Synchronous Generator Power MCQ Quiz in తెలుగు - Objective Question with Answer for Synchronous Generator Power - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

At half full load and 0.8 power factor,

Output power = 500 × 0.5 × 0.8 = 200 kW

Constant losses = Friction and windage loss + core loss + field copper loss

= 1500 + 2500 + 1000 = 5 kW

At half full load, rated phase current

\({{I}_{ph}}=\frac{1}{2}\times \frac{500\times {{10}^{3}}}{\sqrt{3}\times 11\times {{10}^{3}}}\) 

= 13.12 A

Copper losses for all three phases = 3 I²_ph R

= 3 × (13.12)² × 4

= 2.066 kW

Total losses = 5 + 2.066 kW = 7.066 kW

\(Efficiency=\frac{200}{207.066}\times 100=96.58%\)

Angle δ between E and V is the power angle.

\(I = \frac{{1200 \times {{10}^3}}}{{\sqrt 3 \times 8 \times {{10}^3}}}\)

= 86.6 A

\(\begin{array}{l} I{R_a} = 86.6 \times 0.5 = 43.3\ V\\ I{X_S} = 86.6 \times 6 = 519.6\ V \end{array}\)

Voltage per phase \(= \frac{{8000}}{{\sqrt 3 }} = 4618.8\ V\)

\(\begin{array}{l} \phi = {\cos ^{ - 1}}\left( {0.8} \right) = 36.87^\circ \\ \tan \left( {\phi + \delta } \right) = \frac{{V\sin \phi + I{X_S}}}{{V\cos \phi + I{R_a}}}\\ = \frac{{4618.8 \times 0.6 + 519.6}}{{4618.8 \times 0.8 + 43.3}}\\ = \frac{{3290.88}}{{3738.34}}\\ \phi + \delta = 41.36^\circ \\ \delta = 41.36^\circ - 36.87^\circ \end{array}\)

= 4.49°

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

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 1. 

Alternator during short circuit:

• Under solid short circuit, the current is limited by the armature reactance.
• Under short circuit condition, the alternator operates at zero power factor leading by neglecting the armature resistance.
• The armature flux demagnetizes the main field flux and it results into lower resultant flux.

\({V_t} = \frac{{230}}{{\sqrt 3 }} = 132.79\;V\)

\({I_a} = \frac{{30 \times \frac{{{{10}^3}}}{3}}}{{\frac{{230}}{{\sqrt 3 }}}} = \frac{{10 \times {{10}^3}}}{{132.79}} = 75.307\;A\)

ϕ = cos^-1 0.707 = 45°

\(tan\delta = \frac{{{X_q}{I_a}cos\phi - {I_a}{R_a}sin\phi }}{{{V_T} + {X_q}{I_a}sin\phi + {I_a}{R_a}cos\phi }}\)

As given neglect the armature resistance 

\(\tan \delta = \frac{{{I_a}{X_q}\cos \phi }}{{{V_T} + {I_a}{X_q}\sin \phi }}\)

\( = \frac{{75.307 \times 2 \times 0.707}}{{132.79 + 75.307 \times 2 \times 0.707}}\)

= 0.445

δ = tan^-1 0.445 ≃ 24° 

\({\rm{Pe}} = \frac{{{{\rm{V}}_{\rm{t}}}{{\rm{E}}_{\rm{m}}}}}{{0.8}}{\rm{sin}}{{\rm{\delta }}_{\rm{m}}}\)

For maximum E_m, δ_m = 90° limit of stability

\({{\rm{E}}_{{\rm{m}}\left( {{\rm{min}}} \right)}} = \frac{{0.8 \times 0.8}}{1} = 0.64{\rm{PU\;or\;}}256{\rm{V}}\)