Solutions Manual CHAPTER 1 1.1 (a) 100 = ( ) 230 2 R ; Eq (1.21) or R = 529 W (b) I = 230 529 = 0.435 A (c) W = 100 ´ 8 = 800 Wh 1.2 (i) (b) i = 1 2 v dt z Element is inductance; L = 10 2 5 ´ = 4 H (c) i = C dv dt Element is Capacitance; C = 4 F (d) v = Ri Element is resistance; R = 10 5 = 2W (ii) (b) Peak energy = 1 2 Li 2 = 1 2 ´ 4 ´ (5) 2 = 50 J Peak power = 10 ´ 5 = 50 W
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Solutions Manual
CHAPTER 1
1.1 (a) 100 = ( )230 2
R; Eq (1.21)
or R = 529 �
(b) I = 230529
= 0.435 A
(c) W = 100 � 8 = 800 Wh
1.2 (i) (b) i = 12
v dt�
Element is inductance; L = 10 2
5�
= 4 H
(c) i = C dvdt
Element is Capacitance; C = 4 F(d) v = Ri
Element is resistance; R = 105
= 2�
(ii) (b) Peak energy = 12
Li2 = 12
� 4 � (5)2 = 50 J
Peak power = 10 � 5 = 50 W
� ����������������� �����
(c) Peak energy = 12
Cv2 = 12
�� 4 � (5)2 = 50 J
Peak power = 10 � 5 = 50 W(d) Energy storage is zero.
Peak power = 10 � 5 = 50 W(iii) 0 < t < 2s
v = 5t, i = 52
t
W1 = vi dt0
2
� = 252
2
0
2 ����� t dt = 33 1
3 J
2 < t < 6s
v = 10 – 104
(t – 2)
W2 = 50 1 14
2 1 14
20
� � � ���
��
��
��� ( ) ( )t t dt
t = 66 2
3 J
W = 3313
+ 66 23
= 100 J
1.3 (a) i = C dvdt
i = 100 � 10 – 6 � ddt
(200 2 sin 314t) = 8.88 cos 314t
(b) 90°
i
v p
314 t
i vleads by 90°
(c) p = vi (see figure)instantaneous power at double frequency (628 rad/s)
Average power = 0
p = 200 2 � 0.888 sin 314t cos 314t
= 100 2 � 0.888 sin 628t = 125.56 sin 628t W
1.4 (a) i = 1L
v dt�= 103
500 � 200 2 � sin 314t dt = – 1.8 cos 314t
i C
v+ –
i L
+ v –
��������������� �
+ + +
–
–+
–+
– –Vs = 4V V1 V244
4
22V Vs – 1
V V1 2–
21
(b) i lags v by 90°
90°
i
v
p
314 t
(c) p = vi = – 200 2 � 1.8 sin 314t cos 314t
= – 180 2 sin 628t WPower oscillates at double frequency 628 rad/s with zero average value
1.5 (a) v = 200 2 sin 314t V
i = 200 21000
sin 314t A
= 2 /5 sin 314t A
(b)
i
pv
314 t
pav
p = 200 2 � 2 /5 sin2 314t
= 40 (1 – cos 628t)Pav = 40 W; Frequency of oscillating component of power = 628 rad/s
1.6 p = i2 R = I02 R e–2R t/L
W = p dt0
�
� = I02 R e Rt L��
� 2
0
/ dt
= 12
LI02
1.7 (a)(b) At node 1
V V V V V V Vs1 1 1 2 1 2
4 2 4 2�
��
��
� = 0
or 32
V1 – 34
V2 = Vs
2(i)
i R
+ –v
i L
+v
–
� ����������������� �����
At node 2
V V V V V2 2 1 2 1
4 4 2�
��
� = 0
or – 34
V1 + V2 = 0 (ii)
(c) 6V1 – 3V2 = 2Vs = 8 (iii)– 3V1 + 4V2 = 0 (iv)
Solving Eqs. (iii) and (iv)
V1 = 3215
V, V2 = 1.6 V
1.8(a) I1 = 4 – 1 = 3 A
