Alternator is a machine which generates an alternating emf

Alternator runs at constant speed called synchronous speed, hence they have

given the name synchronous Generator

Alternator consists of two parts

Stator

Rotor

It is a static part and outer part

Stator consists of steel frame which encloses

hollow cylindrical core

Core is made up of laminated silicon steel to

reduce Eddy current loss and hysteresis loss

Core has slots on its inner periphery

3 phase star connected winding is placed in

the slots and it is called armature winding.

It is a rotating part it is placed inside the stator

Rotor is mounted on the shaft which is connected to the prime mover

It has field windings which are excited by the pilot exciter which is mounted

on the shaft

There are two types of rotors

Salient pole type rotor

Non salient pole type rotor

Poles are mounted on the steel frame called magnetic

wheel

Field windings are connected in a series such a way

that when they are excited by DC exciter alternative

poles behaves as NORTH-SOUTH poles

Poles are projected

Used for low speed

Large number of poles N=120f/P

Water turbine is prime mover

Rotor is made of smooth solid cylinder having

numerous slots on outer periphery

Field windings are placed in these slots

Regions which are left un slotted form the poles

Poles are not projected

Used for high speed

Less number of poles N=120f/P

Steam turbines are prime movers

It works on the principle of electromagnetic

induction

Rotor field windings are excited by 110V Or

220 V DC voltage generated by pilot exciter

Rotor is rotated by prime mover

Flux produced by the rotor conductor sweeps

across the stator conductor and hence EMF is

induced.

Ns = Synchronous Speed = Rotar Speed in r.p.m

P = number of poles

F = frequency of e.m.f in Hz

No of cycles/revolution = P/2 ……………..(1)

No of revolution / second = Ns/60………….(2)

(1)*(2)

No of cycles / Second = No of cycles/revolution X No of revolution/Second

f = P/2 X Ns/60

f = Ns P/120

OR

Ns = 120 f/P

Winding Terminology

Conductor

Turn

Coil

Coil Side

Pole pitch

n = 180 electrical = slots / pole

Slot angle(𝛽)

𝛽 = 1 slot angle = 180

ƞ

Coil Span

Full pitch

Short pitch

Angle of short pitch (𝛼)

𝛼 = 𝛽 𝑋 𝑁𝑜 𝑜𝑓 𝑠𝑙𝑜𝑡𝑠 𝑏𝑦 𝑤ℎ𝑖𝑐ℎ 𝑐𝑜𝑖𝑙𝑠 𝑎𝑟𝑒 𝑠ℎ𝑜𝑟𝑡 𝑝𝑖𝑡𝑐ℎ Concentrated windings :-

Distributed Windings

Pitch Factor (Kp)

Kp = Cos(𝛼

2)

Distribution Factor (Kd)

Kd = 𝑆𝑖𝑛(

𝑚𝛽

2)

𝑚 𝑆𝑖𝑛(𝛽

2)

Winding factor [Kw]:

Kw = Kp X Kd

EMF equation of an alternator

Let Zph = No of stator conductors/Phase

Zph = 2Tph

Or Tph = Zph/2 = No of turns / Phase

P = No of Poles

f = Frequency of the induced e.m.f in Hz

∅ = flux/pole in weber

The flux cut by the conductor in one revolution is

d∅ = 𝑃∅

The time taken for one revolution is

dt =60

𝑁𝑠

Hence Average e.m.f induced in one conductor

= 𝑑∅

𝑑𝑡

= 𝑃∅ 60

𝑁𝑠

= 𝑁𝑠 𝑃∅

60

Since there are Zph conductors/phase

Hence Average e.m.f / phase = 𝑁𝑠 𝑃∅

60 XZph

Eavg = 𝑍𝑝𝑕 𝑃∅

60 X Ns

Eavg = 𝑍𝑝𝑕 𝑃∅

60 X

120𝑓

𝑃

Eavg/phase = 2 Zphf∅………………………(1)

W . K . T form factor , Kf = 𝐸𝑟𝑚𝑠

𝐸𝑎𝑣𝑔

Erms = Kf Eavg

Erms/phase = Kf Eavg/phase

Erms/phase = 1.11 Eavg/phase ……………….(2)

(1) In (2) we get

Erms/phase = 1.11 X 2 Zph f ∅

Eph = 2.22 Zph f ∅

Due to Kp and Kd the induced e.m.f gets reduced by a small quantity

Eph = 2.22 Kp Kd Zph f ∅

Voltage Regulation: Change in terminal voltage from no load to full load

% Regulation = 𝐸 −𝑉

𝑉 X 100%

E = No load voltage/phase

V = Full load voltage/phase

Important formulae

Ns = 120𝑓

𝑃 OR P =

120𝑓

𝑁

Were , Ns = Synchronous speed or rotor speed in r.p.m

f = frequency of e.m.f in Hz

P = No of poles

Tph = Zph / 2

were , Tph = No of turns conductor /phase

Zph = No of stator conductors / phase

n = 𝑇𝑜𝑡𝑎𝑙 𝑁𝑜 𝑜𝑓 𝑠𝑙𝑜𝑡

𝑝𝑜𝑙𝑒

were , n = Slots per pole

𝛼 = 180

𝑛

were 𝛼 = Slot angle

n = Slots per pole

𝛽 = 𝑛𝑜 𝑜𝑓 𝑠ℎ𝑜𝑟𝑡 𝑝𝑖𝑡𝑐ℎ𝑒𝑑 𝑋 𝛼

Were, 𝛽 =

Kp = cos (𝛽

2)

were , Kp =

𝛽 =

Kd = =sin (𝑞

𝛼

2)

