Synchronous Reactance MCQ Quiz - Objective Question with Answer for Synchronous Reactance - Download Free PDF

Explanation:

Synchronous Reactance in Synchronous Machines

Definition: In the equivalent circuit model of a synchronous machine, the synchronous reactance (denoted by X_s) is the combined reactance that accounts for the effects of both the leakage reactance (X₁) and the magnetizing reactance (X_a). It is a crucial parameter in the analysis and operation of synchronous machines, such as synchronous generators and motors.

Formula: The synchronous reactance is expressed as:

X_s = X₁ + X_a

Here:

• X₁: Leakage reactance, which represents the reactance due to leakage flux that does not contribute to the main magnetic field.
• X_a: Magnetizing reactance, which represents the reactance due to the main magnetic flux linking the stator and rotor.

Working Principle:

The equivalent circuit of a synchronous machine includes the synchronous reactance as a single reactance to simplify the analysis. When an alternating current flows through the stator winding of the synchronous machine, part of the flux produced by the current leaks out and does not link with the rotor. This is accounted for by the leakage reactance (X₁). The remaining flux links with the rotor, and its effect is represented by the magnetizing reactance (X_a). Together, these reactances form the synchronous reactance (X_s).

Importance of Synchronous Reactance:

• Synchronous reactance plays a significant role in determining the voltage regulation and power factor of synchronous machines.
• It impacts the steady-state performance of the machine, particularly under load conditions.
• Higher synchronous reactance leads to reduced fault current levels, which is advantageous for system protection.

Correct Option Analysis:

The correct option is:

Option 1: X_s = X₁ + X_a

This option is correct because it accurately represents the synchronous reactance as the sum of the leakage reactance (X₁) and the magnetizing reactance (X_a). The addition of these two reactances provides a complete representation of the reactance in the equivalent circuit of the synchronous machine. This formula is derived from the fact that both leakage and magnetizing reactances are connected in series within the machine's equivalent circuit.

Additional Information

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

Option 2: X_s = X₁ × X_a

This option is incorrect because it implies that the synchronous reactance is the product of the leakage reactance and the magnetizing reactance. In the equivalent circuit model, the reactances are in series, and their effects are additive, not multiplicative. Therefore, this representation does not hold true for synchronous machines.

Option 3: X_s = X₁ - X_a

This option is also incorrect because it suggests that the synchronous reactance is the difference between the leakage reactance and the magnetizing reactance. However, the synchronous reactance is a sum of these two reactances, as they both contribute positively to the total reactance in the equivalent circuit.

Option 4: X_s = X₁ ÷ X_a

This option is incorrect because it implies that the synchronous reactance is the division of the leakage reactance by the magnetizing reactance. This does not represent the physical or mathematical relationship between these reactances in the equivalent circuit of the synchronous machine.

Conclusion:

Understanding the concept of synchronous reactance is vital for analyzing the performance of synchronous machines. The synchronous reactance (X_s) is the sum of the leakage reactance (X₁) and the magnetizing reactance (X_a). This additive relationship is fundamental to the equivalent circuit representation and helps in understanding the operation of synchronous machines under various load conditions. Incorrect options, such as multiplication, subtraction, or division of reactances, do not accurately describe the concept of synchronous reactance and are not applicable in this context.

Problem Statement: A reactor having an inductive reactance of 4 ohms is connected to the terminals of a 120 V AC generator. We need to calculate the power associated with the reactor.

Solution:

When dealing with an inductive load connected to an AC source, the power associated with the reactor refers to the reactive power (denoted as Q). Reactive power is measured in kVAR (kilovolt-ampere reactive) and is calculated using the formula:

Q = V² / X_L

Where:

• Q = Reactive power (in VAR or kVAR)
• V = Voltage across the reactor (in volts)
• X_L = Inductive reactance of the reactor (in ohms)

Step-by-Step Calculation:

Q = V² / X_L

Q = (120 × 120) / 4

Q = 14400 / 4

Q = 3600 VAR

Q = 3600 / 1000 = 3.6 kVAR

1. Given Data:
   • Voltage across the reactor, V = 120 V
   • Inductive reactance of the reactor, X_L = 4 ohms


2. Substitute the values into the formula:


3. Perform the calculation:


4. Convert VAR to kVAR:

Final Answer:

The reactive power associated with the reactor is 3.6 kVAR.

Correct Option: Option 1

Important Information

To further analyze the other options:

• Option 2 (30 kVAR): This value is incorrect and does not align with the formula for calculating reactive power. A value of 30 kVAR would require much higher voltage or a much lower reactance.
• Option 3 (7.2 kVAR): This value is also incorrect. For the given voltage (120 V) and reactance (4 ohms), the calculated reactive power is 3.6 kVAR, as shown above. The value of 7.2 kVAR is double the correct value, which could result from a misunderstanding of the problem or incorrect input of data.
• Option 4 (4.16 kVAR): This value is incorrect as well. The calculated value from the given data is 3.6 kVAR, not 4.16 kVAR. It does not match the mathematical derivation using the provided formula.

