GATE 2023 Electrical Engineering (EE) Electrical Machines (2023)

SSolution

Understanding Induction Machine Operation Modes:

Key Concept: Slip Definition

For an induction machine:

where:

• \(N_s\) = Synchronous speed
• \(N_r\) = Rotor speed
• \(s\) = Slip

Analysis of Different Operating Modes:

1. Motor Mode (Normal Motoring):

• Rotor speed: \(N_r < N_s\) (rotor runs slower than synchronous speed)
• Slip: \(s = \frac{N_s - N_r}{N_s} > 0\)
• At standstill: \(N_r = 0\), \(s = 1\)
• At synchronous speed: \(N_r = N_s\), \(s = 0\)
• Range: 0 < s < 1 (or 0.0 to 1.0)
• Motor draws power from supply, develops mechanical power
• Torque is positive (in direction of rotation)

Matching: b → p

2. Generator Mode (Generating):

• Rotor speed: \(N_r > N_s\) (rotor runs faster than synchronous speed)
• Slip: \(s = \frac{N_s - N_r}{N_s} < 0\) (negative slip)
• Rotor is driven by prime mover above synchronous speed
• Machine feeds electrical power back to the supply
• Requires external mechanical drive
• Range: -1 < s < 0 (or -1.0 to 0.0)
• Torque is negative (opposes direction of rotation)

Matching: a → r

3. Plugging Mode (Braking):

• Also called "plugged" or "counter-current braking"
• Rotor rotates in opposite direction to rotating magnetic field
• Achieved by reversing phase sequence while motor is running
• Rotor speed: \(N_r < 0\) (negative, opposite to field rotation)
• Slip: \(s = \frac{N_s - N_r}{N_s} = \frac{N_s - (-|N_r|)}{N_s} = \frac{N_s + |N_r|}{N_s} > 1\)
• At maximum negative speed equal to \(-N_s\): \(s = \frac{N_s - (-N_s)}{N_s} = 2\)
• Range: 1 < s < 2 (or 1.0 to 2.0)
• Very high currents, high losses, severe braking torque
• Used for quick stopping or reversing

Matching: c → q

Summary Table:

Physical Interpretation:

Motor Mode (s = 0 to 1):

• Normal operation
• Slip = 0: No-load, no torque
• Slip = 1: Locked rotor, maximum starting torque
• Typical rated slip: 2-5%

Generator Mode (s = -1 to 0):

• Requires external mechanical power
• Used in regenerative braking
• Wind turbines, overhauling loads
• Slip becomes negative

Plugging Mode (s = 1 to 2):

• Emergency braking
• Very high rotor currents (2-7 times rated)
• High thermal stress on rotor
• Rapid deceleration
• Should be used only briefly

Correct answer: A (a-r, b-p, and c-q)

SSolution

Given:

• Number of poles: \(P = 10\)
• Supply frequency: \(f = 50\) Hz
• Rated voltage: \(V = 240\) V
• Operating speed: \(N_r = 540\) RPM
• Find: Frequency of rotor currents due to backward field

Solution:

Relative speed between backward field and rotor:

Magnitude: \(1140\) RPM

Slip magnitude:

Step 5: Calculate rotor current frequency due to backward field

Frequency of rotor currents:

For backward field:

Verification:

The rotor current frequency due to forward field would be:

The backward field frequency is much higher because the backward field and rotor move in opposite directions, creating a much larger relative motion.

Physical Explanation:

• Forward field rotates at +600 RPM
• Rotor rotates at +540 RPM
• Relative speed: \(600 - 540 = 60\) RPM
• This produces 5 Hz rotor currents (forward field effect)

• Backward field rotates at -600 RPM
• Rotor rotates at +540 RPM
• Relative speed: \((-600) - (+540) = -1140\) RPM
• Magnitude: 1140 RPM
• This produces 95 Hz rotor currents (backward field effect)

General Formula:

For single-phase induction motor:

where \(s_f\) is the forward slip.