𝑞 sin (𝛼

2) =

q = no of slots / pole per phase

Eph = 4.44 f ∗ ∅ ∗ 𝑇𝑝ℎ ∗ 𝐾𝑝 ∗ 𝐾𝑑

were , Eph =

f =

Tph =

Kp =

Kd =

1) A six pole dc generator is running and producing a voltage at a frequency of 60 Hz . Calculate the revolutions

per minute of the generator . If the frequency of the generated voltage is required to be decreased to 20 Hz , how

many poles would be needed on the generator ,if it still runs at same speed ?

The speed of rotation is given by

Ns = 120𝑓

𝑃

If the frequency is decreased to 20 Hz , required number of poles ,

P = 120𝑓

𝑁

2) A 4-pole ,3 phase alternator has 36 slots . It has an armature winding which is short-pitched by one slot . Calculate

its coil span factor

Coil span factor as

Kp = cos (𝛽

2)

As the winding is short pitched by one slot , the angle of short pitch is given by

𝛽 = 1 𝑋 𝛼

Since one pole corresponds to 180 degree and there are n slots per pole , the angle corresponds to one slot or its slot

angle is

𝛼 = 180

𝑛

The number of slots per pole

n = 𝑇𝑜𝑡𝑎𝑙 𝑁𝑜 𝑜𝑓 𝑠𝑙𝑜𝑡

𝑝𝑜𝑙𝑒

3) An 8-pole , 3 phase , 120 slots alternator has distributed armature winding . Calculate its distribution

factor

Distribution factor as

Kd = sin (𝑞

𝛼

2)

𝑞 sin (𝛼

2)

Slot angle 𝛼 = 180

𝑛

Number of slots per pole ,

q = 𝑛

3

The number of slot per pole

n = 𝑇𝑜𝑡𝑎𝑙 𝑁𝑜 𝑜𝑓 𝑠𝑙𝑜𝑡

𝑝𝑜𝑙𝑒

4) Determine the distribution factor for a machine having 9 slots per pole for the following cases

a) A three phase winding with 120 degree phase group

b) A three phase winding with 60 degree phase group

The slot angle , 𝛼 = 180

𝑛

a) Since one phase occupies 120 degree , the number of slots in one phase group ,

q = 120

20

Kd = sin(𝑞

𝛼

2)

𝑞 sin(𝛼

2)

b) Since one phase occupies 60 degree , the number of slots in one phase group ,

q = 60

20

Kd = sin (𝑞

𝛼

2)

𝑞 sin (𝛼

2)

5) A 2-pole , 3-phase alternator running at 3000rpm has 42 slots with 2 conductors per slot . Claculate the flux per

pole required to generate line voltage of 2300 v . Assume Kd = 0.952, Kp = 0.956.

Ns = 120𝑓

𝑃

f =

Zph = Z/3

Z = slots X P

Zph =

Tph = Zph / 2

Assuming star connected armature ,

Eph = 𝐸𝐿

3

Eph =

Eph = 4.44 f ∗ ∅ ∗ 𝑇𝑝ℎ ∗ 𝐾𝑝 ∗ 𝐾𝑑

∅ =

6) A 3-phase , 16 pole , star –connected alternator has 144 slots on armature periphery . Each slot contains 10

conductors . It is driven at 375 r.p.m . The line value of e.m.f available across the terminals is observed

to be 2.657 KV . Find the frequency of the induced e.m.f and the flux per pole

Give ,P = 16 , Ns = 375 rpm ,EL = 2.657KV , slots = 144 , conductors/slot = 10 , f = ? Kp = ? Kd = ? ∅ =

Ns = 120𝑓

𝑃 = f = Ns P/120 = ?

Since nothing is mentioned about the pitch , assume full pitch , i.e. ., Kp = 1.

Now , the number of slots/pole , n = slots/pole = ?

The number of slots / pole /phase q = n/phases = ?

The slot angle . 𝛼 = 180

𝑛 = ?

Hence Kd = sin(𝑞

𝛼

2)

𝑞 sin(𝛼

2) = ?

Total number of conductors = Z = Slots X conductors/slot = ?

Conductors / phase , Zph = Z/3 = ?

Turns/phase , Tph = Zph/2

For star connected armature , Eph = 𝐸𝐿

3 = ?

Eph = 4.44 f ∗ ∅ ∗ 𝑇𝑝ℎ ∗ 𝐾𝑝 ∗ 𝐾𝑑

∅ = Eph

4.44 f 𝑇𝑝𝑕 𝐾𝑝 𝐾𝑑