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

Explanation:

Excitation Voltage in Salient Pole Synchronous Machine

Definition: In a salient pole synchronous machine, excitation voltage is the voltage required at the field winding to establish the necessary magnetic flux for the machine's operation. This flux interacts with the armature winding to produce the required electromagnetic torque. The excitation voltage is also influenced by the power factor, load conditions, and machine's reactance.

Expression for Excitation Voltage:

The excitation voltage for generating action in a salient pole synchronous machine is given by:

E_o = V Cos δ + I_qR_a + I_dX_d

Where:

• E_o: Excitation voltage
• V: Terminal voltage
• δ: Power angle
• I_q: Quadrature axis component of armature current
• I_d: Direct axis component of armature current
• R_a: Armature resistance
• X_d: Direct axis synchronous reactance

Derivation:

The excitation voltage in salient pole synchronous machines depends on the phasor relationship between the terminal voltage, armature current, and the synchronous reactance. The machine's synchronous reactance is divided into two components:

• Direct axis reactance (X_d): This is associated with the magnetic flux along the direct axis.
• Quadrature axis reactance (X_q): This is associated with the magnetic flux along the quadrature axis.

The armature current can be resolved into two components:

• I_d: Direct axis component, which contributes to the direct axis magnetic flux.
• I_q: Quadrature axis component, which contributes to the quadrature axis magnetic flux.

The excitation voltage is determined by considering the power angle δ and the phasor addition of the resistive and reactive components of voltage drops across the armature resistance (R_a) and reactances.

Advantages of This Expression:

• Provides a comprehensive understanding of the relationship between excitation voltage and operating parameters such as load current and power factor.
• Helps in designing and analyzing the performance of synchronous machines under different load conditions.

Correct Option Analysis:

The correct option is:

Option 3: E_o = V Cos δ + I_qR_a + I_dX_d

This expression correctly represents the excitation voltage in a salient pole synchronous machine. The positive signs indicate the additive nature of the voltage drops due to the armature resistance and direct axis reactance.

Additional Information

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

Option 1: E_o = V Cos δ + I_qR_a - I_dX_d

This option is incorrect because the sign of the term involving I_dX_d is negative. The excitation voltage expression inherently requires the addition of the direct axis reactance component, as it contributes positively to the net excitation voltage.

Option 2: E_o = V Cos δ - I_qR_a - I_dX_d

This option is incorrect as both I_qR_a and I_dX_d have negative signs. These terms represent the voltage drops due to the resistive and reactive components, which are additive in the excitation voltage equation.

Option 4: E_o = V Cos δ - I_qR_a + I_dX_d

This option is partially correct, as it correctly includes a positive sign for the I_dX_d term. However, the negative sign for I_qR_a is incorrect, as the armature resistance contributes positively to the net excitation voltage.

Conclusion:

The excitation voltage for a salient pole synchronous machine is expressed as:

E_o = V Cos δ + I_qR_a + I_dX_d

This expression accounts for the terminal voltage, power angle, and the contributions of armature resistance and direct axis reactance. Understanding this equation is crucial for analyzing the performance and operation of synchronous machines under various load conditions. Evaluating the other options highlights the importance of correctly interpreting the signs and relationships in the excitation voltage formula.

Concept

The synchronous speed is given by:

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

where, Ns = Synchronous speed

f = Frequency

P = No. of poles

Calculation

Given, Ns = 250 RPM

f = 50 Hz

The no. of poles are given by:

\(250={120\times 50\over P}\)

P = 24 

Concept

The efficiency of a synchronous motor is given by:

\(\eta={P_{out}\over P_{in}}× 100\)

Also, P = S cosϕ 

where, P = Active power

S = Apparent power

cosϕ = Power factor

Calculation

Given, S = 2000 kVA 

cos ϕ = 0.9 leading

P_in = 2000 × 0.9 = 1800 kW

\(0.95={P_{out}\over 1800}\)

P_out = 1800 × 0.95 

Pout = 1710 kW

Voltage regulation of synchronous generators

Voltage regulation of synchronous generators refers to the ability of the generator to maintain a constant output voltage despite variations in load conditions or other external factors. 

The voltage regulation of the alternator can be defined as the change in terminal voltage from no-load to full-load rated value divided by the full-load rated voltage.

\(V.R.={E-V\over V}\)

Direct Load Test of alternators:

• The direct load test involves connecting the synchronous generator to a load and measuring the terminal voltage at different load levels (from no load to full load). By doing so, one can determine how the voltage varies with load and calculate the voltage regulation.
• This method requires the generator to be operated under its rated conditions, making it practical mainly for smaller generators with less than 5 kVA.
• For larger alternators (generators with a power rating greater than 100 kVA), the direct load test may not be practical due to the challenges associated with handling high power loads, safety concerns, and equipment limitations.
• Instead, larger generators often undergo open-circuit tests and short-circuit tests to determine their voltage regulation, as these methods do not require the generator to be connected to a load during the test.

