GATE 2021 Electrical Engineering (EE) Power Systems (2021)

SSolution

Given:

• Total number of buses: \(N\)
• Number of PQ buses: \(N_L\)
• Load flow using Newton-Raphson in polar form
• Find: Dimension of sub-matrix \(M\)

Solution:

Step 1: Understand bus classification

In load flow studies:

• Slack bus: 1 bus (voltage magnitude and angle are specified)
• PV buses: Generator buses where \(P\) and \(|V|\) are specified
• PQ buses: Load buses where \(P\) and \(Q\) are specified (count = \(N_L\))

Number of buses:

• Slack buses: 1
• PV buses: \(N - 1 - N_L\) (remaining buses after slack and PQ)
• PQ buses: \(N_L\)

Step 2: Variables in Newton-Raphson

Unknown variables:

• Voltage angles \(\delta\): All buses except slack = \((N-1)\) unknowns
• Voltage magnitudes \(|V|\): Only for PQ buses = \(N_L\) unknowns

Total unknowns: \((N-1) + N_L\)

Step 3: Equations in Newton-Raphson

Mismatch equations:

• Real power mismatches \(\Delta P\): All buses except slack = \((N-1)\) equations
• Reactive power mismatches \(\Delta Q\): Only for PQ buses = \(N_L\) equations

Total equations: \((N-1) + N_L\)

Step 4: Jacobian structure

Dimensions:

• \(H\): \(\frac{\partial P}{\partial \delta}\) → \((N-1) \times (N-1)\)
• \(S\): \(\frac{\partial P}{\partial V}\) → \((N-1) \times N_L\)
• \(M\): \(\frac{\partial Q}{\partial \delta}\) → \(N_L \times (N-1)\)
• \(R\): \(\frac{\partial Q}{\partial V}\) → \(N_L \times N_L\)

Step 5: Dimension of M

Sub-matrix \(M\) represents:

• Rows: Number of \(Q\) equations = \(N_L\) (PQ buses only)
• Columns: Number of \(\delta\) variables = \((N-1)\) (all buses except slack)

Therefore, dimension of \(M\) is: \(\boxed{N_L \times (N-1)}\)

Verification:

Check Jacobian dimensions:

• \(H\): \((N-1) \times (N-1)\) ✓
• \(S\): \((N-1) \times N_L\) ✓
• \(M\): \(N_L \times (N-1)\) ✓
• \(R\): \(N_L \times N_L\) ✓

Total rows: \((N-1) + N_L\) ✓\\ Total columns: \((N-1) + N_L\) ✓

Correct answer: A (\(N_L \times (N-1)\))

SSolution

Given:

• Generator 1 incremental cost: \(\frac{dF_1}{dP_1} = 40 + 0.2P_1\) (\$/MWh)
• Generator 2 incremental cost: \(\frac{dF_2}{dP_2} = 32 + 0.4P_2\) (\$/MWh)
• Total load demand: \(P_D = 260\) MW
• Network losses ignored
• Find: Optimal generation \(P_1\) and \(P_2\)

Solution:

Step 1: Economic dispatch criterion

For economic operation without transmission losses, the incremental costs of all generators must be equal at the optimal point:

where \(\lambda\) is the system incremental cost (Lagrange multiplier).

Step 2: Set up equations

Equation 1 (Equal incremental costs):

Multiply by 5:

Equation 2 (Power balance):

Step 3: Solve simultaneous equations

From equation (1):

Substitute into equation (2):

From equation (2):

Step 4: Verification

Check power balance:

Check equal incremental costs:

Both are equal ✓

Physical Interpretation:

• Generator 1 has the higher no-load incremental cost (40 vs 32) but the flatter slope (0.2 vs 0.4)
• Its incremental cost rises more slowly with output, so it takes the larger share at this load
• Generator 2 starts cheaper but its incremental cost climbs twice as fast
• Optimal dispatch: Generator 1 produces more (160 MW vs 100 MW)
• At optimum, both generators have same incremental cost (72 \$/MWh)
• Any deviation from this dispatch would increase total cost

Correct answer: B (\(P_1 = 160\) MW, \(P_2 = 100\) MW)

SSolution

Given:

• Bus-1 voltage: \(|E|\angle\delta\)
• Bus-2 voltage: \(|V|\angle 0°\)
• Self-impedance of each line: \(Z_s = 1.5j\) p.u.
• Mutual impedance: \(Z_m = 0.5j\) p.u.
• Two parallel transmission lines with mutual coupling
• Find: Maximum real power transfer

Solution:

Step 1: Model for mutually coupled lines

For two mutually coupled transmission lines, the voltage-current relationships are:

where \(I_1\) and \(I_2\) are currents in lines 1 and 2.

