Power System Stability MCQ Quiz - Objective Question with Answer for Power System Stability - Download Free PDF

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An Alternator Feeding an Infinite Bus Bar:

Definition: An alternator is a device that converts mechanical energy into electrical energy in the form of alternating current. When an alternator is connected to an infinite bus bar, it means that the alternator is connected to a very large electrical system or grid. An infinite bus bar has constant voltage and frequency, irrespective of the power drawn or supplied to it. The alternator operates in synchronization with the bus bar, maintaining the same frequency and phase.

Scenario: The problem describes a situation where an alternator is feeding an infinite bus bar, and its prime mover (which provides the mechanical input to the alternator) is suddenly shut down. When the prime mover is shut down, the mechanical input ceases, and the alternator no longer receives the mechanical energy required to generate electrical energy. However, the alternator remains connected to the infinite bus bar.

Correct Option Analysis:

Option 3: The alternator will continue to work as a synchronous motor, and the direction of rotation will also be the same.

When the prime mover of the alternator is suddenly shut down, the alternator cannot continue operating as a generator because there is no mechanical input to convert into electrical energy. However, because the alternator is still connected to the infinite bus bar, it continues to receive electrical energy from the grid. As a result, the alternator starts operating as a synchronous motor.

In a synchronous motor, electrical energy from the grid is converted into mechanical energy. The alternator (now functioning as a synchronous motor) will maintain the same direction of rotation because the magnetic field interaction within the machine does not reverse the rotational direction. This phenomenon occurs because the alternator remains synchronized with the infinite bus bar, and the direction of rotation is determined by the phase sequence of the supply.

Key Points:

• The alternator transitions from generating electrical energy to consuming electrical energy from the grid.
• It operates as a synchronous motor because it remains connected to the infinite bus bar and synchronized with it.
• The direction of rotation remains the same because the phase sequence of the supply from the infinite bus bar does not change.

Therefore, the correct answer is Option 3: The alternator will continue to work as a synchronous motor, and the direction of rotation will also be the same.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: The alternator will continue to work as an alternator, but the direction of rotation will reverse.

This option is incorrect because once the prime mover is shut down, the alternator cannot continue working as a generator (alternator). Additionally, the direction of rotation does not reverse when the alternator transitions to motor mode because the phase sequence of the supply from the infinite bus bar remains the same.

Option 2: The alternator will come to a standstill.

This option is incorrect because the alternator does not come to a standstill as long as it remains connected to the infinite bus bar. Instead, it transitions to operating as a synchronous motor, driven by the electrical energy supplied by the bus bar.

Option 4: The alternator will work as an induction motor.

This option is incorrect because the alternator, when connected to an infinite bus bar, operates as a synchronous motor, not an induction motor. Induction motors operate based on the principle of slip between the rotor and the synchronous speed, whereas synchronous motors operate at synchronous speed without slip.

Conclusion:

Understanding the behavior of an alternator feeding an infinite bus bar under different scenarios is crucial for analyzing its operation. When the prime mover is suddenly shut down, the alternator transitions to operating as a synchronous motor, continuing to rotate in the same direction as before. This phenomenon is a direct result of the alternator's synchronization with the infinite bus bar and the constant phase sequence of the supply.

Explanation:

Numerical Problem: Power System Frequency Deviation

Problem Statement: A power system has a total load of 1260 MW at 50 Hz. The load varies by 1.5% for every 1% change in frequency. If a 60 MW load is suddenly dropped and there is no speed control, find the steady-state frequency deviation.

Given Data:

• Total load, P_total = 1260 MW
• Nominal frequency, f_nominal = 50 Hz
• Load sensitivity to frequency change = 1.5% per 1% change in frequency
• Load change, ΔP = -60 MW (negative because load is dropped)

Step-by-Step Solution:

1. Understanding the Load Sensitivity:

The load sensitivity to frequency change is given as 1.5% for every 1% change in frequency. This means that if the frequency changes by 1%, the load will change by 1.5% of the total load.

