GATE 2011 Electrical Engineering (EE) Power Systems (2011)

SSolution

Factors for selection:

1. Power transmission capability:

Higher voltage increases capacity quadratically.

2. Losses:

Higher voltage reduces losses significantly.

3. Voltage regulation:

Better regulation at higher voltages.

4. Stability: Long transmission (300 km) requires high voltage for adequate stability margin.

Analysis:

(A) 132 kV double circuit:

• [\(\times\)] Too low for 500 MW at 300 km
• [\(\times\)] Very high losses (\(\propto 1/V^2\))
• [\(\times\)] Poor voltage regulation
• [\(\times\)] Requires very large conductors

(B) 132 kV with 40% compensation:

• [\(\times\)] Improves power transfer but still inadequate
• [\(\times\)] High losses remain
• [\(\times\)] Increased complexity and cost

(C) 400 kV single circuit:

• [\checkmark] Adequate voltage for 500 MW
• [\checkmark] Much lower losses
• [\(\times\)] No redundancy - complete shutdown if line fails
• [\(\times\)] Unacceptable for nuclear plant (base load)

(D) 400 kV double circuit:

• [\checkmark] Optimal voltage level
• [\checkmark] Minimum losses (highest efficiency)
• [\checkmark] Excellent voltage regulation
• [\checkmark] High reliability (redundancy)
• [\checkmark] One circuit can carry load during maintenance
• [\checkmark] Standard for nuclear power evacuation

Nuclear plant requirements:

• Base load generation - must be highly reliable
• 500 MW is large capacity
• 300 km is long distance
• Redundancy essential

Correct answer: D (though answer key shows A)

SSolution

Negative sequence components:

In balanced 3-phase system: \(I_2 = 0\)

Negative sequence appears during:

• Unbalanced faults (LG, LL, LLG)
• Unbalanced loads
• Phase-to-phase faults
• Open conductor conditions

Effect on alternator:

Negative sequence currents have devastating effects on generators:

1. Double-frequency heating:

• \(I_2\) rotates opposite to rotor (relative speed = \(2N_s\))
• Induces currents at \(2f\) in rotor body
• Severe heating in: rotor surface, damper windings, pole faces
• Rotor not designed for this frequency

2. Heating magnitude:

Even small \(I_2\) causes large heating.

3. Permissible limits:

Continuous: \(I_2 < 10%\) of rated

Short time: \(I_2^2 t < 30\) to 40 (for large generators)

4. Protection philosophy:

• Alarm at low \(I_2\) levels
• Trip at high \(I_2\) or sustained conditions
• Inverse time characteristic: \(t = K/I_2^2\)

Other equipment:

Transformer: Better tolerance for unbalance, other protections primary

Transmission line: Uses distance relays, \(I_2\) not main protection

Bus bar: Differential protection is primary

Correct answer: A

SSolution

Power transmission:

where \(X\) is total line reactance.

Methods to enhance power:

(A) Series inductive compensator:

• [\(\times\)] Increases reactance
• [\(\times\)] Reduces power transfer
• [\(\times\)] Worsens stability

(B) Shunt inductive compensator at receiving end:

• [\(\times\)] Absorbs reactive power
• [\(\times\)] Reduces voltage
• [\(\times\)] Decreases power transfer
• Used for controlling overvoltages, not enhancing power

(C) Series capacitive compensator:

• [\checkmark] Reduces effective reactance: \(X_{eff} = X_L - X_C\)
• [\checkmark] Increases power: \(P \propto 1/(X_L - X_C)\)
• [\checkmark] Improves stability margin
• [\checkmark] Better voltage regulation
• [\checkmark] Directly addresses transmission angle limitation
• [\checkmark] Most effective for long lines

(D) Shunt capacitive compensator at sending end:

• Provides reactive power support
• Improves sending end voltage
• Less effective than series compensation
• Doesn't reduce line reactance

Why series capacitive is preferred:

For long EHV lines:

• Line reactance dominates
• Series compensation directly reduces \(X\)
• Typical compensation: 40-70% of line reactance
• Effect: \(P_{new} = P_{old} \times \frac{1}{1-k}\) where \(k\) is compensation degree

Example: 50% compensation doubles power capability!

Correct answer: C

SSolution

Key principle:

Transmission loss \(\propto I^2R \propto P^2 \times \text{length}\)

System configuration:

• G1 connected via 25 km line
• G2, G3 connected via 75 km line each
• Load = 120 MW at center

Loss analysis:

Loss coefficient for each line \(\propto\) length

For minimum loss with identical costs:

• Shorter line (25 km) should carry more power
• Longer lines (75 km) should carry less power

Check options:

Option A: \(P_1 = 80\) MW (25 km), \(P_2 = P_3 = 20\) MW (75 km each)

• Most power through shortest line
• Minimum losses
• Losses \(\propto 80^2(25) + 20^2(75) + 20^2(75)\)
• \(\propto 160,000 + 30,000 + 30,000 = 220,000\) units

Option B: Equal distribution would give higher losses

Option C: Equal distribution - moderate losses

Option D: Less power through short line - highest losses

Economic dispatch with losses:

For identical costs: \(\frac{dC}{dP_i} = \lambda(1 - \partial P_L/\partial P_i)\)

Shorter line has smaller \(\partial P_L/\partial P_i\), so carries more power.

Correct answer: A

SSolution

Given impedances:

• Bus 1-2: \(Z_{12} = j0.1\) pu
• Bus 2-3: \(Z_{23} = j0.08\) pu
• Bus 1: shunt \(Z_1 = j0.2\) pu
• Bus 3: shunt \(Z_3 = j0.1\) pu

Admittance values:

Diagonal elements:

From the answer choices, checking option B:

This suggests different connectivity. The off-diagonal elements are positive (indicating connections), while diagonal elements are negative (sum of connected admittances).

Correct answer: B

Note: Without seeing the exact figure, the pattern suggests Bus 1 connects to Bus 2, and Bus 2 connects to Bus 3, with shunt elements at buses.