GATE 2021 Electrical Engineering (EE) Electrical Machines.

Section 01

0-Mark Questions

QQuestion 1 0 Mark

The power input to a 500 V, 50 Hz, 6-pole, 3-phase induction motor running at 975 RPM is 40 kW. The total stator losses are 1 kW. If the total friction and windage losses are 2.025 kW, then the efficiency is \_\_\_\_\_\_\_%. (Round off to 2 decimal places.)

SSolution

Given:

Supply voltage: $V = 500$ V
Frequency: $f = 50$ Hz
Number of poles: $P = 6$
Speed: $N_r = 975$ RPM
Power input: $P_{in} = 40$ kW
Stator losses: $P_{stator} = 1$ kW
Friction and windage losses: $P_{fw} = 2.025$ kW

Solution:

Step 1: Calculate synchronous speed

\[N_s = \frac{120f}{P} = \frac{120 \times 50}{6} = 1000 \text{ RPM}\]

Step 2: Calculate slip

\[s = \frac{N_s - N_r}{N_s} = \frac{1000 - 975}{1000} = 0.025 = 2.5%\]

Step 3: Power flow in induction motor

Power flow diagram:

\[P_{in} \xrightarrow{-P_{stator}} P_{ag} \xrightarrow{-P_{cu,rotor}} P_{mech} \xrightarrow{-P_{fw}} P_{out}\]

where:

$P_{in}$ = Input power (electrical)
$P_{stator}$ = Stator copper loss + core loss
$P_{ag}$ = Air-gap power
$P_{cu,rotor}$ = Rotor copper loss
$P_{mech}$ = Mechanical power developed
$P_{fw}$ = Friction and windage losses
$P_{out}$ = Output power

Step 4: Calculate air-gap power

\[P_{ag} = P_{in} - P_{stator} = 40 - 1 = 39 \text{ kW}\]

Step 5: Calculate rotor copper loss

For an induction motor:

\[P_{cu,rotor} = s \times P_{ag} = 0.025 \times 39 = 0.975 \text{ kW}\]

Step 6: Calculate mechanical power developed

\[P_{mech} = P_{ag} - P_{cu,rotor} = 39 - 0.975 = 38.025 \text{ kW}\]

Or alternatively:

\[P_{mech} = (1-s) \times P_{ag} = 0.975 \times 39 = 38.025 \text{ kW}\]

Step 7: Calculate output power

\[P_{out} = P_{mech} - P_{fw} = 38.025 - 2.025 = 36 \text{ kW}\]

Step 8: Calculate efficiency

\[\eta = \frac{P_{out}}{P_{in}} \times 100 = \frac{36}{40} \times 100 = 90%\]

Verification:

Total losses:

\[P_{losses} = P_{stator} + P_{cu,rotor} + P_{fw}\]

\[= 1 + 0.975 + 2.025 = 4 \text{ kW}\]

Output power:

\[P_{out} = P_{in} - P_{losses} = 40 - 4 = 36 \text{ kW}\]

Efficiency:

QQuestion 2 0 Mark

An alternator with internal voltage of $1\angle\delta_1$ p.u and synchronous reactance of 0.4 p.u is connected by a transmission line of reactance 0.1 p.u to a synchronous motor having synchronous reactance 0.35 p.u and internal voltage of $0.85\angle\delta_2$ p.u. If the real power supplied by the alternator is 0.866 p.u, then $(\delta_1 - \delta_2)$ is \_\_\_\_ degrees. (Round off to 2 decimal places.)

(Machines are of non-salient type. Neglect resistances.)

SSolution

Given:

Alternator internal voltage: $E_g = 1\angle\delta_1$ p.u.
Alternator synchronous reactance: $X_g = 0.4$ p.u.
Line reactance: $X_{line} = 0.1$ p.u.
Motor synchronous reactance: $X_m = 0.35$ p.u.
Motor internal voltage: $E_m = 0.85\angle\delta_2$ p.u.
Real power supplied: $P = 0.866$ p.u.
Non-salient machines, resistances neglected
Find: $\delta_1 - \delta_2$ in degrees

Solution:

Step 1: Calculate total reactance

Total reactance between the two machines:

\[X_{total} = X_g + X_{line} + X_m\]

\[X_{total} = 0.4 + 0.1 + 0.35 = 0.85 \text{ p.u.}\]

Step 2: Power transfer equation

For two synchronous machines connected through reactance:

\[P = \frac{E_g \cdot E_m}{X_{total}} \sin(\delta_1 - \delta_2)\]

where $\delta_1 - \delta_2$ is the power angle between the two machines.