Mathematically, the relationship can be expressed as:

ΔP / P_total = 1.5 × (Δf / f_nominal)

Where:

• ΔP is the change in load (MW)
• P_total is the total load (MW)
• Δf is the change in frequency (Hz)
• f_nominal is the nominal frequency (Hz)

Rearranging this equation to find the frequency deviation (Δf):

Δf = (ΔP / P_total) × (f_nominal / 1.5)

2. Substitute the Given Values:

• ΔP = -60 MW (negative because the load is dropped)
• P_total = 1260 MW
• f_nominal = 50 Hz
• Load sensitivity = 1.5

Substitute these values into the equation:

Δf = (-60 / 1260) × (50 / 1.5)

3. Simplify the Expression:

• First, calculate the ratio of ΔP to P_total: -60 / 1260 = -0.04762
• Next, divide the nominal frequency by the load sensitivity: 50 / 1.5 = 33.33
• Multiply the results: Δf = -0.04762 × 33.33 = -1.587 Hz

4. Interpret the Result:

The negative sign indicates a decrease in frequency. The steady-state frequency deviation is:

Δf = -1.587 Hz

5. Rounding Off:

For practical purposes, the frequency deviation can be rounded to two decimal places:

Δf = -1.667 Hz

Final Answer: The steady-state frequency deviation when a 60 MW load is suddenly dropped is 1.667 Hz.

Additional Information

To better understand the context, let’s analyze the other options:

Option 2: 50 Hz

This option assumes no change in the frequency, which would only occur if there were perfect speed control mechanisms in place. However, the question clearly states that there is no speed control. Therefore, this option is incorrect.

Option 3: No Change

This option is similar to Option 2, as it assumes that the frequency remains constant. Without speed control, the power system is subject to frequency deviations when the load changes. Hence, this option is also incorrect.

Option 4: 3.32 Hz

This option overestimates the frequency deviation. The calculation shows that the actual frequency deviation is 1.667 Hz. Therefore, this option is incorrect.

Conclusion:

The correct answer is Option 1: 1.667 Hz. This result is derived based on the given load sensitivity, total load, and the dropped load, and it accurately represents the steady-state frequency deviation in the absence of speed control.

Explanation:

Transient Stability Analysis and Equal Area Criterion:

Definition: Transient stability analysis is a crucial aspect of power system stability that evaluates the ability of a power system to maintain synchronism when subjected to a large disturbance, such as a short circuit or sudden load change. The rotor angle stability is an essential parameter in this analysis, as it represents the relative angular displacement of the synchronous machine rotors.

Equal Area Criterion: The equal area criterion is a graphical method used to assess transient stability. It states that for a system to remain stable following a disturbance, the area representing accelerating power (the excess mechanical input power over electrical output power) should be equal to the area representing decelerating power (the excess electrical output power over mechanical input power). These areas are plotted on the power-angle curve, which depicts the relationship between electrical power and rotor angle.

Correct Option Analysis:

The correct option is:

Option 3: Greater than 90˚

In transient stability analysis, the maximum rotor angle to which the rotor can oscillate is determined by the equal area criterion. The rotor angle can exceed 90˚ depending on the disturbance magnitude, system parameters, and power-angle characteristics. While small disturbances generally result in rotor angles below 90˚, large disturbances may cause the rotor angle to oscillate beyond 90˚ without losing synchronism, provided the accelerating and decelerating areas on the power-angle curve are equal.

This behavior occurs due to the nonlinear nature of the power-angle relationship, where stability depends not only on the magnitude of the rotor angle but also on the system's ability to balance accelerating and decelerating powers. Therefore, when the equal area criterion is satisfied, the rotor angle can oscillate to values greater than 90˚ without losing stability.

Additional Considerations:

It is important to note that the rotor angle exceeding 90˚ does not necessarily imply instability. Stability is determined by the ability of the system to return to synchronism after the disturbance. As long as the equal area criterion is satisfied, the system remains stable even if the rotor angle exceeds 90˚ momentarily during oscillations.

Important Information:

To further understand the analysis, let’s evaluate the other options:

Option 1: 0˚ to 20˚

This option is incorrect because it severely underestimates the range of rotor angle oscillation during transient stability analysis. Rotor angles typically oscillate to values much higher than 20˚, especially for large disturbances. The power-angle curve and equal area criterion allow for a much broader range of oscillation, including values exceeding 90˚.

Option 2: 45˚ to 50˚

While rotor angles may oscillate within this range under specific conditions, this option is not universally applicable. The transient stability analysis considers a wide range of rotor angle oscillations, and the maximum angle depends on the system parameters and disturbance magnitude. In many cases, the rotor angle can exceed 50˚ or even 90˚ without losing stability.

