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

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)?

AOptions

Star-Star ($\theta = 0°$)
Star-Delta ($\theta = -30°$)
Delta-Star ($\theta = 30°$)
Star-Zigzag ($\theta = 30°$)

SSolution

Principle:

Two paths that meet at the same pair of terminals must accumulate the same phase shift from the source, otherwise they cannot be paralleled.

Path 1 - through the upper transformer:

The upper transformer is marked $-30°$ in the direction top bus $\rightarrow$ 15 kV bus, so the top bus, and with it terminal A, leads the generator bus by $30°$:

\[\theta_A = +30°\]

Path 2 - through the 400 kV bus:

The generator transformer contributes $+30°$ from 15 kV to 400 kV and the autotransformer contributes $0°$ from 400 kV to 220 kV. Terminal B sits on the 220 kV bus, so

\[\theta_B = +30° + 0° = +30°\]

Required shift:

\[\theta = \theta_B - \theta_A = 0°\]

Star-delta ($-30°$), delta-star ($+30°$) and star-zigzag ($+30°$) all introduce a shift; only a star-star connection gives $\theta = 0°$.

Correct answer: A (Star-Star, 0°)

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:

AOptions

Generator A: 400 MW, Generator B: 300 MW
Generator A: 350 MW, Generator B: 350 MW
Generator A: 450 MW, Generator B: 250 MW
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

AOptions

Point X (on the top sub-conductor, facing the centre of the bundle)
Point Y (on the top sub-conductor, outer surface)
Point Z (a point inside the triangle)
Point W (near the centre of the bundle)

SSolution

Electric field of a bundled conductor:

All three sub-conductors carry charge of the same sign, $q/3$ each. The gradient is largest on a conductor surface, so only X and Y are candidates; Z and W lie inside the bundle, where the three contributions largely cancel.

At Y (outer surface of the top sub-conductor):

The conductor's own field points radially outward, away from the bundle. The two lower sub-conductors lie on the far side, so their fields at Y point in that same outward direction and add.

At X (inner surface of the same sub-conductor):

Here the conductor's own field points towards the centre of the bundle, while the fields of the two lower sub-conductors point away from them, back towards the top conductor. The contributions oppose, so the resultant is smaller than at Y.

This is exactly why bundling lowers the surface gradient: the maximum always occurs on the outer periphery of the bundle.

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

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

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

AOptions

0.87
0.74
0.67
0.54

SSolution

Network reactances:

Each line has reactance $x$, so before the trip the transfer reactance is $x' + x/2 = 0.1 + x/2$ and after the trip it is $x' + x = 0.1 + x$.

Post-trip power-angle curve:

With $E = 1.0$ pu and $V = 1.0$ pu,

\[P_{max} = \frac{EV}{x' + x} = \frac{1}{0.1 + x}\]

State at the instant of tripping:

The oscillation is undamped and $\delta = 130°$ is its maximum, so at that instant $d\delta/dt = 0$. The rotor therefore starts from rest on the post-trip curve at $\delta = 130°$.

Stability condition:

Because $130° > 90°$, the post-trip electrical power falls as $\delta$ grows. If $P_e(130°) < P_m$ the rotor accelerates, $P_e$ drops further and synchronism is lost; if $P_e(130°) \ge P_m$ the rotor decelerates and swings back. The limiting case is therefore

\[\frac{\sin 130°}{0.1 + x} = P_m = 1.0\]

\[0.1 + x = \sin 130° = 0.766\]

\[x = 0.666 \approx 0.67 \text{ pu}\]

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

72.95 mH in phase C and 139.02 $\mu$F in phase B
72.95 mH in phase B and 139.02 $\mu$F in phase C
42.12 mH in phase C and 240.79 $\mu$F in phase B
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