that

  = cos−1 ( 0.8 ) = 36.87 and Q = 60 tan ( 36.87 ) = 45 kVAR ∴The required capacitor’s kVAR is QC = 80 − 45 = 35 kVAR  𝑉2

(1000)2

It follows then 𝑋𝐶 = 𝑆̄ ∗ = 𝑗35000 = −𝑗28.57 𝛺 𝐶

and

106 = 92.85 F  2 ( 60 )( 28.57 ) 𝑆̄ ′∗

The new current is 𝐼 ′ = 𝑉̄ ∗ =

60,000−𝑗45,000 = 60 − 𝑗45 = 75 − 36.87 A 1000∠0°

The supply current, in magnitude, is reduced from 100A to 75A  𝑉 ∠𝛿1 −𝑉2 ∠𝛿2 𝑉 𝑉 = ( 1 ∠𝛿1 − 90°) − 2 ∠𝛿2 − 90° 𝑋∠90° 𝑋 𝑋

2.31 (a) 𝐼̄12 = 1

V V  Complex power S12 = V1 I12* = V11  1 90 − 1 − 2 90 −  2  X X  2 V VV = 1 90 − 1 2 90 + 1 −  2 X X ∴ The real and reactive power at the sending end are

P12 =

Q12 =

V12 VV cos90 − 1 2 cos ( 90 + 1 −  2 ) X X VV = 1 2 sin (1 −  2 )  X

V12 VV sin 90 − 1 2 sin ( 90 + 1 −  2 ) X X V = 1 V1 − V2 cos (1 −  2 )   X

Note: If 𝑉̄1 leads 𝑉̄2 , 𝛿 = 𝛿1 − 𝛿2 is positive and the real power flows from node 1 to node 2. If 𝑉̄1 Lags 𝑉̄2 , 𝛿 is negative and power flows from node 2 to node 1. (b) Maximum power transfer occurs when 𝛿 = 90° = 𝛿1 − 𝛿2

PMAX =

←

V1V2  X

2.32 4 Mvar minimizes the real power line losses, while 4.5 Mvar minimizes the MVA power flow into the feeder.

2.33

2.34

Qcap

MW Losses

Mvar Losses

0.42

0.84

0.5

0.4

0.8

0.383

0.766

1.5

0.369

0.738

0.357

0.714

2.5

0.348

0.696

0.341

0.682

3.5

0.337

0.675

0.336

0.672

4.5

0.337

0.675

0.341

0.682

5.5

0.348

0.696

0.357

0.714

6.5

0.369

0.738

0.383

0.766

7.5

0.4

0.801

0.42

0.84

8.5

0.442

0.885

0.467

0.934

9.5

0.495

0.99

0.525

1.05

7.5 Mvars

2.35

(.3846 + .4950 ) + j (10 − 1.923 − 4.950 )  − (.4950 − j 4.950 ) 

− (.4950 − j 4.950 )

 V10  (.3846 + .4950 ) + j (10 − 1.923 − 4.95 ) V20  1.961 − 48.69 =  1.961 − 78.69 

[

0.8796 + 𝑗3.127 −0.4950 + 𝑗4.950 𝑉̄10 1.961∠ − 48.69° ][ ] = [ ] −0.4950 + 𝑗4.950 0.8796 + 𝑗3.127 𝑉̄20 1.961∠ − 78.69°

2.36 Note that there are two buses plus the reference bus and one line for this problem. After converting the voltage sources in Fig. 2.29 to current sources, the equivalent source impedances are: Z S1 = Z S 2 = ( 0.1 + j 0.5 ) // ( − j 0.1) = =

( 0.1 + j 0.5)( − j 0.1) 0.1 + j 0.5 − j 0.1

( 0.509978.69)( 0.1 − 90 ) = 0.1237 − 87.27

0.412375.96 = 0.005882 − j 0.1235 

The rest is left as an exercise to the student. 2.37 After converting impedance values in Figure 2.30 to admittance values, the bus admittance matrix is: 1

−1 0 0 1 1 1 1 1 −1 (1 + + + − 𝑗1) − ( − 𝑗1) −( ) 2 3 4 3 4 1 1 1 1 1 𝑌̄𝑏𝑢𝑠 = 0 − ( − 𝑗1) ( − 𝑗1 + 𝑗 + 𝑗 ) − (𝑗 ) 3 3 4 2 4 1 1 1 1 1 0 −( ) − (𝑗 ) ( +𝑗 −𝑗 ) [ 4 4 4 4 3 ] Writing nodal equations by inspection: 𝑉̄10 −1 0 0 1 1∠0° −0.25 𝑉̄20 −1 (2.083 − 𝑗1) (−0.3333 + 𝑗1) 0 =[ [ (−0.3333 ] ] −𝑗0.25 + 𝑗1)(0.3333 − 𝑗0.25) 0 0 𝑉̄30 (0.25 − 𝑗0.08333) [𝑉̄ ] −𝑗0.25 (−0.25) 0 2∠30° 40

2.38 The admittance diagram for the system is shown below:

𝑌̄11 𝑌̄ 𝑌̄𝐵𝑈𝑆 = 21 𝑌̄31 [𝑌̄41 where

𝑌̄12 𝑌̄22 𝑌̄32 𝑌̄42

𝑌̄13 𝑌̄23 𝑌̄33 𝑌̄43

𝑌̄14 0 −8.5 2.5 5.0 𝑌̄24 0 2.5 −8.75 5.0 = 𝑗[ ]𝑆 ̄𝑌34 12.5 5.0 5.0 −22.5 0 0 12.5 −12.5 𝑌̄44 ]

𝑌̄11 = 𝑦̄10 + 𝑦̄12 + 𝑦̄13 ; 𝑌̄22 = 𝑦̄ 20 + 𝑦̄12 + 𝑦̄ 23 ; 𝑌̄33 = 𝑦̄13 + 𝑦̄ 23 + 𝑦̄ 34 𝑌̄44 = 𝑦34 ; 𝑌̄12 = 𝑌̄21 = −𝑦̄12 ; 𝑌̄13 = 𝑌̄31 = −𝑦̄13 ; 𝑌̄23 = 𝑌̄32 = −𝑦̄ 23

and

𝑌̄34 = 𝑌̄43 = −𝑦̄ 34

2.39 (a) 𝑌̄𝑐 + 𝑌̄𝑑 + 𝑌̄𝑓 −𝑌̄𝑑 𝑉̄1 𝐼̄1 = 0 −𝑌̄𝑓 −𝑌̄𝑐 ̄ ̄ ̄ ̄ ̄ −𝑌𝑑 𝑌𝑏 + 𝑌𝑑 + 𝑌𝑒 𝑉2 𝐼̄ = 0 −𝑌̄𝑏 −𝑌̄𝑒 = 2 𝑉̄3 𝐼̄3 −𝑌̄𝑐 −𝑌̄𝑏 0 𝑌̄𝑎 + 𝑌̄𝑏 + 𝑌̄𝑐 ̄ ̄ ̄ 𝑌𝑒 + 𝑌𝑓 + 𝑌𝑔 ] [𝑉̄4 ] [ 𝐼̄4 ] −𝑌̄𝑓 −𝑌̄𝑒 0 [ 0 −14.5 8 4 2.5 𝑉̄1 ̄ 𝑉 0 8 −17 4 5 (b) 𝑗 [ ] ̄2 = [ ] 4 −8.8 0 1∠ − 90° 4 𝑉3 0.62∠ − 135° 2.5 5 0 −8.3 [𝑉̄4 ] −1 ̄ −1 ̄ 𝑌̄𝐵𝑈𝑆 𝑉̄ = 𝐼̄; 𝑌̄𝐵𝑈𝑆 𝑌𝐵𝑈𝑆 𝑉̄ = 𝑌̄𝐵𝑈𝑆 𝐼

where

0.71870.66880.63070.6194 0.66880.70450.62420.6258 −1 ̶ 𝑌̄𝐵𝑈𝑆 = 𝑍̄𝐵𝑈𝑆 = 𝑗[ ] 𝛺 0.63070.70450.68400.5660 0.61940.62580.56600.6840

−1 ̄ 𝑉̄ = 𝑌̄𝐵𝑈𝑆 𝐼

𝑉̄1 0 ̄2 𝑉 0 𝑉̄ = and 𝐼̄ = [ ] ̄𝑉3 1∠ − 90° 0.62∠ − 135° [𝑉̄4 ]

where

Then solve for 𝑉̄1, 𝑉̄2 , 𝑉̄3 , and 𝑉̄4 . 2.40 (a) 𝑉̅𝑎𝑛 =

240 ∠0° = 138.56∠0° V (Assumed as reference) √3

𝑉̅𝑎𝑏 = 240∠30°V; 𝑉̅𝑏𝑐 = 240∠ − 90°V; 𝐼𝐴̅ = 12∠ − 90°A ̅

𝑉 138.56∠0° 𝑍̅𝑌 = 𝐼𝐴𝑁 = 12∠−90° = 11.55∠90° = 0 + 𝑗11.55 Ω ̅ 𝐴

𝐼̅

√

√(3)

̅ = 𝐴 ∠30° = (b) 𝐼𝑎𝑏 3

∠(−90° + 30°) = 6.93∠ − 60°A

𝑉 240∠30° 𝑍̅Δ = 𝐼 ̅ 𝐴𝐵 = 6.93∠−60°A = 34.6∠90° = (0 + 𝑗34.6)Ω 𝐴𝐵

Note: 𝑍̄𝑌 = 𝑍̄𝛥̶ /3 2.41

S3 = 3VLL I L  cos −1 ( pf ) = 3 ( 480 )( 20 )  cos −1 0.8 = 16.627  10 3 36.87 = (13.3  10 3 ) + j (9.976  10 3 )

P3 = Re S3 = 13.3 kW Delivered Q3 = I m S3 = 9.976 kVAR Delivered 2.42 (a) With 𝑉̄𝑎𝑏 as reference

𝑍̄𝛥̶ = 4 + 𝑗3 = 5∠36.87° 𝛺 3

208 𝑉̄𝑎𝑛 = 3 ∠ − 30° √

Ia =

Van 120.1 − 30 = = 24.02 − 66.87 A ( Z / 3) 536.87

𝑆̄3𝜑 = 3𝑉̄𝑎𝑛 𝐼̄𝑎∗ = 3(120.1∠ − 30°)(24.02∠ + 66.87°) = 8654∠36.87° = 6923 + 𝑗5192 P3 = 6923 W; Q3 = 5192 VAR; both absorbed by the load

pf = cos ( 36.87 ) = 0.8 Lagging; S3 = S3 = 8654 VA (b)

Vab = 2080 V

I a = 24.02 − 66.87 A

13.87 − 36.87 A

2.43 (a) Transforming the -connected load into an equivalent Y, the impedance per phase of the equivalent Y is 𝑍̄2̶ =

60 − 𝑗45 = (20 − 𝑗15) 𝛺 3

With the phase voltage V1 = 1203 3 = 120 V taken as a reference, the per-phase equivalent circuit is shown below:

Total impedance viewed from the input terminals is Z = 2 + j4 + I=

( 30 + j 40 )( 20 − j15 ) = 2 + j 4 + 22 − j 4 = 24  ( 30 + j 40 ) + ( 20 − j15 )

V1 1200 = = 50 A Z 24

The three-phase complex power supplied = S = 3V1 I * = 1800 W P = 1800 W and Q = 0VAR delivered by the sending-end source

(b) Phase voltage at load terminals 𝑉̄2 = 120∠0° − (2 + 𝑗4)(5∠0°)

= 110 − j 20 = 111.8 − 10.3 V The line voltage magnitude at the load terminal is

(VLOAD )L -L = 3 111.8 = 193.64 V (c) The current per phase in the Y-connected load and in the equiv.Y of the -load: I1 =

V2 = 1 − j 2 = 2.236 − 63.4 A Z1

I2 =

V2 = 4 + j 2 = 4.472 26.56 A Z2

The phase current magnitude in the original -connected load

= 2.582 A ( I ) = I 3 = 4.472 3 2

ph 

(d) The three-phase complex power absorbed by each load is

S1 = 3V2 I1* = 450 W + j 600VAR S2 = 3V2 I 2* = 1200 W − j 900VAR The three-phase complex power absorbed by the line is SL = 3 ( RL + jX L ) I 2 = 3 ( 2 + j 4 ) (5)2 = 150 W + j 300 VAR

The sum of load powers and line losses is equal to the power delivered from the supply:

S1 + S2 + SL = ( 450 + j 600 ) + (1200 − j 900 ) + (150 + j 300 ) = 1800 W+ j 0 VAR 2.44

𝑉𝑎𝑛 = 𝑉∠0° because load is resistive. So 𝑉𝑏𝑛 = 𝑉∠ − 120°, 𝑉𝑐𝑛 = 𝑉∠ + 120°. 𝑉𝑏𝑐 = 𝑉𝑏𝑛 − 𝑉𝑐𝑛 = −√3 𝑉∠180°. So phase angle of 𝑉𝑏𝑐 = 180°.

2.45

𝑋𝑐𝑎𝑝,𝑦 =

𝑋𝑐𝑎𝑝,𝑑𝑒𝑙𝑡𝑎

= −𝑗

1 ; 𝜔𝐶

𝑋𝑖𝑛𝑑1 = 𝜔𝐿1 ; 𝑋𝑖𝑛𝑑2 =

𝜔𝐿2 ; 3

𝑉2

𝑄= 𝑋; unity power factor requires 𝑄𝑡𝑜𝑡𝑎𝑙 = 𝑄𝑐 + 𝑄𝐿1 + 𝑄𝐿2 = 0; 1

𝑐

𝐿1

𝐿2

0=𝑋 +𝑋 +𝑋 1

= −3𝜔𝐶 + 𝜔𝐿 + 𝜔𝐿 ; 1

𝜔 = √(𝐿 + 𝐿 ) (3𝐶 ) = 1500; 𝜔 𝑓 = 2𝜋 = 239 Hz

2.46 (a)

̶̄

𝑍𝛥 /3 Using voltage division: 𝑉̄𝐴𝑁 = 𝑉̄𝑎𝑛 (𝑍 ̶ /3)+𝑍 ̶̄ 𝛥

𝐿𝐼𝑁𝐸

= =

208 3

0

1030 1030 + ( 0.8 + j 0.6 )

(120.09 )(1030) = 1200.930 9.46 + j 5.6

10.9930.62

= 109.3 − 0.62 V Load voltage = VAB = 3 (109.3) = 189.3V Line-to-Line (b)

Zeq = 10 30  || (− j 20) = 11.547 0  VAN = Van

Z eq Z eq + Z LINE

(

= 208 =

) (11.54711.547 + 0.8 + j 0.6 )

1386.7 = 112.2 − 2.78 V 12.3622.78

Load voltage Line-to-Line VAB = 3 (112.2 ) = 194.3 V 2.47

(a) I G1 =

15  103 3 ( 460 )( 0.8 )

 − cos −1 0.8 = 23.53 − 36.87 A

460

0 − (1.4 + j1.6 )( 23.53 − 36.87 ) 3 = 216.9 − 2.73 V Line to Neutral

VL = VG1− Z LINE1 I G1 =

Load Voltage VL = 3 216.9 = 375.7 V Line to line

(b) I L =

30  103 3 ( 375.7 )( 0.8 )

 − 2.73 − cos−1 0.8 = 57.63 − 39.6 A

𝐼̄𝐺2 = 𝐼̄𝐿 − 𝐼̄𝐺1 = 57.63∠ − 39.6° − 23.53∠ − 36.87° = 34.14 − 41.49 A

̶ 𝑉̄𝐺2 = 𝑉̄𝐿 + 𝑍̄𝐿𝐼𝑁𝐸2 𝐼̄𝐺2 = 216.9∠ − 2.73° + (0.8 + 𝑗1)(34.14∠ − 41.49°)

= 259.7 − 0.63 V Generator 2 line-to-line voltage 𝑉𝐺2 = √3(259.7)

= 449.8V (c) 𝑆̄𝐺2 = 3𝑉̄𝐺2 𝐼̄𝐺2∗ = 3(259.7∠ − 0.63°)(34.14∠41.49°) = 20.12 × 103 + 𝑗17.4 × 103

PG 2 = 20.12 kW; QG 2 = 17.4 kVAR; Both delivered

2.48

(a)

(b) pf = cos31.32 = 0.854 Lagging (c) I L =

SL 3VLL

26.93  103 3 ( 480 )

= 32.39 A

(d) QC = QL = 14  103 VAR = 3 (VLL ) / X  2

X =

3 ( 480 )

14  103

= 49.37 

I C = VLL / X  = 480 / 49.37 = 9.72 A

(e)

I LINE =

PL 3 VLL

23  103 3 480

= 27.66 A

2.49 (a) Let 𝑍̄𝑌̶ = 𝑍̄𝐴̶ = 𝑍̄𝐵̶ = 𝑍̄𝐶̶ for a balanced Y-load ̶ = 𝑍̄ ̶ = 𝑍̄ ̶ 𝑍̄𝛥̶ = 𝑍̄𝐴𝐵 𝐵𝐶 𝐶𝐴 Using equations in Fig. 2.27 ̄ 2̶

̄ 2̶

𝑍 +𝑍 +𝑍 𝑍̄𝛥̶ = 𝑌 𝑍̄𝑌̶ 𝑌 = 3𝑍̄𝑌̶ 𝑌

̄ 2̶

̶̄

𝑍 𝑍 𝑍̄𝑌̶ = 𝑍̄ ̶ +𝑍̄𝛥̶ +𝑍̄ ̶ = 3𝛥 𝛥

(b) Z A =

ZB =

𝛥

( j10 )( − j 25) = − j 50 

j10 + j 20 − j 25

( j10 )( j 20 ) = j 40 ; Z = ( j 20 )( − j 25) = − j 100  C

2 2.50 Replace delta by the equivalent WYE: 𝑍̄𝑌̶ = −𝑗 3 𝛺

Per-phase equivalent circuit is shown below:

Noting that (𝑗1.0 ‖−𝑗 3) = −𝑗2, by voltage-divider law, −𝑗2 𝑉̄1 = −𝑗2+𝑗0.1 (100∠0°) = 105∠0°

1 (t ) = 105 2 cos ( t + 0) = 148.5cos  t V 

In order to find 𝑖2 (𝑡) in the original circuit, let us calculate 𝑉̄𝐴′ 𝐵′ 𝑉̄𝐴′𝐵′ = 𝑉̄𝐴′ 𝑁′ − 𝑉̄𝐵′ 𝑁′ = √3𝑒 𝑗30° 𝑉̄𝐴′𝑁′ = 173.2∠30° Then

173.2∠30° 𝐼̄𝐴′ 𝐵′ = = 86.6∠120° −𝑗2

i2 (t ) = 86.6 2 cos (t + 120 ) = 122.5cos ( t + 120 ) A 

1 (150 + j120 ) = ( 50 + j 40 ) kVA 3

2.51 On a per-phase basis S1 =  I1 =

( 50 − j 40 )103 = 25 − j 20 A ( ) 2000

Note: PF Lagging

Load 2: Convert  into an equivalent Y 1 (150 − j 48 ) = ( 50 − j16 )  3 20000  I2 = = 38.117.74 50 − j16 Z 2Y =

= ( 36.29 + j 11.61) A Note: PF Leading

1 S3 per phase = (120  0.6 ) − j 120sin( cos −1 0.6 )  = ( 24 − j32 ) kVA 3  I3 =

( 24 + j32 )103 = 12 + j 16 A ( ) 2000

Note:PF Leading

Total current drawn by the three parallel loads 𝐼̄𝑇 = 𝐼̄1 + 𝐼̄2 + 𝐼̄3

ITOTAL = ( 73.29 + j 7.61) A Note:PF Leading Voltage at the sending end: 𝑉̄𝐴𝑁 = 2000∠0° + (73.29 + 𝑗7.61)(0.2 + 𝑗1.0)

= 2007.05 + j 74.81 = 2008.442.13 V 3( 2008.44 ) = 3478.62 V 

Line-to-line voltage magnitude at the sending end = 2.52 (a) Let 𝑉̄𝐴𝑁 be the reference: VAN =

2160 3

0 24000 V

Total impedance per phase 𝑍̄ ̶ = (4.7 + 𝑗9) + (0.3 + 𝑗1) = (5 + 𝑗10) 𝛺  Line Current =

24000 = 214.7 − 63.4 A = I A  5 + j10

With positive A-B-C phase sequence, I B = 214.7 − 183.4 A; I C = 214.7 − 303.4 = 214.756.6 A 

(b) (VAN ) LOAD = 24000 − ( 214.7 − 63.4 )( 0.3 + j1)  = 24000 − 224.159.9 = 2179.2 − j38.54 = 2179.5 − 1.01 V 

(V ) BN

LOAD

= 2179.5 − 121.01 V , ; (VC N ) LOAD = 2179.5 − 241.01 V ,

(c) S / Phase = (VAN )LOAD I A = ( 2179.5)( 214.7 ) = 467.94 kVA  Total apparent power dissipated in all three phases in the load

 S3  = 3 ( 467.94 ) = 1403.82 kVA  LOAD Active power dissipated per phase in load = (𝑃1𝜑 )

𝐿𝑂𝐴𝐷

= ( 2179.5 )( 214.7 ) cos ( 62.39 ) = 216.87 kW 

  P3 

LOAD

= 3 ( 216.87 ) = 650.61kW 

Reactive power dissipated per phase in load = (𝑄1𝜑 )

𝐿𝑂𝐴𝐷

= ( 2179.5 )( 214.7 ) sin ( 62.39 ) = 414.65 kVAR 

 Q3 

LOAD

= 3 ( 414.65 ) = 1243.95kVAR 

(d) Line losses per phase ( P1 )

( )

Total line loss P3

LOSS

= ( 214.7 ) 0.3 = 13.83 kW  2

LOSS

= 13.83  3 = 41.49 kW 

Chapter 3 Power Transformers 3.1

(a) Impedance as seen from primary is (221 kV / 18 A) = 12.28 kΩ. Impedances convert by the square of the turns ratio, 12.28𝐾

so 𝑎𝑡 = √ 400 = 5.540, 𝑜𝑟 100: 18.05. (b) 𝑆 = 𝑉𝐼 ∗ = (221 kV)(18 A) = 3.98 MVA on the primary side; A

on the secondary side (221 kV ∗ 0.1805) ∗ (18 0.1805) = 39,780 V ∗ 100 A = 3.978 MVA 3.2

𝑁 250 (1000∠0°) = 250∠0° 𝑉 𝑉̄2 = 2 𝑉̄1 = 𝑁1

←

1000

𝑁 1000 (5∠ − 30°) = 20∠ − 30° 𝐴 𝐼̄2 = 1 𝐼̄1 = 𝑁2

←

250

𝑍̄ 2̶ =

𝑉̄2 250∠0° = = 12.5∠30° 𝛺 𝐼̄2 20∠ − 30°

𝑁1 2 1000 2 𝑍̄ 2′̶ = 𝑍2̶ ( ) = (12.5∠30°) ( ) = 200∠30° 𝛺 𝑁2 250 (1000∠0°) Also 𝑍̄ 2′̶ = 𝑉̄1 /𝐼̄1 = (5∠−30°) = 200∠30° 𝛺

←

3.3

3.4

(a) E1 =

N1 2400 E2 = ( 230 ) = 2300 V N2 240 ∗

−1 0.8 ∗

𝑆 75×10 ∠ cos (b) 𝑆2̅ = 𝐸̅2 𝐼2∗̅ ; 𝐼2̅ = ( ̅2 ) = [ 𝐸2

𝐸 𝑍̅2 = 2̅ = 𝐼2

230

] = 326.1∠ − 36.87°

230 = 0.706∠36.87°Ω = 0.565 + 𝑗0.523 326∠−36.87°

𝑁 (c) 𝑧̅1′ = ( 1) 𝑍2̅ = 100𝑍̅2 = 70.6∠36.87°Ω 𝑁2

(d) 𝑃1 = 𝑃2 = 75(. 8) = 60 kW 𝑄1 = 𝑄2 = 60 ∗ tan 36.87° = 45 kVar

3.5

(a) E1 = (b)

N1  2400  E2 =   ( 230 ) = 2300 V N2  240  ∗

𝑆 125×10 ∠ −cos 𝑆2̅ = 𝐸̅2 𝐼2∗̅ ; 𝐼2̅ = (𝐸̅2 ) = [ 230 2

𝐸 𝑍2̅ = ̅2 = 𝐼2

−1 0.8 ∗

] = 543.5∠36.87°

230 = 0.424∠ − 36.87°Ω = 0.339 − 𝑗0.254 543∠−36.87°

(c)

𝑁 𝑧̅1′ = (𝑁1 ) 𝑍2̅ = 100𝑍2̅ = 42.4∠ − 36.87°Ω

(d)

𝑃1 = 𝑃2 = 125(. 8) = 100 kW

𝑄1 = 𝑄2 = 100 ∗ tan −36.87° = −75 kVar supplied to primary winding 3.6

(a) E2 = 2770; E1 = E2 e j 30 = 27730 V 110×103 ) ∠ 𝑐𝑜𝑠 −1 0 . 8 = 397∠36.87°𝐴 277

𝑆

(b) 𝐼̄2 = (𝐸2 ) ∠ 𝑐𝑜𝑠 −1 (𝑃𝐹) = ( 2

𝐼̄1 =

𝐼̄2

∗

(𝑒 𝑗30° )

= 𝐼̄2 𝑒 𝑗30° = 397∠66.87°𝐴

𝐸 277∠0° (c) 𝑍̄ 2̶ = 𝐼̄ 2 = 397 ∠36.87° = 0.698∠ − 36.87° 𝛺 2

̄ 1̶

𝑍 2 = 𝑍̄2̶ = 0.698∠ − 36.87° 𝛺 (d) 𝑆̄1 = 𝑆̄2 = 110∠ − 𝑐𝑜𝑠 −1 0 . 8;

kVA = 110∠ − 36.87°

𝑆̄1 = 88 𝑘𝑊 − 𝑗66𝑘𝑉𝐴𝑅 delivered to primary 3.7

(a)

For maximum power transfer to the load, 𝑅𝐿′ = 𝑎2 𝑅𝐿 = 𝑅𝑆 or 50𝑎2 = 1800 or 𝑎 = 6 = 𝑁1 /𝑁2 (b)

By voltage division,

 L = 5 2 sin 2t V

(VL )RMS = 5V

(VL ) RMS  25 Pa =  = W 13.9mW 1800 1800 2

3.8

1 = (18sin10t ) 2 = 9sin10t V 1 3 1 3 = 2 = 1.5sin10t V 2 out (t ) = −23 = −3sin10t V

2 = 1 = 3sin10t V

3.9

(a) Winding resistance or copper loss (b) Primary side: 0.5Ω, Secondary side: 0.2Ω. Secondary side referred to primary: 0.2Ω ⋅ 32 = 1.8Ω. Represent these losses with a series resistance on the primary side of 2.3 Ω.

3.10 Rated current magnitude on the 66-kV side is given by

𝐼1 =

13,000 = 197.0𝐴 66

𝐼12 𝑅𝑒𝑞 1 = (197.0)2 𝑅𝑒𝑞 1 = 100 × 103 ∴ 𝑅𝑒𝑞 1 = 2.58 𝛺 ̶ = 𝑍𝑒𝑞1

5.5×103 197.0

←

2 2 2 2 ̶ 1 − 𝑅𝑒𝑞 = 27.9 𝛺 Then 𝑋𝑒𝑞 1 = √𝑍𝑒𝑞 1 = √(27.9) − (2.58) = 27.8 𝛺

←

Turns Ratio = a = N1 N 2 = 66 /11.5 = 5.74

3.11

With high-voltage side designated as 1, and L-V side as 2, (11.5 × 103 )2 𝑎2 𝐺𝐶1 = 65 × 103 , based on O.C test. Note: To transfer shunt admittance from H-V side to L-V side, we need to multiply by a2.

GC1 = Y1 =

65  103

(11.5  10 ) ( 5.74 ) 3 2

= 14.9  10 −6 S 

I2 1 30 1  2 =  = 79.2  10 −6 S 3 V2 a 11.5  10 ( 5.74 )2

 Bm1 = Y12 − GC2 1 = 10 −6

( 79.2 ) − (14.9 ) 2

= 77.79  10 −6 S 

Total loss under rated conditions is approximately the sum of short-circuit and open-circuit test losses. Efficiency FL =

10,000  100 = 98.38%  (10,000 ) + (100 + 65)

3.12 (a)

Neglecting series impedance: E1 =

N1 N  2400  E2 = 1 V2 =   2400 = 24000 V N2 N2  240 

N2 240 I2 = ( 4.85) = 0.485A N1 2400 GC = P2 E12 = 173 ( 2400 ) = 3.003  10 −5 S 2

N  YC =  2 I 2  E1 = 0.485 / 2400 = 2.021  10 −4 S  N1  Bm = YC2 − GC2 =

( 2.021  10 −4 ) − ( 3.003  10 −5 ) = 1.998  10 −4 S 2

YC = GC − jBm = 3.003  10 −5 − j1.998  10 −4 S = 2.021  10 −4  − 81.45 S

(b)

𝑃 𝐼1

650

𝑅𝑒𝑞 1 = 21 = (20.8)2 = 1.502 𝛺 ̶ 1= 𝑍𝑒𝑞

𝑉1 52 = = 2.5 𝛺 𝐼1 20.8

2 2 2 𝑋𝑒𝑞 1 = √𝑍 2̶ 𝑒𝑞 1 − 𝑅𝑒𝑞 1 = √(2.5) − (1.502) = 1.998 𝛺

̶ 𝑍̄𝑒𝑞 1 = 𝑅𝑒𝑞 1 + 𝑗𝑋𝑒𝑞 1 = 1.502 + 𝑗1.998 = 2.5∠53.07° 𝛺

(c)

3.13

Using voltage division: 𝐸̅1 = (2400∠0°)

𝑗800 = 2398.5 V 𝑗(800 + 0.5)

𝑉2 = 𝐸̅2 =

𝑁2 𝐸̅ = 239.85 V 𝑁1 1

3.14

(a)

𝑆 50×10 𝐼̄2 = 2𝑉 ∠ − 𝑐𝑜𝑠 −1 (𝑃. 𝐹. ) = 240 ∠ − 𝑐𝑜𝑠 −1 (0.8) = 208.3∠ − 36.87°𝐴 2

I1 =

N2 1 I 2 = ( 208.3 − 36.87 ) = 20.83 − 36.87 A N1 10

E1 =

N1 V2 = 10 ( 2400 ) = 24000 V N2

𝑉̄1 = 2400∠0° + (1 + 𝑗2.5)(20.83∠ −

V1 = E1 + ( Reg1 + jX eg1 ) I1

36.87°) = 2400 + 56.095∠31.329° = 2447.9 + 𝑗29.166 = 2448.1∠0.683°𝑉 29 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

(b) 𝑉̄𝑆 = 𝐸̄1 + (𝑅𝑓𝑒𝑒𝑑 + 𝑗𝑋𝑓𝑒𝑒𝑑 + 𝑅𝑒𝑞1 + 𝑗𝑋𝑒𝑞1 )𝐼̄1 = 2400∠0° + (2.0 + 𝑗4.5)(20.83∠ − 36.87°) = 2400 + 102.59∠29.168° = 2489.6 + 𝑗50.00 = 2490.1∠1.1505°𝑉 (c) 𝑆̄𝑆 = 𝑉̄𝑆 𝐼̄𝑆∗ = (2490∠1.1505°)(20.83∠36.87°) = 51868. ∠38.05°

PS = Re ( SS ) = 40.87 kW

  delivered to the sending end of feeder. QS = I m ( SS ) = 31.95 kVARS

3.15 (a) 𝐼̄1 = 20.83∠0°

V1 = 24000 + (1 + j 2.5 )( 20.830 ) = 2400 + 56.09568.199 = 2420.8 + j 52.08 = 2421.1.232 V

(

)

VS = 24000 + 2.0 + 4.5 ( 20.830 ) = 2400 + 102.5966.04 = 2441.7 + j 93.74 = 2443.2.199 V 𝑆̄𝑆 = 𝑉̄𝑆 𝐼̄𝑆∗ = (2443∠2.199°)(20.83∠0°) = 50896. ∠2.199° = 50,859. +𝑗1953. PS = 50.87 kW

  delivered QS = 1.953kVARS

(b) I1 = 20.8336.87 A V1 = 24000 + (1 + j 2.5 )( 20.8336.87 ) = 2400 + 56.095105.07 = 2385.4 + j 54.17 = 23861.301 V VS = 24000 + ( 2.0 + j 4.5)( 20.8336.87 ) = 2400 + 102.59102.91 = 2377.1 + j100.0 = 2379.2.409 V

𝑆̄𝑆 = 𝑉̄𝑆 𝐼̄𝑆∗ = (2379. ∠2.409°)(20.83∠ − 36.87°) = 49,566. ∠ − 34.46° = 40868. −𝑗28047. PS = 40.87 kW delivered

QS = −28.05 kVARS delivered by source to feeder = +28.04 kVARS absorbed by source

Note: Real and reactive losses, 0.87 kW and 1.95 kVARS, absorbed by the feeder and transformer, are the same in all cases. Highest efficiency occurs for unity P.F ( EFF = Pout / Ps × 100 = ( 50 / 50.87 ) ×100 = 98.29% ).

3.16 (a) 𝑎 = 2400/240 = 10 2400 2

𝑅2′ = 𝑎2 𝑅2 = ( 240 ) 0.0075 = 0.75 𝛺 𝑋2′ = 𝑎2 𝑋2 = (10)2 0.01 = 1.0 𝛺 Referred to the HV-side, the exciting branch conductance and susceptance are given by

(1/ a2 ) 0.003 = (1/100 ) 0.003 = 0.03  10−3 S and

(1/ a2 ) 0.02 = (1/100 ) 0.02 = 0.2  10−3 S

The equivalent circuit referred to the high-voltage side is shown below:

(b) 𝑅1′ = 𝑅1 /𝑎2 = 0.0075𝛺 𝑋1′ = 𝑋1 /𝑎2 = 0.01𝛺 The equivalent circuit referred to the low-voltage side is shown below:

3.17 (a) Neglecting the exciting current of the transformer, the equivalent circuit of the transformer, referred to the high-voltage (primary) side is shown below:

The rated (full) load current, Ref. to HV-side, is given by

With a lagging power factor of 0.8, I1 = 20.8 − cos−1 0.8 = 20.8 − 36.9 A Using KVL, V1 = 24000 + ( 20.8 − 36.9 )(1.5 + j 2 ) = 24500.34 V (b) The corresponding phasor diagram is shown below:

(c)

Using KVL, VS = 24000 + ( 20.8 − 36.9 )( 2 + j 4 ) = 2483.50.96 V pf at the sending end is cos ( 36.9 + 0.96 ) = 0.79 Lagging 3.18

𝑆𝑏𝑎𝑠𝑒 = 50 kVA, 𝑉𝑏𝑎𝑠𝑒1 = 2400 V, 𝑉𝑏𝑎𝑠𝑒2 = 240 V, 𝑍𝑏𝑎𝑠𝑒1 =

(2400)2 = 115.2 Ω 50 × 103

Using voltage division 𝑉̅2𝑝𝑢 = (1.0∠0°)

𝑗6.944 = 0.99938 𝑗(6.944 + 4.3402 × 10−3 )

𝑉̅2 = 𝑉̅2𝑝𝑢 𝑉𝑏𝑎𝑠𝑒2 = (0.99938)(240) = 239.85∠0°V

3.19

Zone 1

Zone 2

Sbase = 50 kVA

VBase 2 = 240 V

Vbase1 = 2400 V Z base1 = ( 2400 ) 50  103 2

= 115.2  (a) 𝑉̄1𝑝𝑢 = 1.0∠0° + (8.6803 × 10−3 + 𝑗2.1701 × 10−2 )(1.0∠ − 36.87°) = 1.0 + 0.02337331.33 = 1.01997 + j 0.012157 = 1.0200.683 pu V1 = V1 puVbase = (1.0200.683 )( 2400 ) = 2448.0.683 V

(b) 𝑉̄𝑠𝑝𝑢 = 1.0∠0° + (1.7361 × 10−2 + 𝑗3.9063 × 10−2 )(1.0∠ − 36.87°) = 1.0 + 0.04274729.168 = 1.03733 + j 0.020833 = 1.037 S1.1505 pu VS = VspuVbase1 = (1.03751.1505 )( 2400 ) = 2490.1.1505 V

(c) 𝑃𝑠𝑝𝑢 + 𝑗𝑄𝑠𝑝𝑢 = 𝑉̄𝑠𝑝𝑢 𝐼̄∗ 𝑠𝑝𝑢 = (1.0375∠1.1505°)(1.0∠36.87°) = 1.037538.02 = 0.8173 + j 0.6390 per unit PS = ( 0.8173 )( 50 ) = 40.87 kW

  delivered QS = ( 0.6390 )( 50 ) = 31.95 kVARS

3.20

Zone 1  240  Vbase1 =   460  480  = 230 V

Zone 2

Zone 3

 460  Vbase 2 =  115 = 460 V  115 

( 460 ) = 10.58  = 2

Z base 2

20,000

Vbase3 = 115V

(115) = 0.6613  = 2

Z base3

20,000 20,000 I base3 = = 173.9 A 115

0.9+𝑗0.2 ̶ 𝑍̄𝐿𝑜𝑎𝑑 = 1.361 + 𝑗0.3025 𝑝𝑢 = 0.6613

XT 2 pu = 0.10 pu

2 = 0.1890 pu 10.58

X Line pu =

 480   20  XT 1 pu = ( 0.10 )     = 0.07259 pu  460   30 

220 = 0.9565 pu 230

Vspu = I Load pu =

Vspu

j ( XT 1 pu + XT 2 pu + X Line ) + Z

Load pu

0.95650 j (.07259 + .1890 + .10 ) + (1.361 + j.3025)

0.95650 0.95650 = 1.361 + j 0.6641 1.51426.010

= 0.6316 − 26.01 pu

I Load = I Load pu I base3 = ( 0.6316 − 26.01 )(173.9 ) = 109.8 − 26.01 A 3.21

( 600 )

Z base =

100  10

= 3.6  3

I base =

Z L pu =

185 = 0.277885 pu 3.6

ZY pu =

1040 = 2.77840 pu 3.6

Ea pu =

480

600

𝐿𝑝𝑢

3 ( 600 )

= 96.23A

 − 30 = 0.8 − 30 pu

𝐸 𝐼̄𝑎𝑝𝑢 = 𝑍̄ ̶ 𝑎𝑝𝑢 +𝑍̄ ̶

100  103

𝑌𝑝𝑢

0.8∠−30°

= 0.2778∠85°+2.778∠40°

0.8∠−30° 0.8∠−30° 𝐼̄𝑎𝑝𝑢 = 2.1521+𝑗2.0622 = 2.9807∠43.78°

I a pu = 0.2684 − 73.78 pu

𝐼̄𝑎 = 𝐼̄𝑎𝑝𝑢 𝐼𝑏𝑎𝑠𝑒 = (0.2684∠ − 73.78°)(96.23) I a = 25.83 − 73.78 A

3.22

Per-unit positive-sequence circuit (2772 )

̶ 𝑍𝑏𝑎𝑠𝑒 =

10,000

= 7.673 𝛺

ZY 1pu = ( 20 + j10 ) 7.673 = ( 2.607 + j1.303) pu = 2.91426.57 pu Z1 pu = ( 30 − j15) 3(7.673) = (1.303 − j 0.6516 ) pu = 1.437 − 26.57 pu

I1 pu =

VS1 pu

1.0 0

 ( 2.914 26.57 )(1.457  − 26.57 )     ( 2.607 + j1.303 ) + (1.303 − j 0.6516 )  1.0 0 1.0 0 = = = 0.9337 9.463pu  4.246 0  1.071 − 9.463    3.91 + j 0.6517  ZY 1 pu Z 1 pu

I base = 1000 277 = 36.1 A

I1 = I1 pu I base = ( 0.93379.463 ) (36.1) = 33.71 9.463 A

3.23 Select a common base of 100MVA and 22kV on the generator side; Base voltage at bus 1 is 22kV; this fixes the voltage bases for the remaining buses in accordance with the transformer turns ratios. Using Eq. 3.3.11, per-unit reactances on the selected base are given by 100 100 ) = 0.2; 𝑇1 : 𝑋 = 0.1 ( ) = 0.2 90 50

𝐺: 𝑋 = 0.18 (

100

𝑇2 : 𝑋 = 0.06 ( 40 ) = 0.15; 𝑇2 : 𝑋 = 0.06 ( 40 ) = 0.15 100 100 ) = 0.16; 𝑇4 : 𝑋 = 0.08 ( ) = 0.2 40 40

𝑇3 : 𝑋 = 0.064 (

100

10.45 2

𝑀: 𝑋 = 0.185 (66.5) ( 11 ) = 0.25 (220)2

̶ For Line 1, 𝑍𝐵𝐴𝑆𝐸 = 100

(110)2

̶ For Line 2, 𝑍𝐵𝐴𝑆𝐸 = 100

48.4

= 484 𝛺 and 𝑋 = 484 = 0.1 65.43

= 121 𝛺 and 𝑋 = 121 = 0.54 35

The load complex power at 0.6 Lagging pf is SL (3 ) = 5753.13 MVA

 The load impedance in OHMS is Z L =

(10.45)

57 − 53.13

VLL2 SL*(3 )

= 1.1495 + 𝑗1.53267 𝛺 The base impedance for the load is  Load Impedance in pu =

(11)2 100

= 1.21 𝛺

1.1495 + j1.53267 = 0.95 + j1.2667 1.21

The per-unit equivalent circuit is shown below:

3.24

𝑉

75 𝑘V

𝑍𝐿

30 Ω

(a) 𝑉𝑙 = 5 ∗ 𝑉𝑠 = 75 kV. 𝐼𝑠 = 5 ∗ 𝐼𝐿 = 5 ∗ ( 𝐿 ) = 5 ∗

15 kV 15 kV

= 1.0, 𝑍𝑝𝑢 =

30Ω 56.3Ω

= 12.5 𝑘A. 𝑃𝐿 = 𝑉𝐼 =

Primary

Secondary

𝑆𝑏

100 MVA

𝑉𝑏

15 kV

75 kV

𝐼𝑏

6667 A

1333 A

𝑍𝑏

2.25 Ω

56.3 Ω

= 0.533, 𝐼𝑝𝑢 =

1 0.533

(75 kV)2 30Ω

= 187 MW (𝑏)

= 1.87, 𝑃𝑝𝑢 = 1.87. 𝐼𝑆 = 𝐼𝑝𝑢 ∗ 𝐼𝑏,𝑝𝑟𝑖𝑚 = 12.5 kA. 𝑃𝐿 = 𝑃𝑝𝑢 ∗

𝑆𝑏 = 187 MW

3.25

3 2

(69×10 ) Base impedance in circuit 𝑍 ̶ = 15×106 = 317.4 𝛺 500

Pre-unit impedance of load in circuit 𝑍 ̶ = 317.4 = 1.575 The impedance diagram in PU is shown below:

3.26 Base impedance on the low-voltage, 3.81kV-side is (3.81)2 = 0.1613 𝛺 90 Note: The rating of the transformer as a 3-phase bank is

3  30 = 90 MVA,

3 ( 38.1) : 3.81 = 66 : 3 . 81kV 

With a base of 66kV on the H-V side, base on L.V side = 3.81 kV so, on L.V. side RL = 66

1 = 6.2 pu 0.1613 2

Resistance referred to H.V. side = 1 (3.81) = 300 𝛺 Per-unit value should be the same as 6.2 pu. Check: base impedance on H.V. side =

(66)2 90

= 48.4 𝛺 and

300 = 6.2pu 48.4

3.27 Transformer reactance on its own base is

0.4 = 0.828 pu (222 )/1000 220 2

100

On the chosen base, reactance becomes 0.828 ∗ (230) (1000) = 0.076 3.28 Eq. 3.3.11 of the text applies. 2

 2400   100  G1 : Z = j 0.2     = j 2 pu  2400   10  2

 2400   100  G2 : Z = j 0.2     = j1pu  2400   20  2

 2400   100  T1 : Z = j 0.1    = j 0.25pu  2400   40  2

 10   100  T2 : Z = j 0.1    = j 0.136 pu  9.6   80 

For the transmission-line zone, base impedance =  Z LINE = ( 50 + j 200 )

M ; kV A =

(9600)2 100×103

100  103

( 9600 )

= ( 0.054 + j 0.217 ) pu

25 4 = 0.25pu; 4 kV = = 0.833pu 100 4.8

The impedance diagram for the system is shown below:

3.29 Since 𝑉̄𝑎′ 𝑛′ = 𝑛𝑉̄𝑎𝑏 = 𝑛(𝑉̄𝑎𝑛 − 𝑉̄𝑏𝑛 ) 𝑉̄𝑎′ 𝑛′ = (√3𝑛𝑒 𝑗30° )𝑉̄𝑎𝑛 = 𝐶̄1 𝑉̄𝑎𝑛 } 𝑉̄𝑏′ 𝑛′ = 𝐶̄1 𝑉̄𝑏𝑛 ; 𝑉̄𝑐 ′𝑛′ = 𝐶̄1 𝑉̄𝑐𝑛

←

(a) Yes ← (b) Since 𝐼̄𝑎 = 𝐼̄𝑎𝑏 − 𝐼̄𝑐𝑎 = 𝑛(𝐼̄𝑎′ − 𝐼̄𝑐′ ) 𝐼̄𝑎 = (√3𝑛𝑒 −𝑗30° )𝐼̄𝑎′ = 𝐶̄1∗ 𝐼̄𝑎′ ; 𝐼̄𝑎′ = 𝐼̄𝑎 /𝐶̄1∗ }← 𝐼̄𝑏′ = 𝐼̄𝑏 /𝐶̄1∗ ; 𝐼̄𝑐′ = 𝐼̄𝑐 /𝐶̄1∗ ∗ (c) 𝑆̄ ′ = 𝑉̄𝑎′𝑛′ (𝐼̄𝑎′ ) = 𝐶̄1 𝑉̄𝑎𝑛 (𝐼̄𝑎 /𝐶̄1∗ )∗ = 𝑉̄𝑎𝑛 𝐼̄𝑎∗ = 𝑆̄

←

3.30 For a negative sequence set, Van =

( 3n e )V = C V = C V  − j 30

* 1

2 an

Vbn = C2Vbn ;Vcn = C2Vcn where C2 = C1*     I a = C2* I a or I a = I a / C2*  * * * I b = I b / C2 ; I c = I c / C2 where C2 = C1 

(a)

  Also C1 = C2* ; Note:Taking the complex conjugate 

C1 = 3 n e j 30 ; C2 = 3 n e − j 30 = C1*

Transforms a positive sequence set into a negative sequence set. 3.31

  n j 30  Van =  e  Van = C3Van  For the positive seq. set  3   Vbn = C3 Vbn ; Vcn = C3Vcn  (a) 𝐶̄4 = 𝐶̄3∗ for the negative sequence set (b) Complex power gain = C (1 C * ) = 1 *

Van Van / C 1 = * = 2 Z L ,where C = C .  Ia C Ia C

3.32 (a)

For positive sequence, 𝑉̄𝐻1 leads 𝑉̄𝑀1 by 90, and 𝑉̄𝐻1 lags 𝑉̄𝑋1 by 90. For negative sequence, 𝑉̄𝐻2 lags 𝑉̄𝑀2 by 90, and 𝑉̄𝐻2 leads 𝑉̄𝑋2 by 90.

(b)

For positive sequence 𝑉̄𝐻1 leads 𝑉̄𝑋1 by 90 and 𝑉̄𝑋1 is in phase with 𝑉̄𝑀1. For negative sequence 𝑉̄𝐻2 lags 𝑉̄𝑋2 by 90 and 𝑉̄𝑋2 is in phase with 𝑉̄𝑀2. Note that a  – zigzag transformer can be used to obtain the advantages of a 𝛥 − 𝑌 transformer without phase shift. (c)

For positive sequence, 𝑉̄𝐻1 lags 𝑉̄𝑋1 by 23.4. For negative sequence, 𝑉̄𝐻2 leads 𝑉̄𝑋2 by 23.4.

3.33

3.34 (a) (b)

3.35 Secondary side: 𝑉𝑏 =

50 ∠(18° − 30°) = 28.9∠ − 12°; 𝑉𝑎 = 28.9∠108°; 𝑉𝑐 = 28.9∠ − 132° then √3 230

primary side must lead secondary by 30°. 𝑉𝑎 = 28.9 ∗ 46 ∠(108° + 30°) = 144.5∠138°; 𝑉𝑏 = 144.5∠18°; 𝑉𝑐 = 144.5∠ − 102°.

3.36

Open  Transformer (a) 𝑉̄𝑏𝑐 and 𝑉̄𝑐𝑎 remain the same after one, single-phase transformer is removed. Therefore, 𝑉̄𝑎𝑏 = −(𝑉̄𝑏𝑐 + 𝑉̄𝑐𝑎 ) remains the same. The load voltages are then balanced, Positive-sequence. Selecting 𝑉̄𝑎𝑛 as reference: Van =

13.8 3

0 = 7.967 0 kV

Vbn = 7.967  − 120 kV Vcn = 7.967 + 120 kV

(b) I a =

S3 3VLL

0 =

43.3  106

3 (13.8  103 )

= 1.8120 kA

I b = 1.812 − 120 kA I c = 1.812 + 120 kA

(c) Vbc = 13.8 − 120 + 30 = 13.8 − 90 kV Transformer bc delivers 𝑆̄𝑏𝑐 = 𝑉̄𝑏𝑐 𝐼̄𝑏∗ Sbc = (13.8 − 90)(1.812 + 120) = 25  30 MVA

𝑆̄𝑏𝑐 = (21.65 + 𝑗12.5) × 106 Transformer ac delivers 𝑆̄𝑎𝑐 = 𝑉̄𝑎𝑐 𝐼̄𝑎∗ where Vac = −Vca = −13.8120 + 30 = 13.8 − 30 kV Sac = (13.8 − 30 )(1.8120 ) = 25   − 30 MVA

𝑆̄𝑎𝑐 = (21.65 − 𝑗12.5) × 106 The open- transformer is not overloaded. Note that transformer bc delivers 12.5 Mvars and transformer ac absorbs 12.5 Mvars. The total reactive power delivered by the open- transformer to the resistive load is therefore zero.

3.37 Noting that √3(54.2) = 94, the rating of the 3-phase transformer bank is 75 MVA, 94Y5.42  kV. Base impedance for the low-voltage side is

5.422 = 0.3917𝛺 75

0.6

On the low-voltage side, 𝑅𝐿 = 0.3917 = 1.53𝑝𝑢 Base impedance on high-voltage side is

(94)2

= 117.8𝛺

The resistance ref. to HV-side is 0.6 (5.42) = 180𝛺 180

or 𝑅𝐿 = 117.8 = 1.5𝑝𝑢 3.38 (a) The single-line diagram and the per-phase equivalent circuit, with all parameters in per unit, are given below:

Current supplied to the load is

240  103 3  230

Base current at the load is 100,000

= 602.45A

( 3  230 ) = 251.02 A

The power-factor angle of the load current is  = cos−1 0.9 = 25.84 Lag . With 𝑉̄𝐴 = 1.0∠0° as reference, the line currents drawn by the load are IA =

602.45  − 25.84 = 2.4 − 25.84 per unit 251.02

I B = 2.4 − 25.84 − 120 = 2.4 − 145.84 per unit

I C = 2.4 − 25.84 + 120 = 2.494.16 per unit

(b) Low-voltage side currents further lag by 30° because of phase shift 𝐼̄𝑎 = 2.4∠ − 55.84°; 𝐼̄𝑏 = 2.4∠ − 175.84°; 𝐼̄𝑐 = 2.4∠64.16° (c) The transformer reactance modified for the chosen base is X = 0.11  (100 / 330 ) =

1 pu 30

The terminal voltage of the generator is then given by

Vt = VA  − 30 + jXI a

= 1.0 − 30 + j (1/ 30 )( 2.4 − 55.34 ) = 0.9322 − j 0.4551 = 1.0374  − 26.02 pu

Terminal voltage of the generator is 23  1.0374 = 23.86 kV The real power supplied by the generator is

Re Vt I a*  = 1.0374  2.4cos ( −26.02 + 55.84 ) = 2.16 pu which corresponds to 216 MW absorbed by the load, since there are no 𝐼 2 𝑅 losses. (d) By omitting the phase shift of the transformer altogether, recalculating 𝑉̄𝑡 with the reactance 1 𝑗 (30) on the high-voltage side, the student will find the same value for 𝑉𝑡 i.e. |𝑉̄𝑡 |.

3.39

Zero Sequence

Positive Sequence

Negative Sequence

3.40

 230   100  X pu new = 0.06     = 0.01837 pu  240   300 

Zero Sequence

Positive Sequence

Negative Sequence 3.41

Per-unit positive-sequence reactance diagram

Sbase = 100 MVA Vbase H = 500 kV in transmission-line Zones Vbase X = 20 kV in motorgenerator Zones 2

 18  X g1 = X g2 = 0.2   = 0.162 pu  20   1000  X m 3 = 0.2   = 0.1333pu  1500  XT 1 = XT 2 = XT 3 = XT 4 = 0.1pu  1000  XTS = 0.1  = 0.06667 pu  1500  Z base H = ( 500 ) 1000 = 250  2

Xline 50 = 50 / 250 = 0.2 pu Xline 25 = 25 / 250 = 0.1pu

3.42

18 0 = 0.90 pu 20 1500 I3 =  cos−1 0.8 = 60.1436.87 kA 3 (18 )( 0.8 )

V3 pu =

I base X =

1000

= 28.87 kA 20 3 60.14 I3 = 36.87 = 2.083 36.87 pu 28.87

1 V1 = V2 = V3 + I 3 ( j T 5 ) + I 3 ( jX Line 25 + jXT 3 ) 2   0.1 + 0.1   = 0.90 + ( 2.083 36.87 )  j  0.06667 +  2    = 0.7454 21.88pu V1 = V2 = 0.7454 ( 20 ) = 14.91 kV

3.43

Sbase = 30 MVA Vbase H = 66.4 3 = 115kV I H1 = I X1 = (a) I base H =

1.00 = 10.0 − 90 pu j 0.1 30

115 3

I base X =

= 0.1506 kA Vbase X = 12.5 3 = 21.65 kV

= 0.8kA 21.65 3 I H = 10 ( 0.1506 ) = 1.506 kA I X = 10 ( 0.8 ) = 8kA

(b) I base H = 0.1506 kA; Vbase X = 12.5kV; 30 = 1.386 kA 12.5 3 I H = 1.506 kA;

I base X =

I X = 10 (1.386 ) = 13.86 kA

3.44

Sbase = 135MVA

VBase H = 115kV

Vbase X = 13.2kV  135  (a) X g1 = 1.5   = 1.558 pu  130  (b)

135 115 3

= 0.6778kA; I H =

150

(115 3 ) (1.0)

= 0.07530 kA = 0.11110pu

VX = 10 + ( 0.005 + j 0.1)( 0.11110 ) = 1.0021 + j 0.04286 = 1.0032.45 pu; VX = 1.003(13.2) = 13.24 kV

Eg = 10 + ( 0.005 + j1.85 )( 0.42860 ) = 1.96123.86 pu; Eg = 1.961  13.2 = 25.89 kV Px + jQx = Vx I x* = (1.0032.45 )( 0.42860 ) = 0.42992.45 pu = 0.4299 ( 35 ) 2.45 MVA = 15.03MW + j 0.643MVAR PF = cos 2.45 = 0.999 Lagging

3.45 Three-phase rating of transformer T2 is 3  100 = 300 MVA and its line-to-line voltage ratio is √3(127): 13.2 or 220 :13.2 kV . Choosing a common base of 300 MVA for the system, and selecting a base of 20 kV in the generator circuit, The voltage base in the transmission line is 230 kV and the voltage base in the motor circuit is 230 (13.2220) = 13.8 kV transformer reactances converted to the proper base are given by 300 13.2 2 𝑇1 : 𝑋 = 0.1 × = 0.0857; 𝑇2 : 0.1 ( ) = 0.0915 350 13.8 Base impedance for the transmission line is (230)2300=176.3  The reactance of the line in per unit is then

0.5×64 = 0.1815 176.3

300

13.2 2

300

13.2 2

Reactance 𝑋𝛼″ of motor 𝑀1 : 0.2 (200) (13.8) = 0.2745 Reactance 𝑋𝛼″ of motor 𝑀2 : 0.2 (100) (13.8) = 0.549 Neglecting transformer phase shifts, the positive-sequence reactance diagram is shown in figure below:

3.46 The motors together draw 180 MW, or

180 = 0.6 pu 300

With phase-a voltage at the motor terminals as reference, V=

13.2 = 0.95650 pu 13.8

The motor current is given by I =

0.6 0 = 0.62730 pu 0.9565

Referring to the reactance diagram in the solution of Problem 3.33, phase-a per-unit voltages at other points of the system are At m : V = 0.9565 + 0.6273 ( j 0.0915) = 0.9582 3.434 pu At l : V = 0.9565 + 0.6273 ( j 0.0915 + j 0.1815) = 0.971710.154 pu At k : V = 0.9565 + 0.6273 ( j 0.0915 + j 0.1815 + j 00857 ) = 0.982613.237 pu

The voltage regulation of the line is then 0.9826 − 0.9582 = 0.0255 0.9582 The magnitude of the voltage at the generator terminals is

0.9826  20 = 19.652 kV Note that the transformer phase shifts have been neglected here. 3.47 (a) For positive sequence operation and standard -Y connection, the per-phase diagram is shown below:

Van =

230 3

0 kV, choosing that as a reference.

100  106  cos−1 0.8 = 41.6736.87 MVA 0.8  3 S  41.67  106 36.87 I a * = = = 313.836.87 A Van 132.8  103

S =

 I a = 313.8 − 36.87 A I a = 10 3 e − j 30 I a = 5435 − 66.87 A

The primary current magnitude is 5435A.   1  Van = V + j 0.08I a =  132.8  103  − 30  + j 0.08 ( 5435 − 66.87 )  10 3  = 7667.4 − 30 + 434.823.13 = 7936 − 27.45 V

Line-to-line primary voltage magnitude = 3 ( 7936 ) = 13.75kV 

Three-phase complex power supplied by the generator is S3 = 3Van I a* = 3 ( 7396 − 27.45 )( 543566.87 ) =12939.4 MVA 

(b) The secondary phase leads the primary by 27.45; this phase shift applies to line-to-neutral (phase) as well as line-to-line voltages.  3.48 (a) For positive sequence operation and standard -Y & Y- connections, the per-phase diagram is drawn below:

(b) V1 =

15 3

0 kV, choosing that as a reference.

𝑍̄ 3̶ = (5 + 𝑗1) + 𝑗0.08 = (5 + 𝑗1.08)𝛺 2

𝑍̄2̶ = 𝑗100 + (10√3) 𝑍̄3̶ = (1500 + 𝑗424)𝛺 𝑍̄1̶ = [

𝑍̄2̶

2 ] + 𝑗0.08 = 5 + 𝑗1.4933 = 5.22∠16.63°

(10√3)

∴ 𝐼̄1 = 𝑉̄1 /𝑍̄1̶ =

8660.5 = 1659.1∠ − 16.63° (5.22∠16.63°)

Generator current magnitude = I1 = 1659.1 A  I2 =

1 10 3

e j 30 I1 = 95.813.37 A

Transmission-line current magnitude = 95.8 A  I3 = 10 3e− j 30 I 2 = 1659.3 − 16.63 A

Load current magnitude = 1659.3 A  V3 = Z LOAD I 3 = ( 5 + j1)(1659.3 − 16.63 ) = 8462.4 − 5.32 V

Line-to-line voltage magnitude at load terminals = 14.66 kV  Three-phase complex power delivered to the load is

S3 = 3V3 I 3* = 3 Z LOAD I 32 = 3 ( 8462.4 − 5.32 )(1659.3 + 16.63 ) = 42.12511.31 MVA  3.49 Base kV in transmission-line circuit = 132 kV Base kV in the generator G1 circuit = 132 

13.2 = 10.56 kV 165

Base kV in the generator G2 circuit = 132 

13.8 = 11.04 kV 165

Impedance diagram of the system with pu values

On the common base of 100 MVA for the entire system, 2

100  13.2  G1 : Z = j 0.15   = j 0.4688 pu 50  10.56  2

G2 : Z = j 0.15 

100  13.8   = j1.1719 pu 20  11.04  2

T1 : Z = j 0.1 

100  13.2   = j 0.1953pu 80  10.56  2

100  13.8  T2 : Z = j 0.1   = j 0.3906 pu 40  11.04 

Base impedance in transmission-line circuit is

(132 ) = 174.24  2

100

50  j 200 = 0.287 + j1.1478 pu 174.24 25  j100 ZTR. LINE 2 = = 0.1435 + j 0.5739 pu 174.24 ZTR. LINE 1 =

LOAD : 40 ( 0.8 + j 0.6 ) = ( 32 + j 24 ) MVA

(150 ) = 703.1 = 703.1 pu = 4.035pu = 2

RLOAD

174.24

(150 ) = 937.5  = 937.5 pu = 5.381pu 2

X LOAD =

24 Z LOAD = ( RLOAD

174.24 jX LOAD )

3.50

3.51 (a) X12 = 0.08 pu; X13 = 0.1 pu;

X 23 = 0.09 (15/10 ) = 0.135 pu

1 ( 0.08 + 0.1 − 0.135) = 0.0225 pu 2 1 X 2 = ( 0.08 + 0.135 − 0.1) = 0.0575 pu 2 1 X3 = ( 0.135 + 0.1 − 0.08 ) = 0.0775 pu 2 X1 =

(b)

Using 𝑃3𝜙 =

2 2 3𝑉𝐿𝑁 𝑉𝐿𝐿 = 𝑅 𝑅

R2 = (13.2 ) 7.5 = 23.23 ; R3 = ( 2.3 ) 5 = 1.058  2

Z 2 base = (13.2 ) 15 = 11.616 ; Z 3 base = ( 2.3 ) 15 = 0.3527  2

R2 pu = R2 Z 2 base

23.23 1.058 = 2 pu; R3 pu = = 3pu 11.616 0.3527

3.52 Voltage ratings indicate the turns ratio. Tertiary voltage is 20 kV according to the turns ratio. Tertiary current is determined by

𝑁1 𝐼1 −𝑁2 𝐼2 138∗10−69∗18 = = 6.9 A 𝑁3 20

3.53 With a base of 15 MVA and 66 kV in the primary circuit, the base for secondary circuit is 15 MVA and 13.2 kV, and the base for tertiary circuit is 15 MVA and 2.3 kV, Note that XPS and XPT need not be changed. XST is modified to the new base as follows: 𝑋𝑆𝑇 = 0.05 ×

15 = 0.075 10

With the bases specified, the per-unit reactances of the per-phase equivalent circuit are given by 1 𝑋𝑃 = (𝑗0.09 + 𝑗0.08 − 𝑗0.075) = 𝑗0.0475 2 1 𝑋𝑆 = (𝑗0.09 + 𝑗0.075 − 𝑗0.08) = 𝑗0.0425 2 53 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1 𝑋𝑇 = (𝑗0.08 + 𝑗0.075 − 𝑗0.09) = 𝑗0.0325 2 3.54 The constant voltage source is represented by a generator having no internal impedance. On a base of 5 MVA, 2.3 kV in the tertiary, the resistance of the load is 1.0 pu. Expressed on a 15 MVA, 2.3 kV base, the load resistance is R = 1.0 

15 = 3.0 pu 5

On a base of 15 MVA, 13.2 kV, the reactance of the motor is X  = 0.2 

15 = 0.4 pu 7.5

The impedance diagram is given below:

Note that the phase shift that occurs between the Y-connected primary and the -connected tertiary has been neglected here.

3.55 (a)

(b) SX = ( 2300) ( 47.83) = 110kVA 𝑆𝐻 = 2530 (43.48) 10 kVA is transformed by magnetic induction. 100 kVA is transformed electrically. (c) At rated voltage, core losses = 70 W 2

 4.348  At full-load current, winding losses =   240 = 224W  4.5 

Total losses = PLoss = 294 W

pout = ( 2530 ) ( 43.48 )( 0.8 ) = 88003W Pin = Pout + PLoss = 88003 + 294 = 88297 W 0

Efficiency = ( Pout Pin )  100 = ( 88003 88297 )100 = 99.67 0 0 54

3.56 (a)

(b) As a normal, single-phase, two-winding transformer, rated: 3 kVA, 220110 V; 2

(220) ̶ Xeq = 0.10 per unit.𝑍𝐵𝑎𝑠𝑒 𝐻𝑜𝑙𝑑 = 3000 = 16.133𝛺

As a single - phase autotransformer rated: 330 (13.64 ) = 4.50 kVA, 330 /110 V, Z Base H new = ( 330 ) / 4500 = 24.2  2

 16.133  X eq = ( 0.10 )   = 0.06667 per unit  24.2 

SBase 3 = 13.5 kVA

VBase X = 110 V

VBase H = 479.5V

I Base X =

I Base H =

13.5  103 110 3

= 70.86 A

13.5  103

479.5 3 = 16.256 A

IX =

6000 − cos−1 .8

(110 3 ) ( 0.8 )

= 39.36 − 36.87 A

39.36  − 36.87 = 0.5555 − 36.87 per unit 70.86 I H = I X = 0.5555 per unit IX =

I H = ( 0.5555 ) (16.256 ) = 9.031A

VH = VX + jX eq I X = 1.00 + ( j 0.06667 ) (.5555 − 36.86 ) VH = 1.0 + 0.0370453.13 = 1.0222 + j 0.02963 VH = 1.02261.66 per unit

VH = (1.0226 )( 479.5 ) = 490.3V

3.57 Rated currents of the two-winding transformer are I1 =

60,000 = 250 A 240

and

I2 =

60,000 = 50 A 1200

The autotransformer connection is shown below:

(a) The autotransformer secondary current is I L = 300 A With windings carrying rated currents, the autotransformer rating is

(1200 )( 300 )10−3 = 360 kVA (b) Operated as a two-winding transformer at full-load, 0.8 pf, Efficiency =

60  0.8 = 0.96 ( 60  0.8 ) + pLoss

From which the total transformer loss PLOSS =

48 (1 − 0.96 ) 0.96

= 2 kW

The total autotransformer loss is same as the two-winding transformer, since the windings are subjected to the same rated voltages and currents as the two-winding transformer. ∴ 𝜂𝐴𝑈𝑇𝑂. 𝑇𝑅. =

360 × 0.8 = 0.9931 (360 × 0.8) + 2

3.58 (a) The autotransformer connection is shown below:

90,000 90,000 = 1125A; I 2 = = 750 A 80 120 V1 = 80 kV; V2 = 120 + 80 = 200 kV I1 =

I in = 1125 + 750 = 1875A (b) Input kVA is calculated as 80  1875 = 150,000 kVA which is same as

Output kVA = 200  750 = 150,000 Permissible kVA rating of the autotransformer is 150,000. The kVA transferred by the magnetic induction is same as the rating of the two-winding transformer, which is 90,000 kVA. 3.59

 PLoad + jQLoad = V I Load = (1.00 )(1.0 − 30 ) = 1.030 = 0.866 + j 0.50 per unit 

(a) No regulating transformer, 𝐶̄ = 1.0 Using current division:  XL 2   0.25  I L1 =   I Load =   (1.0 − 30 ) = 0.5556 − 30 per unit  0.45   X L1 + X L 2 

PL1 + jQL1 = V I L1 = 0.5556 + 30 = 0.4811 + j 0.2778 per unit PL 2 + jQL 2 = ( PLoad + jQLoad ) − ( PL1 + jQL1 ) = ( 0.866 + j 0.5 ) − (.4811 + j.2778 ) = 0.3849 + j 0.2222 per unit

(b) Voltage magnitude regulating transformer, C = 0.9524 Using the admittance parameters from Example 3.13(a) 𝑌̄ 𝐼̄ 𝐼̄ [ ]=[ ̄ ] = [ 11 −𝐼𝐿𝑜𝑎𝑑 −1.0∠ − 30° 𝑌̄21

−𝑗9.0 𝑗8.810 𝑌̄12 𝑉̄ ̄ ][ ] = [ ][ 𝑉 ] ̄𝑌22 𝑉̄ 1 𝑗8.810 −𝑗8.628 1.0∠0°

Solving the second equation above for 𝑉̄: −1.0∠ − 30° = (𝑗8.810)𝑉̄ − (𝑗8.628)(1.0∠0°)

8.62890 − 1.0 − 30 −.866 + j 9.128 = = 1.0415.42 per unit j8.810 j8.810

Then: ̄′

𝑉 −𝑉 1.041∠5.42°−1.0∠0° 0.0361+𝑗0.0983 𝐼̄𝐿1 = = = = 0.5235∠ − 20.14° 𝑗𝑋𝐿1

𝑗0.20

PL1 + jQL1 = V I L1 = 0.523520.14 = 0.4915 + j 0.1802 per unit PL 2 + jQL 2 = ( PLoad + jQLoad ) − ( PL1 + jQL1 ) = 0.3745 + j 0.3198 per unit The voltage magnitude regulating transformer increases the reactive power delivered by line L-2 43.970 (from 0.2222 to 0.3198) with a relatively small change in the real power delivered by line L2. (c) Phase angle regulating transformer, 𝐶̄ = 1.0∠ − 3° Using 𝑌̄21 = −0.2093 + 𝑗8.9945 and 𝑌̄22 = −𝑗9.0 per unit from Example 3.13(b): V= = I L1 =

−Y22V  − I Load ( j 9.0 )(1.00 ) − 1.0 − 30 = Y21 −0.2093 + j8.9945 −0.8660 + j 9.50 9.53995.21 = = 1.0603.89 per unit −0.2093 + j8.9945 8.99791.33 V − V  1.0603.879 − 1.00 0.0578 + j 0.717 = = jX L1 j 0.20 j 0.20

= 0.4606 − 38.87 per unit PL1 + jQL1 = V I L1 = 0.4606 + 38.87 = 0.3586 + j 0.2890 PL 2 + jQL 2 = ( PLoad + jQLoad ) − ( PL1 + jQL1 ) = 0.5074 + j 0.2110

The phase-angle regulating transformer increases the real power delivered by line L2 31.8% (from 0.3849 to 0.5074) with a relatively small change in the reactive power delivered by line L2.