V1 = 1 � 3 = 3 V(b) KVL for mesh 1
3 + 1 � 1 + V2 = 0or V2 = 2V (voltage across R3)
I2 = 21
– 1 = 1 A
Voltage drop across R4 = 1 � 1 = 1 V(c) KVL for mesh 2
– 2 – 1 + Vs = 0or Vs = 3 V
1.9 (a)(b) At node 1
V VR
V VR
V VR
s1
1
1 2
2
1 3
4
��
��
� = 0
or 1 1
1 2R R�
���
���
V1 – 1
2R V2 – 1
4R V3 =
VR
s
1(i)
At node 2
– 1
2R V1 + 1 1 1
2 3 5R R R� �
���
���
V2 – 1
3R V3 = Is (ii)
At node 3
– 1
4R V1 – 1
3R V2 + 1 1 1
3 4 6R R R� �
���
���
V3 = 0 (iii)
Observe the symmetry in these equations
+
–
+
– –
+
–
–
–
+ –
+
+ +
VsIs
R5
R4
R1R6
( – )V V1 3
( – )V V1 2
( – )V V1 s
( – )V V2 3
312
V2 V3
+
–4A
( )R1 ( )R3
( )R4( )R2
1 11 2
1A
1 121
I1
I2
Vs
��������������� �
1.10
v6 W2F1His
iL+
–
iL = e–2t
v = 1 � ddt
e–2t = –2e–2t
iC = 2 dvdt
= 2 ddt
(–2e–2t ) = 8e–2t
iR = v2
= – e–2t
Henceis = iL + iC + iR = e–2t + 8e–2t – e–2t = 8e–2t
1.4
v 2 W 4 W1F
1H
is
i+
–
i = sin 2t
vH = 1 � ddt
sin 2t = 2 cos 2t
vR (4�) = 4 sin 2t� v = VH + vR = 2 cos 2t + 4 sin 2t
H (central limb) = 460 AT/m (from Fig. Q8.6)F (care of central limb) = 460 ¥ 20 ¥ 10–2 = 92 AT
Fcoil = FAB + F (care of central limb)= 495 + 92 = 587 AT
Exciting coil current = 5871000
= 0.587 A
200 400 600 800 1000 1200 14000
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
AT/m
BT(
)
Fig. Q.8.6
8.7 R g = 1 10
4 10 12 10
3
7 4
¥¥ ¥ ¥
-
- -p = 6.63 ¥ 105
R c = ( . )20 01 10
4 10 12 10
2
7 4
- ¥¥ ¥ ¥ ¥
-
- -p m r
= 0 132 10 9. ¥ +
m r
R = 80 10
4 10 6 10
2
7 4
¥¥ ¥ ¥ ¥
-
- -p m r
= 106 109. ¥
m r
f = FR R R
g c+ + 2
F1
R g
R cRR
375 AT0.4 mW b
�� ���� ��� ��� �������������
0.6 ¥ 10–3 = 875
6 630 662 10
104
5..+ ¥F
HGIKJ ¥m r
or mr = 7609
8.8 Rg = 2 10
4 10 4 4 10
3
7 4
¥¥ ¥ ¥ ¥
-
- -p = 9.95 ¥ 105
R C = [ ( ) . ]2 25 20 0 2 10
4 10 4000 4 4 10
2
7 4
+ - ¥¥ ¥ ¥ ¥ ¥
-
- -p = 1.11 ¥ 105
Rg + R C = (9.95 + 1.11) ¥ 105 = 11.06 ¥ 105
f = 400 4
11 06 105
¥¥.
= 1.45 mWb
Wf (air-gap) = 12
Rg f2 = 12
¥ 9.95 ¥ 105 ¥ (1.45)2 ¥ 10–6 = 1.046 J
Wf (core) = 12
R cf2 = 1
2 ¥ 1.11 ¥ 105 ¥ (1.45)2 ¥ 10–6 = 0.036 J
Now mr (core) = • \ R c = 0
f = 400 4
9 95 105
¥¥.
= 1.61 mWb
Wf (air-gap) = 12
¥ 9.95 ¥ 105 ¥ (1.61)2 ¥ 10–6 = 1.29 J
Wf (core) = 0(b) i = F/N = f R total/N
= ( . sin ) .0 4 314 10 11 06 10
400
3 5t ¥ ¥ ¥-
= 1.11 sin 314t A
e = wN fmax cos wt
= 314 ¥ 400 ¥ 0.4 ¥ 10–3 cos 314t
= 50.24 cos 314t(c) L = N2/R total
= ( ).400
1106 10
2
5¥ = 144.7 mH
If mr = • R total = 9.95 ¥ 105, then L = ( )
.
400
9 95 10
2
5¥ = 161 mH.
