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Electrical EngineeringElectrical Machines
WORKBOOKWORKBOOKWORKBOOKWORKBOOKWORKBOOK
2016
Detailed Explanations ofTry Yourself Questions
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T1 : Solution
(c)(c)(c)(c)(c)As shown in figure below,
A A
0.3 Ω
0.2 Ω
0.2 Ω
0.3 Ω
Ia1 Ia2
If1 If2
300 V300 V
500 AA
B
300 – 0.2 ��I = 300 – 0.3
��I
⇒��I (0.2) =
��I (0.3)
⇒ �
�
�
�
II
=� �
����
�� �⇒ =I I
⇒� �� �+I I = 500
� ���� � �+I I = 500
��I =
���
��� = 200
��I = 1.5
��I = 1.5 × 200 = 300 A
and��I (0.3) =
��I (0.2)
�
�
�
�
II
=�
�
��I = 1.5
��I
��I +
��I = 500 A
��I + 1.5
��I = 500
⇒��I =
���
��� = 200
DC Machines1
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T2 : Solution
(34)(34)(34)(34)(34)
��� = 500 – 0.5 × 50 = 475 ...(1)
Let the current when joined in series = I2 AThe resistance = 0.5 × 2 = 1 Ω
∴��
� = 500 – I2 × 1 = 500 – I2 ...(2)
�
�
�
�
�
�= � �
� �
���
��� �
� ��
�
φ= × =
φ...(3)
where, k = �
�
φφ
From (1), (2) and (3)
����
�
− I=
�
�
500 – I2 =�
�
�...(4)
�
�
�
�= � � �
� �
���
�φ= =
φI II
(∵ T2 = 2T1)
100 = kI2
k =�
���
I...(5)
From (5) and (4),
500 – I2 =� �
� ��� �����×=
I I∴ I2 = 34 A
T3 : Solution
(0.85)(0.85)(0.85)(0.85)(0.85)
Input power =�� �
��� ����
×=
∴ Input line current =�
���� ����� �
���
×=
Shunt field current =���
� ���
=
∴ Back emf, ��
� = 500 – 0.4 × 85.7 = 466.5 V
Since the load and the constant losses are constant
� ��� I =� ��� I = 466.5 × 83.7 = 39046
Total resistance after the series field is added= 0.4 + 0.1 = 0.5
4 Electrical Engineering • Electrical Machines
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∴ (500 – 0.5 I2)I2 = 39046
∵ ��I – 1000 I2 + 78092 = 0
This gives, I2 = 85 AThe input current = 85 + 2 = 87 A
∴ The efficiency =������
����� =
�� � �
� ���× = 0.85
T4 : Solution
(403)(403)(403)(403)(403)
T =�
� ��
�
�
⎛ ⎞φ ⎜ ⎟⎝ ⎠πI
209 =� �� ��
� ���� �� �
�− ×× × × ×
πI
= 10.34 Ia
Ia = the line current =���
�� ����
≈
Eb =��
�
�
φ× =
�� � ��
� �φ φ× =
∴ 440 =����� � ���
�
�−× × ×
or, N =� ��
�� ������ ���
×=
×
T5 : Solution
(1989)(1989)(1989)(1989)(1989)
Eg = �
�φ ×
200 =���� �
������ �
φ × × ×
or, φ =��
�� ����� �!� � �!�����
∴ The flux density, B = ���� "�#�
φ×
= �
��
���� �� �� ��−× × × = ��
��$����
∴ The ampere-turns required for 0.8 cm gap
=�
�
βμl =
�
�
���� ��
��� ��
−− × ×
× π ×
= 1989
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T1 : Solution
(d)(d)(d)(d)(d)Maximum allowable flux, φmax = Bmax × core area
= 1.2 × 0.025= 0.03 Wb
Voltage on primary (high voltage) side,V1 = 6000 V
Number of turns of (low voltage) side of main transformer
N2 = � �
���� ���� � ��
�
�=
φ × × = 60
Number of turns on low voltage side of teaser transformer= 0.866 N2 = 0.866 × 60 = 52
Point to be tapped = ��� ��
�× =
T2 : Solution
(c)(c)(c)(c)(c)
pu resistance =���
��������
=
pu reactance =�
�������
=
Voltage regulation will be maximum, when phase angle,
φ = � � �� �
� �� �
$���
$
� �
� �
− II
= � ��"#�"�#��"�
��%�%�"�#�
��
��−
= � ������� �����
�����
− = °
Power factor, cos φ = cos 63.435° = 0.447 lagging (for maximum regulation)
Maximum regulation = pu resistance × cos φ + pu reactance × sin φ= 0.025 × 0.447 + 0.05 × 0.8944
= 0.0559 or 5.59%
Transformers2
