INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY POWER SYSTEM ANALYSIS AND DESIGN FIFTH EDITION PDF

Preview

Summary

INSTRUCTOR'S SOLUTIONS MANUAL     TO ACCOMPANY            POWER SYSTEM  ANALYSIS AND DESIGN        FIFTH EDITION                      J. DUNCAN GLOVER    MULUKUTLA S. SARMA    THOMAS J. OVERBYE      Contents      Chapter 2                  1    Chapter 3                  27    Chapter 4        ...

Description

INSTRUCTOR'S SOLUTIONS MANUAL    

TO ACCOMPANY           

POWER SYSTEM  ANALYSIS AND DESIGN       

FIFTH EDITION                     

J. DUNCAN GLOVER   

MULUKUTLA S. SARMA   

THOMAS J. OVERBYE 

 

 

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

1 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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

2 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(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

3 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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

4 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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

5 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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.

6 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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 ←

7 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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

8 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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

9 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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°

10 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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 ←

11 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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°

12 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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.

13 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2.33 Qcap

MW Losses

Mvar Losses

0

0.42

0.84

0.5

0.4