# A 20-pole alternator is having 180 identical stator slots with 6 conductors in each slot. All the coils of a phase are in series. If the coils are con

140 views
in General
closed
A 20-pole alternator is having 180 identical stator slots with 6 conductors in each slot. All the coils of a phase are in series. If the coils are connected to realize single-phase winding, the generator voltage is V1. If the coils are reconnected to realize three-phase star-connected winding, the generated phase voltage is V2. Assuming full pitch, single-layer winding, the ratio V1/V2 is
1. $\frac{1}{{\sqrt 3 }}$
2. $\frac{1}{2}$
3. $\sqrt 3$
4. 2

by (41.0k points)
selected

Correct Answer - Option 4 : 2

Distribution factor:

• The Distribution Factor is also known as the Breadth Factor or Belt factor or Spread factor.
• It is defined as the ratio of the actual voltage obtained to the possible voltage if all the coils of a polar group were concentrated in a single slot.
• It is denoted by Kd

Kd = A / B

Where,

A is the vector sum of induced EMF

B is the arithmetic sum of induced EMF

Distribution factor Kd

${K_d} = \frac{{\sin \frac{{mβ }}{2}}}{{m\sin \frac{β }{2}\;\;\;}}$

Where,

m is slots per pole per phase

β angular displacement of the slots

Single layer winding: One coil side occupies one slot completely, Number of coils (C) is equal to the half of the number of slots (S)

Double layer winding: Number of coils (C) is equal to the number of slots (S)

Integral slot winding: Number of slots per pole per phase is an integer

Fractional slot winding: Number of slots per pole per phase is not an integer

Full pitched winding: Coil span is equal to slots per pole (S/P)

Short pitched winding: Coil span is less than slots per pole (S/P)

Calculation:

Given that, poles of alternator = 20

Stator slots = 180

Conductors in each slot = 6

For single-phase winding

No. of coils in phase, $n = \frac{{180}}{{20 \times 1}} = 9$

Slots span in electrical degrees, β =180/(no. of slots per pole)

$\Rightarrow \beta = \frac{{180}}{{180/20}} = 20^\circ$

Distribution factor ${k_d} = \frac{{\sin n\beta /2}}{{n\sin \beta /2}}$

$= \frac{{\sin 9 \times 20/2}}{{9\sin 20/2}}$

kd1 = 0.6398

No. of turns N1 = No. of slots × (conductors/2)

$= 180 \times \frac{6}{2} = 540$

For three-phase winding

Slot span, β = 20°

No. of coils in a phase $= \frac{{180}}{{20 \times 3}} = 3 = n$

Distribution factor ${k_d} = \frac{{\sin 3 \times 20/2}}{{3\sin 20/2}}$

kd2 = 0.9597

No. of turns N2 = no. of slots × (conductors/2) × (1/number of phases)

$= 180 \times \frac{6}{2} \times \frac{1}{3}$

N2 = 180

Now, the RMS value of generated emf for single phase is

V1 = 4.44 kd1 ϕ N1

For 3-phase, the RMS value of generated emf is

V2 = 4.44 kd2 ϕ N2

Now, $\frac{{{V_1}}}{{{V_2}}} = \frac{{{k_{d1}} \cdot {N_1}}}{{{k_{d2}} \cdot {N_2}}} = \frac{{0.6398 \times 540}}{{0.9597 \times 180}}$

$\Rightarrow \frac{{{V_1}}}{{{V_2}}} \approx 2$