3.60 Losses are minimum at 0 degrees = 16.606 MW (there can be a +/- 0.1 variation in this value values of the power flow solution tolerance). MW Losses

-10 -9 -8 -7 -6 -5 -4 -3 -2

3.61 Minimum occurs at a tap of 1.0 = 16.604 (there can be a +/- 0.1 variation in this value values of the power flow solution tolerance). MW Losses 0.9 0.90625 0.9125 0.91875 0.925 0.93125 0.9375 0.94375 0.95

3.62 Using (3.8.1) and (3.8.2) 𝑎𝑡 =

13.8 13.8 = 0.03636𝑏 = = 0.04 345(1.1) 345

𝑐 = 𝑎𝑡 /𝑏 = 0.03636/0.04 = 0.90909 From Figure 3.25(a):

 1  cYeq = ( 0.90909 )   = − j18.18 per unit  j 0.05   1  (1 − c ) Yeq = ( 0.0909 )   = − j1.818 per unit  j 0.05  1  ( c − c )Y = ( 0.82645 − 0.90909)  j0.05  = + j1.6529 per unit  2

The per-unit positive-sequence network is:

3.63 A radial line with tap-changing transformers at both ends is shown below:

𝑉̄1′ and 𝑉̄2′ are the supply phase voltage and the load phase voltage, respectively, referred to the high-voltage side. 𝑉̄𝑆 and 𝑉̄𝑅 are the phase voltages at both ends of the line. 𝑡𝑆 and 𝑡𝑅 are the tap settings in per unit. The impedance 𝑍̄ ̶ includes the line impedance plus the referred impedances of the sending end and the receiving end transformers to the high-voltage side. After drawing the voltage phasor diagram for the KVL 𝑉̄𝑆 = 𝑉̄𝑅 + (𝑅 + 𝑗𝑋)𝐼̄, neglecting the phase shift between 𝑉̄𝑆 and 𝑉̄𝑅 as an approximation, and noting that 𝑉̄𝑆 = 𝑡𝑆 𝑉̄1′ and 𝑉̄𝑅 = 𝑡𝑅 𝑉̄2′ , it can be shown that |𝑉̄2′ | |𝑉̄1′ | 𝑡𝑆 = 𝑅𝑃𝜑 + 𝑋𝑄𝜑 1− |𝑉̄1′ ||𝑉̄2′ | √

where 𝑃𝜑 and 𝑄𝜑 are the load real and reactive powers per phase and it is assumed that 𝑡𝑆 𝑡𝑅 = 1. In our problem,

P =

1 (150  0.8 ) = 40 MW 3

and

Q =

1 (150  0.6 ) = 30 MVAR 3

V1 = V2 =

𝑡𝑆 is calculated as

tS =

230 3

(18)( 40 ) + ( 60 )( 30 ) 1−

( 230 / 3 )

and

tR =

= 1.08pu

1 = 0.926 pu 1.08

3.64 With the tap setting t = 1.05, V = t − 1 = 0.05pu The current setup by V = 0.050 circulates around the loop with switch S open; with S closed, only a very small fraction of that current goes through the load impedance, because it is much larger than the transformer impedance; so the superposition principle can be applied to 𝛥𝑉̄ and the source voltage. From 𝛥𝑉̄ Alone, 𝐼𝐶𝐼𝑅𝐶 = 0.05/𝑗0.2 = −𝑗0.25 With 𝛥𝑉̄ shorted, the current in each path is one-half the load current. Load current is

1.0 = 0.8 − 𝑗0.6 0.8+𝑗0.6

Superposition yields: 𝐼̄𝑇𝑎 = 0.4 − 𝑗0.3 − (−𝑗0.25) = 0.4 − 𝑗0.05 𝐼̄𝑇𝐵 = 0.4 − 𝑗0.3 + (−𝑗0.25) = 0.4 − 𝑗0.55 So that STa = 0.4 + j 0.05pu and 𝑆̄𝑇𝐵 = 0.4 + 𝑗0.55 The transformer with the higher tap setting is supplying most of the reactive power to the load. The real power is divided equally between the transformers.

3.65 Same procedure as in Problem 3.64 is followed. Now 𝑡 = 1.0∠4° So 𝑡 − 1 = 1.0∠4° − 1∠0° = 0.07∠91.5° 0.07∠91.5° 𝐼̄𝐶𝐼𝑅𝐶 = 0.2∠90° = 0.350 + 𝑗0.009

Then 𝐼̄𝑇𝑎 = 0.4 − 𝑗0.3 − (0.350 + 𝑗0.009) = 0.050 − 𝑗0.309 𝐼̄𝑋𝑗 = 0.4 − 𝑗0.3 + (0.350 + 𝑗0.009) = 0.750 − 𝑗0.291 So 𝑆̄𝑋𝑗 = 0.050 + 𝑗0.309; 𝑆̄𝑇𝑏 = 0.750 + 𝑗0.291 The phase shifting transformer is useful to control the amount of real power flow; out has less effect on the reactive power flow.

Chapter 4 Transmission Line Parameters

4.1

𝑅𝑑𝑐,20 =

4.2

𝜌=

(10.37 Ω-cmil/ft)(10 ft) 50+234.5 = 1.659 mΩ; 𝑅𝑑𝑐,50 = (1.659 mΩ) ( ) = 1.854 mΩ (250 mil)2 20+234.5

(1 Ω)(500 mil)2 14000 feet

𝜌

= 17.857 Ω-cmil/ft; 𝑇2 = 𝜌2 (𝑇1 + 𝑇) − 𝑇 = 1

17.857 (20 + 228.1) − 228.1 = 17

32.5° C 4.3

l = 1.05  2000 = 2100 m , Allowing for the twist. X-sectional area of all 19 strands = 19 

 4

 (1.5  10 −3 ) = 33.576  10 −6 m 2 . 2

Pl 1.72  10−8  2100 = = 1.076  A 33.576  10−6 2

4.4

  sq  mil   1in   0.0254 m  (a) 795 MCM = ( 795  103 cmil )  4      1 cmil  1000 mil   1in  = 4.0283  10 −4 m 2

 50 + T  (b) R60 HZ, 50°C = R60 HZ, 75°C    75 + T   50+228.1  = 0.0880    75+228.1  = 0.0807 /km

4.5

Ω mi

1mi Ω ) = 0.07365 per conductor (at 75% current 1.609km km

From Table A-4 𝑅60 𝐻𝑧,50°𝐶 = (0.1185 ) ( capacity). For 3 conductors per phase: 𝑅60 𝐻𝑧,50°𝐶 =

4.6

Total transmission line loss PL =

0.07365 Ω = 0.02455 km per phase 3

2.5 (190.5) = 4.7625MW 100

190.5  103

= 500 A

3 ( 220 )

From PL = 3I 2 R , the line resistance per phase is R=

4.7625106 3 ( 500 )

= 6.35 

The conductor cross-sectional area is given by

( 2.84  10−8 )( 63  103 ) = 2.81764  10−4 m 2 6.35

d = 1.894cm = 0.7456in = 745.6 mil A = d 2 = ( 745.6 ) = 556,000 cmil 2

4.7

12 W 50+234.5 = 37.0 mΩ. 𝜌 = 10.37 ( ) = 11.59 Ω-cmil/ft; A)2 20+234.5

𝑅 = (18

(11.59)(4000)

diameter = √

4.8

(a)

0.037

= 1119 mil = 1.119 in

From Eq. (4.4.10) H  1000 m  1000 m H  1 Lint =   10 −7    = 0.05mH/ km Per Conductor 2 m   1km  1H 

(b) From Eq. (4.5.2) DH Lx = Ly = 2  10 −7 Ln      m −1  0.015  −3 D = 0.5m  = e 4   = 5.841  10 m 2  

0.5   H  1000 m  1000 mH  Lx = Ly = 2  10 −7 Ln     −3  H  5.841  10  m  km  

= 0.8899

mH per conductor km

mH per circuit km

(a) Lint = 0.05mH/ km Per Conductor 0.5   6 Lx = Ly = 2  10 −7 ln  10 = 0.8535mH/ km Per Conductor −3   1.2  5.841  10  L = L x + L y = 1.707mH / km Per Circuit

(b) 12( Ic ) = 0.2 I c ln

Dc 2 mWb/km Dc1

0.5   6 Lx = Ly = 2  10 −7 ln  10 = 0.9346 mH/km Per Conductor −3  0.8  5.841  10   L = L x + L y = 1.869mH/km Per Circuit.

Lint is independent of conductor diameter.

The total inductance decreases 4.1% (increases 5%) at the conductor diameter increases 20 % (decreases 20%). 4.10 From Eq. (4.5.10) DH L1 = 2  10 −7 Ln    r  m

D = 4 ft −1  .5   1ft  r = e 4      2   12 in 

4   L1 = 2  10 −7 Ln  −2  1.6225  10   H m

L1 = 1.101  10 −6

r  = 1.6225  10 −2 ft

X1 =  L1 = ( 2 60 ) (1.101 10−6 ) (1000 ) = 0.4153  / km 4.8   4.11 (a) L1 = 2  10 −7 ln  = 1.138  10 −6 H/m −2   1.6225  10 

X1 =  L1 = 2 ( 60 ) (1.138  10−6 ) (1000 ) = 0.4292  /km 3.2   (b) L1 = 2  10 −7 ln  = 1.057  10 −6 H/m −2  1.6225  10  

X1 = 2 ( 60 ) (1.057  10−6 ) (1000 ) = 0.3986  / km L1 and X1 increase by 3.35% (decrease by 4.02%) as the phase spacing increases by 20% (decreases by 20%). 4.12 For this conductor, Table A.4 lists GMR to be 0.0217 ft. For one conductor, 𝐿𝑥 = 2 × 10−7 ln

25 =1.409μH/m; The inductive reactance is 2𝜋(60)(𝐿𝑥 )Ω/m = 0.0217

0.531 Ω/km = 0.855 Ω/mi. For the single-phase line, with two conductors, 2 × 0.855 = 1.710 Ω/mi. 4.13 (a) The total line inductance is given by D  LT =  4  10−4 ln  mH/ m r   3.5 = 4  10−4 ln = 0.002 mH/ m ( 0.7788 )( 0.03)

(b) The total line reactance is given by XT = 2 ( 60 ) 4  10 −4 ln = 0.1508 ln

0.2426 ln

D r

D  / km r

D  / mi r

 X T = 0.755  / km or 1.215  / mi (c) LT = 4  10−4 ln

7 = 0.00228 mH/ m 0.7788 ( 0.03)

Doubling the separation between the conductors causes only about a 14% rise in inductance. 4.14 (a) Eq. (4.5.9): L = 2  10 −7 ln

D H/m Per Phase r

X =  L = 4 f  10 −7 ln ( D / r  )  / m/ Phase = f .  10 −7 (1609.34 ) ln ( D / r  )  / MILE /PH. = f .4 (1609.34 )( 2.3026 )10 −7 log ( D / r  )  /mi /ph. = 4.657  10 −3 f log ( D / r  )  /mi/ph. = 0.2794 log ( D / r  )  / mi / ph. at f = 60 HZ . D 1  X = k log   = k log D + k log   , Where k = 4.657  10 −3 f  r    r 



(b) r  = r.e −1/ 4 = 0.06677 ( 0.7788 ) = 0.052 ft. X a = k log

1  1  = 0.2794 log   = 0.35875 r  0.052 

X d = k log D = 0.2794 log (10 ) = 0.2794 X = X a + X d = 0.63815  / mi/ ph. 

When spacing is doubled, X d = 0.36351 and X = 0.72226  / mi/ph. 

4.15 For each of six outer conductors: −

D11 = r  = e 7 r D12 = D16 = D17 = 2r D13 = D15 = 2 3r D14 = 4r

For the inner conductor: −

D77 = 1 = e 4 r D71 = D72 = D73 = D74 = D75 = D76 = 2r     2 1  − 1      −   3 6  e 4 r  ( 2r ) 2 3r ( 4r )    e 4 r  ( 2r )  Ds = GMR = 49  6        Distances for each Outer conducter  Six outer conductors  Distances for inner conductor 

(

)

(

)

(

)

1 12  −1  18 6 −  6 Ds = GMR = r  e 4  ( 2 ) 2 3 ( 4 )  e 4  ( 2 )     49

12  −1  24 6 Ds = GMR = r  e 4  ( 2 ) 2 3 ( 4 ) = 2.177 r   49

4.16

(

DSL = N b D11 D12  D1N b

) = ( D D  D ) Nb

1N b

1 Nb

 ( n -1)   D11 = DS D1n = 2 A sin   n = 2,3,  N b  Nb  1

       2     3     N b − 1    Nb DSL =  DS 2 A sin  2 A sin 2 A sin  2 A sin              N b     N b     N b     N b      Using the trigonometric identity,



( N −1)

DSL = DS ( A ) b



N b Nb which is the desired result.

Two-conductor bundle, Nb = 2

d A= 2

 d  DSL =  DS   ( 2 )   2 

1/ 2

= DS d Eq (4.6.19) Three-conductor bundle, Nb = 3 A=

  d 2  DSL =  DS   3   3  

d 3

1/ 3

= √𝐷𝑆 𝑑2Eq (4.6.20)

Four-conductor bundle, Nb = 4

  d 3  A= DSL =  DS   4 2   2   d

1/ 4

 4  = 4 DS d 3 4   2 2  = 1.0905 4 DS d 3 Eq (4.6.21) 3

1  − 1   3 − 4.17 (a) GMR = 9  e 4 r  ( 2r )( 2r )  = r 4e 4    = 1.4605 r

     − 1    − 1   (b) GMR = 16  e 4 r  ( 2r )( 4r )( 6r )   e 4 r  ( 2r )( 2r )( 4r )         Distancesfor each outer conductor   Distancesfor each inner conductor 

 −1  6 4 2 GMR = 16  e 4  ( 2 ) ( 4 ) ( 6 ) ( r ) = 2.1554r  

  −  (c) GMR = r  e  ( 2 ) ( 4 )   

    −  32    e  ( 2 )     

( 20 ) ( 8 )(

)

Distances for each corner conductor

( 8) ( 2

  20 ( 4 )   

)

Distances for each outside non-corner conductor

  −    e  ( 2 )    1

  8   

( )

Distances for the center conductor

 −1  24 GMR = r 81  e 4  ( 2 )  

( 8 ) ( 20 ) ( 4 ) ( 32 ) 16

GMR = 2.6374 r

4.18 𝑅 = 0.1128

Ω 3 ; 𝑅𝑒𝑞 = √7.5 ⋅ 7.5 ⋅ 15 = 9.45 m; mi 1m

From Table A.4, 𝐷𝑠 = (0.043 ft) 3.28ft = 0.0123 m; 𝐷𝑒𝑞

9.45

𝐿𝑖 = 2 × 10−7 ln ( 𝐷 ) = 2 × 10−7 ln (0.0123) = 1.3239 μH/m; 𝑠

𝑍 = 𝑅 + 𝑗𝜔𝐿 = 0.1128 ( = 0.0701 + 𝑗0.499

1 1.609

km mi

) + 𝑗(2)(𝜋)(60)(1.3239 × 10−6 )(1000 m/km)

Ω km 𝐷𝑒𝑞

10.39

4.19 (a) 𝑅𝑒𝑞 = √8.25 ⋅ 8.25 ⋅ 16.5 = 10.39 m; 𝐿𝑖 = 2 × 10−7 ln ( 𝐷 ) = 2 × 10−7 ln (0.0123) = 𝑠

1.3478 μH/m; 𝑍 = 𝑅 + 𝑗𝜔𝐿 = 0.1128 (

1 1.609

km mi

) + 𝑗(2)(𝜋)(60)(1.3478 ×

10−6 )(1000 m/km) = 0.0701 + 𝑗0.508 km 𝐷

8.50 )= 0.0123

(b) 𝑅𝑒𝑞 = √6.75 ⋅ 6.75 ⋅ 13.5 = 8.50 m; 𝐿𝑖 = 2 × 10−7 ln ( 𝑒𝑞 ) = 2 × 10−7 ln ( 𝐷𝑠

1.3076 μH/m; 𝑍 = 𝑅 + 𝑗𝜔𝐿 = 0.1128 (

1 1.609

km mi

) + 𝑗(2)(𝜋)(60)(1.3076 ×

10−6 )(1000 m/km) = 0.0701 + 𝑗0.493 km

Ω 1m ; From Table A.4, 𝐷𝑠 = (0.0391 ft) = 0.0119 m; 𝐿𝑖 = 2 × mi 3.28ft

𝐷

9.45

10−7 ln ( 𝐷𝑒𝑞 ) = 2 × 10−7 ln (0.0119) = 1.3354 μH/m; 𝑍 = 𝑅 + 𝑗𝜔𝐿 = 0.1185 ( 𝑠

1 1.609

km mi

𝑗(2)(𝜋)(60)(1.3354 × 10−6 )(1000 m/km) = 0.0736 + 𝑗0.503 km (d) 𝑅 = 0.1035

Ω 1m ; From Table A.4, 𝐷𝑠 = (0.0420 ft) = 0.00128 m; 𝐿𝑖 = mi 3.28ft 𝐷

9.45

2 × 10−7 ln ( 𝐷𝑒𝑞 ) = 2 × 10−7 ln (0.00128) = 1.3208 μH/m; 𝑍 = 𝑅 + 𝑗𝜔𝐿 = 𝑠

0.1035 (

1 1.609

km mi

) + 𝑗(2)(𝜋)(60)(1.3208 × 10−6 )(1000 m/km) = 0.0643 + 𝑗0.498 km

4.20 𝐷𝑒𝑞 = √12 × 12 × 24 = 15.12 m 1m

From table A.4, 𝐷𝑠 = (0.0435 ft) (3.28ft) = 0.0133m DSL = [Dsd2](1/3) = [(0.0133)(0.5)2] (1/3) = 0.1493m 15.12

𝑋1 = 𝜔𝐿1 = 2𝜋(60)2 × 10−7 ln (0.1493) = 0.348 Ω/km 4.21 (a)

From Table A.4:  1  DS = ( 0.0479 )   = 0.0146 m  3.28  DSL = 3 ( 0.0146 )( 0.5 ) = 0.154 m 2

   15.12   X 1 = ( 2 60 )  2  10−7 Ln     1000 = 0.346 km  0.154     1  (b) DS = ( 0.0391)   = 0.0119m  3.28  DSL = 3 ( 0.0119 )(.5 )(.5 ) = 0.144 m    15.12   X 1 = ( 2 60 )  2  10−7 Ln     1000 = 0.351 km  0.144   

Results ACSR Conductor

Aluminum Cross Section

kcmil

km

Canary

900

0.352

Finch

1113

0.348

Martin

1351

0.346

% change 0.9% 0.6%

4.22 Application of Eq. (4.6.6) yields the geometric mean distance that separates the two bundles:

DAB = 9 ( 6.1) ( 6.2 ) ( 6.3) 6 ( 6.05)( 6.15)( 6.25) = 6.15m 2

The geometric mean radius of the equilateral arrangement of line A is calculated using Eq. (4.6.7):

RA = 9 ( 0.015576 ) ( 0.1) = 0.0538m 3

In which the first term beneath the radical is obtained from r  = 0.7788 r = 0.7788 ( 0.02 ) = 0.015576 m

The geometric mean radius of the line B is calculated below as per its configuration:

RB = 9 ( 0.015576 ) ( 0.1) ( 0.2 ) = 0.0628m 3

The actual configuration can now be replaced by the two equivalent hollow conductors each with its own geometric mean radius and separated by the geometric mean distance as shown below:

4.23 (a) The geometric mean radius of each phase is calculated as

R = 4 ( r  ) ( 0.3) where r  = 0.7788  0.0074 2

= 0.0416m The geometric mean distance between the conductors of phases A and B is given by

DAB = 4 62 ( 6.3)( 5.7 ) = 5.996 − 6m Similarly, DBC = 4 62 ( 6.3)( 5.7 ) = 5.996 − 6m and

DCA = 4 122 (12.3)(11.7 ) = 11.998 −12m

The GMD between phases is given by the cube root of the product of the three-phase spacings. Deq = 3 6  6  12 = 7.56m

The inductance per phase is found as L = 0.2 ln

7.56 = 1.041mH/km 0.0416

L = 1.609  1.041 = 1.674 mH/mi

(b) The line reactance for each phase then becomes X = 2 f L = 2 ( 60 )1.674  10 −3 = 0.631 / mi per phase

4.24 From the ACSR Table A.4 of the text, conductor GMR = 0.0244 ft. Conductor diameter = 0.721 in.; since

( 40 ) + (16 ) = 43.08, 2

GMD between phases = ( 43.08 )( 80 )( 43.08 ) 

1/ 3

(a)  X = k log

= 52.95 in.

D  52.95 /12  = 0.2794 log   = 0.6307  /mi  r  0.0244 

 52.95 /12  −7 (b) L = 2  10 −7 ln   = 10.395  10 H/m  0.0244 

X =  L = 2 ( 60 )10.395  10 −7  / m = 2 ( 60 )10.395  10 −7 (1609.34 )

 mi

= 0.6307 /mi  4.25 Resistance per phase =

0.12 = 0.03 / mi 4

GMD = ( 41.76 )( 80 )( 41.76 ) 

1/ 3

, using

402 + 122 = 41.76.

= 51.78ft. 3 GMR for the bundle :1.091 ( 0.0403 )(1.667 )   

1/ 4

by Eq. (4.6.21)

[Note: from Table A.4, conductor diameter. = 1.196 in.; r =

1.196 1  = 0.0498ft.] and conductor 2 12

GMR = 0.0403 ft. GMR for the 4-conductor bundle = 0.7171 ft  51.87   X = 0.2794 log   = 0.5195  / mi   0.7171 

Rated current carrying capacity for each conductor in the bundle, as per Table A.4, is 1010 A; since it is a 4-conductor bundle, rated current carrying capacity of the overhead line is

1010  4 = 4040A  4.26 Bundle radius A is calculated by 0.4572 = 2 A sin( / 8) or A = 0.5974 m GMD = 17m  4.572  Subconductor’s GMR is r  = 0.7788   10 −2  = 1.7803  10 −2 m  2 

  GMD 20   −7 L = 2  10 ln = 2  10 ln   1/8 N -1 1/ N 7 [ Nr( A) ]  8 (1.7803  10−2 ) ( 0.5974 )      −7

which yields L = 7.38  10−7 H / m  4.27 (a) DABeq = [30  30  60  120]1/ 4 = 50.45ft DBCeq = [30  30  60  120]1/ 4 = 50.45ft DACeq = [60  60  150  30]1/ 4 = 63.44 ft  GMD = (50.45  50.45  63.44)1/ 3 = 54.46 ft Equivalent GMR = (0.0588)3 (90)3 

1/ 6

= 2.3ft

 54.46  −6  L = 2  10 −7 ln   = 0.633  10 H/m   2.3 

(b) Inductance of one circuit is calculated below: Deq = [30  30  60]1/ 3 = 37.8ft; r  = 0.0588ft  37.8  −6  L = 2  10 −7 ln   = 1.293  10 H/m  0.0588 

Inductance of the double circuit =

1.293  10 −6 = 0.646  10 −6 H/m  2

 0.633 − 0.646  Error percent =    100 = −2.05%  0.633  

4.28 With N = 3, S = 21 , A =

S 21/12 = = 1.0104 ft 2sin 60 2  0.866

Conductor GMR = 0.0485ft 2 Bundle GMR = 3 ( 0.0485 )(1.0104 )   

Then rA = ( GMRb )( DAA )  rB = ( GMRb  DBB )

1/ 2

rC = ( GMRb  DCC  )

1/ 2

1/ 3

= 0.5296 ft

(

1/ 2

= 0.5296  322 + 362

= ( 0.5296  96 )

1/ 2

1/ 3

1/ 2

= 7.13ft

= 0.5296  322 + 362   

Overall Phase GMR = ( rArBrC )

) = 5.05ft

1/ 2

= 5.05ft

= 5.67ft

DABeq =  32 2 + 36 2  64 2 + 36 2  ( 64 )( 32 )    DBCeq = ( 32 ) 64 2 + 36 2  64  32 2 + 36 2    DACeq = ( 36 )( 32 )( 36 )( 32 ) 

1/ 4

= 51.88ft

= 33.94 ft

 GMD =  51.88  51.88  33.94 

1/ 3

= 45.04 ft

 45.04  Then X L = 0.2794 log   = 0.2515  /mi/phase   5.67 

4.29

rA = 0.5296 ( 32 )  rB = 0.5296 ( 32 ) 

1/ 2

= 4.117ft

1/ 2

= 4.117ft

rC = 0.5296 ( 32 ) 

1/ 2

= 4.117ft

Then GMR phase = 4.117ft





1/ 4

2 2 DABeq =  ( 36 ) + ( 32 ) 36  642 + 362   

DBCeq = ( 64 )( 96 )( 32 )( 64 ) 

1/ 4



DACeq =  ( 32 ) + ( 36 )  2

= 49.76ft

= 59.56ft

 36 + 64  (36) = 49.76ft 1/ 4

Then GMD = ( 49.76  59.56  49.76 )

1/ 3

= 52.83ft

 52.83  X L = 0.2794 log   = 0.3097  /mi/phase   4.117 

4.30

rA = 0.5296 ( 96 )  rB = 0.5296 ( 32 ) 

1/ 2

= 7.13ft

1/ 2

= 4.117ft = rC

From which GMRphase = ( 7.13  4.117  4.117 )

1/ 3

DABeq = ( 32  64  32  64 )

1/ 4



= 4.95ft

= 45.25ft



2 2 DBCeq = ( 36 ) ( 32 ) + ( 36 ) ( 36 )   

DACeq = ( 322 + 362 ) (642 + 362 ) 

1/ 4

1/4

= 41.64 ft

= 59.47ft

Then GMD = ( 45.25  41.64  59.47 )

1/ 3

= 48.21ft

 48.21  X L = 0.2794 log   = 0.2762  /mi/ph.   4.95 

4.31 Flux linkage between conductors 1 & 2 due to current, Ia, is

12( I ) = 0.2 I a ln a

Da 2 mWb/km Da1

Db1 = Db 2 , 12 due to I b is zero.

12( I ) = 0.2 I c ln c

Dc 2 mWb/km Dc1

Total flux linkages between conductors 1 & 2 due to all currents is 74 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

12 = 0.2 I a ln

Da 2 D + 0.2 I c ln c 2 mWb/km Da1 Dc1

For positive sequence, with I a as reference, I c = I a  − 240

 D D   12 = 0.2 I a  ln a 2 + (1 − 240 ) ln c 2  Dc1   Da1 = 0.2 ( 250 )  ln ( 7.21/ 6.4 ) + (1 − 240 ) ln ( 6.4 / 7.21)  = 10.31 − 30 mWb/km  Note: Da1 = Dc 2 = 42 + 52 = 6.4 m    2 2 1/ 2     D = D = 5.2 + 5 = 7.21m a2 c1  ( ) ( )  

with I a as reference, instantaneous flux linkage is

12 ( t ) = 2 12 cos(t +  )  Induced voltage in the telephone line per km is VRMS = 12  + 90 = j12 = j ( 2  60 )(10.31 − 30 )10 −3 = 3.8960 V 

4.32

Cn =

2 ( 8.854  10 −12 ) 2 0 F = = 1.3246  10 −11 to neutral m D  0.5  Ln   Ln     0.015 / 2 

Yn = jCn = j ( 2 60 ) (1.3246  10 −11 ) Yn = j 4.994  10 −6

4.33 (a) Cn =

S m  1000 m km

S to neutral km

2 ( 8.854  10−12 ) = 1.385  10−11 F/ m to neutral  0.5  ln    0.018 / 2 

Yn = j 2 ( 60 )1.385  10 −11 (1000 ) = j 5.221  10 −6 S/km to neutral

(b) Cn =

2 ( 8.854  10 −12 )  0.5  ln    0.012 / 2 

= 1.258  10 −11 F/m to neutral

Yn = j 2 ( 60 )1.258  10 −11 (1000 ) = j 4.742  10 −6 S/km to neutral

Both the capacitance and admittance-to-neutral increase 4.5% (decrease 5.1%) as the conductor diameter increases 20% (decreases 20%).

4.34

2 ( 8.854  10 −12 ) 2 0 F C1 = = = 1.058  10 −11 4 m D   Ln   Ln    0.25 /12     Y1 = jC1 = j ( 2 60 ) (1.058  10 −11 ) (1000 ) = j 3.989  10 −6

4.35 (a) C1 =

S km

2 ( 8.854  10 −12 )  4.8  ln    0.25 /12 

= 1.023  10 −11 F/m

Y1 = j 2 ( 60 )1.023  10 −11 (1000 ) = j3.857  10 −6 S/km

(b) C1 =

2 ( 8.854  10 −12 )  3.2  ln    0.25 /12 

= 1.105  10 −11 F/m

Y1 = j 2 ( 60 )1.105  10 −11 (1000 ) = j 4.167  10 −6 S/km

The positive sequence shunt capacitance and shunt admittance both decrease 3.3% (increase 4.5%) as the phase spacing increases by 20% (decreases by 20%) 4.36 (a) Equations (4.10.4) and (4.10.5) apply. For a 2-conductor bundle, the GMR Dsc = rd = 0.0074  0.3

= 0.0471 The GMD is given by Deq = 3 6  6  12 = 7.56m Hence the line-to-neutral capacitance is given by Can =

2 F/m ln( Deq / Dsc )

55.63 = 10.95nF/km ln ( 7.56 / 0.0471)

( with  =  0 ) or 1.609  10.95 = 17.62 n F/mi (b) The capacitive reactance at 60 HZ is calculated as XC =

Deq 1 = 29.63  103 ln  − mi 2 ( 60 ) Can Dsc

= 29.63  10 3 ln

7.56 = 150,500  − mi 0.0471

4.37 (a) Eq (4.9.15): capacitance to neutral = XC =

2 F/m ln( D / r )

1 ln( D / r ) =   m to neutral 2 fC (2 f )(2 )

with f = 60HZ,  = 8.854  10 −12 F/m . or XC = k  log ( D / r ) ,

where k  =

1 = k  log D + k  log   , r

4.1  106 ,   mile to neutral  f

where k  = 0.06833  106

at f = 60 HZ .

(b) X d = k  log D = 0.06833  106 log (10 ) = 68.33  103 1 1 X a = k  log   = 0.06833  106 log = 80.32  103 0.06677 r

 Xc = X d + X a = 148.65  103   mi to neutral  When spacing is doubled, X d = 0.06833  106 log ( 20 ) = 88.9  103

Then XC = 169.12  103   mi to neutral  4.38

 52.95 /12  C = 0.0389 / log  = 0.018 F/mi/ph .  0.721/ (12  2 )    XC =

1 = 147.366  103   mi  2 ( 60 ) 0.018  10 −6 3

4.39 𝐷𝑒𝑞 = √7.5 ⋅ 7.5 ⋅ 15 = 9.45 m; From Table A.4, 𝑟 = 𝐶1 =

2𝜋𝜖0 𝐷 ln( ) 𝑟

1.196 0.0254m in 1in = 0.01519m; 2

2𝜋 8.854×10−12 ln(

9.45 ) 0.01519

= 8.648 × 10−12 F/m; S

𝑌 = 𝑗𝜔𝐶 = 𝑗2𝜋(60)8.648 = 𝑗3.260 × 10−6 km

4.40 (a) 𝐷𝑒𝑞 = √8.25 ⋅ 8.25 ⋅ 16.5 = 8.50 m; 𝐶1 =

2𝜋𝜖0 𝐷 ln( ) 𝑟

2𝜋 8.854×10−12 ln(

8.50 ) 0.01519

= 8.522 × 10−12 F/m

𝑌 = 𝑗𝜔𝐶 = 𝑗2𝜋(60)8.648 = 𝑗3.213 × 10−6

S km

(b) 𝐷𝑒𝑞 = √6.75 ⋅ 6.75 ⋅ 13.5 = 10.39 m; 𝐶1 =

2𝜋𝜖0 𝐷 ln( ) 𝑟

2𝜋 8.854×10−12 ln(

10.39 ) 0.01519

= 8.792 × 10−12 F/m;

𝑌 = 𝑗𝜔𝐶 = 𝑗2𝜋(60)8.648 = 𝑗3.315 × 10−6 4.41

S km

Deq = 3 10  10  20 = 12.6m

From Table A.4, r =

1.293  0.0254 m  in   = 0.01642 m 2  1in 

DSC = 3 rd 2 = 3 0.01642(0.5)2 = 0.16 m C1 =

2 ( 8.854  10 −12 ) 2 0 = = 1.275  10 −11 F/m Deq  12.6  ln  ln  DSc  0.16 

Y1 = jC1 = j 2 ( 60 )1.275  10 −11 (1000 ) = j 4.807  10 −6 S/km Q1 = VLL 2Y1 = ( 500 ) 4.807  10 −6 = 1.2 MVAR/ km 2

4.42 (a) From Table A.4, r =

1.424 ( 0.0254 ) = 0.0181m 2

DSc = 3 0.0181( 0.5 ) = 0.1654 m 2

C1 =

2 ( 8.854  10 −12 )

= 1.284  10 −11 F/m  12.6  ln    0.1654  Y1 = j 2 ( 60 ) (1.284  10 −11 ) (1000 ) = j 4.842  10 −6 S/km

Q1 = ( 500 ) 4.842  10 −6 = 1.21MVAR/ km 2

(b) r =

1.162 2 ( 0.0254 ) = 0.01476 m; DSc = 3 0.01476 ( 0.5 ) = 0.1546 m 2

C1 =

2 ( 8.854  10 −12 )

= 1.265  10 −11 F/m  12.6  ln    0.1546  Y1 = j 2 ( 60 )1.265  10 −11 (1000 ) = 4.77  10 −6 S/km

Q1 = ( 500 ) 4.77  10 −6 = 1.192 MVAR/ km 2

C1,Y1, and Q1 increase 0.8% (decrease 0.7%) For the larger, 1351 kcmil conductors (smaller, 700 kcmil conductors). 4.43 (a) For Drake, Table A.4 lists the outside diameter as 1.108 in  r=

1.108 = 0.0462 ft 2  12

Deq = 3 20  20  38 = 24.8ft

Can =

2  8.85  10 −12 = 8.8466  10 −12 F/m ln ( 24.8 / 0.0462 )

XC =

1012 = 0.1864  106   mi 2 (60)8.8466  1609

(b) For a length of 175 mi 0.1864  106 = 1065  to neutral 175 220  103 1 0.22 I chg = = = 0.681A/mi XC 3 3  0.1864

Capacitive reactance =

0.681175 = 119A for the line

Total three-phase reactive power supplied by the capacitance is given by

3  220  119  10 −3 = 43.5MVAR 4.44

 51.87  C = 0.0389 log   = 0.0212  F /mi/ph  0.7561 

 Note: Equivalent radius of a 4-cond. bundle is given by    3 1/ 4 3 1/ 4  1.091( 0.0498 d ) = 1.091( 0.0498  1.667 ) = 0.7561ft 

XC =

1 = 125.122  103   mi  −6 2 ( 60 ) 0.0212  10

4.45

From Example 4.8, 2 0

C Xn =

D  HXY  Ln   − Ln     HXX 

C Xn = 1.3247  10 −11

2 ( 8.854  10 −12 )  0.5   20.006  Ln  − Ln     0.0075   20 

F which is 0.01% larger than in Problem 4.32 m

4.46 (a) Deq = 3 12  12  24 = 15.12m

r = 0.0328 / 2 = 0.0164 m XC =

Where

1 2 fCan

2  8.85  10 −12 ln (15.12 / 0.0164 )

Can =

 XC =

2.86 15.12  10 9 ln = 3.254  108   m 60 0.016 

For 125 km, XC =

3.254  108 = 2603  125  1000

(b)

H1 = H2 = H3 = 40m

H12 = H23 = 402 + 122 = 41.761m H31 = 402 + 242 = 46.648m

Deq = 15.12 m and r = 0.0164 m

 XC =

 Deq 1 H12 H 23 H31  2.86  10 9  ln − ln m 60 3 H1 H 2 H3   r

 15.12 1 41.761  41.761  46.648  = 4.77  107  ln − ln  40  40  40  0.0164 3  8 = 3.218 10   m For 125 km, XC =

3.218  108 = 2574  125  103

4.47 D = 10 ft; r = 0.06677 ft; Hxx = 160 ft.; H xy = 1602 + 102

= 160.3ft (See Fig. 4.24 of text) Line-to-line capacitance C xy =

 H D ln − ln xy r H xx

F/ m

(See Ex. 4.8 of the text) C xy =

 ( 8.854  10 −12 )  10   160.3  ln   − ln  160  0.06677    

Neglecting Earth effect, C xy =

= 5.555  10 −12 F/ m

 ( 8.854  10 −12 )  10  ln    0.06677 

= 5.553  10 −12 F/ m Error - percentage =

5.555 − 5.553  100 = 0.036% 5.555

When the phase separation is doubled, D = 20 ft

H xy = 1602 + 202 = 161.245 With effect of Earth,

C xy =

 ( 8.854  10 −12 )  20   161.245  ln  − ln     0.06677   160 

= 4.885  10−12 F/m

Neglecting Earth effect, C xy =

 ( 8.854  10 −12 )  20  ln    0.06677 

= 4.878  10−12 F/ m

Error percentage =

4.885 − 4.878  100 = 0.143% 4.885

4.48 (a) H1 = H2 = H3 = 2  50 = 100 ft H12 = H 23 = 252 + 100 2 = 103.08ft H13 = 50 2 + 100 2 = 111.8ft Deq = 3 ( 25 )( 25 )( 50 ) = 31.5ft r= Can =

1.065 = 0.0444 ft 2  12 2 ( 8.854  10 −12 )  31.5   106  ln  − ln     0.0444   100 

= 9.7695  10 −12 F/ m 

 Note : H m = (103.08  103.08  111.8)1/ 3 = 106ft    H s = 100ft   (b) Neglecting the effect of ground, Can =

2 ( 8.854  10 −12 )  31.5  ln    0.0444 

= 8.4746  10 −12 F/ m

Effect of ground gives a higher value. Error percent =

4.49

9.7695 − 8.4746  100 = 13.25% 9.7695

GMD = ( 60  60  120 )

1/ 3

= 75.6ft

1.16  = 0.0483ft; N = 4; S = 2 A sin 2  12 N A=

18 = 1.0608ft ( 2sin 45 )12

N −1 GMR = rN ( A )   

 Can =

1/ N

3 1/ 4

= 0.0483  4  (1.0608 )    = 0.693ft

2 ( 8.854  10 −12 )  75.6  ln    0.693 

= 11.856  10 −12 F/m 

Next X d = 0.0683log ( 75.6 ) = 0.1283  1  X a = 0.0683log   = 0.0109  0.693  X c = X a + X d = 0.1392 M   mi to neutral = 0.1392  106   mi to neutral 

4.50 From Problem 4.45 1 1 F C xy = C xn = (1.3247  10−11 ) = 6.6235  10−12 2 2 m with Vxy = 20 kV Qx = C xy Vxy = ( 6.6235  10−12 )( 20  103 ) = 1.3247  10−7

C m

From Eq (4.12.1) The conductor surface electric field strength is:

ET =

1.3247  10 −7 ( 2 ) (8.854  10 −12 ) ( 0.0075 )

= 3.1750  10 5 = 3.175

V  kV  m    m  1000 V   100 cm 

kVrms cm

Using Eq (4.12.6), the ground level electric field strength directly under the conductor is:  ( 2 )(10 ) ( 2 )(10 )  1.3247  10 −7  − 2 −12 ( 2 ) (8.854  10 )  (10 ) (10 )2 + ( 0.5)2  V  kV  kV = 1.188   = 0.001188  m  1000 V  m

Ek =

4.51 (a) From Problem 4.30, 2 ( 8.854  10 −12 )

C xn =

 0.5   20.006  Ln  − Ln     0.009375   20  1 F C xy = C xn = 6.995  10 −12 2 m

= 1.3991  10 −11

F m

Qx = C xy Vxy = ( 6.995  10 −12 )( 20  103 ) = 1.399  10 −7 E =

c m

1.399  10 −7  1  1    −12 ( 2 ) (8,854  10 ) ( 0.009375 )  1000   100 

= 2.682

kVrms cm

 ( 2 )(10 ) ( 2 )(10 )  1.399  10 −7  − 2 ( 2 ) (8.854  10 −12 )  (10 ) (10 )2 + ( 0.5 )2  V  kV  kV = 1.254   = 0.001254  m  1000 V  m

Ek =

(b) C xn =

2 ( 8.854  10 −12 )  0.5   20.06  Ln  − Ln     .005625   20 

= 1.2398  10 −11

F m

1 F C xy = C xn = 6.199  10 −12 2 m Qx = C xy Vxy = ( 6.199  10 −12 )( 20  103 ) = 1.2398  10 −7 E =

c m

1.2398  10 −7  1  1    −12 ( 2 ) (8.854  10 ) (.005625)  1000   100 

E = 3.962

kVrms cm

 ( 2 )(10 ) ( 2 )(10 )  1.2398  10 −7  − 2 ( 2 ) (8.854  10 −12 )  (10 ) (10 )2 + ( 0.5)2  V  kV  kV Ek = 1.112   = 0.001112  m  1000 V  m Eq =

The conductor surface electric field strength E decreases 15.5 % (increases 24.8%) as the conductor diameter increases 25 % (decreases 25%). The ground level electric field strength Ek increases 5.6 %( decreases 6.4%).

Chapter 5 Transmission Lines: Steady-State Operation 5.1

A = D = 1.00 pu; C = 0. S

(a)

B = Z = ( 0.19 + j 0.34 ) 30 = 11.68560.8  (b) VR = IR =

33 3

0 = 19.050 kVLN

SR  − cos−1 ( pf ) 3 VR L - L

10 3(33)

 − cos−1 ( 0.9 )

= 0.175 − 25.84 kA

VS = AVR + B I R = (1.0 )(19.05) + (11.68560.8 )( 0.175 − 25.84 ) = 20.73 + j1.17 = 20.763.236 kVL − N

VSL-L = 20.76 3 = 35.96 kV (c) I R = 0.17525.84

VS = AVR + B I R = (1.0 )(19.05) + (11.68560.8 )( 0.17525.84 ) = 19.17 + j 2.04 = 19.286.078 kVL − N

VS L−L = 19.27 3 = 33.4kV 5.2

Y Z 1 = 1 + ( 3.33  10−6  20090 ) ( 0.08 + j 0.48 )( 200 ) 2 2 = 1 + ( 0.0324 170.5 ) = 0.968 + j 0.00533 = 0.9680.315 pu

(a) A = D = 1 +

 Y Z  −4 C = Y 1 +  = ( 6.66  10 90 ) (1 + .0.0162 170.5 ) 4   = 6.553  10 −4 90.155 S B = Z = 97.32 80.54 

(b) VR =

220

0 = 1270 kVL − N 3 P  − cos−1 ( pf ) 250 − cos−1 0.99 IR = R = = 0.6627 − 8.11 kA 3VR L−L ( pf ) 3 ( 220 )( 0.99 )

VS = AVR + B I R = ( 0.9680.315 )(1270 ) + ( 97.3280.54 )( 0.6627 − 8.11 ) = 142.4 + j 62.16 = 155.423.58 kVL− N

VS L−L = 155.4 3 = 269.2 kV

I S = CVR + D I R = ( 6.553  10−4 90.155 ) (127 ) + ( 0.968 − 0.315 )( 0.6627 − 8.11 ) = 0.6353 − j 3.786  10−3 = 0.6353 − 0.34 kA (c) VRNL = VS A = 269.2 0.968 = 278.1kVLL % VR =

5.3

VR NL − VR FL VR FL

 278.1 − 220   100 =  100 = 26.4% 220  

Z base = V 2base Sbase = ( 230 ) 1000 = 52.9  2

Ybase = 1 Z base = 1.89  10−2 S

(a) A = D = 0.9680 0.315 pu B = ( 97.32 80.54 ) 52.9  = 1.8480.54 pu C = ( 6.553  10−4 90.155 S) (1.89  10 −2 S ) = 0.0346790.155 pu

(b) I base =

Sbase 3 3 Vbase L − L

1000 = 2.51 kA 3 (230)

VRpu = ( 220 230 ) 0 = 0.95650 I Rpu = ( 0.6627  − 8.11 ) 2.51 = 0.264 − 8.11

VSPU = APU VRPU + BPU I RPU

= ( 0.968 0.315 )( 0.9565 0 ) + (1.84 80.54 )( 0.264  − 8.11 ) = 1.0725 + j 0.4682 = 1.1723.58 pu; VS = 1.17 pu

I SPU = CPU VRPU + DPU I RPU

= ( 0.03467 90.155 )( 0.95650 ) + ( 0.9680.315 )( 0.264 − 8.11 ) = 0.2531 − j 0.0015 = 0.2531 − 0.34 pu;

I S = 0.2531 pu

%VR =

VRNLpu − VRFLpu VRFLpu

 1.209 − 0.9565   100 =  100 = 26.4% 0.9565  

5.4

VS   A1  =  I S  C1

B1  Vx   A1   =  D1   I x  C1

VS  ( A1 A2 + B1C2 )  =  I S  ( C1 A2 + D1C2 )

B1   A2  D1  C2

B2  VR    D2   I R 

( A B + B D )  V  (C B + D D )  I  1

5.5

KCL : I S = I R + Y (VR + Z 2 I R ) = Y VR + (1 + Y Z 2 ) I R KVL : VS = VR + Z 2 I R + Z1 I S

= VR + Z 2 I R + Z1 Y VR + (1 + Y Z 2 ) I R  = (1 + Y Z1 ) VR + ( Z1 + Z 2 + Y Z1 Z 2 ) I R

In matrix format:

VS  (1 + Y Z1 )  = Y  I S   5.6

( Z + Z + Y Z Z ) V   (1 + Y Z )   I  1

(a)

VR is taken as reference; I = I R + I CR ; I S = I + I CS ; I CR ⊥ VR (Leading); I CS ⊥ VS (Leading); VR + IR + jIX L = VS

( IR) I ; ( jIX L ) ⊥ I

(b) (i)

 R jX  I S = I R + I C ; VC = VR + I R  + L  ; I C ⊥ VC (Leading) 2  2  R jX  VS = VC + I S  + L  ; VR is taken as reference. 2  2

(ii) For nominal T-circuit

1  1  A = 1 + Y Z = D; B = Z  1 + Y Z  ; C = Y  2 4  

For nominal -circuit of part (a) 1  1  A = D = 1 + Y Z ; B = Z ; C = Y 1 + Y Z   2 4  

5.7

VS =

3300

= 1905.3V (Line-to-neutral) 3 0.553.13 = 0.5 ( 0.6 + j 0.8 ) = 0.3 + j 0.4 I=

(1200 / 3)103 = 500  103 A 0.8  VR

From the phasor diagram drawn below with I as reference,

VS2 = (VR cos R + IR ) + (VR sin R + IX ) 2

(1)

 500  103  0.3   500  103  0.4  (1905.3) =  0.8VR +  +  0.6VR +  VR VR    

From which one gets VR = 1769V (a) Line-to-line voltage at receiving end = 1769 3 = 3064 V  = 3.064 kV

(b) Line current is given by

500  103 = 282.6 A  1769

5.8

(a) From phasor diagram of Problem 5.7 solution,

VR cos R + IR = 1769 ( 0.8 ) + ( 282.6  0.3) = 1500 V Sending-End PF =

1500 1500 = = 0.786 Lagging  VS 1905.3

(b) Sending-End 3-Phase Power = PS = 3 (1905.3)( 282.6 ) 0.786 = 1272 kW 

5.9

(a) From Table A.4, 𝑅 = 0.1128 Ω/mi = 0.0701 Ω/km 𝑧̅ = 0.0701 + 𝑗0.497 = 0.502∠81.97°Ω/km; 𝑦̅ = 3.30 × 10−6 ∠90° s/km ̅̅

̅ = 1 + 𝑌𝑍 = 1 + 1 (3.22 × 10−6 × 100∠90°)(0.513 × 100∠82.14°) = 𝐴̅ = 𝐷 2 2 0.9918∠0.0999° 𝐵̅ = 𝑍̅ = 𝑧̅𝑙 = 0.513 × 100∠82.14° = 51∠82.11° Ω ̅̅

𝑌𝑍 𝐶̅ = 𝑌̅ (1 + ) = (3.22 × 10−4 ∠90°)[1 + 0.004130∠172.14°] 4

218 𝑉̅𝑅 = 3 ∠0° = 125.9∠0° kV𝐿𝑁 √

300

𝐼𝑅̅ = 218 3 ∠(− cos −1 0.9) = 0.7945∠ − 25.84° kA √

𝑉̅𝑆 = 𝐴̅𝑉̅𝑅 + 𝐵̅𝐼𝑅̅ = 0.9918∠0.999 (125.9) + 51∠82.11° (0.7945∠-25.84°) 𝑉𝑠 = 151.3√3 = 262 kVLL 𝑉

262

𝑉𝑅𝑁𝐿 = 𝐴𝑠 = 0.9918 = 264.2 kVLL %𝑉𝑅 =

𝑉𝑅𝑁𝐿 −𝑉𝑅𝐹𝐿 264.2−218 = 218 = 21.2% 𝑉𝑅𝐹𝐿

(b) 𝐼𝑅̅ = 0.7945∠0° kA 𝑉̅𝑆 = 𝐴̅𝑉̅𝑅 + 𝐵̅𝐼𝑅̅ = 0.9918∠0.999 (125.9) + 51∠82.11° (0.7945∠0°) 𝑉𝑠 = 136.6√3∠17.2° = 238.6 kVLL %𝑉𝑅 =

𝑉𝑅𝑁𝐿 −𝑉𝑅𝐹𝐿 238.6−218 = 218 = 9.43% 𝑉𝑅𝐹𝐿

(c) 𝐼𝑅̅ = 0.7945∠25.84° kA 𝑉̅𝑆 = 𝐴̅𝑉̅𝑅 + 𝐵̅𝐼𝑅̅ = 0.9918∠0.999 (125.9) + 51∠82.11° (0.7945∠25.84°) = 118.9∠19.1° kVLN 89 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

𝑉𝑠 = 118.9√3 = 205.9 kVLL %𝑉𝑅 =

𝑉𝑅𝑁𝐿 −𝑉𝑅𝐹𝐿 205.9−218 = 218 = −5.6% 𝑉𝑅𝐹𝐿 1

5.10 From Table A.4, R=3 (0.0969) mi = 0.0201 km From problems 4.20 and 4.41, 𝑧̅ = 0.0201 + 𝑗0.531 = 0.531∠87.8° Ω/km 𝑦̅ = 4.807 × 10−6 ∠90° S/km ̅̅

̅ = 1 + 𝑌𝑍 = 1 + 1 (0.531 × 180∠87.8°)(4.807 × 10−6 × 180∠90°) = 0.9587∠0.0936° pu (a) 𝐴̅ = 𝐷 2 2 𝐵̅ = 𝑍̅ = 𝑧̅ℓ = 0.531 (180)∠87.8° = 95.58∠87.8° Ω ̅̅

𝑌𝑍 𝐶̅ = 𝑌̅ (1 + 4 ) = (4.806 × 10−6 × 180∠90°)(1 + 0.0207∠177.8°) = 8.47 × 10−4 ∠90.05° S

(b) 𝑉̅𝑅 = 𝐼𝑅̅ =

475 ∠0° = 274.24∠0°kV𝐿𝑁 √3

𝑃𝑅 ∠ cos−1(𝑝𝑓) 1600∠ cos−1 0.95 = 3 475 (0.95) = 2.047∠18.19° kA (𝑝𝑓) 3 𝑉 √ 𝑅𝐿𝐿 √

𝑉̅𝑆 = 𝐴̅𝑉̅𝑅 + 𝐵̅𝐼𝑅̅ = (0.9587∠0.0936° pu)(274.24∠0° kVLN ) + (95.58∠87.8° Ω)(2.047∠18.19°) = 281.5∠42.05° kVLN = 487.6∠42.05° kVLL ̅ 𝐼𝑅̅ = (8.47 × 10−4 ∠90.05°)(274.24) + (95.58∠87.8°Ω)(2.047∠18.19°) = 𝐼𝑆̅ = 𝐶̅ 𝑉̅𝑅 + 𝐷 2.047∠24.47° kA (c) 𝑆𝑆̅ = 𝑉̅𝑆 𝐼𝑆∗̅ = (487.6∠42.05° kVLL × √3)(2.047∠24.47°)∗ = 1648 + 𝑗522 p.f. = cos[tan-1(522/1648)] = 0.95 lagging (d) Full-load line losses = 𝑃𝑠 − 𝑃𝑅 = 1648 − 1600 = 48 MW 𝑃

1600

Efficiency = ( 𝑃𝑅 ) = 1648 = 97.1% 𝑆

𝑉

487.6

(e) 𝑉𝑅𝑁𝐿 = 𝐴𝑆 = 0.9587 = 508.6 kVLL %VR =

𝑉𝑅𝑁𝐿 −𝑉𝑅𝐹𝐿 508.6−475 = 475 = +7.1% 𝑉𝑅𝐹𝐿

5.11 (a) The series impedance per phase Z = ( r + j L ) l = ( 0.15 + j 2 (60)1.3263  10−3 ) 40 = 6 + j20 

The receiving end voltage per phase VR =

220 3

0 = 1270 kV

Complex power at the receiving end SR(3 ) = 381 cos−1 0.8MVA = 304.8 + j 228.6 MVA

IR =

( 381 − 36.87)103 = 1000 − 36.87 A

3  1270 The sending end voltage, as per KVL, is given by

VS = VR + Z I R = 1270 + ( 6 + j 20 )(1000 − 36.87 )10 −3 = 144.334.93 kV The sending end line-to-line voltage magnitude is then

VS ( L − L ) = 3 (144.33) = 250 kV The sending power is SS (3 ) = 3VS I S* = 3 (144.334.93 )(100036.87 )10 −3 = 322.8MW+ j 288.6 MVAR = 43341.8 MVA

250 − 220 = 0.136 220 P (3 ) 304.8 Transmission line efficiency is  = R = = 0.944 PS (3 ) 322.8 Voltage regulation is

(b) With 0.8 leading power factor, I R = 100036.87 A The sending end voltage is VS = VR + Z I R = 121.399.29 kV The sending end line-to-line voltage magnitude VS ( L − L ) = 3  121.39 = 210.26 kV

The sending end power SS (3 ) = 3V I

* S S

= 3 (121.399.29 )(1 − 36.87 ) = 322.8 MW − j168.6 M var = 364.2 − 27.58 MVA 210.26 − 220 = −0.0443 220 PR (3 ) 304.8 = = 0.944 Transmission line efficiency  = PS (3 ) 322.8 Voltage regulation =

5.12 (a) The nominal  circuit is shown below:

The total line impedance Z = ( 0.1826 + j 0.784 )100 = 18.26 + j 78.4

= 80.576.89  /ph. The line admittance for 100 mi is

(b) VR =

IR =

230 3

1 1 90 = 90 = 0.5391  10 −3 90 S/ph. 185.5  103 XC 100

0 = 132.80 kV

200  103 3 ( 230 )

0 = 5020 A ( Unity Power Factor )

Y  I Z = I R + VR   = 5020 + (132,8000 ) ( 0.27  10 −3 90 ) 2 = 502 + j 35.86 = 503.34.09 A

The sending end voltage VS = 132.80 + ( 0.50334.09 )( 80.576.89 )

= 139.152 + j 40.01 = 144.7916.04 kV The line-to-line voltage magnitude at the sending end is

3 (144.79) = 250.784 kV

Y  I S = I Z + VS   = 502 + j 35.86 + (144.7916.04 )( 0.2790 ) 2 = 491.2 + j 73.46 = 496.78.5 A

Sending end power SS (3 ) = 3 (144.79)(0.4967 ) 16.04 − 8.5

= 213.88 + j 28.31MVA So PS (3 ) = 213.88 MW; QS (3 ) = 28.31MVAR (c) Regulation = 5.13

VS − VR 144.79 − 132.8 = = 0.09 VR 132.8

 l = 0.485 = 0.034862 + j 0.39848 e l = e0.034862 e j 0.39848 = 1.035480.39848 radians = 0.9543 + j 0.40178 e − l = e −0.034862 e − j 0.39848 = 0.965738 − 0.39848 radians = 0.89007 − j 0.37472 cosh  l =

e l + e −  l = ( 0.9543 + j 0.40178 ) + ( 0.89007 − j 0.37472 ) 2 2 = 0.92219 + j 0.01353 = 0.92230.841 pu

Alternatively:

cosh ( 0.034862 + j 0.39848 ) = cosh ( 0.034862 ) cos ( 0.39848 radians ) + j sinh ( 0.034862 ) sin ( 0.39848 radians ) = (1.00060 )( 0.92165 ) + j (0.034869) ( 0.388018 ) = 0.9222 + j 0.01353 = 0.9223 0.841 pu e l − e −  l = ( 0.9543 + j 0.40178 ) − ( 0.89007 − j 0.37472 )  2 2 = 0.03212 + j 0.38825 = 0.38958 85.271 pu

sinh  l =

  l  cosh (  l ) − 1 ( 0.9222 + j 0.01353 ) − 1 tanh   = = sinh (  l ) 0.38958 85.271  2  0.07897 170.13 = 0.2027 84.86 pu 0.3895885.271

5.14 (a)

ZC =

z 0.03 + j 0.35 = = 282.6 − 2.45  y 4.4  10 −6 90

(b)  l = z y (l ) =

( 0.3512885.101 ) ( 4.4  10−6 90 ) ( 500 )

= 0.62287.55 = 0.02657 + j 0.6210 pu

= ( cosh 0.02657 )( cos 0.6210 radians ) + j ( sinh 0.02657 )( sin 0.6210 radians ) = (1.00035 )( 0.8133) + j ( 0.02657 )( 0.5818 ) = 0.8136 + j 0.015458 = 0.8137 1.09 pu sinh  l = sinh ( 0.02657 + j 0.6210 ) = sinh ( 0.02657 ) cos ( 0.6210 radians ) + j ( cosh 0.02657 )( sin 0.6210 radians ) = (0.02657)(0.8133) + j (1.0004)(0.5818) = 0.0216 + j 0.5820 = 0.582487.88

B = ZC sinh (  l ) = ( 282.6 − 2.45 )( 0.582487.88 ) = 164.6 85.43 C=

5.15

(

VR = 475

IR =

1 0.5824 87.88 sinh (  l ) = = 2.061 10−3 90.3 S ZC 282.6  − 2.45

)

3 0 = 274.20 kVL − N

PR  cos −1 ( pf ) 3 VR LL ( pf )

900 cos −1 1.0 3 ( 475 )(1.0 )

= 1.0940 kA

(a) VS = AVR + B I R = ( 0.8794 0.66 )( 274.2 0 ) + (134.885.3 )(1.0940 ) = 241.13 0.66 + 147.47 85.3 = 253.2 + j149.8 = 294.1730.60 kVL − N

VS L − L = 294.17 3 = 509.5 kV (b) I S = CVR + D I R = (1.688 10−3 90.2 ) ( 274.2 0 ) + ( 0.87940.66 )(1.0940 ) = 0.462890.2 + .96210.66 = .9604 + j 0.4739 = 1.07126.26 kA

I S = 1.071 kA (c) PFS = cos ( 32.67 − 24.01) = cos ( 6.41 ) = 0.994 Lagging (d) PS = 3 VS LL I S ( pf S )

= 3 ( 509.5)(1.071)( 0.994) = 939.5 MW Full-load line losses = PS − PR = 939.5 − 900 = 39.5MW (e) VR NL = VS / A = 509.5 / 0.8794 = 579.4 kVL − L

% VR =

VR NL − VR FL VR FL

 100 =

579.4 − 475 = 22% 475

5.16 Table A.4 three ACSR finch conductors per phase

(a)

ZC = z / y =

0.0969   1 mi    = 0.02  / km 3 mi  1.609 km 

0.33686.6 = 264.4 − 1.7  4.807  10 −6 90

(b)  l = z y l = 0.336  4.807  10−6 86.6 + 90 ( 300 )

= 0.0113 + j 0.381 pu (c) A = D = cosh (  l ) = cosh ( 0.0113 + j 0.381)

= cosh 0.0113 cos0.381+ j sinh 0.0113 sin 0.381 rad.

rad.

= 0.9285 + j 0.00418 = 0.92850.258 pu sinh  l = sinh ( 0.0113 + j 0.381) = sinh 0.0113 cos 0.381+ j cosh 0.0113 sin 0.381 rad.

rad.

= 0.01045 + j 0.3715 = 0.371688.39 B = ZC sinh  l = 264.4 − 1.7 ( 0.371688.39 ) = 98.2586.69 C = sinh  l / ZC = 0.371688.39 ( 264.4 − 1.7 ) = 1.405  10−3 90.09 S

5.17

(

)

VR = 480 / 3 0 = 277.10 kVLN

(a) I R =

1500 480 3

 − cos−1 0.9 = 1.804 − 25.84 kA

VS = AVR + B I R = 0.92850.258 ( 277.1) + 98.2586.69 (1.804 − 25.84 )

= 377.424.42 kVLN ; VS = 377.4 3 = 653.7kVLL VRNL = VS / A = 653.7 / 0.9285 = 704 kVLL

% VR =

VR NL − VR FL VR FL

 100 =

704 − 480  100 = 46.7% 480

(b) VS = 0.92850.258 ( 277.1) + 98.2586.69 (1.8040 )

= 321.433.66 kVLN ; VS = 321.4 3 = 556.7kVLL VR NL = VS / A = 556.7 / 0.9285 = 599.5 kVLL % VR =

599.5 − 480  100 = 24.9% 480

VS = 251.2 3 = 435.1kVLL VR NL = VS / A = 435.1/ 0.9285 = 468.6 kVLL % VR =

5.18

468.6 − 480  100 = −2.4% 4.80

 l = l y z = 230

( 0.843  5.105  10 )  ( 79.04 + 90) 2 −6

= 0.477284.52 = 0.0456 + j 0.475 = ( + j  ) l

ZC = VR =

z 0.8431 =  ( 79.04 − 90 ) 2 = 406.4 − 5.48  y 5.105  10 −6 215 3

= 124.130 kV/ph.; I R =

125  103 3  215

0 = 335.70 A

1 1 cosh  l = e0.0456 27.22 + e −0.0456  − 27.22 = 0.89041.34 2 2 1 1 sinh  l = e0.0456 27.22 − e −0.0456  − 27.22 = 0.459784.93 2 2

VS = VR cosh  l + I R Z C sinh  l

= (124.13  0.89041.34 ) + ( 0.3357  406.4 − 5.48  0.459784.93 ) = 137.8627.77 kV

Line-to-Line voltage magnitude at the sending end is I S = I R cosh  l +

3 137.86 = 238.8kV

VR sinh  l = ( 335.7  0.89041.34 ) + ZC 124,130  0.459784.93 406.4 − 5.48 = 332.3126.33 A

Sending-end line current magnitude is 332.31 A

PS (3 ) = 3 ( 238.8 )( 332.31) cos ( 27.77 − 26.33 ) = 137, 433 kW QR ( 3 ) = 3 ( 238.8 )( 332.31) sin ( 27.77 − 26.33 ) = 3454 kVAR Voltage Regulation =

(137.86 / 0.8904 ) − 124.13 = 0.247 124.13

( Note that at no load, I = 0; V = V / cosh  l ) R

since  = 0.475/ 230 = 0.002065 rad/mi

The wavelength  =

2



2 = 3043 mi 0.002065

and the velocity of propagation = f  = 60  3043 = 182,580 mi/s

5.19 Choosing a base of 125 MVA and 215 kV,

( 215) = 370 ; Base Current = 125 103 = 335.7 A 2

Base Impedance =

125

3  215

406.4 − 5.48 = 1.098 − 5.48 pu; VR = 10 pu 370

So Z C =

The load being at unity pf, I R = 1.00 pu  VS = VR cosh  l + I R Z C sinh  l

= (10  0.89041.34 ) + (10  1.098 − 5.48  0.459784.93 ) = 1.1102 27.75 pu

I S = I R cosh  l +

VR sinh  l ZC

1.00   = (10  0.89041.34 ) +   0.459784.93  1.098  − 5.48    = 0.99 26.35 pu 96 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

At the sending end Line to line voltage magnitude = 1.1102  215 = 238.7 kV

Line Current Magnitude = 0.99  335.7 = 332.3A 5.20 (a) Let  =  l = Then A = 1 +

ZY ZY Z 2Y 2 Z 3Y 3 + + + 2 24 720

which iscosh  l

  ( ZY )3/ 2 ( ZY )5/ 2 ( ZY )7/ 2 B = Z C  ( ZY )1/ 2 + + + +  which is Z C sinh  l 6 120 5040   3/ 2 5/ 2  1 1  ( ZY ) ( ZY ) ( ZY )7/ 2 1/ 2 C= sinh  l = ( ZY ) + + + +   ZC Zc  6 120 5040  D=A Considering only the first two terms,    3/ 2   ( ZY )   B = Z C  ( ZY ) 2 +   6   1  ( ZY )3/ 2   2 C= ( ZY ) +   Zc  6   A = D =1+

ZY 2

(b) Refer to Table 5.1 of the text. For Nominal- circuit:

A −1 Y = ;B= Z  B 2

For Equivalent- circuit: 5.21

A −1 Y = ; B = Z  B 2

Vs = AVR + BI R

Eq. (5.1.1):

VS I S = AVR I S + BI R I S Substituting I S = CVR + DI R and A = D

VS I S = A VR I S + BI R (CVR + AI R ) Now adding VR I R on both sides,

VS I S + VR I R = AVR I S + ( BC + 1)VR I R + BAI R2 But A2 − BC = 1 Hence VS I S + VR I R = AVR I S + A 2VR I R + BAI R2 = A (VR I S + VS I R ) 97 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

A = Now B =

VS I S + VR I R  VR I S + VS I R

VS − AVR ; substituting the above result IR

For A , one obtains B = Thus B =

or B =

5.22

VS VR  VS I S + VR I R  −   I R I R  VR I S + VS I R 

VSVR I S + VS2 I R − VRVS I S − VR2 I R I R (VR I S + VS I R )

VS2 − VR2  VR I S + VS I R

1 +X e + e− ; with X = e− , A = X A= 2 2

or X 2 − 2 AX + 1 = 0 ; Substituting X = X1 + jX 2 And A = A1 + jA2 , one gets

X12 − X22 + 2 jX1 X2 − 2  A1 X1 − A2 X2 + j ( A2 X1 + A1 X2 ) + 1 = 0 which implies X12 − X 22 − 2  A1 X1 − A2 X 2  + 1 = 0     and X1 X 2 − ( A2 X1 + A1 X 2 ) = 0   5.23

Equivalent  circuit:

Z  = B = 164.685.43 = (13.1 + j164.1)  = R + jx

Alternatively: Z  = Z F1 = ( zl )

sinh  l l

 0.582487.88  = ( 0.35128 85.1 ) 500    0.622 87.55  = 164.585.43 

  Y Y yl yl  cosh  l − 1  = F2 =  tanh (  l /2 ) (  l /2 )  =   2 2 2 2   l  sinh  l   2    0.8136 + j 0.01548 − 1  4.4  10−6  = 90  ( 400 )   2    ( 0.31187.55 )( 0.5824 87.75 )  G  + jB = 1.136  10−3 89.8 S = (3.96  10 −6 + j1.136  10 −3 )S = 2

R = 13.1  is 13% smaller than R = 15  X  = 164.1 is 6% smaller than X = 175  B / 2 = 1.136 10−3 S is 3% larger than Y / 2 = 1.110−3 S G / 2 = 3.96 10−6 S is introduced into the equivalent  circuit.

