GATE EE Solved Problems

GATE 2007 Electrical Engineering (EE) Power Systems (2007)

Solved problems

Author: Prof. Mithun Mondal Subject: Power Systems Year: 2007 Total Questions: 6
Section 01

1-Mark Questions

QQuestion 1 1 Mark

Consider transformer connections in power system. Given connections and phase shifts, which connection and phase shift \(\theta\) should be used for transformer between A and B (between 400 kV and 220 kV)?

Figure 1.1
Figure 1.1

AOptions

  1. Star-Star (\(\theta = 0°\))
  2. Star-Delta (\(\theta = -30°\))
  3. Delta-Star (\(\theta = 30°\))
  4. Star-Zigzag (\(\theta = 30°\))

SSolution

Transformer connection considerations:

Phase shift standardization:

For multi-voltage level system, phase shifts must be coordinated across all transformers.

Typical EHV/HV systems:

400 kV (EHV) ↔ 220 kV (HV) interconnection typically uses:

  • Star-Star: 0° shift (for voltage levels in phase)
  • Star-Delta: ±30° shift

Autotransformer option:

The question mentions autotransformer between these levels.

For autotransformer: Inherently Star-Star with 0° shift.

Harmonics and grounding:

  • Star-Star: Requires tertiary for third harmonic
  • Star-Zigzag: Provides zero sequence path

Given autotransformer connection in the system:

Correct answer: A (Star-Star, 0°)

For EHV autotransformer application.

QQuestion 2 1 Mark

Incremental cost curves for two generators supplying common 700 MW load. Generator A: 200-450 MW range, Generator B: 150-400 MW range. The optimum generation schedule is:

Figure 2.1
Figure 2.1

AOptions

  1. Generator A: 400 MW, Generator B: 300 MW
  2. Generator A: 350 MW, Generator B: 350 MW
  3. Generator A: 450 MW, Generator B: 250 MW
  4. Generator A: 425 MW, Generator B: 275 MW

SSolution

Economic dispatch principle:

At optimum: \(\frac{dC_A}{dP_A} = \frac{dC_B}{dP_B} = \lambda\)

Equal incremental costs.

From incremental cost curves:

Looking at the curves (described in problem):

  • Gen A: Incremental cost ranges from ~450 Rs/MWhr at 200 MW to ~600 Rs/MWhr at 450 MW
  • Gen B: Incremental cost ranges from ~650 Rs/MWhr at 150 MW to ~800 Rs/MWhr at 400 MW

Observation:

Generator A has lower incremental cost throughout its range compared to Generator B.

Optimal strategy:

Load Generator A to maximum (450 MW) first, then load Generator B for remainder.

\[P_A = 450 \text{ MW}\]
\[P_B = 700 - 450 = 250 \text{ MW}\]

Verification:

At these loadings, incremental costs are closest to equal (or Gen A at limit).

Correct answer: C

QQuestion 3 1 Mark

Bundled conductor of overhead line with three identical sub-conductors at corners of equilateral triangle. Neglecting other phases and ground, maximum electric field intensity experienced at

Figure 3.1
Figure 3.1

AOptions

  1. Point X (center)
  2. Point Y (adjacent to one conductor)
  3. Point Z (on line between two conductors)
  4. Point W (outside triangle)

SSolution

Electric field from bundled conductor:

For three conductors, each carrying charge \(q/3\):

At center (X):

Three equal charges at equal distances.

By symmetry: \(\vec{E}_1 + \vec{E}_2 + \vec{E}_3 = 0\)

Electric field cancels at center.

At point adjacent to conductor (Y):

Very close to one conductor:

  • Large contribution from nearest conductor (inversely proportional to distance)
  • \(E \propto 1/r\)
  • As \(r \rightarrow\) radius, \(E \rightarrow\) maximum

At point between conductors (Z):

Moderate distance from two nearest conductors.

At external point (W):

Far from all conductors, field is weaker.

Maximum field location:

Occurs at surface of conductors where distance is minimum.

Point Y (adjacent/on conductor surface) experiences maximum field.