Concept:

Pitch factor(k_p) - it is the ratio of EMF induced in the short pitch coil to the full pitch coil.

it is given by , k_p = \(cos(\frac{α}{2} n)\) ;

here 'n' represents the n^th order harmonics

Calculation:

To eliminate the nth order harmonics the pitch factor must be equated to 0

⇒ kp = \(cos(\frac{α}{2} n)\) = 0

According to the question we have to eliminate the 5^th harmonic 

⇒ \(cos(\frac{α}{2} *5 ) \) = \(0\), here n = 5

\(\frac{α}{2} *5 = 90\)

⇒ α = 36°

Therefore the coil should be pitched at  α = 36 to eliminate the 5th order harmonic in the induced e.m.f. of a synchronous generator.

Hence the correct option is 5. 

Synchronous motor:

V curves for synchronous motor give the relation between armature current and DC field current. The curves are shown below.

A synchronous motor is capable of operating at all types of power factor i.e. either UPF, leading, or lagging power factor.

• Lagging power factor: If field excitation is such that Eb < V the motor is said to be under excited and it has a lagging power factor.
• Leading power factor: If field excitation is such that Eb > V the motor is said to be over-excited and it draws leading current. So that the power factor improves.
• Unity power factor: If field excitation is such that Eb = V the motor is said to be normally excited.

Synchronous generator:

A synchronous generator or alternator is capable of operating at all types of power factor i.e. either UPF, leading or lagging power factor.

• Leading power factor: If field excitation is such that Eb < V the alternator is said to be under excited and it has a leading power factor.
• Lagging power factor: If field excitation is such that Eb > V the alternator is said to be over-excited and it draws lagging current. 
• Unity power factor: If field excitation is such that Eb = V the alternator is said to be normally excited.
• V -curve for synchronous generator or alternator is shown below

The correct answer is option 1):(one level of AC to another level of AC)

Concept:

• The permamt magnet synchronous generator is mostly used in power plant
• It converts the mechanical power from aerodynamic wind system to electric power
• The wind turbine generator system requires a power conditioning circuit called power converter
• The power converter is capable of adjusting the generator frequency and voltage to the grid. 
• The purpose of power converter is to enable efficient conversion of the variable frequency output of an induction generator, driven by a variable speed wind turbine, to a fixed frequency appropriate for the grid or a load
• In case of Permanent magnet synchronous generator operated wind power plant, the converter used in the generator interface converts one level of AC to another level of AC

In a synchronous generator, the real power is given by

P ∝ VI cos ϕ

It is given that the real power and terminal voltages are constant.

⇒ I cos ϕ = constant

⇒ I_a1 cos ϕ₁ = I_a2 cos ϕ₂

Given that, cos ϕ₁ = 0.9, I_a1 = 100 A and Q₂ = 2Q₁

tan ϕ₁ = 0.4843

We know that, Q = P tan ϕ

At constant real power, Q ∝ tan ϕ

\( \Rightarrow \frac{{{Q_1}}}{{{Q_2}}} = \frac{{\tan {\phi _1}}}{{\tan {\phi _2}}}\)

\( \Rightarrow \frac{{{Q_1}}}{{2{Q_1}}} = \frac{{0.4843}}{{\tan {\phi _2}}}\)

⇒ tan ϕ₂ = 0.9686

⇒ cos ϕ₂ = 0.718

Now, I_a1 cos ϕ₁ = I_a2 cos ϕ₂

⇒ 100 × 0.9 = I_a2 × 0.718

Concept:

\(P = \sqrt 3 \;{V_L}{I_L}\;cos\phi \)

Where,

P = Machine rating in MW

V_L = Line voltage

I_L = Line current

Cosϕ = Power factor

Calculation:

Given-

V_L = 20 kV, cosϕ = 0.8, P = 480 MW

\(480 \times {10^6} = \sqrt 3 \times 20 \times {10^3} \times {I_L} \times 0.8\)

Voltage regulation of synchronous generators

Voltage regulation of synchronous generators refers to the ability of the generator to maintain a constant output voltage despite variations in load conditions or other external factors. 

The voltage regulation of the alternator can be defined as the change in terminal voltage from no-load to full-load rated value divided by the full-load rated voltage.

\(V.R.={E-V\over V}\)

Direct Load Test of alternators:

• The direct load test involves connecting the synchronous generator to a load and measuring the terminal voltage at different load levels (from no load to full load). By doing so, one can determine how the voltage varies with load and calculate the voltage regulation.
• This method requires the generator to be operated under its rated conditions, making it practical mainly for smaller generators with less than 5 kVA.
• For larger alternators (generators with a power rating greater than 100 kVA), the direct load test may not be practical due to the challenges associated with handling high power loads, safety concerns, and equipment limitations.
• Instead, larger generators often undergo open-circuit tests and short-circuit tests to determine their voltage regulation, as these methods do not require the generator to be connected to a load during the test.

