Solutions Manual for Power System Analysis and Design 6th Edition by Glover IBSN 9781305632134
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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
2.15 a
(ii) b
2.16 b
(iii) a
2.17 A. a
(iv) d
B. b
2.31 a
C. a
2.32 a
2.18 c
1
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2.1
(a) A1 = 6∠30° = 6 [ cos30° + j sin 30°] = 5.20 + j 3
5
= 6.40 ∠128.66° = 6.40e j128.66°
−4
(c) A3 = ( 5.20 + j 3) + ( −4 + j 5 ) = 1.20 + j 8 = 8.01∠81.50°
(b) A2 = −4 + j 5 = 16 + 25 ∠ tan −1
(d) A4 = ( 6∠30° )( 6.40 ∠128.66° ) = 38.414∠158.658° = −35.78 + j13.98
(e) A5 = ( 6∠30° ) / ( 6.40∠ − 128.66° ) = 0.94 ∠158.66° = 0.94e j158.66°
2.2
(a) I = 500∠ − 30° = 433.01 − j 250
(b) i(t ) = 4sin (ω t + 30° ) = 4cos (ω t + 30° − 90° ) = 4cos ( ω t − 60° )
I = ( 4 ) ∠ − 60° = 2.83∠ − 60° = 1.42 − j 2.45
(
)
(c) I = 5 / 2 ∠ − 15° + 4∠ − 60° = ( 3.42 − j 0.92 ) + ( 2 − j 3.46 )
= 5.42 − j 4.38 = 6.964∠ − 38.94°
2.3
(a) Vmax = 400 V; I max = 100 A
(b) V = 400
2 = 282.84 V; I = 100
2 = 70.71A
(c) V = 282.84∠30° V; I = 70.71∠ − 80° 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
(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
2
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(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 = 75
)
2 ∠ − 15° = 53.03∠ − 15° ; ω does not appear in the answer.
(b) υ (t ) = 50 2 cos (ω t + 10° ) ; with ω = 377,
υ (t ) = 70.71cos ( 377t + 10° )
(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
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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 ) =  400cos (ω t + 30° )  100cos (ω t − 80° ) 
1
( 400 )(100 ) cos110° + cos ( 2ω t − 50°) 
2
= −6840.4 + 2 × 104 cos ( 2ω t − 50° ) W
=
(b) P = VI cos ( δ − β ) = ( 282.84 )( 70.71) cos ( 30° + 80° )
= −6840 W Absorbed
= +6840 W Delivered
(c) Q = VI sin (δ − β ) = ( 282.84 )( 70.71) sin110°
= 18.79 kVAR Absorbed
(d) The phasor current ( − I ) = 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 ) = 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
Q = VI sin (δ − β ) = 277 × 73.46 sin 90° = 20.35 kVAR
pf = cos (δ − β ) = 0 Lagging
4
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(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 ( 2 cot + 90° )
= −2582sin 2ω t W
2)
(
X = ( 359.3 2 )
2
(c) P = V 2 R = 359.3
(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
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2.14 (a) I = 2∠0° kA
V = Z I = ( 3∠ − 45° )( 2∠0° ) = 6∠ − 45° kV
υ (t ) = 6 2 cos ( ω t − 45° ) kV
p(t ) = υ (t )i (t ) = 6 2 cos (ω t − 45° )   2 2 cos ω t 
1
= 24    cos ( −45° ) + cos ( 2ω t − 45° ) 
2
= 8.49 + 12cos ( 2ω t − 45° ) MW
(b) P = VI cos ( δ − β ) = 6 × 2cos ( −45° − 0° ) = 8.49 MW Delivered
(c) Q = VI sin (δ − β ) = 6 × 2sin ( −45° − 0° )
= −8.49 MVAR Delivered = + 8.49 MVAR 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.
(c) The instantaneous power has an average value of 3.46 W, and the frequency is twice
that of the voltage or current.
6
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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 ←
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 )
(b) υ (t ) = L
= −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
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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.
where QL = ω LI 2 and QC = −
Thus p(t ) = P (1 + cos2ω t ) − QL sin 2ω t − QC sin 2ω t ←
If ω 2 LC = 1,
QL + QC = Q = 0