Conclusion:

The correct answer is Option 1, as the reactive power associated with the reactor, calculated using the formula Q = V² / X_L, is 3.6 kVAR. The other options do not match the results obtained through proper calculations.

Explanation:

Calculation of Maximum Power in a Synchronous Generator:

Problem Statement: A 30 MVA, 15 kV, 1500 rpm, 3-phase synchronous generator connected to a power grid has a synchronous reactance of 9 Ω per phase. The exciting voltage (E_f) is 12 kV (line-to-neutral), and the system voltage (V) is 17.3 kV (line-to-line). We are required to calculate the maximum power that the generator can deliver before it loses synchronism.

Solution:

The maximum power output of a synchronous generator can be determined using the following formula:

P_max = (3 × E_f × V) / X_s

Where:

• P_max = Maximum power output of the generator (in MW)
• E_f = Excitation voltage (line-to-neutral) in kV
• V = System voltage (line-to-neutral) in kV
• X_s = Synchronous reactance per phase (in Ω)

Step 1: Convert the system voltage to line-to-neutral:

The given system voltage is 17.3 kV (line-to-line). To convert it to line-to-neutral, divide by √3:

V_{line-to-neutral} = V_line-to-line / √3 = 17.3 / √3 = 10 kV

Step 2: Substitute the values into the formula:

We now substitute the given data into the power formula:

P_max = (3 × E_f × V) / X_s

Substitute:

• E_f = 12 kV (line-to-neutral)
• V = 10 kV (line-to-neutral, calculated above)
• X_s = 9 Ω

Thus:

P_max = (3 × 12 × 10) / 9

Perform the calculations:

P_max = (360) / 9 = 40 MW (approximately 39.952 MW)

Step 3: Final Answer:

The maximum power output of the generator before it loses synchronism is 39.952 MW.

Important Information:

Let us analyze why the other options are incorrect:

• Option 2: 78.349 MW – This value is much higher than the calculated maximum power. It could arise from incorrectly substituting or misinterpreting the given values in the formula.
• Option 3: 16.67 MW – This value is significantly lower than the calculated maximum power. It might result from using an incorrect voltage or reactance value in the formula.
• Option 4: 112.89 MW – This value is unrealistic and far exceeds the calculated maximum power. Such an error could result from ignoring the synchronous reactance or misrepresenting the excitation voltage.

Conclusion:

The correct answer is Option 1: 39.952 MW. The calculation demonstrates that the maximum power output of the generator before it loses synchronism is approximately 39.952 MW. This result aligns with the standard formula for maximum power in a synchronous generator, taking into account the excitation voltage, system voltage, and synchronous reactance.

Synchronizing Torque of an Alternator

Synchronizing torque (T_sy) is the torque that acts on a synchronous machine (alternator or motor) when there is a small deviation in its rotor angle (δ) from the synchronously rotating magnetic field. This torque helps the rotor return to its steady-state position, maintaining synchronization with the infinite busbar.

The synchronizing torque of an alternator connected to an infinite busbar is given by:

\(T_{s y}=\frac{28.64 α E^{2}}{X_{s} N_{s}} \mathrm{Nm}\)

where, E =  Generated EMF

X_s = Synchronous reactance

α = Slot angle

N_s = Synchronous speed

This equation indicates that:

• Higher EMF (E) leads to a higher synchronizing torque, making synchronization stronger.
• Higher synchronous reactance (X_S) reduces synchronizing torque, making the system more sensitive to disturbances.
• Higher synchronous speed (N_S) reduces synchronizing torque, meaning larger machines (higher N_S) might have weaker synchronization per unit power.

Types of rotor used in alternator

Based on the requirement of speed and application, rotor is classified into following types:

Concept:

Open Circuit Test:

• This test is used for determining the synchronous impedance.
• The alternator is running at the rated synchronous speed, and the load terminals are kept open.
• The excitation current may be increased to get 25% more than the rated voltage i.e. up to 125% of the rated voltage.
• A graph is drawn between the open circuit phase voltage Eg and the field current If. The curve so obtained is called Open Circuit Characteristic (O.C.C).
• The linear portion of the O.C.C is extended to form an air gap line.

The Open Circuit Characteristic (O.C.C) and the air gap line is shown in the figure below

Short circuit test:

• The machine is driven at rated synchronous speed and armature terminals are short-circuited through an ammeter.
• Now, field current gradually increased from zero, until the short-circuit armature current reached its safe maximum value, equal to about 125 to 150% of the rated current.
• Latter readings may be taken in a short time to avoid armature overheating.
• A graph is drawn between short-circuit current Isc and field current if as shown in the figure.