Concept:

Total Active Power across three-phase alternator is given by  

\(P=\sqrt 3V_LI_Lcosϕ =3V_{ph}I_{ph}cosϕ\)

In star connection,

V_L = √3 V_ph ,

I_L = I_ph

Application:

Given:

P = 900 MW, Terminal phase Voltage(V_ph) = 5 kV, cos ϕ = 0.6

\(P=3V_{ph}I_{ph}cosϕ\)

900 × 10⁶  = 3 × (5 × 10³) × I_ph ×  0.6

\(I_{ph}=\dfrac{900\times10^6}{3\times5\times10^3\times0.6}\)

= 100 kA

Concept:

Speed of synchronous generator in rpm is given by

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

Where,

f = supply frequency in Hz

P = Number of poles

Calculation:

Given-

Ns = 100 r/min = 100 rpm,

f = 50 Hz

When hydraulic turbine is connected to synchronous generator, then both are running at same speed

∴ \(100 = \frac{{120 \times 50}}{P}\)

Concept: 

Power Transmission Capability of Synchronous Generator:

Considered a synchronous generator connected to the infinite bus through a transmission line of reactance X_l.

Assumed that the resistance and capacitance of the system are zero.

Where,

V = V∠0 = voltage at infinite bus

E = E∠δ = voltage behind direct axis synchronous reactance of the generator

X_d = synchronous or transient reactance of machine

X_l = line reactance

Complex power delivered by the generator to the system is,

\(S = V{I^*} = V{\left[ {\frac{{E\angle {δ _\;} - V}}{{j\left( {{X_d} + {X_l}} \right)}}} \right]^*}\)

Let, X = X_d + X_l

\(S = P + jQ = \frac{{EV}}{X}sinδ + j\left( {\frac{{EV}}{X}cosδ - \frac{{{V^2}}}{X}} \right)\)

Active power transferred to the system,

\(P = \frac{{EV}}{X}sinδ \)

The maximum steady-state power transfer occurs when,

δ = 90°

Reactive power transferred to the system,

\(Q = \left( {\frac{{EV}}{X}cosδ - \frac{{{V^2}}}{X}} \right)\)

Calculation:

Given,

X_d = 0.4 p.u

X_l = 0.1 p.u

X_m = 0.35 p.u

X = X_d + X_l + X_m = 0.4 + 0.1 + 0.35 = 0.85 p.u

E = 1∠δ1 p.u

V = 0.85 ∠δ2 p.u

P = 0.866 p.u

\(P = \frac{{EV}}{X}{\rm{sin}}\left( {{δ _1} - {δ _2}} \right)\)

\(0.866 = \frac{{1 \times 0.85}}{{0.85}}{\rm{sin}}\left( {{δ _1} - {δ _2}} \right)\)

\(\sin \left( {{δ _1} - {δ _2}} \right) = 0.866\)

(δ₁ - δ₂) = 60° 

Concept:

Speed of synchronous generator in rpm is given by

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

Where,

F = supply frequency in Hz

P = Number of poles

Calculation:

Given-

N_s = 250 rpm,

F = 50 Hz

When hydraulic turbine is connected to synchronous generator, then both are running at same speed

∴ \(250 = \frac{{120 \times 50}}{P}\)

Sending end voltage (V_s) = 17.87 kV

Receiving end voltage (V_r) = 17.32 kV

R = 0.25 Ω

X_L = 3.925 Ω

\(Z = \sqrt {{R^2} + X_L^2} = 3.933\;{\rm{\Omega }}\)

\({P_r} = \frac{{\left| {{V_r}} \right|\left| {{V_s}} \right|}}{Z}\cos \left( {\theta - \delta } \right) - \frac{{V_r^2}}{Z}cos\theta \)

\(= \frac{{\left| {{V_r}} \right|\left| {{V_s}} \right|}}{Z}\cos \left( {\theta - \delta } \right) - \frac{{V_r^2}}{Z}\left( {\frac{R}{Z}} \right)\)

\(\Rightarrow 3 = \frac{{\left( {17.32} \right)\left( {17.87} \right)}}{{3.933}}\cos \left( {\theta - \delta } \right) - \frac{{{{\left( {17.32} \right)}^2} \times 0.25}}{{{{\left( {3.933} \right)}^2}}}\)

⇒ (θ - δ) = 87.276°

\({Q_r} = \frac{{\left| {{V_s}} \right|\left| {{V_r}} \right|}}{Z}\sin \left( {\theta - \delta } \right) - \frac{{V_r^2}}{Z}\sin \theta \)

\(= \frac{{\left| {{V_s}} \right|\left| {{V_r}} \right|}}{Z}\sin \left( {\theta - \delta } \right) - \frac{{V_r^2}}{Z}\left( {\frac{X}{Z}} \right)\)

= \(\frac{{\left( {17.87} \right)\left( {17.32} \right)}}{{3.933}}\sin \left( {84.276} \right) - \frac{{{{\left( {17.32} \right)}^2} \times 3.925}}{{{{\left( {3.933} \right)}^2}}} = 2.185\;MVar\)

\({\rm{Power\;factor}} = \cos \left( {{{\tan }^{ - 1}}\left( {\frac{{{Q_r}}}{{{P_r}}}} \right)} \right) = 0.8\;lag\)