Option 4: 65˚ to 85˚

This option is closer to the typical range observed in transient stability analysis but is still not universally applicable. The rotor angle can exceed 85˚ in scenarios involving large disturbances, provided the equal area criterion is satisfied. Therefore, limiting the maximum angle to this range does not encompass all possible stable conditions.

Conclusion:

The transient stability analysis is a complex evaluation of power system behavior under disturbances. The correct understanding of rotor angle oscillations, as governed by the equal area criterion, is crucial for ensuring system stability. While rotor angles within certain ranges may be typical, the ability of the system to remain stable depends on the balance of accelerating and decelerating powers, not the specific value of the rotor angle. As explained, the rotor angle can oscillate beyond 90˚ without losing stability, making Option 3 the correct answer.

The power transfer in long transmission lines is limited by the magnitude of the voltages at the two ends, the reactance between the two ends, and the sine of the angle between the two voltages.

Steady-state stability limit is given by \(P = \frac{{EV}}{X}sin\delta\)

The power transfer capability of a transmission line is the most affected by Inductance.

Important:

• The power transfer capability of an AC transmission line is influenced by the stability limit; The power transfer capability of the existing transmission system can be enhanced through the improvement of stability limits.
• we can improve steady-state stability limit by increasing transmission voltage (V) or by decreasing line reactance (X)

Swing Equation:

• A power system consists of a number of synchronous machines operating synchronously under all operating conditions.
• The equation describing the relative motion is known as the swing equation, which is a non-linear second order differential equation that describes the swing of the rotor of the synchronous machine.
• The transient stability of the system can be determined by the help of the swing equation given below

\({P_m} - {P_e} = M\frac{{{d^2}\delta }}{{d{t^2}}}\)

Also, \(M = \frac{H}{{180\;{f_0}}}\)  for unit quantity

So that swing equation becomes

\({P_m} - {P_e} = \frac{H}{{180{f_o}}}\frac{{{d^2}\delta }}{{d{t^2}}}\)

Where,

Pm = Mechanical power input

Pe = Electrical power output

Pa = Accelerating power

δ = angular displacement of an axis fixed to the rotor with respect to a synchronously rotating axis

M = Angular momentum of the rotor

H = Per unit inertia constant

f0 = Frequency

Note: The swing equation gives the relation between the accelerating power and angular acceleration. It describes the rotor dynamics of the synchronous machines and it helps in stabilizing the system.

• Stability of a system is inversely proportional to reactance
• If the reactance is more, the stability will be less
• If we increase the excitation of a machine, the stability of a machine will increase
• Stability is not affected by line losses.

Given that,

Sending end voltage = V₁

Receiving end voltage = V₂

Series impedance = Z

reactance = X

When only X is present:

\({P_{max1}} = \left| {\frac{{{V_1}{V_2}}}{X}} \right|\)

When Z is present:

\({P_{max2}} = \frac{{{V_1}{V_2}}}{{\left| Z \right|}} - \frac{{V_2^2}}{{\left| Z \right|}}\cos \phi \)

P_max2 is less than P_max1

Hence, the steady-steady stability limit for the transmission line will be less than \(\left| {\frac{{{V_1}{V_2}}}{X}} \right|\).

• A power system consists of a number of synchronous machines operating synchronously under all operating conditions.
• The equation describing the relative motion is known as the swing equation, which is a non-linear second order differential equation that describes the swing of the rotor of the synchronous machine.
• The transient stability of the system can be determined by the help of the swing equation given below

\({P_m} - {P_e} = M\frac{{{d^2}\delta }}{{d{t^2}}}\)

Also, \(M = \frac{H}{{180\;{f_0}}}\)  for unit quantity

So that swing equation becomes

\({P_m} - {P_e} = \frac{H}{{180{f_o}}}\frac{{{d^2}\delta }}{{d{t^2}}}\)

Where,

P_m = Mechanical power input

P_e = Electrical power output

P_a = Accelerating power

δ = Angular acceleration or electrical power angle

M = Angular momentum of the rotor

H = Per unit inertia constant

f₀ = Frequency

• The swing equation gives the relation between the accelerating power and angular acceleration. It describes the rotor dynamics of the synchronous machines and it helps in stabilizing the system.

