Solutions Manual for Power System Analysis And Design 5th Edition by Glover

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INSTRUCTOR'S SOLUTIONS MANUAL

TO ACCOMPANY

POWER SYSTEM ANALYSIS AND DESIGN

FIFTH EDITION

J. DUNCAN GLOVER

MULUKUTLA S. SARMA

THOMAS J. OVERBYE

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Contents

Chapter 2

1

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

27

71

95

137

175

195

231

303

323

339

353

379

Chapter 12

Chapter 13

Chapter 14

Chapter 2 Fundamentals ANSWERS TO MULTIPLE-CHOICE TYPE QUESTIONS 2.1 b 2.19 a 2.2 a 2.20 A. c 2.3 c B. a 2.4 a C. b 2.5 b 2.21 a 2.6 c 2.22 a 2.7 a 2.23 b 2.8 c 2.24 a 2.9 a 2.25 a 2.10 c 2.26 b 2.11 a 2.27 a 2.12 b 2.28 b 2.13 b 2.29 a 2.14 c 2.30 (i) c (ii) b 2.15 a (iii) a 2.16 b (iv) d 2.17 A. a 2.31 a B. b 2.32 a C. a 2.18 c

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2.1

(a) A1 = 5∠30° = 5 [ cos30° + j sin 30°] = 4.33 + j 2.5 4 = 5 ∠126.87° = 5e j126.87° −3 (c) A3 = ( 4.33 + j 2.5 ) + ( −3 + j 4 ) = 1.33 + j 6.5 = 6.635∠78.44°

(b) A2 = −3 + j 4 = 9 + 16 ∠ tan −1

(d) A4 = ( 5∠30° )( 5 ∠126.87° ) = 25 ∠156.87° = −22.99 + j 9.821 (e) A5 = ( 5∠30° ) / ( 5∠ − 126.87° ) = 1∠156.87° = 1 e j156.87° 2.2

(a) I = 400∠ − 30° = 346.4 − j 200 (b) i(t ) = 5sin (ω t + 15° ) = 5cos (ω t + 15° − 90° ) = 5cos (ω t − 75° )

( 2 ) ∠ − 75° = 3.536∠ − 75° = 0.9151− j3.415 (c) I = ( 4 2 ) ∠ − 30° + 5∠ − 75° = ( 2.449 − j1.414 ) + (1.294 − j 4.83 ) I = 5

= 3.743 − j 6.244 = 7.28∠ − 59.06°

2.3

(a) Vmax = 359.3V; I max = 100 A (b) V = 359.3

2 = 254.1V; I = 100

2 = 70.71A

(c) V = 254.1∠15° V; I = 70.71 ∠ − 85° A 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) υ (t ) = 277 2 cos (ω t + 30° ) = 391.7cos (ω t + 30° ) V (b)

I = V / 20 = 13.85∠30° A i(t ) = 19.58cos (ω t + 30° ) A

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(c) Z = jω L = j ( 2π 60 ) (10 × 10 −3 ) = 3.771∠90° Ω

I = V Z = ( 277 ∠30° ) ( 3.771 ∠90° ) = 73.46 ∠ − 60° A

i(t ) = 73.46 2 cos (ω t − 60° ) =103.9cos (ω t − 60° ) A

(d) Z = − j 25 Ω I = V Z = ( 277∠30° ) ( 25∠ − 90° ) = 11.08∠120° A i(t ) = 11.08 2 cos (ω t + 120° ) = 15.67cos (ω t + 120° ) A

2.6

(

(a) V = 100

)

2 ∠ − 30°= 70.7∠ − 30° ; ω does not appear in the answer.

(b) υ (t ) = 100 2 cos (ω t + 20° ) ; with ω = 377,

υ (t ) = 141.4 cos ( 377t + 20° ) (c) A = A∠α ; B = B∠β ; C = A + B c(t ) = a(t ) + b(t ) = 2 Re Ce jωt 

The resultant has the same frequency ω.