Correct answer: B (95 Hz)

\textit{Note: The backward field in a single-phase induction motor creates a braking torque. The high frequency rotor currents (95 Hz) due to the backward field produce additional losses and reduce motor efficiency.}

SSolution

Given:

• Four stator conductors: A, A', B, B' carrying DC currents
• Current directions shown in figure (i)
• Rotor coils: a-a' and b-b'
• Rotor angular speed: \(\omega\) rad/s (clockwise)
• Find: Relationship between induced EMFs

Solution:

Step 1: Analyze magnetic field produced by stator

The stator conductors carry DC currents:

• Conductors A and A' form one pair
• Conductors B and B' form another pair
• DC currents create a stationary magnetic field

From the figure:

• If A has current going into the page (⊗) and A' has current coming out (⊙)
• Similarly for B and B'
• This creates a stationary flux pattern

Step 2: Determine nature of the magnetic field

With DC currents in stator conductors:

• The magnetic field is stationary in space
• Field does not rotate
• Field magnitude is constant in time
• Field distribution is fixed in space

Step 3: Analyze EMF induction in rotor coils

For EMF to be induced in rotor coils:

where \(\phi\) is the flux linking the coil.

As the rotor rotates:

• Rotor coil a-a' rotates through the stationary field
• The flux linking coil a-a' changes with position
• Similarly for coil b-b'

Step 4: Calculate flux linkage variation

If the stator creates a sinusoidally distributed stationary field:

For rotor coil at angle \(\theta = \omega t\):

where \(\Phi_m\) is the maximum flux linkage.

Induced EMF:

Step 5: Analyze coil b-b' position

From figure (ii), coil b-b' is positioned at 90° (electrical) from coil a-a':

Induced EMF:

Or:

Step 6: Compare the two EMFs

where \(E_m = \Phi_m \omega\)

This shows:

• Both EMFs have finite magnitudes: \(E_m\)
• \(E_{b-b'}\) leads \(E_{a-a'}\) by 90°
• Same frequency: \(\omega\) rad/s

Physical Interpretation:

• DC stator currents create stationary magnetic field
• Rotating rotor coils cut through this stationary field
• This is the principle of a DC generator or AC generator with DC field
• Coil a-a' experiences maximum flux at one instant
• Coil b-b' (positioned 90° away) experiences maximum flux 90° later in time
• Hence \(E_{b-b'}\) leads \(E_{a-a'}\) by 90° (or \(\frac{\pi}{2}\) radians)

Note on phase relationship:

If the rotor rotates clockwise and coil b-b' is ahead of coil a-a' in the direction of rotation, then b-b' reaches any given flux position before a-a', so \(E_{b-b'}\) leads \(E_{a-a'}\).

\textbf{Correct answer: B (\(E_{a-a'}\) and \(E_{b-b'}\) have finite magnitudes with \(E_{b-b'}\) leading \(E_{a-a'}\))}

\textit{Note: This configuration represents the basic principle of AC generation. The stationary DC field on the stator and rotating rotor coils produce AC voltages. The 90° spatial displacement between rotor coils results in 90° phase difference in induced EMFs.}

SSolution

Given:

• Rated voltage: \(V = 400\) V
• Rated current: \(I_a = 15\) A
• Rated speed: \(N_1 = 1500\) RPM
• Armature resistance: \(R_a = 1.2\) \(\Omega\)
• Supply voltage drops by 10%: \(V_2 = 0.9 \times 400 = 360\) V
• Field current: Constant (unchanged)
• Load: Constant torque
• Find: New speed \(N_2\)

Solution:

Step 1: Calculate initial back EMF

At rated conditions:

Step 2: DC motor speed equation

For a separately excited DC motor:

where:

• \(K\) = machine constant
• \(\phi\) = field flux (constant since field current is unchanged)
• \(N\) = speed in RPM

At rated conditions:

Step 3: Analyze new operating condition

Since load torque is constant and field is constant:

For constant torque and constant flux:

The armature current remains the same because:

• Torque is constant: \(T = K\phi I_a\)
• Field flux is constant: \(\phi =\) constant
• Therefore: \(I_a\) must be constant

Step 4: Calculate new back EMF

With reduced supply voltage:

Step 5: Calculate new speed

From equations (1) and (2):

Verification:

Speed drop:

Percentage drop:

This is reasonable given that voltage dropped by 10%.

