GATE 2012 Electrical Engineering (EE) Electrical Machines (2012)

Section 01

1-Mark Questions

QQuestion 1 1 Mark

The slip of an induction motor normally does not depend on

AOptions

rotor speed
synchronous speed
shaft torque
core-loss component

SSolution

The slip of an induction motor is defined as:

\[s = \frac{N_s - N_r}{N_s}\]

where \(N_s\) is synchronous speed and \(N_r\) is rotor speed.

Analysis of each option:

(A) Rotor speed: Slip directly depends on rotor speed (appears in numerator).

(B) Synchronous speed: Slip directly depends on synchronous speed (appears in equation).

(C) Shaft torque:

Torque is related to slip: \(T \propto sE^2R_r/(R_r^2 + (sX_r)^2)\)
As torque changes, rotor speed changes, which changes slip
Higher torque load \(\rightarrow\) lower rotor speed \(\rightarrow\) higher slip

(D) Core-loss component:

Core losses occur in the stator and rotor iron
These are primarily due to hysteresis and eddy currents
Core losses do not directly affect the slip
Slip is determined by the balance between electromagnetic torque and load torque
Core losses only affect efficiency, not the mechanical speed relationship

Correct answer: D

Section 02

2-Mark Questions

QQuestion 2 2 Mark

A 220 V, 15 kW, 1000 rpm shunt motor with armature resistance of 0.25 \(\Omega\) has a rated line current of 68 A and a rated field current of 2.2 A. The change in field flux required to obtain a speed of 1600 rpm while drawing a line current of 52.8 A and a field current of 1.8 A is

AOptions

18.18% increase
18.18% decrease
36.36% increase
36.36% decrease

SSolution

Given:

Rated: \(V_t = 220\) V, \(N_1 = 1000\) rpm, \(I_L = 68\) A, \(I_f = 2.2\) A
New: \(N_2 = 1600\) rpm, \(I_L = 52.8\) A, \(I_f = 1.8\) A
\(R_a = 0.25\) \(\Omega\)

Condition 1 (Rated):

Armature current: \(I_{a1} = I_L - I_f = 68 - 2.2 = 65.8\) A

Back EMF: \(E_{b1} = V_t - I_{a1}R_a = 220 - 65.8(0.25) = 220 - 16.45 = 203.55\) V

Condition 2 (New):

Armature current: \(I_{a2} = 52.8 - 1.8 = 51\) A

Back EMF: \(E_{b2} = 220 - 51(0.25) = 220 - 12.75 = 207.25\) V

Using back EMF equation: \(E_b = K\phi N\)

\[\frac{E_{b1}}{E_{b2}} = \frac{\phi_1 N_1}{\phi_2 N_2}\]

\[\frac{203.55}{207.25} = \frac{\phi_1 \times 1000}{\phi_2 \times 1600}\]

\[\frac{\phi_2}{\phi_1} = \frac{207.25 \times 1000}{203.55 \times 1600} = \frac{207250}{325680} = 0.636\]

Change in flux:

\[\frac{\phi_1 - \phi_2}{\phi_1} = 1 - 0.636 = 0.364 = 36.36%\]

The flux decreased by 36.36%.

Correct answer: D

QQuestion 3 2 Mark

The locked rotor current in a 3-phase, star connected 15 kW, 4-pole, 230 V, 50 Hz induction motor at rated conditions is 50 A. Neglecting losses and magnetizing current, the approximate locked rotor line current drawn when the motor is connected to a 236 V, 57 Hz supply is

AOptions

58.5 A
45.0 A
42.7 A
55.6 A

SSolution

For a locked rotor condition (standstill, \(s = 1\)):

Locked rotor impedance:

\[Z_{LR} = \sqrt{R_r^2 + X_{LR}^2}\]

where \(X_{LR} = 2\pi f L\) (leakage reactance).

At rated conditions (230 V, 50 Hz):

\[I_{LR1} = \frac{V_1}{\sqrt{3} Z_{LR1}} = 50 \text{ A}\]

At new conditions (236 V, 57 Hz):

Since \(X_{LR} \propto f\):

\[X_{LR2} = X_{LR1} \times \frac{57}{50}\]

Assuming resistance is small compared to reactance:

\[Z_{LR2} \approx X_{LR2} = X_{LR1} \times \frac{57}{50}\]

\[I_{LR2} = \frac{V_2}{\sqrt{3} Z_{LR2}} = \frac{V_2}{\sqrt{3} X_{LR1}} \times \frac{50}{57}\]

\[I_{LR2} = I_{LR1} \times \frac{V_2}{V_1} \times \frac{f_1}{f_2} = 50 \times \frac{236}{230} \times \frac{50}{57}\]

\[I_{LR2} = 50 \times 1.026 \times 0.877 = 45.0 \text{ A}\]

Correct answer: B

QQuestion 4 2 Mark

A single phase 10 kVA, 50 Hz transformer with 1 kV primary winding draws 0.5 A and 55 W, at rated voltage and frequency, on no load. A second transformer has a core with all its linear dimensions \(\sqrt{2}\) times the corresponding dimensions of the first transformer. The core material and lamination thickness are the same in both transformers. The primary windings of both the transformers have the same number of turns. If a rated voltage of 2 kV at 50 Hz is applied to the primary of the second transformer, then the no load current and power, respectively, are

AOptions

0.7 A, 77.8 W
0.7 A, 155.6 W
1 A, 110 W
1 A, 220 W

SSolution

Transformer 1:

Rating: 10 kVA, 1 kV, 50 Hz
No-load: \(I_0 = 0.5\) A, \(P_0 = 55\) W

Transformer 2:

All linear dimensions scaled by \(\sqrt{2}\)
Same number of turns, \(V_2 = 2\) kV

Scaling relationships:

Cross-sectional area: \(A_2 = (\sqrt{2})^2 A_1 = 2A_1\)

Core length: \(l_2 = \sqrt{2} l_1\)

Volume: \(Vol_2 = 2\sqrt{2} Vol_1\)

Flux density:

\[B = \frac{V}{4.44 f N A}\]

\[B_2 = \frac{2000}{4.44 \times 50 \times N \times 2A_1} = \frac{1000}{4.44 \times 50 \times N \times A_1} = B_1\]

Same flux density!

No-load current:

Magnetizing current \(I_m \propto \frac{B l}{N}\)

\[I_{m2} = I_{m1} \times \frac{l_2}{l_1} = I_{m1} \times \sqrt{2}\]

Core loss component \(I_c = \frac{P_0}{V}\)

\[I_{c2} = \frac{P_{02}}{V_2}\]

Total no-load current:

\[I_{02} \approx I_{m2} = 0.5 \times \sqrt{2} = 0.707 \approx 0.7 \text{ A}\]

No-load power (iron loss):

Hysteresis loss \(\propto f B^2 \times Volume\)

\[P_h \propto Vol = 2\sqrt{2} P_{h1}\]

Eddy current loss \(\propto f^2 B^2 \times Volume\)

\[P_e \propto Vol = 2\sqrt{2} P_{e1}\]

Total iron loss:

\[P_{02} = 2\sqrt{2} \times P_{01} = 2 \times 1.414 \times 55 = 155.6 \text{ W}\]

Correct answer: B