 ←
p(t ) = P (1 + cos2ω t ) 
Then
*
 150
 5

2.19 (a) S = V I * = 
∠10° 
∠ − 50°  = 375 ∠60°
 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
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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
=
=
= 0.16∠ − 51.34°
Z1 ( 4 + j 5 ) 6.4∠51.34°
= ( 0.1 − j 0.12 ) S
Y2 =
1
1
= = 0.1S
Z 2 10
P = V 2 ( G1 + G2 ) ⇒ V =
P
=
G1 + G2
1000
= 70.71 V
( 0.1 + 0.1)
P1 = V 2 G1 = ( 70.71) 0.1 = 500 W
2
P2 = V 2 G2 = ( 70.71) 0.1 = 500 W
2
(b) Yeq = Y1 + Y2 = ( 0.1 − j 0.12 ) + 0.1 = 0.2 − j 0.12
= 0.233∠ − 30.96° S
I S = V Yeq = 70.71( 0.233) = 16.48A
9
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2.23
S = V I * = (120∠0° )(15∠ − 30° ) = 1800∠ − 30°
= 1558.85 − j 900
P = Re S = 1558.85 W Delivered
Q = Im S = −900 VAR S Delivered = +900 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 = 0.85 Lagging
θ L = cos−1 0.85 = 31.79°
Power triangle for the load:
10
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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
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2.28
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 + S2 + S3 = ( 32.88 + j 5.827 ) 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 =
=
= (1.698 − j 0.301) Ω
S ( 32.88 + j 5.827 )103
2
*
∴ Z = (1.698 + j 0.301) Ω ←
(ii) Let Z be the impedance of a parallel combination of R and X
( 240 )
= 1.7518 Ω
( 32.88)103
2
240 )
(
X=
= 9.885 Ω
( 5.827 )103
2
R=
Then
∴ Z = (1.7518 j 9.885 ) 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 ←
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
(b) ITOTAL =
12
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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 )
It follows then XC = * =
= − j 28.57 Ω
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

V12
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
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2.33
Qcap
Mvar Losses
0
0.42
0.84
0.5
0.4
0.8
1
0.383
0.766
1.5
0.369
0.738
2
0.357
0.714
2.5
0.348
0.696
3
0.341
0.682
3.5
0.337
0.675
4
0.336
0.672
4.5
0.337
0.675
5
0.341
0.682
5.5
0.348
0.696
6
0.357
0.714
6.5
0.369
0.738
7
0.383
0.766
7.5
0.4
0.801
8
0.42
0.84
8.5
0.442
0.885
9
0.467
0.934
9.5
0.495
0.99
0.525
1.05
10
2.34
MW Losses
7.5 Mvars
2.35
14
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(.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 + j 3.127 −0.4950 + j 4.950  V10  1.961∠ − 48.69°

  = 

 −0.4950 + j 4.950 −0.8796 + j 3.127  V20  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 Ω
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:
Ybus
−1
0
0
 1



1

1
 −1  1 + 1 + 1 + 1 − j1 

−
−
j
1
−
 2 3 4

3

4






 


=
1
1
1

1
 1

0
−  − j1 
−
j
1
+
j
+
j
−
j
3

 4


3
4
2




 


1
1 
1
 1
1

− 
− j 
 + j 4 − j 3 
 0
4
 4
4


Writing nodal equations by inspection:
−1
0
 1
 V10   1∠0° 
0


  

−0.25
 −1 ( 2.083 − j1) ( −0.3333 + j1)
 V20  =  0 
 0 ( −0.3333 + j1) ( 0.3333 − j 0.25 )
 V30   0 
− j 0.25