Calculation:

Rated (kVA) = 720 kVA

Base V = 240√3 V

Base impedance \(= \frac{{{{\left( { 240√3} \right)}^2}}}{{720000}} = 0.24\;{\rm{\Omega }}\)

I_{f (O.C.)} = 20 A (field current to get rated O.C. Voltage)

If (S.C.) = 10 A (field current to get rated S.C. Current)

\({V_{OC}} = \frac{{240√3}}{{√ 3 }} = 240 V\)

I_rated =\( {kVA_{rated} \over\sqrt3 V_{L-L} } =\) \({720000 \over {\sqrt3 \times 240\sqrt3}}\) = 1 kA

ISC =\({I_{f(O.C.)} \over I_{f(S.C.)}} \times I_{rated} =\) \({20\over 10}\times{1000}= 2 kA\)

\({X_{s\left( {sat} \right)}} = {V_{OC}\over I_{SC}}= \frac{{240}}{{2000}} = 0.12\;{\rm{\Omega }}\)  (Actual value in ohms)

\({X_{s\left( {sat} \right)}}\left( {pu} \right) ={{X_{s\left( {sat} \right)}}\left( {actual} \right)\over{X_{s\left( {sat} \right)}}\left( {base} \right)}= \frac{{0.12}}{{0.24}}\)

= 0.5 pu

Causes for voltage drop:

(i) drop due to armature resistance i.e. I_aR_a drop

(ii) drop due to armature leakage resistance i.e. I_a X_ℓ drop

(iii) drop due to armature reaction i.e. I_aX_a drop.

Now, ‘I_aX_ℓ’ will be the phase with ‘I_aX_a’

X_ℓ + X_a = X_s

R_a + jX_s = Z_s

In synchronous machine irrespective of its rating always, X_s > X_a > X_ℓ > R_a

We know, Z_s = R_a + j(X_a + X_ℓ)

Since X_s = X_a + X_ℓ

⇒ Z_s = R_s + j(X_s)

Explanation:
The unsaturated synchronous reactance (X_m) in the alternator is equal to the ratio of length of "ad" and the ab.
In general for, it is equal to the ratio of voltage E_ad and current I_ab.

Xsu = Ead / Iab

Slip test in synchronous motor:

• The slip test is a simple no-load test, which is used to determine the direct-axis and quadrature-axis synchronous reactances of a salient-pole synchronous machine.
• The impedance during the slip test is low so the voltage drop will be less and the current will be more.

Circuit Arrangement for Slip Test:

• In this test, a small voltage at a rated frequency is applied to the 3-phase stator winding of the synchronous machine. In this test, a small voltage at a rated frequency is applied to the 3-phase stator winding of the synchronous machine.
• The field winding is unexcited and left open circuited.
• The rotor is driven by an auxiliary motor at a speed slightly less than or more than the synchronous speed.
• The direction of rotation should be the same as that of the rotating magnetic field produced by the stator.
• The rotor is running at a speed (N_R) close to the synchronous speed (N_S) and there will be a small slip between the rotating magnetic field produced by the armature and the actual salient-field poles.
• Thus, the relative speed between the armature MMF and the salient field poles is equal to the slip speed (N_S − N_R).
• Since the armature MMF moves slowly past the actual field poles, hence there will be an instant when the peak of the armature MMF wave is in line with the axis of the actual salient field poles.
• The axis of the field poles is the direct axis or d-axis. In this position, the reluctance offered is known as the direct axis synchronous reactance (X_d).
• After one-quarter of the slip cycle the peak armature wave is in line with the quadrature axis. In this position, the reluctance offered is known as the quadrature axis synchronous reactance (X_q).​

The correct answer is (option 4) i.e. It can take leading and lagging current by over or under-exciting its field winding.

Concept:

V curve is a plot of the stator current (I_a) versus field current (I_f) for different constant loads. Since the shape of these curves is similar to the letter “V”, thus they are called the V curve of a synchronous motor.

• The power factor of the synchronous motor can be controlled by varying the field current I_f.
• The armature current, I_a changes with the change in the field current, I_f.
• Let us assume that the motor is running at NO load. And small field current means an under-excitation field is given to the motor, then armature current (I_a) decreases and becomes up to the minimum value, in this stage motor operates at lagging power factor.
• If the field current is increased from this small value, the armature current Ia decreases until the armature current becomes minimum. At this minimum point, the motor is operating at a unity power factor.
• If now, the field current is increased further means giving over excitation to the motor field, the armature current increases and the motor starts operating as a leading power factor.

The correct answer is option 4.

The main reasons for the voltage drop in the alternator are:

1.) Armature resistance

The voltage drop caused by armature resistance per phase is IR_a, which is in phase with armature current I.

