Alternator Classification

1. Electrical Machines-II
6th Semester, EE and EEE
By
Dr. Binod Kumar Sahu
Associate Professor, Electrical Engg.
Siksha ‘O’ Anusandhan, Deemed to be University,
Bhubaneswar, Odisha, India
Lecturer-2

2. 2
Learning Outcomes: - (Lecturer_01)
Students will be able to:
 Know the basics of a synchronous machine.
 Know different types of alternators from application point of view.
 Understand the concept of power generation in hydro and thermal power plant.
 Know the various sources of power generation in India and Odisha.
 Analyse the basic concept of emf generation.

3. 3
Learning Outcomes: - (Today’s Lecturer)
Students will be able to:
 Classify different types of alternators from construction point of view.
 Know the advantages of stationary armature (rotating field) type construction
over rotating armature (stationary field) type construction.
 Know which type of alternator is suitable for a particular application.
 Determine the generated emf in an alternator.

4. 4
So, the basic requirements for generation of emf in rotating electrical machines are :
i. Magnetic field.
ii. Conductor or coil.
iii. Relative motion between the magnetic field and the coil.
Relative motion between the magnetic field and the coil can be achieved in the following
two ways:
 By rotating the coil in a stationary magnetic field (i.e. rotating armature type,
employed in machines with commutator, e.g. DC Machines) or
 By rotating the magnetic field keeping the coil stationary (stationary armature type
or rotating field type, generally employed in large alternators).
Construction of Alternator: -

5. 5
Rotating Armature type construction: -

6. 6
Rotating Field type construction: -

7. 7
N SS N
x-axis (direction of
induced current/emf)
y-axis (Direction
of Magnetic Field)
z-axis
I
I
-
Motion
+ Motion of the
conductor
-
Motion
of pole
N SS NI
+
x-axis (direction of
induced current/emf)
y-axis (Direction
of Magnetic Field)
z-axis
IMotion of the
conductor
A
B
Motion
of pole
Rotating armature type Stationary armature type
A
B


N
S


Rotating Armature type Stationary Armature type


N
S






N
S




Fleming’s Right Hand Rule

8. 8
 It is easier to insulate stationary winding for high voltages for which the
alternators are usually designed as it is not subjected to centrifugal forces.
 Cooling ducts can also be provided in the stationary armature for efficient
cooling of the machine.
 The stationary 3-phase armature can be directly connected to load without using
unreliable slip rings and brushes.
 Only two slip rings are required for DC supply to the field winding on the rotor.
 For armature we need more insulation, thicker conductor and better
ventilation. All these increases it’s weight. So if the armature is placed in the
rotor, rotor weight and inertia becomes excessively high which is not desirable.
 Simple, robust and light construction of the rotor, permits the rotor to rotate at
high speed there by increasing the output obtainable from the machine from a
given machine dimension.
Advantage of stationary armature or rotating field system: -

9. 9
Types of alternator from construction point of view: -
 Any rotating electrical machine consists of two parts: a stationary part called
stator and a rotating part called rotor.
 Generally, in an alternator stator is the armature and rotor is the field
system, which produces the magnetic field.
Stator: -
 Stator is the stationary part of the alternator and contains 3-phase armature
windings.
 Stator core is built up of silicon steel laminations to reduce iron losses.
 The laminations are provided with slots on its inner periphery and are packed
tightly together by cast iron frame.
 Open slots are used as they allow easy installation of stator coils and easy removal
in case of repair. Coils are insulated before inserting in the slots.
 The three phase windings are placed in these slots and serves as the armature
windings of the alternator. The armature windings are always connected in star and
the neutral is connected to ground.

10. 10
Stator of a turbo alternator: -

11. 11
Stator of a hydro-alternator

12. 12
Rotor: -
Rotor of an alternator can be of two types
i. Salient pole type rotor and
ii. Non-salient pole type or cylindrical pole type rotor.
According to the type of rotor, alternator can be of salient pole type or non-salient pole
type.
Salient pole type alternator Non-salient pole type alternator
Un-slotted
Portion
Projected
Poles

13. 13
Salient pole type Rotor: -
 A salient pole type of rotor consists of large number of projected poles (salient poles)
mounted on a magnetic wheel.
 The projected poles are made up from laminations of steel.
 The rotor winding is provided on these poles and is supported by pole shoes.
 These type of rotors are useful for low speed application generally in the range of 150-
600 rpm (Hydro Alternators) as these are not mechanically that much strong.
 So to have a frequency of 50 Hz,
with a rotor speed of 150 rpm, the number of poles in the rotor must be:
and with a rotor speed 600 rpm, the number of poles must be 10.
 To accommodate large number of poles, the diameter of the rotor and hence the
overall diameter of the alternator has to be more.
 So salient pole alternators have large diameter (3-15 m) but shorter axial length.
 These are generally used for low speed applications i.e. in hydroelectric power plants.
120 120 50
40
150s
f
P
N
 