Step 3: Substitute values

\[0.866 = \frac{1.0 \times 0.85}{0.85} \sin(\delta_1 - \delta_2)\]

\[0.866 = 1.0 \times \sin(\delta_1 - \delta_2)\]

\[\sin(\delta_1 - \delta_2) = 0.866\]

Step 4: Calculate angle

\[\delta_1 - \delta_2 = \sin^{-1}(0.866)\]

\[\delta_1 - \delta_2 = 60°\]

Verification:

Check: $\sin(60°) = \frac{\sqrt{3}}{2} = 0.866$ ✓

This is a standard angle where:

\[\sin(60°) = 0.866025... \approx 0.866\]

Physical Interpretation:

Power angle of 60° indicates moderately heavy loading
Generator internal voltage leads motor internal voltage by 60°
This creates power flow from generator to motor
The system is stable (angle < 90°)

Answer: 60.00 degrees

QQuestion 3 0 Mark

In a single-phase transformer, the total iron loss is 2500 W at nominal voltage of 440 V and frequency 50 Hz. The total iron loss is 850 W at 220 V and 25 Hz. Then, at nominal voltage and frequency, the hysteresis loss and eddy current loss respectively are

AOptions

1600 W and 900 W
900 W and 1600 W
250 W and 600 W
600 W and 250 W

SSolution

Given:

At nominal: $V_1 = 440$ V, $f_1 = 50$ Hz, Total loss $P_{i1} = 2500$ W
At test: $V_2 = 220$ V, $f_2 = 25$ Hz, Total loss $P_{i2} = 850$ W
Find: Hysteresis and eddy current losses at nominal conditions

Solution:

Step 1: Iron loss equations

Total iron loss = Hysteresis loss + Eddy current loss

Hysteresis loss:

\[P_h = K_h f B_m^n\]

where typically $n \approx 1.6$ to $2$ (Steinmetz constant)

For constant flux: $P_h \propto f$

Eddy current loss:

\[P_e = K_e f^2 B_m^2\]

For constant flux: $P_e \propto f^2$

Step 2: Flux density relationships

From transformer voltage equation:

\[V = 4.44 f N B_m A\]

For constant N and A:

\[B_m \propto \frac{V}{f}\]

At nominal conditions:

\[B_{m1} = k \frac{V_1}{f_1} = k \frac{440}{50} = 8.8k\]

At test conditions:

\[B_{m2} = k \frac{V_2}{f_2} = k \frac{220}{25} = 8.8k\]

\[B_{m1} = B_{m2}$$ (Same flux density!) Step 3: Loss relationships with same flux Since flux density is the same: At nominal ($f_1 = 50$ Hz): $$P_{i1} = P_{h1} + P_{e1} = K_h f_1 B_m^n + K_e f_1^2 B_m^2\]

\[2500 = K_h \times 50 + K_e \times 50^2 \text{...(1)}\]

At test ($f_2 = 25$ Hz):

\[P_{i2} = P_{h2} + P_{e2} = K_h f_2 B_m^n + K_e f_2^2 B_m^2\]

\[850 = K_h \times 25 + K_e \times 25^2 \text{...(2)}\]

Step 4: Solve simultaneous equations

From equation (1):

\[2500 = 50K_h + 2500K_e \text{...(1)}\]

From equation (2):

\[850 = 25K_h + 625K_e \text{...(2)}\]

Multiply equation (2) by 2:

\[1700 = 50K_h + 1250K_e \text{...(3)}\]

Subtract (3) from (1):

\[2500 - 1700 = (50K_h + 2500K_e) - (50K_h + 1250K_e)\]

\[800 = 1250K_e\]

\[K_e = 0.64\]