�������������� ��
CHAPTER 9
9.1 (a) I2 = 25 1000
200¥
= 125 A, I1 = 25 1000
600¥
= 41.7 A
(b) (i)( )600
21
2
= 25 ¥ 1000 or Z1 = 14.4 W
Similarity
( )200 2
2Z= 25 ¥ 1000 or Z2 = 1.6 W
(c) fmax = V
f N1
12p = 600
2 50 60¥ ¥ ¥p = 0.045 Wb
The core flux will be the source when the transformer is excited at rated voltage in
secondary side VN
VN
1
1
2
2=
FHG
IKJ
(d)Vf1 =
¢¢
Vf1
or V1¢ = ( f ¢/f ) V1 = 6050
¥ 600 = 720 V
V ¢2 = 7203
= 240 V
(e) fmax = 600
2 40 60¥ ¥ ¥p = 0056 mb
The core flux density has increased. As the core loss is proportional to the square offlux density and directly proportional to frequency, core loss would increase causing thecore to get overheated. Also the magnetizing current would increase which is detrimentalto transformer.
9.2 (a) On 600 – V side:VSC = 600 ¥ 0.052 = 31.2 V
ISC = 25 1000
600¥
= 41.7 A
R1 = 24241 7 2( . )
= 0.139 W
X1 = ( . ) ( . )0 748 0 1392 2- = 0.748 WOn 200 – V side:
R2 = 0 139
9.
= 0.0817 W; X2 = 0 735
9.
= 0.0817 W
PU values:
ZB(HV) = 60041 7.
= 14.4 W
�� ���� ��� ��� �������������
R(pu) = 0 13914 4.
. = 0.0097 or 0.97%
X(pu) = 0 73514 4.
. = 0.051 or 5.1%
(b) Since frequency is same (50 Hz) during SC test
(Core flux) SC(Core flux) Rated Voltage
= 5.2%
(c) Core flux and also core density reduce to 5.2% of rated value. The core losses reduceto (0.052)2 or 0.27% of core loss at rated voltage. Hence core losses during SC are ofnegligible order and the total power input constitutes ohmic losses.
Ia = I Iab ca- = 3 ¥ 7.723 –– 41.3° = 13.376 –– 41.3° kA
Similarly
Ib = 13.376 –– 161.3° kA
Ic = 13.376 –– 281.3° kA
Note: It is easily observed from above that the line voltages and currents on star side leadthose on delta side by 30° the same holds for phase voltages and current.
9.16
22 3
kV
6.6
3 k
V Z Z
Z
I1 I2
IZ
6.6
kV
22 k
V
( )22 3 2
Z= 3 MVA
or Z = 484 W
3 ¥ 22 3 ¥ I2 = 9 ¥ 1600 or I2 = 136.4 A
(transformer current secondary)
�������������� ��
I1 = 136.4 ¥ 226 6.
= 454.5 A (transformer primary current)
IZ (current in leg of delta) = I2
3 = 136 4
3. = 78.75 A
9.17
22 kV
22 3
kV
I1/ 3Z Z
Z
I1
6.6
kV
IZ
As before (Problem 9.16)I2 = 136.4 A (current in transformer secondary)IZ = 78.75 A (current in leg of a connected load)
I1/ 3 = current in transformer primary
= 136.4 ¥ 226 6.
= 454.6 A
I1 (primary line current) = 454.6 3 = 787.4 A
CHAPTER 10
10.1 slots/pole = 3 ¥ 3 = 9
g = 180
9∞
= 20°
Short pitching angle, b = 20°
Kb = sin
sinm
m
g
g2
2
= sin /
sin /3 20 2
3 20 2¥ ∞
∞ = 0.96
Kp = cos 20°/2 = 0.985
10.2 f = 300 20
120¥
= 50 Hz
Total turns = 180 8
2¥
= 720
Nph (series) = 7203
= 240
m = 1803 20¥
= 3
g = 180 20
180∞ ¥
= 20°
�� ���� ��� ��� �������������
Kb (as calculated in Problem 10.1) = 0.96Ep = 4.44 ¥ 0.96 ¥ 50 ¥ 240 ¥ 0.05 = 426 V
El = 426 3 = 738 V
10.3 P = 120 f
ns =
120 50750
¥ = 8
m = 4S = 4 ¥ 3 ¥ 8 = 96
g = 180 8
96∞ ¥
= 15°
Full pitch = 968
= 12 slots
Coil pitch = 11 slotsShort pitching angle, b = 15°
Total armature turns = 2 8 96
2¥ ¥
= 768
Nph (series) = 7683 2¥
= 128
Kb = sin /
sin /4 15 2
4 15 2¥ ∞
∞ = 0.96
Kp = cos 15°/2 = 0.99Ep = 4.44 Kw f Nph (series) f
kVA = 3 ¥ 20.96 ¥ 10 = 6.28.5Observation: For the same winding/same amount of copper) 3-phase winding has 33.5%higher kVA capacity.
