100 MCQ OF Synchronous Generator or Alternator with Explanation

Answer.C. 1400 V

Explanation:-

Given Data

The number of Poles P = 4
No of slots = 48
Flux φ = 0.25 Wb
Number of turns = 10

For double layer winding,

No of coils = No of slots = 48

Total number of Turns = 48 × 10 = 480

Hence turns per phase = 480 ⁄ 3 = 160

Pitch factor K_P is

$\begin{array}{l}{K_P} = \cos \dfrac{\alpha }{2}\\\\{K_P} = \cos \dfrac{{36^\circ }}{2}\\\\{K_P} = \cos 18^\circ \\\\{K_P} = 0.951\end{array}$

Now distribution factor K_d is given as

$\begin{array}{l}{K_d} = \dfrac{{\sin (m\alpha /2)}}{{m\sin \alpha /2}}\\\\{\text{where alpha is slot angle = }}\dfrac{{{\text{180 electrical}}}}{{{\text{No of slots/pole}}}}\\{\rm{\alpha }} = \dfrac{{180 \times 4}}{{48}} = 15^\circ \\\\{\text{m = no of slots per pole per phase}}\\\\m = \dfrac{{48}}{{4 \times 3}} = 4\\\\\therefore {K_d} = \dfrac{{\sin (60^\circ /2)}}{{4\sin (15^\circ /2)}}\\\\{K_d} = 0.9576\end{array}$

∴ EMF per phase E_ph of an alternator is

E_ph = 4.44.K_p.K_d.φ.f.T_ph

E_ph = 4.44 × 0.951 × 0.9576 × 0.25 × 50 × 160

E_ph = 808.68

Line to Line induced EMF

E_L = √3E_PH

E_L = √3 × 808.68

E_L = 1400.67 V

Answer A. 0.9494

Explanation:-

Given Data

The number of Poles P = 4
No of slots = 48
Phase  = 3

Distribution factor K_d is given as

$\begin{array}{l}{K_d} = \dfrac{{\sin (m\alpha /2)}}{{m\sin \alpha /2}}\\\\{\text{where alpha is slot angle = }}\dfrac{{{\text{180 electrical}}}}{{{\text{No of slots/pole}}}}\\{\rm{\alpha }} = \dfrac{{180 \times 4}}{{48}} = 15^\circ \\\\{\text{m = no of slots per pole per phase}}\\\\m = \dfrac{{48}}{{4 \times 3}} = 4\\\\\therefore {K_d} = \dfrac{{\sin (60^\circ /2)}}{{4\sin (15^\circ /2)}}\\\\{K_d} = 0.9576\end{array}$

Pitch factor K_P is

$\begin{array}{l}{K_P} = \cos \dfrac{\alpha }{2}\\\\{K_P} = \cos \dfrac{{15^\circ }}{2}\\\\{K_P} = 0.9914\end{array}$

Sometimes distribution factor (K_d) and pitch factor (K_p) of an alternator are combined into a single factor called winding factor (K_w). The winding factor is the product of K_d and K_p i.e.

K_w = K_p × K_d

K_w = 0.9576 × 0.9914

K_w = 0.9494

Answer 2. Cot 15° ⁄ 4

Explanation:-

Distribution factor K_d is given as

$\begin{array}{l}{K_d} = \dfrac{{\sin (m\alpha /2)}}{{m\sin \alpha /2}}\\\\{\text{where alpha is slot angle = }}\dfrac{{{\text{180 electrical}}}}{{{\text{No of slots/pole}}}}\\{\rm{\alpha }} = \dfrac{{180 \times 6}}{{36}} = 30^\circ \\\\{\rm{m = no of slots per pole per phase}}\\\\m = \dfrac{{36}}{{6 \times 3}} = 2\\\\\therefore {K_d} = \dfrac{{\sin (2 \times 15^\circ /2)}}{{4\sin (30^\circ /2)}}\\\\{K_d} = \dfrac{1}{{4\sin 15^\circ }}\end{array}$

Also the Pitch factor K_P is

$\begin{array}{l}{K_P} = \cos \dfrac{\alpha }{2}\\\\{K_P} = \cos \dfrac{{30^\circ }}{2}\\\\{K_P} = \cos 15^\circ \end{array}$

The winding factor is the product of K_d and Kp i.e.

K_w = K_p × K_d

$\begin{array}{l}{K_w} = \dfrac{1}{{4\sin 15^\circ }} \times \cos 15^\circ \\\\\dfrac{{Cos\theta }}{{Sin\theta }} = Cot\theta \\\\\therefore {K_w} = \dfrac{{Cot15^\circ }}{4}\end{array}$

Answer. 2. π/7

Explanation:-

For eliminating 7th harmonic

K_p7 = 0

$\begin{array}{l}\cos \dfrac{{7\alpha }}{2} = 0\\\\ = \dfrac{{7\alpha }}{2} = \dfrac{\pi }{2}\\\\\alpha = \dfrac{\pi }{7}\end{array}$

Answer C. 265.7 V

Explanation:-

Given

Phase displacement angle α = 30°

No of slots per pole per phase (n) = Number of coil = 8

Distribution factor K_d is

$\begin{array}{l}\therefore {K_d} = \dfrac{{\sin (8 \times 30^\circ /2)}}{{8\sin (30^\circ /2)}}\\\\{K_d} = 0.4182\end{array}$