2.7

(a) The circuit diagram is shown below:

(b) Z = 3 + j8 − j 4 = 3 + j 4 = 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 Ω Z LL = j ( 377 ) ( 5 × 10 −3 ) = j1.885 Ω ZC = − j V=

1 = − j 2.88 Ω ( 377 ) ( 921 × 10−6 )

120 2 2

∠ − 30° V

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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 ) = υ (t )i(t ) = 359.3cos (ω t + 15° )  100 cos (ω t − 85° )  1 ( 359.3)(100 ) cos100° + cos ( 2ω t − 70°)  2 = −3120 + 1.797 × 10 4 cos ( 2ω t − 70° ) W =

(b) P = VI cos (δ − β ) = ( 254.1)( 70.71) cos (15° + 85° ) = −3120 W Absorbed = +3120 W Delivered

(c) Q = VI sin (δ − β ) = ( 254.1)( 70.71) sin100° = 17.69 kVAR Absorbed

(d) The phasor current ( − I ) = 70.71∠ − 85° + 180° = 70.71 ∠ 95° A leaves the positive terminal of the generator.

The generator power factor is then cos (15° − 95° ) = 0.1736 leading 2.11 (a) p(t ) = υ (t )i(t ) = 391.7 × 19.58cos2 (ω t + 30° ) 1 = 0.7669 × 10 4   1 + cos ( 2ω t + 60° )  2 = 3.834 × 103 + 3.834 × 103 cos ( 2ω t + 60° ) W

P = VI cos (δ − β ) = 277 × 13.85cos0° = 3.836 kW

Q = VI sin (δ − β ) = 0 VAR

Source Power Factor = cos (δ − β ) = cos ( 30° − 30° ) = 1.0

(b) p(t ) = υ (t )i(t ) = 391.7 × 103.9cos (ω t + 30° ) cos (ω t − 60° ) 1 = 4.07 × 10 4    cos90° + cos ( 2ω t − 30° )  2 4 = 2.035 × 10 cos ( 2ω t − 30° ) W

P = VI cos (δ − β ) = 277 × 73.46 cos ( 30° + 60° ) = 0 W

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Q = VI sin (δ − β ) = 277 × 73.46 sin 90° = 20.35 kVAR pf = cos (δ − β ) = 0 Lagging

(c) p(t ) = υ (t )i(t ) = 391.7 × 15.67 cos (ω t + 30° ) cos (ω t + 120° ) 1 = 6.138 × 103   cos ( −90° ) + cos ( 2ω t + 150° )  = 3.069 × 103 cos ( 2ω t + 150° ) W 2 P = VI cos (δ − β ) = 277 × 11.08cos ( 30° − 120° ) = 0 W

Q = VI sin (δ − β ) = 277 × 11.08sin ( −90° )

= −3.069 kVAR Absorbed = +3.069 kVAR Delivered pf = cos (δ − β ) = cos ( −90° ) = 0 Leading

2.12 (a) pR (t ) = ( 359.3cos ω t )( 35.93cos ω t ) = 6455 + 6455cos2ω t W

(b) px (t ) = ( 359.3cos ω t ) 14.37cos (ω t + 90° )  = 2582 cos ( 2cot + 90° ) = −2582sin 2ω t W

2) ( X = ( 359.3 2 )

(c) P = V 2 R = 359.3

2

(d) Q = V 2

2

10 = 6455 W Absorbed 25 = 2582 VAR S Delivered

(e) ( β − δ ) = tan −1 ( Q / P ) = tan −1 ( 2582 6455 ) = 21.8°

Power factor = cos (δ − β ) = cos ( 21.8° ) = 0.9285 Leading

2.13

Z = R − jxc = 10 − j 25 = 26.93 ∠ − 68.2° Ω i(t ) = ( 359.3 / 26.93 ) cos (ω t + 68.2° ) = 13.34 cos (ω t + 68.2° ) A

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

(b) px (t ) = 13.34 cos (ω t + 68.2° )  333.5cos (ω t + 68.2° − 90° )  = 2224sin  2 (ω t + 68.2° )  W 2 ) 10 = 889.8 W ( (d) Q = I X = (13.34 2 ) 25 = 2224 VAR S

(c) P = I 2 R = 13.34 2

2

2

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

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2.14 (a) I = 4∠0° kA V = Z I = ( 2∠ − 45° )( 4∠0° ) = 8∠ − 45° kV

υ (t ) = 8 2 cos (ω t − 45° ) kV p(t ) = υ (t )i(t ) = 8 2 cos (ω t − 45° )   4 2 cos ω t     1 = 64    cos ( −45° ) + cos ( 2ω t − 45° )  2 = 22.63 + 32 cos ( 2ω t − 45° ) MW