10.5S = 24
g = 180 2
24∞ ¥
= 15°
f = 2.2 Wb(a) Single-phase winding
m = 242
= 12
Kb = sin /
sin /12 15 2
12 15 2¥ ∞
∞ = 0.638
Nph(series) = 24 2
2¥
= 24
E = 4.44 ¥ 0.638 ¥ 50 ¥ 24 ¥ 2.2= 7478 V
kVA = 7.478 ¥ 8 = 59.8(b) Two-phase winding
m = 242 2¥
= 6
�� ���� ��� ��� �������������
Kb = sin /
sin /6 15 2
6 15 2¥ ∞
∞ = 0.9
Nph (series) = 242
= 12
Ep = 4.44 ¥ 0.9 ¥ 50 ¥ 12 ¥ 2.2= 5272 V
El = 5.272 2 = 7458 VkVA = 2 ¥ 5.272 ¥ 8 = 84.4
(c) Three-phase winding
m = 243 2¥
= 4
Kb = sin /
sin /4 15 2
4 15 2¥ ∞
∞ = 0.96
Nph (series) = 243
= 8
Ep = 4.44 ¥ 0.96 ¥ 50 ¥ 8 ¥ 2.2= 3751 V
El = 3751 3 = 6497 V
kVA = 3 ¥ 6.497 ¥ 8 = 90
10.6 Br = m 0Fr
g
Bav = 2p
Br
fr = p Dl
PFH
IK Bav =
2 0DlP g
¥FHG
IKJ
m Fr
P = 2 0DlP g
¥m
Wb/AT
10.7 Set speed = 120 50
2¥
= 3000 rpm
P (gen) = 120 400
3000¥
= 16
10.8 n(no-load) = 990 rpmn(full-load) = 950 rpm
ns = 1000 rpm
(a) P = 120 50
1000¥
= 6
�������������� ��
(b) s(no-load) = 101000
¥ 100 = 1%
s(full-load) = 50
1000 ¥ 100 = 5%
(c) No-load case
Speed of rator field wrt rotor surface = 120 0 5
6¥ .
= 10 rpmSpeed of rator field wrt stator = 990 + 10 = 1000 rpmSpeed of rator field wrt (rotor field = 0 rpmFull-load case:
f = 0.05 ¥ 50 = 2.5 HzSpeed of rotor field wrt rotor surface = (120 ¥ 0.5)/6 = 50 rpmSpeed of rotor field wrt stator surface = 950 + 50 = 1000 rpm
10.9 ns = 1000 rpm
(a) s = 1000 960
1000-
¥ 100 = 4%
(b) P = 6(c) T = kI2 cos q2 (Eq. (10.45))
q2 = 0\ T = kI2
For same torque I2 should remain the same. If R2 is doubled, the slip will became doublei.e., s = 2 ¥ 0.04 = 0.08 or n = 1000 (1 – 0.8) = 920 rpm. Thus adding external resistancein rotor circuit gives control over motor speed. This is only possible in slip ring inductionmotor and not in squirrel cage motor where rotor bars are premanentaly short circuited.
CHAPTER 11
11.1 VOC for the OCC is now reduced in the ratio of 1600/1000. The revised data are as under:
or n = 1159 rpm11.9 Ea1 = 250 – 85 ¥ (0.12 + 0.1) = 231.3 V
231.3 = K ¢a Iaw = k ¢a ¥ 85 ¥ 2 600
60p ¥
or K ¢a = 0.0433(a) Ia = 100 A
Ea2 = 250 – 100 ¥ (0.12 + 0.1) = 228 V
228 = 0.0433 ¥ 100 ¥ 2
602p ¥ n
or n2 = 503 rpm(b) Ia = 40 A
Ea2 = 250 – 40 ¥ (0.12 + 0.1) = 241.2 V
241.2 = 0.0433 ¥ 40 ¥ 2
602p n
or n2 = 1330 rpm(c) n2 = 800 rpm or 83.78 rad/s
Ea2 = 0.0433 ¥ Ia2 ¥ 83.78 (i)
Ia2 = 250
0 222- Ea
.(ii)
Eliminating Ia2 in Eqs. (i) and (ii)
Ea2 = 0 0433 83 78
0 22. .
.¥
¥ (250 – Ea2)
or Ea2 = 235.6
Ia2 = 250 235 6
0 22- ..