Airthmetic sum of voltage = Total Number of coils x Given voltage

= 8 x 80 = 640 V

The EMF of right coils in series is

Vector sum of voltage = Distribution factor x Airthmetic sum of voltage

= 0.4182 × 640

= 267. 6 V

Answer 4. 6 Poles, 14433.7 V

Explanation:-

Given data

Frequency f = 50 Hz
Line voltage V_L = 25 kV = 25 x 10³ V
Speed N_s = 1000 RPM

The synchronous speed of an alternator is given as

N_s = 120f/P

P = 120f/N

P = 120 x 50/1000

P = 6

In star connection Voltage per phase is given by

V_PH = VL ⁄ √3

V_PH = 25 x 10³ ⁄ √3

V_PH = 14433.7 V

Answer. C. 0.144

Explanation:-

Given Data,

Line voltage V_L = 400 V
Alternator Rating = 15 KVA = 15 x 10³ VA
Resistance R_a = 0.5 Ω
Synchronous Reactance X_L = 10 Ω

Full Load current I_L =  VA ⁄ √3 V_L

I_L = 15 x 10³ ⁄ √3 x 400

I_L = 21.6 A

For star connection Voltage per phase is given by

V_PH = VL ⁄ √3

V_PH = 400 ⁄ √3

V_PH = 230.9 V

Generated voltage per phase of an alternator is given as

E_PH = V_PH + I_LR_a + I_LX_L

= 230 + 21.6 × 0.5 + 21.6 × 10

= 457.7 V

The voltage regulation of an alternator is defined as the change in its terminal voltage when the full load is removed, keeping field excitation and speed constant, divided by the rated terminal Voltage.

Soif V_ph = Rated terminal voltage

Eph = No load induced e.m.f

Voltage Regulation = (E_ph − V_ph)/V_ph

= (457.7 − 400) ⁄ 400

= 0.144

Answer. C. 7.5 Ω

The Nominal impedance of a synchronous machine is also called as the base Impedance. The nominal impedance of an alternator is the ratio of the rated line-to-neutral voltage divided by the rated line current. The nominal impedance is used as a base of comparison for other impedances that the alternator possesses.

Z_base = V_base ⁄ I_base

Apparent Power (S) = V x I

Z_base = V²_base ⁄ S_base

∴ Nominal Impedance

Z_base = (15000)² ⁄ 30 × 10⁶

Z_base = 7.5Ω

Answer. 3. Larger in size than

Explanation:-

The efficiency of alternator increase with the increase in size because losses to heat are reduced. 

The coefficient of efficiency can be increased by designing the machine to have reduced losses windings (copper losses) and magnetic circuit (iron losses). By increasing the cross-section of conductors, the resistance of copper conductors is decreased which leads to reduced copper losses. By increasing the cross-section of the magnetic circuit, the magnetic induction decreases for the same flux. Consequently, the iron losses are smaller. On the other hand, this approach to reducing the current density and magnetic induction leads to an increased volume and mass of the machine. 

The efficiency of an alternator automatically improves as the power rating increases. For example, if an alternator of 1 kVA has an efficiency of 50%, a larger but similar model having a capacity of 10 MVA will have an efficiency of about 90%.

Answer.3. Inter turn Fault protection

Explanation:-

The Merz-Price protection system gives protection against phase to phase faults and earth faults. It does not give protection against inter-turn faults. The inter-turn fault is a short circuit between the turns of the same phase winding. Thus the current produced due to such fault is a local circuit current and it does not affect the currents entering and leaving the winding at the two ends, where C.T.s are located. Hence Merz-Price protection cannot give protection against inter-turn faults.

In a single turn generator, there is no question of inter-turn faults but in multiturn generators, the inter-turn fault protection is necessary. So such inter-turn protection is provided for multiturn generators such as hydroelectric generators. These generators have double winding armatures. This means each phase winding is divided into two halves, due to the very heavy currents which they have to carry. This splitting of the single-phase winding into two is advantageous in providing inter-turn fault protection to such hydroelectric generators.

The scheme uses a cross differential principle. Each phase of the generator is doubly wound and split into two parts S₁ and S₂ as shown in Fig. The current transformers are connected in the two parallel paths of each phase winding.

The secondaries of the current transformers are cross-connected. The current transformers work on circulating the current principle. The relay is connected across the cross-connected secondaries of the current transformers.

Under normal operating conditions, when the two paths are sound then currents In the two parallel paths S₁ and S₂ are equal. Hence currents in the secondaries of the current transformers are also equal. The secondary current flows around the loop and Is the same at all the points. Hence no current flows through the relay and the relay inoperative.

If the short circuit is developed between the adjacent turns of the part Sit of the winding say then currents through S₁ and S₂ no longer remain the same. Thus unequal currents will be induced in the secondaries of the current transformers. The difference of these currents flows through the relay R. Relay then closes its contacts to trip the circuit breaker which isolates the generator from the system.

Such an inter-turn fault protection system is extremely sensitive but it can be applied to the generators having doubly wound armatures.