5.24

Z  = B = 98.2586.69 = 5.673 + j 98.09 

    Y  Y  cosh  l − 1  4.807  =   F2 =   10 −6 90  300     l 2 2  2   sinh  l   2      1.442 0.9285 + j 0.00418 − 1   =  10 −3 90     2   0.3812 88.3 ( 0.371688.39 )   2   −0.0715 + j 0.00418  = 7.21  10 −4 90    0.0708176.7  = 6.37  10 −7 + j 7.294  10 −4 S

R = 5.673  is 5.5% smaller than R = 6  X  = 98.09  is 2.4% smaller than X = 100.5  B / 2 = 7.294  10−4 S is 1.2% larger than Y / 2 = 7.211 10−4 S G / 2 = 6.37  10−7 S is introduced into the equivalent  circuit

5.25 The long line -equivalent circuit is shown below:

z = ( 0.1826 + j 0.784 )  / mi per phase y=

1 1 = = 5.391  10 −6 90 S /mi per phase xc  − 90 185.5  103  − 90

 = y z ; Z = zl = 160.9976.89 ; Y = yl = 1.078  10 −3 90 S F1 = ( sinh  l ) /  l = 0.9720.37; F2 =  Z = Z

tanh (  l / 2 )

 l /2

= 1.0144 − 0.19

sinh  l = 156.4877.26  l

Y  Y  tanh (  l / 2 )  =   = 0.5476  10 −3 89.81 S 2 2   l / 2 

I Z  = I R + VR

Y = 5020 + (132,800 0 ) ( 0.5476  10−3 89.81 ) 2 = 507.5 8.24 A

VS = VR + I Z  Z  = 132,8000 + 507.58.24 (156.4877.26 ) = 160,835 29.45 V (a) Sending end line to line voltage magnitude = 3160.835 = 278.6 kV

Y  (b) I S = I Z  + VS   = 507.58.24 + 160.835 ( 0.5476 ) 29.45 + 89.81 2 = 482.93 18.04 A; I S = 482.93A

= 228.41 MW+ j 46.1 M var (d) Percent voltage regulation =

160.835 − 132.8  100 = 21.1% 132.8

5.26 (a)

z = y

ZC =

j 0.34 = 274.9 0 = 274.9  j 4.5  10−6

(b)  l = z y ( l ) =

( j 0.34) ( j 4.5 10−6 ) ( 350 ) = j 0.4330pu

A = D = cos (  l ) = cos ( 0.4330 radians ) = 0.90770 pu B = j Z C sin (  l ) = j ( 274.9 ) sin ( 0.4330 radians ) = j115.35   1   1  C = j  sin (  l ) = j   sin ( 0.4330 radians )  274.9   ZC  = j1.516  10−3 S (d)  = 0.4330 / 350 = 1.237 10−3 radians/ km

 = 2 / = 5079 km (e) SIL =

2 Vrated L−L

( 500 ) = 909.4 MW 3 = ( ) 2

274.9

0.4×10−6

𝐿

5.27 𝑍𝑐 = √𝐶 = √28.3×10−12 = 118.9 Ω 5.28

√

60⋅√0.4×10−6 ⋅28.3×10−12

𝜆 = 𝑓 𝐿𝐶 =

= 4.95 × 106 m; 𝑣𝑝𝑟𝑜𝑝 = 𝜆𝑓 = (4.95 × 106 m)(60 𝐻𝑧) =

2.97 × 108 m/s. Yes, it is reasonable because it is slightly lower than the speed of light. 5.29

𝑍𝑐 and 𝑣𝑝𝑟𝑜𝑝 have not been affected as they are not frequency dependent. The new value for 𝑣

𝜆 = 𝑝𝑟𝑜𝑝 = 𝑓

5.30

ZC =

2.97×108 = 7.42 × 105 m 400

L1 z = = y C1

0

2

Ln ( Deq / DSL )

D  2 0 / Ln  eq   DSC 

  D   D   Ln  eq  Ln  eq   0   DSL   DSC   ZC =   o  2  Geometric factors Characteristic   impedence of   free space

where

0 4  10 −7 = = 377  0  1  −9  36   10  

1 = L1C1

=

1  Deq  0  Deq  Ln   2 0 / Ln   2  DSL   DSC 

    1  Ln ( Deq / DSC )    =       Ln ( D / D )  eq SL  0 0    Free space  Geometric factors  velocity of propagation

where

0 0

(

1 m = 3.0  108 s  1  4  10 −7   10 −9   36 

)

For the 765 kV line in Example 5.10, Deq = 3 (14)(14)(28) = 17.64 m 3  .0403  DSL = 1.091 4  ( 0.457 ) = 0.202 m   3.28  3  1.196  DSC = 1.091 4  (.0254 )( 0.457 ) = 0.213m   2 

  17.64   17.64    Ln   Ln    0.202   0.213    Z C = 377 = 267    2    

 = 3  108

Ln (17.64 /.213 )

Ln (17.64 /.202 )

= 2.98  108

m s

5.31 (a) For a lossless line,  =  LC = 2 ( 60 ) 0.97  0.0115  10 −9 = 0.001259 rad/km ZC = L / C =

0.97  10 −3 = 290.43  0.0115  10 −6

Velocity of propagation  =

1 LC

and the line wave length is  =  / f = (b) VR =

500 3

1 0.97  0.0115  10 −9

(

= 2.994  105 km/s

)

1 2.994  103 = 4990 km 60

0 kV = 288.6750 kV

800  cos−1 0.8 = 800 + j 600 MVA = 100036.87 MVA 0.8 (1000 − 36.87 )103 = 1154.7 − 36.87 A I R = S * R(3 ) / 3VR* = 3  288.6750

SR(3 ) =

Sending end voltage VS = cos  lVR + jZ c sin  lI R

 l = 0.001259  300 = 0.3777rad = 21.641 VS = 0.9295 ( 288.6750 ) + j ( 290.43) 0.3688 (1154.7 − 36.87 ) (10−3 ) = 356.5316.1 kV Sending end line-to-line voltage magnitude = 3 356.53 = 617.53 kV IS = j

1 sin  lVR + cos  l I R ZC

1 0.3688 ( 288.6750 )103 + 0.9295 (1154.7 − 36.87 ) 290.43 = 902.3 − 17.9 A; Line current = 902.3A =j

SS ( 3 ) = 3VS I S = 3 ( 356.5316.1 )( 902.3 − 17.9 )10 −3 = 800 MW + j 539.672 MVAR Percent voltage regulation =

( 356.53 / 0.9295 ) − 288.675  100 288.675

= 32.87%

5.32 (a) The line phase constant is  l =

2



360

lrad =



360 315 3000

= 22.68

From the practical line loadability,

P3 =

VS puVR pu ( SIL ) sin  l

sin  ; 700 =

(1.0 )( 0.9 )( SIL ) sin 36.87 sin 22.68

 SIL = 499.83MW

( kV Since SIL =

) MW, kV = Z SIL = 2

L rated

(

)

( 320 )( 499.83) = 400 kV

(b) The equivalent line reactance for a lossless line is

X  = Z c sin  l = 320 ( sin 22.68 ) = 123.39  For a lossless line, the maximum power that can be transmitted under steady-state condition occurs for a load angle of 90. 103 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

With VS = 1pu = 400 kV ( L − L ) , VR = 0.9 pu = 0.9 ( 400 ) kV ( L − L )

Theoretical Maximum Power =

( 400 )( 0.9  400 )  1

123.39 = 1167 MW

5.33 (a) V2 = Z C I 2 since the line is terminated in Z C . Then V1 = V2 ( cosh  l + sinh  l ) = V2 e l = V2 e l e j  l

(1)

I1 = I 2 ( cosh  l + sinh  l ) = I 2 e l = I 2 e l e j  l

(2)

V V  1 = 2 = ZC  I1 I 2 (b) V1 = V1 = V2 e l or (c)

( Note : =  + j  ) V2 = e − l V1

From (1) 

I2 = e − l I1

From (2) 

(d) −S21 = V2 I 2 = V1e − l e − j  l I1e− l e j  l = S12 e −2 l

S Thus − 21 = e −2 l  S12 which is ( I 22 / I12 ) . (e) Noting that  is real,

=

− P21 = e−2 l  P12

5.34

For a lossless line, ZC =

L , Eq. (5.4.3) of text which is pure real, i.e. resistive. C

 = j  is pure imaginary;  =  LC ;  = 0 

V2 I 2 −S21 − P21 = = = = = 1 V1 I1 S12 P12

P12 = Re (V1I1 ) = Re ZC I12 = ZC I12 Since ZC is real. Since I1 = V1 / Z C , P12 = V12 / Z C  5.35 Open circuited  I 2 = 0; Lossless   = 0;  = j  . Short line: V1 = V2  (  l )2    ZY  V2  Medium Line: Nominal  : V1 = 1 + V2 = 1 +   2  2     2  (l)   = 1 −  V2   2     Long Line: Equiv.  : V1 = V2 cosh  l = V2 cos  l   

Note: The first two terms in the series expansion of

cos  l are 1 −

(l)

While V1 = V2 in the case of short-line model,   the voltage at the open receiving end is higher    than that at the sending end,for small  l,for   the medium and long-line models.

5.36 From Problem 5.7 solution, see Eq. (1)

VS2 = VR2 + 2VR I ( R cosR + X sin R ) + I 2 ( R2 + X 2 ) Using P = VR I cosR and Q = VR I sin R , one gets −VS2 + VR2 + 2 PR + 2QX +

1 ( P 2 + Q2 )( R 2 + X 2 ) = 0 VR2

(2)

In which only P and Q vary.

For maximum power, dP / dQ = 0 : dP 2 X + 2QC R2 + X 2 =− , Where C = dQ 2 R + 2 PC VR2

and for

V2X dP = 0, Q = − 2 R 2  dQ R +X

Substituting the above in (2), after some algebraic simplification, one gets

PMAX =

 VR2  ZVS − R    Z 2  VR 

where Z = R 2 + X 2 . *

V −V  V 2 V V * 5.37 (a) S12 = V I = V1  1 2  = 1 * − 1 *2 Z Z  Z  * 1 1

V12 j Z V1V2 j Z j12 e − e e  (1) Z Z

which is the power sent by V1 . S21 =

V22 jZ V2V1 jz − j12 e − e e Z Z

and − S21 = −

V22 jz V2V1 jz − j12 e + e e 2 Z Z

which is the power received by V2 . (b) (i)

With V1 = V2 = 1.0     −S21 = −185 + 175 = 0.1717 − j 0.0303  S12 = 185 − 195 = 0.1743

(ii) With V1 = 1.1 and V2 = 0.9  S12 = 1.2185 − 0.9995 = 0.1917 + j 0.2192    −S21 = −0.8185 + 0.9975 = 0.1856 + j 0.1493 

   but Q12 and − Q21 have changed considerably. 

Comparing, P12 has not changed much, 5.38 From Problem 5.14 A = 0.81371.09 pu;

A = 0.8137 and  A = 1.09

B = Z  = 164.685.3 ;

Z  = 164.6 and  Z = 85.3

500  500 ( 0.8137 )( 500 ) PR max = − cos ( 85.3 − 1.09 ) 164.4 164.4 = 1519 − 122 = 1397 MW ( 3 ) 2

For this loading at unity power factor,

IR =

PR max 3 VR LL ( PF )

1397 3 ( 500 )(1.0 )

= 1.61kA / Phase

From Table A.4, the thermal limit for 3 ACSR 1113 kcmil conductors is 3 1.11 = 3.33 kA/ p hase. The current 1.61 kA corresponding to the theoretical steady-state stability limit is well below the thermal limit of 3.33 kA. 5.39 Line Length

200 km

600 km



282.6 − 2.45

282.6  − 2.45

l

0.248687.55

0.745987.55

A=D

0.96940.1544

0.73561.685



69.5485.15

191.885.57

8.7110−4 90.05

2.403 10−3 90.47

PR max

3291

1201

The thermal limit of 3.33 kA/phase corresponds to

3 ( 500) 3.33 = 2884 MW at 500 kV and unity

power factor. (100 kV)(100 kV)

𝑉 𝑉

𝑅 𝑆 5.40 𝑃𝑟𝑚𝑎𝑥 = 𝜔𝐿 sin 𝜃 = (2𝜋60)(12mH) sin 30° = 1105 MW

5.41 (a)

Z = zl = ( 0.088 + j 0.465 )100 = 8.8 + j 46.5  Y yl = = ( j 3.524  10−6 )100 / 2 = j 0.1762mS 2 2

( 230 ) = 529  = 2

Z base = V

2 L base

S3 base

100

 Z = ( 8.8 + j 46.5 ) / 529 = 0.0166 + j 0.088 pu Y = j 0.1762 (1/ 0.529 ) = j 0.09321pu 2

The nominal  circuit for the medium line is shown below:

<<Figure 12>> (b) S3 rated = VL rated I L rated 3 = 230 ( 0.9 ) 3 = 358.5 MVA (c) A = D = 1 +

ZY = 1 + (8.8 + j 46.5 ) ( j 0.1762 10−3 ) = 0.99180.1 2

B = Z = 8.8 + j 46.5 = 47.3279.3 

C =Y +

Z Y2 = 0.351090.04 mS 4

(d) SIL = VL2rated ZC

ZC =

z 0.088 + j 0.465 =  103 = 366.6 − 5.36  y j 3.524

 SIL = ( 230 ) 366.6 = 144.3M W 2



5.42

2 360  360   l = l radians =  l = ( 500 ) = 36     5000 Using Eq. (5.4.29) of the text,

460 =

1.0  0.9 ( SIL )

sin36.87 sin36 1  0.9  SIL = ( 0.6 ) 0.5878

From which SIL = 500.7 MW From Eq. (5.4.21) of the text,

VL − L =

( ZC ) SIL = ( 500.7 ) 500 = 500.3kV

Nominal voltage level for the transmission line is

500 kV  For  loss less line, X  = ZC sin  l

= 500sin36 = 293.9  From Eq. (5.4.27) of the text,

P3 max =

( 500 )( 0.9  500 ) = 765.6 MW  293.9

5.43 The maximum amount of real power that can be transferred to the load at unity pf with a bus voltage greater than 0.9 pu (688.5 kV) is 2950 MW.

5.44 The maximum about of reactive power transfer that can be transferred to the load with a bus voltage greater than 0.9 pu is 650 Mvar.

5.45 (a) Using Eq. (5.5.3) with  = 35

( 500 )( 0.95  500 ) cos 85.3 − 35 − ( 0.8137 )( 0.95  500 ) cos 85.3 − 1.09 ( ) ( ) 2

PR =

164.6 = 919 − 110 = 810 MW (3 )

164.6

PR = 810 MW is the practical line loadability provided that the voltage drop limit and thermal limits are not exceeded. (b) I R FL =

3 VR LL ( PF )

810 3 ( 0.95  500 )( 0.99 )

= 0.99 kA

 = ( 0.81371.09 ) (VR FL 0 ) + (164.685.3 )( 0.998.109 ) 3 288.68 = 0.8137VR FL 1.09 + 162.993.41

500

288.68 = ( 0.8136VR FL − 10.06 ) + j ( 0.0154VR FL + 162.6 )

Taking the squared magnitude of the above equation:

83333 = 0.6622VR2 FL − 11.36VR FL + 26540 Solving the above quadratic equation:

VR FL =

11.36 +

(11.36) + 4 ( 0.6622)( 56793) = 301.6kVL − N 2 ( 0.6622 ) 2

VR FL = 301.6 3 = 522kVL −L = 1.04pu (d) VR NL = VS /A = 500 0.8137 = 614.5kVL − L % VR =

614.5 − 522  100 = 18% 522

(e) From Problem 5.38, thermal limit is 3.33 kA. Since VR FL / VS = 522 / 500 = 1.04  0.95, and the thermal limit of 3.33 kA is greater than 0.99kA, the voltage drop and thermal limits are not exceeded at PR = 810MW. Therefore, loadability is determined by stability. 5.46

A = 0.97390.0912 pu; A = 0.9739,  A = 0.0912 B = Z = 60.4886.6 ; Z = 60.48,  Z = 86.6

(a) Using Eq. (5.5.3) with  = 35 : 500 ( 0.95  500 )

cos ( 86.6 − 35 ) − 60.48 = 2439.2 − 221.2 = 2218 MW ( 3 )

PR =

0.9739 ( 0.95  500 ) 60.48

cos ( 86.6 − 0.0912 )

PR = 2218MW is the line loadability if the voltage drop and thermal limits are not exceeded. (b) I R FL =

3VR LL ( pf )

2218 3 ( 0.95  500 )( 0.99 )

= 2.723kA

 = ( 0.97390.0912 ) VR FL 0 + 60.4886.6 ( 2.7238.11 )

288.68 = ( 0.9739 VR FL − 13.55) + j ( 0.0016VR FL + 164.14 )

Taking the squared magnitude of the above

Solving the above quadratic equation: VR FL =

25.60 +

( 25.60 ) + 4 (.93664 )( 56,207 ) = 250.68 kVLN 2 ( 0.9678 ) 2

VR FL = 250.68 3 = 434.18 kVLL = 0.868 per unit for this load current, 2.723 kA, the voltage drop limit VR / VS = 0.95 is exceeded. The thermal limit, 3.33 kA is not exceeded. Therefore the voltage drop limit determines loadability for this line. Based on VR FL = .95 per unit , IRFL is calculated as follows:

VS = AVR FL + B I R FL  0.95  500   = ( 0.97390.0912 )  0  + 60.4886.6 ( I R FL 8.11 ) 3 3   288.68 = 267.090.0912 + 60.48 I R FL 94.71 500

= ( −4.966 I R FL + 267.09 ) + j ( 60.28 I R FL + 0.4251) Taking squared magnitudes;

83,333 = 3658 I R2 FL − 2601 I R FL + 71,337 Solving the quadratic: I R FL =

2601 +

( 2601) + 4 (3658 )(11,996 ) = 2.2 kA 2 ( 3658 ) 2

At 0.99 pf leading, the practical line loadability for the line is

PR = 3 ( 0.95  500 ) 2.2 ( 0.99 ) = 1792MW which is based on the voltage drop limit VR / VS = 0.95. (d) VR NL = VS / A = 500 / 0.9739 = 513.4 kVLL

% VR =

513.4 ( 500  0.95) 500  0.95

 100 = 8.08%

5.47 (a) l = 200kM; The steady-state limit is: PR = −

500 ( 0.95  500 ) 69.54

cos ( 86.15 − 35 )

0.9694 ( 0.95  500 )

69.54 = 1914 MW I R FL =

3 VR FL ( PF )

cos ( 85.14 − 0.154 ) 1914

3 ( 0.95  500 )( 0.99 )

= 2.35 kA

VS = AVR FL + B I R FL 500 3

 = ( 0.96940.154 ) (VR FL 0 ) + ( 69.5485.15 )( 2.358.109 )

288.675 = ( 0.9694VR FL − 9.29 ) + j ( 0.0026VR FL + 163.1) Taking the squared magnitude:

83333. = 0.9397 VR2FL − 17.16VR FL + 26707 Solving

VR FL =

17.16 +

(17.16) + 4 ( 0.9397 )( 56626) = 254.8kVL − N 2 ( 0.9397 ) 2

VR FL = 254.8 3 = 441.3kVL − L = 0.8825 pu The voltage drop limit VR FL VS  0.95 is not satisfied. At the voltage drop limit: VS = AVR FL + B I R FL  0.95  500   = ( 0.9694 0.154 )  0  + 3 3  

500

( 69.5485.15 ) ( I R FL 8.109 )

288.675 = ( 265.85 − 3.953I R FL ) + j ( 0.715 + 69.4 I R FL ) 83333 = 70677 − 2003 I R FL + 4836I R2 FL

I R FL =

2003 +

( 2003) + 4 ( 4836)(12656) = 1.84kA 2 ( 4836 ) 2

The practical line loadability for this 200-km line is

PR FL = 3 ( 0.95  500 )1.84 ( 0.99 ) = 1499 MW = 1499/885 = 1.69 pu at VR FL VS = 0.95pu and at 0.99pf leading. Note: SIL = (500)2/282.6 = 885 MW 112 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

(b) l = 600 kM

PR =

500 ( 0.95  500 ) 191.8

cos ( 85.57 − 35 ) −

0.7356 ( 0.95  500 ) 191.8

cos ( 85.57 − 1.685 )

= 786.5 − 92.3 = 694.2 MW = 694.2/885 = 0.78 pu The practical line loadability is 694.2 MW corresponding to steady-state stability for the 600km line. 5.48 (a) SIL = ( 345) 300 = 396.8MW 2

Neglecting losses and using Eq. (5.4.29) P=

1  0.95 ( SIL ) sin 35

 2 ( 300 )  sin  radians   5000 

Number of 345-kV lines =

= 1.48 ( SIL ) = 1.48 ( 396.8 ) = 587.3 MW/line

2000 + 1 = 3.4 + 1 5Lines 587.3

( 500 ) = 909.1MW SIL = 2

(b) For 500-kV lines,

275

P = 1.48 ( SIL ) = 1.48  909.1 = 1345.6 MW/ Line Number of 500-kV Lines =

2000 + 1 = 1.49 + 1 3Lines 1345.6

( 765) = 2250.9 MW SIL = 2

260

P = 1.48 ( SIL ) = 1.48  2250.9 = 3331.3MW/ Line Number of 765-kV lines =

2200 + 1 = 0.6 + 1 2 Lines 3331.3

5.49 (a) Using Eq. (5.4.29): 1  0.95 ( SIL ) sin 35

= 1.48 ( SIL )  2 ( 300 )  sin  radians   5000  P = 1.48 ( 396.8 ) = 587.3MW / 345 − kV Line P=

#345-kV Lines =

3200 + 1 = 5.4 + 1 7 Lines 587.3

P = 1.48 ( 909.1) = 1345.5MW / 500 − kV Line #500-kV Lines =

3200 + 1 = 2.4 + 1 4 Lines 1345.5

P = 1.48(2250.9) = 3331.3MW / 765 − kV Line #765-kV Lines = (b) P =

3200 + 1 = 0.96 + 1 2 Lines 3331.3

(1)(.95)( SIL )( sin 35 ) = 1.131 SIL ( ) 2  400   sin  radians   5000 

P = 1.131( 396.8 ) = 448.8 MW / 345 − kV Line #345-kV Lines =

2000 + 1 = 4.5 + 1 = 6 Lines 448.8

P = (1.131)( 909.1) = 1028.3MW / 500 kV Line #500-kV Lines =

2000 + 1 = 1.94 + 1 = 3Lines 1028.3

P = (1.131)( 2250.9 ) = 2545.9 MW / 765 kV Line #765-kV Lines = 5.50

2000 + 1 = 0.79 + 1 = 2 Lines 2545.9

 l = ( 9.46  10−4 ) ( 300 )(180 /  ) = 16.26 Real power for one transmission circuit P = 4000 / 4 = 1000 MW From the practical line loadability, P3 = or

1000 =

VS puVR pu ( SIL ) sin  l

sin 

(1.0)(0.9)( SIL) sin(36.87) sin16.26

From which SIL = 518.511MW 2 Since SIL = ( kVL rated ) Z C  MW  

kVL =

Z C ( SIL ) =

( 343) (518.5) = 422kV

5.51 To show: PR + jQR =

| VR || VS |  −  |B|

−

| A || VR |2  −  |B|

(a) The phasor diagram corresponding to the above equation is shown below:

(b) By shifting the origin from 0 to 0, the power diagram is shown in Fig. (b) above. For a given load and a given value of | VR | , 0 A = | VR | | VS | | B | the loci of point A will be a set of circles of radii 0 A, one for each of the set of values of | VS | . Portions of two such circles (known as receiving-end circles) are shown below:δ

(c) Line 0A in the figure above is the load line whose intersection with the power circle determines the operating point. Thus, for a load (with a lagging power-factor angle  R ) A and C are the operating points corresponding to sending-end voltages | VS1 | and | VS 2 | , respectively. These operating points determine the real and reactive power received for the two sending-end voltages. The reactive power that must be supplied at the receiving end in order to maintain constant | VR | when the sending-end voltage decreases from | VS1 | to | VS 2 | is given by AB, which is

parallel to the reactive-power axis.

5.52 (a) See Problem 5.37(a) solution: Eqs. (1) and (2) with the substitution of Z  for Z , adding the contribution of the complex power consumed by Y  / 2 , using Eq. (1) of Problem 5.37(a) solution, one gets

S12 =

Y * 2 V12 V1V2 j12 V1 + * − * e  2 Z Z

Similarly, subtracting the complex power consumed in

Y (on the right-hand side in Fig. 2

5.17), For the received power, one has

−S21 = −

Y * 2 V22 V1V2 − j12 V2 − * + * e  2 Z Z

Except for the additional constant terms, the equations have the same form as those in Problem 5.37. (b) For a lossless line, ZC = L / C is purely real and  = j  is purely imaginary. Also

Y = Y

tanh (  l / 2 )

 l /2

= j c

tan (  l / 2 )

l / 2

and Z  = Z C sinh( l )

which becomes jZC sinh(  l ) . Note: Y  is now the admittance of a pure capacitance; Z  is now the impedance of a pure inductance.

Active power transmitted, P12 = − P21 And P12 =

V12 sin 12  Z C sin (  l )

Using Eq. (5.4.21) of the text for SIL

P12 = PSIL

sin 12  sin (  l )

P12 1 = 0.707  PSIL sin ( 0.1146l ) 

(d) Thermal limit governs the short lines;    Stability limit prevails for long lines.  5.53 The maximum power that can be delivered to the load is 10,250 MW. 5.54 For 8800 MW at the load the load bus voltage is maintained above 720 kV even if 2 lines are taken out of service (8850 MW may be OK since the voltage is 719.9 kV). 5.55 From Problem 5.23, the shunt admittance of the equivalent  circuit without compensation is

Y  = G + jB = 2 (1.57  10−6 + j8.981 10−4 ) = (3.14  10−6 + j1.796  10 −3 ) S With 65% shunt compensation, the equivalent shunt admittance is

65   −6 −4 Yeq = 3.14  10 −6 + j1.796  1 −  = 3.14  10 + j 6.287  10 100   = 6.287  10 −4 89.71 S Since there is no series compensation

Zeq = Z  = 134.885.3  The equivalent A parameter of the compensated line is

Yeq Z eq

1 ( 6.287  10−4 89.71 ) (134.885.3 ) 2 2 = 0.95780.22 pu

Aeq = 1 +

=1+

The no-load voltage is

VR NL = VS Aeq = 526.14 0.9578 = 549.6 kVL − L where VS is obtained from Problem 5.15.

VR FL = 475kVL − L is the same as given in Problem 5.15, since the shunt reactors are removed at full load. Therefore

% VR =

VR NL − VR FL VR FL

 100 =

549.6 − 475  100 = 15.7% 475

The impedance of each shunt reactor is −1

 B  1  Z reactor = j  ( 0.65 )  = j   1.796  10−3  0.65 2  2 

−1

= j 1713  / phase, at each end of the line. 5.56 (a) VS = 653.7 kVLL (same as Problem 5.17)

Yeq = 2 6.37  10−7 + j 7.294  10 −4 (1 − 0.5 ) from Problem 5.24 = 1.274  10−6 + j 7.294  10−4 = 7.294  10−4 87.5 S

Zeq = Z  = 98.2586.69  Aeq = 1 +

Yeq Z eq 2

=1+

1 ( 7.294  10−4 87.5 ) ( 98.2586.69 ) 2

= 1 + 0.0358174.19 = 0.9644 + j 0.0036 = 0.96440.21

VRNL = VS / Aeq = 653.7 / 0.9644 = 677.8 kVLL % VR =

677.8 − 480  100 = 41.2% 480

(b) VS = 556.7 kVLL (same as Problem 5.17)

VR NL = VS / A = 556.7 / 0.9644 = 577.3 kVLL % VR =

577.3 − 480  100 = 20.3% 480

VR NL = VS / A = 435.1/ 0.9644 = 451.2 kVLL % VR =

451.2 − 480  100 = −6% 480

5.57

From Problem 5.23

Z  = R + jX  = (13.1 + j164.1)  Based on 40% series compensation, half at each end of the line, the impedance of each series capacitor is Z CAP = − jX CAP = − j

1 ( 0.4 )(164.1) = − j 32.82  / phase (at each end) 2

Using the ABCD parameters from Problem 5.14, the equivalent ABCD parameters of the compensated line are  Aeq C  eq

Beq  1 − j 32.82   0.87941.09 134.885.3  =    −3  Deq  0 1  1.688  10 90.2 0.87940.66 

compensated line

Sending-end series capacitors

uncompensated line

1 − j 32.82  0 1   receiving-end series capacitors

 Aeq  Ceq 𝑉 𝑉

Beq  0.92480.64 86.6682.31  = Deq   1.68890.2 0.92480.64  (115 kV)(115 kV)

𝑅 𝑆 5.58 For 115 kV, 𝑃𝑟𝑚𝑎𝑥 = 𝜔𝐿 sin 𝜃 = (2𝜋60)(12 mH) sin 30° = 1462 MW

For 50 HZ, 𝑃𝑟𝑚𝑎𝑥 =

(100 kV)(100 kV) 𝑉𝑅 𝑉𝑆 sin 𝜃 = (2𝜋50)(12 sin 30° = 1327 MW 𝜔𝐿 mH)

Both changes allow more power to be transferred. 5.59 From Problem 5.57: Aeq = 0.9248 pu;  A = 0.64  = 86.66  ;  Zeq = 82.31 Beq = Z eq

From Eq. (5.5.6), with VS = VR = 500 kVL − L PR max =

500  500 0.9248(500) 2 − cos ( 82.31 − 0.64 ) 86.66 86.66

= 2885 − 387 = 2498 MW(3 ) which is 46.7% larger than the value PR max = 1702 MW calculated in Problem 5.38 for the uncompensated line.

5.60 Let X eq be the equivalent series reactance of one 765-kV, 500 km, series compensated line. The equivalent series reactance of four lines with two intermediate substations and one line section outof-service is then: 12  11  X eq  +  X eq  = 0.2778 X eq  43  3 3 

From Eq. (5.4.26) with  = 35, VR = 0.95per unit , and P = 9000MW;

( 765)(.95  765) sin (35) = 9000. .2778 Xeq

Solving for X eq :

N  N    X eq = 127.54  = X   1 − C  = 156.35  1 − C   100   100  Solving: N C = 18.4% series capacitive compensation ( N C = 21.6% including 4% line losses). 5.61

X   NC  164.1  40  =−j   = − j 32.82  2  100  2  100  B N L 2.272  10−3  70  −3 YREAC = − j =−j   = − j1.59  10 S 2 100 2  100  Z CAP = − j

 Aeq  Ceq

0  1 − j 32.82  Beq  1 = −4 1  0 1  Deq   − j 2.92  10 Sending-end shunt compensation

Sending-end series compensation

134.885.3   0.87941.09 1.688  10−3 90.2 0.87940.66    Line

1 1 − j 32.82      1   − j1.59  10−3 0 Receiving end series compensation

0 1 

Receiving end shunt compensation

After multiplying the above 5 matrices, the answer can be obtained.

5.62 See solution of Problem 5.18 for  l, Z C ,cosh  l, and sinh  l . (a) For the uncompensated line: A = D = cosh  l = 0.89041.34 B = Z  = ZC sinh  l = 186.7879.46  C=

sinh  l 0.459684.94 = = 0.00113190.42 S ZC 406.4 − 5.48

(b) Noting that the series compensation only alters the series arm of the equivalent  -circuit, the new series arm impedance is

 = Bnew = 186.7879.46 − j 0.7  230 ( 0.8277 ) = 60.8855.85  Z new In which 0.8277 is the imaginary part of z = 0.843179.04  / mi Noting that A =

Y  1 cosh  l − 1 Z Y  = = 0.00039989.82 S + 1 and 2 ZC sinh  l 2

Anew = ( 60.8855.85  0.00059989.81 ) + 1 = 0.971.24  Z Y   Z Y 2 Cnew = Y   1 +  = Y + 4  4  = 2  0.00059989.81 + 60.8855.85 ( 0.00059989.81 )

= 0.0011890.41 S The series compensation has reduced the parameter B to about one-third of its value for the uncompensated line, without affecting the A and C parameter appreciably. Thus, the maximum power that can be transmitted is increased by about 300%. 5.63 The shunt admittance of the entire line is Y = yl = − j 5.105  10 −6  230 = − j 0.001174S

With 70% compensation, Ynew = 0.7  (− j 0.001174) = − j 0.000822S From Fig. 5.4 of the text, for the case of ‘shunt admittance’, A = D = 1; B = 0; C = Y

 C = Ynew = − j 0.000822S For the uncompensated line, the A, B, C , D parameters are calculated in the solution of Problem 5.49. For ‘series networks’, see Fig.5.4 of the text to modify the parameters. So for the line with a shunt inductor,

Aeq = 0.89041.34 + 186.7879.46 ( 0.000822 − 90 ) = 1.0411 − 0.4 121 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

The voltage regulation with the shunt reactor connected at no load is given by

(137.86 /1.0411) − 124.13 = 0.0667 124.13

which is a considerable reduction compared to 0.247 for the regulation of the uncompensated line. (see solution of Problem 5.18) 5.64 (a) From the solution of Problem 5.31,

ZC = 290.43 ;  l = 21.641 For a lossless line, the equivalent line reactance is given by

X  = ZC sin  l = (290.43)sin 21.641 = 107.11  The receiving end power SR(3 ) = 1000 cos−1 0.8 = 800 + j600MVA Since P3 =

VS ( L − L )VR( L − L ) X

sin  , the power angel  is obtained from

800 = ( 500  500 /107.11) sin  or

 = 20.044

The receiving end reactive power is given by (approximately) QR ( 3 ) =

VS ( L − L )VR ( L − L ) X

cos  −

VR2( L − L ) X

cos  l

( 500 ) cos 21.641 500  500 = cos ( 20.044 ) − ( ) 107.11 107.11 = 23.15MVAR 2

Then the required capacitor MVAR is SC = j 23.15 − j 600 = − j 576.85 The capacitive reactance is given by (see Eq. 2.3.5 in text)

XC =

− jVL2 − j 5002 = = 433.38  SC − j 567.85

106 = 6.1 F 2 ( 60 ) 433.38

(b) For 40% compensation, the series capacitor reactance per phase is

X ser = 0.4 X  = 0.4 (107.1) = 42.84  The new equivalent  -circuit parameters are given by

Z  = j ( X  − X ser ) = j 64.26 ; Y  = j

2  l  tan   = j 0.001316S ZC  2 

Bnew = j 64.26 ; Anew = 1 +

Z Y  = 0.9577 2

The receiving end voltage per phase VR =

500 3

0 kV = 288.6750 kV

The receiving end current is I R = SR*(3 ) / 3VR* Thus I R =

1000 − 36.87 = 1.1547 − 36.87 kA 3  288.6750

The sending end voltage is then given by

VS = AVR + B I R = 326.410.47 kV; VS ( L − L ) = 3 326.4 = 565.4 kV Percent voltage regulation =

( 565.4 / 0.958 ) − 500  100 = 18% 500

5.65 The maximum amount of real power which can be transferred to the load at unity pf with a bus voltage greater than 0.9 pu is 3900 MW.

5.66

Chapter 6 Power Flows 6.1

10 10 10 x1  0  − 25  5 − 10 10 0 x2  2    10 5 − 10 10 x3  = 1       0 0 − 20 x4  − 2  10 There are N − 1 = 3 Gauss elimination steps. Step 1: Use equation 1 to eliminate x1 from equations 2, 3, and 4.

10 10   x1   0   −25 10    0 −40 60 10   x2   10   =  0 22.5 −15 35   x3   2.5      10 10 −40   x4   −5   0 Step 2: Use equation 2 to eliminate x2 from equations 3 and 4. 10 10   −25 10  0 −40   x1   0  60 10    x2   10    0 0 33.33 72.22    =  14.44    x3  −150      0 x4   −10  0 100  9  

Step 3: Use equation 3 to eliminate x3 from equation 4.

10 10  x1   0  − 25 10    0 − 40 60 10  x2   10   =  0 0 33.33 72.22  x3  14.44      0 0 122.22 x4  17.77  0 The matrix is now triangular. Solving via back substitution,

x4 = 0.1455 x3 = 0.1182 x2 = − 0.0364 x1 = 0.0909

6.2

8 2 1   x1   3  4 6 2   x  = 4   2    3 4 14   x3   2 

There are N − 1 = 2 Gauss elimination steps. Step 1: Use equation 1 to eliminate x1 from equations 2 and 3.

2 1   x1  3 8  0 − 10     − 3  x2  = − 5   0 − 8.67 − 36.33 x3  − 2.33 Step 2: Use equation 2 to eliminate x2 from equation 3.

2 1  x1   3  8 0 − 10 − 3  x2  =  − 5   0 0 38.92 x3  −2.31 The matrix is now triangular. Solving via back substitution,

x3 = − 0.0593 x2 = 0.5178 x1 = 0.2530 6.3

4 2 1  x1  3 4 6 2    = 4   x2    3 4 14 x3  2 There are N − 1 = 2 Gauss elimination steps. Step 1: Use equation 1 to eliminate x1 from equations 2 and 3.

1   x1  3 4 2  0 4      1  x2  = 1   0 2.5 13.25 x3  −0.25 Step 2: Use equation 2 to eliminate x2 from equation 3.

1  x1   3  4 2 0 4 1  x2  =  1   0 0 12.625 x3  − 0.875 The matrix is now triangular. Solving via back substitution,

x3 = − 0.0693 x2 = 0.2673 x1 = 0.6337 6.4

− 5 5   x1   5 10 − 10 x  = − 5    2   If we perform Gauss elimination, we arrive at:

− 5 5  x1   5   0 0 x  = 2.5    2   The 2nd row says that

0 x1 + 0 x2 = 2.5 Clearly, this is not possible. The problem in this case is that the system we started with is indeterminate. That is,

−5 5 =0 10 − 10 This system has no solution. 6.5

From Eq. (6.1.6), back substitution is given by: Xk =

yk − Ak ,k +1 X k +1 − Ak ,k + 2 X k + 2

− Ak , N X N

Akk

which requires one division, (N − k) multiplications and (N − k) subtractions for each k = N ,( N − 1) 1 . Summary- Back Substitution Solving for

# Divisions

# Multiplications

# Subtractions

XN−1

XN−2

(N−1)

N −1

Totals

i = i =1

N ( N − 1) 2

N −1

i = i =1

N ( N − 1) 2

6.6

−0.25 −0.125  0  M = D −1 ( D − A) =  −0.667 0 −0.333  −0.2143 −0.2857 0 

8 0 0  D = 0 6 0  0 0 14 

x (i + 1) = Mx (i) + D −1 B x1 = 0.2528 x 2 = 0.5175 x 3 = −0.0595 after 14 iterations

6.7 8 0 0  0 −0.2500 −0.1250    −1 D =  4 6 0  M = D ( D − A) = 0 0.1667 −0.2500   3 4 14  0 0.0060 0.0982  x (i + 1) = Mx (i ) + D −1 B x1 = 0.2531 x 2 = 0.5177 x 3 = −0.0593 after 5 iterations

The Gauss-Seidel method converges over twice as fast as the Jacobi method. 6.8

10 −2 −4  x1  −2 − 2 6 − 2 x  =  3    2   − 4 − 2 10 x3  − 1

1 10  −1 D =0   0 

10 0 0 D =  0 6 0  0 0 10

 0  1 M = D −1 (D − A) =  3 2  5

1 5 0 1 5

 0  1 0 6  1 0  10 0

2 5  1 3  0 

Expressing in the form of Eq. (6.2.6)

 0  x1 (i + 1)    1 x2 (i + 1) = 3 x3 (i + 1)  2  5

1 5 0 1 5

2 1  5  x (i) 10  1  1   x (i) +  0 3  2    x3 (i)  0 0  

 0  − 2 1 0   3  6  1  − 1 0  10 0

Using

1 x(0) = 1 0 we arrive at the following table: i

0.1667

–0.04

–0.0444

–0.0962

–0.1085

–0.1239

–0.1303

–0.1355

–0.1382

0.8333

0.6667

0.5778

0.52

0.485

0.4633

0.45

0.4418

0.4367

0.4336

0.5

0.0667

0.1

–0.0004

–0.0138

–0.0415

–0.0507

–0.0596

–0.0638

–0.0668



Inf

1.24

1.0044

2.0095

0.2235

0.1741

0.0704

0.0484

 for iteration 9 is less than 0.05.  x1 = − 0.14 x2 = 0.43 x3 = − 0.07

6.9

x2 − 3 x1 + 1.9 = 0 x2 + x12 − 3.0 = 0

Rearrange to solve for x1 and x2 :

1 x1 = x2 + 0.6333 3 x2 = 3.0 − x12

Starting with an initial guess of x1 (0) = 1 and x2 (0) = 1

1 x1 (1) = x2 (0) + 0.6333 = 0.9667 3 x2 (1) = 3 − [ x1 (0)]2 = 2 We repeat this procedure using the general equations

1 x1 (n + 1) = x2 (n) + 0.6333 3 x2 (n + 1) = 3 − [ x1 (n)]2 n

...

0.9667 1.3

1.3219

2.0

1.31

2.0656

...

1.1746 1.1735

1.1734

1.1743

...

1.6205 1.6202

1.6229

1.6231

After 50 iterations, x1 and x2 have a precision of 3 significant digits.

x1 = 1.17 x2 = 1.62 6.10

x2 − 4 x + 1 = 0

Rearrange to solve for x:

1 2 1 x + 4 4

Using the general form

1 1 x(n + 1) = [ x(n)]2 + 4 4

and

x(0) = 1

we can obtain the following table: n

5 0.2681

0.5

0.3125

0.2744

0.2688



0.5

0.375

0.1219

0.0204

0.0028

 (4) = 0.0028  0.01  x = 0.27

Using the quadratic formula x=

−b  b2 − 4ac 2a

4  (− 4)2 − 4 2 x = 0.268, 3.732 x=

Note: There does not exist an initial guess which will give us the second solution x = 3.732. 6.11 After 100 iterations the Jacobi method does not converge. After 100 iterations the Gauss-Seidel method does not converge. 6.12 Rewriting the given equations, x1 =

x2 + 0.633; 3

x2 = 1.8 − x12

With an initial guess of x1 (0) = 1 and x2 (0) = 1, update x1 with the first equation above, and x2 with the second equation. x2 (0) 1 + 0.633 = + 0.633 = 0.9663 3 3

Thus

x1 =

and

x2 = 1.8 − x1 (0)2 = 1.8 − 1 = 0.8

In succeeding iterations, compute more generally as x1 ( n + 1) =

x2 ( n ) + 0.633 3

x2 (n + 1) = 1.8 − x12 (n)

and

After several iterations, x1 = 0.938 and x2 = 0.917 . After a few more iterations, x1 = 0.93926 and x2 = 0.9178 . However, note that an “uneducated guess” of initial values, such as x1 (0) = x2 (0) = 100 , would have caused the solution to diverge. 6.13

x1 =

 1  − 1 + 0.5 j − j10  x2 − j10 − 20 j  x1* 

x2 =

 1 −3 + j − j10  x1 − j10  * − 20 j  x2 

Rewriting in the general form:

x1 (n + 1) =

 1  − 1 + 0.5 j − j10  x2 (n) − j10  * − 20 j  [ x1 (n)] 

x2 (n + 1) =

1  −3 + j  − j10  x1 (n + 1) − j10  − 20 j [ x2 (n)]* 

Solving using MATLAB with x1 (0) = 1, x2 (0) = 1,  = 0.05 : x1 = 0.91 − 0.16 j x2 = 0.86 − 0.22 j

 = 0.036 after 3 iterations 6.14

f ( x) = x3 − 6 x 2 + 9 x − 4 = 0 1 2 4 x = − x 3 + x 2 + = g( x ) 9 3 9

Apply Gauss-Seidel algorithm with x(0) = 2. First iteration yields 1 2 4 x (1) = g (2) = − 23 + 22 + = 2.2222 9 3 9

Second iteration: x(2) = g ( 2.2222 ) = 2.5173 Subsequent iterations will result in

2.8966,3.3376, 3.7398, 3.9568, 3.9988, and 4.0000  Note:

There is a repeated root at x = 1 also. With a wrong initial estimate, the solution might diverge. 131

6.15

x2 cos x − x + 0.5 = 0

Rearranging to solve for x, x = x2 cos x + 0.5

In the general form

x(n + 1) = [ x(n)]2 cos x(n) + 0.5 Solving using MATLAB with x(0) = 1,  = 0.01: x = 1.05  = 0.007 after 2 iterations

To test which initial guesses result in convergence, we use MATLAB and try out initial guesses between − 20 and 20, with a resolution of 0.001. In the following plot the curve is the original function we solved. The shaded area is the values of x(0) which resulted in convergence.

6.16 Eq. (6.2.6) is: x (i + 1) = Mx (i ) + D −1 y Taking the Z transform (assume zero initial conditions): zX ( z ) = MX ( z ) + D −1Y ( z )

( zU − M ) X (z) = D −1Y (z) −1 X ( z) = ( zU − M ) D −1Y ( z ) = G( z )Y ( z )

6.17 For Jacobi,  A11 M = D −1 ( D − A) =   0   Z Det ( ZU − M ) = Det   A21  A  22

0  A22 

−1

 0 − A  21

− A12  A11    0  

 0 − A12   = 0   − A21   A22

A12  A11   = Z 2 − A12 A21 = 0  A11 A22 Z   A A  Z =   12 21   A11 A22 

For Gauss-Seidel, − A12   0  A11   A11 0   0 − A12    −1 M = D ( D − A) =  =   0   A12 A21   A21 A22  0 0  A11 A22    A12   Z  A11  = Z  Z − A12 A21  = 0 det ( ZU − M ) = Det     A12 A21  A11 A22   0 Z −  A11 A22    A A Z = 0, 12 21 A11 A22 −1

When N = 2 , both Jacobi and Gauss-Seidel converge if and only if 6.18

A12 A21  1. A11 A22

x3 + 8 x 2 + 2 x − 40 = 0 x(0) = 1 J (i) =

df = [3x 2 + 16 x + 2]x = x (i ) dx x = x (i )

In the general form:

x(i + 1) = x(i) +

1  (− x(i)3 − 8 x(i)2 − 2 x(i) + 40) 3x(i) + 16 x(i) + 2 2

Using x(0) = 1, we arrive at i

4 1.9106

2.381

1.9675

1.9116



1.381

0.1737

0.0284

5.19E-4

After 4 iterations,   0.001.

 x = 1.9106 6.19 Repeating Problem 6.18 with x(0) = −2, we get: i

3 –3.0528

–2

–3.111

–3.0526



0.5556

0.0188

7.58E-5

After 3 iterations,   0.001 . Therefore,

x = − 3.0528 Changing the initial guess from 1 to − 2 caused the Newton-Raphson algorithm to converge to a different solution. 6.20

x 4 + 3 x 3 − 15 x 2 − 19 x + 30 = 7 J (i) =

df = [4 x 3 + 9 x 2 − 30 x − 19]x = x (i ) dx x = x (1)

In general form

x(i + 1) = x(i) +

1  (7 − x(i)4 − 3x(i)3 + 15 x(i) 2 + 19 x(i) − 30) 4 x(i)3 + 9 x(i)2 − 30 x(i) − 19

Using x(0) = 0, we arrive at: i

1.211

0.796

0.804





0.342

0.010

8.7E-6

After 4 iterations,   0.001. Therefore,

x = 0.804

6.21 Repeating Problem 6.20 with x(0) = 4, we arrive at x = 3.082 after 4 iterations. 6.22

For points near x = 2.2, the function is flat, meaning the Jacobian is close to zero. In Newton-Raphson, we iterate using the equation

x(i + 1) = x(i) + J −1 (i)[ y − f ( x(i))] If J (i) is close to zero, J −1 (i) will be a very large positive or negative number. If we use an initial guess of x(0) which is near 2.2, the first iteration will cause x(i) to jump to a point far away from the solution, as illustrated in the following plot:

6.23 (a) e x1 x2 − 1.2 = 0 cos( x1 + x2 ) − 0.5 = 0 x2 e x1 x2  x1e x1x2 df J (i) = = dx x = x (i ) − sin( x1 + x2 ) − sin( x1 + x2 ) x = x (i )

In general form

x(i + 1) = x(i) − J −1 (i) f ( x) Solving using MATLAB with x1 (0) = 1, x2 (0) = 1,  = 0.001: i

0.6836

0.8956

0.8315

0.8267

0.8266

0.5

0.3860

0.1518

0.2157

0.2205

0.2206



0.3164

0.6068

0.4210

0.0226

0.0001

After 5 iterations   0.005. Therefore, x1 = 0.8266 x2 = 0.2206

(b) When the initial conditions are changed, the solution diverges as each iteration results in very large solutions (on the order of 1e83 ). The error is (until the program fails): 1e82 

 6.24

4.9335

NaN

2 x1 + x22 − 8 = 0 x12 − x22 + x1 x2 − 3 = 0 J (i) =

2 df = dx x = x (i ) 2 x1 + x2

2 x2  − 2 x2 + x1 x = x (i )

In general form

x(i + 1) = x(i) − J −1 (i) f ( x) Solving using MATLAB with x1 (0) = 1, x2 (0) = 1,  = 0.001: i

1.9

1.6875

1.6743

1.7

1.5543

1.5471



0.9

0.2125

0.0132

5.037E-5

After 4 iterations   0.001. Therefore, x1 = 1.6743 x2 = 1.5471 136 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

6.25

10 x1 sin x2 + 2 = 0 10 x12 − 10 x1 cos x2 + 1 = 0 J (i) =

10 x1 sin x2  10sin x2 df = dx x = x (i ) 20 x1 − 10 cos x2 10 x1 sin x2  x = x (i )

In the general form

x(i + 1) = x(i) − J −1 (i) f ( x) Solving using MATLAB with x1 (0) = 1, x2 (0) = 0,  = 1E − 4 : i

0.90

0.8587

0.8554

– 0.20

– 0.2333

– 0.2360





0.1667

0.01131

7.0E-5

After 4 iterations, Newton-Raphson converges to x1 = 0.8554 rad x2 = − 0.2360 rad

6.26 Repeating Problem 6.25 with x1 (0) = 0.25 and x2 (0) = 0, Newton-Raphson converges to x1 = 0.2614 rad x2 = − 0.8711 rad

after 6 iterations. 6.27 To determine which values of x1 (0) result in the answer obtained in Problem 6.25, we test out initial guesses for x1 between 0 and 1, with a resolution of 0.0001. The conclusion of this search is that if 0.5  x1 (0)  1 and x2 (0) = 0, Newton-Raphson will converge to x1 = 0.8554 rad and x2 = − 0.2360 rad.

Note: If x1 (0) = 0.5 and x2 (0) = 0. Newton-Raphson does not converge. This is because 4.79 5 J (0) =   0  0

Thus, J −1 does not exist. 6.28 (a) 𝑌11 = 2 − 𝑗4 + 3 − 𝑗6 = 5 − 𝑗10

𝑌22 = 2 − 𝑗4 𝑌33 = 3 − 𝑗6 𝑌12 = 𝑌21 = −2 + 𝑗4 𝑌13 = 𝑌31 = −3 + 𝑗6 137 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

𝑌23 = 𝑌32 = 0 (b) 𝑃𝑘 = 𝑉𝑘 ∑𝑁 𝑛=1 𝑉𝑛 [𝐺𝑘𝑛 cos(𝛿𝑘 − 𝛿𝑛 ) + 𝐵𝑘𝑛 sin(𝛿𝑘 − 𝛿𝑛 )] At bus 2, 𝑃2 = 𝑉2 [𝐺21 𝑉1 cos(𝛿2 − 𝛿1 ) + 𝐵21 𝑉1 sin(𝛿2 − 𝛿1 ) + 𝐺22 𝑉2 cos(𝛿2 − 𝛿2 ) + 𝐵22 𝑉2 sin(𝛿2 − 𝛿2 ) + 𝐺23 𝑉3 cos(𝛿2 − 𝛿3 ) + 𝐵23 𝑉3 sin(𝛿2 − 𝛿3 )] 1.5 = 1.1[(1)(−2) cos(𝛿2 ) + (1)(4) sin 𝛿2 + (2)(1.1)] Solve for 𝛿2 = 15.795° (c) At bus 3 𝑃3 = 𝑉3 [𝐺31 𝑉1 cos(𝛿3 − 𝛿1 ) + 𝐵31 𝑉1 sin(𝛿3 − 𝛿1 ) + 𝐺32 𝑉2 cos(𝛿3 − 𝛿2 ) + 𝐵32 𝑉2 sin(𝛿3 − 𝛿2 ) + 𝐺33 𝑉3 cos(𝛿3 − 𝛿3 ) + 𝐵33 𝑉3 sin(𝛿3 − 𝛿3 )] 𝑄3 = 𝑉3 [𝐺31 𝑉1 sin(𝛿3 − 𝛿1 ) − 𝐵31 𝑉1 cos(𝛿3 − 𝛿1 ) + 𝐺32 𝑉2 sin(𝛿3 − 𝛿2 ) − 𝐵32 𝑉2 cos(𝛿3 − 𝛿2 ) + 𝐺33 𝑉3 sin(𝛿3 − 𝛿3 ) − 𝐵33 𝑉3 cos(𝛿3 − 𝛿3 )] Which lead to the following two equations −1.5 = 𝑉3 (−3 cos 𝛿3 + 6 sin 𝛿3 + 3𝑉3 ) 0.8 = 𝑉3 (−3 sin 𝛿3 − 6 cos 𝛿3 + 6𝑉3 )* Solve using substitution, 𝑉3 = 0.9723, 𝛿3 = −15.10° (d) At bus 1 𝑃1 = 𝑉1 [𝐺11 𝑉1 cos(𝛿1 − 𝛿1 ) + 𝐵11 𝑉1 sin(𝛿1 − 𝛿1 ) + 𝐺12 𝑉2 cos(𝛿1 − 𝛿2 ) + 𝐵12 𝑉2 sin(𝛿1 − 𝛿2 ) + 𝐺13 𝑉3 cos(𝛿1 − 𝛿3 ) + 𝐵13 𝑉3 sin(𝛿1 − 𝛿3 )] = (1)[(5)(1) cos(0) + (−10)(1) sin(0) + (−2)(1.1) cos(0 − 15.795°) + (4)(1.1) sin(0 − 15.695°) + (−3)(0.9723) cos(0 + 15.10°) + (6)(0.9623) sin 0 + 15.10°)] = 0.39 (e) 𝑃𝑙𝑜𝑠𝑠 = 𝑃𝑔𝑒𝑛 − 𝑃𝑙𝑜𝑎𝑑 = 0.39 + 1.5 − 1.5 = 0.39 𝑝𝑒𝑟 − 𝑢𝑛𝑖𝑡

𝑌21 = 0

6.29

1.72

0.88

𝑌22 = 0.009+𝑗0.1 + 0.009+𝑗0.100 + 𝑗 ( 2 ) + 𝑗 ( 2 ) = 1.79 − 𝑗18.54 𝑌23 = 0 1

𝑌24 = 𝑌25 = − 0.009+𝑗0.1 = −0.893 + 𝑗9.92 6.30

6.25 − j18.695

− 5.00 + j15.00

− 1.25 + j3.75

− 5.00 + j15.00

12.9167 − j38.665

− 1.6667 + j 5.00

− 1.25 + j3.75

− 5.00 + j15.00

− 1.25 + j3.75

− 1.6667 + j 5.00

8.7990 − j 32.2294

− 5.8824 + j 23.5294

− 1.25 + j3.75

− 5.8824 + j 23.5294

9.8846 − j 36.4037

− 2.7523 + j 9.1743

− 5.00 + j15.00

− 2.7523 + j 9.1743

7.7523 − j 24.1443

6.31 First, we need to find the per-unit shunt admittance of the added capacitor.

75 Mvar = 0.75 p.u. To find Y, we use the relation S = V 2Y

S 0.75 = 2 = 0.75 p.u. V2 1

Now, we can find Y44 1 1 1 + + 0.08 + j 0.24 0.01 + j 0.04 0.03 + j 0.10 0.05 + 0.01 + 0.04 +j + j 0.75 2 = 9.8846 − j35.6537

Y44 =

6.32

Assume flat start V1 (0) = 1.0  0, V2 (0) = 1.0  0

V2 (i + 1) =

1  P2 − jQ2  − Y21V1 (i) Y22  V2* (i) 

1 = − 4 + j8 0.05 + j 0.1 1 Y22 = = 4 − j8 0.05 + j 0.1 Y21 = −

Using MATLAB, with V1 (0) = 1.0  0 and V2 (0) = 1.0  0 : i

1.0  0

1 0.9264  − 3.0941 

0.9176 − 3.0941 

0.9168  − 3.1263 

After 3 iterations, V2 = 0.9168  − 3.1263 

6.33 If we repeat Problem 6.32 with V2 (0) = 1.0  30 , we get V2 = 0.9172  − 3.1449  after 3 iterations. 6.34

V1 = 1.0  0

V2 (i + 1) =

1  P2 − jQ2  − Y21V1 − Y23V3 (i) *  Y22  V2 (i) 

1  − 1 + j 0.5  − j 5V1 − j 5V3 (i) − j10  V2* (i) 

V3 (i + 1) =

 1  P3 − jQ3 − Y31V1 − Y32V2 (i + 1) Y33  V3* (i) 

 1  − 1.5 + j 0.75 − j5V1 − j 5V2 (i + 1)  * − j10  V3 (i) 

Solving using MATLAB with V2 (0) = 1.0  0 and V3 (0) = 1.0  0 : i

1.0  0

0.9552  − 6.0090 

0.9090  − 12.6225 

1.0  0

0.9220  − 12.5288 

0.8630  − 16.1813 

After 2 iterations, V2 = 0.9090  − 12.6225  V3 = 0.8630  − 16.1813 

6.35 Repeating Problem 6.34 with V1 = 1.05  0 i

1.0  0

0.9801  − 5.8560 

0.9530  − 11.8861 

1.0  0

0.9586  − 120426 

0.9129  − 14.9085 

After 2 iterations, V2 = 0.9530  − 11.8861  V3 = 0.9129  − 14.9085 

6.36

V21 =

 1  P2 − jQ2 − Y21V1 − Y23V30 − Y24V40   0 * Y22  (V2 ) 

 1  0.5 + j 0.2 − 1.04(−2 + j 6) − (−0.666 + j 2) − ( −1 + j3)   Y22  1 − j 0 

4.246 − j11.04 = 1.019 + j 0.046 = 1.022.58 p.u. 3.666 − j11

V22 =

 1  P2 − jQ2 1 1  − Y V − Y V − Y V 21 1 23 3 24 4  Y22  (V21 )*  

Determine V31 and V41 in the same way as V21 . Then

V22 =

1  0.5 + j 0.2 − 1.04(−2 + j 6) − (−0.666 + j 2.0)(1.028 − j 0.087) Y22 1.019 + j 0.046 − (−1 + j 3)(1.025 − j 0.0093) 

4.0862 − j11.6119 = 1.061 + j 0.0179 = 1.06160.97 p.u. 3.666 − j11.0

6.37 Gauss-Seidel iterative scheme: With  2 = 0;V3 = 1 p.u.;  3 = 0 (starting values) Given

Also given

 7 −2 −5  YBus = − j  −2 6 −4  ;  −5 −4 9 

V1 = 1.00 p.u. V2 = 1.0 p.u.; P2 = 60 MW P3 = −80 MW; Q3 = −60 M var (lag)

Iteration 1: Q2 = − Im(Y22V2 + Y21V1 + Y23V3 )V2*

= − Im  − j 6(10) + j 2(10) + j 4(10) 10 = 0

V2 = =

 1  P2 − jQ2 − Y21V1 − Y23V3   * Y22  V2  1  0.6 + j 0  − j 2(10) − j 4(10)   − j 6  10 

= 1 + j 0.1 15.71

Repeat:

V2 = 1 +

0.190 = 0.99005 + j 0.0995 1 − 5.71

= 0.9955.74 V2 = 15.74 = 0.995 + j 0.1 V3 = =

 1  P3 − jQ3 − Y31V1 − Y32V2   * Y33  V3  1  −0.8 + j 0.6  − j 5 (10 ) − j 4 ( 0.995 + j 0.1)   − j 9  10 

0.111153.13 10 = 0.9311 − j 0.0444 = 0.9322 − 2.74 = 0.9978 + j 0.0444 −

Repeat:

V3 = 0.9978 + j 0.0444 −

0.111153.13 0.93222.74

= 0.9218 − j 0.0474 = 0.923 − 2.94 Check:

V2 = (0.995 + j 0.100) − (1 − j 0) = −0.005 + j 0.1

V3 = (0.9218 + j 0.0474) − (1 − j 0) = −0.0782 − j 0.0474 xmax = 0.1

Iteration 2: Q2 = − Im  − j 6(0.995 + j 0.1) + j 2(10) + j 4(0.9218 − j 0.0474) (0.995 − j 0.1) = 0.36 V2 =

1  0.6 − j 0.36  − j 2(10) − j 4(0.9218 − j 0.0474)   − j 6  1 − 5.74 

= 0.9479 + j 0.316 +

0.116659.04 = 14.24 = 0.9973 + j 0.0739 1 − 5.74

V2 = 0.9479 − j 0.0316 +

Repeat:

0.116659.04 = 1.0003 + j 0.0725 1 − 4.24

14.15 = 0.9974 + j 0.0723 V3 =

1  −0.8 + j 0.6  − j 5(10) − j 4(0.9974 + j 0.0723)   j 9  0.92302.94 

= 0.9988 + j 0.0321 −

0.111153.13 = 0.9217 − j 0.0604 0.9232.94

= 0.9237 − 3.75 V3 = 0.9988 + j 0.0321 −

Repeat:

0.111153.13 = 0.9205 − j 0.0592 0.92373.75

= 0.9224 − 3.68 V2 = (0.9974 + j 0.0732) − (0.995 + j 0.1) = 0.0024 − j 0.0277

Check:

V3 = (0.9205 − j 0.0592) − (0.9218 − j 0.0474) = −0.0013 − j 0.0118 xmax = 0.028

Third iteration yields the following results within the desired tolerance:

   V3 = 0.9208 − j 0.0644 = 0.923 − 4 V2 = 0.9981 + j 0.0610 = 13.5 6.38

First, convert all values to per-unit 𝑃

𝑃𝑝𝑢 = 𝑆

−180

= 100 MW = −1.8

𝑏𝑎𝑠𝑒

𝑄𝑝𝑢 = 𝑆

MVA

𝑄

−60 MW

𝑏𝑎𝑠𝑒

= 100

Δ𝑦

MVA

= −0.6

0.1

Δ𝑦max,pu = 𝑆 max = 100 MVA = 10−4 𝑏𝑎𝑠𝑒

MVA

Since there are 2 buses, we need to solve 2(n-1)=2 equations. Therefore, J has dimension 2x2. Using Table 6.5 𝐽122 = 𝑉2 𝑉1 (−𝐺12 sin(𝛿2 − 𝛿1 ) + 𝐵12 cos(𝛿2 − 𝛿1 )) 143 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

𝐽222 = 2𝑉2 𝐺22 + 𝑉2 𝑉1 (𝐺12 cos(𝛿2 − 𝛿1 ) + 𝐵12 sin(𝛿2 − 𝛿1 )) 𝐽322 = 𝑉2 𝑉1 (𝐺12 cos(𝛿2 − 𝛿1 ) + 𝐵12 sin(𝛿2 − 𝛿1 )) 𝐽422 = 2𝑉2 𝐵22 + 𝑉2 𝑉1 (𝐺12 sin(𝛿2 − 𝛿1 ) − 𝐵12 cos(𝛿2 − 𝛿1 )) Also, 𝑌𝑏𝑢𝑠 = [

−𝑗10 𝑗10

𝑗10 ] −𝑗10

Finally, using Eqs. (6.6.2 and 6.6.3), 𝑃2 (𝛿2 , 𝑉2 ) = 𝑉2 (𝑉1 (𝐺21 cos(𝛿2 − 𝛿1 ) + 𝐵21 sin(𝛿2 − 𝛿1 )) + 𝑉2 𝐺22 ) 𝑄2 (𝛿2 , 𝑉2 ) = 𝑉2 (𝑉1 (𝐺21 sin(𝛿2 − 𝛿1 ) − 𝐵21 cos(𝛿2 − 𝛿1 )) − 𝑉2 𝐵22 ) 6.39 If we repeat Problem 6.38 with V2 (0) = 1.0  30 , Newton-Raphson converges to

0.9149  − 11.347  after 5 iterations. 6.40 If we repeat Problem 6.38 with V2 (0) = 0.25 0 , Newton-Raphson converges to V2 = 0.2074  − 1140.2  after 5 iterations.