= 65.6 A
11.10At 600 V1, 80 A, 70 rpm
Ea1 = 600 – 80 ¥ (0.215 + 0.08) = 576.4 V
576.4 = K¢a ¥ 80 ¥ 2 750
60p ¥
or K¢a = 0.092
�������������� ��
At 95A current:Ea2 = 600 – 95 ¥ (0.215 + 0.08) = 572 V
572 = 0.092 ¥ 95 ¥ 2
602p ¥ n
or n2 = 625 rpmT = K ¢a Ia
2
= 0.092 ¥ (95)2 = 830.3 Nm11.11
(a) Ia = 220 AEa = 600 – 220 ¥ 0.15 = 567 V
n = 600 ¥ 507480
= 709 rpm
480 = Kaf ¥ 2 600
60p ¥F
HIK = K ¢a ¥ 220 ¥
2 60060
p ¥FH
IK
or K ¢a = 0.037T = K ¢aIa
2 = 0.0347 ¥ (220)2 = 1681 Nm
(b) Ia = 300 = 600R( )total
or R(total) = 2WR(ext) = 2 – 0.15 = 1.85 W
T(start) = 0.0347 ¥ (300)2 = 3132 NmObservation: With an armature current which is 36% more than the full-load current,the motor give 186% full-load torque. The series motor has therefore excellent startingcharacteristic and is capable of starting high torque loads.
11.12(a) Ia = 220 A
Ea = 600 – 220 ¥ 0.15 = 567 V
n = 600 ¥ 567450
= 756 rpm
450 = Kaf ¥ 2 600
60p ¥
or Kaf = 7.162T = Kaf Ia = 7.162 ¥ 220 = 1576 Nm
(b) R (ext) = 1.85 W (as before)
518 = Kaf ¥ 2 600
60p ¥
Kaf = 8.244T(start) = 8.244 ¥ 300 = 2473 Nm
Remark: Because of saturation torques have reduced
Ra = 0.15 W, Rf = 200 WAt no load; Ea1 ª V = 500 V
n1 = 1000 rpm
If 1 = 500200
= 2.5 A
Ea = kaf n = k ¢e If n
Substituting values500 = K ¢E ¥ 2.5 ¥ 1000
or K ¢E = L/5when had (20° + 40° = 60° C):
R a = 0.5 ¥ 234 5 60234 5 20
.
.++
= 0.579 W
Rf = 200 ¥ 234 5 60234 5 20
.
.++
= 232 W
If 2 = 500232
= 2.18 W
Ea2 = 500 – 70 ¥ 0.579 = 459.5 V
459.5 = 15
¥ 2.16 ¥ n2
or n2 = 1063.6 rpm11.15 Neglecting windage, friction and iron losses the load torque equals developed torque.
With linear magnetisation characteristicEa = K ¢e ¥ If ¥ n (i)T = k ¢T ¥ If ¥ Ia (ii)
�������������� ��
At n1 = 800 rpmEa1 = 220 – 40 ¥ 0.3 = 208 V
If 1 = 220200
= 1.1 A
Substituting in Eqs (i) and (ii)208 = K ¢e ¥ 1.1 ¥ 800 (iii)
T = K ¢T ¥ 1.1 ¥ 40 (iv)At n2 = 1050 rpm
Ea2 = 220 – 0.3 Ia2 = K ¢e ¥ If 2 ¥ 1050 (v)T = K ¢T ¥ If 2 ¥ Ia2 (vi)
Dividing Eq. (v) by (iii) and (vi) by (iv)
1050
2082I f
= 220 0 3
11 8002-
¥.
.Ia (vii)
I If a2 2
11 40. ¥= 1 (viii)
Solving Eqs (vii) and (viii)Ia2
2 – 733.3 Ia2 + 36400 = 0or Ia2 = 53.65 A, 679.7 AThe higher value of current i not an acceptable solution as the motor would operate at toopoor on efficiencyThus Ia2 = 53.65 A
If 2 = 11 4053 65.
.¥
= 0.82 A
Hence
Rf2 = 2200 82.
= 268 W
Rf (ext) = 268 – 200 = 68 W11.16 Since field excitation remains constant
Ea = K¢E n (i)T = K ¢T Ia (ii)
TL = KLn2 (iii)At n1 = 1500 rpm
Ea1 = 250 – 35 ¥ 0.2 = 243 V (iv)T1 = K ¢T ¥ 35 (v)
or 115 – 0.9 Ia2 = K ¢e ¥ If 2 ¥ 1400 (iii)Dividing Eq. (iii) by Eq. (i)
115 0 9107 5
2- ..