Additional energy loss such as Hysteresis loss, Eddy current Loss, and loss due to unequal distribution of currents are part of armature resistance losses. However, these losses are practically negligible.

2.) Armature leakage reactance

When the load current flows through the armature winding it builds up the local flux which cuts the winding and counters the EMF generated. This effect produces armature reactance that is equal to 2πfl.

This armature reactance is called leakage reactance X_L and this leakage flux is proportional to the armature current. Therefore, the voltage drop due to armature leakage reactance increases with increasing load.

3.) Armature Reaction

When the load current flows in the stator conductor, it produces a magnetic field that has a cross-magnetization, de-magnetization, and magnetizing effect upon the main flux due to the field winding.

Such an effect of armature current upon the main flux is known as an armature reaction. The armature reaction depends upon the power factor of the load.

4.) Synchronous Reactance

The combination of leakage reactance along armature reaction is called synchronous reactance.

Concept:

Slip test:

Direct and quadrature axis reactance can be determined by this test.

During this test, the synchronous machine is driven by the separate prime mover at slightly different from synchronous speed.

The field winding is open-circuited and rated frequency and reduced voltage is applied across the terminal.

Under these conditions, the relative velocity between the field pole and the rotating armature m.m.f wave is equal to the difference between synchronous speed and rotor speed.

\({X_q} = \frac{{minimum\;armature\;terminal\;voltage\;per\;phase}}{{maximum\;armature\;current\;per\;phase}}\)

Explanation:

For maximum current, during slip test on a Synchronous machine, the armature MMF aligns along the q-axis.

The correct answer is option 2) : 2.5\(\Omega\)

Concept:

Synchronous reactance is the fictional reactance used as an explanation for the voltage effects in the armature circuit produced by the actual armature leakage reactance and by the change in the air gap flux caused by the armature reaction.

It is said to be the addition of leakage reactance along with armature reaction 

X_s = X_l + Xa

Xl  is the leakage Reactance, X_a is the armature Reactance

Open circuit voltage per phase to the short circuit armature current at the same value of field current (Excitation)

X_s = \(\frac{V_o}{I_s}\)  

V_o is the open circuit voltage

I_s is the short circuit current

Calculation:

Given

 For the same field current 40 A

Vo = 5000 V

Is = 2000 A

Xs = \(\frac{V_o}{I_s}\)  

=\(\frac{5000}{2000}\)

=2.5 \(\Omega\)

Additional Information Short circuit test is used to calculate the synchronous Reactance. In this test, the alternator will run at the rated synchronous speed, and the load terminals are open. The excitation current increased to get 25% more than the rated voltage i.e. up to 125% of the rated voltage.

Armature leakage reactance (X_l):

When the load current flows through the armature winding it builds up the local flux which cuts the winding and counters EMF is generated. This effect produces armature reactance that is equal to 2πfl.

This armature reactance is called leakage reactance X_L and this leakage flux is proportional to the armature current.

Armature reaction (X_a): 

When the load current flows in the stator conductor, it produces a magnetic field which has a cross-magnetization, de-magnetization and magnetizing effect upon the main flux due to the field winding. Such an effect of armature current upon the main flux is known as armature reaction. The armature reaction depends upon the power factor of the load.

Armature reaction reactance is interpreted as inductive reactance which accounts for the effect of armature reaction.

Synchronous reactance (Xs):

It is the imaginary reactance employed to account for the voltage effects in the armature circuit produced by the actual armature leakage reactance and by the change in the air gap flux caused by the armature reaction.

The combination of leakage reactance along with armature reaction is called as synchronous reactance.

\({X_s} = {X_a} + {X_l}\)

Concept:

The % voltage regulation of an alternator is given by:

% V.R. = \( {E_p-V_p \over V_p}× 100\)

where, V_p = Terminal voltage

E_p = Induced EMF

The induced EMF in an alternator is given by:

\(E_p = \sqrt{(V_pcosϕ+I_aR_a)^2+(V_psinϕ+I_aX_a)^2}\)

where, I_a = Armature current

R_a = Armature resistance

X_a = Armature reactance

All the values must be per phase values.

Calculation:

Given, V_L = 440 V

\(V_p = {440 \over \sqrt{3}} = 254.03V\)

Ra = 0.2 Ω 

Xa = 3 Ω 

cosϕ = 1

P = \( { \sqrt{3} }\)V_LI_Lcosϕ 

60 × 10³ = \( { \sqrt{3} }\)(440)(I_L)(1)

I_L = 78.72 A

\(E_p = \sqrt{(254.03+78.72\times0.2)^2+(78.72\times 3)^2}\)

E_P = 358.53 V

% V.R. = \({358.53-254.03\over 254.03}\times 100\)

% V.R. = 41.14%