6.41 The number of equations is 2(n − 1) = 4. Therefore, J will have dimension 4  4. Using Table 6.5

10 − 5 0 0 − 2 7 0 0  J (0) =  0 0 10 − 5    0 0 − 2 13 6.42 First, we convert the mismatch into per-unit 0.1 MVA = 1E-4 p.u. with Sbase = 100 MVA. Solving using MATLAB with V2 (0) = 1.0  0 and V3 (0) = 1.0  0,

V2 V3

1.0  0

0.8833  − 13.3690 

0.8145  − 16.4110 

0.8019  − 16.9610 

0.8015 − 16.9811 

1.0  0

0.8667  − 15.2789 

0.7882  − 19.2756 

0.7731  − 20.1021 

0.7725  − 20.1361 

After 4 iterations, Newton-Raphson converges to V2 = 0.8015  − 16.9811  V3 = 0.7725  − 20.1361 

6.43

First, form the Y-bus −𝑗12.5 𝑗10 −𝑗15 𝑌̅𝑏𝑢𝑠 = [ 𝑗10 𝑗2.5 𝑗5

𝑗2.5 𝑗5 ] −𝑗7.5

𝛿2 (0) = 𝛿3 (0) = 0°; 𝑉2 (0) = 1.0 Compute Δ𝑦(0) 𝑃2 (𝑋) = 𝑉2 (𝑉1 (𝐺21 cos(𝛿2 − 𝛿1 ) + 𝐵21 sin(𝛿2 − 𝛿1 )) + 𝑉2 𝐺22 + 𝑉3 (𝐺23 cos(𝛿2 − 𝛿3 ) + 𝐵23 sin(𝛿2 − 𝛿3 ))) = 0 𝑃3 (𝑋) = 𝑉3 (𝑉1 (𝐺31 cos(𝛿3 − 𝛿1 ) + 𝐵31 sin(𝛿3 − 𝛿1 )) + 𝑉2 (𝐺32 cos(𝛿3 − 𝛿2 ) + 𝐵32 sin(𝛿3 − 𝛿2 )) + 𝑉3 𝐺33 ) = 0 𝑄2 (𝑋) = 𝑉2 (𝑉1 (𝐺21 sin(𝛿2 − 𝛿1 ) − 𝐵21 cos(𝛿2 − 𝛿1 )) − 𝑉2 𝐵22 + 𝑉3 (𝐺23 sin(𝛿2 − 𝛿3 ) − 𝐵23 cos(𝛿2 − 𝛿3 ))) = 0 𝑃2 − 𝑃2 (𝑋) −2.0 − 0 −2 Δ𝑦(0) = [ 𝑃3 − 𝑃3 (𝑋) ] = [ 1.0 − 0 ] = [ 1 ] −0.5 − 0 −0.5 𝑄2 − 𝑄2 (𝑋) Compute 𝐽(0) (see Table 6.5) 𝐽122 = 𝑉2 (𝑉1 (−𝐺12 sin(𝛿2 − 𝛿1 ) + 𝐵12 cos(𝛿2 − 𝛿1 )) + 𝑉3 (−𝐺32 sin(𝛿2 − 𝛿3 ) + 𝐵32 cos(𝛿2 − 𝛿3 ))) = 15 𝐽123 = 𝑉2 𝑉3 (𝐺23 sin(𝛿2 − 𝛿3 ) − 𝐵23 cos(𝛿2 − 𝛿3 )) = −5 𝐽132 = 𝑉3 𝑉2 (𝐺32 sin(𝛿3 − 𝛿2 ) − 𝐵32 cos(𝛿3 − 𝛿2 )) = −5 𝐽133 = 𝑉3 (𝑉1 (−𝐺13 sin(𝛿3 − 𝛿1 ) + 𝐵13 cos(𝛿3 − 𝛿1 )) + 𝑉2 (−𝐺23 sin(𝛿3 − 𝛿2 ) + 𝐵23 cos(𝛿3 − 𝛿3 ))) = 7.5 𝐽222 = 2𝑉2 𝐺22 + 𝑉2 (𝑉1 (𝐺12 cos(𝛿2 − 𝛿1 ) + 𝐵12 sin(𝛿2 − 𝛿1 )) + 𝑉3 (𝐺32 cos(𝛿2 − 𝛿3 ) + 𝐵32 sin(𝛿2 − 𝛿3 )) ) = 0 𝐽232 = 𝑉3 (𝐺32 cos(𝛿3 − 𝛿2 ) + 𝐵32 sin(𝛿3 − 𝛿2 )) = 0 𝐽322 = 𝑉2 (𝑉1 (𝐺12 cos(𝛿2 − 𝛿1 ) + 𝐵12 sin(𝛿2 − 𝛿1 )) + 𝑉3 (𝐺32 cos(𝛿2 − 𝛿3 ) + 𝐵32 sin(𝛿2 − 𝛿3 ))) = 0 𝐽323 = −𝑉2 𝑉3 (𝐺23 cos(𝛿2 − 𝛿3 ) + 𝐵23 sin(𝛿2 − 𝛿3 )) = 0 𝐽422 = 2𝑉2 𝐵22 + 𝑉2 (𝑉1 (𝐺12 sin(𝛿2 − 𝛿1 ) − 𝐵12 cos(𝛿2 − 𝛿1 )) + 𝑉3 (𝐺32 sin(𝛿2 − 𝛿3 ) − 𝐵32 cos(𝛿2 − 𝛿3 )))=15 145 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

15 −5 0 𝐽 = [−5 7.5 0 ] 0 0 15 For the first iteration, solve −0.1142857 𝑥(1) = 𝑥(0) − 𝐽−1 (0) ⋅ Δ𝑦(0) = [ 0.05714285 ] 0.96666 6.44 Repeating Problem 6.43 with PL 2 = 1.0 p.u.

Similar to solution to 6.43. Y-bus is not affected. 𝑃2 − 𝑃2 (𝑋) −1.0 − 0 −1 Δ𝑦(0) = [ 𝑃3 − 𝑃3 (𝑋) ] = [ 1.0 − 0 ] = [ 1 ] −0.5 − 0 −0.5 𝑄2 − 𝑄2 (𝑋) Jacobian is not affected. −0.02857 𝑥(1) = 𝑥(0) − 𝐽−1 (0) ⋅ Δ𝑦(0) = [ 0.11429 ] 0.96667 6.45 After the first three iterations J22 = 104.41, 135.86, 114.91; and with the next iteration it converges to 106.66. 6.46 First, convert all values to per-unit. PG 2 = 80 MW = 0.8 p.u. PL3 = 180 MW = 1.8 p.u.

For convergence,  ymax = 0.1 MVA = 1E-4 p.u. Second, find Ybus .

8 − j16 − 4 + j8 − 4 + j8 Ybus = − 4 + j8 8 − j16 − 4 + j8 − 4 + j8 − 4 + j8 8 − j16  To solve the case, we follow the steps outlined in section 6.6. Iteration 0 Step 1

 2 (0) 0    Start with x(0) =  3 (0) = 0 V3 (0) 1

P ( x(0)) 0 P ( x(0)) =  2 =  P3 ( x(0)) 0

Q ( x(0)) = Q3 ( x(0)) = 0

 P2 − P2 (x(0))   0.8     y(0) =  P3 − P3 ( x(0))  = − 1.8 Q3 − Q3 ( x(0))  0   

Step 2 16 − 8 J1 =   − 8 16 J 3 = 4 − 8

− 4 J2 =   8 J 4 = 16

16 − 8 − 4   J (0) = − 8 16 8   4 − 8 16  

Step 3

J (0)x(0) = y(0) 16 − 8 − 4   2 (0)  0.8   − 8 16 8   3 (0) = − 1.8   4 − 8 16  V3 (0)  0  − 0.0083333  x(0) = − 0.094167  − 0.045 

Step 4

− 0.0083333 x(1) = x(0) +  x(0) = − 0.094167   0.955  Iteration 1 Step 1  0.7825 P ( x(1)) =   − 1.6861 Q ( x(1)) = 0.0610  0.017502    y(1) = − 0.11385  − 0.06104  147 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

Step 2

15.9057 − 7.93935 −3.29945 J (1) = − 7.28439 14.5314 5.8744   4.4609 − 8.98235 15.3439  Step 3

− 0.0038007  x(1) = − 0.0069372 − 0.0069342 Step 4

− 0.012134 x(2) = − 0.1011   0.94807  Iteration 2 Step 1  0.7999 P ( x(2)) =   − 1.7990 Q ( x(2)) = 6.484 E-4  1.3602 E-4    y(2) = − 1.0458 E-4 − 6.484 E-4 

Step 2 −7.89148 −3.27336 15.8424  J (2) = − 7.21758 14.3806 5.68703  4.45117 −8.98958 15.1697 

Step 3 − 3.7748 E-5  x (2) = − 6.412E-5  − 6.9664 E-5

Step 4

Iteration 3 Step 1  0.8000 P ( x(3)) =   − 1.8000 Q ( x(3)) = 6.4875 E-8  1.1294 E-8   y(3) = − 9.7437 E-8 − 6.4875 E-8  y(3)  1E-4

 Newton-Raphson has converged in 3 iterations to V1 = 1.0  0 V2 = 1.0  − 0.012172 rad = 1.0  − 0.6974° V3 = 0.948  − 0.10117 rad = 0.948  − 5.797°

Using the voltages, we can calculate the powers at buses 1 and 2 with Eqs. (6.6.2) and (6.6.3). P1 = 1.0910 p.u. = 109.10 MW Q1 = 0.0237 p.u. = 2.37 Mvar Q2 = 0.1583 p.u. = 15.83 Mvar

Using PowerWorld simulator, we get

P1 = 109.1 MW Q1 = 2.4 Mvar Q2 = 15.8 Mvar V3 = 0.948 p.u. which agrees with our solution. 6.47 First, convert values to per-unit Q2 = 50 Mvar = 0.5 p.u. Ybus stays the same as in Problem 6.46.

Since bus 2 is now a PQ bus, we introduce a fourth equation in order to solve for V2 . Iteration 0 Step 1

 2 (0) 0   (0) 0 Start with x(0) =  3  =   V2 (0) 1    V3 (0) 1

P ( x(0)) 0 P ( x(0)) =  2 =  P3 (x(0)) 0 Q ( x(0)) 0 Q ( x(0)) =  2 =  Q3 ( x(0)) 0

 P2 − P2 (x(0))   0.8   P − P ( x(0))  − 1.8  y(0) =  3 3 =  Q2 − Q2 ( x(0)   0.5     Q3 − Q3 ( x(0))  0

Step 2

16  −8 J (0) =  − 8  4

−8 8

− 4  16 − 4 8  4 16 − 8  − 8 − 8 16 

Step 3 J (0) x(0) =  y(0) − 0.023333 − 0.10167    x(0) =   1.03     0.97 

Step 4

− 0.023333   − 0.10167  x(1) = x(0) +  x(0) =   1.03     0.97  Iteration 1 Step 1  0.8174 P ( x(1)) =   − 1.7299 0.5517 Q ( x(1)) =   0.0727 − 0.01739    − 0.070052   y(1) = − 0.051746   − 0.072698 150 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

Step 2

9.03358 −3.46256 16.4227 −8.28102   − 7.65556 14.9817 −4.47535 5.97655 J (1) =  − 7.66981 3.35868 17.0157 −8.53714    4.60961 −9.25715 −7.43258 15.5949  Step 3

− 0.00057587   − 0.0032484    x(1) = − 0.0077284    − 0.010103  Step 4

− 0.023904   − 0.10492   x(2) =  1.0223     0.9599  Iteration 2 Step 1  0.7999 P ( x(2)) =   − 1.7989 0.5005 Q ( x(2)) =   0.0007  0.00013185   − 0.0011097   y(2) =  − 0.00048034   − 0.00072284

Step 2

 16.2201 − 8.14207 8.96061 −3.41392   − 7.50685 14.7417 − 4.44838 5.80513  J (2) =  − 7.56045 3.27701 16.8459 −8.48223    4.54745 − 9.17011 −7.34331 15.3591  Step 3

 − 6.3502 E-7   − 5.3289E-5    x(2) =   − 7.6468E-5    − 0.00011525 151 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

Step 4

− 0.02391 − 0.10497  x(3) =   1.0222     0.95978  Iteration 3 Step 1  0.8000  P ( x(3)) =   − 1.8000 0.5000 Q ( x(3)) =   0.0000  6.1267 E-8    − 2.0161E-7   y(3) =  − 6.5136 E-8    − 8.9553E-8 

y(3)  1 E-4

 Newton-Raphson has converged in 3 iterations to V1 = 1.0  0 V2 = 1.0222  − 0.02391 rad =1.0222  − 1.370  V3 = 0.95978  − 0.10497 rad = 0.95978  − 6.014°

Using the voltages, we can calculate the powers at bus 1 with Eqs. (6.6.2) and (6.6.3). P1 = 1.0944 p.u. = 109.44 MW Q1 = − 0.3112 p.u. = − 31.12 Mvar

Using PowerWorld Simulator, we get

P1 = 109.4 MW Q1 = − 31.1 Mvar V2 = 1.025 p.u. V3 = 0.960 p.u. Which agrees with our solution.

6.48

6.49 Tap setting

Mvar @ G1

V5 (p.u.)

V2 (p.u.)

P losses

0.975

0.954

0.806

37.64

0.98125

0.961

0.817

36.63

0.9875

0.965

0.823

36.00

0.99375

106

0.97

0.828

35.4

1.0

114

0.974

0.834

34.84

1.00625

123

0.979

0.839

34.31

1.0125

131

0.983

0.845

33.81

1.01875

140

0.987

0.850

33.33

1.025

149

0.992

0.855

32.89

1.03125

158

0.996

0.86

32.47

1.0375

167

1.0

0.865

32.08

1.04375

176

1.004

0.87

31.72

1.05

185

1.008

0.874

31.38

1.05625

195

1.012

0.879

31.06

1.0625

204

1.016

0.884

30.76

1.06875

214

1.02

0.888

30.49

1.075

224

1.024

0.893

30.23

1.08125

233

1.028

0.897

30.00

1.0875

243

1.031

0.901

29.79

1.09375

253

1.035

0.906

29.59

1.1

263

1.039

0.910

29.42

6.50

Shunt Capacitor

Line

Total Power

State

Rating (Mvar)

Output (Mvar)

2-4

2-5

4-5

Losses (MW)

Disconnected

0.834

27%

49%

19%

34.37

Connected

200

184.0

0.959

25%

44%

16%

25.37

Connected

261

261.0

1.000

25%

43%

15%

23.95

V2 (p.u.)

When the capacitor rating is set at 261 Mvar, the voltage at bus 2 becomes 1.0 per unit. As a result of this, the line loadings decrease on all lines, and thus the power loss decrease as well.

6.51

Before new line

After new line

Bus voltage V2 (p.u.)

0.834

0.953

Total real power losses (MW)

34.8

18.3

Branch bw bus 1–5 (% loading)

68.5

63.1

Branch bw bus 2–4 (% loading)

27.3

17.5

49.0

25.4 (both lines)

Branch bw bus 2–5 (% loading) Branch bw bus 3–4 (% loading)

53.1

45.7

Branch bw bus 4–5 (% loading)

18.8

22.1

6.52 With the line connecting REDBUD 69 and PEACH 69 removed, the voltage at REDBUD 69 drops to 0.968 p.u. To raise that voltage to 1.0 p.u., we require 26.0 Mvar from the capacitor bank of REDBUD 69. 6.53

When the generation at PEAR138 is 300 MW, the losses on the system are minimized to 10.24 MW.

6.54

Total System Losses (MW)

16 14 12 10 8 6 4 2 0 0

100

200

300

400

500

P(PEAR138)(MW)

When the generation at PEAR138 is 260 MW or 280 MW, the losses on the system are minimized to 11.48 MW. 6.55 DIAG

= [17 25 9 2 14 15]

OFFDIAG = [−9.1 −2.1 −7.1 −9.1 −8.1 −1.1 −6.1 −8.1 −1.1 −2.1 −6.1 −5.1 −7.1 −5.1] C0L

=[ 2

ROW

=[ 3

6.56 By the process of node elimination and active branch designation, in Fig. 6.9:

The fill-in (dashed) branch after Step 6 is shown below:

Note that two fill-ins are unavoidable. When the bus numbers are assigned to Fig. 6.9 in accordance with the step numbers above, the rows and columns of YBUS will be optimally ordered 155 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

for Gaussian elimination, and as a result, the triangular factors L and U will require minimum storage and computing time for solving the nodal equations. 6.57 Table 6.6 w DC Approximation Generation Bus #

Load

Voltage

Phase Angle

Magnitude

(degrees)

(per unit)

(per unit) 1

1.000

0.000

3.600

0.000

1.000

−18.695

0.000

8.000

0.000

1.000

0.524

5.200

0.000

0.800

0.000

1.000

−1.997

0.000

1.000

−4.125

0.000

TOTAL

8.800

0.000

8.800

0.000

When comparing these results to Table 6.6 in the book, voltage magnitudes are all constant. Most phase angles are close to the NR algorithm except bus 3 has a positive angle in DC and a negative value in NR. Total generation is less since losses are not taken into account and reactive power is completely ignored in DC power flow.

Table 6.7 w DC Approximation Line # 1 2 3

Bus to Bus

−2.914

0.000

2.914

0.000

2.914

−5.086

0.000

5.086

0.000

5.086

1.486

0.000

1.486

−1.486

0.000

1.486

With the DC power flow, all reactive power flows are ignored. Real power flows are close to the NR algorithm except losses are not taken into account, so each end of the line has the same flow.

Table 6.8 w DC Approximation Tran. # 1 2

Bus to Bus

3.600

0.000

3.600

−3.600

0.000

3.600

4.400

0.000

4.400

−4.400

0.000

4.400

With DC approximation the reactive power flows in transformers are also ignored and losses are also assumed to be zero. 6.58 With the reactance on the line from bus 2 to bus 5 changed, the B matrix becomes:

0 − 43.333  0 − 100 B=  10 100  0  33.333

10 100 − 150 40

33.333  0   40  − 123.333

P remains the same:

− 8.0   4.4   P=  0     0  − 0.2443 −13.995°      0.0255 1.4638°  = − B−1 P =  rad =  − 0.0185  −1.0572°     − 0.0720  −4.1253°

6.59 (a) Since bus 7 is the slack bus, we neglect row 7 and column 7 in our B matrix and P vector.

16.667 4.167 0 0 0 − 20.833    5.556 5.556 8.333 16.667  16.667 − 52.778  4.167  5.556 −43.056 33.333 0 0 B=  5.556 33.333 −43.056 4.167 0  0   0  8.333 0 4.167 −29.167 0   16.667 0 0 0 −25  0   160  1.6      110  1.1 − 110 − 1.1 P=  MW=   p.u. − 30  − 0.3 − 130 − 1.3      0   0  (b)  = −B−1P

 7.6758     4.2020  − 0.4315  =  − 0.3265  − 1.3999     2.8014  This agrees with the angles shown in PowerWorld. 6.60 First, we find the angle of bus 1 required to give a 59 MW line loading to the line between buses 1 and 3. P13 = 0.59 =

1 −  3 x13

1 −  3 0.24

1 =  3 + 0.142 Now, we solve for the angles:

P = − B

1.6 +  P1   1.1       − 1.1     = −  − 0.3    − 1.3       0  

  3 + 0.142   2     3   4     5   6   

This is a system of 6 unknowns ( 2 ,3

6 , P1) and 6 equations. Solving yields

 P1 = 0.0115 p.u. =1.15 MW Therefore, if the generator at bus 1 increases output by 1.115 MW, the flow on the line between buses 1 and 3 will reach 100%. 6.61 A type 1 or 2 wind turbine is one which maintains a constant real and reactive power output. Hence, the bus it is connected to will be a PQ bus. A type 3 or 4 wind turbine is one which maintains a constant voltage and real power output; the bus it is connected to will be a PV bus. In a contingency such as a line outage, the voltage in the PQ bus will change due to the redistribution of power flows on the remaining lines. However, the voltage on a PV bus remains constant in a contingency, due to the fact that the type 3 or 4 wind turbine changes its reactive power output to maintain that voltage. In this regard, a type 3 or 4 wind turbine is more favorable. From Eq. (6.6.3), we can see that the reactive power injection at a bus is directly proportional to the voltage magnitude at that bus.

Chapter 7 Power System Economics and Optimization 7.1

dC1 = 15 + 0.1 P1 =  dP1 dC2 = 20 + 0.08 P2 =  dP2 P1 + P2 = 1000

−1  P1   −15  0.1 0 0 0.08 −1  P2  =  −20   1 1 0     1000  P1 = 472.2 MW, P2 = 527.8 MW,  = 62.2 $/MWh Total cost = $41,230/hr

7.2

Since the unlimited solution violates the P2 limit constraint, P2 = 500 MW, which means

P1 = 500 MW. The system lambda is determined by the unlimited unit = 15 + 0.1* 500 =

$65/MWh. Total cost is $41,300/hr. 7.3

dC1 = 15 + 0.1 P1 =  dP1 dC2 = 0.95 ( 20 + 0.08P2 ) =  dP2 P1 + P2 = 1000 0 −1  P1   −15  0.1  0 0.076 −1  P  =  −19    2    1 1 0     1000

P1 = 454.5 MW, P2 = 545.5 MW,  = 60.5 $/MWh Total cost = $41,259/hr

7.4

  dC1  1  15 + 0.1P1  = = 60.0 dP1  1 − PL  1 − 4  10−4 P1 − 4  10−4 P2  P1  

(

)

  dC2  1  20 + 0.08P2  = = 60.0 dP2  1 − PL  1 − 6  10−4 P2 − 4  10−4 P1  P2  

(

)

Rearranging the above two equations gives

( 0.1 + 0.024 ) P1 − (0.024)P2 = 45 ( 0.08 + 0.036 ) P2 − (0.024)P1 = 40 Solving gives P1 = 447.6, P2 = 437.4, Plosses = 19.15 MW, Pload = 865.85 MW, Total cost = $34,431/hr

7.5

For N = 2, (7.12.14) becomes: 2

i =1 j =1

i =1

PL =   Pi Bij Pj =  Pi ( Bi1 P1 + Bi 2 P2 ) = B P + B12 P1 P2 + B21 P1P2 + B22 P2 2 2 11 1

Assuming B12 = B21 , PL = B11P12 + 2 B12 P1P2 + B22 P2 2 PL = 2 ( B11P1 + B12 P2 ) P1

PL = 2 ( B12 P1 + B22 P2 ) P2

Also, from (7.12.15):

i =1

2 PL = 2  B1 j Pj = 2 ( B11P1 + B12 P2 ) j =1 P1

i=2

2 PL = 2  B2 j Pj = 2 ( B21P1 + B22 P2 ) j =1 P2

= 2 ( B12 P1 + B22 P2 )

which checks. 7.6

Choosing Sbase as 100 MVA (3-Phase),

1 = ( S3 base ) 0.01 = 100;  2 = 40 2

1 = ( S3 base ) 2.00 = 200;  2 = 260  1 = 100

;  2 = 80

In per unit, 0.25  PG1  1.5;0.3  PG 2  2.0;0.55  PL  3.5

1 =

 1  2 = 200 PG1 + 200; 2 = = 80 PG 2 + 260 PG1 PG 2

Calculate 1 and 2 for minimum generation conditions (Point 1, in Figure shown below). Since

2  1 , in order to make  ’s equal, load unit 1 first until 1 = 284 which occurs at PG1 =

284 − 200 = 0.42 (Point 2 in Figure) 200

Chapter 4 Transmission Line Parameters

4.1

𝑅𝑑𝑐,20 =

4.2

𝜌=

(10.37 Ω-cmil/ft)(10 ft) 50+234.5 = 1.659 mΩ; 𝑅𝑑𝑐,50 = (1.659 mΩ) ( ) = 1.854 mΩ (250 mil)2 20+234.5

(1 Ω)(500 mil)2 14000 feet

𝜌

= 17.857 Ω-cmil/ft; 𝑇2 = 𝜌2 (𝑇1 + 𝑇) − 𝑇 = 1

17.857 (20 + 228.1) − 228.1 = 17

32.5° C 4.3

l = 1.05  2000 = 2100 m , Allowing for the twist. X-sectional area of all 19 strands = 19 

 4

 (1.5  10 −3 ) = 33.576  10 −6 m 2 . 2

Pl 1.72  10−8  2100 = = 1.076  A 33.576  10−6 2

4.4

  sq  mil   1in   0.0254 m  (a) 795 MCM = ( 795  103 cmil )  4      1 cmil  1000 mil   1in  = 4.0283  10 −4 m 2

 50 + T  (b) R60 HZ, 50°C = R60 HZ, 75°C    75 + T   50+228.1  = 0.0880    75+228.1  = 0.0807 /km

4.5

Ω mi

1mi Ω ) = 0.07365 per conductor (at 75% current 1.609km km

From Table A-4 𝑅60 𝐻𝑧,50°𝐶 = (0.1185 ) ( capacity). For 3 conductors per phase: 𝑅60 𝐻𝑧,50°𝐶 =

4.6

Total transmission line loss PL =

0.07365 Ω = 0.02455 km per phase 3

2.5 (190.5) = 4.7625MW 100

190.5  103

= 500 A

3 ( 220 )

From PL = 3I 2 R , the line resistance per phase is R=

4.7625106 3 ( 500 )

= 6.35 

The conductor cross-sectional area is given by

( 2.84  10−8 )( 63  103 ) = 2.81764  10−4 m 2 6.35

d = 1.894cm = 0.7456in = 745.6 mil A = d 2 = ( 745.6 ) = 556,000 cmil 2

4.7

12 W 50+234.5 = 37.0 mΩ. 𝜌 = 10.37 ( ) = 11.59 Ω-cmil/ft; A)2 20+234.5

𝑅 = (18

(11.59)(4000)

diameter = √

4.8

(a)

0.037

= 1119 mil = 1.119 in

From Eq. (4.4.10) H  1000 m  1000 m H  1 Lint =   10 −7    = 0.05mH/ km Per Conductor 2 m   1km  1H 

(b) From Eq. (4.5.2) DH Lx = Ly = 2  10 −7 Ln      m −1  0.015  −3 D = 0.5m  = e 4   = 5.841  10 m 2  

0.5   H  1000 m  1000 mH  Lx = Ly = 2  10 −7 Ln     −3  H  5.841  10  m  km  

= 0.8899

mH per conductor km

mH per circuit km

(a) Lint = 0.05mH/ km Per Conductor 0.5   6 Lx = Ly = 2  10 −7 ln  10 = 0.8535mH/ km Per Conductor −3   1.2  5.841  10  L = L x + L y = 1.707mH / km Per Circuit

(b) 12( Ic ) = 0.2 I c ln

Dc 2 mWb/km Dc1

0.5   6 Lx = Ly = 2  10 −7 ln  10 = 0.9346 mH/km Per Conductor −3  0.8  5.841  10   L = L x + L y = 1.869mH/km Per Circuit.

Lint is independent of conductor diameter.

The total inductance decreases 4.1% (increases 5%) at the conductor diameter increases 20 % (decreases 20%). 4.10 From Eq. (4.5.10) DH L1 = 2  10 −7 Ln    r  m

D = 4 ft −1  .5   1ft  r = e 4      2   12 in 

4   L1 = 2  10 −7 Ln  −2  1.6225  10   H m

L1 = 1.101  10 −6

r  = 1.6225  10 −2 ft

X1 =  L1 = ( 2 60 ) (1.101 10−6 ) (1000 ) = 0.4153  / km 4.8   4.11 (a) L1 = 2  10 −7 ln  = 1.138  10 −6 H/m −2   1.6225  10 

X1 =  L1 = 2 ( 60 ) (1.138  10−6 ) (1000 ) = 0.4292  /km 3.2   (b) L1 = 2  10 −7 ln  = 1.057  10 −6 H/m −2  1.6225  10  

25 =1.409μH/m; The inductive reactance is 2𝜋(60)(𝐿𝑥 )Ω/m = 0.0217

(b) The total line reactance is given by XT = 2 ( 60 ) 4  10 −4 ln = 0.1508 ln

0.2426 ln

D r

D  / km r

D  / mi r

 X T = 0.755  / km or 1.215  / mi (c) LT = 4  10−4 ln

7 = 0.00228 mH/ m 0.7788 ( 0.03)

Doubling the separation between the conductors causes only about a 14% rise in inductance. 4.14 (a) Eq. (4.5.9): L = 2  10 −7 ln

D H/m Per Phase r



(b) r  = r.e −1/ 4 = 0.06677 ( 0.7788 ) = 0.052 ft. X a = k log

1  1  = 0.2794 log   = 0.35875 r  0.052 

X d = k log D = 0.2794 log (10 ) = 0.2794 X = X a + X d = 0.63815  / mi/ ph. 

When spacing is doubled, X d = 0.36351 and X = 0.72226  / mi/ph. 

4.15 For each of six outer conductors: −

D11 = r  = e 7 r D12 = D16 = D17 = 2r D13 = D15 = 2 3r D14 = 4r

For the inner conductor: −

(

)

(

)

(

)

1 12  −1  18 6 −  6 Ds = GMR = r  e 4  ( 2 ) 2 3 ( 4 )  e 4  ( 2 )     49

12  −1  24 6 Ds = GMR = r  e 4  ( 2 ) 2 3 ( 4 ) = 2.177 r   49

4.16

(

DSL = N b D11 D12  D1N b

) = ( D D  D ) Nb

1N b

1 Nb

 ( n -1)   D11 = DS D1n = 2 A sin   n = 2,3,  N b  Nb  1



( N −1)

DSL = DS ( A ) b



N b Nb which is the desired result.

Two-conductor bundle, Nb = 2

d A= 2

 d  DSL =  DS   ( 2 )   2 

1/ 2

= DS d Eq (4.6.19) Three-conductor bundle, Nb = 3 A=

  d 2  DSL =  DS   3   3  

d 3

1/ 3

= √𝐷𝑆 𝑑2Eq (4.6.20)

Four-conductor bundle, Nb = 4

  d 3  A= DSL =  DS   4 2   2   d

1/ 4

 4  = 4 DS d 3 4   2 2  = 1.0905 4 DS d 3 Eq (4.6.21) 3

1  − 1   3 − 4.17 (a) GMR = 9  e 4 r  ( 2r )( 2r )  = r 4e 4    = 1.4605 r

 −1  6 4 2 GMR = 16  e 4  ( 2 ) ( 4 ) ( 6 ) ( r ) = 2.1554r  

  −  (c) GMR = r  e  ( 2 ) ( 4 )   

    −  32    e  ( 2 )     

( 20 ) ( 8 )(

)

Distances for each corner conductor

( 8) ( 2

  20 ( 4 )   

)

Distances for each outside non-corner conductor

  −    e  ( 2 )    1

  8   

( )

Distances for the center conductor

 −1  24 GMR = r 81  e 4  ( 2 )  

( 8 ) ( 20 ) ( 4 ) ( 32 ) 16

GMR = 2.6374 r

4.18 𝑅 = 0.1128

Ω 3 ; 𝑅𝑒𝑞 = √7.5 ⋅ 7.5 ⋅ 15 = 9.45 m; mi 1m

From Table A.4, 𝐷𝑠 = (0.043 ft) 3.28ft = 0.0123 m; 𝐷𝑒𝑞

9.45

𝐿𝑖 = 2 × 10−7 ln ( 𝐷 ) = 2 × 10−7 ln (0.0123) = 1.3239 μH/m; 𝑠

𝑍 = 𝑅 + 𝑗𝜔𝐿 = 0.1128 ( = 0.0701 + 𝑗0.499

1 1.609

km mi

) + 𝑗(2)(𝜋)(60)(1.3239 × 10−6 )(1000 m/km)

Ω km 𝐷𝑒𝑞

10.39

4.19 (a) 𝑅𝑒𝑞 = √8.25 ⋅ 8.25 ⋅ 16.5 = 10.39 m; 𝐿𝑖 = 2 × 10−7 ln ( 𝐷 ) = 2 × 10−7 ln (0.0123) = 𝑠

1.3478 μH/m; 𝑍 = 𝑅 + 𝑗𝜔𝐿 = 0.1128 (

1 1.609

km mi

) + 𝑗(2)(𝜋)(60)(1.3478 ×

10−6 )(1000 m/km) = 0.0701 + 𝑗0.508 km 𝐷

8.50 )= 0.0123

(b) 𝑅𝑒𝑞 = √6.75 ⋅ 6.75 ⋅ 13.5 = 8.50 m; 𝐿𝑖 = 2 × 10−7 ln ( 𝑒𝑞 ) = 2 × 10−7 ln ( 𝐷𝑠

1.3076 μH/m; 𝑍 = 𝑅 + 𝑗𝜔𝐿 = 0.1128 (

1 1.609

km mi

) + 𝑗(2)(𝜋)(60)(1.3076 ×

10−6 )(1000 m/km) = 0.0701 + 𝑗0.493 km

Ω 1m ; From Table A.4, 𝐷𝑠 = (0.0391 ft) = 0.0119 m; 𝐿𝑖 = 2 × mi 3.28ft

𝐷

9.45

10−7 ln ( 𝐷𝑒𝑞 ) = 2 × 10−7 ln (0.0119) = 1.3354 μH/m; 𝑍 = 𝑅 + 𝑗𝜔𝐿 = 0.1185 ( 𝑠

1 1.609

km mi

𝑗(2)(𝜋)(60)(1.3354 × 10−6 )(1000 m/km) = 0.0736 + 𝑗0.503 km (d) 𝑅 = 0.1035

Ω 1m ; From Table A.4, 𝐷𝑠 = (0.0420 ft) = 0.00128 m; 𝐿𝑖 = mi 3.28ft 𝐷

9.45

2 × 10−7 ln ( 𝐷𝑒𝑞 ) = 2 × 10−7 ln (0.00128) = 1.3208 μH/m; 𝑍 = 𝑅 + 𝑗𝜔𝐿 = 𝑠

0.1035 (

1 1.609

km mi

) + 𝑗(2)(𝜋)(60)(1.3208 × 10−6 )(1000 m/km) = 0.0643 + 𝑗0.498 km

4.20 𝐷𝑒𝑞 = √12 × 12 × 24 = 15.12 m 1m

From table A.4, 𝐷𝑠 = (0.0435 ft) (3.28ft) = 0.0133m DSL = [Dsd2](1/3) = [(0.0133)(0.5)2] (1/3) = 0.1493m 15.12

𝑋1 = 𝜔𝐿1 = 2𝜋(60)2 × 10−7 ln (0.1493) = 0.348 Ω/km 4.21 (a)

From Table A.4:  1  DS = ( 0.0479 )   = 0.0146 m  3.28  DSL = 3 ( 0.0146 )( 0.5 ) = 0.154 m 2

Results ACSR Conductor

Aluminum Cross Section

kcmil

km

Canary

900

0.352

Finch

1113

0.348

Martin

1351

0.346

% change 0.9% 0.6%

4.22 Application of Eq. (4.6.6) yields the geometric mean distance that separates the two bundles:

DAB = 9 ( 6.1) ( 6.2 ) ( 6.3) 6 ( 6.05)( 6.15)( 6.25) = 6.15m 2

The geometric mean radius of the equilateral arrangement of line A is calculated using Eq. (4.6.7):

RA = 9 ( 0.015576 ) ( 0.1) = 0.0538m 3

In which the first term beneath the radical is obtained from r  = 0.7788 r = 0.7788 ( 0.02 ) = 0.015576 m

The geometric mean radius of the line B is calculated below as per its configuration:

RB = 9 ( 0.015576 ) ( 0.1) ( 0.2 ) = 0.0628m 3

The actual configuration can now be replaced by the two equivalent hollow conductors each with its own geometric mean radius and separated by the geometric mean distance as shown below:

4.23 (a) The geometric mean radius of each phase is calculated as

R = 4 ( r  ) ( 0.3) where r  = 0.7788  0.0074 2

= 0.0416m The geometric mean distance between the conductors of phases A and B is given by

DAB = 4 62 ( 6.3)( 5.7 ) = 5.996 − 6m Similarly, DBC = 4 62 ( 6.3)( 5.7 ) = 5.996 − 6m and

DCA = 4 122 (12.3)(11.7 ) = 11.998 −12m

The GMD between phases is given by the cube root of the product of the three-phase spacings. Deq = 3 6  6  12 = 7.56m

The inductance per phase is found as L = 0.2 ln

7.56 = 1.041mH/km 0.0416

L = 1.609  1.041 = 1.674 mH/mi

(b) The line reactance for each phase then becomes X = 2 f L = 2 ( 60 )1.674  10 −3 = 0.631 / mi per phase

4.24 From the ACSR Table A.4 of the text, conductor GMR = 0.0244 ft. Conductor diameter = 0.721 in.; since

( 40 ) + (16 ) = 43.08, 2

GMD between phases = ( 43.08 )( 80 )( 43.08 ) 

1/ 3

(a)  X = k log

= 52.95 in.

D  52.95 /12  = 0.2794 log   = 0.6307  /mi  r  0.0244 

 52.95 /12  −7 (b) L = 2  10 −7 ln   = 10.395  10 H/m  0.0244 

X =  L = 2 ( 60 )10.395  10 −7  / m = 2 ( 60 )10.395  10 −7 (1609.34 )

 mi

= 0.6307 /mi  4.25 Resistance per phase =

0.12 = 0.03 / mi 4

GMD = ( 41.76 )( 80 )( 41.76 ) 

1/ 3

, using

402 + 122 = 41.76.

= 51.78ft. 3 GMR for the bundle :1.091 ( 0.0403 )(1.667 )   

1/ 4

by Eq. (4.6.21)

[Note: from Table A.4, conductor diameter. = 1.196 in.; r =

1.196 1  = 0.0498ft.] and conductor 2 12

GMR = 0.0403 ft. GMR for the 4-conductor bundle = 0.7171 ft  51.87   X = 0.2794 log   = 0.5195  / mi   0.7171 

Rated current carrying capacity for each conductor in the bundle, as per Table A.4, is 1010 A; since it is a 4-conductor bundle, rated current carrying capacity of the overhead line is

  GMD 20   −7 L = 2  10 ln = 2  10 ln   1/8 N -1 1/ N 7 [ Nr( A) ]  8 (1.7803  10−2 ) ( 0.5974 )      −7

1/ 6

= 2.3ft

 54.46  −6  L = 2  10 −7 ln   = 0.633  10 H/m   2.3 

(b) Inductance of one circuit is calculated below: Deq = [30  30  60]1/ 3 = 37.8ft; r  = 0.0588ft  37.8  −6  L = 2  10 −7 ln   = 1.293  10 H/m  0.0588 

Inductance of the double circuit =

1.293  10 −6 = 0.646  10 −6 H/m  2

 0.633 − 0.646  Error percent =    100 = −2.05%  0.633  

4.28 With N = 3, S = 21 , A =

S 21/12 = = 1.0104 ft 2sin 60 2  0.866

Conductor GMR = 0.0485ft 2 Bundle GMR = 3 ( 0.0485 )(1.0104 )   

Then rA = ( GMRb )( DAA )  rB = ( GMRb  DBB )

1/ 2

rC = ( GMRb  DCC  )

1/ 2

1/ 3

= 0.5296 ft

(

1/ 2

= 0.5296  322 + 362

= ( 0.5296  96 )

1/ 2

1/ 3

1/ 2

= 7.13ft

= 0.5296  322 + 362   

Overall Phase GMR = ( rArBrC )

) = 5.05ft

1/ 2

= 5.05ft

= 5.67ft

DABeq =  32 2 + 36 2  64 2 + 36 2  ( 64 )( 32 )    DBCeq = ( 32 ) 64 2 + 36 2  64  32 2 + 36 2    DACeq = ( 36 )( 32 )( 36 )( 32 ) 

1/ 4

= 51.88ft

= 33.94 ft

 GMD =  51.88  51.88  33.94 

1/ 3

= 45.04 ft

 45.04  Then X L = 0.2794 log   = 0.2515  /mi/phase   5.67 

4.29

rA = 0.5296 ( 32 )  rB = 0.5296 ( 32 ) 

1/ 2

= 4.117ft

1/ 2

= 4.117ft

rC = 0.5296 ( 32 ) 

1/ 2

= 4.117ft

Then GMR phase = 4.117ft





1/ 4

2 2 DABeq =  ( 36 ) + ( 32 ) 36  642 + 362   

DBCeq = ( 64 )( 96 )( 32 )( 64 ) 

1/ 4



DACeq =  ( 32 ) + ( 36 )  2

= 49.76ft

= 59.56ft

 36 + 64  (36) = 49.76ft 1/ 4

Then GMD = ( 49.76  59.56  49.76 )

1/ 3

= 52.83ft

 52.83  X L = 0.2794 log   = 0.3097  /mi/phase   4.117 

4.30

rA = 0.5296 ( 96 )  rB = 0.5296 ( 32 ) 

1/ 2

= 7.13ft

1/ 2

= 4.117ft = rC

From which GMRphase = ( 7.13  4.117  4.117 )

1/ 3

DABeq = ( 32  64  32  64 )

1/ 4



= 4.95ft

= 45.25ft



2 2 DBCeq = ( 36 ) ( 32 ) + ( 36 ) ( 36 )   

DACeq = ( 322 + 362 ) (642 + 362 ) 

1/ 4

1/4

= 41.64 ft

= 59.47ft

Then GMD = ( 45.25  41.64  59.47 )

1/ 3

= 48.21ft

 48.21  X L = 0.2794 log   = 0.2762  /mi/ph.   4.95 

4.31 Flux linkage between conductors 1 & 2 due to current, Ia, is

12( I ) = 0.2 I a ln a

Da 2 mWb/km Da1

Db1 = Db 2 , 12 due to I b is zero.

12( I ) = 0.2 I c ln c

Dc 2 mWb/km Dc1

12 = 0.2 I a ln

Da 2 D + 0.2 I c ln c 2 mWb/km Da1 Dc1

For positive sequence, with I a as reference, I c = I a  − 240

with I a as reference, instantaneous flux linkage is

4.32

Cn =

2 ( 8.854  10 −12 ) 2 0 F = = 1.3246  10 −11 to neutral m D  0.5  Ln   Ln     0.015 / 2 

Yn = jCn = j ( 2 60 ) (1.3246  10 −11 ) Yn = j 4.994  10 −6

4.33 (a) Cn =

S m  1000 m km

S to neutral km

2 ( 8.854  10−12 ) = 1.385  10−11 F/ m to neutral  0.5  ln    0.018 / 2 

Yn = j 2 ( 60 )1.385  10 −11 (1000 ) = j 5.221  10 −6 S/km to neutral

(b) Cn =

2 ( 8.854  10 −12 )  0.5  ln    0.012 / 2 

= 1.258  10 −11 F/m to neutral

Yn = j 2 ( 60 )1.258  10 −11 (1000 ) = j 4.742  10 −6 S/km to neutral

Both the capacitance and admittance-to-neutral increase 4.5% (decrease 5.1%) as the conductor diameter increases 20% (decreases 20%).

4.34

4.35 (a) C1 =

S km

2 ( 8.854  10 −12 )  4.8  ln    0.25 /12 

= 1.023  10 −11 F/m

Y1 = j 2 ( 60 )1.023  10 −11 (1000 ) = j3.857  10 −6 S/km

(b) C1 =

2 ( 8.854  10 −12 )  3.2  ln    0.25 /12 

= 1.105  10 −11 F/m

Y1 = j 2 ( 60 )1.105  10 −11 (1000 ) = j 4.167  10 −6 S/km

= 0.0471 The GMD is given by Deq = 3 6  6  12 = 7.56m Hence the line-to-neutral capacitance is given by Can =

2 F/m ln( Deq / Dsc )

55.63 = 10.95nF/km ln ( 7.56 / 0.0471)

( with  =  0 ) or 1.609  10.95 = 17.62 n F/mi (b) The capacitive reactance at 60 HZ is calculated as XC =

Deq 1 = 29.63  103 ln  − mi 2 ( 60 ) Can Dsc

= 29.63  10 3 ln

7.56 = 150,500  − mi 0.0471

4.37 (a) Eq (4.9.15): capacitance to neutral = XC =

2 F/m ln( D / r )

1 ln( D / r ) =   m to neutral 2 fC (2 f )(2 )

with f = 60HZ,  = 8.854  10 −12 F/m . or XC = k  log ( D / r ) ,

where k  =

1 = k  log D + k  log   , r

4.1  106 ,   mile to neutral  f

where k  = 0.06833  106

at f = 60 HZ .

(b) X d = k  log D = 0.06833  106 log (10 ) = 68.33  103 1 1 X a = k  log   = 0.06833  106 log = 80.32  103 0.06677 r

 Xc = X d + X a = 148.65  103   mi to neutral  When spacing is doubled, X d = 0.06833  106 log ( 20 ) = 88.9  103

Then XC = 169.12  103   mi to neutral  4.38

 52.95 /12  C = 0.0389 / log  = 0.018 F/mi/ph .  0.721/ (12  2 )    XC =

1 = 147.366  103   mi  2 ( 60 ) 0.018  10 −6 3

4.39 𝐷𝑒𝑞 = √7.5 ⋅ 7.5 ⋅ 15 = 9.45 m; From Table A.4, 𝑟 = 𝐶1 =

2𝜋𝜖0 𝐷 ln( ) 𝑟

1.196 0.0254m in 1in = 0.01519m; 2

2𝜋 8.854×10−12 ln(

9.45 ) 0.01519

= 8.648 × 10−12 F/m; S

𝑌 = 𝑗𝜔𝐶 = 𝑗2𝜋(60)8.648 = 𝑗3.260 × 10−6 km

4.40 (a) 𝐷𝑒𝑞 = √8.25 ⋅ 8.25 ⋅ 16.5 = 8.50 m; 𝐶1 =

2𝜋𝜖0 𝐷 ln( ) 𝑟

2𝜋 8.854×10−12 ln(

8.50 ) 0.01519

= 8.522 × 10−12 F/m

𝑌 = 𝑗𝜔𝐶 = 𝑗2𝜋(60)8.648 = 𝑗3.213 × 10−6

S km

(b) 𝐷𝑒𝑞 = √6.75 ⋅ 6.75 ⋅ 13.5 = 10.39 m; 𝐶1 =

2𝜋𝜖0 𝐷 ln( ) 𝑟

2𝜋 8.854×10−12 ln(

10.39 ) 0.01519

= 8.792 × 10−12 F/m;

𝑌 = 𝑗𝜔𝐶 = 𝑗2𝜋(60)8.648 = 𝑗3.315 × 10−6 4.41

S km

Deq = 3 10  10  20 = 12.6m

From Table A.4, r =

1.293  0.0254 m  in   = 0.01642 m 2  1in 

DSC = 3 rd 2 = 3 0.01642(0.5)2 = 0.16 m C1 =

2 ( 8.854  10 −12 ) 2 0 = = 1.275  10 −11 F/m Deq  12.6  ln  ln  DSc  0.16 

Y1 = jC1 = j 2 ( 60 )1.275  10 −11 (1000 ) = j 4.807  10 −6 S/km Q1 = VLL 2Y1 = ( 500 ) 4.807  10 −6 = 1.2 MVAR/ km 2

4.42 (a) From Table A.4, r =

1.424 ( 0.0254 ) = 0.0181m 2

DSc = 3 0.0181( 0.5 ) = 0.1654 m 2

C1 =

2 ( 8.854  10 −12 )

= 1.284  10 −11 F/m  12.6  ln    0.1654  Y1 = j 2 ( 60 ) (1.284  10 −11 ) (1000 ) = j 4.842  10 −6 S/km

Q1 = ( 500 ) 4.842  10 −6 = 1.21MVAR/ km 2

(b) r =

1.162 2 ( 0.0254 ) = 0.01476 m; DSc = 3 0.01476 ( 0.5 ) = 0.1546 m 2

C1 =

2 ( 8.854  10 −12 )

= 1.265  10 −11 F/m  12.6  ln    0.1546  Y1 = j 2 ( 60 )1.265  10 −11 (1000 ) = 4.77  10 −6 S/km

Q1 = ( 500 ) 4.77  10 −6 = 1.192 MVAR/ km 2

C1,Y1, and Q1 increase 0.8% (decrease 0.7%) For the larger, 1351 kcmil conductors (smaller, 700 kcmil conductors). 4.43 (a) For Drake, Table A.4 lists the outside diameter as 1.108 in  r=

1.108 = 0.0462 ft 2  12

Deq = 3 20  20  38 = 24.8ft

Can =

2  8.85  10 −12 = 8.8466  10 −12 F/m ln ( 24.8 / 0.0462 )

XC =

1012 = 0.1864  106   mi 2 (60)8.8466  1609

(b) For a length of 175 mi 0.1864  106 = 1065  to neutral 175 220  103 1 0.22 I chg = = = 0.681A/mi XC 3 3  0.1864

Capacitive reactance =

0.681175 = 119A for the line

Total three-phase reactive power supplied by the capacitance is given by

3  220  119  10 −3 = 43.5MVAR 4.44

 51.87  C = 0.0389 log   = 0.0212  F /mi/ph  0.7561 

 Note: Equivalent radius of a 4-cond. bundle is given by    3 1/ 4 3 1/ 4  1.091( 0.0498 d ) = 1.091( 0.0498  1.667 ) = 0.7561ft 

XC =

1 = 125.122  103   mi  −6 2 ( 60 ) 0.0212  10

4.45

From Example 4.8, 2 0

C Xn =

D  HXY  Ln   − Ln     HXX 

C Xn = 1.3247  10 −11

2 ( 8.854  10 −12 )  0.5   20.006  Ln  − Ln     0.0075   20 

F which is 0.01% larger than in Problem 4.32 m

4.46 (a) Deq = 3 12  12  24 = 15.12m

r = 0.0328 / 2 = 0.0164 m XC =

Where

1 2 fCan

2  8.85  10 −12 ln (15.12 / 0.0164 )

Can =

 XC =

2.86 15.12  10 9 ln = 3.254  108   m 60 0.016 

For 125 km, XC =

3.254  108 = 2603  125  1000

(b)

H1 = H2 = H3 = 40m

H12 = H23 = 402 + 122 = 41.761m H31 = 402 + 242 = 46.648m

Deq = 15.12 m and r = 0.0164 m

 XC =

 Deq 1 H12 H 23 H31  2.86  10 9  ln − ln m 60 3 H1 H 2 H3   r

 15.12 1 41.761  41.761  46.648  = 4.77  107  ln − ln  40  40  40  0.0164 3  8 = 3.218 10   m For 125 km, XC =

3.218  108 = 2574  125  103

4.47 D = 10 ft; r = 0.06677 ft; Hxx = 160 ft.; H xy = 1602 + 102

= 160.3ft (See Fig. 4.24 of text) Line-to-line capacitance C xy =

 H D ln − ln xy r H xx

F/ m

(See Ex. 4.8 of the text) C xy =

 ( 8.854  10 −12 )  10   160.3  ln   − ln  160  0.06677    

Neglecting Earth effect, C xy =

= 5.555  10 −12 F/ m

 ( 8.854  10 −12 )  10  ln    0.06677 

= 5.553  10 −12 F/ m Error - percentage =

5.555 − 5.553  100 = 0.036% 5.555

When the phase separation is doubled, D = 20 ft

H xy = 1602 + 202 = 161.245 With effect of Earth,

C xy =

 ( 8.854  10 −12 )  20   161.245  ln  − ln     0.06677   160 

= 4.885  10−12 F/m

Neglecting Earth effect, C xy =

 ( 8.854  10 −12 )  20  ln    0.06677 

= 4.878  10−12 F/ m

Error percentage =

4.885 − 4.878  100 = 0.143% 4.885

4.48 (a) H1 = H2 = H3 = 2  50 = 100 ft H12 = H 23 = 252 + 100 2 = 103.08ft H13 = 50 2 + 100 2 = 111.8ft Deq = 3 ( 25 )( 25 )( 50 ) = 31.5ft r= Can =

1.065 = 0.0444 ft 2  12 2 ( 8.854  10 −12 )  31.5   106  ln  − ln     0.0444   100 

= 9.7695  10 −12 F/ m 

 Note : H m = (103.08  103.08  111.8)1/ 3 = 106ft    H s = 100ft   (b) Neglecting the effect of ground, Can =

2 ( 8.854  10 −12 )  31.5  ln    0.0444 

= 8.4746  10 −12 F/ m

Effect of ground gives a higher value. Error percent =

4.49

9.7695 − 8.4746  100 = 13.25% 9.7695

GMD = ( 60  60  120 )

1/ 3

= 75.6ft

1.16  = 0.0483ft; N = 4; S = 2 A sin 2  12 N A=

18 = 1.0608ft ( 2sin 45 )12

N −1 GMR = rN ( A )   

 Can =

1/ N

3 1/ 4

= 0.0483  4  (1.0608 )    = 0.693ft

2 ( 8.854  10 −12 )  75.6  ln    0.693 

= 11.856  10 −12 F/m 

Next X d = 0.0683log ( 75.6 ) = 0.1283  1  X a = 0.0683log   = 0.0109  0.693  X c = X a + X d = 0.1392 M   mi to neutral = 0.1392  106   mi to neutral 

4.50 From Problem 4.45 1 1 F C xy = C xn = (1.3247  10−11 ) = 6.6235  10−12 2 2 m with Vxy = 20 kV Qx = C xy Vxy = ( 6.6235  10−12 )( 20  103 ) = 1.3247  10−7

C m

From Eq (4.12.1) The conductor surface electric field strength is:

ET =

1.3247  10 −7 ( 2 ) (8.854  10 −12 ) ( 0.0075 )

= 3.1750  10 5 = 3.175

V  kV  m    m  1000 V   100 cm 

kVrms cm

Ek =

4.51 (a) From Problem 4.30, 2 ( 8.854  10 −12 )

C xn =

 0.5   20.006  Ln  − Ln     0.009375   20  1 F C xy = C xn = 6.995  10 −12 2 m

= 1.3991  10 −11

F m

Qx = C xy Vxy = ( 6.995  10 −12 )( 20  103 ) = 1.399  10 −7 E =

c m

1.399  10 −7  1  1    −12 ( 2 ) (8,854  10 ) ( 0.009375 )  1000   100 

= 2.682

kVrms cm

 ( 2 )(10 ) ( 2 )(10 )  1.399  10 −7  − 2 ( 2 ) (8.854  10 −12 )  (10 ) (10 )2 + ( 0.5 )2  V  kV  kV = 1.254   = 0.001254  m  1000 V  m

Ek =

(b) C xn =

2 ( 8.854  10 −12 )  0.5   20.06  Ln  − Ln     .005625   20 

= 1.2398  10 −11

F m

1 F C xy = C xn = 6.199  10 −12 2 m Qx = C xy Vxy = ( 6.199  10 −12 )( 20  103 ) = 1.2398  10 −7 E =

c m

1.2398  10 −7  1  1    −12 ( 2 ) (8.854  10 ) (.005625)  1000   100 

E = 3.962

kVrms cm

Now, calculate 1 and 2 at the maximum generation conditions: Point 3 in Figure, now that 1  2 , unload unit 1 first until 1 is brought down to 1 = 420 which occurs at PG1 =

420 − 200 = 1.10 (Point 4 in Figure) 200

Notice that, for 0.72  PL  3.1 , it is possible to maintain equal  ’s. Equations are given by

1 = 2 ; 200 PG1 + 200 = 80 PG 2 + 260; and PG1 + PG 2 = PL These linear relationships are depicted in the Figure below: For PL = 282 MW = 2.82 pu, PG 2 = 2.82 − PG1 ; PG1 = 0.4 PG 2 + 0.3 = 1.128 − 0.4 PG1 + 0.3 1.4 PG1 = 1.428 or PG1 = 1.02 = 102 MW PG 2 = 2.82 − 1.02 = 1.8 = 180 MW

Results are tabulated in the table given below:

Table of Results Point

PG1

PG 2

1

2

0.25

0.30

0.55

250

284

0.42

0.30

0.72

284

1.50

2.00

3.50

500

420

1.10

2.00

3.10

420

7.7

(To solve the problem change the Min MW field for generator 2 to 0 MW). The minimum value in the plot above occurs when the generation at bus 2 is equal to 180MW. This value corresponds to the value found in Example 7.20 for economic dispatch at generator 2 (181MW).

7.8

To achieve loss sensitivity values that are equal the generation at bus 2 should be about 159 MW and the generation at bus 4 should be about 215 MW. Minimum losses are 7.79 MW. The operating cost in Example 7.22 is lower than that found in this problem indicating that minimizing losses does not usually result in a minimum cost dispatch. The minimum loss one-line is given below. 20 MW 1.05 pu 1

20 MW

MVA

5 MW slack

103 MW

0.99 pu

25.8 MW 0.0000 24 MW AGC ON

104 MW

MVA

78 MW 29 Mvar

147 MW 39 Mvar

1.00 pu

7 MW

215.0 MW -0.0003 OFF AGC

MVA

5 MW 1.04 pu

MVA

38 MW A

7 MW 93 MW

2 39 MW 20 Mvar

MVA

25 MW

MVA

89 MW 5

159.0 MW -0.0001 OFF AGC

0.96 pu

127 MW 39 Mvar

Total Hourly Cost:

6437.97 $/h

Load Scalar: 1.00

Total Area Load:

392.0 MW

MW Losses:

Marginal Cost ($/MWh):

At Min Gen $/MWh

7.79 MW

7.9

To achieve loss sensitivity values that are equal the generation at bus 2 should be about 230 MW and the generation at bus 4 should be at its maximum limit of 300 MW. Minimum losses are 15.38 MW. The minimum loss one-line is given below. 29 MW

28 MW

1.05 pu 1

MVA

143 MW

0.98 pu

33.9 MW 0.0000 35 MW AGC ON

5 MW slack

145 MW

MVA

206 MW 55 Mvar

1.00 pu

11 MW

MVA

5 MW 1.04 pu

97%

300.0 MW -0.0055 OFF AGC

MVA

36 MW

110 MW 41 Mvar

53 MW A

11 MW 133 MW

2 55 MW 27 Mvar

MVA

125 MW 5

230.3 MW 0.0000 OFF AGC

0.92 pu

178 MW 55 Mvar

Total Hourly Cost:

9404.56 $/h

Load Scalar: 1.40

Total Area Load:

548.8 MW

MW Losses:

Marginal Cost ($/MWh):

At Min Gen $/MWh

15.38 MW

7.10 The line between buses 2 and 5 reaches 100% at a load scalar of about 1.68; above this value the flow on this line is constrained to 100%, causing the additional load into bus 5 to increase the flow on the line between buses 4 and 5. This line reaches its limit at a load scalar of 1.81. Since both lines into bus 5 are loaded at 100%, no additional load can be brought into the bus without causing at least one of these lines to be overloaded. The plot of the bus 5 MW marginal cost is given below.

7.11 The OPF solution for this case is given below, with an operating cost of 16,091.6 $/hr. The marginal cost at bus UIUC69 is 41.82 $/hr. Increasing the UIUC69 MW load from 61 to 62 MW increases the operating cost to $16,132.80, an increase of $41.2, verifying the marginal cost. Similarly at the DEMAR69 bus the marginal cost is 17.74 $/hr; increasing the load from 44 to 45 MW increases the operating cost by $17.8. Note, because of convergence tolerances the manually calculated changes may differ slightly from the marginal cost values. SLA CK345

A MVA A MVA

Total Cost: 16091.6 $/h 1.02 pu

1.02 pu A

MVA

SLA CK138

1.01 pu

MVA

RA Y138

1.03 pu

A MVA

33 MW 13 Mvar

15.9 Mvar

1.02 pu

TIM69

1.02 pu

PA I69

1.01 pu

GROSS69

20 MW 8 Mvar

1.00 pu A

PETE69

MVA

HA NNA H69 51 MW 15 Mvar

58 MW 40 Mvar

DEMA R69

44.0 MW

UIUC69

1.00 pu

0.99 pu

A MA NDA 69

12.6 Mvar

A MVA

HA LE69

MVA

60 MW 12 Mvar

BLT138 A

99% MVA

1.01 pu

MVA

LA UF69

49 MW 16 Mvar 1.00 pu

A MVA

23 MW 6 Mvar

22 MW 15 Mvar

BUCKY138

ROGER69

2 Mvar A MVA

1.02 pu

SA VOY69 A

38 MW 2 Mvar

JO138

MVA

A MVA

7.12

14 MW

14 MW 3 Mvar 1.02 pu

MVA

45 MW 0 Mvar

A MVA

10 MW 5 Mvar

MVA

1.01 pu

PA TTEN69

WEBER69

LA UF138 1.01 pu

MVA

1.00 pu

1.01 pu

A MVA

MVA

7.3 Mvar

1.00 pu

1.02 pu

MVA

15 MW 5 Mvar

15 MW 60 Mvar

1.02 pu

SHIMKO69 7.4 Mvar

BLT69

1.01 pu

MVA

36 MW 10 Mvar

MVA

LYNN138

14 MW 4 Mvar

MVA

1.00 pu

0.0 Mvar

13 Mvar

0 MW 0 Mvar A

MVA

BOB69 56 MW

0.99 pu

36 Mvar

33 MW 10 Mvar

MVA

1.02 pu

24 MW 45 Mvar

MVA

BOB138

61.0 MW

MVA

15 MW 3 Mvar 1.00 pu

0.99 pu

MVA

0.99 pu

A A A

0.996 pu

MVA

A A

MVA

12 Mvar

28.8 Mvar

14.3 Mvar

HOMER69

WOLEN69

4.8 Mvar

MVA

1.00 pu

1.01 pu

21 MW 7 Mvar

MVA

HISKY69

MVA

12 MW 5 Mvar

13 Mvar

MVA

FERNA 69

MORO138

A MVA

MVA

37 MW

17 MW 3 Mvar

MVA

1.01 pu

MVA

1.02 pu

RA Y69

23 MW 7 Mvar

MVA

18 MW 5 Mvar

MVA

TIM138

1.00 pu

398 MW 72 Mvar

RA Y345 slack

TIM345

JO345

SA VOY138

MVA

150 MW 1 Mvar

MVA

The economic dispatch solution would be 𝑃𝐺1 = 400 MW, 𝑃𝐺2 = 370 MW, 𝑃𝐺3 =

50 MW. With this solution, the angles would be 𝛿1 = 0, 𝛿2 = 0.8°, 𝛿3 = −8.0°. The line flows would be 𝐹12 = −20 MW, 𝐹13 = 280 MW, 𝑎𝑛𝑑 𝐹23 = 85 MW This is not a valid solution because the line flow limits F13 of 120 MW is violated. 7.13

In example 7.6, the solution is 𝑃𝐺1 = 394 MW, 𝑃𝐺2 = 150 MW, 𝑃𝐺3 = 276 MW. The

resulting LMPs are 𝜆1 = 18 $/MWh, 𝜆2 = 9.72 $/MWh, 𝜆3 = 31 $/MWh. Why is there this variation in LMPs?

Chapter 8 Symmetrical Faults 8.1

(a)

Z = R + j L = 0.4 + j ( 2. .60 ) 2  10 −3 = 0.4 + j 0.754 = 0.853562.05  Z = 0.8535  and  = 62.05 I ac =

V 277 = = 324.5A Z 0.8535

(b) Irms (0) = I ac k (0) = 324.5 3 = 562.1 A (c) Using Eq. (8.1.12), with

X 0.754 = = 1.885 R 0.4

k ( = 5cycles) = 1 − 2 e−4 (5) 1.885 1.0 I rms ( = 5cycles) = I ac k (5cycles) = 324.5A (d) From Eq. (8.1.1)  300   300  V (0) = 2 V sin  = 300;  = sin −1   = sin −1   = 49.98  2V   2 277 

From Eq. (8.1.4) 2V sin ( −  ) e − t / T Z 2 (277) =− sin ( 49.98 − 62.05 ) e − t / T 0.8535 = 95.98e − t / T

idc (t ) = −

where T =

L 2  10−3 = = 5  10−3 s R 0.4

idc (t ) = 95.98 e−t (510 ) A −3

8.2

(a)

Z = 1 + j 2 = 2.23663.43

I ac = V Z = 4000 2.236 = 1789 A (b) I rms (0) = 1789 3 = 3098 A (c) with X R = 2, k (5) = 1 + 2e−4 (5)/2

1.0

I rms (5) = k (5)I ac = (1.0 )(1789) = 1789 A

(

)

(d)  = sin −1 300 4000 2 = 3.04;

X 2 = = 5.305  10−3 s  R (2 .60)(1)

idc (t ) = −

2(4000) sin ( 3.04 − 63.43 ) e − t / T 2.236

= 2200e − t ( 5.30510 ) A −3

8.3

Z = 0.125 + j 2 (60)0.01 = 0.125 + j3.77 = 3.772 88.1 1 40 = A 2 3.772 2 T = L/R = 0.08Sec.

I ac rms =

151

The response is then given by i( ) = 40sin ( t +  − 88.1 ) − 40e − t / 0.08 sin ( − 88.1 )

(a) No dc offset, if switch is closed when  = 88.1 . (b) Maximum dc offset, when  = 88.1 − 90 = −1.9 Current waveforms with no dc offset (a), and with Max. dc offset (b) are shown below:

8.4

(a) For X / R = 0 , i(t ) = 2 I rms sin (t −  Z )  (b) The wave form represents a sine wave, with  no dc offset. For X / R = , i (t ) = 2 I rms sin (t −  Z ) + sin  Z   The dc offset is maximum for (X/R) equal to infinity. 

(c) For ( X/R ) = 0, Asymmetrical Factor = 1.4141   For ( X/R ) = , Asymmetrical Factor = 2.8283   The time of peak, t p , For X / R = 0, is 4.2 ms.  and For X / R = ,is 8.3 ms.  Note: The multiplying factor that is used to determine the maximum peak instantaneous fault current can be calculated by taking the derivative of the bracketed term of the given equation for i(t) in PR.8.4 with respect to time and equating to zero, and then solving for the time of maximum peak tp; substituting tp into the equation, the appropriate multiplying factor can be determined.

8.5

VL − N =

VL − L 3

13.2  103 3

= 7621V

RMS symmetrical fault current, I rms =

7621

( 0.52 + 1.52 ) 2 1

= 4820 A  X/R Ratio of the system is

1.5 = 3 , for which the asymmetrical factor is 1.9495. 0.5

 The maximum peak instantaneous value of fault current is I max peak = 1.9495 ( 4820 )

= 9397 A    able to withstand a peak current of approximately    9400 A.  All substation electrical equipment must be

8.6

(a) Neglecting the transformer winding resistance, I  =

Eq 1.0 1 = = = 3.704 pu X d + XTR 0.17 + 0.1 0.27

The base current on the HV side of the transformer is: Srated

I base H =

3 Vrated H

1000 3 ( 345 )

= 1.673 kA

I  = 3.704  1.673 = 6.198 kA

(b) Using Eq (8.2.1) at t = 3 cycles = 0.05 S with the transformer reactance included,  1 1  −0.05 0.05  1 1  −0.05 1.0 1  I ac ( 0.05) = 1.0  − e + − +  e 1.6   0.4 1.6   0.27 0.4  = 2.851 Pu

From Eq (8.2.5)

idc (t ) = 2 ( 3.704 ) e−t / 0.1 = 5.238e−t / 0.1 pu The rms asymmetrical current that the breaker interrupts is I rms ( 0.05S ) = I ac2 ( 0.05 ) + idc2 ( 0.05 )

( 2.851) + ( 5.238) e−2( 0.05) 0.1 = 4.269 pu = 4.269 (1.673) = 7.144 kA =

8.7

(a) Using Eq. (8.2.1) with the transformer reactance included, and with  = 0 for maximum dc offset,  1 1  − t / 0.05  1 1  − t /1.0 1    iac (t ) = 2 (1.0)  − + − + sin   t −  e e  1.6   2  0.4 1.6   0.27 0.4 

  = 2 1.204 e − t / 0.05 + 1.875 e − t /1.0 + 0.625sin   t −  pu 2  The generator base current is I base L =

srated

3 Vrated L − L

1000 3 ( 20 )

= 28.87 kA

   iac (t ) = 40.83 1.204 e − t / 0.05 + 1.875e − t /1.0 + 0.625  sin  t −  kA 2  where the effect of the transformer on the time constants has been neglected. (b) From Eq. (8.2.5) and the results of Problem 8.6, idc (t ) = 3 I e− t / TA = 2 ( 3.704 ) e − t / 0.1 = 5.238 e− t 0.1 pu = 151.2 e− t 0.1 kA

8.8

(a) iac (t ) = 10 (1 + e− t / 200 + 6 e−t /15 ) , t in ms and i in kA. The plot is show below:

(b)

I base =

300

= 12551A;

Z base =

0.0138 3 i(k ) = 0.797 (1 + e − t / 1 + 6e − t /  2 ) pu

(13.8)2 = 0.635  300

1 = 1.255pu t → 0.797 1 lim i(t ) = 8  0.797;  X d = = 0.157 pu t →0 8  0.797

lim i(k ) = 0.797;  X d =

X d = 1.255  0.635 = 0.797  X d = 0.157  0.635 = 0.996 

8.9

The prefault and postfault per-phase equivalent circuits are shown below:

Vf =

14.5 = 0.9670 pu , taken as reference 15

60  106

Base Current =

= 2309.5A 3  15  103 40  103 36.9 I MOTOR = = 199136.9 A 0.8  3  14.5 = 0.862136.9 pu = ( 0.69 + j 0.52 ) pu

For the generator, Vt = 0.967 + j 0.1( 0.69 + j 0.52 ) = ( 0.915 + j 0.069 ) pu Eg = 0.915 + j 0.069 + j 0.1( 0.69 + j 0.52 ) = ( 0.863 + j 0.138 ) pu I g =

0.863 + j 0.138 = ( 0.69 − j 4.315 ) pu j 0.2

= 2309.5 ( 0.69 − j 4.315 ) = (1593.6 − j 9965.5) A 

For the motor: Vt = V f = 0.967 0 Em = 0.967 − j 0.1( 0.69 + j 0.52 ) = (1.019 − j 0.069 ) pu I m =

1.019 − j 0.069 = ( −0.69 − j10.19 ) pu j 0.1

= 2309.5 ( −0.69 − j10.19 ) = −1593.6 − j 23533.8 A 

In the fault: I f = I g + I m = 0.69 − j 4.315 − 0.69 − j10.19

= − j14.505pu = − j14.505  2309.5 = − j 33,499 A  Note: The fault current is very high since the subtransient reactance of synchronous machines and the external line reactance are low.

