Distribution Factor MCQ – Objective Question with Answer for Distribution Factor

Answer.4. Both 2 and 3

Explanation

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 K.

Answer.1. Actual voltage to Possible voltage

Explanation

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 K.

Answer.4. $ \frac{{sin30^\circ }}{{3\;sin10^\circ }}$

Explanation

Distribution factor ${k_d} = \frac{{sin\frac{{m\gamma }}{2}}}{{m\sin \frac{\gamma }{2}}}$

Where $m = \frac{{slots}}{{pole \times phase}}$

$\gamma = \frac{{\pi \times p}}{s}$

P = no of poles

S = no of slots

Calculation

No of slots = 18

No of poles = 2

No of phase = 3

$m = \frac{{18}}{{2 \times 3}} = 3$

$\gamma = \frac{{\pi \times 2}}{{18}} = 20^\circ$

${k_d} = \frac{{sin\frac{{3 \times 20}}{2}}}{{3\;sin\frac{{20}}{2}}} = \frac{{sin30^\circ }}{{3\;sin10^\circ }}$

Answer.4. 1.273T volts

Explanation

For a uniformly distributed 1-phase alternator, the distribution factor,

${k_d} = \frac{{\sin \frac{{m\gamma }}{2}}}{{\frac{{m\gamma }}{2}{\rm{\;}}\frac{\pi }{2}}} = \frac{2}{\pi }$

Phase Spread mγ = 180°

Total induced emf,

E = T × EMF in each turn × k_p × k_d

= 2T × 1 × 2/π = 1.273 T volts

Answer.3. 0.67

Explanation

The breadth factor or a distribution factor is given by

${k_d} = \frac{{\sin n \cdot \frac{{m\beta }}{2}}}{{n \cdot \sin \frac{{m\beta }}{2}}}$ (for n^th harmonics)

m = no. of slots per pole per phase

β = slots angle

Calculation:

3 – phase, 4 – pole synchronous machine.

No. of slots = 36 (stator slots)

$m = \frac{{36}}{{3\; \times \;4}} = 3$

$\beta = \frac{{180^\circ }}{{slots\;per\;pole}}$

$\beta = \frac{{180^\circ }}{{36/4}} = 20^\circ $

So, ${k_d} = \frac{{\sin \frac{{\left( 3 \right)\left( 3 \right)\; \times \;20^\circ }}{2}}}{{3\sin \frac{{3\; \times \;20}}{2}}}$

$\Rightarrow {k_d} = \frac{{\sin 90^\circ }}{{3\sin 30^\circ }} = \frac{2}{3}$

Answer.3. Number of distributed slots under a given pole

Explanation

• The Distribution Factor is also known as the Breadth Factor or Belt factor or Spread factor.
• In alternators, the distribution factor for the given number of phases is dependent only on the number of distributed slots under a given pole.
• 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 K_d

K_d = A / B

Where,

A is the vector sum of induced EMF

B is the arithmetic sum of induced EM

• In a concentrated winding, each phase of a coil is concentrated in a single slot.
• The individual coil voltages induced are in phase with each other.
• These voltages must be added arithmetically.
• In order to determine the induced voltage per phase, a given coil voltage is multiplied by the number of series-connected coils per phase.
• In actual practice, in each phase, coils are not concentrated in a single slot.
• They are distributed in a number of slots in space to form a polar group under each pole.
• The voltages induced in coil sides are not in phase, but they differ by an angle β which is known as the angular displacement of the slots.
• The phasor sum of the individual coil voltages is equal to the total voltage induced in any phase of the coil

Distribution factor K_d

${K_d} = \frac{{\sin \frac{{m\beta }}{2}}}{{m\sin \frac{\beta }{2}\:\:\:}}$

Where,

m is slots per pole per phase

β angular displacement of the slots

Answer.2. 200 kVA

Explanation

According to Distribution factor

kVA ∝ KD

${\dfrac{{kV{A_{{{60}^0}\left( {3 – \phi } \right)}}}}{{kV{A_{{{180}^0}\left( {1 – \phi } \right)}}}} = \dfrac{{\sin \dfrac{{60}}{2}}}{{\sin \dfrac{{180}}{2}}}{\rm{x}}\dfrac{{{{180}^0}}}{{{{60}^0}}}}$

0.5 × 3/1 = 1.5

${ \Rightarrow kV{A_{{{180}^0}\left( {1 – \phi } \right)}} = \frac{{300}}{{1.5}} = 200\;kVA}$

Answer.2. $\frac{\sin 30°}{10\sin 3°}$

Explanation

The mathematical expression of the distribution factor is given as,

${k_d} = \frac{{\sin \frac{{m\beta }}{2}}}{{m\sin \frac{\beta }{2}\;}}$

Where m is the number of slots per pole per phase

β is the slot angle in degrees

Calculation:

3 – phase, 8 – pole synchronous machine.