(b) P = VI cos (δ − β ) = 8 × 4 cos ( −45° − 0° ) = 22.63MW Delivered (c) Q = VI sin (δ − β ) = 8 × 4sin ( −45° − 0° ) = −22.63 MVAR Delivered = + 22.63MVAR Absorbed

(d) pf = cos (δ − β ) = cos ( −45° − 0° ) = 0.707 Leading

(

2.15 (a) I =  4 

)

2 ∠60° 

( 2∠30°) =

2 ∠30° A

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

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

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2.16 (a) Z = 10 + j 120 π × 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 × 6.63 × sin 36.4° = 663VAR (c) Peak Magnetic Energy = W = LI 2 = 0.04 ( 6.63 ) = 1.76 J 2

Q = ωW = 377 × 1.76 = 663VAR is satisfied.

2.17 (a) S = V I * = Z I I * = Z I

2

= jω LI 2

Q = Im[ S ] = ω LI 2 ←

(b) υ (t ) = L

di = − 2ω L I sin (ω t + θ ) dt

p(t ) = υ (t ) ⋅ i(t ) = −2ω L I 2 sin (ω t + θ ) cos (ω t + θ ) = −ω L I 2 sin 2 (ω t + θ ) ← = − Q sin 2 (ω t + θ ) ←

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 2 cos ∠Z ; Q = Z I 2 sin ∠Z ←

(b) Choosing i(t ) = 2 I cos ω t , Then υ (t ) = 2 Z I cos (ω t + ∠Z ) ∴ p(t ) = υ (t ) ⋅ i(t ) = Z I 2 cos (ω t + ∠Z ) ⋅ cos ω t = Z I 2 cos ∠Z + cos ( 2ω t + ∠Z )  = Z I 2 [ cos ∠Z + cos2ω t cos ∠Z − sin 2ω t sin ∠Z ] = P (1 + cos2ω t ) − Q sin 2ω t ←

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1 jωC

(c) Z = R + jω L +

From part (a), P = RI 2 and Q = QL + QC 1 2 I ωC which are the reactive powers into L and C, respectively.

Thus p(t ) = P (1 + cos2ω t ) − QL sin 2ω t − QC sin 2ω t ←

where QL = ω LI 2 and QC = −

If ω 2 LC = 1,

  ← p(t ) = P (1 + cos2ω t ) 

QL + QC = Q = 0

Then

*

 150  5  ∠10°  ∠ − 50°  = 375 ∠60° 2.19 (a) S = V I =   2  2  = 187.5 + j 324.8 *

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 = 187.5 tan 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

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2.21 (a)

φL = cos−1 0.6 = 53.13° QL = P tan φL = 500 tan 53.13° = 666.7 kVAR φS = cos−1 0.9 = 25.84° QS = P tan φS = 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 0.9

m

= 0 kVAR

Source PF = cos  tan −1 ( 666.7 914.4 )  = 0.808 Lagging 2.22 (a) Y1 =

1 1 1 = = = 0.2∠ − 53.13° Z1 ( 3 + j 4 ) 5∠53.13° = ( 0.12 − j 0.16 ) S

Y2 =

1 1 = = 0.1S Z 2 10

P = V 2 ( G1 + G2 )  V =

P = G1 + G2

1100 = 70.71 V ( 0.12 + 0.1)

P1 = V 2 G1 = ( 70.71) 0.12 = 600 W 2

P2 = V 2 G2 = ( 70.71) 0.1 = 500 W 2

(b) Yeq = Y1 + Y2 = ( 0.12 − j 0.16 ) + 0.1 = 0.22 − j 0.16 = 0.272∠ − 36.03° S I S = V Yeq = 70.71( 0.272 ) = 19.23 A

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2.23

S = V I * = (120∠0° )(10∠ − 30° ) = 1200∠ − 30° = 1039.2 − j 600 P = Re S = 1039.2 W Delivered Q = Im S = −600 VAR S Delivered = +600 VAR SAbsorbed

2.24

S1 = P1 + jQ1 = 10 + j 0; S2 = 10∠ cos−1 0.9 = 9 + j 4.359 10 × 0.746 ∠ − cos−1 0.95 = 9.238∠ − 18.19° = 8.776 − j 2.885 0.85 × 0.95 SS = S1 + S2 + S3 = 27.78 + j1.474 = 27.82 ∠3.04° S3 =

PS = Re(SS ) = 27.78 kW QS = Im(SS ) = 1.474 kVAR SS = SS = 27.82 kVA

2.25

SR = VR I * = RI I * = I 2 R = (20)2 3 = 1200 + j 0 SL = VL I * = ( jX L I )I * = jX L I 2 = j8(20)2 = 0 + j 3200 SC = VC I * = (− jIXC )I * = − jX C I 2 = − j 4(20)2 = 0 − j1600

Complex power absorbed by the total load SLOAD = SR + SL + SC = 2000∠53.1° Power Triangle:

Complex power delivered by the source is * SSOURCE = V I * = (100 ∠0° )( 20∠ − 53.1° ) = 2000∠53.1° The complex power delivered by the source is equal to the total complex power absorbed by the load.