Substitute back into equation (2):

\[850 = 25K_h + 625 \times 0.64\]

\[850 = 25K_h + 400\]

\[25K_h = 450\]

\[K_h = 18\]

Step 5: Calculate losses at nominal conditions

Hysteresis loss at 50 Hz:

\[P_h = K_h \times f_1 = 18 \times 50 = 900 \text{ W}\]

Eddy current loss at 50 Hz:

\[P_e = K_e \times f_1^2 = 0.64 \times 50^2 = 0.64 \times 2500 = 1600 \text{ W}\]

Verification:

\[P_{total} = P_h + P_e = 900 + 1600 = 2500 \text{ W}$$ ✓ At 25 Hz: $$P_h = 18 \times 25 = 450 \text{ W}\]

\[P_e = 0.64 \times 625 = 400 \text{ W}\]

QQuestion 4 0 Mark

A belt-driven DC shunt generator running at 300 RPM delivers 100 kW to a 200 V DC grid. It continues to run as a motor when the belt breaks, taking 10 kW from the DC grid. The armature resistance is 0.025 $\Omega$, field resistance is 50 $\Omega$, and brush drop is 2 V. Ignoring armature reaction, the speed of the motor is \_\_\_\_\_\_\_\_\_\_\_\_ RPM. (Round off to 2 decimal places.)

SSolution

Given:

As generator: Speed $N_g = 300$ RPM, Output power $P_g = 100$ kW
DC grid voltage: $V = 200$ V
As motor: Input power $P_m = 10$ kW
Armature resistance: $R_a = 0.025$ $\Omega$
Field resistance: $R_f = 50$ $\Omega$
Brush drop: $V_b = 2$ V
Ignore armature reaction
Find: Motor speed $N_m$

Solution:

Step 1: Calculate field current (constant for both modes)

\[I_f = \frac{V}{R_f} = \frac{200}{50} = 4 \text{ A}\]

Field current remains constant in both generator and motor modes.

Step 2: Generator mode analysis

For a shunt generator the grid receives the load current:

\[P_g = V \times I_L\]

where $I_L$ is the load current.

\[I_L = \frac{100000}{200} = 500 \text{ A}\]

Armature current:

\[I_{a,g} = I_L + I_f = 500 + 4 = 504 \text{ A}\]

Generated EMF:

\[E_g = V + I_{a,g}R_a + V_b\]

\[E_g = 200 + 504 \times 0.025 + 2\]

\[E_g = 200 + 12.6 + 2 = 214.6 \text{ V}\]

Step 3: Motor mode analysis

Total input power:

\[P_m = V \times I_{L,m} = 10000 \text{ W}\]

\[I_{L,m} = \frac{10000}{200} = 50 \text{ A}\]

For shunt motor:

\[I_{a,m} = I_{L,m} - I_f = 50 - 4 = 46 \text{ A}\]

Back EMF (motor):

\[E_m = V - I_{a,m}R_a - V_b\]

\[E_m = 200 - 46 \times 0.025 - 2\]

\[E_m = 200 - 1.15 - 2 = 196.85 \text{ V}\]

Step 4: Speed relationship

For DC machine with constant flux (since $I_f$ is constant):

\[E \propto \phi N\]

Since $\phi$ is constant:

\[\frac{E_m}{E_g} = \frac{N_m}{N_g}\]

\[N_m = N_g \times \frac{E_m}{E_g}\]

\[N_m = 300 \times \frac{196.85}{214.6}\]

\[N_m = 300 \times 0.917241 = 275.17 \text{ RPM}\]

Verification:

Check power balance:

Generator mode: - Generated power: $E_g \times I_{a,g} = 214.6 \times 504 = 108.16$ kW - Armature copper loss: $I_{a,g}^2 R_a = 504^2 \times 0.025 = 6.35$ kW - Brush loss: $V_b \times I_{a,g} = 2 \times 504 = 1.01$ kW - Output: $108.16 - 6.35 - 1.01 = 100.8$ kW ≈ 100 kW ✓

