E.M.F. Equation of an Alternator

E.M.F. Equation of an Alternator

AU : Nov. 04, April-99, Dec.-07, 08, 09, 12, 13, 15, 16, May-04, 05, 06, 07, 09,10, 11, 12, De. 17

Let     ϕ = Flux per pole, in Wb,        P = Number of poles

N_s = Synchronous speed in r.p.m., f = Frequency of induced e.m.f. in Hz

Z = Total number of conductors

Z_ph = Conductors per phase connected in series 

Z_ph = Z/3

Consider a single conductor placed in a slot.

The average value of e.m.f. induced in a conductor = dϕ / dt

For one revolution of a conductor

e_avg per conductor = Flux cut in one revolution / Time taken one revolution

Total flux cut in one revolution is ϕ × P

Time taken for one revolution is 60/N_S seconds.

e_avg per conductor = ϕP/(60/N_S) = ϕ PN_S/60 …(1.12.1)

But f = PN_S/120

Hence PN_S/60 = 2f

Substituting in equation (1.12.1),

eavg per conductor = 2f ϕ volts

Assume full pitch winding for simplicity i.e. this conductor is connected to a conductor which is 180° electrical apart. So these two e.m.f.s will try to set up a current in the same direction i.e. the two e.m.f. are helping each other and hence resultant e.m.f. per turn will be twice the e.m.f. induced in a conductor.

E.M.F. per turn = 2 × (e.m.f. per conductor) = 2 × (2f ϕ) = 4f ϕ volts.

Let T_ph be the total number of turns per phase connected in series. Assuming concentrated winding, we can say that all are placed in single slot per pole per phase. So induced e.m.f.s in all turns will be in phase as placed in single slot. Hence net e.m.f. per phase will be algebraic sum of the e.m.f.s per turn.

Average E_ph = T_ph × (Average e.m.f. per turn)

.-. Average E_ph = T_ph × 4 f ϕ

But in a.c. circuits R.M.S. value of an alternating quantity is used for the analysis. The form factor is 1.11 of sinusoidal e.m.f. 

K_f = R.M.S./ Average = 1.11

R.M.S. value of E_ph = K_f × Average value

E_ph = 1.11 × 4 f ϕ T_ph

E_ph = 4.44 f ϕ T_ph volts

Key Point : This is the basic e.m.f. equation for an induced e.m.f. per phase for full pitch, concentrated type of winding.

Where T_ph = Number of turns per phase

T_ph = Z_ph/2 …. As 2 conductor constitute 1 turn

But as mentioned earlier, the winding used for the alternators is distributed and short pitch hence e.m.f. induced slightly gets affected. Let us see now the effect of distributed and short pitch type of winding on the e.m.f. equation.

 

1. Pitch Factor or Coil Span Factor (K_c)

In practice short pitch coils are preferred. So coil is form by connecting one coil side to another which is less than one pole pitch away. So actual coil span is less than 180°. The coil is generally shorted by one or two slots.

Key Point : The angle bp which coils are short pitched is called angle of short pitch denoted as ‘ɑ’.

ɑ = Angle by which coils are short pitched. As coils are shorted in terms of number of slots i.e. either by one slot, two slots and so on and slot angle is 3 then angle of short pitch is always a multiple of the slot angle β.

ɑ = β × Number of slots by which coils are short pitched.

or ɑ = 180° - Actual coil span of the coils.

This is shown in the Fig. 1.12.2.

Now let E be the induced e.m.f. in each coil side. If coil is full pitch coil, the induced e.m.f. in each coil side help each other. Coil connections are such that both will try to set up a current in the same direction in the external circuit Hence the resultant e.m.f. across a coil will be algebraic sum of the two.

E_R = E + E = 2E  ... For full pitch

Now the coil is short pitched by angle 'a', the two e.m.f. in two coil sides no longer remains in phase from external circuit point of view. Hence the resultant e.m.f. is also no longer remains algebraic sum of the two but becomes a phasor sum of the two as shown in the Fig. 1.12.4.

Obviously ER in such a case will be less than what it is in case of full pitch coil.

From the geometry of the Fig. 1.12.4, we can write,

AC is perpendicular drawn on OB bisecting OB.

This is the resultant e.m.f. in case of a short pitch coil which depends on the angle of short pitch 'ɑ'.

Key Point : The factor by which, induced e.m.f gets reduced due to short pitching is called pitch factor or coil span factor denoted by K_c.