Alternative Approach - Using Speed-Voltage Relationship:

For constant torque load with constant field:

Since \(I_a\) is constant:

Physical Explanation:

• When supply voltage drops, back EMF must also drop
• Since \(E = K\phi N\) and flux is constant, speed must drop
• For constant torque, current remains same
• The voltage drop \((I_a R_a = 18\) V) remains constant
• The back EMF (and hence speed) is proportional to \((V - I_a R_a)\)

Answer: 1343 RPM

SSolution

Given:

• Synchronous impedance: \(Z_s = 0.1 + j0.3\) pu
• Static stability limit: \(P_{max} = 2.5\) pu
• Find: Excitation voltage \(E_f\) (pu)

Solution:

For synchronous motor with impedance \(Z_s = R_a + jX_s\):

The power transfer is maximum when the load angle satisfies certain conditions. For the maximum power:

\(P_{max} = \frac{VE_f}{|Z_s|} - \frac{V^2 R_a}{|Z_s|^2}\)

With \(V = 1\) pu: \(2.5 = \frac{E_f}{0.3162} - \frac{1 \times 0.1}{0.1}\)

\(2.5 = \frac{E_f}{0.3162} - 1\)

\(\frac{E_f}{0.3162} = 3.5\)

\(E_f = 3.5 \times 0.3162 = 1.11 \text{ pu}\)

Answer: 1.11 pu

\textit{Note: The static stability limit represents the maximum power that can be transferred before the motor loses synchronism. It depends on both the excitation voltage and the synchronous impedance.}

SSolution

Given:

• Rated voltage: \(V_1 = 415\) V (line)
• Rated frequency: \(f_1 = 50\) Hz
• Number of poles: \(P = 6\)
• Rated speed: \(N_1 = 960\) RPM
• Rated power: 4 HP
• Load: Constant torque
• New voltage: \(V_2 = 0.8 \times 415 = 332\) V
• New frequency: \(f_2 = 0.8 \times 50 = 40\) Hz
• Neglect: Stator leakage impedance and rotational losses
• Find: New speed \(N_2\) (RPM)

Solution:

Step 1: Calculate rated synchronous speed and slip

Synchronous speed at rated frequency: \(N_{s1} = \frac{120 f_1}{P} = \frac{120 \times 50}{6} = 1000 \text{ RPM}\)

Rated slip: \(s_1 = \frac{N_{s1} - N_1}{N_{s1}} = \frac{1000 - 960}{1000} = 0.04 = 4%\)

Step 2: Synchronous speed at new frequency

\(N_{s2} = \frac{120 f_2}{P} = \frac{120 \times 40}{6} = 800 \text{ RPM}\)

Step 3: Torque equation for induction motor

For an induction motor, the torque is: \(T = \frac{3V^2 s R_r'}{\omega_s[(R_r')^2 + (sX_r')^2]}\)

where \(\omega_s = \frac{2\pi f}{P/2}\) is the synchronous angular velocity.

Simplifying: \(T \propto \frac{V^2 s}{f}\)

For constant torque and constant \(\frac{V}{f}\) ratio:

Step 4: V/f control analysis

When both voltage and frequency are reduced by the same percentage (20%): \(\frac{V_2}{f_2} = \frac{0.8V_1}{0.8f_1} = \frac{V_1}{f_1}\)

The V/f ratio is maintained constant.