8.10

The pre-fault load current in pu is IL =

S pu Vpu

 − cos(PF) =

1.0  − cos−1 ( 0.8 ) = 1.0  − 36.86pu 1.0

The initial generator voltage behind the subtransient reactance is Eg = V + j ( X + XTR ) I L = 1.0 0 + ( j 0.27 )(1.0 − 36.86 ) = 1.162 + j 0.216 = 1.182 10.53 pu

The subtransient fault current is

I  = Eg j ( X d + XTR ) = 1.182 10.53 j 0.27 = 4.378  − 79.47pu I  = 4.378 (1.673 )  − 79.47 = 7.326  − 79.47kA Alternatively, using superposition, I  = I1 + I 2 = I1 + I L = 3.704  − 90 + 1.0 − 36.86

[From Pr. 8.6 (a)] I  = 0.8 − j 4.304 = 4.378  − 79.47 pu = 7.326  − 79.47kA

8.11

The prefault load current in per unit is:

IL =

S 1.0  − cos−1 (P.F.) =  − cos−1 0.95 = 0.9524  − 18.195 per unit V 1.05

The internal machine voltages are: Eg = V + jX g I L = 1.050 + ( j 0.15 )( 0.9524 − 18.195 ) = 1.05 + 0.142971.81 = 1.0946 + j 0.1358 = 1.10307.072 per unit Em = V − j ( XT 1 + X Line + XT 2 + X m ) I L = 1.050 − ( j 0.505 )( 0.9524  − 18.195 ) = 1.05 + 0.48095 − 18.195 = 0.8998 − j 0.4569 = 1.0092 − 26.92 per unit

The short circuit currents are: I g = I m =

Eg jX g

1.10307.072 = 7.353  − 82.93 per unit j 0.15

Em 1.0092  − 26.92 = = 1.998 243.1 per unit j ( X1 + X Line + X 2 + X m ) j 0.505

I F = I g + I m = 7.353 − 82.93 + 1.998243.1 = − j 9.079 per unit

8.12

Per Unit Positive Sequence Reactance Diagram  1000  X g1 = ( 0.20 )   = 0.40 per unit  500  Xg 2 = ( 0.18 )(1000 / 750 ) = 0.24 Per unit

XT 2 = ( 0.10 )(1000 / 750 ) = 0.1333per unit

X j 3 = 0.17 per unit

XT 3 = 0.10 per unit

Vbase 4 ( 500 ) = = 250  Sbase 1000

XT 1 = ( 0.12 )(1000 / 500 ) = 0.24 per unit

Z base 4 =

X Line1.2 = X Line 2.3 = X Line 2.4 =

50 = 0.20 per unit 250

(a) XTh = ( 0.40 + 0.24 ) // 0.20 + ( 0.20 + 0.10 + 0.17 ) // ( 0.20 + 0.1333 + 0.24 )

= 0.64 //  0.20 + 0.47 // 0.5733 = 0.64 // 0.4583 = 0.2670 per unit 173 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

(b) VF =

I F =

525 = 1.050 per unit 500

I base 4 =

Sbase 3base 4

1000

( 3 ) (500)

= 1.155 kA

VF 1.050 = = − j 3.933per unit = ( − j 3.933)(1.155 ) = − j 4.541 kA ZTh j 0.2670

0.4583 = ( − j 3.933)( 0.4173) = − j1.641per unit = − j1.896 kA ( 0.4583 + 0.64 )

0.64   1.2 = I F  I Line  = ( − j 3.933)( 0.5827 ) = − j 2.292 per unit = − j 2.647 kA  0.4583 + 0.64 

8.13 (a) XTh = ( 0.20 + 0.24 + 0.40 ) // ( 0.20 + 0.10 + 0.17 ) // ( 0.20 + 0.1333 + 0.24 )

XTh = 0.84 // 0.47 // 0.5733 =

(b) I F =

1 = 0.1975per unit 1 1 1 + + 0.84 0.47 0.5733

VF 1.050 = = − j 5.3155per unit Z Th j 0.1975

I F = ( − j 5.3155 )(1.155 ) = − j 6.1379 kA

1.050 = − j1.25per unit = ( − j1.25 )(1.155 ) = − j1.443kA j 0.84 1.050  = I 32 = − j 2.234 per unit = ( − j 2.234 )(1.155) = − j 2.580 kA j 0.47 1.050  = I 42 = − j1.8315per unit = ( − j1.8315 )(1.155 ) = − j 2.115kA j 0.5733

8.14 (a) Zone 1 Vbase1 = 10 kV

Zone 2

Zone 3

Vbase 2 = 15 kV

Vbase3 = 138 kV

(138 ) = 190.44  = 2

Z base 3

100

 12   100  X g1 = ( 0.20 )     = 0.576 per unit  10   50  X g2 = 0.20 per unit  100  XT 1 = ( 0.10 )   = 0.20 per unit  50  XT 2 = 0.10 per unit X Line = 40 /190.44 = 0.21per unit (b) XTh = ( 0.20 ) // ( 0.576 + 0.20 + 0.21 + 0.21 + 0.10 ) = 0.20 //1.296 = 0.1733per unit VF = 1.0 per unit

I F =

(c)

I base 2 =

VF 1.00 = = − j 5.772 per unit ZTh j 0.1733 100

= 3.849 kA 15 3 I F = ( − j 5.772 )( 3.849 ) = − j 22.21kA

(d)

1.00 = − j 5.0 per unit = ( − j 5.0 )( 3.849 ) = − j19.245kA j 0.20 1.00 I T2 = = − j 0.7716 per unit = ( − j 0.7716 )( 3.849 ) = − j 2.970 kA j1.296 I g2 =

8.15 (a) XTh = ( 0.20 + 0.10 ) // ( 0.576 + 0.20 + 0.21 + 0.21)

= 0.30 //1.196 = 0.2398 per unit (b)

I F = I base3 =

1.00 = − j 4.1695per unit j 0.2398 100

= 0.4184 kA 138 3 I F = ( − j 4.1695 )( 0.4184 ) = − j1.744 kA

1.00 = − j 3.333per unit = ( − j 3.333)( 0.4184 ) = − j1.395 kA j 0.30

 = I 34

1.00 = − j 0.836 per unit = ( − j 0.836 )( 0.4184 ) = − j 0.350 kA j1.196

8.16 Choosing base MVA as 30 MVA and the base line voltage at the HV-side of the transformer to be 33kV, 30 30 30  0.15 = 0.225pu; X G 2 =  0.1 = 0.3pu; XTRANS =  0.05 = 0.05pu 20 10 30 30 Z LINE = ( 3 + j15 ) 2 = ( 0.0826 + j 0.4132 ) pu 33 X G1 =

The system with pu-values is shown below:

ZTOTAL = 0.0826 + j 0.5918 = 0.597582 pu

Then

IF = I base =

1.0 = 1.674 − 82 pu 0.597582 30  106 3 33  10

= 524.8 A

I F = 1.674  524.8 = 878.6 A

8.17 Choosing base values of 25 MVA and 13.8 kV on the generator side Generator reactance = 0.15 pu 2

Transformer reactance =

25  13.2    0.11 = 0.101 pu 25  13.8 

Base voltage at the transmission line is 13.8  Per-unit line reactance:

69 = 72.136 kV 13.2

( 72.136 )

= 0.312 2

25  13  X M = 0.15     = 0.222 pu 15  13.8 

The reactance diagram is shown below; Switch SW simulates the short circuit, and Eg and Em are the machine prefault internal voltages.

Choose VF to be equal to the voltage at the fault point prior to the occurrence of the fault; then

VF = Em = Eg ; prefault currents are neglected; I F 2 = 0; so VF may be open circuited as shown below:

Equivalent impedance between

Neglecting prefault currents,

terminals a & b is

Eg = Em = VF = 10 pu

j 0.15  j 0.736 = j 0.1246 j 0.15 + j 0.736  I F = I F1 =

10 = − j8.025pu j 0.1246

8.18 (A) (a)

(b) I F 2 =

VF 1.00 = = − j8.333per unit Z 22 j 0.12

Using Eq (8.4.7):  Z   0.08  E1 = 1 − 12  VF = 1 − 1.00 = 0.33330 per unit Z 22   0.12    Z  E2 = 1 − 22  VF = 0 Z 22  

 Z   0.06  E3 =  1 − 23  VF =  1 − 1.00 = 0.500 per unit Z 22   0.12  

(B) (a)

(b)

I F2 = VF Z 22 = 1.00 j 0.8 = − j1.25 pu

Using Eq (8.4.7)  Z  0.1   E1 =  1 − 12  VF =  1. − (10 ) = 0.875 0 pu Z 22  0.8   

 Z  E2 =  1 − 22  VF = 0 Z 33    Z   0.5  E3 =  1 − 23  VF =  1 − 10 = 0.375 0 pu Z33   0.8  

8.19

6.5625  −5 YBUS = − j   0   0

−5 0 0  15 −5 −5  per unit −5 8.7037 0   −5 0 7.6786 

Using the personal computer subroutine

0.2671 0.1505 Z bus = j  0.0865  0.098

8.20

0.1505 0.0865 0.098  0.1975 0.1135 0.1286  per unit 0.1135 0.1801 0.0739   0.1286 0.0739 0.214 

−5 0 0 0  6.7361  −5 9.7619 −4.7619 0 0   Ybus = − j  0 −4.7619 9.5238 −4.7619 0  per unit   0 −4.7619 14.7619 −10   0  0 0 0 −10 15 

Using the personal computer subroutine 0.3542 0.2772  Z bus = j 0.1964  0.1155 0.077

0.2772 0.1964 0.1155 0.077  0.3735 0.2645 0.1556 0.1037  0.2645 0.3361 0.1977 0.1318  per unit  0.1556 0.1977 0.2398 0.1599  0.1037 0.1318 0.1599 0.1733 

8.21 (a) The admittance diagram is shown below:

(b) Y11 = − j 0.5 − j 5 − j 5 = − j10.5;Y22 = − j 0.5 − j 2.5 − j 5 = − j8.0 Y33 = − j 0.5 − j 5 − j10 − j 2.5 = − j18.0;Y44 = − j 5 − j10 − j 5 = − j 20.0 Y12 = Y21 = 0; Y13 = Y31 = j 5.0; Y14 = Y41 = j 5.0 Y23 = Y32 = j 2.5; Y24 = Y42 = j 5; Y34 = Y43 = j10.0

Hence the bus admittance matrix is given by

0 j 5.0 j 5.0   − j10.5  0 − j8.0 j 2.5 j 5.0   YBUS =  j 5.0 j 2.5 − j18.0 j10.0    j 5.0 j10.0 − j 20.0   j 5.0 −1 (c) The bus impedance matrix Z BUS = YBUS is given by

 j 0.724  j 0.620 Z BUS =   j 0.656   j 0.644 (1)

j 0.620

j 0.656

j 0.738

j 0.642

j 0.702

j 0.660

j 0.676

j 0.644  j 0.660  j 0.676   j 0.719

(2)

(3)

(4)

(1)  j1.5 − j 0.25 0 0   (2)  − j 0.25 j 0.775 − j 0.4 − j 0.125 8.22 (a) YBUS = (3)  0 − j 0.4 j1.85 − j 0.2    (4)  0 − j 0.125 − j 0.2 j 0.325  (1) (1)  j 0.71660 (2)  j 0.60992 (b) Z BUS = (3)  j 0.53340  (4)  j 0.58049

(2)

(3)

(4)

j 0.60992

j 0.53340

j 0.73190

j 0.64008

j 0.71660

j 0.69659

j 0.66951

j 0.58049  j 0.69659  j 0.66951   j 0.76310 

Note:

Z BUS may be formulated directly (instead of inverting YBUS ) by adding the branches in the

order of their labels; and numbered subscripts on Z BUS will indicate the intermediate steps of the solution. For details of this step-by-step method of formulating Z BUS , please refer to the 2nd edition of the text. 8.23 (a) I f =

1.0 1.0 = = − j 4.348 pu  Due to the fault Z 22 j 0.23

Note: Because load currents are neglected, the prefault voltage at each bus is 1.00 pu , the same as V f at bus 2. (b) Voltages during the fault are calculated below: j 0.2    1 − j 0.23    0.134  V1     0      =  0  pu  V2  =  j 0.15  V3   1 − 0.3478  j 0.23         0.3435 V4   1 − j 0.151   j 0.23 

V3 − V1 0.3478 − 0.1304 = = − j 0.8696 pu  Z 3−1 j 0.25

(d) Fault currents contributed to bus 2 by the adjacent unfaulted buses are calculated below: From Bus 1:

V1 0.1304 = = − j1.0432  Z 2 −1 j 0.125

From Bus 3:

V3 0.3478 = = − j1.3912  Z 2 −3 j 0.25

From Bus 4:

V4 0.3435 = = − j1.7175  Z2−4 j 0.2

Sum of these current contributions = −j4.1519 Which is approximately same as I f .

8.24 For a fault at bus 3, generator 5 supplies 0.789 per unit current, generator 6 supplies 1.154 pu, and generator 7 supplies 2.625 pu. The per unit fault-on voltages are bus 1 = 0.546, bus 2 = 0.388, bus 3 = 0, bus 4 = 0.619, bus 5 = 0.735, bus 6 = 0.773, and bus 7 = 0.263. 8.25 For a fault at bus 4, generator 5 supplies 0.795 per unit current, generator 6 supplies 2.813 pu, and generator 7 supplies 1.114 pu. The per unit fault-on voltages are bus 1 = 0.541, bus 2 = 0.382, bus 3 = 0.605, bus 4 = 0, bus 5 = 0.732, bus 6 = 0.375, and bus 7 = 0.716. 8.26 For a fault midway on the line between buses 2 and 3, generator 5 supplies 0.966 per unit current, generator 6 supplies 1.416 pu, and generator 7 supplies 2.100 pu. The per unit fault-on voltages are bus 1 = 0.432, bus 2 = 0.238, bus 3 = 0.210, bus 4 = 0.521, bus 5 = 0.663, bus 6 = 0.710, and bus 7 = 0.420. 8.27 Generator G2 (at bus 6) supplies the most fault current for a bus 6 fault, during which it supplies 4.375 per unit fault current. This value can be limited to 2.5 per unit by increasing the G3 positive sequence impedance from 0.24 to 1.05/2.5 = 0.42, which would require an addition of 0.18 per unit reactance. 8.28 For a three phase fault at POPULAR69, the generators supply the following per unit fault currents: REDUBUD69 = 0.000 (off-line), ELM345 G1 = 1.971, ELM345, G2=1.971, SLACK345 = 3.198, PEACH69 = 3.000, CEDAR69 = 0.000 (off-line), BIRCH69 = 0.952, PEAR138 = 2.871, PEAR69 = 2.871. During the fault a total of 5 of the 37 buses have voltages at or below 0.75 per unit (one bus has a voltage of 0.75324 so some students may round this to 0.75 and say 6 buses are at or BELOW 0.75). 8.29 For a three phase fault at bus REDBUD69 the generators supply the following per unit fault currents: REDBUD69 = 0.000 (off-line), ELM345 G1 = 1.939, ELM345 G2 = 1.939, SLACK345 = 3.010, PEACH69 = 2.907, CEDAR69 = 0.000 (off-line), BIRCH69 = 1.626, PEAR138 = 3.152, PEAR69 = 2.228. During the fault a total of 5 of the 37 buses have voltages at or below 0.75 per unit. 8.30 With the line connecting buses 4 and 5 open in the Example 7.5 case, the generator at bus 3 supplies 7.925 per unit fault current. 8.31 (a) The symmetrical interrupting capability is:  15.5  at 10 kV: ( 9.0 )   = 13.95 kA  10  Vmin =

Vmax 15.5 = = 6.20 kV k 2.50

at 5kV: I max = k  = ( 2.50 )( 9.0 ) = 22.5kA (b) The symmetrical interrupting capability at 13.8 kV is: 182 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

 15.5  9.0   = 10.11kA  13.8 

Since the interrupting capability of 10.11 kA is greater than the 10 kA symmetrical fault current and the (X/R) ratio is less than 15, the answer is yes. This breaker can be safely installed at the bus. 8.32

From Table 8.10, select the 345 kV (nominal voltage class) breaker with a 40 kA rated short circuit current. This breaker has a 3 kA rated continuous current.

8.33

The maximum symmetrical interrupting capability is k  rated short-circuit current 1.25  19,000 = 23,750 A which must not be exceeded. Rated maximum voltage K 72.5 = = 58kV 1.25

Lower limit of operating voltage =

Hence, in the operating voltage range 72.5–58 kV, the symmetrical interrupting current may exceed the rated short-circuit current of 19,000 A, but is limited to 23,750 A. For example, at 66 kV the interrupting current can be 72.5  19,000 = 20,871A 66

8.34

(a) For a base of 25 MVA, 13.8 kV in the generator circuit, the base for motors is 25 MVA, 6.9 kV. For each of the motors, X d = 0.2

25000 = 1.0 pu 5000

The reactance diagram is shown below:

For a fault at P, VF = 10 pu; ZTh = j 0.125pu

I f = 10 j 0.125 = − j8.0 pu The base current in the 6.9 kV circuit is

25000 3  6.9

= 2090 A

so, subtransient fault current = 8  2090 = 16,720A

(b) Contributions from the generator and three of the four motors come through breaker A. The generator contributes a current of − j8.0 

0.25 = − j 4.0 pu 0.50

Each motor contributes 25% of the remaining fault current, or − j1.0 pu amperes each. For breaker A I  = − j 4.0 + 3 ( − j1.0 ) = − j 7.0 pu or 7  2090 = 14,630 A

(c) To compute the current to be interrupted by breaker A, let us replace the sub transient reactance of j1.0 by the transient reactance, say j1.5, in the motor circuit. Then Z Th = j

0.375  0.25 = j 0.15pu 0.375 + 0.25

The generator contributes a current of 1.0 0.375  = − j 4.0 pu j 0.15 0.625

Each motor contributes a current of

1 1 0.25   = − j 0.67 pu 4 j 0.15 0.625

The symmetrical short-circuit current to be interrupted is

( 4.0 + 3  0.67 )  2090 = 12,560 A Supposing that all the breakers connected to the bus are rated on the basis of the current into a fault on the bus, the short-circuit current interrupting rating of the breakers connected to the 6.9 kV bus must be at least

4 + 4  0.67 = 6.67pu, or 6.67  2090 = 13,940 A . A 14.4-kV circuit breaker has a rated maximum voltage of 15.5kV and a k of 2.67. At 15.5kV its rated short-circuits interrupting current is 8900 A. This breaker is rated for a symmetrical short-circuit interrupting current of 2.67  8900 = 23,760A , at a voltage of 15.5 / 2.67 = 5.8 kV .

This current is the maximum that can be interrupted even though the breaker may be in a circuit of lower voltage. The short-circuit interrupting current rating at 6.9 kV is 15.5  8900 = 20,000 A 6.9

The required capability of 13,940 A is well below 80% of 20,000 A, and the breaker is suitable with respect to short-circuit current.

Chapter 9 Symmetrical Components 9.1

Using the identities shown in Table 9.1 (a)

a −1 −(1 − a) (−1) 3  − 30 3 = = =  90 2 2 2 1+ a − a (1 + a + a ) − 2a (−1) 2  240 2

(b)

( a 2 − a ) + j = −1 + j = ja + a 2 a( j + a) =

2135  1 3 (1120 )  j − + j  2 2  

215 215 = = 0.7321  270  3  1.932  105 1 − + j  + 1  2  2 

(c)

(1 + a ) (1 + a2 ) = ( −a2 ) ( −a ) = a3 = 1  0

(d)

( a − a 2 )( a 2 − 1) = ( a 2 − a )(1 − a 2 ) = ( 3270 )( 330 ) = 3  300 𝑗

(e) − 𝑎 = (1∠180°)(1∠90°)(1∠ − 120°) = 1∠150°

9.2

1 2

3 2

(a)

( a ) = a ( a3 ) = a = − + j

(b)

( ja ) = ( j ) ( a ) = ( j ) ( j ) ( j ) ( a ) = −a =

(c)

(1 − a ) = ( 3 − 30 ) = ( 3 )  − 90 = 0 − j3 3

1 3 −j 2 2

= 0 − j 5.196 1

3 2

3 radians 2 = 0.606549.62 = 0.3929 + j 0.4620 −

(d) ea = e 2

(e)

𝑗 10

−( ) 𝑎

−

= e 2

𝑗 10

(−1)5

𝑎

= − 10 = −

1 𝑎

√3

= 𝑎2 = − − 𝑗

9.3

(a)

I 0  1 1    I1  = 1 1 a I 2  3 1 a 2

1   6  90   1 90  + 1  320  + 1220    6 2  a  6  320  = 1 90  + 1 80  + 1 100   3 1  90  + 1 200  + 1  340  a  6  220 

0 − j 0.2856 0.571  − 90  6  0 + j 2.97  =  5.94  90   A = 3  0 + j 0.316   0.632  90  

(b) I 0  1 1    I1  = 1 1 a I 2  3 1 a 2 9.4

Van  1 1    2 Vbn  = 1 a Vcn  1 a

1  40  90   1  90  + 1  0    18.86  45    2  a   40  0   40 1 90  + 1120   = 25.76  105  A = a   0  3 1  90  + 1240   6.90  165  

1  45  80  1 80  + 2  0  + 1 90       a   90  0   = 45 1  80  + 2  240  + 1 210  1  80  + 2 120  + 1 330   a 2  45  90 

 2.174 + j1.985   132.48 42.4  = 45 − 1.692 − j1.247 = 94.58  216.4  V 0.0397 + j 2.217 99.78  88.97 

9.5

Eq 9.1.12 of text: 1 1 𝐼0̅ = (𝐼𝑎̅ + 𝐼𝑏̅ + 𝐼𝑐̅ ) = (9∠0° + 8∠ − 90° + 6∠150°) = 1.268 − 𝑗1.667 3 3 = 2.094∠ − 52.74° A 1 1 𝐼1̅ = (𝐼𝑎̅ + 𝑎𝐼𝑏̅ + 𝑎2 𝐼𝑐̅ ) = (9∠0° + (1∠120°)8∠ − 90° + (1∠ − 120°)6∠150°) 3 3 = 7.04 − 𝑗2.33 = 7.42∠18.33° A 1 1 𝐼2̅ = (𝐼𝑎̅ + 𝑎2 𝐼𝑏̅ + 𝑎𝐼𝑐̅ ) = (9∠0° + (1∠ − 120°)8∠ − 90° + (1∠120°)6∠150°) 3 3 = 0.691 − 𝑗0.667 = 0.960∠ − 43.99° A

9.6

(a)

Eq. (9.1.9) of text:

Va = (V0 + V1 + V2 ) = (100 + 8030 + 40 − 30 ) = 114 + j 20 = 1169.9 V  Vb = V0 + a 2V1 + aV2 = 100 + 1240 ( 8030 ) + 1120 ( 40 − 30 ) 

= (100 + 80 − 90 + 4090 ) = 10 − j 40 = 41.3 − 76 

Vc = V0 + aV1 + a 2V2 = 100 + 1120 ( 8030 ) + 1240 ( 40 − 30 ) 

= (100 + 80150 + 40 − 150 ) = −94 + j 20 = 96.1168 

(b) Vab = Va − Vb = (114 + j 20 ) − (10 − j 40 ) = 104 + j 60 = 12030 V 

Vbc = Vb − Vc = (10 − j 40 ) − ( −94 + j 20 ) = 104 − j60 = 120 − 30 V  Vca = Vc − Va = ( −94 + j 20 ) − (114 + j 20 ) = −208 + j 0 = 208180 V 

(V ) = 13 (V + V + V ) = 13 (12030 + 120 − 30 + 208180 ) = 0  ab 0

12030 + 1120 (120 − 30 )   +1240 ( 208180 )  

(V ) = 13 (V + aV + a V ) = 13  ab 1

1 (12030 + 12090 + 20860 ) = 69.33 + j120 = 138.660 V  3 12030 + 1240 (120 − 30 )   +1120 ( 208180 )  

(V ) = 13 (V + a V + a (V ) ) = 13  ab 2

1 (12030 + 120210 + 208 − 60 ) = 34.67 − j60 = 69.3 − 60 V  3

Since (Vab )0 = Va 0 − Vb 0 = 0 And (Vab )1 = Va1 − Vb1 ; (Vab )2 = Va 2 − Vb 2 , we have VL − L 0 = 0; VL − L 1 =

( 330°)V ; V 1

L −L 2

( 3 − 30°)V

 1   1   − 30  VL1 and V2 =  30  VL 2 Or V1 =   3   3  Applying the above, one gets

 1  V1 =   − 30  (138.660 ) = 8030 = 69.3 + j 40   3   1  V2 =  30  ( 69.3 − 60 ) = 40 − 30 = 34.6 − j 20   3 

Phase voltages are then given by

Va = V1 + V2 = 103.9 + j 20 = 105.910.9 V  Vb = a 2V1 + aV2 = 1240 ( 8030 ) + 1120 ( 40 − 30 ) = ( 80 − 90 + 4090 ) = − j 40 = 40 − 90 V 

Vc = aV1 + a 2V2 = 1120 (8030 ) + 1240 ( 40 − 30 ) = 80150 + 40210 = −104 + j 20 = 105.9169 V 

The above are not the same as in part (a)  However, either set will result in the same line voltages. Note that the zero-sequence line voltage is always zero, even though zero-sequence phase voltage may exist. So it is not possible to construct the complete set of symmetrical components of phase voltages even when the unbalanced system of line voltages is known. But we can obtain a set with no zerosequence voltage to represent the unbalanced system. 9.7 𝐼0̅ 1 1 1 [𝐼1̅ ] = [1 𝑎 3 1 𝑎2 𝐼2̅

1 0 500∠90° 1500 1∠150° + 1∠30° [1∠270° + 1∠270°] = [1000∠270°] A 𝑎2 ] [1500∠150°] = 3 𝑎 1500∠30° 1∠30° + 1∠150° 500∠90°

Current to ground, 𝐼𝑛̅ = 3𝐼0̅ = 1500∠90° A 9.8

Vab = Va − Vb ; Vbc = Vb − Vc ; Vca = Vc − Va Vab + Vbc + Vca = 0, Vab 0 = Vbc 0 = Vca 0 = 0 Choosing Vab as the reference, 1 (Vab + aVbc + a 2Vca ) 3 1 = (Va − Vb ) + a (Vb − Vc ) + a 2 (Vc − Va ) 3 1 = (Va + aVb + a 2Vc ) − ( a 2Va + Vb + aVc )  3 1 = (Va + aVb + a 2Vc ) − a 2 (Va + aVb + a 2Vc )  3 1 = (1 − a 2 ) (Va + aVb + a 2Vc )  = (1 − a 2 ) Va1 3

Vab1 =

(

)

= 3 Va1 e j 30 

1 (Vab + a 2Vbc + aVca ) 3 1 = (Va − Vb ) + a 2 (Vb − Vc ) + a (Vc − Va ) 3 1 = (Va + a 2Vb + aVc ) − ( aVa + Vb + a 2Vc )  3 1 = (Va + a 2Vb + aVc ) − a (Va + a 2Vb + aVc )  3 1 = (1 − a ) (Va + a 2Vb + aVc )  = (1 − a ) Va 2 3

Vab 2 =

(

)

= 3 Va 2 e − j 30 

9.9

Choosing Vbc as reference and following similar steps as in Problem 9.8 solution, one can get Vbc 0 = 0; Vbc1 = 3Va1e − j 90 = − j 3Va1 ;     j 90 and Vbc 2 = 3Va 2 e = j 3Va 2  

9.10 (a) VLg 0  1 1   1 VLg1  = 1 a   3 1 a 2 VLg 2 

1  2800  2 a  250 − 110  a  290130 

2800 + 250 − 110 + 290130  2.696 − j 4.257  1  = 270.6 + j31.26  = 2800 + 250 10 + 29010    3 2800 + 250130 + 290250   6.706 − j 27  5.039  − 57.65 =  272.4  6.59  V  27.82 − 76.05 

(b) Vab  Vag − Vbg  2800 − 250 − 110        Vbc  = Vbg − Vcg  = 250 − 110 − 290130  Vca  V − V  290130 − 2800   cg ag    365.5 + j 234.9   434.532.73  =  100.9 − j 457.1  =  468.1 − 77.55 V  −466.4 + j 222.2   516.6154.5 

1   434.532.73  a 2   468.1 − 77.55 a   516.6154.5 

 434.5  32.73 + 468.1  − 77.55 + 516.6 154.5   0 + j 0  1   =  434.5  32.73 + 468.1  42.55 + 516.6  34.5  = 378.9 + j 281.2  3  434.5  32.73 + 468.1 162.5  + 516.6  274.5   −13.46 − j 46.44    0 0     =  471.8  36.58  V =  3 VLg1  + 30     48.35  − 106.2  3VLg 2  − 30 

9.11 The circuit is shown below:

1 (100 + 10180 + 0 ) = 0 3 1 I a1 = (100 + 10180 + 120 + 0 ) = 5 − j 2.89 = 5.78 − 30 A 3 1 I a 2 = (100 + 10180 + 240 + 0 ) = 5 + j 2.89 = 5.7830 A 3 Ia0 =

Then I b 0 = I a 0 = 0 A;

Ic0 = Ia0 = 0 A

I b1 = a 2 I a1 = 5.78 − 150 A; I c1 = aI a1 = 5.7890 A I b 2 = aI a 2 = 5.78150 A;

I c 2 = a 2 I a 2 = 5.78 − 90 A

9.12 Selecting a base of 2300 V and 500 kVA, each resistor has an impedance of 10 pu ; Vab = 0.8 ; Vbc = 1.2 ; Vca = 1.0

The symmetrical components of the line voltages are:

Vab1 =

Vab 2 =

1 ( 0.882.8 + 1.2120 − 41.4 + 1.0240 + 180 ) = 0.2792 + j 0.9453 3 = 0.985773.6 1 ( 0.882.8 + 1.2240 − 41.4 + 1.0120 + 180 ) = −0.1790 − j0.1517 3 = 0.2346220.3

(These are in pu on line-to-line voltage base.) Phase voltages in pu on the base of voltage to neutral are given by Van1 = 0.985773.6 − 30 = 0.985743.6 Van 2 = 0.2346220.3 + 30 = 0.2346250.3

[Note: An angle of 180 is assigned to Vca ]

Zero-sequence currents are not present due to the absence of a neutral connection.

I a1 = Va1 /10 = 0.985743.6 pu I a 2 = Va 2 /10 = 0.2346250.3 pu

The positive direction of current is from the supply toward the load. 9.13 (a) I  0  1 1 I  1 1 a  1  =  I  2  3 1 a 2

1   10  0   a 2  12  − 90  a   15  90  

10  0  + 12 − 90  + 15  90    3.333 + j1   3.48 16.7   1 =  10  0  + 12  30  + 15  330   = 11.13 − j0.5  = 11.14 − 2.58  A 3 10  0  + 12 150  + 15  210  −4.46 − j0.5 4.49 − 173.6 

(b) I a  I ab − I ca   10  0  − 15  90    10 − j15   18.03  − 56.3             Ib  = Ibc − I ab  =  12  − 90  − 10  0   = − 10 − j12 = 15.62  − 129.81  A I c  I ca − Ibc  15  90  − 12  − 90   j 27    27  90  (c) I L 0  1 1   1 I L1  = 1 a I L 2  3 1 a 2

1   18.03  − 56.3   a 2  15.62  − 129.81   a   27  90 

 18.03  − 56.3  + 15.62  − 129.81  + 27  90   1 =  18.03  − 56.3  + 15.62  − 9.81  + 27 330   3 18.03  − 56.3  + 15.62  − 249.81  + 27  210  0  0 + j 0  0        = 16.26 − j10.39  = 19.29  − 32.57   A =  3 I 1  − 30     − 6.257 − j 4.61 7.77  − 143.59   3 I  2  + 30 

9.14

I0 = I1 = I2 =

VLg 0 Z0 VLg1 Z1 VLg 2 Z2

5.039 − 57.65  = 0.252  − 110.78  A 20 53.13 

272.4 6.59  = 13.62  − 46.54  A 20  53.13 

27.82  − 76.05  = 1.391  − 129.18  A 20  53.13 

I a  1 1    2 I b  = 1 a I c  1 a  

1  0.252  − 110.78    a   13.62  − 46.54   a 2  1.391  −129.18  

0.2520  − 110.78  + 13.62  − 46.54  + 1.391 − 129.18  = 0.2520  − 110.78  + 13.62 193.46  + 1.391  − 9.18   0.2520  − 110.78  + 13.62  73 .46  + 1.391 110.82    8.4 − j11.2   14  − 53.13       = − 11.96 − j3.628 = 12.5  − 163.1  A  3.294 + j14.12   14.5  76.37  

Note: The source and load neutrals are connected with a zero-ohm wire. I a  Vag zY   280  0  20  53.13    14  − 53.13           I b  = Vbg zY  = 250  − 110  20 53.13  = 12.5  − 163.1  I c  V z   290 130  20 53.13    14.5 76.87      cg Y 

Which agrees with the above result. 9.15

I 0 = 0 ; From Problem 9.14, I1 = 13.62 − 46.54 A; I 2 = 1.391 − 129.18 A  I a  1 1    2  I b  = 1 a  I c  1 a  

1  0    a   13.62 − 46.54  a 2  1.391 − 129.18

13.62 − 46.54 + 1.391 − 129.18  =  13.62193.46 + 1.391 − 9.18   13.6273.46 + 1.391110.82   8.49 − j10.96  13.86 − 52.24 =  −11.87 − j 3.392  = 12.35195.9  A  3.383 + j14.36  14.7576.74 

9.16

272.46.59 = 40.86 − 46.54 A  20   53.13   3    27.82 − 76.05 I2 = = 4.173 − 129.18 A  20   3  53.13   I 0 = 0; I1 =

 I a  1 1    2  I b  = 1 a  I c  1 a  

1  0   41.58 − 52.24   a   40.86 − 46.54  = 37.05195.9  A a 2   4.173 − 129.18  44.2576.74 

Note: These currents are 3 times those in Problem 9.15. 9.17

I0 = I1 = I2 =

VLg 0 Z0 VLg1 Z1 VLg 2 Z2

5.039 − 57.65 = 0.4826 − 131 A 3 + j10

272.46.59 = 36.54 − 19.98 A 7.45426.57

27.82 − 76.05 = 3.732 − 102.72 A 7.45426.57

 I a  1 1    2  I b  = 1 a  I c  1 a  

1  0.4826 − 131  a  36.54 − 19.98  a 2  3.732 − 102.72

0.4826 − 131 + 36.54 − 19.98 + 3.732 − 102.72  =  0.4826 − 131 + 36.54220.02 + 3.73217.28   0.4826 − 131 + 36.54100.02 + 3.732137.28   33.2 − j16.49  37.07 − 26.41 =  −24.74 − j 22.75 = 33.61222.6  A  −9.416 + j 38.15 39.29103.9 

9.18

 Z0 Z  10  Z 20

Z 01 Z1 Z 21

Z 02  1 1  Z12  1 1 a = Z 2  3 1 a 2

1 a 2  a 

( Z aa + Z ab + Z ac ) ( Z aa + a 2 Z ab + aZ ac )( Z aa + aZ ab + a 2 Z ac )    2 2  ( Z ab + Z bb + Z bc ) ( Z ab + a Z bb + aZ bc )( Z ab + aZ bb + a Z bc )   ( Z ac + Z bc + Z cc ) ( Z ac + a 2 Z bc + aZ cc )( Z ac + aZ bc + a 2 Z cc )   

( Z aa + Z bb + Z cc ) + 2 ( Z ab + Z ac + Z bc ) 1 = ( Z aa + aZ bb + a 2 Z cc ) + Z ab (1 + a ) + Z ac (1 + a 2 ) + Z bc ( a + a 2 ) 3 ( Z aa + a 2 Z bb + aZ cc ) + Z ab (1 + a 2 ) + Z ac (1 + a ) + Z bc ( a 2 + a )  ( Z aa + a 2 Z bb + aZ cc ) + Z ab ( a 2 + 1) + Z ac ( a + 1) + Z bc ( a + a 2 ) ( Z aa + a3 Z bb + a3 Z cc ) + Z ab ( a 2 + a ) + Z ac ( a + a 2 ) + Z bc ( a 2 + a 4 )

( Z aa + a 4 Z bb + a 2 Z cc ) + Z ab ( a 2 + a 2 ) + Z ac ( 2a ) + Z bc ( a 2 + a 4 ) ( Z aa + aZ bb + a 2 Z cc ) + Z ab (1 + a ) + Z ac (1 + a 2 ) + Z bc ( a + a 2 )   ( Z aa + a 2 Z bb + a 4 Z cc ) + Z ab ( 2a ) + Z ac ( 2a 2 ) + Z bc ( 2 )  3 3 2 2 2  Z + a Z + a Z + Z a + a + Z a + a + Z a + a ( aa ) ac ( ) bc ( ) bb cc ) ab (

 Z aa + Z bb + Z cc + 2 Z ab + 2 Z ac + 2 Z bc 1 =  Z aa + aZ bb + a 2 Z cc − a 2 Z ab − aZ ac − Z bc 3  Z aa + a 2 Z bb + aZ cc − aZ ab − a 2 Z ac − Z bc Z aa + a 2 Z bb + aZ cc − aZ ab − a 2 Z ac − Z bc Z aa + Z bb + Z cc − Z ab − Z ac − Z bc Z aa + aZ bb + a 2 Z cc + 2 a 2 Z ab + 2 aZ ac + 2 Z bc Z aa + aZ bb + a 2 Z cc − a 2 Z ab − aZ ac − Z bc

  Z aa + a Z bb + aZ cc + 2 aZ ab + 2 a Z ac + 2 Z bc   Z aa + Z bb + Z cc − Z ab − Z ac − Z bc 2

9.19 (a)

writing KVL equations [see Eqs (9.2.1) – (9.2.3)]:

Vag = Z y I a + Z n ( I a + I b + I c ) Vbg = Z y I b + Z n ( I a + I b + I c ) Vcg = Z y I c + Z n ( I a + I b + I c ) In matrix format [see Eq (9.2.4)] ( Z y + Z n )   Zn   Zn  ( 3 + j 5 )   j1  j 1 

 V    I a   ag     I b = Vbg  Zn     ( Z y + Z n )  I c  Vcg 

(Z + Z ) y

Zn j1

( 3 + j 5) j1

 I a  ( 3 + j 5 )     I b  =  j1  I c   j1   

  I a   1000    j1   I b  = 75180 ( 3 + j 5)  I c   5090  j1

( 3 + j5) j1

  j1  ( 3 + j 5) j1

−1

 1000  75180    5090 

Performing the indicated matrix inverse (a computer solution is suggested):  I a  0.1763 − 56.50 0.02618150.2 0.02618150.2   1000        I b  =  0.02618150.2 0.1763 − 56.50 0.02618150.2  75180   I c   0.02618150.2 0.02618150.2 0.1763 − 56.50   5090   

Finally, performing the indicated matrix multiplication:

 I a  17.63 − 56.50 + 1.964330.2 + 1.309240.2       I b  = 2.618150.2 + 13.22123.5 + 1.309240.2   I c  2.618150.2 + 1.964330.2 + 8.81533.5     I a   10.78 − j16.81  19.97 − 57.32        I b  =  −10.22 + j11.19  = 15.15132.4  A  I c   6.783 + j 5.191  8.54137.43   

(b) Step (1): Calculate the sequence components of the applied voltage:

Vg 0  1 1   1 Vg1  = 1 a   3 1 a 2  Vg 2 

1   1000  a 2  =  75180 a   5090 

 1000 + 75180 + 5090  1 = 1000 + 75300 + 50330  3  1000 + 7560 + 50210  8.333 + j16.667  18.6363.43  =  60.27 − j 29.98  = 67.32 − 26.45 V  31.40 + j13.32  34.1122.99  Step (2): Draw sequence networks:

Step (3): Solve sequence networks I0 =

Vg0 Z0

Vg0 Z y + 3 Zn

18.6363.43 18.6363.43 = 3 + j7 7.61666.80

I 0 = 2.446 − 3.37 A I1 = I2 =

Vg1 Z1 Vg 2 Z2

67.32 − 26.45 67.32 − 26.45 = = 13.46 − 79.58 A 3 + j4 553.13

34.1122.99 = 6.822 − 30.14 553.13

Step (4): Calculate the line currents (phase components):  I a  1 1    2  I b  = 1 a  I c  1 a  

1   2.446 − 3.37  a  13.46 − 79.58  a 2  6.822 − 30.14 

2.446 − 3.37 + 13.46 − 79.58 + 6.822 − 30.14  = 2.446 − 3.37 + 13.46160.42 + 6.82289.86  2.446 − 3.37 + 13.4640.42 + 6.822209.86   10.78 − j16.81  19.97 − 57.32  =  −10.22 + j11.19  = 15.15132.4  A  6.773 + j 5.187  8.53137.45 

9.20 (a) The line-to-line voltages are related to the  currents by Vab   j 27    Vbc  =  0 Vca   0  

0   I ab    0   I bc  j 27   I ca 

0 j 27 0

Transforming to symmetrical components, Vab 0   j 27   A Vab1  =  0   Vab 2   0

0 j 27 0

0   I ab 0    0  A  I ab1  j 27   I a b 2   

Premultiplying each side by A−1 , Vab 0   I ab 0   j 27      −1 Vab1  = j 27 A A  I ab1  =  0     Vab 2   I ab 2   0

0 j 27 0

0   I ab 0    0   I ab1  j 27   I ab 2   

As shown in Fig. 9.5 of the text, sequence networks for an equivalent Y representation of a balanced- load are given below:

(b) With a mutual impedance of (j6)  between phases, Vab 0   j 27   A Vab1  =  j 6    j6 Vab 2  

j6 j 27 j6

j 6   I ab 0    j 6  A  I ab1  j 27   I ab 2   

Rewriting the coefficient matrix into two parts,  j 27  j6   j 6

j6 j 27 j6

j6  1 0 0  1 1 1    j 6  = j 21 0 1 0  + j6 1 1 1 0 0 1  1 1 1 j 27 

and substituting into the previous equation, Vab 0   1 1 1   I ab 0        −1 −1  Vab1  =  j 21A A + jGA 1 1 1 A   I ab1     1 1 1   I  Vab 2    ab 2    j 21  =   0  0   j 39 =  0  0

0 j 21 0 0 j 21 0

0   3 0 0    I ab 0     0  + j 6 0 0 0    I ab1   0 0 0    I  j 21  ab 2  0   I ab 0    0   I ab1  j 21  I ab 2   

Then the sequence networks are given by:

9.21 From Eq (9.2.28) and (9.2.29), the load is symmetrical. Using Eq. (9.2.31) and (9.2.32): ̅ = 3 + 𝑗7 𝑍0̅ = 𝑍̅𝑎𝑎 2 𝑍𝑎𝑏 ̅ − 𝑍̅𝑎𝑏 = 3 + 𝑗7 𝑍1̅ = 𝑍̅2 = 𝑍𝑎𝑎 3 + 𝑗7 𝑍𝑆 = [ 0 0

0 0 3 + 𝑗7 0 ]Ω 0 3 + 𝑗7

9.22 Since Z s is diagonal, the load is symmetrical. Using Eq. (9.2.31) and (9.2.32): Z 0 = 6 + j10 = Z aa + 2 Z ab Z1 = 5 = Z aa − Z ab

Solving the above two equations 1 1 1 10 ( 6 + j10 − 5) = (1 + j10 ) = + j  3 3 3 3 16 10 Z aa = Z ab + 5 = + j  3 3 Z aa =

10 1 10 1 10  16 +j   3 + j 3 3+ j 3 3 3   1 10 16 10 1 10   Zp = +j +j +j  3 3 3 3 3 3    1 10 16 10  1 + j 10 +j +j  3 3 3 3 3   3

9.23 The line-to-ground voltages are Va = Z s I a + Z m I b + Z m I c + Z n I n Vb = Z m I a + Z s I b + Z m I c + Z n I n Vc = Z m I a + Z m I b + Z s I c + Z n I n

Since I n = I a + I b + I c , it follows

Va   Z s + Z n    Vb  =  Z m + Z n Vc   Z m + Z n   

Zm + Zn  Ia    Zm + Zn   Ib  Z s + Z n   I c 

Zm + Zn Zs + Zn Zm + Zn

phase impedance matrix Z p

or in compact form Vp = Z p I p From Eq. (9.2.9) Z s = A−1 Z p A 1 1 1  Z s = 1 a 3 1 a 2

1   Zs + Zn  a 2   Z m + Z n a   Z m + Z n

 Zs + 3 Zn + 2 Zm  = 0  0 

Zm + Zn Zs + Zn Zm + Zn

0 Zs − Zm 0

Z m + Z n  1 1  Z m + Z n  1 a 2 Z s + Z n  1 a

1 a  a 2 

  0  Z s − Z m  0

Sequence impedance matrix

When there is no mutual coupling, Z m = 0  Zs + 3 Zn   Zs =  0  0 

0 Zs 0

0  0 Z s 

9.24 (a) The circuit is shown below:

KVL:

( j12 ) I a + ( j 4 ) I b − ( j12 ) I b − j 4 ( I a ) = Va − Vb = VLINE 30 ( j12 ) I b + ( j 4 ) I c − ( j12 ) I c − ( j 4 ) I b = Vb − Vc = VLINE  − 90

KCL: I a + I b + I c = 0

In matrix form:  j12 − j 4 − ( j12 − j 4 )   I a  VL 30  0  ( j12 − j 4 ) − ( j12 − j 4 )  I b  = VL 90  0  1   I c   0  1 1    where VL = 100 3

Solving for I a , I b , I c , one gets I a = 12.5 − 90; I b = 12.5150; I c = 12.530A (b) Using symmetrical components,  j12 + 2 ( j 4 )  0     Vs = 100  ; Z s =  0   0  0 

  0  j12 − j 4  0

j12 − j 4 0

From the solution of Problem 9.18 upon substituting the values Is = Zs

−1

1 1 Vs and I p = A I s where A = 1 a 2 1 a

1 a  a 2 

which result in

I a = 12.5 − 90; I b = 12.5150; I c = 12.530 A which is same as in (a)

9.25 (a)

1 1 Z s = A−1 Z p A; A = 1 a 2 1 a

1 1 1 1  a  ; A−1 = 1 a 3 1 a 2 a 2 

1 a 2  a 

The load sequence impedance matrix comes out as 0 0  6 + j 30  Zs =  0 6 + j 21 0   0 6 + j 21  0

see the result of Problem 9.18 20025  1 1 1   −1 −1 (b) Vp = 100 − 155 ; Vs = A Vp ; A = 1 a 3 80100  1 a 2

1 a 2  a 

Symmetrical components of the line-to-neutral voltages are given by:

V0 = 47.773957.6268; V1 = 112.7841 − 0.0331; V2 = 61.623145.8825 V (c) Vs = Z s I s ; I s = Z s −1 Vs , which results in I 0 = 1.562 − 21.06; I1 = 5.164 − 74.0877; I 2 = 2.822 − 28.1722 A

1 1 (d) I p = A I s ; A = 1 a 2 1 a

1 a  a 2 

The result is: I a = 8.706 − 51.99; I b = 5.062136; I c = 3.12932.04 A

9.26

I0 = I1 = I2 =

VLg 0 3 + j 4 + Z0 VLg1 3 + j 4 + Z1 VLg 2 3 + j 4 + Z2

5.039 − 57.65 5.039 − 57.65 = = 0.2016 − 110.78 A 2553.13 ( 3 + j 4 ) + (12 + j16 )

272.46.59 = 10.896 − 46.54 A 2553.13

27.82 − 76.05 = 1.1128 − 129.18 A 2553.13

 I a  1 1    2  I b  = 1 a  I c  1 a  

1  0.2016 − 110.78  a   10.896 − 46.54  a 2  1.1128 − 129.18 

0.2016 − 110.78 + 10.896 − 46.54 + 1.1128 − 129.18  = 0.2016 − 110.78 + 10.896193.46 + 1.1128 − 9.18  0.2016 − 110.78 + 10.89673.46 + 1.1128110.82   6.72 − j8.96  11.2  − 53.13  =  − 9.57 − j 2.902  =  10  − 163.1   A 2.635 + j11.297  11.676.87  

Also, since the source and load neutrals are connected with a zero-ohm neutral wire,  I a  Vag / ( 3 + j 4 + ZY )   2800 / 2553.13  11.2 − 53.13           I b  = Vbg / ( 3 + j 4 + ZY )  = 250 − 110 / 2553.13  =  10 − 163.1  A  I c  V / ( 3 + j 4 + Z )   290130 / 2553.13   11.676.87     cg  Y 

Which checks. 9.27 (a) KVL : Van = Z aa I a + Z ab I b + Z ab I c + Z an I n + Van

− ( Z nn I n + Z an I c + Z an I b + Z an I a ) Voltage drop across the line section is given by

Van − Van = ( Z aa − Z an ) I a + ( Z ab − Z an )( I b + I c ) + ( Z an − Z nn ) I n Similarly for phases b and c

Vbn − Vbn = ( Z aa − Z an ) I b + ( Z ab − Z an )( I a + I c ) + ( Z an − Z nn ) I n Vcn − Vcn = ( Z aa − Z an ) I c + ( Z ab − Z an )( I a + I b ) + ( Z an − Z nn ) I n KCL :

In = − ( Ia + Ib + Ic )

Upon substitution

Van − Van = ( Z aa + Z nn − 2 Z an ) I a + ( Z ab + Z nn − 2 Z an ) I b + ( Z ab + Z nn − 2 Z an ) I c Vbn − Vbn = ( Z ab + Z nn − 2 Z an ) I a + ( Z aa + Z nn − 2 Z an ) I b + ( Z ab + Z nn − 2 Z an ) I c Vcn − Vcn = ( Z ab + Z nn − 2 Z an ) I a + ( Z ab + Z nn − 2 Z an ) I b + ( Z aa + Z nn − 2 Z an ) I c The presence of the neutral conductor changes the self- and mutual impedances of the phase conductors to the following effective values:

Z s  Z aa + Z nn − 2 Z an ; Z m  Z ab + Z nn − 2 Z an

Using the above definitions Vaa  Van − Van   Z s      Vbb  = Vbn − Vbn  =  Z m Vcc  Vcn − Vcn   Z m     

Zm Zs Zm

Zm  Ia    Zm  Ib  Z s   I c 

Where the voltage drops across the phase conductors are denoted by Vaa , Vbb , and Vcc . (b) The a-b-c voltage drops and currents of the line section can be written in terms of their symmetrical components according to Eq. (9.1.9); with phase a as the reference phase, one gets Vaa0    Z s − Z m    A Vaa1  =   Vaa2      

Zs − Zm

  Zm    +  Zm Z s − Z m   Z m

Zm Zm Zm

Zm   Ia0     Z m   A  I a1  Z m    I a 2 

Multiplying across by A−1,  Vaa0  1  1 1 1   I a 0          −1  Vaa1  = A ( Z s − Z m )  1  + Z m 1 1 1  A  I a1   Vaa2   1 1 1   I a 2  1   

Vaa0   Z s − 2 Z m V  =   aa1   Vaa2  

Zs − Zm

 Ia0      I a1  Z s − Z m   I a 2 

Now define zero-, positive-, and negative-sequence impedances in terms of Z s and Z m as Z 0 = Z s + 2 Z m = Z aa + 2 Z ab + 3 Z nn − 6 Z an Z1 = Z s − Z m = Z aa − Z ab Z 2 = Z s − Z m = Z aa − Z ab

Now, the sequence components of the voltage drops between the two ends of the line section can be written as three uncoupled equations: Vaa0 = Van 0 − Van0 = Z 0 I a 0 Vaa1 = Van1 − Van1 = Z1 I a1 Vaa2 = Van 2 − Van2 = Z 2 I a 2

9.28 (a) The sequence impedances are given by Z 0 = Z aa + 2 Z ab + 3 Z nn − 6 Z an = j 60 + j 40 + j 240 − j180 = j160  Z1 = Z 2 = Z aa − Z ab = j 60 − j 20 = j 40 

The sequence components of the voltage drops in the line are Vaa 0  (182.0 − 154.0 ) + j ( 70.0 − 28.0 )  Van − Van      −1  −1  Vaa1  = A Vbn − Vbn  = A ( 72.24 − 44.24 ) − j ( 32.62 − 74.62 )       V − V − 170.24 − 198.24 + j 88.62 − 46.62 ) ( ) c n   cn  ( Vaa 2  28.0 + j 42.0  28.0 + j 42.0   kV = A 28.0 + j 42.0  =  0  28.0 + j 42.0    0 −1

From Problem 9.22 result, it follows that

Vaa0 = 28,000 + j 42,000 = j160 I a 0 ; Vaa1 = 0 = j 40 I a1 ; Vaa2 = 0 = j 40 I a 2 From which the symmetrical components of the currents in phase a are

I a 0 = ( 262.5 − j175) A; I a1 = I a 2 = 0 The line currents are then given by

I a = I b = I c = ( 262.5 − j175 ) A (b) Without using symmetrical components: The self- and mutual impedances [see solution of Problem 9.22(a)] are Z s = Z aa + Z nn − 2 Z an = j 60 + j80 − j 60 = j80  Z m = Z ab + Z nn − 2 Z an = j 20 + j80 − j 60 = j 40 

So, line currents can be calculated as [see solution of Problem 9.22(a)] Vaa  28 + j 42   j80      3 Vbb  = 28 + j 42   10 =  j 40 Vcc  28 + j 42   j 40    I a   j80     I b  =  j 40  I c   j 40  

j 40 j80 j 40

j 40  j 40  j80 

−1

j 40 j80 j 40

j 40   I a    j 40   I b  j80   I c 

28 + j 42   28 + j 42   103    28 + j 42 

262.5 − j175 = 262.5 − j175  A 262.5 − j175 

9.29

Z1 = Z 2 = j 0.5  200 = j100 

Z 0 = j 2  200 = j 400  Y1 = Y2 = j 3  10 −9  200  103 = j 6  10 −4 S Y0 = j1  10 −9  200  103 = j 2  10 −4 S

Nominal-  sequence circuits are shown below:

9.30 (a) I AB =

VAB 4800° = = 23.31 − 29.05° A (18 + j10 ) 20.5929.05°

I BC =

VBC 480120° = = 23.31 − 149.05° A 18 + j 10 20.59 29.05° ( )

(b) I A = I AB = 23.31 − 29.05° A I B = I BC − I AB = 23.31 − 149.05° - 23.31 − 29.05° I B = −40.37 − j 0.6693 = 40.38180.95° A I C = − I BC = 23.3130.95° A IL0  1 1 1  23.31 − 29.05°   1 (c)  I L1  = 1 a a 2   40.38180.95°  3 IL2  1 a 2 a   23.3130.95°     23.31 − 29.05 + 40.38180.95 + 23.3130.95  1 = 23.31 − 29.05 + 40.38300.95 + 23.31270.95  3  23.31 − 29.05 + 40.3860.95 + 23.31150.95 

0 + j0 0        = 13.84 − j 23.09 =  26.92 − 59.06 A 6.536 + j11.77  13.4660.96 

9.31

I Line 0 = I Line 2 = 0  200    90 3  I Line1 = =  Z   10  Z Line1 +   0.580 +  3  40    3  Vg1

115.4790 115.4790 = = 30.9645.05 A 2.64 + j 2.635 3.7344.95

9.32

Vgan = Vg1 = Vm1 + Z Line1 I m1 I m1 =

5000 cos−1 0.8 200 3 ( 0.8 )

= 18.0436.87 A

200

0 + ( 0.580 )(18.0436.87 ) 3 = 115.47 + ( −4.077 + j8.046 )

Vgan =

= 111.39 + j8.046 = 111.74.131 V Vg = 3 (111.7 ) = 193.5V (Line to Line)

9.33 Converting the  load to an equivalent Y, and then writing two loop equations:

 2 ( Z L + ZY ) − ( Z L + ZY )   I  Vcg − Vag    c  =   − Z + Z 2 Z + Z  ( L ( L Y )   − I b  Vag − Vbg  Y )   21.4643.78 −10.7343.78   I c  295115 − 2770   −10.7343.78 21.4643.78    =      − I b  2770 − 260 − 120  −1

 I c   21.4643.78 −10.7343.78  482.5146.35   =  −10.7343.78 21.4643.78     − I b     465.128.96   I c   0.06213 − 43.78 0.03107 − 43.78   482.5146.35   =  0.03107 − 43.78 0.06213 − 43.78      465.128.96   − I b    I c   29.98102.57 + 14.45 − 14.82  7.445 + j 25.57   = =   − I b  14.99102.57 + 28.90 − 14.82  24.68 + j 7.239   I c   26.6273.77  = A  − I b   25.7116.34 

Also, I a = − I b − I c I a = ( 24.68 + j 7.239 ) − ( 7.445 + j 25.57 ) I a = 17.23 − j18.33 = 25.15 − 46.76  I a   25.15 − 46.76      I b  =  25.71196.34  A  I c   26.6273.77 

which agrees with Ex. 9.6. The symmetrical components method is easier because it avoids the need to invert a matrix. 9.34 The line-to-ground fault on phase a of the machine is shown below, along with the corresponding sequence networks:

With the base voltage to neutral

13.8 3

kV ,

Va = 0;

Vb = 1.013 − 102.25;

with Z base

(13.8 ) = 9.52 , Z = j 2.38 = j 0.25; Z = j3.33 = j 0.35; =

= ( −0.215 − j 0.99 ) pu

Vc = 1.013102.25 pu.

= ( −0.215 + j 0.99 ) pu

Zg0 =

9.52

j 0.95 = j 0.1; Z n = 0; Z 0 = j 0.1pu 9.52

The symmetrical components of the voltages at the fault point are Va 0  1 1   1 Va1  = 3 1 a Va 2  1 a 2  

Ia0 = − I a1 =

1  0   −0.143 + j 0    a 2   −0.215 − j 0.99  =  0.643 + j 0  pu a   −0.215 + j 0.99  −0.500 + j 0 

( −0.143 + j 0 ) = − j1.43pu Va 0 =− Zg0 j 0.1

Ean − Va1 (1 + j 0 ) − ( 0.643 + j 0 ) = = − j1.43pu Z1 j 0.25

Ia2 = −

( −0.5 + j 0 ) = − j1.43pu Va 2 =− Z2 j 0.35

 Fault current into the ground I a = I a 0 + I a1 + I a 2 = 3I a 0 = − j 4.29 pu

With base current

20,000 3  13.8

= 837 A , the subtransient current in line a is

I a = 4.29  837 = 3590 A

Line-to-line voltages during the fault are: (on base voltage to neutral) Vab = Va − Vb = 0.215 + j 0.99 = 1.0177.7 pu = 8.0577.7 kV Vbc = Vb − Vc = 0 − j1.98 = 1.98270 pu = 15.78270 kV Vca = Vc − Va = −0.215 + j 0.99 = 1.01102.3 pu = 8.05102.3 kV

Phasor diagrams of line voltages before and after the fault are shown below:

Vab = 13.830 kV

Vab = 8.0577.7 kV

Vbc = 13.8270 kV

Vbc = 15.78270 kV

Vca = 13.8150 kV

Vca = 8.05102.3 kV

(Balanced)

(Unbalanced)

9.35 Base MVA = 100 100 100 100 = 0.5; G2 : X = 0.15  = 0.375; G3 : X = 0.15  = 0.375 20 40 40 100 100 Reactors: X1 = 0.05  = 0.25; X 2 = 0.04  = 0.25pu. 20 16

G1 : X = 0.1 

Per-phase reactance diagram is shown below: (Excluding the source) [in pu]

 j 0.5 j ( 0.25 + 0.2344 ) with respect to f = j 0.246 100 = 406.5MVA  0.246 406.5  106 Fault Current = = 17,780 A 3  13.2  103 = 17.78 kA

 Fault MVA =

9.36 Line-to-ground fault: Let Va = 0; I b = I c = 0  Then I a 0 =

1 1 Ia + Ib + Ic ) = Ia ( 3 3

1 1 I a + aI b + a 2 I c ) = I a ( 3 3 1 1 I a 2 = ( I a + a 2 I b + aI c ) = I a 3 3 I a1 =

1 so that I a 0 = I a1 = I a 2 = I a ; Va 0 + Va1 + Va 2 = 0  3 Sequence network interconnection is shown below:

I 0 = I1 = I 2 =

Ea Z 0 + Z1 + Z 2

V0 + V1 + V2 = 0 Ia =



3Ea Z 0 + Z1 + Z 2

9.37 (a) Short circuit between phases b and c: I b + I c = 0; I a = 0 (Open Line); Vb = Vc then I a 0 = 0; 1 1 0 + aI b + a 2 I c ) = ( aI b − a 2 I b ) ( 3 3 1 = ( a − a2 ) Ib 3 1 1 1 I a 2 = ( 0 + a 2 I b + aI c ) = ( a 2 I b − aI b ) = ( a 2 − a ) I b 3 3 3 I a1 =

so that I a1 = − I a 2 From Vb = Vc , one gets Va 0 + a 2Va1 + aVa 2 = Va 0 + aVa1 + a 2Va 2 so that Va1 = Va 2 Sequence network interconnection is shown below:

(b) Double line-to-ground fault: Fault conditions in phase domain are represented by I a = 0; Vb = Vc = 0 1 Sequence components: Va 0 = Va1 = Va 2 = Va 3 I a 0 + I a1 + I a 2 = 0

Sequence network interconnection is shown below:

9.38 (a)

(b)

9.39

9.40

 345   100  X pu new = ( 0.09 )     = 0.02755 per unit  360   300 

9.41

 18   100  X g1− 0 = ( 0.05 )     = 0.0054 = X g 2 −0  20   750   100  X m 3− 0 = ( 0.05 )   = 0.003333  1500  2

 18  X n 2 = ( 0.06 )   = 0.0486  20  9.42

3 X n 2 = 0.1458

VA1 = 145 + 30 = 175 = 0.2588 + j 0.9659

VA 2 = 0.5250 − 30 = 0.5220 = −0.383 − j 0.321 VA = VA1 + VA 2 = 0.12422 + j 0.645 = 0.656100.91 VB1 = a 2VA1 = 1315 = 1 − 45 = 0.7071 − j 0.7071 VB 2 = aVA 2 = 0.5340 = 0.5 − 20 = 0.469 − j 0.171 VB = VB1 + VB 2 = 1.177 − j 0.878 = 1.47 − 36.7 VC1 = aVA1 = 1195 = −0.9659 − j 0.2588 VC 2 = a 2VA2 = 0.5100 = −0.0868 + j 0.492 VC = VC1 + VC 2 = −1.053 − j 0.234 = 1.078167.49

Line-to-line voltages are given by: [in pu on line-neutral voltage base]

VAB = VA − VB = −1.301 + j1.523 = 2130.5  VBC = VB − VC = 2.32 − j1.112 = 2.57 − 25.6  VCA = VC − VA = −1.177 − j 0.411

 Note: Divide by 3  if the base is line-to-line voltage 

= 1.25 − 160.75  Load impedance in each phase is 10 pu.

 I a1 = Va1 in pu; I a 2 = Va 2 in pu Thus I A = VA in pu I A = 0.656100.91 pu   I B = 1.47 − 36.7 pu    I C = 1.078167.49 pu 

9.43 (a)

e j 90 : i

Per Unit Zero Sequence

X1 = X 2 = X3 =

1 ( 0.1 + 0.1 − 0.1) 2

e j 90 : i

(e+ j 90 : i) (e j 90 : i)

= 0.05 per unit

Per Unit Positive Sequence (Per Unit Negative Sequence)

(b)

Per Unit Zero Sequence

Per Unit Positive Sequence (Per Unit Negative Sequence)

(c)

Per unit Zero Sequence

e j 23.40 : i (e− j 23.40 : i) Per Unit Positive Sequence (Per Unit Negative Sequence)

9.44

Per Unit Zero Sequence Network

Per Unit Positive (or Negative) Sequence Network (Phase Shift Not Shown) p-primary s-secondary t-tertiary

9.45

S3 = Van I a* + Vbn I b* + Vcn I c* = ( 2800 )(14 53.13 ) + ( 250 − 110 )(12.5163.1 ) + ( 290130 )(14.5 − 76.87 ) = 392053.13 + 312553.1 + 420553.13 = 6751 + j8999

P3 = Re S3 = 6751W   delivered to the load. Q3 = Im S3 = 8999vars 9.46 (a)

Z0 = Z y + 3 Zn = 2 + j 2 + j3 = 2 + j 5 = 5.38568.20

(b) I 0 =

V0 1060 = Z 0 5.38568.20

I 0 = 1.857 − 8.199 A

Z1 = ZY // I1 =

Z = ( 2 + j 2 ) // ( 2 + j 2 ) = 1 + j = 245. 3

V1 1000 = = 70.71 − 45 A Z1 245

Z 2 = Z1 = 245 I2 =

V2 15200 = = 10.61155 A Z2 2 45

S0 = V0 I 0* = (1060 )(1.8578.199 ) = 6.897 + j17.24 S0 = 18.5768.199 S1 = V1 I1* = (1000 )( 70.7145 ) = 5000 + j 5000 S1 = 707145

S2 = V2 I 2* = (15200 )(10.61 − 155 ) = 112.5 + j112.5

S2 = 159.45

S3 = 15,358. + j15,389. S3 = 21.74  103 45.06 VA 9.47

S3 = Va 0 I a*0 + Va1 I a*1 + Va 2 I a*2 Substituting values of voltages and currents from the solution of Problem 9.12, S3 = 0 + ( 0.985743.6 )( 0.9857 − 43.6 ) + ( 0.2346250.3 )( 0.2346 − 250.3 ) = ( 0.9857 ) + ( 0.2346 ) 2

= 1.02664 pu

With the three-phase 500-kVA base,

S3 = 513.32 kW To compute directly: The Equivalent -Connected resistors are

R = 3RY = 3  10.58 = 31.74  From the given line-to-line voltages

S3 =

Vab

R

Vbc R

Vca R

(1840 ) + ( 2760 ) + ( 2300 ) 2

31.74

= 513.33 kW 9.48 The complex power delivered to the load in terms of symmetrical components:

S3 = 3 (Va 0 I a*0 + Va1I a*1 + Va 2 I a*2 ) Substituting values from the solution of Problem 9.20, S3 = 3  47.773957.6268 (1.56221.06 ) + 112.7841 − 0.0331 ( 5.16474.0877 ) +61.623145.8825 ( 2.82228.1722 )  = 1154.7 + j1388.4 VA

The complex power delivered to the load by summing up the power in each phase:

S3 = Va I a* + Vb I b* + Vc I c* ; with values from Problem 9.25 solution,

= 20025 ( 8.706 − 51.99 ) + 100 − 155 ( 5.062136 ) +80100 ( 3.12932.04 )  = 1154.7 + j1388.4 VA 9.49 From Problem 9.6(a) solution:

Va = 1169.9 V; Vb = 41.3 − 76 V; Vc = 96.1168 V From Problem 9.5, I a = 120 A; I b = 6 − 90 A; I c = 8150 A (a) In terms of phase values S = Va I a* + Vb I b* + Vc I c*

= 1169.9 (120 ) + 41.3 − 76 ( 690 ) + 96.1168 ( 8 − 150 ) = ( 2339.4 + j 537.4 ) VA 

(b) In terms of symmetrical components:

V0 = 100 V; V1 = 8030 V; V2 = 40 − 30 V From Problem 9.6(a) I 0 = 1.82 − 21.5 A; I1 = 8.3716.2 A; I 2 = 2.81 − 36.3 From Problem 9.5 Soln.  S = 3 (V0 I 0* + V1 I1* + V2 I 2* ) = 3 100 (1.8221.5 ) + 8030 (8.37 − 16.2 ) + 40 − 30 ( 2.8136.3 )  = 3 ( 779.8 + j179.2 )

= ( 2339.4 + j 537.4 ) VA 

Chapter 9 Symmetrical Components 10.1 Calculation of per unit reactances Synchronous generators: G1

X1 = X d = 0.18

X 2 = X d = 0.18

X 0 = 0.07

X1 = X d = 0.20

X 2 = X d = 0.20

X 0 = 0.10

 13.8   1000  X1 = X d = 0.15      15   500  = 0.2539 X 2 = X d = 0.2539

 13.8   1000  X 0 = 0.05      15   500  = 0.08464 3 X n = 3 X 0 = 0.2539

 13.8   1000  X1 = X d = 0.30      15   750  = 0.3386 2

 13.8   1000  X 2 = 0.40      15   750  = 0.4514

 13.8   1000  X 0 = 0.10      15   750  = 0.1129

Transformers: XT 1 = 0.10

XT 2 = 0.10

 1000  XT 4 = 0.11  = 0.1467  750 

 1000  XT 3 = 0.12    500  = 0.24

Transmission Lines:

( 765) = 585.23  2

Z base H =

1000

PositiveNegative Sequence

Zero Sequence

50 X12 = = 0.08544 585.23

X12 =

150 585.23 = 0.2563

40 585.23 = 0.06835

100 585.23 = 0.1709

X13 = X 23 =

10.1

Per Unit Sequence Networks

j 0.1709

j 0.2563

j 0.1709

j 0.06835 j 0.08544

j 02539

j 0.06835

j 0.18 j 0.20

j 0.06835

j 0.08544

j 0.06835

j 0.2539

j 0.18 j 0.20

10.2

n = 1 (Bus 1 = Fault Bus)

Thévenin equivalents as viewed from Bus 1:

Per unit zero sequence

VF = 1.00 per unit

Per unit negative sequence

Zero sequence Thévenin Equivalent:

X0 = .10 //.5063 // (.1427 + .3376 ) X0 = 0.07114 per unit

Negative Sequence Thévenin Equivalent:

X2 = .28 //.7605 // (.04902 + .1872 ) X2 = 0.1097per unit

Similarly, X1 = .28 //.7605 // (.04902 + .1745 ) = 0.1068

10.3

Three-phase fault at bus 1. Using the positive-sequence Thévenin equivalent from Problem 10.2:

I base H = =

I1 =

Sbase 3 3 Vbase H 1000 3 ( 765 )

= 0.7547 kA

VF 1.00 = = − j 9.363per unit Z1 j 0.1068

I0 = I2 = 0

 I A  1 1     2  I B  = 1 a  I C  1 a  

1  0  9.363 − 90   a   − j 9.363 = 9.363150  per unit  9.36330  a 2  0

 I A  9.363 − 90  7.067 − 90          I B  = 9.363150   0.7547 = 7.067150  kA  I C  9.36330  7.06730   

10.4 (a) Calculation of Per Unit Line Reactances Note: The generator and transformer reactances are the same as given in the solution to Problem 10.1. Only the transmission line reactances change. Transmission Lines: ZbaseH = (500)2/1000 = 250 ohms PositiveNegative Sequence

Zero Sequence

50 X12 = = 0.08544 0.20 per unit 585.23 250

X12 =

40 585.23 250 = 0.06835

X13 = X 23 =

150 585.23 250 = 0.2563

= 0.60 per unit

100 585.23 250 = 0.1709

X13 = X 23 =

= 0.16 per unit

= 0.40 per unit

10.4 (a) Per Unit Sequence Networks

j 0.2539 j 0.18 j 0.20

j 0.2539

j 0.18 j 0.20

Note: only one impedance in the negative sequence network is different than in the positive sequence network.