It is defined as the ratio of resultant e.m.f. when coil is short pitch to the resultant e.m.f. when coil is full pitched. It is always less than one. It is also defined as the ratio of phasor sum of e.m.f.s to algebraic sum of e.m.f.s.

2. Distribution Factor (K_d)

Similar to full pitch coils, concentrated winding is also rare in practice. Attempt is made to use all the slots available under a pole for the winding which makes the nature of the induced e.m.f. more sinusoidal. Such a winding is called distributed winding. 

Consider 18 slots, 2 pole alternator. So slots per pole i.e. n = 9.

m = Slots per pole per phase = 3 and β = 180°/ 9 = 20°

Let E = Induced e.m.f. per coil and there are 3 coils per phase.

In concentrated type all the coil sides will be placed in one slot under a pole. So induced e.m.f. in all the coils will achieve maxima and minima at the same time i.e. all of them will be in phase. Hence resultant e.m.f. after connecting coils in series will be algebraic sum of all the e.m.f.s as all are in phase.

As against this, in distributed type, coil sides will be distributed, one each in the 3 slots per phase available under a pole as shown in the Fig. 1.12.5 (a).

Fig. 1.12.5

Though the magnitude of e.m.f. in each coil will be same as 'E', as each slot contributes phase difference of 3° i-e. 20° in this case, there will exist a phase difference of 3° with respect to each other as shown in the Fig. 1.12.6 (b).

Hence resultant e.m.f. will be phasor sum of all of them as shown in the Fig. 1.12.6. So due to distributed winding resultant e.m.f. decreases.

Key Point : The factor by which there is a reduction in the e.m.f. due to distribution of coils is called distribution factor denoted as K_d.

a. Derivation of Distribution Factor

In general let there be 'n' slots per pole and zm' slots per pole per phase. So there will be 'm' coils distributed under a pole per phase, connected in series. Let E be the induced e.m.f. per coil. Then all the'm' e.m.f.s induced in the coils will have successive phase angle difference of β = 180^o/n. While finding out the phasor sum of all of them,  phasor diagram will approach a shape of a 'm' equal sided polygon circumscribed by a semicircle of radius 'R'.

This is shown in the Fig. 1.12.7 (a) AB, BC, CD etc., represent e.m.f. per coil. All the ends are joined at 'O' which is centre of the circumscribing semicircle of radius 'R'.

Fig. 1.12.7 (a) Phasor sum of ‘m’ e.m.f.s

Angle subtended by each phasor at the origin 'O' is 3°. This can be proved as below.

All the triangles OAB, OBC ... are similar and isosceles, as AB = BC = CD = ... = E.

Let the base angles be 'x'.

If 'M' is the last point of the last phasor,

∠ AOM = m × β = mβ and

AM = E_R = Resultant of all the e.m.f.s.

Consider a ∆OAB separately as shown in the Fig. 1.12.7 (b).

Let OF be the perpendicular drawn on AB bisecting angle at apex 'O' as β /2.

Now consider ∆OAM as shown in the Fig. 1.12.7 (a) and OG is the perpendicular drawn from 'O' on its base bisecting ∠OAM.

This is the resultant e.m.f. when coils are distributed. If all zm' coils are concentrated, all would have been in phase giving E_R as algebraic sum of all the e.m.f.s.

E_R = m × E ... For concentrated

 From equation (1.12.4), E = 2R sin (β/2)

E_R = 2 m R sin (β/2)

 This is resultant e.m.f. when coils are concentrated.

The distribution factor is define as the resultant e.m.f. when coils are disturbed to the resultant e.m.f. when coils are concentrated. It is always less than one.

This factor is also called winding factor or breadth factor.

 

3. Generalized Expression for E.M.F. Equation of an Alternator

Considering full pitch, concentrated winding,

E_ph = 4.44 f ϕ T_ph volts.

But due to short pitch, distributed winding used in practice, this Eph will reduce by factors K_c and K_d. So generalized expression for e.m.f. equation can be written as

E_ph = 4.44 K_c K_d f ϕ T_ph volts.

For full pitch coil, K_c, = 1.

For concentrated winding K_d = 1. 

Key Point : For short pitch and distributed winding Kc and Kd are always less than unity.

 

Example : 1.12.1 Calculate the distribution factor for a 36 slots, 4 poles, single layer three phase winding.

