[Solved] Power System Stability MCQ [Free PDF] - Objective Question Answer for Power System Stability Quiz - Download Now! (2024)

1. 6/π
2. 24/π
3. 12/π
4. 2/π
5. None of these

Concept:

Accelerating power can be calculated as

Pa= Pm– Pe

Accelerating torque is given as

\({T_a} = \frac{{{P_a}}}{{{\omega _s}\;}}\)

Where,

Pm= Mechanical input power

Pe= Electrical output power

ωs= Synchronous speed in radian/sec

P = Number of poles

F = Supply frequency

Synchronous speed in rpm can be calculated as

\({N_s} = \frac{{120f}}{P}\)

Also,\({\omega _s} = \frac{{2\pi {N_s}}}{{60}}\)

Calculation:

Given-

f = 50 Hz, P = 6

Before fault

Pm= Pe= 500 cosϕ = 1000 x 0.8

Pm= Pe= 800 MW

After fault Pmremains same, while Pereduces by 50%

So electrical output at the time of fault is

Pe’ = (1 – 0.5) = 0.5x 800

Pe’= 400 MW

Now Accelerating power at the time of fault is

Pa’ = 800 – 400 = 400 MW

Now synchronous speed is

\({N_s} = \frac{{120 \times 50}}{6} = 1000\;rpm\)

\({\omega _s} = \frac{{2\pi \times 1000}}{{60}} = 104.71\;radian/sec\)

Hence accelerating torque is

\({T_a} = \frac{{400 \times {{10}^6}}}{{104.71}} = 3.82\; \times {10^6}\;N - m ={12\over\pi}\;MN - m\)

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1. A, B, C only
2. A, B, D only
3. C and D only
4. A, B, C and D
5. A and D only

Steady-state stability limit =\(\frac{{{V_1}\;{V_2}}}{X}\)

More steady-state stability represents more transient stability of the system.

Steady-state stability limit of the system can be increased by the following methods:

• Upgrading voltage on the existing transmission system or opting for higher voltages on the new transmission system.
• The use of an additional parallel transmission line reduces transfer reactance X, thereby increases it.
• The use of a transformer with lower leakage reactance improves a steady-state stability limit.
• Series capacitive compensation of transmission line reduces transfer reactance X and hence steady-state stability limit increases.
• The use of bundled phase conductors reduces the reactance thereby increasing the stability limit.

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1. greater than\(\left| {\frac{{{V_{1{V_2}}}}}{X}} \right|\)
2. less than\(\left| {\frac{{{V_1}{V_2}}}{X}} \right|\)
3. equal to\(\left| {\frac{{{V_1}{V_2}}}{X}} \right|\)
4. equal to\(\left| {\frac{{{V_1}{V_2}}}{Z}} \right|\)
5. none of these

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 \(\left| {\frac{{{V_1}{V_2}}}{X}} \right|\).

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1. Excitation of generating system
2. Reactance of transmission line
3. Reactance of generating system
4. Losses

Concept:

Steady-state stabilityof a power system is

\({P_{max}} = \dfrac{{EV}}{X}\)

• We can increase thesteady-state stability by decreasing the reactance X.
• In adouble circuit line where two transmission lines are connected in parallel,the reactance is less than the single line circuit and hence the stability can be improved.
• The stability willnot only be changed by changing the reactance ofa transmission line but also can be changed by changing the reactance of the generator.
• The disturbance in the system occursif the speed of the rotor is other than synchronous speed and there will be a change in torque angle.
  

Hence,Power system stability is mostly affected and is dependent on the reactance of transmission line, generator, and input torque

Therefore,Power system stability is least affected by losses

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1. 480 MJ
2. 480 KJ
3. 30 MJ
4. 410 MJ

Concept:

TheInertia constant(H) of agenerator-turbine unit is defined as the ratio of kinetic energy stored at thesynchronousspeedto thegenerator'skVA or MVA rating.

Importance of Inertia:

• Inertia is the stored rotating energy in the system
• Following a System loss, the higher the System Inertia (assuming no frequency response) the longer it takes to reach a new steady-state operating frequency.
• Directly connected synchronous generators and Induction Generators will contribute directly to System Inertia.
• Modern Generator technologies such as Wind Turbines or wave and tidal generators which decouple the prime mover from the electrical generator will not necessarily contribute directly to System Inertia

We shall define the inertia constant H,

GH = KE in MJ

Where,

G = three-phase MVA rating (base) of machine

H = inertia constant in MJ/MVA or MW.s/MVA

Inertia constant,\(H = \frac{{kinetic\;energy\;stored}}{{rating}}\)

K.E. stored = H× rating

Calculation:

Given,

K.E. stored = 4× 120 = 480 MJ

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Power System Stability Question 6

1. Shunt conductance
2. Capacitance
3. Resistance
4. Inductance

Power System Stability Question 6 Detailed Solution

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 byInductance.

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)

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Power System Stability Question 7

1. Reactance of line
2. Losses
3. Reactance of generator
4. Excitation of generator

Power System Stability Question 7 Detailed Solution

• 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.

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Power System Stability Question 8

1. greater than\(\left| {\frac{{{V_{1{V_2}}}}}{X}} \right|\)
2. less than\(\left| {\frac{{{V_1}{V_2}}}{X}} \right|\)
3. equal to\(\left| {\frac{{{V_1}{V_2}}}{X}} \right|\)
4. equal to\(\left| {\frac{{{V_1}{V_2}}}{Z}} \right|\)

Power System Stability Question 8 Detailed Solution

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 \(\left| {\frac{{{V_1}{V_2}}}{X}} \right|\).

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Power System Stability Question 9

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

Power System Stability Question 9 Detailed Solution

• 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.

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Power System Stability Question 10

1. angle between stator voltage and current.
2. angular displacement of the rotor with respect to the stator
3. angular displacement of the stator mmf with respect to a synchronously rotating axis.
4. angular displacement of an axis fixed to the rotor with respect to a synchronously rotating axis.

Power System Stability Question 10 Detailed Solution

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.

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Power System Stability Question 11

1. 60°/s²
2. 65°/s²
3. 70°/s²
4. 75°/s²

Power System Stability Question 11 Detailed Solution

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

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

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

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

= 75°/s²

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Power System Stability Question 12

1. 100 MJ
2. 400 MJ
3. 800 MJ
4. 12.5 MJ

Power System Stability Question 12 Detailed Solution

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

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Power System Stability Question 13

1. Red - Purple
2. Black - Green
3. Blue - Black
4. Green-Yellow

Power System Stability Question 13 Detailed Solution

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 protectsappliances from electrical fires

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Power System Stability Question 14

1. \(P = \frac{{EV}}{X}\)
2. \(P = \frac{{EV}}{X}sin\delta\)
3. \(P = \frac{{EV}}{X}cos\delta\)
4. \(P = \frac{{EV}}{X}sin30 {°} \)

Power System Stability Question 14 Detailed Solution

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 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)

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Power System Stability Question 15

Power System Stability Question 15 Detailed Solution

Analysis:

In the given line diagram

Line 1 is tripped at time t1by opening the circuit breakers at the two ends(although there is no fault)

As the line is disconnected the reactance of the system increases

The maximum power transfer increases (as power is inversely proportional to the reactance of the line\(P_{max}=\frac{EV}{X}\))

It is also given thatmechanical power input to the generator and the voltage magnitude E is constant.

Therefore the new electrical power is to be the same as the old power, soδ value should be increased (given in the line reactance diagram the steady-state value ofδbefore the transientis positive)

The rotor angle transient will be

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