Step 2: Equivalent impedance

For two identical mutually coupled lines in parallel, the equivalent impedance is:

Or alternatively, consider the impedance matrix:

Since both equations give the same voltage drop, adding them: \(2(V_1 - V_2) = (Z_s + Z_m)(I_1 + I_2)\)

\(V_1 - V_2 = \frac{Z_s + Z_m}{2} \cdot I_{total}\)

where \(I_{total} = I_1 + I_2\)

\(Z_{eq} = \frac{Z_s + Z_m}{2} = \frac{1.5j + 0.5j}{2} = \frac{2j}{2} = j \text{ p.u.}\)

Step 3: Power transfer equation

For two buses connected through impedance \(Z_{eq} = jX_{eq}\):

\(P = \frac{|E||V|}{X_{eq}} \sin\delta\)

Maximum power occurs when \(\sin\delta = 1\) (i.e., \(\delta = 90°\)):

\(P_{max} = \frac{|E||V|}{X_{eq}} = \frac{|E||V|}{1} = |E||V| \text{ p.u.}\)

Check by symmetry of the two lines

The two lines are identical and are tied to the same pair of buses, so they must carry equal currents, \(I_1 = I_2 = \frac{I_{total}}{2}\). Substituting in either row of the impedance matrix:

\(V_1 - V_2 = Z_s I_1 + Z_m I_2 = (Z_s + Z_m)\frac{I_{total}}{2}\)

\(Z_{eq} = \frac{Z_s + Z_m}{2} = \frac{1.5j + 0.5j}{2} = j1 \text{ p.u.}\)

\(P_{max} = \frac{|E||V|}{X_{eq}} = \frac{|E||V|}{1} = |E||V|\)

Effect of the coupling:

Without mutual coupling the two \(1.5j\) lines in parallel would give \(X_{eq} = 0.75\) p.u. and \(P_{max} = 1.33|E||V|\). The positive mutual impedance raises \(X_{eq}\) to \(1\) p.u. and so lowers the transfer limit to \(|E||V|\).

Correct answer: A (\(|E||V|\))

SSolution

Given:

• Buses 1, 2 and 3 are each joined to one common internal node through \(Z_1 = Z_2 = Z_3 = j\ \Omega\)
• That internal node is earthed through \(Z_4 = j\ \Omega\)
• Generators are ideal voltage sources, so they add no shunt admittance
• Find: \(Y_{Bus}\) referred to buses 1, 2 and 3

Solution:

Step 1: Read the topology

The three buses are not connected to one another directly. Each reaches a common internal node (call it node 4) through \(j\ \Omega\), and node 4 is earthed through \(j\ \Omega\). Every branch admittance is

Step 2: Write the admittance matrix of the four-node network

Node 4 carries four branches, so \(Y_{44} = 4(-j) = -4j\); each connection between node \(i\) and node 4 gives \(Y_{i4} = -y = j\); buses 1, 2 and 3 have no branch between them, so their mutual terms are zero.

Step 3: Eliminate the internal node (Kron reduction)

Node 4 has no injection, so it is removed by

Every \(Y_{i4} = Y_{4k} = j\) and \(Y_{44} = -4j\), so the correction term is the same for all entries:

Diagonal entries:

Off-diagonal entries:

Step 4: Result

Check: each row sums to \(-\frac{3}{4}j + \frac{1}{4}j + \frac{1}{4}j = -\frac{1}{4}j\), which is non-zero exactly because the reduced network keeps a path to earth through \(Z_4\).

