• Number Of Slots Per Pole Per Phase Formula Cost
• Number Of Slots Per Pole Per Phase Formula Per
• Number Of Slots Per Pole Per Phase Formula Calculator
• Number Of Slots Per Pole Per Phase Formula Sheet

Definition

Distribution factor is defined as the ratio of phasor sum of coil emfs to the arithmetic sum of coil emfs. It is also known as Belt or Breadth factor and denoted by k_d.

A 200 KW, 600 V DC, 6 pole dc shunt generator has an armature with 18 slots per pole. It has a simple lap winding with 4 conductors per slot. The armature and shunt field winding has resistance of 0.1 Ω and 50 Ω respectively. The flux per pole is 0.07 Wb. Calculate the speed at which it is to be driven for rated load operation. If the number of slots per pole per phase 𝒒𝒒= 𝑺𝑺 𝒎𝒎𝑷𝑷. Is an integer, then the winding is called an. Integral-slot winding in case the number of slots per pole per phase, q is not an integer, the winding is called. Fractional-slot winding. For example a 3-phase winding with 36 slots and 4 poles is an integral slot. A 30 amp two pole breaker allows up to 30 amps per leg or phase to conduct in fact provided a 'total' of 60 amps. However, 220 volt devices require the rated amperage for each leg. The number of slots should be selected to give an integral number of slots per pole per phase. The stator slot pitch at the air gap surface should be between 1.5 to 2.5 cm. Stator slot pitch at the air gap surface = τ.

In electrical machine, armature winding is distributed in the slots. It is not concentrated in a single slot. Therefore the total emf generated across the armature terminal is the phasor sum of emfs induced in individual coils. If these coils were concentrated, then the total induced emf would have been arithmetic sum of induced emf in a single turn of coil. Thus to find the factor of distribution or the degree of distribution of coils, distribution factor has been defined as the ratio of emf induced in distributed winding to the emf induced in concentrated winding. This definition of distribution factor is same stated earlier in the post.

Calculation of Distribution Factor

To get a general expression or formula of distribution factor, let us assume a 2 pole, 3 phase electrical machine having a total of 18 slots in its stator. These slots are distributed along the periphery of stator in 2 pole pitches. Therefore the angle between any two consecutive slots will be equal to 20 degree (2×180 / 18 = 20). This angle between two consecutive slots is called Angular Slot Pitch. It is in electrical degree not in mechanical degree.

Thus angular slot pitch for our example is 20 degree. The number of slots per pole per phase will be 3 [(18/2)/3 = 3]. This means that a particular phase will consists of three coils distributed in three slots under a single pole. This is shown in figure below.

In the above figure it can be seen that, coils are distributed in slot number 1,2,3 and 10,11,12. It can also be observed that, coil sides are connected in series in such a way that emf induced in coil side at 1 and 10 are additive. This means, if E₁ and E₁₀ are the emf induced in respective coil sides then the total emf induced will be E₁+E₁₀. Similarly, if the emf induced in coil sides 2 and 11 are E₂ & E₁₁, then the total emf in coil 2-11 is given as arithmetic sum of E₂ and E₁₁.

Thus the total emf at the coil terminal A & B is the phasor sum of (E₁+E₁₀), (E₂+E₁₁) and (E₃+E₁₂). Let us assume (E₁+E₁₀) as reference phasor and draw the phasor diagram of the above three emf to get the total emf available at terminal AB.

In the above phasor, ab represents (E₁+E₁₀). Since the slots are displaced from each other by angular slot pitch of γ (in our example, it is 20 degree), this means the emf induced in coil 2-11 will be displace by an angle of γ from reference phasor ab. This is shown by ‘bc’ in above phasor. Similarly, ‘cd’ represents the emf induced in coil 3-12 and is displaced by reference phasor by and angle of 2γ.

Thus the total emf across terminals AB of coil will be the phasor sum of ab, bd and cd which is equal to ‘ad’. Let us now calculate this phasor sum.

Since ab, bc and cd are lying in stator slot, therefore their perpendicular bisector must meet at the centre of the circle. Let us now draw perpendicular bisector oe and of on ab and ad respectively as shown in above figure. In general, the angle aob should be equal to angular slot pitch γ, angle aod equal to qγ and angle aof equal to (qγ/2) where q is number of slots per pole per phase.

In right angled triangle aoe,

Sin (γ/2) = ae / oa

ae = oaSin(γ/2)

Thus, ab = 2oaSin(γ/2) as oe is perpendicular bisector of ab.

Therefore, if the coil were concentrated then the emf across coil terminal AB would have been arithmetic sum of emf. This is given as

Emf per coil = Number of slots per pole per phase x 2oeSin(γ/2)

= 2q(oa)Sin(γ/2)

Now, in right angled triangle aof,

Sin(qγ/2) = af / oa

af = (oa)Sin(qγ/2)

Hence the resultant emf ad equal to the phasor sum of ab, bc and cd is given as

ad = 2(oa)Sin(qγ/2)

Therefore as per the definition of distribution factor,

Kd = Phasor Sum of Coil emf / Arithmetic sum of coil emf

= ad / q ab

= [2(oa)Sin(qγ/2)] / [2q(oa)Sin(γ/2)]

= Sin(qγ/2) / qSin(γ/2)

Thus the formula for distribution factor is given as below.

