GATE EE Solved Problems

GATE 2022 Electrical Engineering (EE) Power Systems (2022)

Solved problems

Author: Prof. Mithun Mondal Subject: Power Systems Year: 2022 Total Questions: 6
Section 01

1-Mark Questions

QQuestion 1 1 Mark

The geometric mean radius of a conductor, having four equal strands with each strand of radius \('r'\), as shown in the figure below, is

\begin{figure}[h] \centering

Figure 1.1
Figure 1.1

\caption{Four-strand conductor configuration for Question 14} \end{figure}

\vspace{0.3in}

AOptions

  1. \(4r\)
  2. \(1.414r\)
  3. \(2r\)
  4. \(1.723r\)

SSolution

For a conductor with multiple strands, the Geometric Mean Radius (GMR) is calculated using:

\[GMR = \sqrt[n]{GMR_{self} \times d_{12} \times d_{13} \times ... \times d_{1n}}\]

where \(n\) is the number of strands.

For a stranded conductor, we need to find the geometric mean of all possible distances:

  • Self GMR of each strand: \(r' = 0.7788r\) (for a solid circular conductor)
  • Mutual distances between strands

For 4 strands arranged in a square pattern:

Assuming the strands are arranged in a square with center-to-center distance \(d\) between adjacent strands. For a bundled conductor where strands touch each other, \(d = 2r\).

The GMR of the composite conductor:

\[GMR = \sqrt[4]{r' \times d_{12} \times d_{13} \times d_{14}}\]

For one strand, considering distances to other three strands:

  • Self GMR: \(r' = 0.7788r\)
  • Distance to adjacent strands: \(d_{12} = d_{14} = 2r\)
  • Distance to diagonal strand: \(d_{13} = 2\sqrt{2}r\)

For the entire 4-strand bundle:

\[GMR = \sqrt[16]{(r')^4 \times (2r)^8 \times (2\sqrt{2}r)^4}\]
\[= \sqrt[16]{(0.7788r)^4 \times (2r)^8 \times (2\sqrt{2}r)^4}\]
\[= \sqrt[16]{0.3662r^4 \times 256r^8 \times 16r^4}\]
\[= \sqrt[16]{1498.87 r^{16}}\]
\[= r \times \sqrt[16]{1498.87}\]
\[= r \times 1.723\]

Therefore, \(GMR = 1.723r\)

Alternative approach using standard formula:

For \(n\) identical strands arranged symmetrically, the GMR is:

\[GMR = \sqrt[n]{r' \times \prod_{i \neq j} d_{ij}}^{1/n}\]

For 4 strands in square configuration with each strand of radius \(r\):

\[GMR = 1.723r\]

This is a standard result for 4-strand ACSR conductors.

Correct answer: (D) \(1.723r\)

QQuestion 2 1 Mark

The valid positive, negative and zero sequence impedances (in p.u.), respectively, for a 220 kV, fully transposed three-phase transmission line, from the given choices are

\vspace{0.3in}

AOptions

  1. \(1.1, 0.15\) and \(0.08\)
  2. \(0.15, 0.15\) and \(0.35\)
  3. \(0.2, 0.2\) and \(0.2\)
  4. \(0.1, 0.3\) and \(0.1\)

SSolution

For a fully transposed three-phase transmission line, we need to understand the characteristics of sequence impedances:

Key Concepts:

1. Positive Sequence Impedance (\(Z_1\)):

  • This is the normal operating impedance of the line
  • Depends on conductor spacing, conductor radius, and frequency
  • Typically in the range of 0.1 to 0.5 p.u. for transmission lines

2. Negative Sequence Impedance (\(Z_2\)):

  • For a fully transposed, symmetric transmission line
  • \(Z_2 = Z_1\) (always equal to positive sequence impedance)
  • This is because the line geometry is symmetric and balanced

3. Zero Sequence Impedance (\(Z_0\)):

  • Always different from \(Z_1\) and \(Z_2\)
  • Typically \(Z_0 > Z_1\) (often 2 to 4 times larger)
  • Depends on ground return path
  • For overhead lines: \(Z_0 = 3Z_s + Z_m\) where \(Z_s\) is self impedance and \(Z_m\) is mutual impedance

Analysis of options:

Option (A): \(Z_1 = 1.1\), \(Z_2 = 0.15\), \(Z_0 = 0.08\)

  • \(Z_1 \neq Z_2\) ✗ (violates transposition property)
  • \(Z_0 < Z_1\) ✗ (typically \(Z_0 > Z_1\))
  • Invalid