Correct answer: C

\(\begin{bmatrix} -\frac{3}{4}j & \frac{1}{4}j & \frac{1}{4}j \\ \frac{1}{4}j & -\frac{3}{4}j & \frac{1}{4}j \\ \frac{1}{4}j & \frac{1}{4}j & -\frac{3}{4}j \end{bmatrix}\)

SSolution

Given:

• Phase-A current: \(I_A = 1.1\angle 0°\) p.u.
• Phase-C current: \(I_C = (1\angle 120° + 0.1)\) p.u.
• Phase-B zero-sequence current: \(I_{B0} = 0.1\angle 0°\) p.u.
• Find: Phase-B current \(I_B\)

Solution:

Step 1: Symmetrical components review

For three-phase currents, the symmetrical components are:

\(\begin{bmatrix} I_{A0} \\ I_{A1} \\ I_{A2} \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & a & a^2 \\ 1 & a^2 & a \end{bmatrix} \begin{bmatrix} I_A \\ I_B \\ I_C \end{bmatrix}\)

where \(a = 1\angle 120° = e^{j2\pi/3}\) and \(a^2 = 1\angle 240° = 1\angle -120°\)

The zero-sequence component is: \(I_0 = \frac{1}{3}(I_A + I_B + I_C)\)

For all phases, the zero-sequence current is the same: \(I_{A0} = I_{B0} = I_{C0} = I_0 = 0.1\angle 0°\)

Step 2: Use zero-sequence relationship

\(I_0 = \frac{1}{3}(I_A + I_B + I_C)\)

\(0.1 = \frac{1}{3}(I_A + I_B + I_C)\)

\(I_A + I_B + I_C = 0.3\angle 0°\)

Step 3: Calculate \(I_B\)

\(I_B = 0.3 - I_A - I_C\)

Substitute given values: \(I_B = 0.3 - 1.1\angle 0° - (1\angle 120° + 0.1)\)

\(I_B = 0.3 - 1.1 - 1\angle 120° - 0.1\)

\(I_B = -0.9 - 1\angle 120°\)

\(I_B = -1\angle 120° - 0.9\)

Now, \(-1\angle 120° = 1\angle(120° + 180°) = 1\angle 300° = 1\angle -60°\)

\(1\angle 120° = -0.5 + j0.866\)

\(I_C = 1\angle 120° + 0.1 = (-0.5 + j0.866) + 0.1 = -0.4 + j0.866\)

\(I_A = 1.1\)

\(I_A + I_C = 1.1 + (-0.4 + j0.866) = 0.7 + j0.866\)

\(I_B = 0.3 - (0.7 + j0.866) = -0.4 - j0.866\)

Step 4: Write \(I_B\) in the form used by the options

Use the identity \(1 + 1\angle 120° + 1\angle -120° = 0\), that is \(-1\angle 120° = 1 + 1\angle -120°\). Substituting in \(I_B = -0.9 - 1\angle 120°\):

Step 5: Numerical check against the options

\(1\angle -120° + 0.1 = (-0.5 - j0.866) + 0.1 = -0.4 - j0.866\), which is exactly the rectangular value obtained in Step 3, with \(|I_B| = \sqrt{0.4^2 + 0.866^2} = 0.954\) p.u.

The other options evaluate to \(-0.6 - j0.866\) (A), \(-0.65 - j0.953\) (B) and \(-0.45 - j0.953\) (C); none of them matches.

Correct answer: D (\(1\angle -120° + 0.1\angle 0°\))

SSolution

Given:

• Three-phase balanced supply, phase sequence RYB, three-wire connection (no neutral)
• From the figure the three lines meet at a common star point through
• line R: capacitor \(Z_R = -j10~\Omega\)
• line B: a plain conductor, \(Z_B = 0\)
• line Y: inductor \(Z_Y = j10~\Omega\)
• Find: \(\frac{|I_B|}{|I_R|}\)

Solution:

Step 1: Locate the star point

Line B reaches the star point through zero impedance, so the star point sits at the potential of line B. Each branch therefore has a line voltage referred to B across it:

Step 2: Line voltages

With \(V_R = V\angle 0°\), \(V_Y = V\angle -120°\), \(V_B = V\angle 120°\) for the RYB sequence:

Step 3: Branch currents

Step 4: Current in line B

KCL at the star point gives

Step 5: Ratio

Correct answer: B (1)