(b) n = 1 (Bus 1 = fault bus) Thévenin equivalents as viewed from bus 1:

j 0.1247

j 0.1266

Zero sequence Thévenin equivalent:

X0 = 0.1 0.85

( 0.3579 + 0.5667 )

X0 = 0.08158 per unit

Negative sequence Thévenin equivalent:

0.4667

0.30

0.875

X2 = 0.25 0.9875

0.1192

0.28

0.875

0.4939 =

0.28

0.1951

0.70

0.2705

( 0.1181 + 0.2143)

X2 = 0.28// 0.875//(0.1192+0.1951) 0.1247 = 0.1266= per unit per unit

Similarly, for Positive sequence: X1 = 0.28// 0.875//(0.1192+0.1836) = 0.1247 per unit

Using the positive sequence Thévenin equivalent from Problem 10.4(b):

I base H = j 0.1247

Sbase 3 3 Vbase H

1000 3 ( 500 )

= 1.155 kA

I0 = I2 = 0 I1 =

VF 1.0 0 8.0192 unit = = − j-j8.177 perper unit Z1 jj0.1223 0.1247

 I A  1 1 1  0 8.0192 8.177  − 90          2 -j 8.0192 8.0192  I B  = 1 a a   − j8.177 = 8.177 150  per unit  1.155  I C  1 a a 2  0 8.0192 8.177 30    9.26 9.441 − 90 





= 9.441150  kA 9.26 9.26 9.44130





10.5 Calculation of per unit reactances Synchronous generators:

 1000  X1 = X d = ( 0.2 )   = 0.4  500  X 2 = X d = 0.4

 1000  X 0 = ( 0.10 )    500  = 0.20

 1000  X1 = X d = 0.18   = 0.24  750  X 2 = X d = 0.24

 1000  X 0 = 0.09    750  = 0.12

X1 = 0.17

X 0 = 0.09

X base3

( 20 ) = 0.4  =

X 2 = 0.20

1000

3Xn =

3 ( 0.028 ) 0.4

= 0.21per unit

Transformers:  1000  XT 1 = 0.12   = 0.24  500   1000  XT 2 = 0.10   = 0.1333  750 

XT 3 = 0.10

Each Line:

( 500 ) = 250  = 2

X base H

1000

X1 = X 2 = X0 =

50 = 0.20 per unit 250

150 = 0.60 per unit 250

10.6 n = 1 (Bus 1 = Fault Bus) Thévenin equivalents as viewed from Bus 1:

Zero Sequence Thévenin equivalents:

X 0 = 0.24 // .6 + (.7333 //.7 )  X 0 = .24 //.9581 = 0.1919 per unit

Positive Sequence Thévenin equivalent:

X1 = .64 // .2 + (.5733 //.47 )  X1 = .64 //.4583 X1 = 0.2670 per unit

Negative Sequence Thévenin equivalent:

X 2 = .64 // .2 + (.5733 //.50 )  X 2 = .64 //.4671 = 0.270 per unit

10.7 Three-phase fault at bus 1. Using the positive-sequence Thévenin equivalent from Problem 10.5:

I base H =

1000

500 3 = 1.155 kA

I1 =

VF 1.00 = = 3.745 − 90 per unit Z1 j 0.267

I0 = I2 = 0

 I A  1 1     2  I B  = 1 a  I C  1 a  

1  0  3.745 − 90    a  3.745 − 90 = 3.745150  per unit  3.74530  a 2  0

 I A  3.745 − 90   4.325 − 90          I B  = 3.745150   1.155 =  4.325150  kA  I C  3.74530   4.32530   

10.8 Calculation of per unit reactances Synchronous generators: 2

 12   100  X1 = X d = 0.2      10   50  X1 = 0.576 per unit

 12   100  X 0 = ( 0.1)      10   50  X 0 = 0.288 per unit

X 2 = X1 = .576 per unit

X1 = X d = 0.2

X 0 = 0.1

X 2 = 0.23

Transformers:

 100  XT 1 = 0.1  = 0.2 per unit  50  XT 2 = 0.1per unit

Each line:

(138 ) = 190.44  = 2

Z base H

100

40 = 0.210 per unit 190.44 100 X0 = = 0.5251per unit 190.44 X1 = X 2 =

10.9 n = 1 (Bus 1 = Fault Bus) Thévenin equivalents as viewed from Bus 1:

Zero Sequence Thévenin equivalent:

Z 0 = j1.3502 per unit

Positive Sequence Thévenin equivalent:

Z1 = j.576 // j.92 = j 0.3542 per unit

Negative Sequence Thévenin equivalent:

Z 2 = j.576 // j.95 = j 0.3586 per unit

10.10 Three-phase fault at bus 1. Using the positive-sequence Thévenin equivalent from Problem 10.9:

I base1 =

100

= 5.774 kA 10 3 V 1.00 I1 = F = = 2.823 − 90 per unit Z1 j.3542

I0 = I2 = 0

 I A  1 1  I   = 1 a 2  B   I C  1 a

1  0  2.823 − 90   a  2.823 − 90 = 2.823150  per unit  2.82330  a 2  0

 I A  2.823 − 90 16.30 − 90  I   = 2.823150   5.774 = 16.30150  kA  B      IC  2.82330  16.3030 

10.11 (a) The positive sequence network and steps in its reduction to its Thévenin equivalent are shown below:

The negative sequence network and its reduction is shown below:

The zero-sequence network and its reduction are shown below:

(b)

For a balanced 3-phase fault, only positive sequence network comes into picture. I sc = I a = I a1 =

10 = 3.23 − 90 j ( 0.26 + 0.05 )

I sc = 3.23pu

10.12 The zero, positive, and negative sequence networks are shown below:

Using delta-wye transformation and series-parallel combinations, Thévenin equivalents looking into bus 3 are shown below:

For a bolted 3-phase fault at bus 3, V1 = 0; Also V2 = V0 = 0 I 0 = 0; I 2 = 0

I1 =

1  0 = − j 5.71 j 0.175

The fault current is 5.71 pu.  I a  1 1    2  I b  = 1 a  I c  1 a  

1  0  5.71 − 90   a   − j 5.71 = 5.71150   5.7130  a 2  0

10.13 For a balanced 3-phase fault at bus 3, we need the positive sequence impedance network reduced to its Thévenin equivalent viewed from bus 3. The development is shown below:

Convert the  formed by buses 1, 2 and 3 to an equivalent Y Z1X =

( j 0.125)( j 0.15) = j 0.0357143

Z2X =

( j 0.125)( j 0.25) = j 0.0595238

Z3 X =

( j 0.15)( j 0.25) = j 0.0714286

j 0.525 j 0.525

j 0.525

Using series-parallel combinations, the positive sequence Thévenin impedance is given by, viewed from bus 3:

( j 0.2857143 )( j 0.3095238 ) + j 0.0714286 j 0.5952381

= j 0.1485714 + j 0.0714286 = j 0.22

With the no-load generated EMF to be 10 pu, the fault current is given by (with Z F = j 0.1 ) I a = I a1 =

1.00 j 0.22 + j 0.1

= − j 3.125pu = 820.1 − 90 A

The fault current is 820.1 A. 10.14 Bolted single-line-to-ground fault at bus 1. I 0 = I1 = I 2 =

 I A  1 1     2  I B  = 1 a  I C  1 a  

VF Z 0 + Z1 + Z 2

1.00 j ( 0.07114 + 0.1068 + 0.1097 )

1.00 = − j 3.4766 per unit j 0.2876

1   − j 3.4766   − j10.43  − j 7.871      a   − j 3.4766  =  0  per unit =  0  kA  0  a 2   − j 3.4766   0 

Using Eq. (10.1): V0   0   j 0.07114      0 V1  = 1.00 −  V2   0   0  

0 j 0.1068 0

  − j 3.4766  0   − j 3.4766  j 0.1097   − j 3.4766  0

V0   −0.2473      V1  =  0.6287  per unit V2   −0.3814    VAg  1 1    2 VBg  = 1 a   1 a VCg  

1   −0.2473   0      a   0.6287  = 0.9502247.0 per unit a 2   −0.3814   0.9502113.0 

10.15 Single-Line-to Ground Arcing Fault at Bus 1.

( 765) = 585.2  = 2

Z base H

1000 150 ZF = = 0.025630 per unit 585.2 VF I 0 = I1 = I 2 = Z 0 + Z1 + Z 2 + 3 Z F =

1.00 = 3.359 − 75.03 per unit 0.0769 + j 0.2876

 I A  1 1    2  I B  = 1 a  I C  1 a  

1  3.359 − 75.03  10.0771 − 75.03   per unit a  3.359 − 75.03  =  0   a 2  3.359 − 75.03   0

7.605 − 75.03   kA =  0    0

V0   0   j 0.07114      0 V1  = 1.00  −  V2   0   0  

0 j 0.1068 0

0  3.359 − 75.03  0  3.359 − 75.03  j 0.1097  3.359 − 75.03 

V0   0.2390 − 165.03      V1  = 0.6600 − 8.0718  per unit V2   0.3685 − 165.03    VAg  1 1    2 VBG  = 1 a VCG  1 a  

1   0.2390 − 165.03   0.2583 − 75.06  a  0.6600 − 8.0718  =  0.9227 − 114.14  per unit a 2   0.3685 − 165.03   0.9835112.84 

10.16 Bolted Line-to-Line Fault at Bus 1. VF = 1.0 0

I1 = − I 2 = =

VF Z1 + Z 2

1.00 j.2165

= − j 4.619 per unit  I A  1 1     2  I B  = 1 a  I C  1 a  

1  0   0 0          a   − j 4.619  = 8.000180 per unit = 6.038180 kA  6.0380  a 2   + j 4.619   8.0000 

V0   0   j.07114      V1  = 10 −  0 V2   0   0  

VAg  1 1    2 VBg  = 1 a   1 a VCg  

0 j 0.1068 0

 0  0  0   − j 4.619  0.5067 per unit j 0.1097   − j 4.619  0.5067 0

1   0  1.0130      a  0.5067  = 0.5067180 per unit a 2  0.5067  0.5067180

10.17 Bolted double-line-to-ground fault at bus 1. VF = 1.00

I1 =

VF Z1 + Z 2 // Z 0

1.00 j (.1068 + .1097 //.0711)

1.00 j 0.14995

= 6.669 − 90 per unit  Z0   .07114  I 2 = − I1   = j 6.669    .18084   Z0 + Z2  I 2 = j 2.623per unit  Z2   .1097  I 0 = − I1   = j 6.669   = j 4.046 per unit  .18084   Z0 + Z2   I A  1 1     2  I B  = 1 a  I C  1 a  

1   j 4.046   0 0          a   − j 6.669  = 10.08143.0  per unit = 7.607143 kA  7.60737  a 2   j 2.623  10.0837.02

V0   0   j 0.07114      0 V1  = 10 −  V2   0   0  

VAg  1 1    2 VBg  = 1 a   1 a VCg  

0 j 0.1068 0

  j 4.046   0.2878 0   − j 6.669 =  0.2878 per unit j 0.1097   j 2.623   0.2878 0

1  0.2878  0.8633 a  0.2878  =  0  per unit a 2  0.2878   0 

10.18

The zero sequence network is the same as in Problem 10.1. The  − Y transformer phase shifts have no effect on the fault currents and no effect on the voltages at the voltages at the fault bus. Therefore, from the results of Problem 10.10:  I A   − j10.43  − j 7.871         I B  =  0  per unit =  0  kA  I C   0   0   

VAg   0      VBg  = 0.9502247.0 per unit    0.9502113.0   VCg  

Contributions to the fault from generator 1: From the zero-sequence network: I G1− 0 = 0 From the positive sequence network, using current division:  .1727  I G1−1 = ( − j 3.4766 )    − 30  .28 + .1727  = 1.326 − 120 per unit

From the negative sequence network, using current division:

 .1802  I G1− 2 = ( − j 3.4766 )    + 30  .28 + .1802  = 1.3615 − 60 per unit

Transforming to the phase domain:  I G1− A  1 1     2  I G1− B  = 1 a  I G1−C  1 a  

1  0  2.328 − 89.6   a  1.326 − 120 = 2.32889.6  per unit a 2   1.326 − 60  0.036180  1.757 − 89.6 = 1.75789.6  kA 0.027180 

10.19 (a) Bolted single-line-to-ground fault at bus 1.

I 0 = I1 = I 2 = =

VF Z 0 + Z1 + Z 2

1.0 0 1.0 0 = j (.08158 + .1223 + .1247 ) j 0.3286

= − j 3.043 per unit  I A  1 1     2  I B  = 1 a  I C  1 a  

1   − j 3.043  − j 9.130  a   − j 3.043 =  0  per unit a 2   − j 3.043  0   − j10.54  =  0  kA  0 

Using Eq. (10.1): V0   0   j.08158       V1  = 1.0 0  −  0 V2   0   0  −0.2483 =  0.6278  per unit  −0.3795 VAg  1 1    2 VBg  = 1 a VCg  1 a  

0 j.1223 0

0   − j 3.043 0   − j 3.043 j.1247   − j 3.043

1   −.2483  0       a   .6278  = 0.9485 246.9 per unit a 2   −.3795  0.9485113.1 

(b) Single-line-to-ground arcing fault at bus 1.

( 500 ) = 250  = 2

Z base H

1000 15 ZF = = 0.06 0 per unit 250 VF I 0 = I1 = I 2 = Z 0 + Z1 + Z 2 + 3 Z F =

1.0 0 1.0 0 = 0.18 + j 0.3286 0.374761.29

= 2.669  − 61.29 per unit  I A  1 1     2  I B  = 1 a  I C  1 a  

1  2.669  − 61.29 8.007  − 61.29  per unit 1.155 a   2.669  − 61.29 =  0   a 2   2.669  − 61.29  0 9.246  − 61.29   kA =  0    0

V0   0   j.08158      V1  = 1.0 0  −  0 V2   0   0  

0 j.1223 0

 2.669  − 61.29 0   2.669  − 61.29  j.1247   2.669  − 61.29 0

0.2177 208.71 =  0.7307  − 12.9  per unit 0.3328 208.71

VAg  1 1    2 VBg  = 1 a   1 a VCg  

1  .2177 208.71 0.4804  − 61.29 a   .7307  − 12.4  = 0.9094 244.0  per unit  a 2   .3328208.71  1.009 113.6

(c)

I1 = − I 2 = =

VF Z1 + Z 2

1.0 0 = − j 4.0486 j ( 0.1223 + 0.1247 )

 I A  1 1     2  I B  = 1 a  I C  1 a  

1  0 0        a   − j 4.0486  =  7.012  180 per unit 1.155 a 2   + j 4.0486   7.012 0   0  =  −8.097  kA  8.097 

V0   0   j.08158       V1  = 1.0 0 −  0 V2   0   0

0 j.1223 0

0  0    0   − j 4.0486  j.1247   + j 4.0486

 0  = 0.5049  per unit 0.5049 

VAg  1 1    2 VBg  = 1 a VCg  1 a  

1   0  1.010 0      a  .5049  =  0.5049 180 per unit a 2  .5049   0.5049 180

(d)

I1 =

VF Z1 + Z 2 // Z 0

1.0 0 j (.1223 + .1247//.0816 )

1.0 0 j 0.1716

= − j 5.827 per unit  Z0   .08158  I 2 = − I1   = j 5.827    .2063   Z0 + Z2  I 2 = j 2.304 per unit  Z2   .1247  I 0 = − I1   = j 5.827   = j 3.523 per unit  .2063   Z0 + Z2 

 I A  1 1     2  I B  = 1 a  I C  1 a  

1   j 3.523   0      a   − j 5.827 = 8.804 143.1 per unit 1.155 a 2   j 2.304   8.30436.9  0    = 10.17143.1 kA 10.1736.9

V0   0   j 0.08158      0 V1  = 1.00  −  V2   0   0  

0 j.1223 0

  j3.523    0  − j 5.827 j.1247  j 2.304  0

0.2874   = 0.2874 per unit 0.2874 VAg  1 1    VBg  = 1 a 2   1 a VCg  

1  .2874 0.86220      a  .2874 =  0  per unit 2   a  .2874  0

(e)

Per unit positive sequence network

Per unit negative sequence network The per unit zero sequence network is the same as in Problem 10.1 244 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

The  − Y transformer phase shifts have no effect on the fault currents and no effect on the voltages at the fault bus. Therefore, from the results of Problem 10.19(a), VAg   I A  − j 9.130 0 − j10.54            I B =  0  per unit=  0  kA VBg  = 0.9485246.90 per unit   0.9485113.1  IC   0   0     VCg  

Contribution to the fault current from generator 1: From the zero-sequence circuit: I G1− 0 = 0 From the positive sequence circuit, using current division:  .2393  IG1−1 = (− j 3.043)    − 30   .25 + .2393  = 1.488 − 120  per unit Transformer phase shift

From the negative sequence circuit, using current division:  .2487  I G1− 2 = (− j 3.043)    + 30   .25 + .2487  = 1.518 − 60  per unit Transformer phase shift

Transforming to the phase domain: I G1− A  1 1     2 IG1− B  = 1 a I G1−C  1 a

1 0  2.603 − 89.7    a  1.488 − 120  =  2.60389.7   per unit a 2   1.518 − 60    0.03180   3.00687.7  = 3.00687.7  kA  0.035180  

10.20

I base H =

Sbase 3 Vbase H

500 3 ( 500 )

= 0.5774 kA

Three- phase fault: I 0 = I 2 = 0 I1 =

VF 1.00 = Z1 j 0.30

= − j 3.333per unit

I a = I1 = − j 3.333per unit = − j1.925 kA

Single line-to-ground fault: I 0 = I1 = I 2 =

VF 1.00 = = − j1.429 per unit Z 0 + Z1 + Z 2 j ( 0.1 + 0.3 + 0.3)

I a = 3I 0 = − j 4.286 per unit = − j 2.474 kA

Line-to-line fault: I 0 = 0 I1 = − I 2 =

VF 1.00 = = − j1.667 per unit Z1 + Z 2 j ( 0.3 + 0.3 )

I b = ( a 2 − a ) I1 = ( a 2 − a ) ( − j1.667 ) = 2.887180 per unit I b = 1.667180 kA

Double line-to-ground fault: I1 =

VF 1.00 1.0 = = = − j 2.667 per unit Z1 + Z 2 Z 0 j ( 0.3 + 0.3 0.1) j 0.375

 Z0   .1  I 2 = − I1   = ( j 2.667 )   = j 0.667 per unit  .4   Z0 + Z2   Z2   .3  I 0 = − I1   = ( j 2.667 )   = j 2.0 per unit  .4   Z0 + Z2  I B = I c + a 2 I1 + aI 2 = 2.090 + 2.667150 + .667210 = 4.163134 per unit = 2.404134 kA

10.21 Bolted single-line-to-ground fault at bus 1. I 0 = I1 = I 2 = =

VF Z 0 + Z1 + Z 2

1.00 = − j1.372 per unit j ( 0.1919 + 0.267 + 0.27 )

 I A  1 1     2  I B  = 1 a  I C  1 a  

1   − j1.372  a   − j1.372  a 2   − j1.372 

 I A   − j 4.116   − j 4.753         I B  =  0  per unit =  0  kA  I C   0   0   

Contributions to the fault current Zero sequence:

 .9581  Transformer: I T − 0 = − j1.372    .24 + .9581  = − j1.097 per unit .24   Line: I L −0 = − j1.372   = − j 0.2748  .24 + .9581 

Positive sequence:

 .4583  Transformer: I T −1 = − j1.372    .64 + .4583  = − j 0.5725 .64   Line: I L −1 = − j1.372   = − j 0.7994  .64 + .4583 

Negative sequence:

 .4671  Transformer: I T − 2 = − j1.372    .64 + .4671  = − j 0.5789 per unit .64   Line: I L − 2 = j1.372   = − j 0.7931per unit  .64 + .4671 

Transformer:  IT− A  1 1  I   = 1 a 2  T −B    IT−C  1 a

1   − j1.097   − j 2.248   − j 2.596       a   − j 0.5725 = .521 − 89.4  per unit = .602 − 89.4  kA .602 − 90.6 a 2   − j 0.5789 .521 − 90.6

Line:  I L− A  1 1  I   = 1 a 2  L−B    I L−C  1 a

1   − j 0.2748   − j1.8673   − j 2.156       a   − j 0.7994  = .52190.6 per unit = .60290.6 kA .60289.4  a 2   − j 0.7931 .52189.4 

10.22 Bolted line-to-line fault at bus 1. I1 = − I 2 = =

VF Z1 + Z 2

1.00 j (.2670 + .270 )

= − j1.862 per unit

 I A  1 1     2  I B  = 1 a  I C  1 a  

1  0   0      a   − j1.862  = 3.225180 per unit a 2   + j1.862   3.2250 

 I A   0 0            I B  = 3.225180  1.155 = 3.724180  kA  I C   3.2250   3.7240   

Contributions to the fault current Zero sequence:

IT −0 = I L −0 = 0 Positive sequence:

 .4583  Transformer: I T −1 = − j1.862    .64 + .4583  = − j 0.7770 per unit .64   Line: I L −1 = − j1.862   = − j1.085per unit  .64 + .4583 

Negative sequence:

 .4671  Transformer I T − 2 = j1.862    .64 + .4671  = j 0.7856 per unit

.64   Line I L − 2 = j1.862   .64 + .4671   = j1.076 per unit Contribution to fault from transformer:  I T− A  1 1     2  I T − B  = 1 a  I T−C  1 a  

1  0   j.0086   0.009990       a   − j.777 = 1.353180.2 per unit = 1.562180.2 kA 1.562 − 0.2  a 2   j.7856  1.353 − 0.2 

Contribution to fault from line:  I L− A  1 1     2  I L − B  = 1 a  I L−C  1 a  

1   0   − j 0.0086   0.0099 − 90      a   − j1.085 = 1.871179.9 per unit =  2.160179.9  kA  2.1600.1  a 2   j1.076   1.8710.1 

10.23 Bolted double-line-to-ground fault at bus 1. VF = 1.00

I1 =

VF Z1 + Z 2 Z 0

I1 =

1.00 j (.267 + .27 .1919 )

1.00 j 0.3792

= − j 2.637 per unit  Z0   .1919  I 2 = − I1   = j 2.637    .27 + .1919   Z0 + Z2  I 2 = j1.096 per unit  Z2  .27   I 0 = − I1   = j 2.637    .27 + .1919   Z0 + Z2  I 0 = j1.541per unit  I A  1 1     2  I B  = 1 a  I C  1 a  

1   j1.541   0      a   − j 2.637  =  3.975144.3  per unit 1.155 a 2   j1.096  3.97535.57 0    =  4.590144.3  kA  4.59035.57

Contributions to the fault current Zero sequence:

 .9581  Transformer: I T − 0 = j1.541   .24 + .9581  = j1.232 per unit .24   Line: I L −0 = j1.541  = j 0.3087 per unit .24 + .9581  

Positive sequence:  .4583  I T −1 = ( − j 2.637 )   = − j1.100  .64 + .4583  .64   I L −1 = ( − j 2.637 )   = − j1.537 per unit  .64 + .4583 

Negative Sequence:  .4671  I T − 2 = ( j1.096 )   = j 0.4624  .64 + .4671  .64   I L − 2 = ( j1.096 )   = j 0.6336 .64 + .4671  

Contribution to fault from transformer:  I T− A  1 1     2  I T − B  = 1 a  I T−C  1 a  

1   j1.232   j 0.5944      a   − j1.10  =  2.058131.1  per unit 1.155 a 2   j 0.4624   2.05848.90  0.686490  =  2.376131.1  kA 2.37648.90

Contributions to fault from line:  I L− A  1 1     2  I L − B  = 1 a  I L−C  1 a  

1   j.3087   − j.594  a   − j1.537  = 2.028158.0 per unit 1.155 a 2   j.6336   2.02822.0  0.686 − 90  =  2.342158  kA  2.34222 

10.24 Bolted single-line-to-ground fault at bus 1. VF = 1.00

I 0 = I1 = I 2 = =

VF Z 0 + Z1 + Z 2

1.00 j (1.3502 + 0.3542 + 0.3586 )

= − j 0.4847 per unit I base 1 =

 I A  1 1     2  I B  = 1 a  I C  1 a  

100 10 3

= 5.774 kA

1   − j.4847  − j1.454   − j8.396       a   − j.4847 =  0  per unit 5.774 =  0  kA  0  a 2   − j.4847  0 

Contributions to fault current Zero sequence: Generator G1 I G1− 0 = 0 Transformer T1 I T 1− 0 = − j 0.4847 per unit Positive sequence:

 .92  Generator G1 I G1−1 = − j.4847    .92 + .576  = − j 0.2981per unit  .576  Transformer T1 IT 1−1 = − j.4847   = − j 0.1866 per unit  .92 + .576 

Negative sequence:

 .95  Generator G1 I G1− 2 = − j.4847    .95 + .576  = − j 0.3017 per unit  .576  Transformer T1 I T 1− 2 = − j.4847    .576 + .95  = − j.1830 Contribution to fault from generator G1:  I G1− A  1 1     2  I G1− B  = 1 a  I G1−C  1 a  

1  0   − j 0.5998      a   − j.2981  =  0.299989.4  per unit 5.774 a 2   − j.3017   0.299990.6 3.463 − 90  =  1.73189.4  kA  1.73190.6 

Contribution to fault from transformer T1:  I T1− A  1 1     2  I T 1− B  = 1 a  I T1−C  1 a  

1   − j.4847   − j.8543      a   − j.1866  = 0.2999 − 90.6 per unit 5.774 a 2   − j.1830   0.2999 − 89.4   4.932 − 90  = 1.731 − 90.6  kA 1.731 − 89.4 

10.25 Arcing single-line-to-ground fault at bus 1. VF = 1.00

VF Z 0 + Z1 + Z 2 + 3 Z F

I 0 = I1 = I 2 = =

1.00 j ( 0.3 + 1.3502 + .3542 + .3586 )

= 0.4232 − 90 per unit  I A  1 1    2  I B  = 1 a  I C  1 a  

1  0.4232 − 90 1.270 − 90  per unit 5.774 a  0.4232 − 90 =  0   a 2  0.4232 − 90  0 7.331 − 90   kA =  0    0

Contributions to fault current Zero sequence: Generator G1: I G1− 0 = 0 Transformer T1: IT 1−0 = 0.4232 − 90 per unit Positive sequence:

 .92  Generator G1: I G1−1 = .4232 − 90    .92 + .576  = 0.2603 − 90  .576  Transformer T1: IT 1−1 = .4232 − 90    .92 + .576  = 0.1629 − 90 per unit Negative sequence: 252 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

 .95  Generator G1: I G1− 2 = .4232 − 90    .576 + .95  = 0.2635 − 90  .576  Transformer T1: IT 1− 2 = .4232 − 90    .576 + .95  = 0.1597 − 90 per unit Contribution to fault from generator G1:  I G1− A  1 1    2  I G1− B  = 1 a  I G1−C  1 a  

1  0   .5238 − 90    a   0.2603 − 90  = .261989.40  per unit 5.774 a 2  0.2635 − 90 .261990.61  3.0244 − 90  = 1.512389.40  kA 1.512390.61

Contribution to fault from generator T1:  IT1− A  1 1    2  IT1− B  = 1 a  IT1−C  1 a  

1  .4232 − 90   0.7458 − 90  a  .1629 − 90  = 0.2916 − 90.61  per unit 5.774 a 2  .1597 − 90  0.2916 − 89.40   4.3063 − 90  = 1.5123 − 90.61  kA 1.5123 − 89.40 

10.26 Bolted line-to-line fault at bus 1. I1 = − I 2 = =

VF Z1 + Z 2

1.00 j (.3542 + .3586 )

= − j1.403per unit

 I A  1 1     2  I B  = 1 a  I C  1 a  

1  0   0      a   − j1.403 = 2.430180 per unit 5.744 a 2   + j1.403  2.4300  0    = 14.03180  kA  14.030 

Contributions to fault current. Zero sequence: Generator G1: I G1− 0 = 0 Transformer T1: I T 1− 0 = 0 Positive sequence:

 .92  Generator G1: I G1−1 = − j1.403    .92 + .576  = − j 0.8628 per unit  .576  Transformer T1: I T 1−1 = − j1.403   = − j 0.5402  .576 + .92 

Negative sequence:

 .95  Generator G1: I G1− 2 = ( j1.403 )    .95 + .576  = j 0.8734 per unit  .576  Transformer T1: I T 1− 2 = ( j1.403 )    .576 + .95  = j 0.5296 per unit Contribution to fault from generator G1:  I G1− A  1 1     2  I G1− B  = 1 a  I G1−C  1 a  

1  0   j 0.0106      a   − j.8628  = 1.504 − 179.8 per unit 5.774 a 2   j.8734   1.504 − 0.2   0.061290  = 8.683 − 179.8  kA  8.683 − 0.2 

Contribution to fault from transformer T1:  I T1− A  1 1     2  I T 1− B  = 1 a  I T1−C  1 a  

1   0   − j 0.0106  a   − j.5402  = 0.9265179.7 per unit 5.774 a 2   j.5296   0.92650.33  0.0612 − 90 =  5.349179.7  kA  5.3490.33 

I1 =

VF Z1 + Z 2 Z 0

1.00 j (.3542 + .3586 1.3502 )

1.00 j 0.6375

= − j1.569 per unit  Z2  .3586   I 0 = − I1   = j1.569    .3586 + 1.3502   Z2 + Z0  = j 0.3292 per unit  Z0  1.3502   I 2 = − I1   = j1.569    1.3502 + .3586   Z0 + Z2  = j1.2394 per unit  I A  1 1     2  I B  = 1 a  I C  1 a  

1   j 0.3292   0      a   − j1.569  =  2.482168.5 per unit 5.774 a 2   j1.2394   2.48211.48 0    = 14.33168.5  kA 14.3311.48

Contributions to fault current. Zero sequence: Generator G1: I G1− 0 = 0 Transformer T1: I T 1−0 = j 0.3292 per unit Positive sequence:

 .92  Generator G1: I G1−1 = − j1.569    .92 + .576  = − j 0.9646 per unit  .576  Transformer T1: I T 1−1 = − j1.569    .576 + .92  = − j 0.6039 per unit

 .95  Generator G1: I G1− 2 = j1.2394    .95 + .576  = j 0.7716 per unit  .576  Transformer T1: IT 1−2 = j1.2394   = j 0.4678 per unit  .576 + .95 

Contribution to fault from generator G1:  I G1− A  1 1     2  I G1− B  = 1 a  I G1−C  1 a  

1   0   − j 0.193  a   − j.9646  = 1.507176.3 per unit 5.774 a 2   j.7716   1.5073.67   1.114 − 90  = 8.701176.3 kA  8.7013.67 

Contribution to fault from transformer T1:  I T1− A  1 1     2  I T 1− B  = 1 a  I T1−C  1 a  

1   j 0.3292   j 0.1931  a   − j 0.6039  = 1.010156.8  per unit 5.774 a 2   j 0.4678  1.01023.17  1.11490  =  5.831156.8  kA 5.83123.17

10.28 I a = ( I 0 + I1 + I 2 ) = 0 I a = ( I 0 + I1 + I 2 ) = 0 Also Vbb = Vcc = 0, or V00  1 1 V  = 1 1 a  11  3  V22  1 a 2

1  Vaa  Vaa / 3     a 2   0  = Vaa / 3 a   0  Vaa / 3

Which gives V00 = V11 = V22 = 0

10.29 I b = I c = 0, or I0  1 1   1  I1  = 3 1 a I2  1 a 2  

1   I a   I a / 3     a 2   0  =  I a / 3  I 0 = I1 = I 2 a   0   I a / 3

Similarly I 0 = I1 = I 2 Also Vaa = (V00 + V11 + V22 ) = 0

10.30 (a) For a single line-to-ground fault, the sequence networks from the solution of Problem 10.10 are to be connected in series.

The sequence currents are given by I 0 = I1 = I 2 =

1 = 1.65 − 90 pu j ( 0.26 + 0.2085 + 0.14 )

The subtransient fault current is

I a = 3 (1.65 − 90 ) = 4.95 − 90 pu Ib = Ic = 0 The sequence voltages are given by Eq. (10.1.1): V1 = 10 − I1 Z1 = 10 − (1.65 − 90 )( 0.2690 ) = 0.57 pu V2 = − I 2 Z 2 = − (1.65 − 90 )( 0.208590 ) = −0.34 pu V0 = − I 0 Z 0 = − (1.65 − 90 )( 0.1490 ) = −0.23pu

The line-to-ground (phase) voltages at the faulted bus are Vag  1 1    2 Vbg  = 1 a   1 a Vcg  

1   −0.23   0      a   0.57  = 0.86 − 113.64 pu a 2   −0.34   0.86113.64 

(b) For a line-to-line fault through a fault impedance Z F = j 0.05, the sequence network connection is shown below:

I1 = − I 2 =

10 = 1.93 − 90 pu 0.518590

I0 = 0

The phase currents are given by (Eq. 8.1.20  8.1.22)

I a = 0; I b = −I c = ( a2 − a ) I1 = 3.34 − 180 pu The sequence voltages are

V1 = 10 − I1 Z1 = 10 − (1.93 − 90 )( 0.2690 ) = 0.5pu V2 = − I 2 Z 2 = − ( −1.93 − 90 )( 0.208590 ) = 0.4 pu V0 = − I 0 Z 0 = 0 The phase voltages are then given by Va = V1 + V2 + V0 = 0.9 pu Vb = a 2V1 + aV2 + V0 = 0.46 − 169.11 pu Vc = aV1 + a 2V2 + V0 = 0.46169.11 pu Check: Vb − Vc = I b Z F = 0.17 − 90

The reductions are shown below:

 I1 = 10 / 0.4590 = 2.24 − 90 0.29   I 2 = − I1   = −1.18 − 90  0.29 + 0.2585  I 0 = −1.06 − 90

The sequence voltages are given by V1 = 10 − I1 Z1 = 10 − ( 2.24 − 90 )( 0.2690 ) = 0.42 V2 = − I 2 Z 2 = − ( −1.18 − 90 )( 0.208590 ) = 0.25 V0 = − I 0 Z 0 = − ( −1.06 − 90 )( 0.1490 ) = 0.15

The phase currents are calculated as I a = 0; I b = a 2 I1 + aI 2 + I 0 = 3.36151.77; I c = aI1 + a 2 I 2 + I 0 = 3.3628.23.

The neutral fault current is I b + I c = 3I 0 = −3.18 − 90. The phase voltages are obtained as Va = V1 + V2 + V0 = 0.82 Vb = a 2V1 + aV2 + V0 = 0.24 − 141.49 Vc = aV1 + a 2V2 + V0 = 0.24141.49

10.31 (a) For a single line-to-ground fault at bus 3, the interconnection of the sequence networks is shown below. (See solution of Problem 10.12)

I 0 = I1 = I 2 =

 I a  1 1    2  I b  = 1 a  I c  1 a  

10 = − j1.82 j ( 0.199 + 0.175 + 0.175 )

1   − j1.82   − j 5.46  a   − j1.82  =  0  a 2   − j1.82   0 

Sequence voltages are given by V0 = − j 0.199 ( − j1.82 ) = −0.362; V1 = 1 − j 0.175 ( − j1.82 ) = 0.681; V2 = − j 0.175 ( − j1.82 ) = −0.319

The phase voltages are calculated as Va  1 1    2 Vb  = 1 a Vc  1 a  

1   −0.362   0      a   0.681  = 1.022238 a 2   −0.319  1.022122 

(b) For a line-to-line fault at bus 3, the sequence networks are interconnected as shown below:

I1 = − I 2 =

10 = − j 2.86 j 0.175 + j 0.175

Phase currents are then  I a  1 1    2  I b  = 1 a  I c  1 a  

1  0   0  a   − j 2.86  =  −4.95 a 2   j 2.86   4.95 

The sequence voltages are

V0 = 0; V1 = V2 = I1 ( j 0.175 ) = 0.5 Phase voltages are calculated as Va  1 1    2 Vb  = 1 a Vc  1 a  

1   0   1.0  a   0.5 =  −0.5 a 2   0.5  −0.5

Sequence currents are calculated as I1 =

10 = − j 3.73 j 0.175 +  j 0.175 ( j 0.199 ) / ( j 0.175 + j 0.199 ) 

0.199 ( j3.73) = j1.99 0.175 + 0.199 0.175 I0 = ( j3.73) = j1.75 0.175 + 0.199 I2 =

Phase currents are given by  I a  1 1    2  I b  = 1 a  I c  1 a  

1   j1.75   0      a   − j 3.73 =  5.6152.1 a 2   j1.99   5.627.9 

The neutral fault current is I b + I c = 3I 0 = j 5.25 Sequence voltages are obtained as

V0 = V1 = V2 = − ( j1.75 )( j 0.199 ) = 0.348 Phase voltages are then Va  1 1    2 Vb  = 1 a Vc  1 a  

1  0.348  1.044  a   0.348  =  0  a 2   0.348   0 

(d) In order to compute currents and voltages at the terminals of generators G1 and G2, we need to return to the original sequence circuits in the solution of Prob. 10.12. Generator G1 (Bus 4): For a single line-to-ground fault, sequence network interconnection is shown below:

From the solution of Prob. 10.31(a), I f = − j1.82 From the circuit above, I1 = I 2 =

1 I f = − j 0.91 2

Transforming the  of (j0.3) in the zero-sequence network into an equivalent Y of (j0.1), and using the current divider, I0 =

0.15 ( − j1.82 ) = − j 0.62 0.29 + 0.15

Phase currents are then  I a  1 1    2  I b  = 1 a  I c  1 a  

1   − j 0.62   2.44 − 90 a   − j 0.91 =  0.2990  a 2   − j 0.91  0.2990 

Sequence voltages are calculated as V0 = − ( − j 0.62 )( j 0.14 ) = −0.087; V1 = 1 − j 0.2 ( − j 0.91) = 0.818; V2 = − j 0.2 ( − j 0.91) = −0.182

Phase voltages are then Va  1 1    2 Vb  = 1 a Vc  1 a  

1   −0.087   0.5490  a   0.818  = 0.956245 a 2   −0.182   0.956115 

Generator G2 (Bus 5): From the interconnected sequence networks and solution of Prob. 10.31, I f = − j1.182; I1 = I 2 =

1 I f = − j 0.91; I 0 = 0 2

Recall that Y –  transformer connections produce 30 phase shifts in sequence quantities. The HV quantities are to be shifted 30 ahead of the corresponding LV quantities for positive sequence, and vice versa for negative sequence. One may however neglect phase shifts. Since bus 5 is the LV side, considering phase shifts, I1 = 0.91 − 90 − 30 = 0.91 − 120; I 2 = 0.91 − 90 + 30 = 0.91 − 60

Phase currents are then given by  I a  1 1    2  I b  = 1 a  I c  1 a  

1  0  1.58 − 90    a  0.91 − 120 = 1.58 + 90  a 2   0.91 − 60   0

Positive and negative sequence voltages are the same as on the G1 side:

V1 = 0.818; V2 = −0.182; V0 = 0; With phase shift V1 = 0.818 − 30 V2 = 0.182210

Phase voltages are calculated as Va  1 1    2 Vb  = 1 a Vc  1 a  

1  0  0.744 − 42.2   a  0.818 − 30 =  0.744222.2  a 2   0.182210   1.0090 

10.32 (a) Refer to the solution of Problem 10.13. The negative sequence network is the same as the positive sequence network without the source.

The zero-sequence network is shown below considering the transformer winding connections:

For the single line-to-ground fault At bus 3 through a fault impedance Z F = j 0.1 , I 0 = I1 = I 2 =

10 j ( 0.22 + 0.22 + 0.35 + 0.3 )

= − j 0.9174

Fault currents are  I a  1 1    2  I b  = 1 a  I c  1 a  

1   I 0   − j 2.7523    a   I1  =  0   a 2   I 2   0

(b) For a line-to-fault at bus 3 through a fault impedance of j0.1,

I0 = 0 I1 = − I 2 =

1 = − j1.8519 j ( 0.22 + 0.22 + 0.1)

Fault currents are then  I a  1 1    2  I b  = 1 a  I c  1 a  

1  0   0    a   − j1.8519 =  −3.2075 a 2   j1.8519   3.2075 

I1 =

10 j 0.22 ( j 0.35 + j 0.3 ) j 0.22 + j 0.22 + j 0.35 + j 0.3

= − j 2.6017

I2 =

1 − ( j 0.22 )( − j 2.6017 )

I0 = −

j 0.22

= j1.9438

1 − ( j 0.22 )( − j 2.6017 ) j 0.35 + j 0.3

= j 0.6579

Fault phase currents are then  I a  1 1    2  I b  = 1 a  I c  1 a  

1   j 0.6579   0      a   − j 2.6017  =  4.058165.93° a 2   j1.9438   4.05814.07° 

Neutral fault current at bus 3 = I b + I c = 3I 0 = 1.973290°

10.33 Positive sequence network of the system is shown below:

Negative sequence network is same as above without sources. The zero sequence network is shown below:

Using any one of the methodsalgorithms, sequence Z BUS can be obtained. (1) (1)  j 0.1437 (2)  j 0.1211 Z BUS1 = Z BUS 2 = (3)  j 0.0789  (4)  j 0.0563 (1) (1)  j 0.16 (2)  0 Z BUS 0 = (3)  0  (4)  0

(2)

(3)

(4)

j 0.1211

j 0.0789

j 0.1696

j 0.1104

j 0.1696

j 0.0789

j 0.1211

j 0.0563  j 0789  j 0.1211  j 0.1437 

(2)

(3)

(4)

j 0.08

j 0.58

0  0  0   j 0.16 

Choosing the voltage at bus 3 as 10° , the prefault current in line (2) – (3) is I 23 =

P − jQ 0.5 ( 0.8 − j 0.6 ) = = 0.4 − j 0.3pu V3 10°

Line (2) – (3) has parameters given by

Z1 = Z 2 = j 0.15; Z 0 = j 0.5

Denoting the open-circuit points of the line as p and p , to simulate opening, we need to develop the Thévenin-equivalent sequence networks looking into the system between points p and p .

Before any conductor opens, the current I mn in phase a of the line (m) – (n) is positive sequence, given by I mn =

Vm − Vn Z1

Z pp1 = −

Z12 ZTh ,mn,1 − Z1

; Z pp 2 =

− Z 02 − Z 22 ; Z pp 0 = ZTh ,mn,2 − Z 2 Z Th ,mn,0 − Z 0

To simulate opening phase a between points p and p , the sequence network connection is shown below:

To simulate opening phases b and c between points p and p , the sequence network connection is shown below:

In this Problem

− j ( 0.15 ) − Z12 Z pp1 = Z pp 2 = = Z 22 1 + Z 33 1 − 2 Z 23 1 − Z1 j 0.1696 + j 0.1696 − 2 ( j 0.1104 ) − j 0.15 2

= j 0.7120 − ( j 0.5 ) − Z 02 Z pp 0 = = Z 22 0 + Z 330 − 2 Z 230 − Z 0 j 0.08 + j 0.58 − 2 ( j 0.08 ) − j 0.5 2

= Note that an infinite impedance is seen looking into the zero sequence network between points p and p of the opening, if the line from bus (2) to bus (3) is opened. Also bus (3) would be isolated from the reference by opening the connection between bus (2) and bus (3). (a) One open conductor: Va 0 = Va1 = Va 2 = I 23

Z pp1 Z pp 2 Z pp1 + Z pp 2

= ( 0.4 − j 0.3 )

( j 0.712 )( j 0.712 ) j ( 0.712 + 0.712 )

= 0.1068 + j 0.1424

V3 1 = V3 2 =

Z 32 1 − Z33 1 Z1

Va1 =

j 0.1104 − j 0.1696 ( 0.1068 + j 0.1424 ) j 0.15

= −0.0422 − j 0.0562 V30 =

Z 32 0 − Z 330 Z0

Va 0 =

j 0.08 − j 0.58 ( 0.1068 + j 0.1424 ) j 0.5 = −0.1068 − j 0.1424

V3 = V30 + V31 + V32 = −0.1068 − j 0.1424 − 2 ( 0.0422 + j 0.0562 ) = −0.1912 − j 0.2548 Since the prefault voltage at bus (3) is 10° , the new voltage at bus (3) is

V3 + V3 = (1 + j 0 ) + ( −0.1912 − j 0.2548 ) = 0.8088 − j 0.2548 = 0.848 − 17.5° pu (b) Two open conductors: Inserting an infinite impedance of the zero sequence network in series between points p and

p of the positive-sequence network causes an open circuit in the latter. No power transfer can occur in the system. Obviously, power cannot be transferred by only one phase conductor of the transmission line, since the zero sequence network offers no return path for current.

10.34

Va = 0; Vb = Vc ; I b + I c = 0

Sequence currents are given by 1 1 1 I 0 = I a ; I1 =  I a + ( a − a 2 ) I b  ; I 2 =  I a + ( a 2 − a ) I b  3 3 3

One can conclude that I1 + I 2 = 2 I 0 Sequence network connection to satisfy the above:

Sequence voltages are obtained below: 1 1 V1 = ( a + a 2 ) Vb  = − Vb 3 3 1 1 V2 = ( a + a 2 ) Vb  = − Vb 3 3 1 2 V0 = ( 2Vb ) = Vb 3 3

Thus V1 = V2 and V1 + V2 + V0 = 0 The sequence network interconnection is then given by:

10.35 (a) Following Ex. 10.2 of the text:

I1 =

10° = − j 6.67; I 2 = 0; I 0 = 0 j 0.15 V0 = V1 = V2 = 0

I a = 6.67 − 90°; I b = 6.67150°; I c = 6.6730° 

Va = Vb = Vc = 0 ( Bolted 3-phase fault )  (b) Following Ex. 10.3 of the text I 0 = I1 = I 2 =

10 j ( 0.15 + 0.15 + 0.2 )

= − j2

I a = 3 ( − j 2 ) = − j 6; I b = I c = 0  V0   0   j 0.2      V1  = 10° −  0 V2   0   0  

0 j 0.15 0

0  − j2 0   − j 2  j 0.15  − j 2 

 −0.4  =  0.7   −0.3  Va  1 1    2 Vb  = 1 a Vc  1 a  

1   −0.4   0      a   0.7  = 1.11233°  a 2   −0.3  1.11125° 

10 j ( 0.15 + 0.15 )

= − j 3.333 I0 = 0

(

)

I b = − j 3 ( − j 3.333 ) = −5.773   I c = 5.773; I a = 0 

V1 = V2 = I1 ( j 0.15 ) = 0.5; V0 = 0 Va  1 1    2 Vb  = 1 a Vc  1 a  

1   0   1.0  a  0.5 =  −0.5  a 2  0.5  −0.5

(d) Following Ex. 10.5 of the text: 10 0.15  0.2   j 0.15 + 0.15 + 0.2   = − j 4.242

I1 =

0.2   I 2 = ( j 4.242 )    0.2 + 0.15  = j 2.424  0.15  I 0 = ( j 4.242 )   = j1.818  0.2 + 0.15 

 I a  1 1    2  I b  = 1 a  I c  1 a  

1   j1.818   0      a   − j 4.242  = 5.5118.2°  a 2   j 2.424   5.561.8° 

V0 = V1 = V2 = − j1.818 ( j 0.2 ) = 0.364 Va  1 1    2 Vb  = 1 a Vc  1 a  

1  0.364  1.092  a  0.364  =  0   a 2  0.364   0 

Worst fault: 3-phase fault with a fault current of 6.67 pu  10.36 The positive-sequence per-phase circuit is shown below:

 V3 I LOAD = 1; I LOAD = 1.02 − 10° prior to the fault

E = 10 + j 0.1(1.02 − 10 ) = 1.0235.64

With a short from bus 2 to ground, i.e. with switch closed, I FAULT =

1.0235.64 = 5.115 − 84.36  j 0.2

10.37 (a) E1 = 10 + (10 )( j 0.1) = 1 + j 0.1

E2 = 10 − (10 )( j 0.15 ) = 1 − j 0.15 With switch closed, I1 =

E1 1 + j 0.1 = = 1 − j10 j 0.1 j 0.1

I2 =

E2 1 − j 0.15 = = −1 − j 6.67 j 0.15 j 0.15

I = I1 + I 2 = − j16.67 

(b) Superposition: Ignoring prefault currents E1 = E2 = 10 I1 =

10 10 = − j10; I 2 = = − j 6.67 j 0.1 j 0.15

I = I1 + I 2 = − j16.67

Now load currents are superimposed:

I1 = I1 FAULT + I1 LOAD = − j10 + 1 = 1 − j10 I 2 = I 2 FAULT + I 2 LOAD = − j 6.67 + (−1) = −1 − j 6.67 I = I FAULT + I LOAD = − j16.67  Same as in part (a)  10.38 I1−1 =

VF 1.00 = = − j8.333per unit Z11−1 j 0.12

 I1a  1 1     2  I1b  = 1 a  I1c  1 a  

1  0  8.333 − 90   a  8.333 − 90 = 8.333150  per unit  8.33330  a 2  0

Using Eq. (10.5.9) with k = 2 and n = 1: V2 − 0   0  0      V2 −1  = 10 − 0 V2 − 2   0  0  

0 j 0.08 0

0  0   0  0   − j8.333 =  0.3333 per unit j 0.08  0   0 

V2 ag  1 1    2 V2 bg  = 1 a   1 a V2 cg  

1   0  0.33330  a  0.3333 =  0.3333240 per unit a 2   0  0.3333120 

10.39 I1−0 = I1−1 = I1−2 =

VF 1.00 = Z11−0 + Z11−1 + Z11−2 j ( 0.10 + 0.12 + 0.12 )

= − j 2.941per unit  I1a  1 1     2  I1b  = 1 a  I1c  1 a  

1   − j 2.941  − j8.824  a   − j 2.941 =  0  per unit a 2   − j 2.941  0 

Using Eq. (10.5.9) with k = 2 and n = 1: V2 − 0   0  0      V2 −1  = 10 − 0 V2 − 2   0  0  

V2 ag  1 1    2 V2 bg  = 1 a   V2 cg  1 a

10.40 I1−1 = − I1−2 =

0 j.08 0

0   − j 2.941  0  0   − j 2.941 =  0.7647  per unit j.08   − j 2.941  −0.2353

1   0  0.52940      a   .7647  = 0.9056253.0 per unit a 2   −.2353 0.9056107.0 

VF 1.00 = Z11−1 + Z11−2 j (.12 + .12 )

= − j 4.167 per unit  I1a  1 1     2  I1b  = 1 a  I1c  1 a  

1  0   0      a   − j 4.167  = 7.217180 per unit a 2   + j 4.167   7.2170 

Using Eq. (10.5.9) with k = 2 and n = 1: V2 − 0   0  0      V2 −1  = 10 − 0 V2 − 2   0  0  

V2 ag  1 1    2 V2 bg  = 1 a   1 a V2 cg  

0 j 0.08 0

0  0   0  0   − j 4.167  = 0.6667  per unit j 0.08   j 4.167   0.3333 

1  0   1.0      a  .6667  = 0.5774210 per unit a 2  .3333  0.5774150 

10.41 I1−1 =

VF 1.00 = Z11−1 + Z11−2 // Z11−0 j ( 0.12 + 0.12 // 0.10 )

= − j 5.729 per unit  0.10  I1− 2 = ( + j 5.729 )   = j 2.604 per unit  0.22 

 0.12  I1−0 = ( + j 5.729 )   = j 3.125per unit  0.22 

 I1a  1 1     2  I1b  = 1 a  I1c  1 a  

1   j 3.125   0      a   − j 5.729  = 8.605147.0 per unit a 2   + j 2.604   8.60533.0 

Using Eq. (10.5.9) with k = 2 and n = 1: V2 −0   0  0      V2 −1  = 1.00 − 0 V2 −2   0  0  

V2 ag  1 1    2 V2 bg  = 1 a   V2 cg  1 a

0 j.08 0

0   j 3.125   0  0   − j 5.729  = 0.5417  per unit j.08   j 2.604   0.2083

1  0   0.750      a  − .5417  =  0.4733217.6 per unit a 2  .2083  0.4733142.4 

[Note: For details on “formation of Z bus one step at a time”, please refer to edition 1 or 2 of the text.] 10.42 (a) Zero sequence bus impedance matrix: Step (1): Add Z b = j 0.10 from the reference to bus 1 (type 1)

Z bus -0 = j 0.10 per unit Step (2): Add Z b = j 0.2563 from bus 1 to bus 2 (type 2) 0.10 0.10  Z bus −0 = j   per unit 0.10 0.3563

Step (3): Add Z b = j 0.10 from the reference to bus 2 (type 3) 0.1  0.1 .07808 .02192  j  .10  Z bus − 0 = j  − .10.3563 = j       0.1 0.3563 .4563 .3563 .02192 .07808 

Step (4): Add Z b = j 0.1709 from bus 2 to bus 3 (type 2)  0.07808 0.02192 0.02192 Z bus −0 = j 0.02192 0.07808 0.07808 per unit 0.02192 0.07808 0.24898

Step (5): Add Z b = j 0.1709 from bus 1 to bus 3 (type 4) .07808+  Z bus − 0 = j  .02192  .02192

.02192 .02192−   .07808 .07808  .07808 .24898   .05616  j  − −.05616  .05616 − .05616 − .22706 .45412   −.22706 

0.05  0.07114 0.02887  Z bus −0 = j  0.02887 0.07114 0.05  per unit  0.05 0.05 0.13545

Positive sequence bus impedance matrix: Step (1): Add Z b = j 0.28 from the reference to bus 1 (type 1)

Zbus−1 = j 0.28 per unit Step (2): Add Z b = j 0.08544 from bus 1 to bus 2 (type 2) .28  .28 Z bus −1 = j   per unit .28 .36544 

Step (3): Add Z b = j 0.3 from the reference to bus 2 (type 3) .16218 .12623 .28  .28 j  .28  Z bus −1 = j  − .28 .36544  = j       .12623 .16475 .28 .36544  .66544 .36544 

Step (4): Add Z b = j.06835 from bus 2 to bus 3 (type 2) 0.16218 0.12623 0.12623 Z bus −1 = j  0.12623 0.16475 0.16475 per unit  0.12623 0.16475 0.2331 

Step (5): Add Z b = j.06835 from bus 1 to bus 3 (type 4)

.16218 .12623 .12623  +.03595 j    Z bus −1 = j .12623 .16475 .16475 − −.03852  .21117  .12623 .16475 .2331   −.10687  .03595 − .03852 − .10687 .15606 .13279 .14442  Z bus −1 = j .13279 .15772 .14526  per unit .14442 .14526 .17901

Step (6): Add Z b = j (.4853 //.4939) = j 0.2448 from the reference to bus 3 (type 3) .15606 .13279 .14442  .14442  j    Z bus −1 = j .13279 .15772 .14526  − .14526  .42379  .14442 .14526 .17901 .17901 .14442 .14526 .17901

 0.1068 0.08329 0.08342  Z bus −1 = j  0.08329 0.1079 0.08390  per unit 0.08342 0.08390 0.10340 

Negative sequence bus impedance matrix: Steps (1) – (5) are the same as for Z bus −1 . Step (6): Add Z b = j (.5981//.4939) = j 0.2705 from the reference bus to bus 3 (type 3) .15606 .13279 .14442 .14442 j    Z bus − 2 = j .13279 .15772 .14526 − .14526 .14442 14526 17901  .44951 .14442 .14526 .17901  .17901  0.1097 0.08612 0.08691 Z bus −2 = j 0.08612 0.11078 0.08741 per unit  0.08691 0.08741 0.10772

(b) From the results of Problem 10.42(a), Z11−0 = j 0.07114, Z11−1 = j 0.1068 , and Z11−2 = j 0.1097 per unit are the same as the Thévenin equivalent sequence impedances at bus

1, as calculated in Problem 10.2. Therefore, the fault currents calculated from the sequence impedance matrices will be the same as those calculated in Problems 10.3 and 10.14–10.17. 10.43 I1−1 =

VF 1.030 = = 5.15 − 90 per unit Z11−1 j 0.20

 I1a  1 1 1  0  5.15  − 90         2  I1b  = 1 a a  5.15  − 90 = 5.15 150  per unit  I1c  1 a a 2  0  5.15 30   

Using (10.5.9) with k = 2 and n = 1: V2− 0  0 0  0 0  0  0            V2−1  = 1.03 0 − 0 j 0.10 0   − j 5.15 = 0.515 per unit V2− 2  0  0 0  0  j 0.10  0  

V2 ag  1 1 1  0  0.515 0          2 V1bg  = 1 a a   0.515 =  0.515 240 per unit   1 a a 2   0   0.515 120   V2 cg  

10.44 I1−0 = I1−1 = I1− 2 =

VF 1.03  0 = = − j 2.06 per unit Z11−0 + Z11−1 + Z11− 2 j ( 0.10 + 0.2 + 0.2 )

 I1a  1 1 1   − j 2.06  6.18  − 90        per unit 2 0  I1b  = 1 a a   − j 2.06  =    I1c  1 a a 2   − j 2.06    0   V2 −0  0 0  0 0   − j 2.06  0            V2 −1  = 1.03 0 − 0 j 0.10 0   − j 2.06  = 0.824  per unit V2 − 2  0  0 0 j 0.10   − j 2.06   −0.206   

V2 ag  1 1 1  0  0.618 + j 0          2 V2 bg  = 1 a a  0.824  =  −0.309 − j 0.892  per unit V2 cg  1 a a 2   −0.206   −0.309 + j 0.892    10.45 I1−1 = − I1− 2 =

VF 1.03 0 = = 2.575  − 90 per unit Z11−1 + Z11− 2 j ( 0.2 + 0.2 )

0   I1a  1 1 1  0           2  I1b  = 1 a a   − j 2.575  =  −2.575 3  per unit  I1c  1 a a 2   + j 2.575  +2.575 3      V2− 0  0 0  0 0  0   0         − j 2.575 = 0.7725 per unit V2−1  = 1.03 0 − 0 j 0.10 0     V2− 2  0        0 0 j 0.10 + j 2.575 0.2575        V2 ag  1 1 1  0   1.03          2 V2bg  = 1 a a  0.7725 =  −0.515 − j 0.446  per unit   1 a a 2  0.2575  −0.515 + j 0.446     V2 cg  

10.46 I1−1 =

VF 1.03 0 = = 3.86 − 90 per unit Z11−1 + Z11− 2 // Z11−0  0.20 )( 0.10 )  ( j 0.20 +  0.30  

 0.10  I1− 2 = ( 3.86 + 90 )   = 1.29 90 per unit  0.30   0.20  I1− 0 = ( 3.86 + 90 )   = 2.58 90  0.30 

 I1a  1 1 1   j 2.58   0          2  I1b  = 1 a a   − j 3.86  =  −4.460 + j 3.86  per unit  I1c  1 a a 2   j1.29   +4.460 + j 3.86    V2− 0  0 0  0 0   j 2.58  0            V2−1  = 1.03 0 − 0 j 0.10 0   − j 3.86  = 0.644  per unit V2− 2  0  0 0 j 0.10   j1.29  0.129    V2 ag  1 1 1  0   0.77          2 V2bg  = 1 a a  0.644  =  −0.386 − j 0.446  per unit   1 a a 2  0.129   −0.386 + j 0.446      V2 cg  

10.47 Zero sequence bus impedance matrix: Step (1): Add Z b = j 0.10 from the reference to bus 1 (type 1) Z bus − 0 = j 0.10 per unit Step (2): Add Z 6 = j 0.60 from bus 1 to bus 2 (type 2) 0.10 0.10  Z bus − 0 = j   per unit 0.10 0.70 

Step (3): Add Z b = j 0.10 from the reference to bus 2 (type 3) 0.10 0.10  0.0875 0.0125 j 0.10   0.10 0.70 Z bus − 0 = j  − = j     0.10 0.70  0.80 0.70  0.0125 0.0875

Step (4): Add Z b = j 0.40 from bus 2 to bus 3 (type 2) 0.0875 0.0125 0.0125 Z bus −0 = j 0.0125 0.0875 0.0875 per unit 0.0125 0.0875 0.4875

Step (5): Add Z b = j 0.40 from bus 1 to bus 3 (type 4)  0.0875 0.0125 0.0125  0.075   0.075 −0.075 −0.475 j    Z bus − 0 = j  0.0125 0.0875 0.0875 − −0.075 0.95   0.0125 0.0875 0.4875  −0.475  0.08158 0.01842 0.050  = j  0.01842 0.08158 0.050  per unit  0.050 0.050 0.250 

Positive sequence Bus Impedance Matrix: Step (1): Add Z b = j 0.25 from the reference to bus 1 (type 1)

Z bus −1 = j 0.25 per unit Step (2): Add Z b = j 0.20 from bus 1 to bus 2 (type 2) 278 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

0.25 0.25 Z bus −1 = j   per unit 0.25 0.45

Step (3): Add Z b = j 0.35 from the reference to bus 2 (types) 0.25 0.25 0.1719 0.1094 j 0.25 0.25 0.45 Z bus −1 = j  − = j     0.25 0.45 0.8 0.45 0.1094 0.1969

Step (4): Add Z b = j 0.16 from bus 2 to bus 3 (type 2) 0.1719 0.1094 0.1094 Zbus −1 = j 0.1094 0.1969 0.1969 per unit 0.1094 0.1969 0.3569

Step (5): Add Z b = j 0.16 from bus 1 to bus 3 (type 4) 0.1719 0.1094 0.1094  0.0625  0.0625 −0.0875 −0.2475 −j    Z bus −1 = j 0.1094 0.1969 0.1969 −0.0875 0.47  0.1094 0.1969 0.3569 −0.2475 0.1636 0.1210 0.1423 = j 0.1210 0.1806 0.1508 per u nit 0.1423 0.1508 0.2266

Step (6): Add Zb = j (0.5786 0.4853) = j0.2639 from the reference to bus 3 (type 3) 0.1636 0.1210 0.1423 0.1423 0.1423 0.1508 0.2266 j   0.1508 Z bus −1 = j 0.1210 0.1806 0.1508 −  0.4905  0.1423 0.1508 0.2266 0.2266 0.1223 0.0773 0.0766 = j 0.0773 0.1342 0.0811 per uni t 0.0766 0.0811 0.1219

Negative Sequence Bus Impedance Matrix: Steps (1)–(5) are the same as for Z bus −1 . Step (6): Add Z b = j (0.5786 .5981) = j 0.2941 from the reference to bus 3 (types 3). 0.1636 0.1210 0.1423 0.1423 0.1423 0.1508 0.2266 j   0.1508 Z bus −1 = j 0.1210 0.1806 0.1508 −  0.5207  0.1423 0.1508 0.2266 0.2266 0.1247 0.0798 0.0804 = j 0.0798 0.1369 0.0852 0.0804 0.0852 0.1280

10.48 From the results of Problem 10.47, Z11− 0 = j 0.08158,

Z11−1 = j 0.1223 and Z11− 2 = j 0.1247 per

unit are the same as the Thévenin equivalent sequence impedances at bus 1 as calculated in Problem 10.4(b). Therefore the fault currents calculated from the sequence impedance matrices will be the same as those calculated in problems 10.4(c) and 10.19(a) through 10.19(d). 10.49 Zero sequence bus impedance matrix: Step (1): Add Z b = j 0.24 from the reference to bus 1 (type 1)

Zbus −0 = j 0.24 per unit Step (2): Add Z b = j 0.6 from bus 1 to bus 2 (type 2) 0.24 0.24  Z bus − 0 = j   per unit 0.24 0.84 

Step (3): Add Z b = j 0.6 from bus 2 to bus 3 (type 2) 0.24 0.24 0.24  Z bus −0 = j 0.24 0.84 0.84  per unit 0.24 0.84 1.44 

Step (4): Add Z b = j 0.10 from the reference to bus 3 (type 3) .24 .24 .24   .24  j    Z bus −0 = j .24 .84 .84  − .84  .24 .84 1.44  1.54 .24 .84 1.44 1.44  0.2026 0.1091 0.01558 Z bus −0 = j  0.1091 0.3818 0.05455 per unit 0.01558 0.05455 0.09351

Step (5): Add Z b = j 0.6 from bus 2 to bus 4 (type 2)

 0.2026 0.1091 0.01558 0.1091   0.1091 0.3818 0.05455 0.3818   per unit Z bus −0 = j  0.01558 0.05455 0.09351 0.05455    0.1091 0.3818 0.05455 0.9818  Step (6): Add Z b = j 0.1333 from the reference bus to bus 4 (type 3)

.2026 .1091 .01558 .1091   .1091  .1091 .3818. .05455 .3818     − j  .3818  .1091 .3818 .05455 .9818 Z bus −0 = j  .01558 .05455 .09351 .05455 1.1151 .05455      .9818  .1091 .3818 .05455 .9818 

 0.1919 0.07175 0.01024 0.01304   0.07175 0.2511 0.03587 0.04564   per unit Z bus −0 = j  0.01024 0.03587 0.09084 0.006521   0.01304 0.04564 0.006521 0.1174  Positive sequence bus impedance matrix can be obtained as:

 0.2671  0.1505 Z bus −1 = j   0.0865  0.0980

0.1505 0.0865 0.0980  0.1975 0.1135 0.1286  per unit 0.1135 0.1801 0.0739   0.1286 0.0739 0.2140 

Negative sequence bus impedance matrix: Steps (1)–(4) are the same as for Z bus −1 Step (5): Add Z b = j 0.3 from the reference bus to bus 3 (type 3)

.64 .64 Z bus − 2 = j  .64  .64

.64 .64 .64   .64     .84 .84 .84  j  .84  − .64 .84 1.04 .84 .84 1.04 .84  1.34 1.04    .84 .84 1.04  .84 

.3343 .2388 Z bus −2 = j  .1433  .2388

.2388 .1433 .2388 .3134 .1881 .3134  per unit .1881 .2328 .1881  .3134 .1881 .5134 

Step (6): Add Z b = j 0.3733 from the reference to bus 4 (type 3)

.3343 .2388 Z bus − 2 = j  .1433  .2388

.2388 .3134 .1881 .3134

 0.2700  0.1544 Z bus −2 = j  0.09264   0.1005

.1433 .1881 .2328 .1881

.2388 .2388    .3134  j .3134 − .2388 .3134 .1881 .5134 .1881 .8867 .1881    .5134  .5134

0.1005  0.2026 0.1216 0.1319  per unit 0.1216 0.1929 0.07919  0.1319 0.07919 0.2161  0.1544 0.09264

10.50 From the results of Problem 10.49, Z11−0 = j 0.1919, Z11−1 = j 0.2671 , and Z11−2 = j 0.2700 per unit are the same as the Thévenin equivalent sequence impedances at bus 1, as calculated in Problem 10.6. Therefore, the fault currents calculated from the sequence impedance matrices are the same as those calculated in Problems 10.7, 10.21–10.23.