2.26 (a) The problem is modeled as shown in figure below: PL = 120 kW pfL = 0.85Lagging

θ L = cos−1 0.85 = 31.79°

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Power triangle for the load: QL = PL tan ( 31.79° )

SL = PL + jQL = 141.18∠31.79° kVA

= 74.364 kVAR

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

Real power loss in the line is zero.

Reactive power loss in the line is QLINE = I 2 X LINE = ( 294.13 ) 1 2

= 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, VS = 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° ; θOLD = 36.87°; Qold = Pold tan (θ old ) = 37.5 kVAR ∴ Sold = 50,000 + j 37,500

θ 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 =

21,070

( 377 )( 220 )

2

= 1155μ F ←

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2.28

S1 = 12 + j 6.667 S2 = 4 ( 0.96 ) − j 4 sin ( cos −1 0.96 )  = 3.84 − j1.12 S3 = 15 + j 0 STOTAL = S1 + S2 + S3 = ( 30.84 + j 5.547 ) kVA

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

V  V2 Since S = V I * = V   = * , it follows that Z Z 

( 240 ) V2 Z = = = (1.809 − j 0.3254) Ω S ( 30.84 + j 5.547 )103 2

*

∴ Z = (1.809 + j 0.3254 ) Ω ←

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

( 240 ) ( 30.84 )103 2 240 ) ( X= ( 5.547 )103 2

R=

Then

= 1.8677 Ω = 10.3838 Ω

∴ Z = (1.8677 j10.3838 ) Ω ←

2.29 Since complex powers satisfy KCL at each bus, it follows that S13 = (1 + j1) − (1 − j1) − ( 0.4 + j 0.2 ) = −0.4 + j1.8 ← S31 = −S13* = 0.4 + j1.8 ←

Similarly, S23 = ( 0.5 + j 0.5 ) − (1 + j1) − ( −0.4 + j 0.2 ) = −0.1 − j 0.7 ← S32 = −S23* = 0.1 − j 0.7 ←

At Bus 3, SG 3 = S31 + S32 = ( 0.4 + j1.8 ) + ( 0.1 − j 0.7 ) = 0.5 + j1.1 ← 2.30 (a) For load 1: θ1 = 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°

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

θ ′ = 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 ← V 2 (1000 ) = − j 28.57 Ω It follows then XC = * = SC j 35000 2

and

C=

106 = 92.85μ F ← 2π ( 60 )( 28.57 )

S ′* 60,000 − j 45,000 = = 60 − j 45 = 75∠ − 36.87° A V* 1000∠0° The supply current, in magnitude, is reduced from 100A to 75A ←

The new current is I ′ =

2.31 (a) I12 =

V1∠δ1 − V2 ∠δ 2  V1  V =  ∠δ1 − 90°  − 2 ∠δ 2 − 90° X ∠90° X  X

V V  Complex power S12 = V1 I12* = V1∠δ1  1 ∠90° − δ1 − 2 ∠90° − δ 2  X X  2 V1 V1V2 = ∠90° − ∠90° + δ1 − δ 2 X X ∴ The real and reactive power at the sending end are

P12 =

Q12 =

V12 VV cos90° − 1 2 cos ( 90° + δ1 − δ 2 ) X X V1V2 = sin (δ1 − δ 2 ) ← X

V12 VV sin 90° − 1 2 sin ( 90° + δ1 − δ 2 ) X X V = 1 V1 − V2 cos (δ1 − δ 2 )  ← X

Note: If V1 leads V2 , δ = δ1 − δ 2 is positive and the real power flows from node 1 to node 2. If V1 Lags V2 , δ is negative and power flows from node 2 to node 1. (b) Maximum power transfer occurs when δ = 90° = δ1 − δ 2 ← 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.

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