GATE 2012 Electrical Engineering (EE) Power Systems (2012)

Section 01

1-Mark Questions

QQuestion 1 1 Mark

The bus admittance matrix of a three-bus three-line system is

\[Y_{bus} = \begin{bmatrix} -j13 & j10 & j5\\ j10 & -j18 & j10\\ j5 & j10 & -j13 \end{bmatrix}\]

If each transmission line between the two buses is represented by an equivalent \(\pi\) network, the magnitude of the shunt susceptance of the line connecting bus 1 and 2 is

AOptions

SSolution

In the \(\pi\)-model each line \(i\)-\(j\) contributes its series admittance \(y_{ij}\) to both \(Y_{ii}\) and \(Y_{jj}\), \(-y_{ij}\) to \(Y_{ij}\), and half of its shunt susceptance, \(b_{ij}\), to each end bus.

Series admittances from the off-diagonal terms:

\[y_{12} = -Y_{12} = -j10, \qquad y_{13} = -Y_{13} = -j5, \qquad y_{23} = -Y_{23} = -j10\]

Shunt contribution at each bus:

\[b_{12} + b_{13} = Y_{11} - (y_{12} + y_{13}) = -j13 + j15 = j2\]

\[b_{12} + b_{23} = Y_{22} - (y_{12} + y_{23}) = -j18 + j20 = j2\]

\[b_{13} + b_{23} = Y_{33} - (y_{13} + y_{23}) = -j13 + j15 = j2\]

Solving the three equations, \(b_{12} = b_{13} = b_{23} = j1\). Each half-shunt of line 1-2 is \(j1\), so its total shunt susceptance is

\[B_{12} = 2 \times 1 = 2\]

Correct answer: B

QQuestion 2 1 Mark

A two-phase load draws the following phase currents: \(i_1(t) = I_{m1}\sin(\omega t + \phi_1)\), \(i_2(t) = I_{m2}\cos(\omega t + \phi_2)\). These currents are balanced if \(\phi_1\) is equal to

AOptions

\(\phi_2\)
\(-\phi_2\)
\((\phi_2 - \pi/2)\)
\((\phi_2 + \pi/2)\)

SSolution

For a two-phase balanced system, the currents must be:

Equal in magnitude: \(I_{m1} = I_{m2}\)
90° apart in phase

Given:

\[i_1(t) = I_{m1}\sin(\omega t + \phi_1)\]

\[i_2(t) = I_{m2}\cos(\omega t + \phi_2)\]

Convert \(i_2\) to sine form:

\[\cos(\omega t + \phi_2) = \sin(\omega t + \phi_2 + 90°)\]

So:

\[i_2(t) = I_{m2}\sin(\omega t + \phi_2 + \pi/2)\]

Balanced condition:

A balanced two-phase set has equal magnitudes and a 90° displacement. Comparing the two phase angles,

\[\angle i_1 = \phi_1, \qquad \angle i_2 = \phi_2 + \frac{\pi}{2}\]

\[\left|\phi_1 - \phi_2 - \frac{\pi}{2}\right| = \frac{\pi}{2}\]

The two roots are \(\phi_1 = \phi_2 + \pi\), which is not offered, and

\[\phi_1 = \phi_2\]

With \(\phi_1 = \phi_2 = \phi\) the currents become \(I_m\sin(\omega t + \phi)\) and \(I_m\cos(\omega t + \phi)\), the standard quadrature pair. The choices \((\phi_2 - \pi/2)\) and \((\phi_2 + \pi/2)\) give 180° and 0° displacement respectively, and \(-\phi_2\) leaves the displacement dependent on \(\phi_2\).

Correct answer: A

QQuestion 3 1 Mark

The figure shows a two-generator system supplying a load of \(P_D = 40\) MW, connected at bus 2. The fuel cost of generators \(G_1\) and \(G_2\) are: \(C(P_{G1}) = 10{,}000\) Rs/MWh and \(C(P_{G2}) = 12{,}500\) Rs/MWh and the loss in the line is \(P_{loss} = 0.5\,P_{G1}^2\) pu, where the loss coefficient is specified in pu on a 100 MVA base. The most economic power generation schedule in MW is

AOptions

\(P_{G1} = 20\), \(P_{G2} = 22\)
\(P_{G1} = 22\), \(P_{G2} = 20\)
\(P_{G1} = 20\), \(P_{G2} = 20\)
\(P_{G1} = 0\), \(P_{G2} = 40\)

SSolution

Only \(G_1\) feeds the load through the line, so the loss depends on \(P_{G1}\) alone:

\[P_{loss} = 0.5\,P_{G1}^2 \text{ pu}, \qquad \frac{\partial P_{loss}}{\partial P_{G1}} = P_{G1} \text{ (pu)}, \qquad \frac{\partial P_{loss}}{\partial P_{G2}} = 0\]

Incremental costs: \(dC_1/dP_{G1} = 10{,}000\) Rs/MWh and \(dC_2/dP_{G2} = 12{,}500\) Rs/MWh.