Solution :  

 

Example : 1.12.2 Calculate the pitch factor for the given winding : 36 stator slots, 4 poles, coil

Solution :  

36 slots, P = 4, Coil span 1-8.

n = Slots / Pole = 36 / 4

For full pitch, coil span must be 1-9 but actually it is 1- 8. Thus coils are short pitched by 1 slot. Hence short pitch angle is one slot angle.

β  = slot angle = 180° / n = 20°

ɑ = Angle of short pitch = 1 slot angle = 20°

K_c = Pitch factor = cos ɑ /2 = cos 10° = 0.9848

 

4. Line Value of Induced E.M.F.

If the armature winding of three phase alternator is star connected, then the value of induced e.m.f. across the terminals is √3 E_ph where E_ph is induced e.m.f. per phase.

While if it is delta connected line value of e.m.f. is same as E_ph.

This is shown in the Fig. 1.12.8 (a) and (b).

Advantages of star connection :

Practically most of the alternators are star connected due to following reasons :

1. Neutral point can be earthed from safety point of view.

2. For the same phase voltage, voltage available across the terminal is more than delta connection. 

3. For the same terminal voltage, the phase voltage in star is 1/√3 times line value.

This reduces strain on the insulation of the armature winding.

 

Example 1.12.3 A 3 phase, 50 Hz, 16 pole star connected alternator has a stator winding with 144 slots with 10 conductors/slot. The magnetic flux/pole is 0.03 webers and is sinusoidally distributed in space. The coil pitch of the winding is 8 slots. Estimate the e.m.f induced between the lines of the alternator.

Solution :   P = 16, f = 50 Hz, 144 slots, 10 conductors/slot

ϕ = 0.03 Wb, Coil pitch = 8 slots.

n = Slots / pole = 144/16 = 9   and m = Slots / pole / phase

= n/3 = 9/3 = 3

β = 180° / n = 180°/9  = 20^o

For full pitch, coil pitch = n = 9 slots but actual coil pitch is 8 slots. So coils are short pitched by 1 slot.

ɑ = 1 slot angle = β = 20°

 

Example 1.12.4 A 3-phase 16 pole alternator has star connected winding with 144 slots and10 conductors per slot. The flux per pole is 0.04 Wb and is distributed sinusoidally. The speed is 375 rpm. Find the frequency, phase emf and line emf. The coil span is 120° electrical AU : Dec.-13, Marks 8

Solution :

 

Example 1.12.5 A three phase, star-connected 16 pole alternator has 192 slots with 8 conductors/slot, coil spain = 160 electrical degrees, speed of alternator = 375 r.p.m., flux/pole = 55 mWb. Calculate the phase and line voltages. AU : Nov.-04, Marks 8

Solution : Refer Ex. 1.12.4 for the procedure and verify the answer as E_line = 5 kv.

 

Example 1.12.6 A 3 ph, Y connected alternator has the following data : Voltage required to be generated on O.C. is 4000 V at 50 Hz, speed is 500 r.p.m., stator slots/pole/ph is 3, conductors/slot is 12. Calculate the number of poles and useful flux/pole. Assume all conductors/ph to be connected in series and coil to be full pitched. AU : May-05, Marks 8

Solution :

 

Example 1.12.7 A 3 phase, 8 pole, 50 Hz, star connected alternator has 96 slots with 4 conductors per slot. The coil pitch is 10 slots. If the flux per pole is 60 mWb find

i) The phase voltage

ii) The line voltage

iii) If each phase is capable of carrying 650 A, what is the kVA rating of the machine ?

Solution : Given values are,

P = 8, f = 50 Hz,   ϕ = 60 mWb,

Coil pitch = 10 slots, No. of slots = 96

 

Example 1.12.8 A 3 phase, 6 pole, star-connected alternator revolves at 1000 r.p.m. The stator has 90 slots and 8 conductors per slot. The flux per pole is 0.05 Wb (sinusoidally distributed). Calculate the voltage generated by the machine if the winding factor is 0.96. AU : May-06, Marks 6

Solution :  Refer Ex. 1.12.4 for the procedure and verify the answer as

E_line = 2214.808 V

 

Example 1.12.9 Find the no load phase and line voltage of a star connected 3 phase, 6 pole alternator which runs at 1200 r.p.m., having flux per pole of 0.1 Wb sinusoidally distributed. It’s stator has 54 solts having double layer winding. Ecah coil has 8 turns and the coil chorded by 1 slot.