6 Electrical Engineering • Electrical Machines
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T3 : Solution
(b)(b)(b)(b)(b)At 3/4 Full load and unity power factor output
=�
��� ��� �� �
× × =
Maximum efficiency, ηmax = 97%
Input =������ �
� � �����
= =η
Total loss = Input – output = 386.6 – 375 = 11.6 kWThis loss is equally divided between iron and copper loss, so
�
���
⎛ ⎞⎜ ⎟⎝ ⎠ = Pi =
������� �
�=
or, Pc =��
��� ���� ��
× =
Percentage resistance =��
�
�����
�×
I
= �
��
� �
�����
×
II
= ������� ����&
���× =
Percentage impedance = 10%
Percentage reactance = � ��� ����− = 9.786%
Percentage regulation = � �� �
�
#�% %��� ��
�
φ + φI I
= 2.06 × 0.8 + 9.786 × 0.6 = 7.52%
T4 : Solution
(c)(c)(c)(c)(c)Rated output, S = 25 kVASecondary voltage, V2 = 220 VFull-load (or rated) secondary current,
I2 =�
���� �� ����
���
�
× ×= = 113.64 A
Total resistance referred to secondary,
��� = K2R1 + R2 =
����
� ��������
⎛ ⎞ × +⎜ ⎟⎝ ⎠
= 0.02 Ω
Full-load copper loss, Pc =�
�� ��I = (113.64)2 × 0.02
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= 258 W or 0.258 kWIron loss, Pi = 80% of full load copper loss
=��
������
× = 206.4 W or 0.2064 kW
Full-load output at 0.8 p.f. = 25 × cos φ = 25 × 0.8= 20 kW
Full-load efficiency at 0.8 p.f., η =������
��������� ' �� �
×+ i
= ��
����� ����� �����
×+ +
= 97.93%
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T1 : Solution
(a)(a)(a)(a)(a)Since the stator resistance is neglected
Stator out-put = Rotor out-put
=(���� ����!� ����
�
=�
��
� ��
� �
⎛ ⎞× ×⎜ ⎟⎝ ⎠ =
����
�� ��� �
�−× s
if E2 = rotor stand-still voltage per phase
rotor current at slips = �� �
�
if s1 and s2 are the slips with two given stator out-puts we have�
�� ����
� �
�−× = 9 ...(i)
and�
�� ����
� �
�−× = 4 ...(ii)
From (i) and (ii)
∴ �
�
�
�= 2.25
since, s1 = � �
���)� �
� �
� � � ��
� �
− −=
so, �
�
�
�= 2.25 = �
�
�
�
� �
� �
−−
⇒ 2.25 (ns – n2) = ns – n1
⇒ 2.25 ns – ns = 2.25 n2 – n1
⇒ 1.25 × 750 = 2.25 n2 – n1 ...(iii)
as �
�
�
�= 1.043 (given)
from equation (iii)1.25 × 750 = 2.25 n2 – 1.043 n2
n2 = 776.719 ≈ 777 rpmn1 = 1.043 × 776.719 ≈ 810 rpm
Induction Machine3
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T2 : Solution
(c)(c)(c)(c)(c)
Torque at any slip, T =� � �
���
� � �+...(1)
This will be maximum,when R = sXSince, maximum torque occurs at s = 0.2
R = 0.2 X⇒ X = 5 R ...(2)At starting, s = 1
Tst =� �
��
� �+
with R increased to R + 0.5,
��� ′ =� �
� ����
� ���� ��
� �
� �
++ +
(∴ X = 5R) ...(3)
putting, S =�
� in (1)
Tmax =�
�
$
� ���
�� � � �
� ��= = (∴ X = 5R) ...(4)
Tst is 75% of Tmax, from (3) and (4)
� �
� ����
� ���� ��
�
� �
++ +
=���
�� ��� �=
⇒ 38 R2 – 17R + 0.75 = 0⇒ R = 0.395 Ω or 0.05 Ωand X = (5 × 0.395)Ω or (5 × 0.05)Ωor X = 1.97 Ω or 0.25 Ω
T3 : Solution
(b)(b)(b)(b)(b)
Synchronous speed, Ns =��� ��
�� ����
×=
Slip, s =�� ���
������
−=
Since the torque in maximum (stalling)
�
�
�
�= s = 0.16
X 2 = � ���
���� ����
�= = 0.44 Ω
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If the total rotor resistance =�� ′ (in the second case)
�
�
�
�
′= s at starting = 1
�� ′ = X 2 = 0.44 Ω
∴ The resistance to be added = 0.44 – 0.07 = 0.37 Ω
T4 : Solution
(b)(b)(b)(b)(b)
Ns =���
���� ���
=
N1 = 1425 rpm ⇒ S1 = ���� ���
��������
−=
N2 = 1275 rpm ⇒ S2 = ���� ���
��������
−=
Slip =����� #����� ��%%
����� ����� =
�
� ��
(���� �����
�I
⇒ slip ∝ R2 ⇒ �
�
�
�= �
�
����
����
� � �
�
+ +=
⇒ R =����
���� ��������
× − = 0.5 Ω/ph
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T1 : Solution
(a)(a)(a)(a)(a)Given data,Given data,Given data,Given data,Given data, z s = (0.5 + j4)Ω = 4.031∠82.8°Vt = 400 V, Ef = 500 V, α2 = 90 – θ2 = (90 – 82.8) = 7.12°
shaft power output = maximum output power – friction windage and iron losses
maximum output power =��
�
��� �� ������
���� ����
� � �
�� �
� � ��
× ⎛ ⎞− × = − ×⎜ ⎟⎝ ⎠
= 41.922 kWshaft power output = (3 × 42.922 kW) – 0.9 kW = 124.867 kW
T2 : Solution
(0.98)(0.98)(0.98)(0.98)(0.98)Taking 11 kV, 1600 KVA as base
The pu synchronous reactance = �
��
����
���
= 0.4
if V1 = 1∠0° = terminal voltage and I = 1∠φThen the generated voltage
�� = �� + I
or, �� = � �� ���� �� � � ��� ��+ ∠φ ∠ ° = + ∠φ + °
= 1 + 0.4(–sinφ + jcosφ) = 1 – 0.04 sinφ + j0.4cosφFor, zero regulation, magnitude of this is 1∴ (1 – 0.4 sin φ)2 + (0.4cosφ)2 = 1⇒ 1 + 0.42 – 0.8 sin φ = 1
⇒ sinφ =����
���
⇒ φ = 11.5°∴ power factor, cosφ = cos11.5° = 0.98
Synchronous Machine4
12 Electrical Engineering • Electrical Machines
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T3 : Solution
(b)(b)(b)(b)(b)
Vt =� �
= 265.58 V
Ef =���
= 317.54 V
zs =�
��� �� ����� ����
� �+ = + = 1.675∠85.44° Ω
Gross Pm(output)⎥maximum =�
�
� � � �
��
� � � �
−+
=�
�
������ ��� ���� � ����
�� ���� �
− × ×+ = 45.53 kW
net Pm (output)⎥ maximum = 45.53 × 3 – 1.4 = 135.19 kW
T4 : Solution
(c)(c)(c)(c)(c)
P =� � �
%�� %�� ��
� � �
� � �
� � �
� � �
⎛ ⎞⋅δ + − δ⎜ ⎟
⎝ ⎠
=������ ����� ����� � �
��� ��� ���� � ��� ���
⎛ ⎞δ + − δ⎜ ⎟⎝ ⎠
=��� �
��� ��� ���� ���
δ + δ
�
�δ= 0 for P = Pmax
0 =��� �
��� ������� ���
δ + δ
⇒ 1.1 cos δ + cos 2δ = 0⇒ 2cos2 δ –1 + 1.1 cos δ = 0
⇒ cos δ =���� ����� ������
�����
− ± +=
δ = 61.12°
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T5 : Solution
(b)(b)(b)(b)(b)
Phase voltage = ��
��� *
=
I =����
��� �
=×
I�
as reference,
drop = �I�
= 131(0.4 + j6) = 52 + j786The terminal voltage = 3810(0.8 + j 0.6)
= 3048 + j 2286 V∴ The generated voltage = 3100 + j3072 V
= 4.364∠44.7° kV ...(i )with I as reference, if x kV is the load phase voltage
Terminal voltage = 0.8 x – j 0.6 x kV
�I�
= 0.052 + j 0.786 kV∴ Generated voltage = (0.8 x + 0.052) + j (0.786 – 0.6 x) ...(ii )From equation (i) and (ii),∴ (0.8 x + 0.052)2 + (0.786 – 0.6x)2 = 4.3642
∴ x 2 – 0.86 x – 18.424 = 0x = 4.744
line voltage = 4.744 × � × 1000 = 8220 V
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Single-Phase Motor& Special Machine,
Energy Conversion Principles5T1 : Solution
(5.22)(5.22)(5.22)(5.22)(5.22)
S =���� ���
����
− = 0.06
The circuit model is given below,
Im Im′ 1.8 Ω 7.8 Ω
1.8 Ω 7.8 Ω
Im′′
j48
j48
220 V
3.4 s = 5.67 Ω
3.4(2 – s) = 1.75 Ω
� �� ����� �
= j48⎥⎥ (1.8 + 56.7 + j7.8) = j48⎥⎥ (58.5 + j7.8)
= 35∠54° = 20.6 + j28.3
� �� ����� �
= j48⎥⎥ (1.8 + 1.75 + j7.8) = j48⎥⎥ (3.55 + j7.8)
= 7.36∠69.1° = 2.63 + j6.88
��������
= (20.6 + j 28.3) + (2.63 + j6.88) = 42.16∠56.6°
I�
=��� �
���� �������� ����
∠ °= ∠ − °
∠ °
⎥ IL⎥ = ⎥ IM⎥ = 5.22 A
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