10.51 Zero sequence bus impedance matrix: Working backwards from bus 4: Step (1): Add Z b = j 0.1 from the reference bus to bus 4 (type 1) 4 Z bus −0 =  j 0.1 4 per unit

Step (2): Add Z b = j 0.5251 from bus 4 to bus 3 (type 2) 3 4 0.6251 0.1 3 Z bus − 0 = j  per unit 0.1 4  0.1

Step (3): Add Z b = j 0.5251 from bus 3 to bus 2 (type 2)

2 3 4 1.1502 0.6251 0.1 2 Z bus −0 = j 0.6251 0.6251 0.1 3 per unit  0.1 0.1 0.1 4 Step (4): Add Z b = j 0.2 from bus 2 to bus 1 (type 2) 1 2 3 4 1.3502 1.1502 0.6251 0.1 1 1.1502 1.1502 0.6251 0.1 2  per unit Z bus − 0 = j  0.6251 0.6251 0.6251 0.1 3   0.1 0.1 0.1 4  0.1

Step (5): Add Z b = j 0.1 from the reference to bus 5 (type 1) 1.3502 1.1502 0.6251 0.1 0  1.1502 1.1502 0.6251 0.1 0    Z bus −0 = j 0.6251 0.6251 0.6251 0.1 0  per unit   0.1 0.1 0.1 0   0.1  0 0 0 0 0.1

Positive sequence bus impedance matrix can be obtained as: 0.3542 0.2772  Z bus −1 = j 0.1964   0.1155 0.0770

0.2772 0.1964 0.1155 0.0770  0.3735 0.2645 0.1556 0.1037  0.2645 0.3361 0.1977 0.1318  per unit  0.1556 0.1977 0.2398 0.1599  0.1037 0.1318 0.1599 0.1733 

Negative sequence bus impedance matrix: Steps (1) – (5) are the same as for Z bus −1 Step (6): Add Z b = j 0.23 from the reference bus to bus 5 (type 3)

.576 .576 .576 .576 .576   .576  .576 .776 .776 .776 .776   .776     j   .986  .576 .776 .986 1.196 1.296 Z bus − 2 = j .576 .776 .986 .986 .986  −   1.526   .576 .776 .986 1.196 1.196  1.196  .576 .776 .986 1.196 1.296  1.296   0.3586  0.2831  Z bus −2 = j  0.2038   0.1246 0.08682

0.2831 0.2038 0.1246 0.08682  0.3814 0.2746 0.1678 0.1170  0.2746 0.3489 0.2132 0.1486  per unit  0.1678 0.2132 0.2586 0.1803  0.1170 0.1486 0.1803 0.1953 

10.52 From the results of Problem 10.51 Z11−0 = j1.3502, Z11−1 = j 0.3542, and Z11−2 = j 0.3586 per unit are the same as the Thévenin equivalent sequence impedances at bus 1, as calculated in Problem 10.9. Therefore, the fault currents calculated from the sequence impedance matrices are the same as those calculated in Problem 10.10, 10.24–10.27. 10.53 (a) & (b)

Either by inverting YBUS or by the building algorithm Z BUS can be obtained as (1) (1)  j 0.0793 (2)  j 0.0558 Z BUS = (3)  j 0.0382  (4)  j 0.0511 (5)  j 0.0608

(2)

(3)

(4)

(5)

j 0.0558

j 0.0382

j 0.0511

j 0.1338

j 0.0664

j 0.0630

j 0.0664

j 0.0875

j 0.0720

j 0.0630

j 0.0720

j 0.2321

j 0.0605

j 0.0603

j 0.1002

j 0.0608 j 0.0605 j 0.0603  j 0.1002  j 0.1301

Simply by closing 8, the subtransient current in the 3-phase fault at bus (4) is given by I f =

1.0 = − j 4.308 j 0.2321

The voltage at bus (3) during the fault is

V3 = Vf − I f Z34 = 1 − ( − j 4.308 )( j 0.0720 ) = 0.6898 The voltage at bus (5) during the fault is

V5 = Vf − I f Z54 = 1 − ( − j 4.308)( j 0.1002 ) = 0.5683 Currents into the fault at bus (4) over the line impedances are From bus (3):

0.6898 = − j 2.053 j 0.336

From bus (5):

0.5683 = − j 2.255 j 0.252

Hence, total fault current at bus (4) = − j 4.308pu 10.54 The impedance of line (1)–(2) is Z b = j 0.168 .

Z BUS is given in the solution of Problem 10.53. The Thévenin equivalent circuit looking into the system between buses (1) and (2) is shown below:

Z kk ,Th = j 0.168 +

( j 0.0235)( − j 0.09 ) + j 0.0558 = j 0.2556 j ( 0.0235 − 0.09 ) 284

Subtransient current into line-end fault I f = 1/ j 0.2556 = − j3.912pu 10.55 (a)

Negative-sequence network is same as above without sources.

(1) (1)  j 0.1437 (2)  j 0.1211 Z BUS1 = Z BUS 2 = (3)  j 0.0789  (4)  j 0.0563 (1) (1)  j 0.19 (2)  0 Z BUS 0 = (3)  0  (4)  0

(2)

(3)

(4)

j 0.1211

j 0.0789

j 0.1696

j 0.1104

j 0.1696

j 0.0789

j 0.1211

j 0.0563  j 0.0789  j 0.1211   j 0.1437 

(2)

(3)

(4)

j 0.08

j 0.58

0  0  0   j 0.19 

(b) For the line-to-line fault, Thévenin equivalent circuit:

Upper case A is used because fault is in the HV-transmission line circuit. I fA 1 = − I fA 2 =

10 = − j 2.9481 j 0.1696 + j 0.1696

I fA = I fA 1 + I fA 2 = 0 I f B = a 2 I fA 1 + aI fA 2 = −5.1061 + j 0 = 855180 A I f C = − I f B = 5.1061 + j 0 = 8550 A

Base current in HV transmission line is

100,000 3  345

= 167.35A

Symmetrical components of phase-A voltage to ground at bus (3) are

V3 A 0 = 0; V3 A 1 = V3 A 2 = 1 − ( j 0.1696 )( − j 2.9481) = 0.5 + j 0 Line-to-Ground voltages at fault bus (3) are

V3 A = V3 A 0 + V3 A 1 + V3 A 2 = 0 + 0.5 + 0.5 = 10 V3 B = V3 A 0 + a2V3 A 1 + aV3 A 2 = 0.5180 V3C = V3 B = 0.5180 Line-to-line voltages at fault bus (3) are

V3, AB = V3 A − V3 B = 1.50 = 1.5 

345 3

= 2990 kV

V3, BC = V3 B − V3C = 0

V3,CA = V3C − V3 A = 1.5180 = 299180 kV Avoiding, for the moment, phase shifts due to  − Y transformer connected to machine 2, sequence voltages of phase A at bus (4) using the bus-impedance matrix are calculated as

V4 A0 = − Z 430 I fA0 = 0

V4 A1 = Vf − Z431 I fA1 = 1 − ( j 0.1211)( − j 2.9481) = 0.643 V4 A2 = − Z432 I fA2 = − ( j 0.1211)( j 2.9481) = 0.357 Accounting for phase shifts

V4 a 1 = V4 A 1 − 30 = 0.643 − 30 = 0.5569 − j 0.3215

V4 a 2 = V4 A 2 30 = 0.35730 = 0.3092 + j 0.1785 V4 a = V4 a 0 + V4 a 1 + V4 a 2 = 0.8661 − j 0.143 = 0.8778 − 9.4 Phase-b voltages at terminals of machine 2 are

V4 b 0 = V4 a 0 = 0 V4b 1 = a2V4 a 1 = 0.643240 − 30 = −0.3569 − j 0.3215 V4b 2 = aV4 a 2 = 0.357120 + 30 = −0.3092 + j 0.1785

V4b = V4b 0 + V4b 1 + V4b 2 = −0.8661 − j 0.143 = 0.8778 − 170.6 For phase C of machine 2,

V4 c 1 = aV4 a 1 = 0.64390; V4 c 2 = a2V4 a 2 = 0.357 − 90

V4 c = V4 c 0 + V4 c 1 + V4 c 2 = j 0.286 Line-to-line voltages at terminals of machine 2 are given by

V4,ab = V4 a − V4 b = 1.7322 + j 0 = 1.7322 

20 3

= 200 kV

V4,bc = V4 b − V4 c = −0.8661 − j 0.429 = 0.9665 − 153.65 = 11.2 kV  − 153.65

V4,ca = V4 c − V4 a = −0.8661 + j 0.429 = 0.9665153.65 = 11.2 kV 153.65 (c) For the double line-to-line fault, connection of Thévenin equivalents of sequence networks is shown below:

I fa1 =

10 = − j 4.4342 j 0.1437 ( j 0.19 ) j 0.1437 + j ( 0.1437 + 0.19 )

Sequence voltages at the fault are

V4 a1 = V4 a 2 = V4 a 0 = 1 − ( − j 4.4342 )( j 0.1437 ) = 0.3628 I fa 2 = j 4.4342

j 0.19 = j 2.5247 j ( 0.1437 + 0.19 )

I fa 0 = j 4.4342

j 0.1437 = j1.9095 j ( 0.1437 + 0.19 )

Currents out of the system at the fault point are

I fa = I fa 0 + I fa 1 + I fa 2 = 0 I fb = I fa 0 + a2 I fa 1 + aI fa 2 = −6.0266 + j 2.8642 = 6.6726154.6

I fc = I fa 0 + aI fa 1 + a2 I fa 2 = 6.0266 + j 2.8642 = 6.672625.4 Current I f into the ground is

I f = I fb + I fc = 3I fa 0 = j 5.7285 a-b-c voltages at the fault bus are 287 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

V4 a = V4 a 0 + V4 a 1 + V4 a 2 = 3V4 a 1 = 3 ( 0.3628) = 1.0884 V4 b = V4 c = 0 V4,ab = V4 a − V4 b = 1.0884; V4,bc = V4 b − V4 c = 0; V4,ca = V4 c − V4 a = −1.0884 Base current =

100  103 3  20

= 2887 A

 I fa = 0; I fb = 19.262154.6 kA; I fc = 19.262 25.4 kA I f = 16.53890 kA Base line-to-neutral voltage in machine 2 is 20 / 3 kV

V4,ab = 12.5680 kV; V4,bc = 0; V4,ca = 12.568180 kV 10.56 (a)

Z BUS 1 and Z BUS 2 are same as in the solution of Problem 10.51. However, because the transformers are solidly grounded on both sides, the zero-sequence network is changed as shown below:

For the single line-to-ground fault, series connection of the Thévenin equivalents of the sequence networks is shown below: (1) (1)  j 0.1553 (2)  j 0.01407 Z BUS 0 = (3)  j 0.0493  (4)  j 0.0347

(2)

(3)

(4)

j 0.1407

j 0.0493

j 0.1999

j 0.0701

j 0.1999

j 0.0493

j 0.1407

j 0.0347  j 0.0493  j 0.1407   j 0.1553 

(b) I fA 0 = I fA1 = I fA 2 =

10 j ( 0.1696 + 0.1696 + 0.1999 )

= − j1.8549 I fA = 3I fA 0 = − j 5.5648 = 931270 A

Base current in HV trans. line is

100,000 3  345

= 167.35A

Phase-a sequence voltages at bus (4), terminals of machine 2, are

V4 a 0 = − Z 430 I fA0 = − ( j 0.1407 )( − j1.8549 ) = −0.2610 V4 a1 = 1 − ( j 0.1211)( − j1.8549 ) = 0.7754  = V f − Z 431 I fA1  V4 a 2 = − ( j 0.1211)( − j1.8549 ) = −0.2246  = − Z 432 I fA2 

Note: Subscripts A and a denote HV and LV circuits, respectively, of the Y-Y connected transformer. No phase shift is involved. V4 a  1 1    2 V4 b  = 1 a V4 c  1 a  

1   −0.2610  0.2898 + j 0  0.28980      a   +0.7754  =  −0.5364 − j 0.866  = 1.0187 − 121.8 a 2   −0.2246   −0.5364 + j 0.866  1.0187121.8 

Line-to-ground voltages of machine 2 in kV are: (Multiply by 20

V4 a = 3.3460 kV; V4 b = 11.763  − 121.8 kV; V4 c = 11.763121.8 kV Symmetrical components of phase-a current are Ia0 = −

I a1 =

V4 a 0 0.2610 = = − j 6.525 jX 0 j 0.04

V f − V4 a1

Ia2 = −

jX 

1.0 − 0.7754 = − j1.123 j 0.2

V4 a 2 0.2246 = = − j1.123 jX 2 j 0.2

The phase-c currents in machine 2 are calculated as I c = I a 0 + aI a1 + a 2 I a 2

= − j 6.525 + a ( − j1.123 ) + a 2 ( − j1.123 ) = − j 5.402

Base current in the machine circuit is

100  103 3 ( 20 )

= 2886.751A

 I c = 15,594 A

10.57 Using equations (10.5.9) in (8.1.3), the phase “a” voltage at bus k for a fault at bus n is:

Vka = Vk -0 + Vk -1 + Vk -2

= VF − ( Z kn-0 I n −0 + Z kn −1I n −1 + Z kn −2 I n 2 )

For a single line-to-ground fault, (9.5.3),

I n −0 = I n −1 = I n −2 =

VF Z nn −0 + Z nn −1 + Z nn −2 + 3Z F

Therefore,

  Z kn −0 + Z kn −1 + Z kn −2 Vka = VF 1 −   Z nn −0 + Z nn −1 + Z nn −2 + 3Z F  The results in Table 10.5 for Example 10.8 neglect resistance of all components (machines, transformers, transmission lines). Also the fault impedance Z F is zero. As such, the impedance in the above equation all have the same phase angle (90), and the phase “a” voltage Vka therefore has the same angle as the pre-fault voltage VF , which is zero degrees. Note also that pre-fault load currents are neglected. Note, the PowerWorld problems in Chapter 10 were solved ignoring the effect of the  − Y transformer phase shift [see Example 10.6]. An upgraded version of PowerWorld Simulator is available from www.powerworld.com/gloversarma that (optionally) allows inclusion of this phase shift. 10.58 During a double line-to-ground fault the bus 2 voltage on the unfaulted phase rises to 1.121 per unit and the fault point voltage raises to 1.252 per unit. 10.59 Bus 2 Phase A Voltage Bus 2 Phase B Voltage

10.60 For a line-to-line fault the magnitude of the per unit fault currents are 125,121 A for a bus 1 fault, 2,672 A for a bus 2 fault, 8,341 A for a bus 3 fault, 6,443 A for a bus 4 fault, and 5,163 A for a bus 5 fault. All values are lower than for the single line-to-ground case, except at bus 2 it is slightly higher.

10.61 For a double line-to-ground fault the magnitude of the per unit fault currents are 228,878 A for a bus 1 fault, 1,918 A for a bus 2 fault, 12,190 A for a bus 3 fault, 12,703 A for a bus 4 fault, and 8,943 A for a bus 5 fault. All values are higher than for the single line-to-ground case, except at bus 2 the DLG current is somewhat lower. 10.62 For the Example 10.8 case with a second line between buses 2 and 4, the magnitude of the per unit fault currents for a single line-to-ground fault are 46.137 for a bus 1 fault, 19.458 for a bus 2 fault, 64.360 for a bus 3 fault, 56.183 for a bus 4 fault, and 42.414 for a bus 5 fault. All values are higher than for the Example 10.8 case since the extra line decreases the overall system impedance. However, the change is really only significant for the bus 2 fault. 10.63 For the Example 10.8 case with a second generator added at bus 3, the magnitude of the per unit fault currents for a single line-to-ground fault are 48.545 for a bus 1 fault, 14.74 for a bus 2 fault, 119.322 for a bus 3 fault, 73.167 for a bus 4 fault, and 46.817 for a bus 5 fault. All values are higher than for the Example 10.8 case since the extra generator provides an additional source of fault current. The change is most dramatic for the bus 3 fault since the fault current is not limited by the step-up transformer. Of course, if a second generator were added at a location, undoubtedly a second step-up transformer would also be added. 10.64 For a fault at the ORANGE69 bus the magnitude of the per unit fault current is 15.122. The amount supplied by each of the generators is given below. During the fault 16 buses have their “a” phase voltages below 0.75 per unit. None of the “b” or “c” phases are below 0.75 per unit. Number of

Name of

Phase Cur

Phase

Bus

Ang A

Ang B

Ang C

REDBUD69

0.00000

0.00

BIRCH69

0.89944

0.22944

0.48395

-73.41

-152.86

80.20

CEDAR69

1.59080

0.66251

0.32882

-94.16

-87.52

-100.19

ELM345

1.62440

1.08826

1.82367

-44.42

-142.50

99.37

ELM345

1.62440

1.08826

1.82367

-44.42

-142.50

99.37

PEAR138

2.52518

1.09055

2.28376

-76.13

168.61

78.29

PEACH69

2.71598

0.33630

0.47639

-96.92

-124.25

-19.25

SLACK345

3.01057

1.62359

3.04381

-62.65

-167.07

86.25

PEAR69

5.35004

2.79977

1.27050

-95.91

-115.38

-101.55

10.65 For a fault at the POPULAR69 bus the magnitude of the per unit fault current is 8.272. The amount supplied by each of the generators is given below. During the fault 6 buses have their “a” phase voltages below 0.75 per unit. None of the “b” or “c” phases are below 0.75 per unit. Number

Name of Bus

of Bus

Phase Cur

Phase

Ang A

Ang B

Ang C

REDBUD69

0.00000

0.00

CEDAR69

0.43570

0.25204

0.14654

-72.62

-91.08

141.29

BIRCH69

0.60286

0.28865

0.46673

-61.51

-157.17

86.60

ELM345

1.53183

1.22198

1.69349

-33.65

-138.69

102.18

ELM345

1.53183

1.22198

1.69349

-33.65

-138.69

102.18

PEAR69

1.73205

1.10874

0.57668

-72.16

-139.16

84.36

PEAR138

1.78585

1.30925

1.76086

-55.50

-168.17

81.18

SLACK345

2.60138

1.87442

2.68988

-49.93

-158.12

88.62

Chapter 11 System Protection 11.1 Using Eq. (11.2.1):

1 345  103 V = V = = 115 V (line-to-line) n 3000 S3 700.  106 I= = = 1171.44 A 3 VLL 3 ( 345  103 )

( )

From Eq. (11.2.2), Ie = 0 for zero CT error. Then, from Figure 11.7: I  + Ie = I  + 0 =

1  5  I =  (1171.44 ) n  1200 

I  = 4.88 A

11.2 (a) Step (1) – I = 10 A Step (2) – From Figure 11.7,

E  = ( Z  + Z B ) I  = ( 0.082 + 1)(10 ) = 10.82 V Step (3) – From Figure 11.8, Ie = 0.6 A Step (4) – From Figure 11.7,  100  I =  (10 + 0.6 ) = 212 A  5 

(b) Step (1) – I = 13 A Step (2) – From Figure 11.7, E  = ( Z  + Z B ) I  = ( 0.082 + 1.3 )(13 ) = 18.0 V

Step (3) – From Figure 11.8, Ie = 1.8 A Step (4) – From Figure 11.7,  100  I =  (13 + 1.8) = 296 A  5 

10.

13.

15.

105.

168.

212.

296.

700.

(d) With a 5-A tap setting and a minimum fault-to-pickup ratio of 2, the minimum relay trip current for reliable operation is Imin = 2  5 = 10 A. From (a) above with Imin = 10 A,

I min = 212 A . That is, the relay will trip reliably for fault currents exceeding 212 A. 11.3 From Figure 11.8, the secondary resistance Z  = 0.125  for the 200:5 CT. (a) Step (1) – I = 10 A Step (2) – E = ( Z  + Z B ) I  = ( 0.146 + 1)(10 ) = 11.46 V Step (3) – From Figure 11.8, Ie = 0.11 A  250  Step (4) – I =   (10 + 0.11) = 505.5A  5 

(b) Step (1) – I = 10 A Step (2) – E = ( Z  + Z B ) I  = ( 0.146 + 4 )(10 ) = 41.46 V Step (3) – From Figure 11.8, Ie = 0.55 A  250  Step (4) – I =   (10 + 0.55 ) = 527.5 A  5 

(c) Step (1) – I = 10. A Step (2) – E = ( Z  + Z B ) I  = ( 0.146 + 5 )(10 ) = 51.46 V Step (3) – From Figure 11.8, Ie = 0.7 A  250  Step (4) – I =   (10 + 0.7 ) = 535 A  5 

11.4 (a)

N1 N 2 = 240,000 120 = 2000 1 Vab = 230,0000 2000 = 1150 Vbc = 230,000 − 120 2000 = 115 − 120

Vca = − (Vab + Vbc ) = − (115 − 60 ) = 115 + 120

(b) Vab = 1150; but now Vbc = −115 − 120 = 11560

Vca = − (Vab + Vbc ) = 199150 The output of the PT bank is not balanced three phase. 11.5 Designating secondary voltage as E2, read two points on the magnetization curve (Ie, E2) = (1, 63) and (10, 100). The nonlinear characteristic can be represented by the so-called Frohlich equation

E2 = ( AI e ) ( B + I e ) , using that 63 =

A 10 A and 100 = B +1 B + 10

Solve for A and B: A = 107 and B = 0.698 For parts (a) and (b), ZT = ( 4.9 + 0.1) + j ( 0.5 + 0.5 ) = 5 + j1

= 5.09911.3  (a) The CT error is the percentage of mismatch between the input current (in secondary terms) denoted by I 2 and the output current I 2 in terms of their magnitudes:

CT error =

I 2 − I 2 I 2

 100

ET = I 2 ZT = 4 ( 5.099 ) = 20.4 I e = 20.4

25 + 1 + 107 ( 0.698 + I e )  = 0.163 (by iteration) 2

From Frohlich’s equation E2 =

0.163 (107 )

= 20.3 0.698 − 0.163 E 20.3 I2 = 2 = = 3.97 Z T 5.099 CT error =

0.03 = 0.7% 4

(b) For the faulted case ET = 12 ( 5.099 ) = 61.2 V; I e = 0.894 A (by iteration) E2 = 60.1 V; I 2 = 60.1 5.099 = 11.78 A

CT error =

0.22  100 = 1.8% 12

(c) For the higher burden, ZT = 15 + j 2 = 15.137.6  For the given load condition, ET = 4 (13.13) = 60.5 V I e = 0.814 A; E2 = 57.6 V; I 2 =  CT error =

57.6 = 3.81 A 15.13

0.19  100 = 4.8% 4

(d) For the fault condition, ET = 181.6 V; I e = 9.21 A; E2 = 99.5 V; I 2 =

99.5 = 6.58 A 15.13

 CT error =

5.42  100 = 45.2% 12

Thus, CT error increases with increasing CT current and is further increased by the high terminating impedance. 11.6 Assuming the CT to be ideal, I2 would be 12 A; the device would detect the 1300–A primary current (or any fault current down to 800 A) independent of Z L . (a) In the solution of Problem 11.5(b), I2 = 11.78 A. Therefore, the fault is detected. (b) In Problem 11.5(d), I2 = 6.58 A. The fault is then not detected. The assumption that the CT was ideal in this case would have resulted in failing to detect a faulted system. 11.7 (a) The current tap setting (pickup current) is Ip = 1.0 A. I  10 = = 10. From curve ½ in Figure 11.12 Ip 1

toperating = 0.08 s (b)

I  10 = = 5. Interpolating between curve 1 and curve 2 in Figure 11.12, toperating = 0.55 s Ip 2

(c)

I  10 = = 5. From curve 7, toperating = 3 s Ip 2

(d)

I  10 = = 3.33 From curve 7, toperating = 5.2 s Ip 3

(e)

I  10 =  1 . The relay does not operate. It remains in the blocking position. I p 12

11.8 From the plot of I vs I in Problem 11.2(c), I  = 13 A.

I  13 = = 3.25 Ip 4

From curve 4 in Figure 11.12, toperating = 2.8 s 11.9 (a)  = RC = 0.1 s

v0 = 2 (1 − e−t /0.1 ) ; at t = Tdelay, V0 = 1

1 − e−Tdelay /0.1 = 0.5 or eTdelay /0.1 = 2 Thus Tdelay = 0.1  ln 2 = 0.0693 s (b)  = RC = 1s

v0 = 2 (1 − e−t ) ; at t = Tdelay, V0 = 1  eTdelay = 2 or Tdelay = ln 2 Thus Tdelay = 0.693 s The circuit time response is sketched below:

11.10 From the solution of Prob. 11.5(b), I2 = Irelay = 11.78 A I relay I pickup

11.78 = 2.945 corresponding to which, from curve 2, Toperating = 0.9 s 4

11.11 (a) For the 700. A fault current at bus 3, fault-to-pickup current ratios and relay operating times are:

B3:

I 3 fault TS3

700 ( 200 / 5) 3

17.5 = 5.83 3

From curve ½ of Figure 11.12, toperating3 = 0.10 seconds. Adding the breaker operating time, primary protection clears this fault in (0.10 + 0.083) = 0.183 seconds.

B2 :

I 2 fault TS 2

700 ( 200 / 5 )

17.5 = 3.5 5

From curve 2 in Figure 11.12, toperating2 = 1.3 seconds. The coordination time interval between B3 and B2 is (1.3 − 0.183) = 1.12 seconds. (b) For the 1500-A fault current at bus 2:

B2:

I 2 fault TS 2

1500 ( 200 / 5) 5

37.5 = 7.5 5

From curve 2 of Figure 11.12, toperating2 = 0.55 seconds. Adding the breaker operating time, primary protection clears this fault in (0.55 + 0.083) = 0.633 seconds.

I1fault

B1:

TS1

1500 ( 400 / 5 ) 5

18.75 = 3.75 5

From curve 3 of Figure 11.12, toperating1 = 1.8 seconds. The coordination time interval between B2 and B1 is (1.8 − 0.633) = 1.17 seconds. Fault-to-pickup ratios are all  2.0 Coordination time intervals are all  0.3 seconds. 11.12 First select current Tap settings (TSs). Starting at B3, the primary and secondary CT currents for maximum load L3 are:

IL3 = I L 3 =

SL 3 V3 3

9  106 34.5  103 3

= 150.6 A

150.6 = 3.77 A ( 200 / 5)

From Figure 11.12, select 4-A TS3, which is the lowest TS above 3.77 A.

IL2 = I L 2 =

( SL 2 + SL 3 ) = ( 9.0 + 9.0 )  106 = 301.2 A 34.5  103 3

V2 3

301.2 = 3.77 A ( 400 / 5)

Again, select 4-A TS2 for B2.

I L1 = I L 1 =

SL 1 + SL 2 + SL 3 V1 3

( 9 + 9 + 9 )  106 = 451.8 A 34.5  103 3

451.8 = 3.77 ( 600 / 5)

Again select a 4-A TS1 for B1. Next select Time Dial Settings (TDSs). Starting at B3, the largest fault current through B3 is 3000 A, for the maximum fault at bus 2 (just to the right of B3). The fault to pickup ratio at B3 for this fault is

I 3 fault

TS3

3000 ( 200 / 5 ) 4

= 18.75

Select TDS = 1 2 at B3, in order to clear this fault as rapidly as possible. Then from curve ½ in Fig. 11.12, toperating3 = 0.05 s. Adding the breaker operating time (5 cycles = 0.083 s), primary protection clears this fault in 0.05 + 0.083 = 0.133 s. For this same fault, the fault-to-pickup ratio at B2 is

I 2 fault TS 2

3000 ( 400 / 5 )

37.5 = 9.4 4

Adding B3 relay operating time, breaker operating time, and 0.3 s coordination interval, (0.05 + 0.083 + 0.3) = 0.433 s, which is the desired B2 relay operating time. From Figure 11.12, select TDS2 = 2. Next select the TDS at B1. The largest fault current through B2 is 5000 A, for the maximum fault at bus 1 (just to the right of B2). The fault-to-pickup ratio at B2 for this fault is

I 2 fault TS 2

5000 ( 400 / 5 ) 4

62.5 = 15.6 4

From curve 2 in Fig. 11.12, the relay operating time is 0.38 s. Adding the 0.083 s breaker operating time and 0.3 s coordination time interval, we want a B1 relay operating time of (0.38 + 0.083 + 0.3) = 0.763 s. Also, for this same fault,

I1 fault TS1

5000 ( 600 / 5 ) 4

41.66 = 10.4 4

From Fig. 11.12, select TDS1 = 3.5. Breaker

Relay

TDS

CO-8

3.5

CO-8

Solution Problem 11.7 11.13 For the 1500-A fault current at bus 3, fault-to-pickup current ratios and relay operating times are:

B3:

I 3 fault TS3

1500 ( 200 / 5) 4

37.5 = 9.4 4

From curve ½ of Figure 11.12, toperating3 = 0.08 s. Adding breaker operating time, primary relaying clears this fault in 0.08 + 0.083 = 0.163 s. 299 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

B2:

I 2 fault TS 2

1500 ( 400 / 5 )

18.75 = 4.7 4

From curve 2 in Fig. 11.12, toperating2 = 0.85 s. The coordination time interval between B3 and B2 is (0.85 − 0.163) = 0.69 s.

B1:

I1fault TS1

1500 ( 600 / 5 )

12.5 = 3.1 4

From curve 3.5 in Fig. 11.12, toperating1 = 2.8 s. The coordination time interval between B3 and B1 is (2.8 − 0.163) = 2.6 s. For the 2250-A fault current at bus 2, fault-to-pickup current ratios and relay operating times are:

B2:

I 2 fault TS 2

2250 ( 400 / 5 ) 4

28.13 = 7.0 4

From curve 2 in Fig. 11.12, toperating2 = 0.6 s. Adding breaker operating time, primary protection clears this fault in (0.6 + 0.083) = 0.683 s.

B1:

I1fault TS1

2250 ( 600 / 5) 4

18.75 = 4.7 4

From curve 3.5 in Fig. 11.12, toperating1 = 1.5 s. The coordination time interval between B2 and B1 is (1.5 − 0.683) = 0.82 s. Fault-to-pickup ratios are all  2.0 Coordination time intervals are all  0.3 s 11.14 The load currents are calculated as

I1 =

4  106

3 (11  103 )

= 209.95A; I 2 =

= 131.22 A; I 3 =

6.75  106

3 (11  103 )

2.5  106

3 (11  103 )

= 354.28 A

The normal currents through the sections are then given by

I 21 = I1 = 209.95 A; I 32 = I 21 + I 2 = 341.16 A; I S = I 32 + I 3 = 695.44 A With the given CT ratios, the normal relay currents are i21 =

209.95 341.16 695.44 = 5.25 A; i32 = = 8.53 A; iS = = 8.69 A ( 200 / 5) ( 200 / 5 ) ( 400 / 5 )

Now obtain CTS (Current Tap Settings) or pickup current in such a way that the relay does not trip under normal currents. For this type of relay, CTS available are 4, 5, 6, 7, 8, 10, and 12 A. For position 1, the normal current in the relay is 5.25 A; so choose (CTS)1 = 6 A. Choosing the nearest setting higher than the normal current. For position 2, normal current being 8.53 A, choose (CTS)2 = 10 A. For position 3, normal current being 8.69 A, choose (CTS)3 = 10 A. Next, select the intentional delay indicated by TDS, time dial setting. Utilize the short-circuit currents to coordinate the relays. The current in the relay at 1 on short circuit is iSC1 =

2500 = 62.5A expressed as a multiple of ( 200 / 5)

the CTS or pickup value, iSC 1 62.5 = = 10.42 (CTS )1 6

Choose the lowest TDS for this relay for fastest action. Thus ( TDS )1 =

1 2

Referring to the relay characteristic, the operating time for relay 1 for a fault at 1 is obtained as T11 = 0.15 s. To set the relay at 2 responding to a fault at 1, allow 0.15 for breaker operation and an error margin of 0.3 s in addition to T11. Thus T21 = T11 + 0.1 + 0.3 = 0.55 s Short circuit for a fault at 1 as a multiple of the CTS at 2 is iSC 1 62.5 = = 6.25 CTS ( )2 10

From the characteristics for 0.55 s operating time and 6.25 ratio, (TDS)2 = 2 Now, setting the relay at 3: For a fault at bus 2, the short-circuit current is 3000 A, for which relay 2 responds in a time T22 calculated as iSC 2 3000 = = 7.5 (CTS )2 ( 200 / 5 )10

For (TDS)2 = 2, from the relay characteristic, T22 = 0.5 s. Allowing the same margin for relay 3 to respond for a fault at 2, as for relay 2 responding to a fault at 1, T32 = T22 + 0.1 + 0.3 = 0.9s

The current in the relay expressed as a multiple of pickup is iSC 2 3000 = = 3.75 (CTS )3 ( 400 / 5 )10

Thus, for T3 = 0.9 s, and the above ratio, from the relay characteristic (TDS)3 = 2.5 Note: Calculations here did not account for higher load starting currents that can be as high as 5 to 7 times rated values. 11.15 (a) Three-phase permanent fault on the load side of bus 3. From Table 11.7, the three-phase fault current at bus 3 is 2000 A. From Figure 11.19, the 560 A fast recloser opens 0.04 s after the 2000 A fault occurs, then recloses ½ s later into the permanent fault, opens again after 0.04 s, and recloses into the fault a second time after a 2 s delay. Then the 560 A delayed recloser opens 1.5 s later. During this time interval, the 100 T fuse clears the fault. The delayed recloser then recloses 5 to 10 s later, restoring service to loads 1 and 2. (b) Single line-to-ground permanent fault at bus 4 on the load side of the recloser. From Table 11.7, the IL-G fault current at bus 4 is 2600 A. From Figure 11.19, the 280 A fast recloser (ground unit) opens after 0.034 s, recloses ½ s later into the permanent fault, opens again after 0.034 s, and recloses a second time after a 2 s delay. Then the 280 A delayed recloser (ground unit) opens 0.7 s later, recloses 5 to 10 s later, then opens again after 0.7 s and permanently locks out. (c) Three-phase permanent fault at bus 4 on the source side of the recloser. From Table 11.7, the three-phase fault at bus 4 is 3000 A. From Figure 11.19, the phase overcurrent relay trips after 0.95 s, thereby energizing the circuit breaker trip coil, causing the breaker to open. 11.16 Load current =

3500 3 ( 34.5 )

= 58.57 A ; max. fault current = 1000 A;

min. fault current = 500 A. Note that from Figure 11.19, reclosers have both a fast (assuming no intentional time delay) recloser clearing curve and a time-delayed recloser clearing curve. (a) For this condition, the recloser must open before the fuse melts. The maximum clearing time of the recloser should be less than the minimum melting time of the fuse for all fault currents between 500 and 1000 A. Referring to Figure 11.44, the fuse melt time is about 0.135 s for the 500 A fault, and about 0.029 s for the 1000 A fault. Hence, select a clearing time of 0.029 s for the recloser fast curve.

(b) For this condition, the fuse must open before the recloser operates. The minimum clearing time of the recloser should be greater than the maximum clearing time of the fuse for all fault currents between 500 and 1000 A. Referring to Figure 11.44, the fuse clear time is about 0.4 s for the 500 A fault, and about 0.066 s for the 1000 A fault. Hence, select a clearing time of 0.4 s for the recloser time-delay curve. (c) The fast and time-delayed recloser clearing curves are drawn on the 80E fuse curve as follows;

Recloser Time-Delay Clearing Curve

Recloser Fast Clearing Curve

Note that for the load current of 58.75 A, neither the recloser nor the fuse operates. 11.17 (a) For a fault of P1, only breakers B34 and B43 operate; the other breakers do not operate. B23 should coordinate with B34 so that B34 operates before B23 (and before B12, and before B1). Also, B4 should coordinate with B43 so that B43 operates before B4. (b) For a fault of P2, only breakers B23 and B32 operate; the other breakers do not operate. B12 should coordinate with B23 so that B23 operates before B12 (and before B1). Also B43 should coordinate with B32 so that B32 operates before B43 (and before B4).

(c) For a fault of P3, only breakers B12 and B21 operate; the other breakers do not operate. B32 should coordinate with B21 so that B21 operates before B32 (and before B43, and before B4). Also, B1 should coordinate with B12 so that B12 operates before B1. (d) Fault Bus

Operating Breakers

B1 and B21

B12 and B32

B23 and B43

B4 and B34

11.18

(a) For a fault at P1, breakers in zone 3 operate (B12a and B21a). (b) For a fault at P2, breakers in zone 7 operate (B21a, B21b, B23, B24a, B24b). (c) For a fault at P3, breakers in zone 6 and zone 7 operate (B23, B32, B21a, B21b, B24a, and B24b). 304 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

11.19 (a)

(b) Scheme

Breakers that open for fault on Line 1

Ring Bus

B12 and B14

Breaker- and ½, Double Bus

B11 and B12

Double Breaker, Double Bus

B11 and B21

Scheme

Lines Removed for a Fault at P1

Ring Bus

Line 1 and Line 2

Breaker- and ½, Double Bus

None

Double Breaker, Double Bus

None

Scheme

Breakers that open for a Fault on Line 1

(c)

(d) with Stuck Breaker Ring Bus

B12, B14 and either B23 or B34

Breaker- and ½, Double Bus

B11, B12 and either B13 or B22

Double Breaker, Double Bus

B11, B21 and all other breakers on bus 1 or bus 2.

11.20 (a)

Z =

V ( 4500 / 1)  VLN  1  VLN = LN =  I L I L ( 2500 / 5 )  I L  9

Z =

Z 9

Set the B12 zone 1 relay for 80% reach of line 1–2: Z r1 = 0.8 ( 6 + j 60 ) 9 = 0.53 + j5.33  secondary

Set the B12 zone 2 relay for 120% reach of line 1–2: Z r 2 = 1.2 ( 6 + j 60 ) 9 = 0.8 + j 0.8  secondary

Set the B12 zone 3 relay for 100% reach of line 1–2 and 120% reach of line 2–3: Z r 3 = 1.0 ( 6 + j 60 ) 9 + 1.2 ( 5 + j50 ) 9 = 1.3 + j13.33  secondary

(b) The secondary impedance viewed by B12 during emergency loading is:

500   0  1  VLN   1  3  = 22.925.8  Z =    =  −1  I L   9   1.4 − cos 0.9  9    

Z  exceeds the zone 3 setting of (1.3 + j13.33) = 13.399 84.3  for B12. Hence, the

impedance during emergency loading lies outside the trip region of this 3-zone mho relay (see Figure 11.29 (b)). 11.21 (a) For the bolted three-phase fault at bus 4, the apparent primary impedance seen by the B12 relay is:

Z apparent =

V1 V1 − V2 + V2 (V1 − V2 ) V2 = = + I12 I12 I12 I12 Z12

Z apparent = Z12 +

V2 I12

Using V2 = Z 24 I 24 and I 24 = I12 + I 32 :

Z apparent = Z12 +

Z 24 ( I12 + I 32 ) I12

I  = Z12 + Z 24 +  32  Z 24 which is the desired result.  I12 

(b) The apparent secondary impedance seen by the B12 relay. For the bolted three-phase fault at bus 4 is:

 Z apparent =

Z apparent

( nV / nI )

I   I12 

( 3 + j 40 ) + ( 6 + j80 ) +  32  ( 6 + j80 ) =

( nV / nI )

   I 32    I 32   9 + 6    + j 120. + 80    I12    I12     Z apparent =  ( nV / nI )

where nV is the VT ratio and nI is the CT ratio. Also, the B12 zone 3 relay is set with a secondary impedance: Zr 3 =

( 3 + j 40 ) + 1.2 ( 6 + j80 ) = 10.2 + j136  secondary ( nV / nI ) ( nV / nI )

   Comparing Z r 3 with Z apparent , Z apparent exceeds Z r 3 when ( I 32 / I12 )  0.2 . Hence Z apparent lies outside the trip region for the three-phase fault at bus 4 when ( I 32 / I12 )  0.2 ; remote backup of line 2–4 at B12 is then ineffective.

(1) 2 2

11.22 Rr =

( 22 + 0.82 )

(1) 0.8 2

= 0.431 p.u.; Xr =

( 22 + 0.82 )

= 0.1724 p.u.

The X–R diagram is given below:

Based on the diagram, Z S can be obtained analytically or graphically:

Z S = Z L + Z r = ( 0.1 + 0.431) + j ( 0.3 + 0.1724 ) = 0.710741.66  0.1724    0.431 

 =  S − r = 41.66 − tan −1  = 41.66 − 22 = 19.66 11.23 (a) Given the reaches, Zone 1: Z r = 0.1  80% = 0.08; Zone 2: 0.1  120% = 0.12; Zone 3: 0.1  230% = 0.25

In the view of the system symmetry, all six sets of relays have identical settings. (b) It should be given in the problem statement that the system is the same as Problem 9.11.

VLN base =

230 3

= 133 kV; I L base =

100 0.23 3

= 251 A

The equivalent instrument transformer’s secondary quantities are  115   5  Vbase = 133  = 115V; I base = 251   = 3.14 A  133   400   Z base = 115 / 3.14 = 36.7 

The settings are (by multiplying by 36.7) Zone 1: 2.93 ; Zone 2: 4.40 ; Zone 3: 9.16  (c) The operating region for three zone distance relay with directional restraint as per the arrangement of Fig. 11.50 is shown below:

Locate Point X on the diagram. Comment on line breaker operations: B31: Fault in Zone 1; instantaneous operation B32: Directional unit should block operation B23: Fault in Zone 2; delayed operation B31 should trip first, preventing B23 from tripping. B21: Fault duty is light. Fault in Zone 3, if detected at all. B12: Directional unit should block operation. B13: Fault in Zone 2; just outside of Zone 1; delayed operation Line breakers B13 and B31 clear the fault as desired. In addition, breakers B1 and B4 must be coordinated with B13 so that the trip sequence is B13, B1, and B4 from fastest to slowest. Likewise, B13, B31, and B23 should be faster than B2 and B5. 11.24 For a 20% mismatch between I1 and I 2 , select a 1.20 upper slope in Figure 11.34. That is: 2+k = 1.20 2−k

Solving, k = 0.1818

11.25 (a) Output voltages are given by V1 = jX m I1 = j 5 ( − j16 ) = 80 V V2 = jX m I 2 = j 5 ( − j 7 ) = 35 V V3 = jX m I 3 = j 5 ( j 36 ) = −180 V V4 = jX m I 4 = j 5 ( − j13) = 65 V

V0 = V1 + V2 + V3 + V4 = 80 + 35 − 180 + 65 = 0 Thus there is no voltage to operate the voltage relay VR. For the external fault on line 3, voltages and currents are shown below:

(b) Moving the fault location to the bus, as shown below, the fault currents and corresponding voltages are indicated. Now V0 = 80 + 35 + 50 + 65 = 230 V and the voltage relay VR will trip all four line breakers to clear the fault.

On Line 2

Here V0 = 80 − 195 + 50 + 65 = 0 VR would not operate. (ii) On Line 4 This case is displayed below:

Here V0 = 80 + 35 + 50 − 165 = 0 VR would not operate. 11.26 First select CT ratios. The transformer rated primary (20 kV side) current is: I1 rated =

5  106 = 250 A 20  103

From Table 11.2, select a 300 : 5 CT ratio on the 20 kV (primary) side to give

I1 = ( 250 )( 5 / 300 ) = 4.167 A at rated conditions. Similarly: I 2 rated =

5  106 = 577.4 A 8.66  103

Select a 600 : 5 secondary CT ratio so that I 2 = ( 577.4 )( 5 / 600 ) = 4.811 A at rated conditions.

Next, select relay taps to balance currents in the restraining windings. The ratio of currents in the restraining windings is: I 2 4.811 = = 1.155 I1 4.167

The closest relay tap ratio is T2 / T1 = 1.10 . The percentage mismatch for this tap setting is:

 4.167   4.811  −     I / T − I / T ( ) ( 2 2 )  100 =  5   5.5   100 %Mismatch = 1 1  4.811  ( I2 / T2 )  5    = 4.7% 11.27 Connect CTs in  on the 500 kV Y side, and in Y on the 345 kV  side of the transformer. Rated current on the 345 kV  side is

I a rated =

500.  106 345.  103 3

= 836.7 A

Select a 900 : 5 CT ratio on the 345 kV  side to give Ia = ( 836.7 )( 5 / 900 ) = 4.649 A at rated conditions in the CT secondaries and in the restraining windings. Similarly, rated current on the 500 kV Y side is

I A rated =

500  106 500  103 3

= 577.4 A

Select a 600 : 5 CT ratio on the 500 kV Y side to give I A = ( 577.4 )( 5 / 600 ) = 4.811 A in the 500

 = 4.811 3 = 8.333 A in the restraining windings. kV CT secondaries and I AB Next, select relay taps to balance currents in the restraining windings.  I AB 8.333 = = 1.79 I a 4.649

 Ta = 1.8 The closest tap ratio is TAB For a tap setting of 5 : 9 . The percentage mismatch for this relay tap setting is: %Mismatch =

( I AB / TAB ) − ( Ia / Ta )  100 = (8.333 / 9 ) − ( 4.649 / 5)  100 ( Ia / Ta ) ( 4.649 / 5)

= 0.4%

11.28 The primary line current is

15  106 3 ( 33  103 )

= 262.43 A (say I p )

The secondary line current is 262.43  3 = 787.3 A (say I r )  5  The CT current on the primary side is ip = 262.43   = 4.37 A  300   5  The CT current on the secondary side is is = 787.3   3 = 3.41 A  2000 

[Note:

3 is applied to get the value on the line side of -connected CTs.]

The relay current under normal load is

ir = i p − is = 4.37 − 3.41 = 0.96 A With 1.25 overload ratio, the relay setting should be

ir = 1.25 ( 0.96 ) = 1.2 A 11.29 The primary line current is I p =

30  106 3 ( 33  103 )

= 524.88 A

Secondary line current is I s = 3I p = 1574.64 A  5  The CT current on the primary side is I1 = 524.88   = 5.25 A  500   5  And that on the secondary side is I 2 = 1574.64   3 = 6.82 A  2000 

Relay current at 200% of the rated current is then

2 ( I 2 − I1 ) = 2 ( 6.82 − 5.25 ) = 3.14 A 11.30 Line currents are: I  =

10  106 3 ( 33  103 )

IY =

= 174.95 A

10  106 3 (11  103 )

= 524.86 A

If the CT’s on HV-side are connected in Y, then the CT ratio on the HV-side is 524.86 / 5 = 104.97

(

)

Similarly, the CT ratio on the LV-side is 174.95 5 / 3 = 505.04

Chapter 12 Transient Stability 12.1 (a) syn = 2 60 = 377 rad/s 2 p

msyn = syn =

2 (377) = 188.5rad/s 4

(b) KE=H Srated=4(500×106) =2×109 joules (c) Using (12.1.16)

𝛼=

𝑃𝑎𝑝.𝑢. 𝜔𝑠𝑦𝑛 2𝐻𝜔𝑝𝑢 2

𝛼𝑚 = 𝑃 𝛼 =

syn

 pu (t )  (t ) = Pa p.u. (t )

500 2𝜋60 500

= (2)(4)(1) = 47.1 𝑟𝑎𝑑/𝑠 2

2 (47.1) = 23.55 𝑟𝑎𝑑/𝑠 2 4

12.2 Using (12.1.17) 𝐽=

2𝐻 𝑆𝑟𝑎𝑡𝑒𝑑 (2)(4)(500×106 ) = = 1.1258 × 105 𝑘𝑔𝑚2 𝜔2 𝑚𝑠𝑦𝑛 (188.5)2

12.3 (a) The kinetic energy in ft-lb is: KE =

1  WR 2  2  m ft- lb 2  32.2 

 2  (b) Using m =   ( rpm )  60 

1  WR 2   2  KE =  (rpm)   2  32.2   60 

ft- lb 

1.356 joules ft- lb

KE = 2.31  10−4 (WR2 ) (rpm)2 joules

( 2.31  10−4 )(WR2 ) ( rpm ) per unit-seconds H= 2

Srated

( 2.31 10−4 )( 4  106 ) ( 3600 ) = 15.97 per unit − seconds H= 2

750  106

12.4 Per unit swing equations 2H

 pu (t ) d 2 (t ) = Pmpu (t ) − Pepu (t ) = Papu (t ) syn dt 2

Assuming  pu (t )  1 →

2 H d 2 ( t ) = Papu (t ) wsyn dt 2

2(4) 𝑑 2 𝛿(𝑡) = 0.7 ∗ (0.4)(0.7) = 0.42 (i.e. a 60% reduction to 0.4) 2𝜋60 𝑑𝑡 2

Initial Conditions: δ(0) = 12° = 0.2094 𝑟𝑎𝑑;

𝑑𝛿(0) =0 𝑑𝑡

Integrating twice and using the above initial conditions: 𝑑𝛿(𝑡) = 19.78𝑡 + 0 𝑑𝑡

𝛿(𝑡) = 9.89𝑡 2 + 0.2094 At t=5 cycles=0.08333 seconds 𝛿(5 𝑐𝑦𝑐𝑙𝑒𝑠) = 9.89(0.08333)2 + 0.2094 𝛿(5 𝑐𝑦𝑐𝑙𝑒𝑠) = 0.2781 𝑟𝑎𝑑 = 15.93°

12.5

1 J 2

m2 syn

Srated

; msyn =

2 (2 f ) p

562

𝐻56 = 𝐻60 602

12.6 (a)𝜔𝑠𝑦𝑛 = 2 ∙ 𝜋 ∙ 60 = 377 𝑟𝑎𝑑/𝑠 2

𝜔𝑚𝑠𝑦𝑛 = 𝑝 𝜔𝑠𝑦𝑛 = 16 377 = 47.125 𝑟𝑎𝑑/𝑠 (b) H=1.5 p.u. -s; KE=H Srated=1.5(500×106) =7.5×108 joules (c) δ(0) = 10° = 0.1745 𝑟𝑎𝑑;

𝑑𝛿(0) =0 𝑑𝑡

2(1.5) 𝑑2 𝛿(𝑡) = 0.5 𝑓𝑜𝑟 𝑡 ≥ 0 2𝜋60 𝑑𝑡 2 𝑑𝛿(𝑡) 2𝜋60 =( )𝑡 + 0 𝑑𝑡 6 2𝜋60 2 𝛿(𝑡) = ( ) 𝑡 + 0.1745 12 At t=5 cycles=0.08333 seconds 𝛿(5 𝑐𝑦𝑐𝑙𝑒𝑠) = 31.4(0.08333)2 + 0.2094 = 0.1274 𝑟𝑎𝑑 = 7.3° 12.7 K𝐸𝑔𝑒𝑛 = 𝐻𝑆𝑟𝑎𝑡𝑒𝑑 = (4)(100 × 106 ) = 4 × 108 𝑗𝑜𝑢𝑙𝑒𝑠 1

For a moving mass 𝑊𝑘𝑖𝑛𝑒𝑡𝑖𝑐 = 2 𝑀𝑉 2 To equal the energy stored in the generator K𝐸𝑔𝑒𝑛 = 𝑊𝑘𝑖𝑛𝑒𝑡𝑖𝑐 2(K𝐸𝑔𝑒𝑛 )

𝑉=√

𝑀 2(4×108 𝑗𝑜𝑢𝑙𝑒𝑠) 𝑚 𝑚𝑖𝑙𝑒𝑠 = 81.6 𝑠 = 182.53 ℎ𝑜𝑢𝑟 120,000𝑘𝑔

𝑉=√

12.8

(a) P =

Vt Vbus X

sin  t



sin  t =

( 0.8 )( 0.22 ) = 0.167619 ( P )( X ) = (Vt )(Vbus ) (1.05)(1.0 )

 t = sin −1 ( 0.167619 ) = 9.65 Vt − Vbus 1.05 9.65 − 1.0 0 = jX j 0.22 0.03514 + j 0.1760 I = = 0.81579 − 11.291 j 0.22 I =

S = Vt I * = (1.05 9.65 )( 0.81579 11.291 ) = 0.8565820.941 S = 0.80 + j 0.306 Q = Im S = 0.306 per unit

(b) E  = Vbus + j ( X d + X ) I = 1.00 + j ( 0.3 + 0.22 )( 0.81579 − 11.291 )

E  = 1.0 0 + 0.424278.709 E  = 1.16021.01 per unit (c)

(1.160 )(1.0 ) sin  E  Vbus sin  = ( X d + X ) 0.3 + 0.22

P = 2.231sin  per unit 12.9 Circuit during the fault at bus 3:

where E = 1.160 is determined in Problem 12.8. The Thévenin equivalent, as viewed from the generator internal voltage source, shown here, is the same as in Figure 12.9

(1.160 )( 0.333) sin  = 0.8279sin  per unit E  VTh sin  = xTh 0.4666

12.10

(X42 + X25 ) + X15 = 0.01 + 0.025 (0.01 + 0.05) + 0.02

XTh = X34 + X 45 = 0.0514 

12.11

(X42 + X25 ) = 0.01 + 0.025 (0.05 + 0.10)

XTh = X34 + X 45

= 0.03143 

12.12

(1.2812 )(1.0 ) sin  = 1.8303sin  E  VBUS sin  = Xeq 0.70

 1   = 0.4179 rad  2.4638   1  1 = sin −1   = 0.5780 rad  1.8303 

 0 = sin −1 

A1 = 

1 = 0.5780

 0 = 0.4179

2

(1.0 − 1.8303 sin  ) d =  =0.5780 (1.8303sin  − 1) d = A2 1

( 0.5780 − 0.4179 ) + 1.8303 ( cos0.5780 − cos0.4179 ) = 1.8303 ( cos.5780 − cos  2 ) − ( 2 − 0.5780 ) 1.8303cos  2 +  2 = 2.0907

Solving iteratively (Newton-Raphson)  2 = 0.7439 rad = 42.62

12.13

 0 = 0.4179 rad 1 = 0.4964 rad (From Example 11.4) A1 = 

1 = 0.4964

 0 = 0.4179

1.0d = 

2

1 = 0.4964

( 2.1353sin  − 1.0 ) d

( 0.4964 − 0.4179) = 2.1353 ( cos0.4964 − cos  2 ) − ( 2 − 0.4964) 2.1353cos  2 +  2 = 2.2955

Solving iteratively using Newton Raphson with  2 (0) = 0.60 rad

 2 ( i + 1) =  2 (i) +  −2.1353sin  2 (i) + 1

−1

2.2955 − 2.1353cos  2 (i) −  2 (i)

2

0.60

0.925

0.804

0.785

0.7850

 2 = 0.7850 rad = 44.98 12.14 Referring to Figure 12.8, the critical clearing time occurs when the accelerating area, A1 , is equal to the decelerating area, A3, in which the maximum value of d is determined from the post-fault curve. 1 − 2.1353sin  max = 0 →  max = 2.654rad = 152.1

To determine the clearing time, first solve for the critical clearing angle. The post fault curve is the same as in Problem 12.13, so we have A1 = 

 cr

 0 = 0.4179

1.0d = 

 max = 2.654

 cr

( 2.1353sin  − 1.0 ) d

( cr − 0.4179 ) = 2.1353 ( cos  cr − cos (2.654) ) − ( 2.654 −  cr ) 0.3496 = 2.1353cos  cr →  cr = 1.4063rad = 80.58

Then we can solve for tcr as in Example 12.5 tcr =

4H (1.4063 − 0.4179 ) = 0.1774 sec = 10.6 cycles (2 60)(1.0)

12.15 If gen 3 X d = 0.24 on an 800 MVA base, then it is 0.03 on a 100 MVA base.

(a) Xeq = 0.02 + 0.025 0.15 + 0.04 = 0.0814 *

 3.00  I =  = 3  0 E  = 1.0 + j 0.0814.3  0  1 0  E  = 1 + j 0.2442 = 1.029413.7 

(b) The faulted network is



Xeq = 0.03 + 0.01 + 0.1 0.025 + (0.02 0.05) = 0.0682 Veq = 10 



0.0357 0.1  = 0.51300 0.0557 0.125 1.0294  0.513 sin  0.0682 = 7.742sin 

Pe2 =

Once the fault is cleared, the network becomes

1.0294  1 sin  0.085 = 12.11sin 

Pe3 =

 0 = 13.7 = 0.239rad  4  d 2 (t )  2 60  dt 2 = pmp.u. − pep.u. = 1 − 12.11sin  (t )    4  d  (t )  2 60  dt = t + 12.11cos  (t ) + 0   1 2  4   2 60   (t ) = 2 t + 12.11sin  (t ) + 0.239   4 1   2  2 60   (0.05) = 2 (0.05) + 12.11sin  (0.05) + 0.239   0.011   (0. 05) = 0.0013 + 12.11sin  (0.05) + 0.239 0.011   (0.05) = 0.2402 + 12.11sin  (0.05)

Solving iteratively, the power angle when the fault is cleared is: 0.011   (0.05) = 0.2402 + 12.11sin  (0.05)

 (0.05) == 0.020rad = −1.15 (c) Using the equal area criteria, we can solve for the maximum value of the power angle.  2 H  d 2 (t )  2 60  dt 2 = pmp.u.    (0) =  0 = 0.239rad d  (0) =0 dt  d  (t )  2 60  pmp.u.  t +0  dt  = 2H   2 60  pmp.u.  t 2 + 0 4H 2 60 1 = (0.05) 2 + 0.239 = 0.3175 12 A1 = 10 pmp.u.d  = 10 1d  = 0.3175 − 0.239 = 0.0785

 (t ) =

2 A2 = 12 pmax sin  − pmp.u.d  = 0.3175 12.11sin  − 1d  = A1 = 0.0785

12.11cos(0.3175) − cos  2  − ( 2 − 0.3175) = 0.0785 12.11cos  2 +  2 = 11.744

Solving iteratively, the maximum power angle is:

 2 = − 0.177 rad = − 10.148

12.16 The system has no critical clearing angle because 

 P − P sin  = Accelerating area m

0

 0 = 13.7  = 0.239 rad =3 ( − 0.239) − Pe2 cos 

 0.239

with Pe2 = 7.742 → 3 (2.9026) + 7.742  (−1 − 0.971) = −6.551

is less than zero 12.17

dx = − x, dt

x (0) = 10,  t = 0.1

(a) x (t +  t) = x(t ) +

dx(t ) t dt

x (0.1) = 10 + (− 10)(0.1) = 9 x (0.2) = 9 + (− 9)(0.1) = 8.1 x (0.3) = 8.1 + (− 8.1)(0.1) = 7.29 x (0.4) = 7.29 + (− 7.29)(0.1) = 6.561 x (0.5) = 6.561 + (− 6.561)(0.1) = 5.9049

(b) x(0) = x(0) +

dx(0)  t = 10 + (− 10)(0.1) = 9 dt

dx(0) = −9 dt

(

)

 dx (t ) + dx ( t )  dt dt  x ( t +  t ) = x (t ) +  t   2   (−10 + (− 9)) 0.1 = 9.05 x(0.1) = 10 + ( ) 2

0.1

0.2

0.3

0.4

0.5

8.145

7.371225

6.670959

6.037218

5.463682

9.05

8.19025

7.412176

6.7080195

6.0707577

12.18

dx1 = x2 dt dx2 = − x1 dt

x1 (0) = 1.0  t = 0.1 x2 (0) = 0

dx1 (0)  t = 1.0 + 0 (0.1) = 1.0 dt dx (0) x2 (0.1) = x2 (0) + 2  t = 0 − 1.0 (0.1) = − 0.1 dt x1 (0.2) = 1 + (− 0.1)(0.1) = 0.99

(a) x1 (0.1) = x1 (0) +

x2 (0.2) = (− 0.1) + (− 1) (0.1) = −0.2 x1 (0.3) = 0.99 + (−0.2) (0.1) = 0.97 x2 (0.3) = − 0.2 + (− 0.99)(0.1) = −0.299

1.0

0.99

0.97

− 0.1

− 0.2

− 0.299

 dx  (b) x1 = x1 +  1   t  dt   dx  x2 = x2 +  2   t  dt 

dx1 = x2 dt dx2 = − x1 dt

 dx1 (t ) + dx1 (t )  dt  t x1 (t + t) = x1 (t ) +  dt   2    dx2 (t ) + dx2 (1)  dt  t x2 (t +  t) = x2 (t ) +  dt   2  

x1 (0) = 1.0 + (0) (0.1) = 1.0 x2 (0) = 0 + (− 1.0)(0.1) = − 0.1  dx1 (0) + dx1 (0)  dt  t x1 (0.1) = x1 (0) +  dt   2    0 + (− 0.1)  = 1.0 +   (0.1) = 0.995 2    − 1.0 + (− 1.0)  x2 (0.1) = 0 +   (0.1) = − 0.1 2  

dx1 (0) = − 0.1 dt dx2 (0) = − 1.0 dt

0.1

0.2

0.3

0.994

0.978428

0.956936

−0.1

−0.1995

−0.25773

−0.31753

0.995

0.980025

0.95915

− 0.1

−0.159725

−0.22161

12.19

𝐻 = 20 𝑠𝑒𝑐 𝑝. 𝑢.

𝐷 = 0.1 𝑝. 𝑢.

100 MVA base

(a) 𝑋𝑒𝑞 = 0.0375 + (0.08||0.08) = 0.0775 500 (100) 𝑃 −1 𝐼= ∠ −cos (𝑝𝑓) = ∠0° = 5.0∠0° 𝑉𝑏𝑢𝑠 (𝑝𝑓) (1.0)(1.0) ̅̅̅ 𝐸 ′ = 𝐸 ′ ∠δ = V𝑏𝑢𝑠 + 𝑗𝑋𝑒𝑞 𝐼 = 1.0∠0° + j0.0775(5.0∠0°) = 1 + j0.388 = 1.0726∠21.2063° (b) 𝑆𝑎𝑚𝑒 𝑎𝑠 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 (c) From the fifth line 𝛿0 = ∠21.2063° = 0.3701 rad from (a) 𝛿0.01667 = 0.3701 + (377 − 377)(0.01667) = 0.3701 377 − 0.1(0) 𝜔0.01667 = 377 + (0.01667) [ ] = 377.15711 2(20)(1) 𝛿0.0333 = 0.3701 + (0.01667)(377.15711 − 377) = 0.3727 rad 377 − 0.1(0.15711) 𝜔0.0333 = 377.15711 + 0.01667 [ ] = 377.3142 𝑟𝑎𝑑 2(20)(1)

12.20

The critical clearing time for a sold fault midway between buses 1 and 2 is about 0.07 seconds. Below are

plots of the generator rotor angle with the fault cleared at 0.07 seconds and 0.08 seconds, respectively.

12.21

The critical clearing time for a sold fault midway between buses 1 and 2 is about 0.37 seconds. Below are

plots of the generator rotor angle with the fault cleared at 0.37 seconds and 0.38 seconds, respectively.

12.22 (a) By inspection: 0 0 0  −30 20 10  20 −30 0 10 0 0    10 0 −50 0 40 0 Ybus = j   0 −50 40 0  0 10  0 0 40 40 −100 20    0 0 0 20 −20   0  1    jX d1  (b) Y22 =  0    0 

0 1 jX d 2 0

 0   − j5 0 0     0  =  0 − j10 0  p.u.   0 0 − j10  1  jX d3 

 −1  jX   d1   0  Y12 =  0   0   0   0 

0 −1 jX d2 0 0 0 0

 0     j 5.0 0   0    0 0  =  0 0   0  0    0 −1   jX d3 

0 j10 0 0 0 0

0  3.0 − j 2.0 Yload3 = = 3 − j2 0  12 0  2 − j 0.9 = 2 − j 0.9  p.u. Yload4 = 12 0  Yload5 = 1 − j 0.3 0   j10 

j 20 j10 0 0 0 − j 35   0 j10 0 0  j 20 − j 40  j10 0 (3 − j 2) 0 j 40 0 Y11 =   j10 0 (2 − j 50.9) j 40 0  0  0 0 j 40 j 40 (1 − j100.3) j 20   0 0 0 j 20 − j30  0

12.23

(a) By inspection with line 4-5 open: −30 20 10 0 0 0 20 −30 0 10 0 0 10 0 −50 0 40 0 𝑌𝑏𝑢𝑠 = 0 10 0 −10 0 0 0 0 40 0 −60 20 [ 0 0 0 0 20 −20] 𝑌𝑙𝑜𝑎𝑑3 =

3.0−𝑗2.0 = 3.0 − 𝑗2.0 𝑝. 𝑢. (1.0)2

𝑌𝑙𝑜𝑎𝑑4 =

2.0−𝑗0.9 = 2.0 − 𝑗0.9 𝑝. 𝑢. (1.0)2

𝑌𝑙𝑜𝑎𝑑5 =

1.0−𝑗0.3 = 1.0 − 𝑗0.3 𝑝. 𝑢. (1.0)2

−𝑗35 𝑗20 𝑗10 0 0 0 𝑗20 −𝑗40 0 𝑗10 0 0 𝑗10 0 3 − 𝑗52 0 𝑗40 0 𝑌11 = 0 𝑗10 0 2 − 𝑗10.9 0 0 0 0 𝑗40 0 1 − 𝑗60.3 𝑗20 0 0 0 𝑗20 −𝑗30] [ 0 1 𝑗𝑋′𝑑1

1 𝑗𝑋′𝑑2

1 𝑗𝑋′𝑑3 ]

𝑌22 = [

𝑗5 0 0 𝑗10 0 0 𝑌12 = 0 0 0 0 0 [0

−𝑗5 0 0 −𝑗10 0 ] 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 =[ 0 0 0 −𝑗10

0 0 0 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 0 0 𝑗10]

(b) Y12 and Y22 are the same as Problem 12.22/12.23 (a). −𝑗35 𝑗20 𝑗10 0 0 0 𝑗20 −𝑗40 0 𝑗10 0 0 𝑗10 0 3 − 𝑗52 0 𝑗40 0 𝑌11 = 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 0 𝑗10 0 −𝑗50 𝑗40 0 0 0 𝑗40 𝑗40 1 − 𝑗100.3 𝑗20 0 0 0 𝑗20 −𝑗30] [ 0 12.24

The critical clearing time for a sold fault occurring on Line 1 close to Bus 2 is about 0.17 seconds. Below

are plots of the generator rotor angle with the fault cleared at 0.17 seconds and 0.18 seconds, respectively.

12.25 The critical clearing time for a fault on the line between buses 44 and 14 for the Example 12.9 case with the fault occurring near bus 14 is about 0.584 seconds. Below is a plot of the generator angles with this clearing time.

12.26

The critical clearing time is 0.68 seconds using a time step of 0.01 seconds for the simulation. Note that the

system is stable if the fault is cleared after 0.9 seconds, 1.0 seconds, etc. Part of this is due to this case not having exciter models. Below are plots of the generator rotor angle with the fault cleared at 0.68 seconds and 0.69 seconds, respectively.