Now winding is double layer means in each slot there are two coil sides and each coil has 8 turns. Thus each coil side has 8 conductors as shown in the Fig. 1.12.9.

 

Example 1.12.10 A 12 pole, three phase, 600 r.p.m., star connected alternator has 180 slots. There are 2 coil sides per slot and total 10 conductors per slot. If flux per pole is 0.05 Wb determine from first principles,

i) R.M.S. value of e.m.f. in a conductor

ii) R.M.S. value of e.m.f. in a turn

iii) R.M.S. value of e.m.f. in a coil iv) Per phase induced e.m.f.

Assume full pitch coils. 

Solution :

 

Example 1,12,11 A single phase 1500 r.p.m., 4 pole alternator has 8 conductors per slot with total of 24 slots. The winding is short pitched by th of full pitch. Assume distributed winding with flux per pole as 0.05 Wb. Calculate the induced e.m.f.

Solution : N_S = 1500 r.p.m., P = 4, Slots = 24; Conductors/slot = 8

Note that the alternator is single phase and not the three phase.

 

Example 1.12.12 A 4-pole alternator has an armatuure with 25 slots and 8 conductors per slot and rotates at 1500 r.p.m. and the flux per pole is 0.05 Wb. Calculate the e.m.f. generated, if winding factor is 0.96 and all the conductors are in series. AU : Dec.-12, Marks 4

Solution : P = 4, N₅ = 1500 r.p.m., 25 slots, 8 conductors/slot,

 

 Examples for practice

Example 1.12.13 A 3 phase, 16 pole alternator has a star connected winding with 144 slots and 10 conductors per slot. The flux per pole is 0.03 Wb, sinusoidally distributed and the speed is 375 r.p.m. Find the frequency and line voltage.

[Ans.: f = 50 Hz, E_line = 2656.94 V]

Example 1.12.14 In a 3 phase, star connected alternator, there are 2 coil sides per slot and 16 turns per coil. Armature has 288 slots on its periphery. When driven at 250 r.p.m. it produces 6600 V between the lines at 50 Hz. The pitch of the coil is 2 slots less than the full pitch. Calculate the flux per pole.

[Ans.: 0 = 12 mWb]

Example 1.12.15 Calculate the distribution factor of a 3-phase winding with 120° phase spread when the winding is i) Uniformly distributed ii) Occupies 6 slots per pole.  

[Ans.: i) 0.8269, ii) 0.8365]

Example 1.12.16 Find the value of Kd for an alternator with 9 slots per pole for the following cases i) One winding in all the slots ii) One winding using only the first 2/3 of the slots / pole iii) Three equal windings placed sequentially in 60° group.

[Ans.: i) 0.6398, ii) 0.8312, iii) 0.9597]

Example 1.12.17 The armature of a single phase alternator is wound completely with T single turn coils, which are uniformly distributed. The induced e.m.f. in each turn is 2 V (r.m.s.). Calculate the e.m.f. of the whole winding with T number of coils connected in series.    

[Ans . E _ph = 4 T/π V]

Example 1.12.18 A 3-phase, 16 pole, star connected alternator has 144 slots having 10 conductors in each slot. The flux per pole is 30 mWb and distributed sinusoidally and the speed is 375 r.p.m. Find the induced e.m.f. for ; i) Full pitch winding ; ii) Coils short pitched by 1 slot ; iii) Coils short chorded by 2 slots.

[Ans.: 1533.98 V, 2656.94 V, 1510.68 V, 2616.574 V, 1441.3318 V, 2496.46 V]

Example 1.12.19 A 3 - 0, 50 Hz, 10 pole alternator has 90 slots. The flux per pole is 0.15 Wb. If the winding is to be star-connected to give a line voltage of 11000 V, find the number of armature conductors to be connected in series/phase.

[Ans.: 396] 

Example 1.12.20 A 3 phase, 4 pole, 50 Hz star connected alternator has flux per pole of 0.12 Wb. The slots per pole per phase is 4 and the number of conductors per slot are 4. If the winding coil span is 150°, calculate the emf generated

[Ans.: 788.5472 V, 1365.8038 V]

Review Questions

1. Derive the e.m.f. equation of an alternator. Explain the pitch factor and distribution factor.

AU : May-04, 06, 07, 10, 11, Dec.-04, 06, 10, 11, 12, 13, 14, 15, 16, Dec.-17 Marks 12

2. Derive the expression for the distribution factor and coil span factor.