GATE 2019 Electrical Engineering (EE) Power Systems (2019)

Section 01

0-Mark Questions

QQuestion 1 0 Mark

In the circuit shown below, the switch is closed at $t = 0$. The value of $\theta$ in degrees which will give the maximum value of DC offset of the current at the time of switching is

AOptions

60 \CorrectChoice $-45$
90
$-30$

SSolution

For maximum DC offset in an R-L circuit, the switch should be closed when the instantaneous voltage is zero and about to go negative (or positive, depending on convention).

Circuit impedance angle:

\[\phi = \tan^{-1}\left(\frac{\omega L}{R}\right) = \tan^{-1}\left(\frac{377 \times 10 \times 10^{-3}}{3.77}\right)\]

\[\phi = \tan^{-1}(1) = 45Ã‚Â°\]

For maximum DC offset, the voltage should be at zero crossing with appropriate polarity. This occurs when:

\[\theta + 90Ã‚Â° = 45Ã‚Â°\]

\[\theta = 45Ã‚Â° - 90Ã‚Â° = \boxed{-45Ã‚Â°}\]

Correct answer: B

QQuestion 2 0 Mark

The $Y_{bus}$ matrix of a two-bus power system having two identical parallel lines connected between them in pu is given as

\[Y_{bus} = \begin{bmatrix} -j8 & j20 \\ j20 & -j8 \end{bmatrix}\]

The magnitude of the series reactance of each line in pu (rounded off to one decimal place) is \_\_\_\_\_\_\_\_.

SSolution

For a two-bus system with parallel lines:

The off-diagonal elements represent: $Y_{12} = Y_{21} = j20$ pu

For two identical parallel lines, each with admittance $y$:

\[Y_{12} = -(y + y) = -2y = j20\]

Therefore: $-2y = j20$

\[y = -j10 = \frac{1}{jx} = \frac{-j}{x}\]

\[-j10 = \frac{-j}{x}\]

\[x = \frac{1}{10} = 0.1 \text{ pu}\]

The magnitude of series reactance of each line: $\boxed{0.1 \text{ pu}}$

QQuestion 3 0 Mark

Five alternators each rated 5 MVA, 13.2 kV with 25% of reactance on its own base are connected in parallel to a busbar. The short-circuit level in MVA at the busbar is \_\_\_\_\_\_\_\_.

SSolution

Given: 5 alternators, each rated 5 MVA with $X = 0.25$ pu on own base

For parallel connection of identical machines:

Equivalent reactance: $X_{eq} = \frac{X}{n} = \frac{0.25}{5} = 0.05$ pu

Short-circuit MVA on combined base (5 ÃƒÆ’Ã¢â‚¬â€ 5 = 25 MVA):

\[\text{SC MVA} = \frac{\text{Base MVA}}{X_{eq}} = \frac{25}{0.05} = 500 \text{ MVA}\]

Alternatively, each machine contributes:

\[\text{SC MVA per machine} = \frac{5}{0.25} = 20 \text{ MVA}\]

Total SC MVA: $5 \times 20 = \boxed{100 \text{ MVA}}$

Correct approach: Each machine on its own base:

\[\text{SC per machine} = \frac{5}{0.25} = 20 \text{ MVA}\]

\[\text{Total SC} = 5 \times 20 = \boxed{100 \text{ MVA}}\]

QQuestion 4 0 Mark

The total impedance of the secondary winding, leads, and burden of a 5 A CT is 0.01 $\Omega$. If the fault current is 20 times the rated primary current of the CT, the VA output of the CT is \_\_\_\_\_\_\_\_.

SSolution

Given: CT ratio with secondary rated current $I_{s,rated} = 5$ A, Secondary burden $Z_b = 0.01$ $\Omega$

Fault current is 20 times rated primary current, so secondary current:

\[I_s = 20 \times 5 = 100 \text{ A}\]

Voltage across burden:

\[V_s = I_s \times Z_b = 100 \times 0.01 = 1 \text{ V}\]

VA output:

\[\text{VA} = V_s \times I_s = 1 \times 100 = \boxed{100 \text{ VA}}\]

QQuestion 5 0 Mark

The line currents of a three-phase four wire system are square waves with amplitude of 100 A. These three currents are phase shifted by 120Ã‚Â° with respect to each other. The rms value of neutral current is

AOptions

0 A
$\frac{100}{\sqrt{3}}$ A \CorrectChoice 100 A
300 A

SSolution

For square wave currents with amplitude 100 A:

RMS value of each phase current:

\[I_{rms} = 100 \text{ A}$$ (for square wave, RMS = amplitude) For three-phase currents (square waves) with 120Ã‚Â° phase shift, the neutral current is: $$I_n = I_a + I_b + I_c\]

At any instant, the square waves can be: - One phase at +100 A - Two phases at -100 A (or vice versa)

Or: - Two phases at +100 A - One phase at -100 A

Maximum neutral current magnitude: $|+100 - 100 - 100| = 100$ A or $|+100 + 100 - 100| = 100$ A

For square wave balanced system with 120Ã‚Â° phase shift, the RMS neutral current:

\[I_{n,rms} = \boxed{100 \text{ A}}\]

Correct answer: C

QQuestion 6 0 Mark

A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line inductance is 1 mH/km per phase, and the capacitance is 0.01 $\mu$F/km per phase. The line is under open circuit condition at the receiving end and energized with 400 kV at the sending end, the receiving end line voltage in kV (rounded off to two decimal places) will be \_\_\_\_\_\_\_\_.

