Synchronising of Alternators MCQ Quiz in বাংলা - Objective Question with Answer for Synchronising of Alternators - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

• The power factor of the alternator defines the phase angle between the current drawn and applied voltage.
• The power factor of the alternator depends upon the type of load. In case if it is an infinite bus, then the power factor of the alternator depends upon the excitation of the alternator.
• According to load, the alternator provides both reactive (kVAR) and real power (kW).
• When the load is RL type it requires lagging VAR and that power has to be provided by the alternator.
• When the load is of RC type, it requires leading VAR, and that power is provided by the alternator.

Important Points

Armature reaction:

The effect of armature (stator) flux on the flux produced by the rotor field poles is called armature reaction. It depends on the power factor i.e. the phase relationship between the terminal voltage and armature current.

• The armature reaction flux is constant in magnitude and rotates at synchronous speed
• At unity power factor load, the armature reaction is cross-magnetizing (distortional)
• At lagging power factor load, the armature reaction is partly demagnetizing and partly cross-magnetizing; At zero power factor lagging load, it is purely demagnetizing and the induced emf will get decreased
• When the generator supplies a load at the leading power factor the armature reaction is partly magnetizing and partly cross-magnetizing; At zero power factor leading load, it is purely magnetizing and the induced emf will get increased.

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