Machines Ty s

Electrical EngineeringElectrical Machines

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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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3Try yourself

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


∴ (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


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



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


7Try yourself

= 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%


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


9Try yourself

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



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


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



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°


13Try yourself

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


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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