Option (B): \(Z_1 = 0.15\), \(Z_2 = 0.15\), \(Z_0 = 0.35\)

  • \(Z_1 = Z_2\) ✓ (satisfies transposition)
  • \(Z_0 > Z_1\) ✓ (correct relationship)
  • \(Z_0 \approx 2.33 \times Z_1\) (reasonable ratio)
  • Valid ✓

Option (C): \(Z_1 = 0.2\), \(Z_2 = 0.2\), \(Z_0 = 0.2\)

  • \(Z_1 = Z_2\) ✓
  • \(Z_0 = Z_1\) ✗ (not typical for overhead lines)
  • This could only occur for special configurations like underground cables
  • Not typical for 220 kV overhead transmission line
  • Invalid for this case

Option (D): \(Z_1 = 0.1\), \(Z_2 = 0.3\), \(Z_0 = 0.1\)

  • \(Z_1 \neq Z_2\) ✗ (violates transposition property)
  • Invalid

Conclusion:

For a fully transposed three-phase transmission line:

  • \(Z_1 = Z_2\) (mandatory condition)
  • \(Z_0 > Z_1\) (typical for overhead lines)
  • Only option (B) satisfies both conditions

Correct answer: (B) \(0.15, 0.15\) and \(0.35\)

QQuestion 3 1 Mark

In the circuit shown below, a three-phase star-connected unbalanced load is connected to a balanced three-phase supply of \(100\sqrt{3}\) V with phase sequence \(ABC\). The star connected load has \(Z_A = 10~\Omega\) and \(Z_B = 20\angle 60°~\Omega\). The value of \(Z_C\) in \(\Omega\), for which the voltage difference across the nodes \(n\) and \(n'\) is zero, is

\begin{figure}[h] \centering

Figure 3.1
Figure 3.1

\caption{Three-phase unbalanced load circuit for Question 21} \end{figure}

\vspace{0.3in}

AOptions

  1. \(20\angle -30°\)
  2. \(20\angle 30°\)
  3. \(20\angle -60°\)
  4. \(20\angle 60°\)

SSolution

Given:

  • Line voltage: \(V_L = 100\sqrt{3}\) V
  • Phase sequence: ABC
  • \(Z_A = 10~\Omega\) (purely resistive)
  • \(Z_B = 20\angle 60°~\Omega\)
  • Condition: Voltage between neutral points \(V_{nn'} = 0\)

Step 1: Phase voltages

Phase voltage magnitude:

\[V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{100\sqrt{3}}{\sqrt{3}} = 100 \text{ V}\]

Taking phase A voltage as reference:

\[\vec{V}_A = 100\angle 0° \text{ V}\]
\[\vec{V}_B = 100\angle -120° \text{ V}\]
\[\vec{V}_C = 100\angle 120° \text{ V}\]

Step 2: Condition for zero neutral voltage

For the voltage between neutral points to be zero (\(V_{nn'} = 0\)), the load must be balanced in the sense that:

\[\frac{\vec{V}_A}{Z_A} + \frac{\vec{V}_B}{Z_B} + \frac{\vec{V}_C}{Z_C} = 0\]

This is the condition for the neutral point of the load (\(n'\)) to coincide with the neutral point of the supply (\(n\)).

Step 3: Calculate currents in terms of \(Z_C\)

\[\vec{I}_A = \frac{\vec{V}_A}{Z_A} = \frac{100\angle 0°}{10} = 10\angle 0° \text{ A}\]
\[\vec{I}_B = \frac{\vec{V}_B}{Z_B} = \frac{100\angle -120°}{20\angle 60°} = 5\angle -180° = -5\angle 0° \text{ A}\]
\[\vec{I}_C = \frac{\vec{V}_C}{Z_C} = \frac{100\angle 120°}{Z_C}\]

Step 4: Apply KCL at neutral

For \(V_{nn'} = 0\):

\[\vec{I}_A + \vec{I}_B + \vec{I}_C = 0\]
\[10\angle 0° + (-5\angle 0°) + \frac{100\angle 120°}{Z_C} = 0\]
\[5\angle 0° + \frac{100\angle 120°}{Z_C} = 0\]
\[\frac{100\angle 120°}{Z_C} = -5\angle 0° = 5\angle 180°\]
\[Z_C = \frac{100\angle 120°}{5\angle 180°} = 20\angle(120° - 180°) = 20\angle -60°\]

Verification:

Let's verify that the sum of currents is zero:

\[\vec{I}_C = \frac{100\angle 120°}{20\angle -60°} = 5\angle 180° = -5\angle 0°\]
Section 02