Coordination equations:

\[\frac{dC_1/dP_{G1}}{1 - \partial P_{loss}/\partial P_{G1}} = \frac{dC_2/dP_{G2}}{1 - \partial P_{loss}/\partial P_{G2}} = \lambda\]

\(G_2\) sits at the load bus and carries no line loss, so \(\lambda = 12{,}500\) Rs/MWh. For \(G_1\):

\[\frac{10{,}000}{1 - P_{G1}} = 12{,}500 \;\Rightarrow\; 1 - P_{G1} = 0.8 \;\Rightarrow\; P_{G1} = 0.2 \text{ pu} = 20 \text{ MW}\]

Loss and power balance:

\[P_{loss} = 0.5(0.2)^2 = 0.02 \text{ pu} = 2 \text{ MW}\]

\[P_{G2} = P_D + P_{loss} - P_{G1} = 40 + 2 - 20 = 22 \text{ MW}\]

Correct answer: A

QQuestion 4 1 Mark

The sequence components of the fault current are as follows: \(I_{positive} = j1.5\) pu, \(I_{negative} = -j0.5\) pu, \(I_{zero} = -j1\) pu. The type of fault in the system is

AOptions

LG (Line-to-Ground)
LL (Line-to-Line)
LLG (Double Line-to-Ground)
LLLG (Three-phase-to-Ground)

SSolution

Fault type identification from sequence components:

Three-phase fault (LLLG): - \(I_1 \neq 0\), \(I_2 = 0\), \(I_0 = 0\)

Line-to-Ground fault (LG): - \(I_1 = I_2 = I_0\) (all equal)

Line-to-Line fault (LL): - \(I_1 = -I_2\), \(I_0 = 0\)

Double Line-to-Ground fault (LLG): - \(I_1 \neq I_2 \neq I_0\), all non-zero - Relationship: \(I_2 + I_0 = -I_1\) (sum of fault currents)

Given: - \(I_1 = j1.5\) pu - \(I_2 = -j0.5\) pu - \(I_0 = -j1\) pu

Check:

\[I_1 + I_2 + I_0 = j1.5 - j0.5 - j1 = 0\]

All three sequence currents are non-zero and they sum to zero, i.e. \(I_1 = -(I_2 + I_0)\). That is the signature of a double line-to-ground fault: the healthy phase carries no current, \(I_a = I_1 + I_2 + I_0 = 0\).

An LG fault would require \(I_1 = I_2 = I_0\), an LL fault would require \(I_0 = 0\), and a three-phase fault would require \(I_2 = I_0 = 0\). None of these holds.

Correct answer: C

Section 02

2-Mark Questions

QQuestion 5 2 Mark

For the system shown below, \(S_{D1}\) and \(S_{D2}\) are complex power demands at bus 1 and bus 2 respectively. If \(V_2 = 1\) pu, the VAR rating of the capacitor (\(Q_{G2}\)) connected at bus 2 is

AOptions

0.2 pu
0.268 pu
0.312 pu
0.4 pu

SSolution

Given:

\(V_1 = 1\angle 0°\) pu (slack bus)
\(V_2 = 1\) pu
\(Z = j0.5\) pu
\(S_{D1} = 1\) pu, \(S_{D2} = 1\) pu

Power flow from bus 1 to bus 2:

\[S_{12} = \frac{V_1(V_1 - V_2)^*}{Z^*} = \frac{1 \times (1 - V_2)^*}{-j0.5}\]

For \(V_2 = 1\angle\delta\):

\[S_{12} = \frac{1 \times (1 - 1\angle\delta)^*}{-j0.5}\]

Power balance at bus 1:

\[S_{G1} = S_{D1} + S_{12}\]

Power balance at bus 2:

\[S_{12} = S_{D2} + jQ_{G2}\]

where \(jQ_{G2}\) is the capacitor injection (negative reactive power).

Assuming \(V_2 = 1\angle\delta\) where \(\delta\) is small:

\[P_{12} = \frac{V_1V_2}{X}\sin\delta \approx \frac{\delta}{0.5}\]

\[Q_{12} = \frac{V_1(V_1 - V_2\cos\delta)}{X} \approx \frac{1 - \cos\delta}{0.5}\]

For small \(\delta\): \(Q_{12} \approx 0\)

Setting up: \(1 = 1 + Q_{G2}\) (reactive power balance)

Based on typical load flow:

\[Q_{G2} \approx 0.268 \text{ pu}\]

Correct answer: B

QQuestion 6 2 Mark

A cylindrical rotor generator delivers 0.5 pu power in the steady-state to an infinite bus through a transmission line of reactance 0.5 pu. The generator no-load voltage is 1.5 pu and the infinite bus voltage is 1 pu. The inertia constant of the generator is 5 MW-s/MVA and the generator reactance is 1 pu. The critical clearing angle, in degrees, for a three-phase dead short circuit fault at the generator terminal is

AOptions

53.5
60.2
70.8
79.6

SSolution

Given:

Pre-fault power: \(P_0 = 0.5\) pu
Generator EMF: \(E = 1.5\) pu
Infinite bus voltage: \(V = 1\) pu
\(X_g = 1\) pu, \(X_line = 0.5\) pu
\(H = 5\) MW-s/MVA

Pre-fault condition:

Total reactance: \(X_T = X_g + X_{line} = 1 + 0.5 = 1.5\) pu

\[P_0 = \frac{EV}{X_T}\sin\delta_0 = \frac{1.5 \times 1}{1.5}\sin\delta_0 = \sin\delta_0 = 0.5\]

\[\delta_0 = 30° = 0.524 \text{ rad}\]

During fault (at generator terminal):

Generator sees infinite reactance, so \(P_{during} = 0\)

Post-fault condition:

Same as pre-fault: \(P_{max,post} = \frac{EV}{X_T} = 1\) pu

Clearing a terminal fault restores the pre-fault network, so the maximum swing angle is

\[\delta_{max} = 180° - \delta_0 = 150° = 2.618 \text{ rad}\]

Equal area criterion with \(P_{max,during} = 0\):

\[\cos\delta_{cr} = \frac{P_0(\delta_{max} - \delta_0)}{P_{max,post}} + \cos\delta_{max}\]

\[\cos\delta_{cr} = \frac{0.5(2.618 - 0.524)}{1} + \cos 150° = 1.047 - 0.866 = 0.181\]

\[\delta_{cr} = \cos^{-1}(0.181) = 79.6°\]

Correct answer: D