Ia = I
If
f
2
1
FHG
IKJ =
14001500
LetI
If
f
2
1
= k
\ 115 – 0.9 Ia2 = 100.33 k (iv)
T2 = 12
Tf l = 12
¥ K ¢T If 1 ¥ 25
K ¢T If2 Ia2 = 12
K ¢T If 1 ¥ 25
or Ia2 = 12.5/k (v)
�������������� ��
Substituting Eq. (v) in Eq. (iii)
115 – 0.9 ¥ 12 5.
k= 100.33 k
100.33 k2 – 115 k + 11.25 = 0or k = 1.037, 0.108Smaller value of k is rejected because it would mean a very large reduction in fieldcurrent resulting in heavy current in the armature circuit which is practically notacceptable (efficiency would be too low)
k = I
If
f
2
1 = 1.037 =
R
Rf
f
1
2
orR
Rf
f
2
1= 0.965
Thus a reduction of 3.5% is needed in the shunt field resistance11.18 For maximum efficiency of 215 A input current
(2/5)2 ¥ Ra = 3300or Ra = 0.071 WNote: Less resistance would mean larger copper conductor cross-section and therefore morevolume of copper to be used in the machineNow PL = 2 ¥ 3300 = 6600 W
Pn = 230 ¥ 215 = 49.45 kW
h = 49 45 6 6
49 45. .
.-
= 86.65%
11.19IL = 79.8 A, If = 2.6 A
Ia = 79.8 – 2.6 = 77.2 A
Ea = 220 – 77.2 ¥ 0.18 = 206 V
206 = f ¥ ¥ ¥ FH IK
1200 62060
42
(a) or f = 8.31 ¥ 10–3 or 8.31 mWb(b) EaIa = Tw
or T = 206 77 22 1200
60
¥¥
.p = 126.6 Nm
(c) Net output = 15000 WMech power developed = 206 ¥ 77.2 = 15903 W
Rotational loss = 15903 – 15000 = 903 W
�� ���� ��� ��� �������������
(d) Armature copper loss= (77.2)2 ¥ 0.18 = 1073 W
Field copper loss = 220 ¥ 2.6 = 572 WTotal copper loss = 1073 + 572 = 1645 W
Total loss = 1645 + 903 = 2548 W
h = 1515 2 548+ .
¥ 100 = 85.5%
CHAPTER 12
12.3No-load voltage = 400 V
If (nl ) = 5 AAt 125% full-load, 0.8 pf lagging
Ia = 62 1000 1 25
3 400
¥ ¥¥
. = 112 A
Ia = 112 –– 36.9° A
Ef = 4003
+ 112 –– 36.9° ¥ j1.08
= 303.65 + j96.76Ef = 318.6 V or 552 V (line)
By linear interpolation
If = 7.5 + 2 560. ¥ 32 = 8.83 A
12.4(i) Unity pf
Ia = 25 –0° A
Ef = 231 – j6.1 ¥ 25 –0° = 231 – j152.5
Ef = 276.8 V or 479.4 V(line)
Pe = 3 ¥ 400 ¥ 23 ¥ 1 = 17.32 kWQe = 0
(ii) 0.8 pf leading
Ia = 25 –36.9°
Ef = 231 – j6.1 ¥ 25 –36.9° = 322.6 – j122
Ef = 345 V or 579 V (line)
Ia
Vt
xs
Ef–
– ++
�������������� �
Pe = 3 ¥ 400 ¥ 25 ¥ 0.8/1000
= 13.86 kW
Qe = + 3 ¥ 400 ¥ 25 ¥ 0.6/1000
= + 10.39 kVAR12.5
(a) Vt = Ef = 113
kV
3 113
4 22
¥ FHG
IKJ
FHG
IKJ
. sin d = 15
or d = 31.4°
(b) Ia xs = 2 ¥ 113
sin 15.7° kV = 3.44 kV
Ia = 34404 2.