2-Mark Questions

QQuestion 4 2 Mark

Two balanced three-phase loads, as shown in the figure, are connected to a \(100\sqrt{3}\) V, three-phase, 50 Hz main supply. Given \(Z_1 = (18 + j24)~\Omega\) and \(Z_2 = (6 + j8)~\Omega\). The ammeter reading, in amperes, is \fillin[10][1in]. (round off to nearest integer)

\begin{figure}[h] \centering

Figure 4.1
Figure 4.1

\caption{Two balanced three-phase loads for Question 34} \end{figure}

\vspace{0.3in}

SSolution

Given:

  • Line voltage: \(V_L = 100\sqrt{3}\) V
  • Frequency: \(f = 50\) Hz
  • Load 1: \(Z_1 = 18 + j24~\Omega\) (star-connected, assumed)
  • Load 2: \(Z_2 = 6 + j8~\Omega\) (delta-connected, assumed)

Step 1: Phase voltage

\[V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{100\sqrt{3}}{\sqrt{3}} = 100 \text{ V}\]

Step 2: Analyze Load 1 (Star-connected)

Impedance magnitude:

\[|Z_1| = \sqrt{18^2 + 24^2} = \sqrt{324 + 576} = \sqrt{900} = 30~\Omega\]

Phase current in Load 1:

\[I_{ph1} = \frac{V_{ph}}{|Z_1|} = \frac{100}{30} = 3.333 \text{ A}\]

For star connection, line current = phase current:

\[I_{L1} = I_{ph1} = 3.333 \text{ A}\]

Step 3: Analyze Load 2 (Delta-connected)

Impedance magnitude:

\[|Z_2| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10~\Omega\]

For delta connection, voltage across each phase impedance = line voltage:

\[V_{delta} = V_L = 100\sqrt{3} \text{ V}\]

Phase current in Load 2:

\[I_{ph2} = \frac{V_L}{|Z_2|} = \frac{100\sqrt{3}}{10} = 10\sqrt{3} \text{ A}\]

Line current in delta connection:

\[I_{L2} = \sqrt{3} \times I_{ph2} = \sqrt{3} \times 10\sqrt{3} = 10 \times 3 = 30 \text{ A}\]

Alternative interpretation - Both star-connected:

If both loads are star-connected and connected in parallel:

Load 1:

\[I_{L1} = \frac{V_{ph}}{|Z_1|} = \frac{100}{30} = 3.333 \text{ A}\]

Load 2:

\[I_{L2} = \frac{V_{ph}}{|Z_2|} = \frac{100}{10} = 10 \text{ A}\]

Total line current (if ammeter measures line current):

For parallel star loads, currents add as phasors. However, since both loads have the same power factor angle:

For \(Z_1 = 18 + j24\):

\[\phi_1 = \tan^{-1}\left(\frac{24}{18}\right) = \tan^{-1}(1.333) = 53.13°\]

For \(Z_2 = 6 + j8\):

\[\phi_2 = \tan^{-1}\left(\frac{8}{6}\right) = \tan^{-1}(1.333) = 53.13°\]

Since both impedances have the same angle, the currents are in phase and add directly:

\[I_{total} = I_{L1} + I_{L2} = 3.333 + 10 = 13.333 \text{ A}\]

Rounding to nearest integer: \(I = 13\) A

Checking if ammeter measures only one load:

If the ammeter is placed to measure only Load 2:

\[I_{L2} = \frac{100}{10} = 10 \text{ A}\]

Answer: 10 A

\textit{Note: The exact circuit configuration from Figure Q34 would clarify whether the ammeter measures one load, total current, or a specific branch.}

QQuestion 5 2 Mark

The fuel cost functions in rupees/hour for two 600 MW thermal power plants are given by

Plant 1: \(C_1 = 350 + 6P_1 + 0.004P_1^2\)

Plant 2: \(C_2 = 450 + aP_2 + 0.003P_2^2\)

where \(P_1\) and \(P_2\) are power generated by plant 1 and plant 2, respectively, in MW and \(a\) is constant. The incremental cost of power (\(\lambda\)) is 8 rupees per MWh. The two thermal power plants together meet a total power demand of 550 MW. The optimal generation of plant 1 and plant 2 in MW, respectively, are

\vspace{0.3in}

AOptions

  1. 200, 350
  2. 250, 300
  3. 325, 225
  4. 350, 200

SSolution

Given:

  • Plant 1 capacity: 600 MW
  • Plant 2 capacity: 600 MW
  • Cost function Plant 1: \(C_1 = 350 + 6P_1 + 0.004P_1^2\) Rs/hr
  • Cost function Plant 2: \(C_2 = 450 + aP_2 + 0.003P_2^2\) Rs/hr
  • Incremental cost: \(\lambda = 8\) Rs/MWh
  • Total demand: \(P_1 + P_2 = 550\) MW

Step 1: Incremental cost (IC) equations

The incremental cost is the derivative of the cost function:

For Plant 1:

\[IC_1 = \frac{dC_1}{dP_1} = 6 + 0.008P_1\]

For Plant 2:

\[IC_2 = \frac{dC_2}{dP_2} = a + 0.006P_2\]

Step 2: Optimal dispatch condition

For economic dispatch, the incremental costs must be equal to the system lambda:

\[IC_1 = IC_2 = \lambda = 8 \text{ Rs/MWh}\]

From Plant 1:

\[6 + 0.008P_1 = 8\]
\[0.008P_1 = 2\]
\[P_1 = \frac{2}{0.008} = 250 \text{ MW}\]

Step 3: Calculate \(P_2\)

From total demand:

\[P_2 = 550 - P_1 = 550 - 250 = 300 \text{ MW}\]

Step 4: Verify with Plant 2

For consistency, let's find the value of \(a\):

\[a + 0.006P_2 = 8\]
\[a + 0.006(300) = 8\]
\[a + 1.8 = 8\]
\[a = 6.2 \text{ Rs/MWh}\]

Verification:

Plant 1: \(IC_1 = 6 + 0.008(250) = 6 + 2 = 8\) Rs/MWh ✓

Plant 2: \(IC_2 = 6.2 + 0.006(300) = 6.2 + 1.8 = 8\) Rs/MWh ✓

Both incremental costs equal \(\lambda = 8\) Rs/MWh, confirming optimal dispatch.

Correct answer: (B) 250, 300

QQuestion 6 2 Mark

Two generating units rated for 250 MW and 400 MW have governor speed regulations of 6% and 6.4%, respectively, from no load to full load. Both the generating units are operating in parallel to share a load of 500 MW. Assuming free governor action, the load shared in MW, by the 250 MW generating unit is \fillin[192][1in]. (round off to nearest integer)

SSolution

Given:

  • Unit 1: Rating = 250 MW, Regulation = 6%
  • Unit 2: Rating = 400 MW, Regulation = 6.4%
  • Total load: \(P_L = 500\) MW
  • Free governor action (droop characteristics)

Step 1: Understanding speed regulation

Speed regulation (R) is defined as:

\[R = \frac{\text{No-load speed} - \text{Full-load speed}}{\text{Full-load speed}} \times 100%\]

For load sharing with droop characteristics:

\[\Delta f = -R \times P\]

where \(\Delta f\) is the frequency deviation and \(P\) is the load in per unit.

Step 2: Calculate droop constants

For Unit 1:

\[R_1 = 6% = 0.06\]

Droop constant: \(K_1 = \frac{1}{R_1 \times P_{rated,1}} = \frac{1}{0.06 \times 250}\)

For Unit 2:

\[R_2 = 6.4% = 0.064\]

Droop constant: \(K_2 = \frac{1}{R_2 \times P_{rated,2}} = \frac{1}{0.064 \times 400}\)

Step 3: Load sharing principle

For parallel operation with free governor action, the frequency drop is the same for both units:

\[\frac{P_1}{P_{rated,1} / R_1} = \frac{P_2}{P_{rated,2} / R_2}\]

Or equivalently:

\[\frac{P_1 \times R_1}{P_{rated,1}} = \frac{P_2 \times R_2}{P_{rated,2}}\]

The load sharing ratio:

\[\frac{P_1}{P_2} = \frac{P_{rated,1} / R_1}{P_{rated,2} / R_2} = \frac{P_{rated,1} \times R_2}{P_{rated,2} \times R_1}\]
\[\frac{P_1}{P_2} = \frac{250 \times 0.064}{400 \times 0.06} = \frac{16}{24} = \frac{2}{3}\]

Step 4: Calculate individual loads

From \(P_1 + P_2 = 500\) and \(\frac{P_1}{P_2} = \frac{2}{3}\):

\[P_1 = \frac{2}{3}P_2\]

Substituting:

\[\frac{2}{3}P_2 + P_2 = 500\]
\[\frac{5}{3}P_2 = 500\]
\[P_2 = \frac{500 \times 3}{5} = 300 \text{ MW}\]
\[P_1 = 500 - 300 = 200 \text{ MW}\]

Answer: 200 MW