 
− j 0.25
( 0.25 − j 0.08333)  V40  2∠30°
 0 ( −0.25 )
15
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2.38 The admittance diagram for the system is shown below:
YBUS
where
Y11

Y
=  21
Y31

Y41
Y12
Y13
Y22
Y23
Y32
Y42
Y33
Y43
Y14 

Y24 
=
Y34 

Y44 
0 
 −8.5 2.5 5.0
 2.5 −8.75 5.0
0 

j
S
 5.0 5.0 −22.5 12.5 


0 12.5 −12.5
 0
Y11 = y10 + y12 + y13 ; Y22 = y20 + y12 + y23 ; Y23 = y13 + y23 + y34
Y44 = y34 ; Y12 = Y21 = − y12 ; Y13 = Y31 = − y13 ; Y23 = Y32 = − y23
Y34 = Y43 = − y34
and
2.39 (a)
Yc





+ Yd + Y f
−Yd
−Yc
−Y f
−Yd
Yb + Yd + Ye
−Yb
−Ye
  V1   I1 = 0 
  

−Ye
 V2  =  I 2 = 0 
 V   I

0
 3  3

Ye + Y f + Yg  V4   I 4

−Y f
−Yc
−Yb
Ya + Yb + Yc
0
4 2.5   V1   0
 −14.5 8

  
 8 −17 4


5  V2   0


(b) j
=
 4

4 −8.8 0  V3  1∠ − 90°

  

0 −8.3 V4   0.62∠ − 135°
 2.5 5
−1
−1
YBUS V = I ; YBUS
YBUS V = YBUS
I
16
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
−1
YBUS
= Z BUS
where
 0.7187
 0.6688
= j
 0.6307

 0.6194
0.6688
0.7045
0.7045
0.6307
0.6242
0.6840
0.6258
0.5660
0.6194 
0.6258 
Ω
0.5660 

0.6840 
−1
V = YBUS
I
 V1 
0

 
0

V

V =  2  and I = 
V3 
1∠ − 90°

 