(a)𝐼 =

12.27

0.8 (1.0)(1.0)

0 = 0.80

𝑉𝑇 = 1.00 + j0.22(0.80) = 1 + j0.176 = 1.02 ∠ 9.98 𝐸̅ = 1.0239∠12.407 + j(2.0)(0.80)

= 1 + j1.82 = 2.0766 ∠ 61.2134 then δ = 61.2134 𝑉𝑑 0.8764 [𝑉 ] = [ 0.4815 𝑞

0.7917 −0.4815 1.0 ][ ]=[ ] 0.6357 0.8764 0.176

𝐼𝑑 0.8764 [𝐼 ] = [ 0.4815 𝑞

0.7011 −0.4815 0.8 ][ ] = [ ] 0 0.3852 0.8764

𝐸𝑞 ′ = 𝑉𝑞 + 𝑅𝑎 𝐼𝑞 + 𝑋𝑑 𝐼𝑑 = 0.6357 + (0)(0.3852) + (0.3)(0.7011) = 0.8460 𝐸𝑑 ′ = 𝑉𝑑 + 𝑅𝑑 𝐼𝑑 − 𝑋𝑞 𝐼𝑞 = 0.7917 + 0 − (0.5)(0.3852) = 0.5991 𝑑𝐸𝑞 𝐸𝑓𝑑 = 𝑇𝑑 ( ) + (𝑋𝑑 − 𝑋𝑑′ )𝐼𝑑 + 𝐸𝑞 𝑑𝑡 = 0 + (2.1 − 0.3)0.7011 + 0.8460 = 2.1080 (b) The critical clearing time is 0.35 seconds using a time step of 0.01 seconds for the simulation. Below are plots of the generator rotor angle with the fault cleared at 0.35 seconds and 0.36 seconds, respectively.

12.28 (a) X  = X a +

X1 X m (0.17)(3.8) = 0.2297 p.u. = 0.067 + X1 + X m (0.17 + 3.8)

X = X a + X m = 3.867 p.u. T0 = I=

X1 + X m (0.17 + 3.8) = 0.85 s =  0 R1 (2 60)(0.0124)

1.0  0 = 1  0 (1.0)(1.0)

Vt = 1  0  + ( j 0.22)(16 ) = 1 + j 0.22 Vr = Er − Ra I r + X I i

= 0.8475 − (0.013)(0.8433) + (0.2297)(0.7119) = 1.0

Vi = E1 − Ra Ii − X Ir

= 0.4230 − (0.013)(0.7119) + (0.2297)(0.8433) = 0.22

dEr 1 = (2 60)(− 0.0129)(0.4230) − (0.8475 − (3.867 − 0.2297)(0.7119) = 0 dt 0.85 dEi 1 = − (2 60)(− 0.0129)(0.8475) − (0.4230 + (3.867 − 0.2297)(0.8433)) = 0 dt 0.85

(b) 1

12.29 pf =

= 0.9766 12 + 0.222 P 1.0 I=  − cos−1 ( pf ) =  − cos−1 (0.9766) Vbus ( pf ) (1.0)(0.9766)

VT = 1.0 + ( j 0.22)(1 − j 0.215) = 1.0473 + j 0.22 = 1.070211.863 

I sore = I −

VT (1.0473 + j022) = (1 − j 0.215) + jXeq j 0.8 = 1.275 − j1.524

I p + jI q = (1.275 − j1.524)(1 − 11.863 ) = 1.987 (− 50.084 − 11.863) = 1.987 − 61.947  = 0.934 − j1.754

Eq = −I q Xeq = (1.754)(0.8) = 1.403

12.32 𝑃𝐿𝑜𝑎𝑑 = 𝑃𝐵𝑎𝑠𝑒𝐿𝑜𝑎𝑑 (𝑃𝑧 |𝑉̅ |2 + 𝑃𝑖 |𝑉̅ | + 𝑃𝑝 ) = (100MW)(0.3 × 0.852 + 0.4 × 0.85 + 0.3) = 21.675 + 34 + 30 = 85.675 𝑀𝑊 𝑄𝐿𝑜𝑎𝑑 = 𝑄𝐵𝑎𝑠𝑒𝐿𝑜𝑎𝑑 (𝑄𝑧 |𝑉̅ |2 + 𝑄𝑖 |𝑉̅ | + 𝑄𝑝 ) = 50 𝑀𝑣𝑎𝑟(0.5 × 0.852 + 0.4 × 0.85 + 0.1) = 18.0625 + 17 + 5 = 40.0625 𝑀𝑣𝑎𝑟

12.33 (a) The input impedance and current are 𝑅 𝑗𝑋𝑚 ( 1 + 𝑗𝑋1 ) 𝑠 𝑍𝑖𝑛 = (𝑅𝑎 + 𝑗𝑋𝑎 ) + = 0.1278 + 𝑗0.2338 𝑅1 + 𝑗(𝑋1 + 𝑋𝑚 ) 𝑠 𝑉̅ 1.00 𝐼̅ = = 𝑍𝑖𝑛 𝑍𝑖𝑛 Then with 𝑠 = 0.01 we get 𝐼̅ =

𝑉̅ 1.00 1.00 = = = 1.8 − 𝑗3.293 𝑍𝑖𝑛 𝑍𝑖𝑛 0.1278 + 𝑗0.2338 𝑆 = 1.8 + 𝑗3.293

Real power comsuption:1.8×100=180 MW Reactive power consumption: 3.293×100=329.3 Mvar (b) if s=0.02 𝑅 𝑗𝑋𝑚 ( 1 + 𝑗𝑋1 ) 𝑠 𝑍𝑖𝑛 = (𝑅𝑎 + 𝑗𝑋𝑎 ) + = 0.064 + 𝑗0.2304 𝑅1 + 𝑗(𝑋1 + 𝑋𝑚 ) 𝑠 ̅ 𝑉 1.00 1.00 𝐼̅ = = = = 1.119 − 𝑗4.029 𝑍𝑖𝑛 𝑍𝑖𝑛 0.064 + 𝑗0.2304 𝑆 = 1.119 + 𝑗4.029 Real power comsuption:1.119×100=111.9 MW Reactive power consumption: 4.029×100=402.9 Mvar

12.34 𝑅 0.054138 𝑗𝑋𝑚 ( 1 + 𝑗𝑋1 ) 𝑗 − 0.9126 𝑠 𝑠 𝑍𝑖𝑛 = (𝑅𝑎 + 𝑗𝑋𝑎 ) + = 𝑅1 0.014 + 𝑗(𝑋1 + 𝑋𝑚 ) + 𝑗3.98 𝑠 𝑠 0.014 + 𝑗3.98 𝑉̅ 1.00 𝑠 ̅𝐼 = = = 0.054138 𝑍𝑖𝑛 𝑍𝑖𝑛 𝑗 − 0.9126 𝑠 S = 0.292

12.35 The fault time duration at which the motor stalls is 0.24 seconds. Below are plots of the bus voltage magnitude and the motor speed with the fault cleared at 0.23 seconds and 0.24 seconds, respectively. If the fault is cleared after 0.23s:

If the fault is cleared after 0.24s:

Chapter 13 Power System Controls 13.1 (a) The open-loop transfer function G(S) is given by G (S) =

(b)

k a ke k f

(1 + Ta S )(1 + Te S ) (1 + T f S )

(1 + Ta S )(1 + Te S ) (1 + T f S ) e 1 = = Vref 1 + G ( S ) (1 + Ta S )(1 + Te S ) (1 + T f S ) + ka ke k f For steady state, setting S = 0 eSS =

or 1 + k = ( Vref )

( V ) , where k = k k k ref

a e

1+ k

eSS

For the condition stipulated, 1 + k  100 or k  99  G (S)  (c) Vt ( t ) = ‹ −1  Vref ( S )  1 + G ( S ) 

The response of the system will depend on the characteristic roots of the equation 1 + G (S) = 0

(i)

If the roots S1, S2, and S3 are real and distinct, the response will then include the transient components A1eS1t , A2 eS2t , and A3e S3t .

(ii) If there are a pair of complex conjugate roots S1, S2 ( = a  j ) , then the dynamic response will be of the form Ae at sin ( t +  ) . 13.2 (a) The open-loop transfer function of the AVR system is K G (S) H (S) = =

KA (1 + 0.1S )(1 + 0.4S )(1 + S )(1 + 0.05S )

500 K A S + 33.5 S + 307.5 S 2 + 775 S + 500 4

The closed-loop transfer function of the system is

Vt ( S )

Vref ( S )

25K A ( S + 20 ) S 4 + 33.5 S 3 + 307.5 S 2 + 775 S + 500 + 500 K A

(b) The characteristic equation is given by

1 + K G(S ) H (S ) = 1 +

500 K A =0 S + 33.5S + 307.5S 2 + 775S + 500 4

Which results in the characteristic Polynomial Equation S 4 + 33.5S 3 + 307.5S 2 + 775S + 500 + 500 K A = 0

The Routh-Hurwitz array for this polynomial is shown below: S4

307.5

500 + 500KA

33.5

775

284.365

500 + 500KA

58.9KA − 716.1

500 + 500KA

From the S1 row, it is seen that KA must be less than 13.16 for control system stability; also from the S0 row, KA must be greater than −1. Thus, with positive values of KA, for control system stability, the amplifier gain must be K A  12.16 .

For K = 13.16, the Auxiliary Equation from the S2 row is 284.365 S 2 + 6580 = 0 or S =  j 4.81

That is, for K = 13.16, there are a pair of conjugate poles on the j axis, and the control system is marginally stable. (c) From the closed-loop transfer function of the system, the steady-state response is

SVt ( S ) = (Vt )SS = lim S →0

KA 1 + KA

For the amplifier gain of KA = 10, the steady-state response is

(Vt )SS =

10 = 0.909 1 + 10

And the steady-state error is

(Ve )SS = 1.0 − 0.909 = 0.091 13.3 System becomes unstable at about K a = 277. Anything with +/– about 10 of this answer should be considered correct.

13.4

13.5

Initial Values

Ka = 100

Ka = 300

Ka = 30

Ka = 10

Vt = 1.0946 pu

Vt = 1.0958 pu

Vt = 1.1033 pu

Vt = 1.0983 pu

Vt = 1.1033 pu

Efd = 2.913

Efd = 2.794

Efd = 3.311

Efd = 2.805

Efd = 2.827

(a) Converting constraints to a 100 MVA base: 100 𝑀𝑉𝐴 ) = 0.02 𝑝. 𝑢. 200 𝑀𝑉𝐴 100 𝑀𝑉𝐴 ) = 0.01667 𝑝. 𝑢. 𝑅2,𝑛𝑒𝑤 = 0.05 ( 300 𝑀𝑉𝐴 100 𝑀𝑉𝐴 ) = 0.012 𝑝. 𝑢. 𝑅3,𝑛𝑒𝑤 = 0.06 ( 500 𝑀𝑉𝐴 𝑅1,𝑛𝑒𝑤 = 0.04 (

From Equation 13.2.3, the area frequency response characteristic 𝛽 is: 𝛽=( ⇒ 𝛽=(

1 𝑅1,𝑛𝑒𝑤

1 𝑅2,𝑛𝑒𝑤

1 1 + +⋯) 𝑅1 𝑅2

1 1 1 ) = 𝟏𝟗𝟑. 𝟑𝟑𝟑 𝒑. 𝒖. )=( + + 𝑅3,𝑛𝑒𝑤 0.02 0.01667 0.012 150

(b) The load decreases by 150 MW, so Δ𝑝𝑚 = − 100 = −1.5 𝑝. 𝑢. Since the reference power setting does not change, Δ𝑝𝑟𝑒𝑓 = 0 Using equation 13.2.4: Δ𝑝𝑚 = Δ𝑝𝑟𝑒𝑓 − 𝛽Δ𝑓 ⇒ Δ𝑓 = − ⇒ Δ𝑓 = −

Δ𝑝𝑚 − Δ𝑝𝑟𝑒𝑓 𝛽

−1.5 − 0 = 0.00776 𝑝. 𝑢. = 𝟎. 𝟒𝟔𝟔 𝑯𝒛 193.333

(c) For each individual turbine, the change in mechanical power is from Equation 13.2.1: Δ𝑝𝑚 = Δ𝑝𝑟𝑒𝑓 − ⇒ Δ𝑝𝑚1 = Δ𝑝𝑟𝑒𝑓 − ⇒ Δ𝑝𝑚2 = Δ𝑝𝑟𝑒𝑓 −

1 𝑅1,𝑛𝑒𝑤 1

𝑅2,𝑛𝑒𝑤

⇒ Δ𝑝𝑚3 = Δ𝑝𝑟𝑒𝑓 −

Δ𝑓 = −

1 𝑅3,𝑛𝑒𝑤

1 Δ𝑓 𝑅

1 0.00776 = −0.3879 𝑝. 𝑢. = −𝟑𝟖. 𝟕𝟗 𝑴𝑾 0.02

1 0.00776 = −0.4655 𝑝. 𝑢. = −𝟒𝟔. 𝟓𝟓 𝑴𝑾 0.01667

Δ𝑓 = −

1 0.00776 = −0.6466 𝑝. 𝑢. = −𝟔𝟒. 𝟔𝟔 𝑴𝑾 0.012

13.6 100

Now, Δ𝑝𝑚 = 100 = 1 𝑝. 𝑢. and Δ𝑝𝑟𝑒𝑓 = 0 (a) Using Equation 13.2.4: Δ𝑓 = −

Δ𝑝𝑚 − Δ𝑝𝑟𝑒𝑓 1−0 =− = −5.172 × 10−3 𝑝. 𝑢. = −𝟎. 𝟑𝟏𝟎 𝑯𝒛 𝛽 193.333

(b) Using Equation 13.2.1 with the new Δ𝑓: ⇒ Δ𝑝𝑚1 = Δ𝑝𝑟𝑒𝑓 − ⇒ Δ𝑝𝑚2 = Δ𝑝𝑟𝑒𝑓 −

1 𝑅1,𝑛𝑒𝑤 1

𝑅2,𝑛𝑒𝑤

⇒ Δ𝑝𝑚3 = Δ𝑝𝑟𝑒𝑓 −

Δ𝑓 = −

1 𝑅3,𝑛𝑒𝑤

1 (−5.172 × 10−3 ) = 0.2586 𝑝. 𝑢. = 𝟐𝟓. 𝟖𝟔 𝑴𝑾 0.02

1 (−5.172 × 10−3 ) = 0.3103 𝑝. 𝑢. = 𝟑𝟏. 𝟎𝟑 𝑴𝑾 0.01667

Δ𝑓 = −

1 (−5.172 × 10−3 ) = 0.4310 𝑝. 𝑢. = 𝟒𝟑. 𝟏𝟎 𝑴𝑾 0.012

13.7 Now, Δ𝑓 = 0.005 𝑝. 𝑢. and Δ𝑝𝑟𝑒𝑓 = 0

Using Equation 13.2.1 with the new Δ𝑓: ⇒ Δ𝑝𝑚1 = Δ𝑝𝑟𝑒𝑓 − ⇒ Δ𝑝𝑚2 = Δ𝑝𝑟𝑒𝑓 −

1 𝑅1,𝑛𝑒𝑤

1 𝑅2,𝑛𝑒𝑤

⇒ Δ𝑝𝑚3 = Δ𝑝𝑟𝑒𝑓 −

Δ𝑓 = −

1 (0.005) = −0.25 𝑝. 𝑢. = −𝟐𝟓 𝑴𝑾 0.02

Δ𝑓 = −

1 (0.005) = −0.3750 𝑝. 𝑢. = −𝟑𝟕. 𝟓𝟎 𝑴𝑾 0.01667

Δ𝑓 = −

1 (0.005) = −0.41667 𝑝. 𝑢. = −𝟒𝟏. 𝟔𝟔𝟕 𝑴𝑾 0.012

1 𝑅3,𝑛𝑒𝑤

13.8 Now, Δ𝑓 = −0.005 𝑝. 𝑢.

Using Equation 13.2.1 with the new Δ𝑓: ⇒ Δ𝑝𝑚1 = Δ𝑝𝑟𝑒𝑓 − ⇒ Δ𝑝𝑚2 = Δ𝑝𝑟𝑒𝑓 − ⇒ Δ𝑝𝑚3 = Δ𝑝𝑟𝑒𝑓 −

1 𝑅1,𝑛𝑒𝑤

1 𝑅2,𝑛𝑒𝑤 1 𝑅3,𝑛𝑒𝑤

Δ𝑓 = −

1 (−0.005) = 0.25 𝑝. 𝑢. = 𝟐𝟓 𝑴𝑾 0.02

Δ𝑓 = −

1 (−0.005) = 0.3750 𝑝. 𝑢. = 𝟑𝟕. 𝟓𝟎 𝑴𝑾 0.01667

Δ𝑓 = −

1 (−0.005) = 0.41667 𝑝. 𝑢. = 𝟒𝟏. 𝟔𝟔𝟕 𝑴𝑾 0.012

13.9 (a)

Using Rnew = Rold

Sbase( new ) Sbase( old )

R1( new) = 0.04

1000 1000 = 0.08pu; R2( new) = 0.06 = 0.08pu 500 750

The area frequency-response characteristic is given by n

1 1 1 1 1 = + = + = 25pu R1 R2 0.08 0.08 k =1 Rk

 =

(b) The per-unit increase in load is 250 /1000 = 0.25 n

k =1

 n 1   f = Pref ( total ) − f  k =1 Rk 

( Pm )TOTAL =  Pmk =  Pref k −   With Pref ( total ) = 0 for steady-state conditions, f = −



Pm = −

1 ( 0.25) = −10  10−3 pu 25

or f = −10  10−3  60 = −0.6 Hz . 13.10

(a) First the regulation constants must be converted to a common 1000 MVA base: 1000 𝑀𝑉𝐴 ) = 0.08 𝑝. 𝑢. 𝑅1,𝑛𝑒𝑤 = 0.04 ( 500 𝑀𝑉𝐴 1000 𝑀𝑉𝐴 ) = 0.05 𝑝. 𝑢. 𝑅2,𝑛𝑒𝑤 = 0.05 ( 1000 𝑀𝑉𝐴 1000 𝑀𝑉𝐴 ) = 0.0667 𝑝. 𝑢. 𝑅3,𝑛𝑒𝑤 = 0.05 ( 750 𝑀𝑉𝐴 To solve for 𝛽, we use Equation 13.2.3: 𝛽=( ⇒ 𝛽=(

1 𝑅1,𝑛𝑒𝑤

1 𝑅2,𝑛𝑒𝑤

1 1 + +⋯) 𝑅1 𝑅2

1 1 1 ) = 𝟒𝟕. 𝟓 𝒑. 𝒖. )=( + + 𝑅3,𝑛𝑒𝑤 0.08 0.05 0.0667

(b) Given Δ𝑝𝑟𝑒𝑓 = 0 and using Equation 13.2.4: Δ𝑓 = −

Δ𝑝𝑚 − Δ𝑝𝑟𝑒𝑓 𝛽

Δ𝑝𝑚 = 200 𝑀𝑊 = 0.2 𝑀𝑉𝐴 ⇒ Δ𝑓 = −

0.2 𝑝. 𝑢. = −4.211 × 10−3 𝑝. 𝑢. = −𝟎. 𝟐𝟓𝟑 𝑯𝒛 47.5 𝑝. 𝑢. 337

1 𝑅1,𝑛𝑒𝑤 1

𝑅2,𝑛𝑒𝑤 1

𝑅3,𝑛𝑒𝑤

Δ𝑓 = −

1 (−0.253) = 0.0526 𝑝. 𝑢. = 𝟓𝟐. 𝟔𝟑𝟐 𝑴𝑾 0.08

Δ𝑓 = −

1 (−0.253) = 0.0842 𝑝. 𝑢. = 𝟖𝟒. 𝟐𝟏𝟏 𝑴𝑾 0.05

Δ𝑓 = −

1 (−0.253) = 0.0632 𝑝. 𝑢. = 𝟔𝟑. 𝟏𝟓𝟖 𝑴𝑾 0.0667

13.11 Given the contingency is a loss of 65 MW, Δ𝑝𝑚 = 65 𝑀𝑊 = 0.65 𝑝. 𝑢. for a 100 MVA

base. Assuming Δ𝑝𝑟𝑒𝑓 = 0, from Equation 13.2.4: Δ𝑓 = −

Δ𝑝𝑚 𝛽

In order to solve for 𝛽, all the regulation constraints R are the same, and the new base power is the same (100 MVA). 𝑆𝑏𝑎𝑠𝑒,𝑛𝑒𝑤 𝑅𝑛𝑒𝑤 = 𝑅𝑜𝑙𝑑 ( ) 𝑆𝑏𝑎𝑠𝑒,𝑜𝑙𝑑 1 1 1 𝛽=( + + ⋯+ ) 𝑅1,𝑛𝑒𝑤 𝑅2,𝑛𝑒𝑤 𝑅𝑛,𝑛𝑒𝑤 =(

𝑆𝑏𝑎𝑠𝑒,𝑜𝑙𝑑,1 𝑆𝑏𝑎𝑠𝑒,𝑜𝑙𝑑,2 𝑆𝑏𝑎𝑠𝑒,𝑜𝑙𝑑,𝑛 + +⋯+ ) 𝑆𝑏𝑎𝑠𝑒,𝑛𝑒𝑤 ∙ 𝑅𝑜𝑙𝑑 𝑆𝑏𝑎𝑠𝑒,𝑛𝑒𝑤 ∙ 𝑅𝑜𝑙𝑑 𝑆𝑏𝑎𝑠𝑒,𝑛𝑒𝑤 ∙ 𝑅𝑜𝑙𝑑 ⇒𝛽=

∑𝑆𝑏𝑎𝑠𝑒,𝑜𝑙𝑑 𝑆𝑏𝑎𝑠𝑒,𝑛𝑒𝑤 ∙ 𝑅𝑜𝑙𝑑

Neglecting the PEAR69 generator, which is the contingency generator, we get: 𝛽= ⇒ Δ𝑓 = −

1102 𝑀𝑉𝐴 = 220.4 𝑝. 𝑢. 100 𝑀𝑉𝐴 (0.05 𝑝. 𝑢. )

Δ𝑝𝑚 0.65 =− = −0.00295 𝑝. 𝑢. = −0.177 𝐻𝑧 𝛽 220.4

Looking at the plot, it’s noted that the final frequency is approximately 59.831 Hz, which is close to the calculated frequency drop of 0.177 Hz. The largest frequency deviation is at 3.708 seconds, with a frequency drop of approximately 0.416 Hz, from 60 Hz to 59.584 Hz. 13.12

The final frequency is approximately 59.832 Hz, indicating that doubling the H values does not impact the final frequency significantly. However, the largest bus frequency deviation occurs much later, at 5.125 seconds, with a deviation of 0.353 Hz, from 60 Hz to 59.647 Hz.

13.13

Using a 60 Hz base, the per unit frequency change is − 0.12/60 = − 0.002 p.u. Since all the

governor R values are identical (on their own base), the total MVA of the generation is given by  P * R /f = (2800*0.04)/0.02 = 5600 MW. 339 © 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

13.14 Adding (13.2.4) for each area with Pref = 0 : Pm1 + Pm 2 = − ( 1 +  2 ) f 400 = − ( 400 + 600 ) f  f = −0.4 H z

Ptie 2 = Pm 2 = −2 f = −600 ( −0.4 ) = 240MW

Ptie1 = −Ptie 2 = −240 MW

13.15 In steady-state, ACE2 = Ptie 2 + B f 2 f = 0

and

Ptie 2 = Pm 2

Pm 2 = Ptie 2 = − B f 2 f

and Pm1 = − 1f Also Pm1 + Pm 2 = 400 MW Solving the equations: − ( 1 + B f 2 ) f = 400

f =

−400 = −0.4 Hz 400 + 600

Ptie 2 = − ( 600 )( −0.4 ) = 240MW Ptie1 = −Ptie 2 = −240MW

Note: The results are the same as those in Problem 13.14. That is, LFC is not effective when employed in only one area. 13.16 In steady-state: ACE1 = Ptie1 + B f 1f = 0

ACE2 = Ptie2 + B f 2 f = 0 Adding ( Ptie1 + Ptie 2 ) + ( B f 1 + B f 2 ) f = 0 Therefore, f = 0 ; Ptie 1 = 0 and Ptie 2 = 0 . That is, in steady-state the frequency error is returned to zero, area 1 picks up its own 400 MW load increase.

13.17 (a) (13.15) LFC in area 2 alone. In steady-state ACE2 = Ptie 2 + B f 2 f = 0 Pm 2 = Ptie 2 = − B f 2 f

and Pm1 = − 1f Also Pm1 + Pm 2 = −300 Solving: − ( 1 + B f 2 ) f = −300

f =

300 = 0.3Hz 400 + 600

Ptie 1 = − ( 400 )( 0.3) = −120MW

Ptie 2 = −Ptie 1 = 120 MW

(b) (13.16) LFC employed in both areas 1 and 2. ACE1 = Ptie1 + B f 1f = 0

ACE2 = Ptie2 + B f 2 f = 0

(

)

Adding: ( Ptie1 + Ptie 2 ) + B f1 + B f 2 f = 0 Thus f = 0 Ptie1 = 0 and Ptie 2 = 0 13.18 (a) The per-unit load change in area 1 is

PL1 =

187.5 = 0.1875 1000

The per-unit steady-state frequency deviation is WSS =

−PL1 −0.1875 = = −0.005  1   1  ( 20 + 0.6 ) + (16 + 0.9 ) + D2   + D1  +   R1   R2 

Thus, the steady-state frequency deviation in Hz is f = ( −0.005 )( 60 ) = −0.3Hz

and the new frequency is f = f0 + f = 60 − 0.3 = 59.7 Hz . The change in mechanical power in each area is

Pm1 = −

W −0.005 =− = 0.1pu = 100 MW R1 0.05

Pm 2 = −

W −0.005 =− = 0.08pu = 80 MW R2 0.0625 341

Thus area 1 increases the generation by 100 MW and area 2 by 80 MW at the new operating frequency of 59.7 Hz. The total change in generation is 180 MW, which is 7.5 MW less than the 187.5 MW load change because of the change in the area loads due to frequency drop. The change in area 1 load is W  D1 = ( −0.005 )( 0.6 ) = −0.003pu or −3.0 MW , and the change in area 2 load is W  D2 = ( −0.005 )( 0.9 ) = −0.0045pu or −4.5 MW. Thus, the change in the total area load is −7.5MW. The tie-line power flow is  1  P12 = W  + D2  = −0.005 (16.9 ) = −0.0845pu  R2  = −84.5MW

That is, 84.5 MW flows from area 2 to area 1. 80 MW comes from the increased generation in area 2, and 4.3MW comes from the reduction in area 2 load due to frequency drop. (b) With the inclusion of the ACE5 , the frequency deviation returns to zero (with a settling time of about 20 seconds). Also, the tie-line power change reduces to zero, and the increase in area 1 load is met by the increase in generation Pm1 .

13.19 Given that the goal is to provide a total of 60° of phase lead compensation, each lead-lag block 𝑟𝑎𝑑

provides 30°. Considering that this is provided at 2 Hz, 𝜔𝑑 = 2𝜋𝑓 = 2𝜋(2) = 4𝜋 𝑠𝑒𝑐 . Using Equation 13.4.3: 𝛼=

1 − sin 𝜙 1 − sin 30° 1 = = 1 + sin 𝜙 1 + sin 30° 3

Using Equation 13.4.4: 𝑇1 =

1 𝜔 𝑑 √𝛼

= 0.138 ⇒ 𝑇2 = 𝛼𝑇1 = 0.0459 ⇒ 𝑻𝟏 = 𝟎. 𝟏𝟑𝟖 𝒔 ⇒ 𝑻𝟐 = 𝟎. 𝟎𝟒𝟓𝟗 𝒔

13.20 Given T1 = T3 = 0.4s and T2 = T4 = 0.1s, we can use Equations 13.4.4 to solve a simple system of equations: 0.4 =

0.1 = 𝛼0.4

𝜔 𝑑 √𝛼

⇒ 𝛼 = 0.25 ⇒ 𝜔𝑑 = 5 Then we use Equation 13.4.3 to solve for 𝜙: 0.25 =

1 − sin 𝜙 ⇒ 𝝓 = 𝟑𝟔. 𝟖𝟔𝟗𝟗° 1 + sin 𝜙

𝜔𝑑 = 2𝜋𝑓 ⇒ 𝑓 =

5 ⇒ 𝒇 = 𝟎. 𝟕𝟗𝟔 𝑯𝒛 2𝜋

PW13.21 Given the total phase lead is 0° at 1.36 Hz, using Equations 13.4.3 and 13.4.4: 𝜔𝑑 = 2 𝜋 × 1.36 𝐻𝑧 𝛼= 𝑇1 =

1 − sin 𝜙 1 − sin 0° = =1 1 + sin 𝜙 1 + sin 0°

1 𝜔 𝑑 √𝛼

= 0.117 ⇒ 𝑇2 = 𝛼𝑇1 = 0.117 ⇒ 𝑇1 = 0.117 𝑠 ⇒ 𝑇2 = 0.117 𝑠

Without a stabilizer enabled, the frequency oscillations are as seen above, with a damping of 5.825% for a 1.36 Hz mode. Setting the values for T1, T2, T3, and T4 to the ones above, a Ks value of 45 results in almost instability, as seen in the Generator Vs plot:

Thus, when we tune Ks value to 15, we get a damping of 4.225% at 1.421 Hz. PW13.22 Given the total phase lead is 30° at 1.36 Hz, using Equations 13.4.3 and 13.4.4: 𝜔𝑑 = 2 𝜋 × 1.36 𝐻𝑧 𝛼= 𝑇1 =

1 − sin 𝜙 1 − sin 15° = = 0.589 1 + sin 𝜙 1 + sin 15° 1

𝜔 𝑑 √𝛼

= 0.153 ⇒ 𝑇2 = 𝛼𝑇1 = 0.0898 ⇒ 𝑇1 = 0.153 𝑠 ⇒ 𝑇2 = 0.0898 𝑠

Similar to PW13.21, there is a 1.36 Hz mode with damping of 5.825% to be worked on. A value of Ks of 120 gives the following plot:

Thus, when we tune the Ks value to 40, we get a damping of 7.32% at 1.492 Hz.

Chapter 14 Transmission Lines: Transient Operation 14.1 From the results of Example 14.2: V ( x, t ) = i ( x, t ) =

E x  x E   U −1  t −  + U −1  t + − 2  2 v  v 2  

E x  x E   U −1  t −  − U −1  t + − 2  2 Zc v 2 Z v     c

For t =  / 2 =

l : 2v

3  l    2−x E  x − 2l    E V  x,  = U −1   + U −1    2 2  v  2  v      3  l    2−x  x − 2l  E E   i  x,  = U −1  U −1  −   2  2 Zc  v  2 Z c  v     

l For t =  = : v V ( x, ) =

E E E l−x E  x −l  l −x  x −l  U −1  U −1  U −1   + U −1   i ( x, ) = −  2 2 Zc  v  2  v   v  2 Zc  v 

For t = 2 = V ( x, 2 ) =

2l : w

E E E  2l − x  E x  2l − x  x U −1  U −1  U −1    + U −1   i ( x, 2 ) = − 2 2 Zc  v  2 v  v  2 Zc v

14.2 From Example 14.2  R = 1 and  S = 0 For a ramp voltage source, EG ( S ) =

E S2

Then from Eqs (14.2.10) and (14.2.11), x  S  − 2    E  1   −vSx V ( x, S ) =  2   e + e  v    S  3    x  S  − 2    E   1   −vSx v  I ( x, S ) =  2     e − e   S   3 Zc   

Taking the inverse Laplace transform: V ( x, t ) = i ( x, t ) =

E x  x E   U −2  t −  + U −2  t + − 2  3  v 3  v 

E  x E  x  U −2  t −  − U −2  t + − 2  3 Zc  v  3 Zc  v 

At the center of the line, where x = l / 2 , l  E   E  3  V  , t  = U −2  t −  + U −2  t −  2 2  3  2 3  E l    E  3  i  ,t  = U −2  t −  − U −2  t −  2  2  3 Zc  2  3 Zc 

14.3 From Eq (14.2.12) with Z R = S ( 2 LR ) and Z G = Z c : S ( 2 LR ) Z −1 S − c Zc 2 LR R ( S ) = = Zc S ( 2 LR ) +1 S + 2 LR Zc

S ( S ) = 0

Then from Eq (14.2.10) with EG ( S ) =

E S

  Z   S − c  x    − Sx − 2  2 LR S  v  E1  e V ( x, S ) =    e v +  S  2    S + Zc      2 L R    

Using partial-fraction expansion  Sx    − x       S  − 2  E e v −1 2 e  v   V ( x, S ) =  + + 2 S   S S + Ze      2 LR    

Taking the inverse Laplace transform: V ( x, t ) =

−1  x   t + − 2   E x E  x    U −1  t −  +  −1 + 2e 2 LR | Zc  v   U −1  t + − 2  2  v  2   v  

At the center of the line, where x = l : 2 3     t − 2   −    2 LR | Z c  l  E   E  3    U V  , t  = U −1  t −  + −1 + 2e −1  t −   2 2  2  2 2      

14.4

 R = 0; EG ( S ) =

E S 348

From Eq (14.2.10)    E Z / 2 LG   − Sxv   e  V ( x, S ) =  c Z  S  S + c   2 LG 

Using partial fraction expansion:   1  − Sx 1 e v V ( x, S ) = E  −  S S + Zc   2 LG 

Taking the inverse Laplace transform,  t−x/v    −    x V ( x, t ) = E 1 − e  2 LG / Zc   U −1  t −   v  

At the center of the line, where x = l / 2 : t − / 2 ) − (   l  2 LG / Z c V  , t  = E 1 − e  U −1 ( t −  / 2 ) 2   

14.5

R =

4 −1 = 0.6 4 +1

EG ( S ) =

1 −1 S = 3 = −0.5 1 +1 3

E S

x   − Sx S − 2    e v + 0.6e  v   E 1    V ( x, S ) =   −2 S 1 S  + 1  1 − ( 0.6 )( −0.5 ) e  3 

x   − Sx S  − 2   3E  e v + 0.6e  v   V ( x, S ) =  4S  1 + 0.3e−2 S  

V ( x, S ) =

x  S  − 2   3E  − Sxv 2 v  1 − 0.3e −2 S + 0.3 e + 0.6 e ( ) e−4 S   4S  

V ( x, S ) =

x  x  x  s  − 2  − S  + 2  S  − 4  3E  − Sxv v  − 0.3e  v  − 0.18e  v  e + 0.6 e  4S 

+0.09e

V ( x, t ) =

x  − S  + 4  v 

x  S  − 6  

+ 0.054e  v

 

  

3E   x x   U −1  t −  + 0.6U −1  t + − 2   4   v v  

x x     −0.3U −1  t − − 2  − 0.18U −1  t + − 4  v  v    x x      +0.09U −1  t − − 4  + 0.054U −1  t + − 6   v  v    

At the center of the line, where x =

l : 2

 l  3E     3   5  V  ,t  = U −1  t −  + 0.6U −1  t −  − 0.3U −1  t −    2  2  4   2   2   7   9   11  −0.18U −1  t −  + 0.09U −1  t −  + 0.054U −1  t − 2  2  2    

  

1  10 −6 L 3 14.6 (a) Z c = = =100.  1 C  10 −10 3

=

1 LC

1 1 −6  1 −10   3  10  3  10    

= 3.0  108 m/s

l 30  103 = = 1  10−4 S = 0.1 ms v 3  108

ZG −1 Zc (b)  S = =0 ZG +1 Zc

ZR (S ) =

R ( SL ) SL + R

EG ( S ) =

100 S

RS 100S = R S + 50,000. S+ L

ZR (S ) S −1 −1 Zc −50,000 S + 50,000 R ( S ) = = = S ZR (S ) 2 S + 50,000 +1 +1 S + 50,000 Zc

R ( S ) =

−25,000. per unit S + 25,000.

(c) Using (14.2.11) with x = l (receiving end) 25000 − S  100 / S   − S I R ( S ) = I ( l, S ) =   e + S + 25000 e  200    IR (S) =

 − S 1  1 1 1 1 25000 −1  − S e  + e =  + + 2  S S ( S + 25000 )  2  S S S + 25000 

IR (S ) =

1 2 −1  − S + e  2  5 S + 25000 

− ( t − )  1 (d) iR ( t ) = 2 − e 0.0410−3  U −1 ( t −  ) A 2 

14.7 (a)

L 2  10 −6 = = 400  C 1.25  10 −11

Zc =

1 LC

=

( 2  10 )(1.25  10 ) −6

−11

= 2.0  108

m s

l 100.  103 = = 5  10−4 S = 0.5ms v 2  108

ZG −1 Zc 100 (b)  S = = 0 EG ( S ) = ZG S +1 Zc

Z R ( S ) = SLR +

1 SC R

LR = 100  10 −3 H C R = 1  10 −6 F

ZR (S ) L 1 −1 S R + −1 Zc Z c SC R Z c R (S ) = = L 1 ZR (S ) S R + +1 +1 Z c SC R Z c Zc

Zc 1 S+ LR LRC R S 2 − 4  103 S + 1  107 R ( S ) = = 2 Z 1 S + 4  103 S + 1  107 S2 + c S + LR LRC R S2 −

VR ( S ) =

100  400   − S  S 2 − 4  103 S + 1  107  − S  e  e +  2 3 7  S  400 + 400    S + 4  10 S + 1  10  

 1 ( S − 2000 + j 2449.5 )( S − 2000 − j 2449.5 )  − S VR ( S ) = 50  + e  S S ( S + 2000 + j 2449.5 )( S + 2000 − j 2449.5 ) 

1 1  − S − j1.633 + j1.633 VR ( S ) = 50  + + + e  S S S + 2000 + j 2449.5 S + 2000 − j 2449.5 

2  −3.266 ( 2449.5 ) VR ( S ) = 50  +  e− S 2 2  S ( S + 2000 ) + ( 2449.5 )  − ( t − )     VR ( t ) = 50 2 − 3.266e 0.510−3 sin ( 2449.5)( t −  )  U −1 ( t −  )    

(d)

14.8 (a)

Zc =

=

L 0.999  10 −6 = = 299.73  C 1.112  10 −11

1 LC

( 0.999  10−6 )(1.112  10−11 )

= 3.0  108

m s

l 60.  103 = = 1.9998  10−4 S = 0.2 ms v 3.0  108

ZG −1 Z (b)  S = c =0 ZG +1 Zc

EG ( S ) =

 1  RR    SC R  = (1/ C R ) ZR = 1 1 RR + S+ SC R RR C R

E S2

RR = 150.  C R = 1  10 −6 F

 1     Z cC R  − 1 ZR 1 −1 S+ Z RR C R R = c = ZR  1  +1   Zc  Z cC R  + 1 S + 1/ Rx C R

 1 1  −S −  −  RRC R Z cC R  −S − 3.330  103  R = = per unit  1 1  S + 1.0003  10 4 S+ +   RRC R Z cC R 

V ( 0, S ) = VS ( S ) =

(d)

E  1    −S − 3.33  103  −2 S  e  1 + S 2  2    S + 1.0003  10 4  

VS ( S ) =

 E 1 −S − 3.33  103 e −2 S   2 + 2 4  2S S ( S + 1.0003  10 ) 

VS ( S ) =

E  1  −0.333 −6.67  10 −5 6.67  10 −5  −2 S  + + e   2 + 2  S  S2 S S + 1.0003  10 4  

VS ( t ) =

− ( t − 2 )    E  −5 −5 0.110 −3 tU t − 0.333 t − 2  + 6.69  10 − 6.67  10 e ( ) ( )  −1   U −1 ( t − 2 )  2    

14.9

14.10

14.12

For a voltage wave VA + arriving at the junction:

VA + + VA− = VB+

KVL :

KCL : I A+ + I A− = I B+ +

(1)

VB+ RJ

 1 VA+ VA− VB+ VB+ 1  VB+ − = + = VB+  + = Zc Z Z c RJ  Z c RJ  Z eq

where Z eq =

(2)

RJ Z c RJ + Z c

Solving (1) and (2):  Z eq  −1  Z V + =  V + VA− =  c AA A  Z eq  A + 1    Zc 

  Z eq    2  Zc   +   + VB = V =  BAVA+  Z eq  A +1    Zc   

Since Line Sections A and B have the same characteristic impedance Z c ,  BB =  AA and  AB =  BA .

=

100  103 1 = ms 3  108 3

(b)

14.14

For a voltage wave VA+ arriving at the junction from line A,

KVL VA+ + VA− = VB+

(1)

VB+ = VC+

(2)

VB+ = VD+

(3)

KCL

I A+ + I A− = I B+ + I C+ + I D+ VA+ VA− VB+ VC+ VD+ − = + + Z A Z A Z B ZC Z D

(4)

Using Eqs (2) and (3) in Eq (4):  1 VA+ VA− 1 1  VB+ − = VB+  + + = ZA ZA  Z B ZC Z D  Z eq

Where Z eq = Z B // ZC // Z D =

(5)

1 1 1 1 + + Z B ZC Z D

Solving Eqs (1) and (5): Z eq   −1   ZA  V + =  AAVA + VA − =   ( Z eq / Z A ) + 1  A    

 2 ( Z eq / Z A )   VA + =  BAVA + VB + =   ( Z eq / Z A ) + 1 

Also VC + = CAVA +

CA =  DA =  BA

VD + =  DAVA +

14.15 (a)

SLG −1 Zc S − Z c / LG S ( S ) = = SLG S + Z c / LG +1 Zc

1 −1 −3 R = 4 = 1 +1 5 4

     Z E 1 1  c  V1 ( S ) =  = E −  Z  S  SLG + Z c  S S+ c   LG  

(b) For 0  t  5 :  3  −3 9  V ( l, S ) =  1 −  V1 ( S ) e − S +  +   S ( S ) V1 ( S ) e − S (3 ) 5 5 25        2E  1 1  − S 6 E  1   S − Z c / LG  Z c / LG  − S (3 ) V ( l, S ) = − e −   e Zc  5 S 25  S   S + Z c / LG  S + Z c / LG  S +  LG  

        2 Z c / LG  − S (3 ) 2E  1 1  − S 6 E  1 1 V ( l, S ) = − e + − − e 2  Zc  Zc  5 S 25  S  Z  S+ S+ c  LG  LG  S + L     G     Taking the inverse Laplace transform:

V ( l, t ) =

−( t − ) − ( t − 3 ) − t − 3  2 ( t − 3 ) L( / Z )  2E  6E   1 − e LG / Zc U −1 ( t −  ) +  1 − e LG / Zc − e G c U −1 ( t − 3 )   5  25  LG / Z c  

14.16 (a) Z G = 100  Z c = 400  E = 100 kV    Zc / 2   200  V1 ( t ) = E U −1 ( t )   = 100   U −1 ( t ) Z  200 + 100   c + ZG   2 

V1 ( t ) = 66.67 U −1 ( t ) kV

(b)

ZG − 1 100 − 1 Z ( 1 c / 2) S = = 200 =− ZG 100 3 +1 +1 200 ( Zc / 2 )

 R = −1

(c)

(d)

400 −1 1 = 14.17 (a)  S = 200 400 3 +1 200

100 −1 1  R = 300 =− 100 2 +1 300

300 −1 1  AA = 200 = 300 5 +1 200

 300  2 200  6  BA =  = 300 5 +1 200

200 −1 1  BB = 300 =− 200 5 +1 300

(b)

 200  2 300  4  AB =  = 200 5 +1 300

Voltage S =

1 1 1  AA =  BB = − 3 5 5 6 4  BA =  AB = 5 5

R = −

1 2

At t = 0, the 10 kA pulsed current source at the junction encounters 200 // 300 = 120 . Therefore the first voltage waves, which travel on both the cable and overhead line, are pulses of width 50 s and magnitude 10 kA 120  = 1200 kV ,

V1 ( S ) =

E (1 − e−TS ) E = 1200. kV T = 50. s S

(c)

14.18

Nodal Equations: Note that  = 0.2 ms and t = 0.02 ms 0.02 Vk ( t ) = 10 − I k ( t − 0.2 ) 0.011Vm ( t ) = I m ( t − 0.2 ) − I L ( t − 0.02 )

Solving:

Vk ( t ) = 50.0 10 − I k ( t − 0.2 )

(a)

Vm ( t ) = 90.909  I m ( t − 0.2 ) − I L ( t − 0.02 )

(b)

Dependent current sources: Eq (12.4.10)

I k ( t ) = I m ( t − 0.2 ) −

2 Vm ( t ) 100

(c)

Eq (12.4.9)

I m ( t ) = I k ( t − 0.2 ) +

2 Vk ( t ) 100

(d)

Eq (12.4.14)

I L ( t ) = I L ( t − 0.02 ) +

Vm ( t )

(e)

500

Equations (a) – (e) can now be solved iteratively by digital computer for time

t = 0, 0.02, 0.04 ms . Note that I k ( ) and I m ( ) on the right hand side of Eqs (a) – (e) are zero during the first 10 iterations while their arguments ( ) are negative. 14.19

Nodal Equations: 1 1  V1 ( t ) = 200  − U −1 ( t − 0.1) − I1 ( t − .1667 )  4 4  

(a)

V2 ( t ) = 133.33  I 2 ( t − .1667 ) − I 3 ( t − .1667 )

(b)

V3 ( t ) = V2 ( t )

(c)

V4 ( t ) = 0

(d)

Dependent current sources:  2  Eq (12.4.10) I1 ( t ) = I 2 ( t − .1667 ) −   V2 ( t )  400 

(e)

 2  Eq (12.4.9) I 2 ( t ) = I1 ( t − .1667 ) +   V1 ( t )  400 

(f)

 2  Eq (12.4.10) I 3 ( t ) = I 4 ( t − .1667 ) −   V4 ( t )  400 

(g)

 2  Eq (12.4.9) I 4 ( t ) = I 3 ( t − .1667 ) +   V3 ( t )  400 

(h)

Equations (a) – (h) can be solved iteratively for t = 0, t,2t

where t = 0.03333ms .

I1 ( ) , I 2 ( ) , I 3 ( ) and I 4 ( ) on the right hand side of Eqs (a) – (h) are zero for the first 5 iterations. 14.20

E = 100.kV

Z G = Z c = 400. 

= 500. s

t = 100. s

( 2LR / t ) = 2000. 

 t    = 50.   2C R 

 1  1  + 0 0    1   − I k ( t − 500 )  400 400   V t    ( ) 4   k    1 1  −1    Vm ( t )  =  I m ( t − 500 ) − I L ( t − 100 )  0  400 + 2000  2000          Vc ( t )   I L ( t − 100 ) + I C ( t − 100 )  −1 1   1    0 +    2000 50 2000    

Solving: 1  Vk ( t ) = 200  − I k ( t − 500 )  4  Vm ( t )  334.7 8.136   I m ( t − 500 ) − I L ( t − 100 )    =   Vc ( t )  8.136 48.98  I L ( t − 100 ) + I C ( t − 100 ) 

Current sources:  2  (12.4.9) I m ( t ) = I k ( t − 500 ) +   Vk ( t )  400   2  (12.4.10) I k ( t ) = I m ( t − 500 ) −   Vm ( t )  400 

(12.4.14) I L ( t ) = I L ( t − 100 ) +

1 Vm ( t ) − VC ( t )  1000 

 1  (12.4.18) IC ( t ) = − IC ( t − 100 ) +   VC ( t )  25 

s

50.0

0.25

100

50.0

0.25

200

50.0

0.25

300

50.0

0.25

400

50.0

0.25

500

50.0

83.68

2.034

0.25

0.2398

0.0816

0.0814

600

50.0

57.69

0.25

14.21

E = 100. kV Z G = Z c = 299.73  RR = 150   = 25.  t = 50. s = 200. s  t   2C R 

Writing nodal equations:  1  1  0 t    299.73 + 299.73       Vk ( t )  =  2.9973 − I k ( t − 200 )      1 1   Vm ( t )    1 0 + +  I m ( t − 200 ) + I C ( t − 50 )     299.73 150. 25.     

Solving:

Vk ( t ) = 50.t − 149.865 I k ( t − 200 ) Vm ( t ) = 19.999  I m ( t − 200 ) + I C ( t − 50 )  Current sources:  2  I m ( t ) = I k ( t − 200 ) +   Vk ( t )  299.73   2  I k ( t ) = I m ( t − 200 ) −   Vm ( t )  299.73 

(12.4.9) (12.4.10)

 1  I C ( t ) = − I C ( t − 50 ) +   Vm ( t )  12.5 

(12.4.18)

s

0.0

0.0025

1.66  10−5

100

0.0050

3.33  10−5

150

0.0075

5.0  10−5

200

0.0100

6.67  10−5

250

0.0125

3.32  10−4

8.34  10−5

1.43  10−5

2.66  10−5

300

0.0150

1.20  10−3

10.0  10−5

2.53  10−5

6.94  10−5

14.22

Nodal Equations: −1

10 V1 ( t )   0.1767 −0.1667     =  −0.1667 0.1767      − I 2 ( t − 0.2 )  V2 ( t )   −1

V3 ( t )   0.1767 −0.1667   I 3 ( t − 0.2 )     =  V4 ( t )   −0.1667 0.1677   − I L ( t − 0.02 )

(a) (b) (c ) (d )

Dependent current sources: Eq (14.4.10)

 2  I 2 ( t ) = I 3 ( t − 0.2 ) −   V3 ( t )  100 

(e)

Eq (14.4.9)

 2  I 3 ( t ) = I 2 ( t − 0.2 ) +   V2 ( t )  100 

(f)

Eq (14.4.14)

I L ( t ) = I L ( t − 0.02 ) +

V4 ( t ) 500

Equations (a) – (g) can be solved iteratively for t = 0, t,2t

(g) where t = 0.02 ms . I 2 ( ) and

I 3 ( ) on the right hand side of Eqs (a) – (g) are zero for the first 10 iterations. 14.23 (a) The maximum 60-Hz voltage operating voltage under normal operating conditions is

(

)

1.08 115/ 3 = 71.7 kV From Table 14.2, select a station-class surge arrester with 84-kV

MCOV. This is the station-class arrester with the lowest MCOV that exceeds 71.7kV, providing the greatest protective margin and economy. (Note: where additional economy is required, an intermediate-class surge arrester with an 84-kV MCOV may be selected.)

(b) From Table 14.2 for the selected station-class arrester, the maximum discharge voltage (also called Front-of-Wave Protective Level) for a 10-kA impulse current cresting in 0.5 s ranges from 2.19 to 2.39 in per unit of MCOV, or 184 to 201 kV, depending on arrester manufacturer. Therefore, the protective margin varies from ( 450 − 201) = 249kV to ( 450 − 184 ) = 266kV . Note. From Table 3 of the Case Study for Chapter 14, select a VariSTAR Type AZE stationclass surge arrester, manufactured by Cooper Power Systems, rated at 108kV with an 84-kV MCOV. From Table 3 for the selected arrester, the Front-of-Wave Protective Level is 313kV, and the protective margin is therefore (450–313) = 137kV or 13784 = 1.63 per unit of MCOV. 14.24 The maximum 60-Hz line-to-neutral voltage under normal operating conditions on the HV side of

(

)

the transformer is 1.1 345 / 3 = 219.1kV . From Table 3 of the Case Study for Chapter 14, select a VariSTAR Type AZE station-class surge arrester, manufactured by Cooper Power Systems, with a 276-kV rating and a 220-kV MCOV. This is the type AZE station-class arrester with the lowest MCOV that exceeds 219.1 kV, providing the greatest protective margin and economy. For this arrester, the maximum discharge voltage (also called Front-of-Wave Protective Level) for a 10-kA impulse current cresting in 0.5 s is 720kV. The protective margin is (1300 – 720) = 580 kV = 580220 = 2.64 per unit of MCOV.

Chapter 15 Power Distribution 15.1 See Figure 15.2. Yes, laterals on primary radial systems are typically protected from short circuits. Fuses are typically used for short-circuit protection on the laterals. 15.2 The three-phase, four-wire multigrounded primary system is the most widely used primary distribution configuration. The fourth wire in these Y-connected systems is used as a neutral for the primaries, or as a common neutral when both primaries and secondaries are present. Usually the windings of distribution substation transformers are Y-connected on the primary distribution side, with the neutral point grounded and connected to the common neutral wire. The neutral is also grounded at frequent intervals along the primary, at distribution transformers, and at customers’ service entrances. Sometimes distribution substation transformers are grounded through an impedance (approximately one ohm) to limit short circuit currents and improve coordination of protective devices. 15.3 See Table 15.1 for a list of typical primary distribution voltages in the United States [1–9]. The most common primary distribution voltage is 15-kV Class, which includes 12.47, 13.2, and 13.8 kV. 15.4 Reclosers are typically used on overhead primary radial systems (see Figure 15.2) and on overhead primary loop systems (see Figure 15.7). Studies have shown that the large majority of faults on overhead primaries are temporary, caused by lightning flashover of line insulators, momentary contact of two conductors, momentary bird or animal contact, or momentary tree limb contact. As such, reclosers are used on overhead primary systems to reduce the duration of interruptions for these temporary faults. Reclosers are not used on primary systems that are primarily underground, because faults on underground systems are usually permanent. 15.5 See Table 15.2. Typical secondary distribution voltages in the United States are 120/240 V, singlephase, three-wire for Residential applications; 208Y/120V, three-phase, four-wire for Residential/Commercial applications; and 480Y/277V, three-phase, four-wire for Commercial/Industrial/High-Rise applications. 15.6 Secondary Network Advantages Secondary networks provide a high degree of service reliability and operating flexibility. They can be used to supply high-density load areas in downtown sections of cities. In secondary network systems, a forced or scheduled outage of a primary feeder does not result in customer outages. Because the secondary mains provide parallel paths to customer loads, secondary cable failures usually do not result in customer outages. Also, each secondary network is designed to share the load equally among transformers and to handle large motor starting and other abrupt load changes without severe voltage drops. Secondary Network Disadvantage Secondary networks are expensive. They are typically used in high-density load areas where revenues justify grid costs. Another disadvantage is that network protectors that are used to prevent “power back-feeding” do not protect against the secondary network cable from overload that could be caused by faults. The network protector does not protect the (secondary) network cable from overload

15.7 As stated in Section 15.3, more than 260 cities in the United States have secondary networks. If one Googles “secondary electric power distribution networks United States,” one of the web sites that appears in the Google list is: Interconnection of Distributed Energy Resources in Secondary... File Format: PDF/Adobe Acrobat. Electric Power Research Institute and EPRI are registered service marks of the Electric ... SECONDARY DISTRIBUTION NETWORK OVERVIEW. Power System Design and Operation ... Printed on recycled paper in the United States of America ... mydocs.epri.com/docs/public/000000000001012922.pdf From this publication, three cities in the Western Interconnection that have secondary distribution systems are: Portland, Seattle, and San Francisco. 15.8 (a) At the OA rating of 40 MVA, I OA,L =

(13.8   )

= 1.673 kA per phase

Similarly, I FA, L =

(13.8   3)

I FOA L =

= 2.092 kA per phase

(13.8   3)

= 2.719 kA per phase

(b) The transformer impedance is 9% or 0.09 per unit based on the OA rating of 40 MVA. Using (3.3.11), the transformer per unit impedance on a 100 MVA system base is:  100  Z transformerPUSystem Base = 0.09   = 0.225 per unit  40 

VF Z transformerPU

1.0 = 11.11 per unit = (11.11)(1.673) ( 0.09 )

= 18.59 kA/phase

Note that in (c) above, the OA rating is used to calculate the short-circuit current, because the transformer manufacturer gives the per unit transformer impedance using the OA rating as the base quantity. 15.9 (a) During normal operations, all four transformers are in service. Using a 5% reduction to account for unequal transformer loadings, the summer normal substation rating is 1.20  (30 + 33.3 + 33.3 + 33.3)  0.95 = 148 MVA. With all four transformers in service, the substation can operate as high as 148 MVA without exceeding the summer normal rating of 120% of each transformer rating.

(b) The summer allowable substation rating, based on the single-contingency loss of one of the 33.3 MVA transformers, would be 1.5  (30 + 33.3 + 33.3)  0.95 = 137 MVA. Each of the three transformers that remain in service is each allowed to operate at 150% of its nameplate rating for two hours (reduced by 5% for unequal transformer loadings), which gives time to perform switching operations to reduce the transformer loading to its 30-day summer emergency rating. However, from the solution to (c), the 30-day summer emergency rating of the substation is 119.3 MVA. Since it is assumed that a maximum reduction of 10% in the total substation load can be achieved through switching operations, the summer allowable substation rating is limited to 119.3/0.9 = 132.6 MVA. Note that, even though the normal summer substation rating is 148 MVA, it is only allowed to operate up to 132.6 MVA, so that in case one transformer has a permanent outage: (1) the remaining in-service transformers do not exceed their two-hour emergency ratings; and (2) switching can be performed to reduce the total substation load to its 30-day emergency rating. (c) Based on the permanent loss of one 33.3-MVA transformer, the 30-day summer emergency rating of the substation is 1.3  (30 + 33.3 + 33.3)  0.95 = 119.3 MVA. When one transformer has a permanent failure, each of the other three transformers can operate at 130% of their rating for 30 days (reduced by 5% for unequal transformer loadings), which gives time to replace the failed transformer with a spare that is in stock. 15.10 Based on a maximum continuous current of 2.0 kA per phase, the maximum power through each circuit breaker at 12.5 kV is 12.5  2.0  3 = 43.3 MVA. The summer allowable substation rating under the single-contingency loss of one transformer, based on not exceeding the maximum continuous current of the circuit breakers, is 43.3  3  0.95 = 123.4 MVA. Comparing this result with the summer allowable substation rating of 132.6 MVA determined in Problem 15.9(b), we conclude that the circuit breakers limit the summer allowable substation rating to 123.4 MVA. 15.11 (a) The power factor angle of the load is L = cos–1(0.85) = 31.79°. Using Figure 2.5, the load reactive power is: QL = SLsin(L) = 10  sin (31.79°) = 5.268 Mvar absorbed Also, the real power absorbed by the load is SL  p.f. = 10  0.85 = 8.5 MW Similarly, the power factor angle of the source is S = cos–1(0.90) = 25.84°. The real power delivered by the source to the load is 8.5 MW, which is unchanged by the shunt capacitors. Using Figure 2.5, the source reactive power is: P   8.5  QS = SS sin (S ) =  S  sin ( S ) =   sin (25.84°)  0.90   p.f.  = 4.117 Mvar delivered

The reactive power delivered by the shunt capacitors is the load reactive power minus the source reactive power: QC = QL − QS = 5.268 − 4.117 = 1.151 Mvar (b) The power factor angle of the load is L = cos–1(0.90) = 25.84°. Using Figure 2.5, the load reactive power is: QL = SL sin ( L ) = 10  sin ( 25.84° ) = 4.359 Mvar absorbed

Also, the real power absorbed by the load is SL  p.f. = 10  0.90 = 9.0 MW Similarly, the power factor angle of the source is S = cos–1(0.95) = 18.19°. The real power delivered by the source to the load is 9.0 MW, which is unchanged by the shunt capacitors. Using Figure 2.5, the source reactive power is: P   9.0  QS = SS sin ( S ) =  S   sin (S ) =   sin(18.19°)  0.95   p.f.  = 2.958 Mvar delivered

The reactive power delivered by the shunt capacitors is the load reactive power minus the source reactive power: QC = QL − QS = 4.359 − 2.958 = 1.401 Mvar (c) Comparing the results of (a) which requires 1.151 Mvar to increase the power factor from 0.85 to 0.90 lagging; and (b) which requires 1.401 Mvar (22% higher than 1.151 Mvar) to increase the power factor from 0.90 to 0.95 lagging; it is concluded that improving the high powerfactor load requires more reactive power. 15.12 (a) Without the capacitor bank, the total impedance seen by the source is: Z TOTAL = RLINE + jX LINE +

1 1 RLOAD

Z TOTAL = 3 + j 6 +

1 jX LOAD

1 1 + 40 j 60 1 33.333 Z TOTAL = 3 + j 6 + = 3 + j6 + 0.03 33.69 −33.69 Z TOTAL = 3 + j 6 + 27.74 + j18.49 = 30.74 + j 24.49 =

39.30 /phase 38.54°

(a1) The line current is:  13.8  1.05   /0° 0.2129 VSLN 3   I LINE = = = kA/phase Z TOTAL −38.54°  39.30   38.54    (a2) The voltage drop across the line is: Z  0.2129  VDROP = LINE = (3 + j 6)   I LINE  −38.54°   6.708  0.2129  =    63.43°  −38.54°  1.428 kV 24.89° | VDROP | = 1.428 kV

(a3) The load voltage is:

   13.8  1.428  − VLOAD = VSLN − Z LINE I LINE = 1.05    3  24.89°      = 8.366 − (1.295 + j 0.60010 ) = 7.071 − j 0.6010 =

7.096 kVLN −4.86°

|VLOAD | = 7.096  3 = 12.29 kVLL (a4) The real and reactive power delivered to the three-phase load is: 3(VLOADLN )2 3(7.096)2 PLOAD3 = = = 3.776 MW RLOAD 40 QLOAD3 =

3 (VLOADLN )2 3(7.096)2 = = 2.518 Mvar X LOAD 60

(a5) The load power factor is:

(a6)

   Q   2.518   p.f. = cos  tan −1    = cos  tan −1    = 0.83 lagging  P   3.776     The real and reactive line losses are: PLINELOSS3 = 3I LINE 2 RLINE = 3(0.2129)2 (3) = 0.408 MW QLINELOSS3 = 3 I LINE 2 X LINE = 3(0.2129)2 (6) = 0.816 Mvar

(a7) The real power, reactive power, and apparent power delivered by the distribution substation are: PSOURCE3 = PLOAD3 + PLINELOSS3 = 3.776 + 0.408 = 4.184 MW

QSOURCE3 = QLOAD3 + QLINELOSS3 = 2.518 + 0.816 = 3.334 Mvar SSOURCE3 =  (4.1842 + 3.334 2 ) = 5.350 MVA (b) With the capacitor bank in service, the total impedance seen by the source is: 1 Z TOTAL = RLINE + jX LINE + 1 1 1 + − RLOAD jX LOAD jXC Z TOTAL = 3 + j6 +

1 1 1 1 + − 40 j 60 j60

1 43.42 = 43 + j6 = /phase 0.025 7.94° (b1) The line current is:  13.8  1.05   0° VSLN 0.1927 3   I LINE = = = kA/phase Z TOTAL 43.42 − 7.94° 7.94  (b2) The voltage drop across the line is: Z TOTAL = 3 + j6 +

1.293  6.708   0.1927  VDROP = Z LINE I LINE =    − 7.94°  = 55.49° kV 63.43°     | VDROP | = 1.293 kV

(b3) The load voltage is: 1.293  13.8  VLOAD = VSLN − Z LINE I LINE = 1.05   /0° − 55.49°  3  = 8.366 − (0.7325 + j1.065) = 7.633 − j1.065 =

7.707 kVLN −7.94°

|VLOAD | = 7.707  3 = 13.35 kVLL (b4) The real and reactive power delivered to the three-phase load is: 3(VLOADLN )2 3(7.707)2 PLOAD3 = = = 4.455 MW RLOAD 40 QLOAD3 =

3(VLOADLN )2 3(7.707)2 = = 2.970 Mvar X LOAD 60

(b5) The load power factor is:    Q   2.970  p.f. = cos tan −1   = cos tan −1   = 0.83 lagging  P   4.455    (b6) The real and reactive line losses are: PLINELOSS3 = 3I LINE 2 RLINE = 3(0.1927)2 (3) = 0.3342 MW QLINELOSS3 = 3 I LINE 2 X LINE = 3(0.1927)2 (6) = 0.6684 Mvar

(b7) The reactive power delivered by the shunt capacitor bank is: 3(VLOADLN )2 3(7.707)2 QC = = = 2.970 Mvar XC 60 (b8) The real power, reactive power, and apparent power delivered by the distribution substation are: PSOURCE3 = PLOAD3 + PLINELOSS3 = 4.455 + 0.3342 = 4.789 MW QSOURCE3 = QLOAD3 + QLINELOSS3 − QC = 2.970 + 0.6684 − 2.97 = 0.6684 Mvar SSOURCE3 =  (4.7892 + 0.66842) = 4.835 MVA

(c) Comparing the results of (a) and (b), with the shunt capacitor bank in service, the real power delivered to the load increases by 18% (from 3.776 to 4.445 MW) while at the same time:

• The line current decreases from 0.2129 to 0.1927 kA/phase • The real line losses decrease from 0.408 to 0.334 MW • The reactive line losses decrease from 0.816 to 0.668 Mvar • The voltage drop across the line decreases from 1.428 to 1.293 kV • The reactive power delivered by the source decreases from 3.334 to 0.6684 Mvar • The load voltage increases from 12.29 to 13.35 kVLL The above benefits are achieved by having the shunt capacitor bank (instead of the distribution substation) deliver reactive power to the load. 15.13 (a) Using all the outage data from the Table in (15.7.1) – (15.7.4):

342 + 950 + 125 + 15 + 2200 + 4000 + 370 4500 = 1.7782 interruptions/year

SAIFI =

(14.4  342) + (151.2  950) + (89.8  125) + (654.6  15) + (32.7  2200) + (10053  4000) + (40  370) 4500 SAIDI = 8992.966 minutes/year = 149.88 hours/year

SAIDI =

SAIDI 8992.966 = = 5057.34 minutes/year = 84.289 hours/year SAIFI 1.7782 8760  4500 − [(14.4  342) + (151.2  950) + (89.8  125) + (654.6  15) + (32.7  2200) + (10053  4000) + (40  370)]/60 ASAI = 8760  4500 ASAI = 0.98289 = 98.289%

CAIDI =

(b) Omitting the major event on 11/04 2010 and using the remaining outage data from the table in (15.7.1) – (15.7.4): 342 + 950 + 125 + 15 + 2200 + 370 SAIFI = = 0.889 interruptions/year 4500 (14.4  342) + (151.2  950) + (89.8  125) + (654.6  15) + (32.7  2200) + (40  370) SAIDI = 4500 SAIDI = 56.966 minutes/year

SAIDI 56.966 = = 64.079 minutes/year SAIFI 0.889 8760  4500 − [(14.4  342) + (151.2  950) + (89.8  125) + (654.6  15) + (32.7  2200) + (40  370)]/60 ASAI = 8760  4500 ASAI = 0.99989 = 99.989%

CAIDI =

Comparing the results of (a) and (b), the CAIDI is 84.289 hours/year including the major event versus 64.079 minutes/year or 1.068 hours/year excluding the major event.

15.14 Using the outage data from the two tables, and omitting the major event: 200 + 600 + 25 + 90 + 700 + 1500 + 100 + 342 + 950 + 125 + 15 + 2200 + 370 SAIFI = 2000 + 4000 SAIFI = 1.203 interruptions/year SAIDI = [(8.17  200) + (71.3  600) + (30.3  25) + (267.2  90) + (120  700) + (10  1500) + (40  100) + (14.4  342) + (151.2  950) + (89.8  125) + (654.6  15) + (32.7  2200) + (40  370)]/[2000 + 4000]

SAIDI = 428,568.3/6000 = 71.43 minutes/year SAIDI 71.43 = = 59.375 minutes/year SAIFI 1.203 ASAI = [8760  6000 − 428,568.3/60]/[8760  6000] = 0.9999 = 99.99%

CAIDI =

15.15 The optimal solution space is rather flat around a value of 0.066 MW for total system losses. The below system solution represents an optimal (or near optimal) solution.

15.16 The losses can be reduced by balancing the taps; opening some of the capacitors can also help reduce losses. The below system solution has reduced the losses to 0.063 MW. Again the solution space is flat so there are a number of near optimal solutions that would be acceptable.

15.17 Because of the load voltage dependence, the total load plus losses are minimized by reducing the voltages to close to the minimum constraint of 0.97 per unit. Again, the solution space is quite flat, with an optimal (or near optimal) solution shown below with case losses of 0.086 MW (versus a starting value of 0.129 MW).

15.18

Initially, the Total Load+Losses is 10.640 MW. By adjusting the tap ratios and the status of the capacitor banks as seen below, while maintaining the constraint that bus voltages must remain above 0.97 p.u., the minimum Total Load+Losses is 9.900 MW.

15.19 Initially, the Total Load+Losses is 7.748 MW. By adjusting the tap ratios and the status of the capacitor banks as seen below, while maintaining the constraint that bus voltages must remain above 0.97 p.u., the minimum Total Load+Losses is 6.812 MW.