Motor mode: - Input power: 10 kW - Field loss: $V \times I_f = 200 \times 4 = 0.8$ kW - Armature input: $10 - 0.8 = 9.2$ kW - Armature copper loss: $46^2 \times 0.025 = 0.053$ kW - Brush loss: $2 \times 46 = 0.092$ kW - Mechanical power: $9.2 - 0.053 - 0.092 = 9.055$ kW ✓

Answer: 275.17 RPM

QQuestion 5 0 Mark

An 8-pole, 50 Hz, three-phase, slip-ring induction motor has an effective rotor resistance of 0.08 $\Omega$ per phase. Its speed at maximum torque is 650 RPM. The additional resistance per phase that must be inserted in the rotor to achieve maximum torque at start is \_\_\_\_\_\_\_\_\_ $\Omega$. (Round off to 2 decimal places.) Neglect magnetizing current and stator leakage impedance. Consider equivalent circuit parameters referred to stator.

SSolution

Given:

Number of poles: $P = 8$
Frequency: $f = 50$ Hz
Effective rotor resistance: $R_r' = 0.08$ $\Omega$ (referred to stator)
Speed at maximum torque: $N_{mt} = 650$ RPM
Find: Additional rotor resistance for maximum torque at start
Neglect magnetizing current and stator leakage impedance

Solution:

Step 1: Calculate synchronous speed

\[N_s = \frac{120f}{P} = \frac{120 \times 50}{8} = 750 \text{ RPM}\]

Step 2: Calculate slip at maximum torque

\[s_{mt} = \frac{N_s - N_{mt}}{N_s} = \frac{750 - 650}{750} = \frac{100}{750} = 0.1333\]

Step 3: Condition for maximum torque

For maximum torque in an induction motor:

\[s_{mt} = \frac{R_r'}{X_r'}\]

where:

$s_{mt}$ = slip at maximum torque
$R_r'$ = rotor resistance (referred to stator)
$X_r'$ = rotor leakage reactance (referred to stator)

From the given condition:

\[0.1333 = \frac{0.08}{X_r'}\]

\[X_r' = \frac{0.08}{0.1333} = 0.6 \text{ } \Omega\]

Step 4: Condition for maximum torque at starting

For maximum torque at starting (when $s = 1$):

\[s_{start} = 1 = \frac{R_{r,new}'}{X_r'}\]

\[R_{r,new}' = X_r' = 0.6 \text{ } \Omega\]

Step 5: Calculate additional resistance

Additional resistance required:

\[R_{additional}' = R_{r,new}' - R_r'\]

\[R_{additional}' = 0.6 - 0.08 = 0.52 \text{ } \Omega\]

Verification:

Original condition (maximum torque at 650 RPM):

QQuestion 6 0 Mark

An air-core radio-frequency transformer as shown has a primary winding and a secondary winding. The mutual inductance M between the windings of the transformer is \_\_\_\_\_\_ μH. (Round off to 2 decimal places.)

SSolution

Given:

Frequency: $f = 100$ kHz
Secondary open circuit voltage: $V_{oc} = 7.3$ V (peak-to-peak)
Voltage across the 22 $\Omega$ resistor in series with the primary: $V_{22} = 5.0$ V (peak-to-peak)
Air-core transformer (loose coupling)
Find: Mutual inductance M (μH)

Solution:

Step 1: Primary current from the 22 $\Omega$ resistor

The 22 $\Omega$ resistor sits in series with the primary winding, so it carries the primary current, and 5.0 V$_{p-p}$ is measured across it:

\[I_1 = \frac{5.0}{22} = 0.2273 \text{ A}_{p-p}\]

Step 2: Voltage induced in the open secondary

The secondary is open-circuited, so it carries no current and the only voltage across it is the mutually induced one:

\[V_2 = \omega M I_1\]

Both readings are peak-to-peak, so the conversion to rms cancels in the ratio.

Step 3: Mutual inductance

\[M = \frac{V_2}{\omega I_1} = \frac{7.3}{2\pi \times 100 \times 10^3 \times 0.2273}\]

\[M = \frac{7.3 \times 22}{2\pi \times 10^5 \times 5.0} = 51.12 \times 10^{-6} \text{ H}\]

Answer: 51.12 μH