 0.62∠ − 135°
V4 
where
Then solve for V1 , V2 , V3 , and V4 .
240
∠0° = 138.56∠0° V (Assumed as Reference)
3
2.40 (a) VAN =
VAB = 240∠30° V; VBC = 240∠ − 90° V; I A = 15∠ − 90° A
VAN 138.56 ∠0°
=
= 9.24∠90° = ( 0 + j 9.24 ) Ω
IA
15∠ − 90°
ZY =
(b) I AB =
IA
3
Z∆ =
∠30° =
15
∠ − 90° + 30° = 8.66∠ − 60°A
3
VAB
240∠30°
=
= 27.71∠90° = ( 0 + j 27.71) Ω
I AB 8.66∠ − 60°
Note: ZY = Z ∆ / 3
2.41
S3φ = 3VLL I L ∠ cos−1 ( pf )
= 3 ( 480 )( 20 ) ∠ cos−1 0.8
= 16.627 × 103 ∠36.87°
= (13.3 × 103 ) + j (9.976 × 103 )
P3φ = Re S3φ = 13.3 kW
Delivered
Q3φ = I m S3φ = 9.976 kVAR
Delivered
2.42 (a) With Vab as reference
Van =
Ia =
208
3
Z∆
= 4 + j 3 = 5∠36.87° Ω
3
∠ − 30°
Van
120.1∠ − 30°
=
= 24.02∠ − 66.87° A
( Z ∆ / 3) 5∠36.87°
17
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S3φ = 3Van I a* = 3 (120.1∠ − 30° )( 24.02∠ + 66.87° )
= 8654∠36.87° = 6923 + j 5192
P3φ = 6923 W; Q3φ = 5192 VAR; both absorbed by the load
pf = cos ( 36.87° ) = 0.8 Lagging; S3φ = S3φ = 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
Z2 =
60 − j 45
= ( 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 Ω
( 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 = 0 VAR delivered by the sending-end source
18
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(b) Phase voltage at load terminals V2 = 120∠0° − ( 2 + j 4 )( 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
(I )
ph ∆
=
I2
3
=
4.472
3
= 2.582 A
(d) The three-phase complex power absorbed by each load is
S1 = 3V2 I1* = 430 W + j 600 VAR
S2 = 3V2 I 2* = 1200 W − j 900 VAR
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 + j300 VAR
The sum of load powers and line losses is equal to the power delivered from the supply:
S1 + S2 + SL = ( 450 + j600 ) + (1200 − j 900 ) + (150 + j 300 )
= 1800 W + j 0 VAR
2.44 (a) The per-phase equivalent circuit for the problem is shown below:
Phase voltage at the load terminals is V2 =
2200 3
= 2200 V taken as Ref.
3
Total complex power at the load end or receiving end is
SR( 3φ ) = 560.1( 0.707 + j 0.707 ) + 132 = 528 + j 396 = 660∠36.87° kVA
19
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With phase voltage V2 as reference,
I =
SR*( 3φ )
*
2
3V
=
660,000∠ − 36.87°
= 100∠ − 36.87° A
3 ( 2200∠0° )
Phase voltage at sending end is given by
V1 = 2200∠0° + ( 0.4 + j 2.7 )(100∠ − 36.87° ) = 2401.7∠4.58° V
The magnitude of the line to line voltage at the sending end of the line is
(V1 ) L -L =
3V1 = 3 ( 2401.7 ) = 4160 V
(b) The three-phase complex-power loss in the line is given by
SL (3φ ) = 3 RI 2 + j 3 × I 2 = 3 ( 0.4 ) (100 2 ) + j 3 ( 2.7 )(100 )
2
= 12 kW + j81kVAR
(c) The three-phase sending power is
SS (3φ ) = 3V1 I * = 3 ( 2401.7∠4.58° )(100∠36.87° )
= 540 kW + j 477 kVAR
Note that SS (3φ ) = SR(3φ ) + SL ( 3φ )
2.45 (a)
IS =
SS
3VLL
=
25.001 × 103
3 ( 480 )
= 30.07 A
(b) The ammeter reads zero, because in a balanced three-phase system, there is no neutral
current.
2.46 (a)
20
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Using voltage division: VAN = Van
=
=
208
3
∠0°
Z∆ / 3
( Z ∆ / 3) + Z LINE
10∠30°
10∠30° + ( 0.8 + j 0.6 )
(120.09 )(10∠30°) =
9.46 + j 5.6
= 109.3∠ − 0.62° V
1200.9∠30°
10.99∠30.62°
Load voltage = VAB = 3 (109.3 ) = 189.3V Line-to-Line
(b)
Z eq = 10 ∠30 ° || (− j 20)
= 11.547 ∠0 °Ω
VAN = Van
Z eq
Z eq + Z LINE
(
= 208
=
3
) (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
∠ − cos−1 0.8 = 23.53∠ − 36.87° A
8 ( 460 )( 0.8 )
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
21
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30 × 103
(b) I L =
3 ( 375.7 )( 0.8 )
∠ − 2.73° − cos−1 0.8 = 57.63∠ − 39.6° A
I G 2 = I L − I G1 = 57.63 ∠ − 39.6° − 23.53∠ − 36.87°
= 34.14∠ − 41.49° A
VG 2 = VL + Z LINE 2 I G 2 = 216.9∠ − 2.73° + ( 0.8 + j1)( 34.14∠ − 41.49° )
= 259.7∠ − 0.63° V
Generator 2 line-to-line voltage VG 2 = 3 ( 259.7 )
= 449.8 V
(c) SG 2 = 3VG 2 I G* = 3 ( 259.7∠ − 0.63° )( 34.14∠41.49° )
2
= 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
SL
(c) I L =
3VLL
=
26.93 × 103
3 ( 480 )
= 32.39 A
(d) QC = QL = 14 × 103 VAR = 3 (VLL ) / X ∆
2
X∆ =
(e)
3 ( 480 )
2
= 49.37 Ω
14 × 103
I C = VLL / X ∆ = 480 / 49.37 = 9.72 A
I LINE =
PL
3 VLL
=
23 × 103
3 480
= 27.66 A
22
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2.49 (a) Let ZY = Z A = Z B = ZC for a balanced Y-load
Z ∆ = Z AB = Z BC = Z CA
Using equations in Fig. 2.27
ZY2 + ZY2 + ZY2
= 3 ZY
ZY
Z∆ =
and
ZY =
(b) Z A =
ZB =
( j10 )( − j 25)
j10 + j 20 − j 25
Z ∆2
Z
= ∆
3
Z∆ + Z∆ + Z∆
= − j 50 Ω
( j10 )( j 20 ) = j 40 Ω; Z
j5
C
=
( j 20 )( − j 25) = − j 100 Ω
j5
2.50 Replace delta by the equivalent WYE: ZY = − j
2
Ω
3
Per-phase equivalent circuit is shown below:

2
Noting that  j 1.0 − j  = − j 2 , by voltage-divider law,
3

V1 =
− j2
(100∠0°) = 105∠0°
− j 2 + j 0.1
∴υ1 (t ) = 105 2 cos (ω t + 0°) = 148.5cos ω t V ←
In order to find i2 (t ) in the original circuit, let us calculate VA′B′
VA′B′ = VA′N ′ − VB′N ′ = 3 e j 30°VA′N ′ = 173.2∠30°
Then
I A′B′ =
173.2∠30°
= 86.6∠120°
− j2
∴ i2 (t ) = 86.6 2 cos (ω t + 120° )
= 122.5cos (ω t + 120° ) A ←
23
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2.51 On a per-phase basis S1 =
∴ I1 =
1
(150 + j120 ) = ( 50 + j 40 ) kVA
3
( 50 − j 40 )103 =
2000
( 25 − j 20 ) A
Note: PF Lagging
Load 2: Convert ∆ into an equivalent Y
1
(150 − j 48 ) = ( 50 − j16 ) Ω
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 120 sin( cos−1 0.6 )  = ( 24 − j 32 ) kVA
3
∴ I3 =
( 24 + j32 )103 =
2000
(12 + j 16 ) A
Note:PF Leading
Total current drawn by the three parallel loads IT = I1 + I 2 + I 3
ITOTAL = ( 73.29 + j 7.61) A
Note:PF Leading
Voltage at the sending end: VAN = 2000∠0° + ( 73.29 + j 7.61)( 0.2 + j 1.0 )
= 2007.05 + j 74.81 = 2008.44∠2.13° V
Line-to-line voltage magnitude at the sending end =
2.52 (a) Let VAN be the reference: VAN =
2160
3
3( 2008.44 ) = 3478.62 V ←
∠0° ≃ 2400∠0° V
Total impedance per phase Z = ( 4.7 + j 9 ) + ( 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 ←
24
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(b) (VA′N ) LOAD = 2400∠0° − ( 214.7∠ − 63.4° )( 0.3 + j1) 
= 2400∠0° − 224.15∠9.9° = 2179.2 − j 38.54
= 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
 S3φ 
= 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 ←
∴  P3φ 
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 ←
∴ Q3φ 
LOAD
= 3 ( 414.65 ) = 1243.95 kVAR ←
(d) Line losses per phase ( P1φ )
Total line loss ( P3φ )
LOSS
= ( 214.7 ) 0.3 = 13.83 kW ←
2
LOSS
= 13.83 × 3 = 41.49 kW ←
25
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Solutions Manual for Power System Analysis and Design 6th Edition by Glover IBSN 9781305632134
Full Download: http://downloadlink.org/product/solutions-manual-for-power-system-analysis-and-design-6th-edition-by-glover-ibs
26
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Full all chapters instant download please go to Solutions Manual, Test Bank site: downloadlink.org