Synchronising of Alternators MCQ Quiz in తెలుగు - Objective Question with Answer for Synchronising of Alternators - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Synchronizing power of a synchronous machine is given by,

\({P_{syn}} = 3\frac{{{E_f}{V_t}}}{{{X_s}}}\sin \delta \)

Where E_f is the internally generated emf

V_t is the terminal voltage

X_S is the synchronous reactance

δ is the torque angle

Calculation:

Given that,

\({E_f} = {V_t} = \frac{{6600}}{{\sqrt 3 }}V\)

X_S = 1.65 Ω

Mechanical angle (θ_m) = 1°

Number of poles (P) = 12

Electrical angle, \({\theta _e} = \frac{{{\theta _m}}}{2} \times P = \frac{1}{2} \times 12 = 6^\circ = \frac{\pi }{{30}}rad\)

\(P = 3 \times \frac{{\frac{{6600}}{{\sqrt 3 }} \times \frac{{6600}}{{\sqrt 3 }}\; \times \;\cos 0^\circ }}{{1.65}}\; \times \;\frac{\pi }{{30}}\) 

= 2764.6 kW

= 921.53 kW per phase

Per phase voltage, \({V_t} = \frac{{6600}}{{\sqrt 3 }} = 3810.5\;V\)

Per phase armature current,

\({I_a} = \frac{{\left( {3000} \right)\left( {{{10}^3}} \right)}}{{\sqrt 3 \left( {6600} \right)}} = 262.43\;A\)

Percentage reactance, \({X_s} = \frac{{{X_s}\;in\;ohms}}{{{V_t}/{I_a}}} \times 100\)

\(= \frac{{20}}{{100}} \times \frac{{3810.5}}{{262.43}} = 2.9\)

At no load, synchronizing power per mechanical degree,

\({P_s} = m\frac{{dp}}{{d\delta }} \cdot \frac{{\pi p}}{{360}} = \frac{{3\;{V_t}{E_f}}}{{{X_s}}}\cos \delta \cdot \frac{{\pi p}}{{360}}\)

\(= \frac{{3 \times {{\left( {3810} \right)}^2}}}{{2.90}} \times \frac{{\pi \left( 8 \right)}}{{360}} = 1048.36\;kW\)

Synchronizing torque,

\({T_s} = \frac{{60}}{{2\pi \times {N_s}}} \cdot {P_s}\)

\(= \frac{8}{{2\pi \left( {100} \right)}}\;\left( {1048.36 \times {{10}^3}} \right)\)

= 13, 348.13 N-m

Concept:

Maximum power output by the synchronous generator:

The output power delivered by the synchronous generator is given by

\({{\bf{P}}_{{\bf{g}}}} = \frac{{{\bf{EV}}}}{{{Z_s}}}\;{\bf{cos}}\left( {{\bf{θ }} - {\bf{δ }}} \right) - \frac{{{{\bf{V}}^2}}}{{{{\bf{Z}}_{\bf{s}}}}}\;{\bf{cos}}\;{\bf{θ }}\)  -------    (1)

Where

E is the excitation voltage line to line

V is the terminal voltage line to line

Zs is the synchronous impedance

θ is the internal angle and δ is the load angle

Condition for maximum power output:

For maximum power developed \(\frac{{{\bf{d}}{{\bf{P}}_{{\bf{g}}}}}}{{{\bf{dδ }}}} = 0\)

⇒ \(\frac{{{\bf{EV}}}}{{{Z_s}}}\;{\bf{sin}}\left( {{\bf{θ }} - {\bf{δ }}} \right) - 0 = 0\)

⇒ sin (θ - δ) = 0

⇒ θ = δ 

⇒ Maximum power output of alternator is given by 

From equation (1)

⇒ \({{\bf{P}}_{{\bf{gmax}}}} = \frac{{{\bf{EV}}}}{{{Z_s}}}\ - \frac{{{{\bf{V}}^2}}}{{{{\bf{Z}}_{\bf{s}}}}}\;{\bf{cos}}\;{\bf{θ }}\)

Calculation:

Given that,

Z_s = 10 Ω, r_a = 1 Ω, E_f = V_t = 500 V

⇒ cos θ = r_a / Z_s =  1/10 = 0.1

The maximum power output is:

\({{\bf{P}}_{{\bf{gmax}}}} = \frac{{{\bf{500^2}}}}{{{10}}}\ - \frac{{{{\bf{500}}^2}}}{{{{10}}}}\times0.1\)  = 22.5 kW

Concept:

• Synchronizing Power is defined as the varying of the synchronous power P on varying in the load angle δ.
• It is also called Stiffness of Coupling, Stability or Rigidity factor. It is represented as P_syn.
• A synchronous machine, whether a generator or a motor, when synchronized to infinite Busbars has an inherent tendency to remain in Synchronism.

Power of salient power: 

\(P = \frac{{EV}}{{{X_d}}}sinδ + \frac{{{V^2}}}{2}\left( {\frac{1}{{{X_q}}} - \frac{1}{{{X_d}}}} \right)sin2δ \)

Where

δ = Power angle

Xd = Direct axis synchronous reactance

Xq = Quadrature axis synchronous reactance

Synchronizing power of salient pole machine:

\({P_{syn}} = \frac{{dP}}{{dδ }}\)

\(P_{syn} = \frac{{EV}}{{{X_d}}}cosδ + {V^2}\left( {\frac{1}{{{X_q}}} - \frac{1}{{{X_d}}}} \right)cos2δ \)

Explanation:

\(P_{syn} = \frac{{EV}}{{{X_d}}}cosδ + {V^2}\left( {\frac{1}{{{X_q}}} - \frac{1}{{{X_d}}}} \right)cos2δ \)

Z_s = (0.7 + 3j) = 3.08 ∠76.87

V_t = 400 V, E_f = 500 V

\(Maximum\;output\;power = \frac{{{E_f}{V_t}}}{{{Z_s}}} - \frac{{E_f^2}}{{Z_s^2}} \times {r_a}\)

\( = \frac{{500 \times 400}}{{3.08}} - {\left( {\frac{{500}}{{3.08}}} \right)^2} \times 0.7 = 46487.60\;W\)

Shaft power = [3 × 46.49 – 1.8]kW = 137.66 kW

Voltage per phase, \(V = \frac{{10,000}}{{\sqrt 3 }}\)

= 5773.5 V

Full load current, \(I = \frac{{500 \times 1000}}{{\sqrt 3 \times 10 \times 1000}}\)

= 28.87 A

I X_S = 30% of V = 0.3 × 5773.5

= 1732.05 V

\({X_S} = \frac{{1732.05}}{{28.87}} = 60\ {\rm{\Omega}}\)

\(I = 28.87\angle 0^\circ\), then V = 5773.5(0.8 + j0.6)

V = 4618.8 + j3464.1 V

\({E_0} = V + I{X_S}\) (R_a is neglegible)

= 4618.8 + j3464.1 + (j 1732.05)

= 4618.8 + j 5196.15

= 6952.2 ∠ 48.36° 

Power angle \(\delta = 48.36^\circ - {\cos ^{ - 1}}\left( {0.8} \right)\)

= 48.36° - 36.87°

= 11.5°

1° of mechanical displacement \(= \frac{6}{2} \times 1^\circ\) Electrical displacement

= 3° of electrical displacement.

Synchronising power per phase, \({P_{SY}} = \frac{{EV}}{{{X_S}}}\cos \delta \sin \alpha\)

\(= \frac{{6952.2 \times 5773.5}}{{60}} \times \cos 11.5^\circ \sin 3^\circ\)

= 34308.6 W

= 34.31 kW