= 819 A
q = 15.7°pf = cos q = 0.96 leading
(c) Qe = + 3 ¥ 11 ¥ 819 sin 15.7°= + 4222 kVAR
12.6
(a) 3 ¥ 12.5 ¥ 60 cos q = 1050
cos q = 0.81 pf leading
(b) Ia = 60 –36.1°
Ef = 12 53. – (2 + j48) ¥ 60 –36.1°/1000
= 8.86 – j2.4Ef = 9.18 kV or 15.9 kV(line)
(c) Mechanical power developed
= 1050 – 3 60 2
1000
2¥ ¥( )
= 1028.4 kWns = 1000 rpm, ws = 157 rad/s
T = 1028 4 1000
157. ¥
= 6650 Nm
(d) Vt = 7.22 kV, Ef = 9.18 kV
15.7°
31.4°q
Ia
I xa s
VL =
Ef =
113
113
Ia
VtEf
–
+
–
+
2 W 48 W
���� ��� ��� �������������
Iaxs = 9.18 – 7.22 = 1.96 kV
Ia = 196 1000
48. ¥
= 40.8 A
pf = cos 90° = 0° leading
Ia
VtEf
Ef
Vt
Ia
–
– ++
48 W
jI xa s
jI xa s
12.7
Ef = 13 23. = 7.62 kV (= OC Voltage)
Vt = 11 53. = 6.64 kV
At max load d = 90°
Pe(max) = 3 ¥ 7 62 6 64
120. .¥
= 1.264 MW
ns = 1500 rpm, ws = 157 rad/s
T = 1264 10
157
6. ¥ = 8051 Nm
From phase diagram
Iaxs = ( . ) ( . )7 62 6 602 2+
= 10.09 kV
Ia = 10 09 1000
120. ¥
= 84.1 A
pf = cos q = 6 6210 09
..
= 0.656 lag
12.8
Vt = Ef = 223
= 12.7 kV
Ia = 200 1003 12 7
¥¥ .
= 5249 –0°
Ef = 12.7 + 52491000
¥ j1.5 = 12.7 + j7.87
= 14.94 –31.8° kV
q7.62
6.64
Ia
I xa s
Ia
VtEf–
– ++
1.5 W
Ia
V Lt 0°E Lf –d–
– ++
120 W
�������������� �
(i) Ef is increased toE ¢f = 14.94 ¥ 1.15 = 17.18 kW
3 ¥ 1718 12 7
1 5. .
.¥
sin d = 200
d = 27.3°From the phase diagram
y = 17.18 sin 27.3° = 7.88x = 17.18 cos 27.3° – 12.7
= 2.57
Iaxs = ( . ) ( . )2 57 17 882 2+ = 8.28 kV
Ia = 8 29 1000
15.
.¥
= 5527 A
cos q = 7 888 29..
= 0.95 lag
(ii) At E ¢f = 17.18, turbine power increased to 250 MW
3 ¥ 1718 12 7
1 5. .
.¥
sin d = 250 MW
d = 35°From the phasor diagram
y = 17.18 sin 35° = 9.85x = 17.18 cos 35° – 12.7 = 1.37
Remark: At high values of slip h drops off very sharply. The induction motor should beoperated at two steps (2 to 8%)
12.15No-load test
ri = ( )
.400444 5
2
= 360 W (inclusive of windage and friction loss)
cos q0 = 444 53 400 3 5
..¥ ¥
= 0.183
xm = ri / tan q0 = 67 WBlocked rotor test
r1 + r ¢2 = 2220
3 16 7 2¥ ( . ) = 2.65 W
�������������� �
r1 = 1.25 W, r ¢2 = 1.4 W
z = 200 3
16 7/.
= 6.91 W
x1 + x ¢2 = ( . ) ( . )6 91 2 652 2- = 6.38 W
Running
ns = 1000 rpm, ws = 2 1000
60p ¥
= 104.7 rad/s
n = 935 rpm s = 0.065
r1 + r ¢2/s = 1.25 + 1 4
0 065.
. = 22.79 W
¢I2 = 400 3 022 79 6 38
/. .
– ∞+ j
= 9.76 –– 15.6° = 9.40 – j2.62
I0 = 3.5 –– cos–1 0.18.3 = 3.5 –– 79.5°
= 0.64 – j3.44 A
I1 = I I0 2+ ¢ = 10.04 – j6.06 A
= 11.73 –– 31.1° AI1 = 11.73 A, pf = 0.856 lag
Pi(input) = 3 ¥ 400 ¥ 11.73 ¥ 0.856 = 6.956 kW
P0 (output) = 3I ¢22 r ¢2
11
s-FH IK
= 3 ¥ (9.76)2 ¥ 1.4 10 65
1.
-FH IK = 5.75 kW
h = 5 7556 956..
= 82.7%
Breakdown torque
smax, T = ¢
+ + ¢
r
r x x
2
12
1 22( )
; (Eq. (12.46))
= 1 4
125 6 382 2
.