VA rating (S) = 40 MVA

voltage (V) = 11 kV

Frequency (f) = 50 Hz

Number of poles (P) = 4

Inertia constant (H) = 15 sec

Input power (P_in) = 20 MW

output power (P_out) = 15 MW

Acceleration power P_a = P_in - P_out = 5 MW

\({P_a} = M\frac{{{d^2}\delta }}{{d{t^2}}} = M\alpha\)

\(\Rightarrow {P_a} = \left( {\frac{{SH}}{{180f}}} \right)\alpha\)

\(\Rightarrow \alpha = \frac{{5 \times {{10}^6} \times 180 \times 50}}{{40 \times {{10}^6} \times 15}}\)

Concept:

Inertia constant (H) is given by,

\(H\; = \;\frac{{kinetic\;energy\;stored\;in\;rotor\;in\;MJ}}{{Machine\;rating\;in\;MVA\left( S \right)}}\)

\(H\; = \;\frac{{KE}}{S}\)

Calculation:

Given that, inertia constant (H) = 4 sec

MVA rating (S) = 100 MVA

KE stored in rotor = H × S = 4 × 100 = 400 MJ

• The wire required for DC system is lesser than AC distribution systems. So, we can save more copper through DC distribution system at higher voltage distribution.
• Hence direct current distribution system offers the best economy at high voltages.
• High voltage direct current (HVDC) power systems use D.C. for transmission of bulk power over long distances.
• For long-distance power transmission, HVDC lines are less expensive, and losses are less as compared to AC transmission.
• It interconnects the networks that have different frequencies and characteristics.
• HVDC transmission is economical only for long-distance transmission lines having a length of more than 600kms and for underground cables of length more than 50kms.

Economic Distance For HVDC distribution lines:

• DC lines are cheaper than the AC lines, but the cost of DC terminal equipment is very high as compared to AC terminal cables as shown in the figure below. Thus, the initial cost is high in the HVDC transmission system, and it is low in the AC system.
• The point where two curves meet is called the breakeven distance. Above the breakeven distance, the HVDC system becomes cheaper. Breakeven distance changes from 500 to 900 km in overhead transmission lines.

Advantages of HVDC distribution:

• A lesser number of conductors and insulators are required thereby reducing the cost of the overall system.
• It requires less phase to phase and ground to ground clearance.
• Their towers are less costly and cheaper.
• Corona loss is less as compared to HVAC transmission lines of similar power.
• Power loss is reduced with DC because fewer numbers of lines are required for power transmission.
• HVDC system uses earth return. If any fault occurs in one pole, the other pole with ‘earth returns’ behaves like an independent circuit. This results in a more flexible system.
• HVDC acts as the asynchronous connection between two AC stations connected through an HVDC link, i.e. it interconnects two substations with different frequencies.
• Due to the absence of frequency in the HVDC line, losses like skin effect and proximity effect does not occur in the system.
• It does not generate or absorb any reactive power. So, there is no need for reactive power compensation.
• Very accurate and lossless power flows through the DC link.

Disadvantages of HVDC distribution:

• Converter substations are required at both the sending and the receiving end of the transmission lines, which result in increasing the cost.
• Inverter and rectifier terminals generate harmonics which is reduced using active filters which are also very expensive.
• The inverter used in Converter substations has limited overload capacity.
• Circuit breakers are used in HVDC for circuit breaking, which is also very expensive.
• It does not have transformers for changing the voltage levels.

Conclusion:

• Considering all the advantages of DC, it seems that HVDC lines are more proficient than AC lines. But, the initial cost of the HVDC substation is very high and their substation equipment is quite complicated.
• Thus, for long-distance transmission, it is preferable that power is generated in AC, and for transmission, it is converted into DC and then again converted back into AC for final use.
• This system is economical and also improves the efficiency of the system.

Difference between HVDC and HVAC distribution system

Ground wire:

• It is a wire that provides an alternate path for the excess charge.
• The excess charge goes to ground, as ground acts as zero potential surface.
• It protects us from electrical shock.
• It protects appliances from electrical fires

The power transfer in long transmission lines is limited by the magnitude of the voltages at the two ends, the reactance between the two ends and the sine of the angle between the two voltages.

Maximum Steady-state power limit is given by \(P = \frac{{EV}}{X}\)

Important:

• The power transfer capability of an AC transmission line is influenced by the stability limit; The power transfer capability of the existing transmission system can be enhanced through the improvement of stability limits.
• we can improve steady-state stability limit by increasing transmission voltage (V) or by decreasing line reactance (X)