SSolution

Given: $f = 50$ Hz, $V_s = 400$ kV, $l = 300$ km, $L = 1$ mH/km, $C = 0.01$ $\mu$F/km

Total inductance: $L_{total} = 1 \times 10^{-3} \times 300 = 0.3$ H

Total capacitance: $C_{total} = 0.01 \times 10^{-6} \times 300 = 3 \times 10^{-6}$ F

Inductive reactance: $X_L = 2\pi f L = 2\pi \times 50 \times 0.3 = 94.25$ $\Omega$

Capacitive reactance: $X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 3 \times 10^{-6}} = 1061.03$ $\Omega$

For open circuit (Ferranti effect), using nominal-$\pi$ model:

Sending end voltage (phase): $V_s = \frac{400}{\sqrt{3}} = 230.94$ kV

\[V_r = V_s \left(1 + \frac{X_L}{2X_C}\right) \approx V_s \left(1 + \frac{\omega^2 LC}{4}\right)\]

\[\omega^2 LC = (2\pi \times 50)^2 \times 0.3 \times 3 \times 10^{-6} = 0.0888\]

\[V_r = 230.94 \times (1 + 0.0222) = 236.06 \text{ kV (phase)}\]

\[V_{r,line} = \sqrt{3} \times 236.06 = \boxed{408.87 \text{ kV}}\]

QQuestion 7 0 Mark

A 30 kV, 50 Hz, 50 MVA generator has the positive, negative, and zero sequence reactances of 0.25 pu, 0.15 pu, and 0.05 pu, respectively. The neutral of the generator is grounded with a reactance so that the fault current for a bolted LG fault and that of a bolted three-phase fault at the generator terminal are equal. The value of grounding reactance in ohms (rounded off to one decimal place) is \_\_\_\_\_\_\_\_.

SSolution

Given: $X_1 = 0.25$ pu, $X_2 = 0.15$ pu, $X_0 = 0.05$ pu, Base: 50 MVA, 30 kV

For three-phase fault:

\[I_{f,3\phi} = \frac{V}{X_1} = \frac{1}{0.25} = 4 \text{ pu}\]

For LG fault with grounding reactance $X_n$:

\[I_{f,LG} = \frac{3V}{X_1 + X_2 + X_0 + 3X_n}\]

For equal fault currents:

\[\frac{3}{X_1 + X_2 + X_0 + 3X_n} = \frac{1}{X_1}\]

\[\frac{3}{0.25 + 0.15 + 0.05 + 3X_n} = \frac{1}{0.25}\]

\[\frac{3}{0.45 + 3X_n} = 4\]

\[3 = 4(0.45 + 3X_n)\]

\[3 = 1.8 + 12X_n\]

\[X_n = \frac{1.2}{12} = 0.1 \text{ pu}\]

Base impedance: $Z_{base} = \frac{(30)^2}{50} = 18$ $\Omega$

\[X_n = 0.1 \times 18 = \boxed{1.8 \text{ } \Omega}\]

QQuestion 8 0 Mark

In the single machine infinite bus system shown, the generator is delivering the real power of 0.8 pu at 0.8 power factor lagging to the infinite bus. The power angle of the generator in degrees (rounded off to one decimal place) is \_\_\_\_\_\_\_\_.

Given: $X_d' = 0.25$ pu, $X_t = 0.2$ pu, $X_{L1} = 0.4$ pu, $X_{L2} = 0.4$ pu, $V = 1\angle 0Ã‚Â°$

SSolution

Total reactance: $X_{total} = X_d' + X_t + X_{L1} + X_{L2} = 0.25 + 0.2 + 0.4 + 0.4 = 1.25$ pu

Given: $P = 0.8$ pu, $\cos\phi = 0.8$ lagging, $V = 1$ pu

Apparent power: $S = \frac{P}{\cos\phi} = \frac{0.8}{0.8} = 1$ pu

Reactive power: $Q = S\sin\phi = 1 \times 0.6 = 0.6$ pu

Current: $I = \frac{S}{V} = \frac{1}{1} = 1$ pu at angle $\phi = \cos^{-1}(0.8) = 36.87Ã‚Â°$ lagging

Taking terminal voltage as reference: $\vec{V} = 1\angle 0Ã‚Â°$

\[\vec{I} = 1\angle -36.87Ã‚Â° = 0.8 - j0.6\]

Internal voltage:

\[\vec{E} = \vec{V} + j\vec{I}X_{total} = 1 + j(0.8 - j0.6)(1.25)\]

\[= 1 + j(1.25)(0.8 - j0.6) = 1 + j1 + 0.75 = 1.75 + j1\]

Power angle: $\delta = \tan^{-1}\left(\frac{1}{1.75}\right) = \tan^{-1}(0.571) = \boxed{29.7Ã‚Â°}$$