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

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Concept

The maximum power that can be transferred through a power system is:

\(P_m={EV\over X}\)

where, E = Induced EMF in an alternator

V = Infinite bus voltage

X = Total reactance

Calculation

Given, E = 1.6 pu

V = 1 pu

X = 0.6 + 0.2 pu

\(P_m={1.6\times 1\over 0.8}\)

P_m = 2 pu

Synchronous Machine Dynamics During Disturbances:

In synchronous machines, the rotor and stator magnetic fields lock together to maintain a constant speed called synchronous speed. During disturbances such as sudden changes in load, the rotor can experience oscillations before settling back to a steady state. These oscillations are characterized by changes in rotor angle and speed.

Explanation of the Concept:

• The power angle curve represents the relationship between the electrical power output of the synchronous machine and the rotor angle. Points A, B, and C on this curve indicate different states during the machine's dynamic response to a disturbance.
• Point A: The initial operating point before the disturbance.
• Point B: The maximum rotor angle deviation during oscillation. At this point, the machine's speed is not synchronous due to the inertia of the rotor, leading to acceleration or deceleration.
• Point C: The final steady-state operating point after the oscillation settles.

Detailed Solution:

• During the oscillation, the rotor speed will vary above and below the synchronous speed. The rotor accelerates when the electrical power is greater than the mechanical power and decelerates when the electrical power is less than the mechanical power.
• At points A and C, the machine is in a steady state, meaning the rotor speed is synchronous. This is because there are no unbalanced forces acting on the rotor, and it is locked with the stator magnetic field.
• At point B, the rotor speed is not synchronous as it represents the maximum deviation from the steady state due to the disturbance. The rotor will have either accelerated or decelerated during its swing from A to B.

Conclusion:

• The correct answer is option 3: A and B. At points A and C, the rotor speed is synchronous, but at point B, it is not.

Explanation:

Reactive Power in Power Systems

Definition: Reactive power is the component of electrical power that oscillates between the source and the load. Unlike active power, which performs useful work and is measured in watts (W), reactive power does not perform any real work and is measured in volt-amperes reactive (VAR). It is primarily associated with the energy storage in the inductive and capacitive elements of the power system.

Understanding Reactive Power:

In an AC power system, the voltage and current waveforms are sinusoidal but may not be in phase. When there is a phase difference between voltage and current, the power system has both active and reactive power components. The active power (P) is responsible for performing useful work, such as turning a motor or lighting a bulb. The reactive power (Q) is associated with the energy storage in the electric and magnetic fields of capacitors and inductors, respectively.

Reactive power is crucial for maintaining the voltage levels across the power system. Proper voltage levels ensure efficient and reliable operation of electrical equipment. Reactive power, however, does not result in actual work being done but is essential for the magnetization of inductive loads and maintaining electric fields in capacitive loads.

Correct Option Analysis:

The correct option is:

Option 3: Inductive loads

Reactive power is primarily due to inductive loads in a power system. Inductive loads, such as motors, transformers, and inductors, require reactive power to create magnetic fields necessary for their operation. When current flows through an inductive load, it lags behind the voltage, causing a phase difference. This phase difference results in the generation of reactive power.

Inductive loads consume reactive power to establish and maintain their magnetic fields. This consumption of reactive power does not result in actual work but is necessary for the proper functioning of inductive components. Without reactive power, inductive loads cannot maintain their magnetic fields, leading to inefficient operation and potential equipment failure.

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

Option 1: Capacitive loads

While capacitive loads also generate reactive power, they are not the primary source in most power systems. Capacitive loads, such as capacitors, store energy in the form of electric fields. The current through a capacitive load leads the voltage, creating a phase difference that results in the generation of reactive power. However, in typical power systems, inductive loads are more prevalent and significant in generating reactive power compared to capacitive loads.

Option 2: Purely active loads

Purely active loads consume only active power and do not generate reactive power. These loads, such as resistive heaters and incandescent lamps, convert electrical energy directly into useful work or heat without creating any phase difference between voltage and current. As a result, purely active loads do not contribute to reactive power in a power system.

Option 4: Resistive loads

Resistive loads, similar to purely active loads, consume only active power. They convert electrical energy into heat or work without causing any phase difference between voltage and current. Therefore, resistive loads do not generate reactive power and are not a primary source of reactive power in a power system.

Conclusion:

In conclusion, reactive power is primarily due to inductive loads in a power system. Inductive loads require reactive power to establish and maintain their magnetic fields, which are essential for their operation. While capacitive loads also generate reactive power, they are not the primary source in most power systems. Understanding the role of reactive power and its sources is crucial for maintaining the efficiency and reliability of electrical power systems. Proper management of reactive power ensures stable voltage levels and efficient operation of electrical equipment, ultimately contributing to the overall performance of the power system.

Additional Information

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

Option 1: Capacitive loads

While capacitive loads also generate reactive power, they are not the primary source in most power systems. Capacitive loads, such as capacitors, store energy in the form of electric fields. The current through a capacitive load leads the voltage, creating a phase difference that results in the generation of reactive power. However, in typical power systems, inductive loads are more prevalent and significant in generating reactive power compared to capacitive loads.

Option 2: Purely active loads

Purely active loads consume only active power and do not generate reactive power. These loads, such as resistive heaters and incandescent lamps, convert electrical energy directly into useful work or heat without creating any phase difference between voltage and current. As a result, purely active loads do not contribute to reactive power in a power system.

Option 4: Resistive loads

Resistive loads, similar to purely active loads, consume only active power. They convert electrical energy into heat or work without causing any phase difference between voltage and current. Therefore, resistive loads do not generate reactive power and are not a primary source of reactive power in a power system.

Conclusion:

In conclusion, reactive power is primarily due to inductive loads in a power system. Inductive loads require reactive power to establish and maintain their magnetic fields, which are essential for their operation. While capacitive loads also generate reactive power, they are not the primary source in most power systems. Understanding the role of reactive power and its sources is crucial for maintaining the efficiency and reliability of electrical power systems. Proper management of reactive power ensures stable voltage levels and efficient operation of electrical equipment, ultimately contributing to the overall performance of the power system.

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)