Number Of Slots Per Pole Per Phase Formula Cost

where q is number of slots per pole per phase and γ is angular slot pitch.

Effect of Harmonics on Distribution Factor

In electrical machine, every effort is made to make the magnetic flux density wave sinusoidal in space. But practically, it is never distributed sinusoidally in space. Rather it contains various harmonics. Out of various harmonics, third harmonic is the most dominant. A third harmonic component of flux density wave may be assumed to be produced by 3 poles as compared to the one pole for fundamental component. In view of this, the angular slot pitch for third harmonic will be 3γ and hence nγ for n^th harmonics.

Thus the distribution factor for nth harmonics is given as below.

Normally distribution factor for n^th harmonics K_dn is less than K_d. This directly means that the emf generated due to nth harmonic component of flux density wave is less than that generated by fundamental component. Thus distribution of armature winding over the slots reduces the harmonics in generated emf and thus making it to approach the sine wave. This is one of the advantages of distributed winding.

As we know that principle of induction motor is very similar to the transformer with some differences. An induction motor at standstill condition is similar to a transformer at no load condition. Therefore, the method of drawing the induction motor phasor diagram is also same as that of a transformer phasor diagram. In this post we will discuss the phasor diagram of induction motor at standstill condition and at full load slip.

Induction Motor Phasor Diagram at Standstill Condition:

Before going into the phasor diagram, there are some important points to be taken care:

• Per phase value of induced emf E₁ in the stator winding is given as below

E₁ = √2πf₁k_w1N₁Ø

where f1 = supply frequency

N1 = Number of series turns per phase

Ø = resultant air gap flux per pole

k_w1 = Stator winding factor

• Per phae value of induced emf E2 in rotor winding is given as

E₂ = √2πf₂k_w2N₂Ø

where f₂ = frequency of induced emf in rotor = sf₁

N₂ = Number of series turns per phase

Ø = resultant air gap flux per pole

k_w2 = Rotor winding factor

• Total air gap mmf F_r of induction motor is the sum of stator mmf (F₁) and rotor mmf (F₂).
• Magnetizing current I_m taken by stator winding from the supply always remains in phase with the resultant flux Ø.
• The induced emf always lags behind the resultant flux Ø by 90°.

Now we are at a stage to draw the induction motor phasor diagram. Let us take the resultant air gap flux Ø as the reference. This flux Ø will be in phase with the resultant mmf F_r. Also, the induced emf E₁ and E₂ in stator and rotor winding will lag behind the Ø by 90°. This is shown in the below phasor diagram of induction motor.

Since the rotor mmf counteracts the stator produced mmf as per lenz’s law, therefore stator takes extra current from supply to counterbalance the effect of rotor current. Therefore under normal condition,

Stator mmf = Rotor mmf

N₁’I₂^’ = N₂’I₂

where N₁’ and N₂’ are the effective stator and rotor turns per phase.

This component of stator current is called the load component. In addition to load component, stator also takes magnetizing current I_m to build magnetic flux in the air gap. Thus the total stator current I₁ = I₂’ + I_m. This is shown in the above phasor diagram. I₂’ is shown opposite to the rotor current I₂ for the reason discussed above.

At standstill condition, E₂ = I₂ (r₂ + jx₂). The core loss component of stator current I_c is in phase phase with V₁’ or –E₁. At standstill condition, the friction and windage loss is zero, therefore the stator no load current is given as

I₀ = I_m + I_c

Since stator applied voltage V₁ must balance the stator back emf V₁’ or -E1, stator impedance drop I₁(r₁ + jx₁), therefore we can write

V₁ = V₁’ + I₁(r₁ + jx₁) ……(1)

Similar equation exists for rotor circuit and can be written as

Number Of Slots Per Pole Per Phase Formula Per

E₂ = I₂ (r₂ + jx₂) ……..(2)

The above equations are applied for drawing induction motor phasor diagram as shown in above figure. It can be easily seen from the above phasor diagram that, the power factor (cosɵ) of induction motor at starting is very poor as ɵ is large.

Induction Motor Phasor Diagram at Full Load Slip:

Number Of Slots Per Pole Per Phase Formula Calculator

At full load, the slip s of induction is low. The stator voltage equation (1) do not changes when the motor is loaded. But the rotor voltage equation changes with slip. The rotor induced voltage at any slip s becomes sE₂ and the rotor circuit reactance becomes sx₂. Therefore,

Number Of Slots Per Pole Per Phase Formula Sheet

sE₂ = I₂ (r₂ + jsx₂)

The above rotor equation when implemented, the induction motor phasor diagram will becomes different from the phasor at standstill condition. The induction motor phasor at full load slip s is shown below.

Since at full load condition, some friction and windage loss will exists. This means that stator will have to draw some extra current from the supply main to provide this additional loss. Therefore the total no load current I₀ taken by stator is the sum current I_fc and I_m. It can be seen from the above phasor diagram that power factor of induction motor improves. generally at full load the power factor ranges in between 0.8 to 0.9.