No. of slots = 240 (stator slots)

$m = \frac{{240}}{{3\; × \;8}} = 10$

$\beta = \frac{{180^\circ }}{{slots\;per\;pole}}$

$\beta = \frac{{180^\circ }}{{240/8}} = 6^\circ $

${K_d} = \frac{{\sin \frac{{10 × 6}}{2}}}{{10\sin \frac{{6}}{2}}} = \frac{{\sin 30^\circ }}{{10\sin 3^\circ }}$

Answer.3. 0.67

Explanation

The breadth factor or a distribution factor is given by

${k_d} = \frac{{\sin n \cdot \frac{{m\beta }}{2}}}{{n \cdot \sin \frac{{m\beta }}{2}}}$ (for n^th harmonics)

m = no. of slots per pole per phase

β = slots angle

Calculation:

3 – phase, 4 – pole synchronous machine.

No. of slots = 36 (stator slots)

$m = \frac{{36}}{{3\; \times \;4}} = 3$

$\beta = \frac{{180^\circ }}{{slots\;per\;pole}}$

$\beta = \frac{{180^\circ }}{{36/4}} = 20^\circ $

So, ${k_d} = \frac{{\sin \frac{{\left( 3 \right)\left( 3 \right)\; \times \;20^\circ }}{2}}}{{3\sin \frac{{3\; \times \;20}}{2}}}$

$\Rightarrow {k_d} = \frac{{\sin 90^\circ }}{{3\sin 30^\circ }} = \frac{2}{3}$

⇒ k_d = 0.67

Answer.1. $\frac{1}{{3\sqrt2}}\cot \left( {{{15}^0}} \right)$

Explanation

Pitch factor:

Pitch factor is defined as the ratio of the induced emf per coil when the winding is short-pitched to the induced emf per coil when the winding is full pitched.

It is given by Kp = cos (α/2)

Where α is the short pitch or chording angle

Distribution factor:

It is defined as the ratio induced emf with distributed winding to the induced emf with a concentrated winding.

It is given by ${k_d} = \frac{{\sin \frac{{mγ }}{2}}}{{m\sin \frac{γ }{2}\;}}$

Where,

mγ = phase spread

γ = slot angle in degrees = (P × 180°) / S

m = slots / pole / phase

Winding factor:

Winding factor Kw = Kp × Kd

Calculation:

Slot angle γ = (180° × 6) / 36 = 30°

m = 36 / (6×2) = 3

${k_d} = \frac{{\sin \frac{{mγ }}{2}}}{{m\sin \frac{γ }{2}\;}}$ = $ \frac{{\sin \frac{{3 ×30 }}{2}}}{{3\sin \frac{30 }{2}\;}}$

⇒ Kd = (sin 45° / 3 sin 15°) = 1/ (3√2) sin 15°

Winding is short pitched by 1 slot

Slots/pole = 36/6 = 6

6 slots = 180 ° ⇒ 1 slot = 30 °

⇒ Slot pitch angle α = 30°

Pitch factor Kp = cos α /2 = cos (30°/2) = cos 15°

Winding factor Kw = (1/ (3√2) sin 15°) × cos 15° = $\frac{1}{{3\sqrt2}}\cot \left( {{{15}^0}} \right)$

Answer.1. Concentrated winding

Explanation

A winding with only one slot per pole per phase is called a concentrated winding. In this type of winding, the emf generated/phase is equal to the arithmetic sum of the individual coil EMFs in that phase.

However, if the coils/phase are distributed over several slots in space (distributed winding), the EMFs in the coils are not in phase (i.e. phase difference is not zero) but are displaced from each by the slot angle α. (The angular displacement in electrical agrees between the adjacent slots is called slot angle).

Answer.2. Distributed winding

Explanation

If the coils/phase are distributed over several slots in space (distributed winding), the EMFs in the coils are not in phase (i.e. phase difference is not zero) but are displaced from each by the slot angle α. (The angular displacement in electrical agrees between the adjacent slots is called slot angle).

A winding with only one slot per pole per phase is called a concentrated winding. In this type of winding, the emf generated/phase is equal to the arithmetic sum of the individual coil EMFs in that phase.

Answer.3. Distributed winding EMF/Concentrated winding EMF

Explanation

The distribution factor kd is defined as the ratio of Distributed winding EMF to the Concentrated winding EMF

Mathematically it is expressed as

k_d =  Distributed winding EMF/Concentrated winding EMF