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

Explanation:

Equal Area Criterion in Power-Angle Curve

Definition: The equal area criterion is a method used to analyze the stability of a synchronous machine under transient conditions. It is based on the principle that the area of acceleration (A₁) and the area of deceleration (A₂) in the power-angle curve should be equal for the system to remain stable after a disturbance. The power-angle curve represents the relationship between the electrical power output of a synchronous machine and the rotor angle (δ).

Working Principle: When a disturbance occurs, the rotor angle changes, causing a temporary imbalance between the mechanical input power (P_m) and the electrical output power (P_e). The rotor accelerates or decelerates depending on whether P_m is greater than or less than P_e. The equal area criterion ensures that the energy gained during acceleration is equal to the energy lost during deceleration, allowing the rotor to settle at a new equilibrium point.

Correct Option Analysis:

The correct option is:

Option 3: \(\rm A_{1}=\int_{\delta_{o}}^{\delta_{c}}\left(P_{m}-P_{e}\right) d \delta=0\)

This option accurately defines the area of acceleration (A₁) in the power-angle curve. The integral represents the energy gained by the rotor during acceleration, which occurs when the mechanical input power (P_m) exceeds the electrical output power (P_e). The limits of integration are δ_o (the initial rotor angle at the start of the disturbance) and δ_c (the critical rotor angle where the system transitions to deceleration). The condition for stability is that the total energy gain (A₁) during acceleration equals the total energy loss (A₂) during deceleration.

Mathematical Representation:

The energy gained during acceleration is given by:

\(A_{1} = \int_{\delta_{o}}^{\delta_{c}} (P_{m} - P_{e}) \, d\delta\)

The energy lost during deceleration is given by:

\(A_{2} = \int_{\delta_{c}}^{\delta_{max}} (P_{e} - P_{m}) \, d\delta\)

For stability:

\(A_{1} = A_{2}\)

This ensures that the rotor angle returns to a stable operating point after the disturbance.

Additional Information

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

Option 1: \(\rm A_{1}=\int_{\delta_{c}}^{\delta_{o}}\left(P_{m}-P_{e}\right) d \delta=0\)

This option incorrectly reverses the limits of integration. The area of acceleration (A₁) is always defined from the initial rotor angle (δ_o) to the critical rotor angle (δ_c), not vice versa. Reversing the limits would result in a negative value for A₁, which is physically incorrect in the context of energy gained during acceleration.

Option 2: \(\rm A_{1}=\int_{\delta_{o}}^{\delta_{c}}\left(P_{e}-P_{m}\right) d \delta=0\)

This option swaps the terms inside the integral, using (P_e - P_m) instead of (P_m - P_e). This representation corresponds to the area of deceleration (A₂), not acceleration (A₁). The area of acceleration must be defined by the difference (P_m - P_e), which represents the net power causing the rotor to accelerate.

Option 4: \(\rm A_{1}=\int_{\delta_{c}}^{\delta_{o}}\left(P_{e}-P_{m}\right) d \delta=0\)

This option combines the errors from options 1 and 2. It reverses the limits of integration and swaps the terms inside the integral. As explained earlier, the limits of integration for A₁ must be from δ_o to δ_c, and the integrand must be (P_m - P_e). This option is therefore doubly incorrect.

Conclusion:

The equal area criterion is a vital tool for analyzing the transient stability of synchronous machines. By ensuring that the areas of acceleration (A₁) and deceleration (A₂) are equal, we can determine whether the system will return to a stable operating point after a disturbance. The correct representation of A₁ is:

\(A_{1} = \int_{\delta_{o}}^{\delta_{c}} (P_{m} - P_{e}) \, d\delta\)

This integral accurately describes the energy gained during acceleration, and its equality with the energy lost during deceleration ensures stability. Understanding the limits of integration and the terms inside the integral is crucial for correctly applying the equal area criterion in power system analysis.

Concept:

The swing equation is a second-order differential equation that describes the rotor angle dynamics of a synchronous machine. It is derived from Newton’s second law applied to rotational motion. The general form of the swing equation is:

\[ \frac{H}{180f} \frac{d^2 \delta}{dt^2} = P_m - P_e \]

Where:

• H = Inertia constant (in MJ/MVA)
• f = System frequency (Hz)
• δ = Rotor angle in electrical degrees
• P_m = Mechanical input power
• P_e = Electrical output power

The equation reflects the accelerating power acting on the rotor as the difference between mechanical and electrical power.

सही उत्तर है- भयानक और घना

सही उत्तर है- पशु और पेड़-पौधे

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.

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)