For constant V/f operation with constant torque: - The flux remains approximately constant: \(\phi \propto \frac{V}{f}\) - For constant torque: \(T = K\phi I_r\)

Step 5: Slip relationship

For constant torque operation with constant V/f:

The torque equation simplifies to: \(T \propto \frac{V^2 s}{f \cdot X_r}\)

Since \(X_r \propto f\): \(T \propto \frac{V^2 s}{f \cdot f} = \frac{V^2 s}{f^2}\)

For constant torque: \(\frac{V_1^2 s_1}{f_1^2} = \frac{V_2^2 s_2}{f_2^2}\)

With \(V_2 = 0.8V_1\) and \(f_2 = 0.8f_1\): \(\frac{V_1^2 s_1}{f_1^2} = \frac{(0.8V_1)^2 s_2}{(0.8f_1)^2}\)

\(\frac{V_1^2 s_1}{f_1^2} = \frac{0.64V_1^2 s_2}{0.64f_1^2}\)

\(s_1 = s_2\)

The slip remains the same!

Step 6: Calculate new speed

\(s_2 = s_1 = 0.04\)

\(N_2 = N_{s2}(1 - s_2) = 800(1 - 0.04) = 800 \times 0.96 = 768 \text{ RPM}\)

Verification:

Speed ratio: \(\frac{N_2}{N_1} = \frac{768}{960} = 0.8\)

Frequency ratio: \(\frac{f_2}{f_1} = \frac{40}{50} = 0.8\) ✓

This confirms that under constant V/f control with constant torque: \(N_2 = N_1 \times \frac{f_2}{f_1}\)

Physical Explanation:

When V/f ratio is maintained constant:

• Flux remains constant
• For constant torque, slip remains constant
• Speed is proportional to frequency: \(N = N_s(1-s) \propto f\)
• Reducing frequency by 20% reduces speed by 20%

This is the principle of V/f control (scalar control) used in variable frequency drives.

Answer: 768 RPM

SSolution

Given:

• Frequency: \(f = 50\) Hz
• Winding a-b (primary): Open circuit voltage \(V_{ab} = 200\) V (rms)
• Winding c-d (secondary): Supplied from AC current source
• Core cross-sectional area: \(A = 5000\) mm\(^2 = 5 \times 10^{-3}\) m\(^2\)
• Average core length: \(l = 500\) mm \(= 0.5\) m
• Maximum flux density: \(B_{max} = 1\) Wb/m\(^2\)
• Relative permeability: \(\mu_r = 5000\)
• Leakage impedance of a-b: \(Z_{ab} = 5 + j16\) \(\Omega\)
• Leakage impedance of c-d: \(Z_{cd} = 11.25 + j36\) \(\Omega\)
• Find: Self-inductance of winding a-b (\(L_{ab}\)) in mH

Solution: \(L_m = \frac{N_{ab}^2}{\mathcal{R}} = \frac{32,400}{159.15} = 203.6 \text{ H}\)

\(\mu_0 \mu_r A = 4\pi \times 10^{-7} \times 5000 \times 5 \times 10^{-3}\) \(= 4\pi \times 10^{-7} \times 25 = 31.416 \times 10^{-6} \text{ Hâ‹…m}\)

\(\mathcal{R} = \frac{0.5}{31.416 \times 10^{-6}} = 15,915 \text{ AT/Wb}\)

\(L_m = \frac{32400}{15915} = 2.036 \text{ H} = 2036 \text{ mH}\)

\(L_{ab} = 2036 + 50.9 = 2086.9 \text{ mH}\)

Rounded: \(\mathbf{2086.9 \text{ mH}}\) or approximately \(\mathbf{2087.0 \text{ mH}}\)

Answer: 2087.0 mH

\textit{Note: The high inductance value is due to the high permeability (5000) and relatively large number of turns (180) in the transformer winding.}