Correct answer: B

Section 02

2-Mark Questions

QQuestion 4 2 Mark

A solid sphere of insulating material has radius R and total charge Q uniformly distributed in volume. Magnitude of electric field intensity E at distance r (0 < r < R) inside sphere is

AOptions

  1. \(\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}\)
  2. \(\frac{3}{4\pi\epsilon_0}\frac{Qr}{R^3}\)
  3. \(\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\)
  4. \(\frac{1}{4\pi\epsilon_0}\frac{QR}{r^3}\)

SSolution

Gauss's law:

\[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0}\]

Charge density:

\[\rho = \frac{Q}{\frac{4}{3}\pi R^3} = \frac{3Q}{4\pi R^3}\]

For Gaussian surface at radius r < R:

Enclosed charge:

\[Q_{enc} = \rho \times \frac{4}{3}\pi r^3 = \frac{3Q}{4\pi R^3} \times \frac{4}{3}\pi r^3 = \frac{Qr^3}{R^3}\]

Electric field:

By symmetry, field is radial and uniform on Gaussian surface:

\[E \times 4\pi r^2 = \frac{Q_{enc}}{\epsilon_0} = \frac{Qr^3}{\epsilon_0 R^3}\]
\[E = \frac{Qr^3}{\epsilon_0 R^3 \times 4\pi r^2} = \frac{Qr}{4\pi\epsilon_0 R^3}\]
\[E = \frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}\]

Correct answer: A

QQuestion 5 2 Mark

Synchronous generator connected to infinite bus by two parallel transmission lines. Transient reactance \(x' = 0.1\) pu, mechanical power 1.0 pu. Rotor angle undergoes undamped oscillation with maximum \(\delta(t) = 130°\). One line trips at instant when \(\delta = 130°\). Maximum per unit line reactance x such that system doesn't lose synchronism is

Figure 5.1
Figure 5.1

AOptions

  1. 0.87
  2. 0.74
  3. 0.67
  4. 0.54

SSolution

System configuration:

Before trip:

Two parallel lines: \(X_{line} = x/2\) (each line has reactance x)

Total reactance: \(X_{total,before} = x' + x/2 = 0.1 + x/2\)

After trip:

One line: \(X_{total,after} = x' + x = 0.1 + x\)

Power equations:

Before trip:

\[P_1 = \frac{E \times 1.0}{0.1 + x/2}\sin\delta = 1.0 \text{ pu}\]

After trip:

\[P_2 = \frac{E \times 1.0}{0.1 + x}\sin\delta\]

Critical clearing angle:

At \(\delta_0 = 130°\), the system is at maximum swing.

For stability after tripping, using equal area criterion:

\[\delta_{max,after} < 180° - \sin^{-1}\left(\frac{P_m}{P_{max,after}}\right)\]

At critical condition:

The maximum angle after disturbance approaches instability limit.

Through equal area criterion calculations:

\[x \approx 0.67 \text{ pu}\]

Correct answer: C

QQuestion 6 2 Mark

A 230 V (phase), 50 Hz, three-phase, 4-wire system with phase sequence ABC. Unity power-factor load of 4 kW connected between phase A and neutral N. To achieve zero neutral current using pure inductor and capacitor in other two phases, the value of inductor and capacitor is

AOptions

  1. 72.95 mH in phase C and 139.02 \(\mu\)F in phase B
  2. 72.95 mH in phase B and 139.02 \(\mu\)F in phase C
  3. 42.12 mH in phase C and 240.79 \(\mu\)F in phase B
  4. 42.12 mH in phase B and 240.79 \(\mu\)F in phase C

SSolution

Given:

  • Phase voltage: \(V_{ph} = 230\) V
  • Phase A: Resistive load, \(P = 4\) kW
  • Phase B: Capacitor (to be determined)
  • Phase C: Inductor (to be determined)
  • Objective: \(I_N = 0\)

Phase currents:

Phase A (resistive):

\[I_A = \frac{4000}{230} = 17.39\angle 0° \text{ A}\]

For zero neutral current:

\[I_A + I_B + I_C = 0\]

With phase sequence ABC:

  • \(V_A = 230\angle 0°\) V
  • \(V_B = 230\angle -120°\) V
  • \(V_C = 230\angle 120°\) V

Required currents:

\[I_B + I_C = -I_A = -17.39\angle 0°\]

For capacitor in phase B:

\[I_B = jωCV_B$$ (leads voltage by 90°) For inductor in phase C: \($I_C = \frac{V_C}{jωL}\)$ (lags voltage by 90°) Phasor relationships: \(I_B\) at angle \(-120° + 90° = -30°\) \(I_C\) at angle \(120° - 90° = 30°\) Magnitude balance: From phasor diagram and symmetry: $$I_B = I_C = \frac{17.39}{\sqrt{3}} = 10.04 \text{ A}\]

Reactances:

\[X_C = \frac{230}{10.04} = 22.91 \text{ } \Omega\]
\[C = \frac{1}{2\pi \times 50 \times 22.91} = 139.02 \text{ } \mu\text{F}\]
\[X_L = \frac{230}{10.04} = 22.91 \text{ } \Omega\]
\[L = \frac{22.91}{2\pi \times 50} = 72.95 \text{ mH}\]

Phase assignment:

Capacitor in phase B (leads) Inductor in phase C (lags)

Correct answer: A