( . ) ( . )+ = 0.215
Now for s = 0.215z = (1.25 + 0.14/0.215) + j6.38
= 10.05 –39.4°
� ���� ��� ��� �������������
I ¢2 = 400 310 05
/.
= 23 A
Tmax = 3 23 1 4 0 215
104 7
2¥ ¥( ) . / ..
= 98.7 NmSpeed = 1000 (1 – 0.215) = 785 rpm
12.16
(a) Ts = 3 12
2
1 22
1 22w s
V r
r r x x◊ ¢
+ ¢ + + ¢( ) ( ); Eq. (12.41)
with s = 1 (i)Parameter values as calculated in Example 12.8 are:
r1 = 0.42 W, r ¢2 = 0.463 W; r1 + r ¢2 = 0.883 Wx1 + x ¢2 = 2.25 W
Substituting the values in Eq. (i)
250 = 378 54
400 3
0 42 2 25
22
22 2.
( / ) ( )
( . ( )) ( . )◊ ¢
+ ¢ +R
R
total
total
= 2037
0 1764 0 84 5 062
2 22
¢+ ¢ + ¢ +
R
R R. . .
or 250 R ¢22 – 2097 R ¢2 + 1265 = 0
or R ¢22 – 8.39 R ¢2 + 5.06 = 0
or R ¢2 = 0.655 W, 7.735 WIt is seen from the figure that with 7.735 W themotor cannot start though it has a Ts of 250 Nm.Thus
R ¢2 = 0.655 WResistance to be added in rotor circuit
= (0.655 – 0.463)/(2.45)2 = 0.032 W(a) (i) Added resistance left in circuit
T = 30 42
12
2
22
1 2w s
V R s
R s x x◊ ¢
+ ¢ + + ¢( ) ( / )
( . / ) ( )
or 250 = 378 54
400 3
0 42
22
22
1 22.
( / ) ( / )
( . / ) ( )◊ ¢
+ ¢ + + ¢R s
R s x x
From the solution of part (a)R ¢2/s = 0.655 or 7.735
But R ¢2 = 0.655 W
0.655 W
7.735 W
250 Nm
�������������� �
\ s = 1 or 0 6557 735..
= 0.085
n = 750 (1 – 0.085) = 686.25 rpm(ii) Added resistance cut and
Again r¢2/s = 0.655 or 7.735But r ¢2 = 0.463 W
s = 0.707 W or 0.06n = 750 (1 – 0.06) = 705 rpm
This illustrates the speed control action of added rotor resistance
(c) T = 3 12
2
1 22
1 22w s
V r s
r r s x x◊ ¢
+ ¢ + + ¢( ) ( / )
( / ) ( ), r ¢2 = 0.463 W
s = 0.085 shift
or 250 = 378 54
3 0 463 0 085
0 42 0 463 0 085 2 25
2 2
2 2.( / ) ( . / . )
( . . / . ) ( . )◊
+ +V
V = 377.3 V(d) Efficiency comparison:
With external resistance in rotor circuit
P0 = 250 ¥ 2 686 25
60p ¥ .
= 17.966 kW
P0 = 3I ¢22 R ¢2 1
1s
-FH IK
17966 = 3I ¢22R ¢2 1
0 0851
.-FH IK
or 3I ¢22 R ¢2 = 1.669 kW
Power lost in iron loss, windage and friction
= Vri
2
= ( )
.400132 2
2
= 1.21 kW
Stator loss cannot be estimated and is ignored in this comparison.
h = 17 96617 966 1 669 1 21
.. . .+ +
= 86.2%
With reduced stator voltagePower lost in iron loss, windage and friction
= ( . )
.377 3132 2
2
= 1.077 kW
P0 = 17.966 kW3I ¢2
2R ¢2 = 1.669 kW
� ���� ��� ��� �������������
h = 17 96617 966 1 669 1 077
.. . .+ +
= 86.74%
Remark: Observe that efficiency is slightly higher for the reduced stator voltage case.However, the equipment to reduce voltage would be more expensive than the resistanceto be added in rotor circuit and would have associated loss not considered above.
The diode will rectify the signal and the meter will measure the average value.\ Imax = 800/4 ¥ 106 = 2 ¥ 10–4 AThe peak value of the voltage across the meter = 400 Vand the meter will read an average value of 200 V.
14.13 Self capacitance, Cd = (C1 – 4C2)/3C1 = 450 pF and C2 = 100 pF
\ Cd = 50/3 pF.14.14 Cd = (C1 – 4C2)/3 C1 = 500 pF and C2 = 60 pF