  

14. 14
Giant shaft connecting turbine and generator
Generator
Turbine

15. 15
Non-salient or cylindrical pole type Rotor: -
 Non-salient pole rotors are cylindrical in shape having parallel slots on it’s outer surface to
carry the rotor windings.
 It is made up of solid steel.
 These are used for high speed applications 1500-3000 rpm for 50 Hz operation. So the
number of rotor poles are either 2 or 4.
 Since the number of poles are less, diameter of the rotor and hence the overall diameter
of the machine is less as compared to salient pole machines.
 So these machines have smaller diameter but large axial length.
 Flux distribution is almost sinusoidal and hence gives better emf waveform.
 Windage loss as well as noise is less as compared to salient pole rotors.
 These are used in Thermal, Nuclear and Gas Power Plants.

16. 16
Non-salient or cylindrical pole type Rotor
Un-slotted
Portion
Slotted
Portion
carrying the
field coils

17. 17
EMF equation of a three phase Alternator: -
Let,
φ = Average value of flux/pole in the air gap in Wb,
Ns = Synchronous speed of the Alternator in rpm,
P = Number of Poles,
Zph = Total number of armature conductors/phase,
Tph = Total number of armature turns/phase =
𝑍 𝑝ℎ
2
f = Frequency of the induced emf =
𝑃𝑁𝑠
120
in Hz.
So, average value of induced emf/conductor,
e = Blv, where, ‘B’ is the average flux density in the air-gap in Wb/m2, ‘l’ is the
length of the conductor in meter and ‘v’ is the velocity in m/sec.
B =
𝑃𝜑
𝜋𝐷𝑙
, and 𝑣 =
𝜋𝐷
60
𝑁 𝑠
=
𝜋𝐷𝑁𝑠
60
.
So, 𝑒 = 𝐵𝑙𝑣 =
𝑃𝜑
𝜋𝐷𝑙
× 𝑙 ×
𝜋𝐷𝑁𝑠
60
=
𝑃𝜑𝑁𝑠
60
= 2𝑓𝜑 volts. (as 𝑓 =
𝑃𝑁𝑠
120
)
RMS value of induced emf/conductor = 𝑓𝑜𝑟𝑚 𝑓𝑎𝑐𝑡𝑜𝑟 × 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑒𝑚𝑓/
𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟


N
S





18. 18
= 1.11 × 2𝑓𝜑 = 2.22𝑓𝜑 volts.
If there are Zph number of conductors connected in series/phase, then RMS value of induced
emf/phase,
= 2.22𝑓𝜑𝑍 𝑝ℎ
= 2.22𝑓𝜑 × 2𝑇𝑝ℎ (as conductors/phase = twice the turns/phase)
= 4.44𝑓𝜑𝑇𝑝ℎ
EMF equation can also be derived as follows:
Average value of induced emf/conductor, 𝑒 =
𝑑𝜑
𝑑𝑡
One revolution of the conductor cuts 𝑃𝜑 amount of flux in
60
𝑁 𝑠
seconds.
So, 𝑑𝜑 = 𝑃𝜑, and 𝑑𝑡 =
60
𝑁 𝑠
.
So, the average induced emf/conductor,
𝑒 =
𝑃𝜑
60
𝑁 𝑠
=
𝑃𝜑𝑁𝑠
60
= 2𝑓𝜑 volt.
N SS N

19. 19
RMS value of emf/conductor = form factor × average value
𝐸 = 1.11 × 2𝑓𝜑 = 2.22𝑓𝜑 volts.
If there are Zph is the total number of conductors/phase, rms value of induced emf/phase
𝐸 𝑝ℎ = 2.22𝑓𝜑𝑍 𝑝ℎ = 2.22𝑓𝜑 × 2𝑇𝑝ℎ = 4.44𝑓𝜑𝑇𝑝ℎ volts.
If the three phase armature windings are connected in star, line value of induced voltage
would be, 𝐸𝐿 = 3𝐸 𝑝ℎ.

20. 20
Numerical related to EMF equation: -
1. Find the number of armature conductors in series for a 11 kV, 10 pole, 3 phase, 50 Hz
alternator with 90 slots. Flux per pole is 0.1016 Wb.
Solution: -
Assuming the alternator to be connected in star, phase value of induced emf/phase,
𝐸 𝑝ℎ =
11×103
3
= 6350.85 𝑉
We know that,
𝐸 𝑝ℎ = 4.44𝑓𝜑𝑇𝑝ℎ
So, the number of turns/phase,
𝑇𝑝ℎ =
𝐸 𝑝ℎ
4.44𝑓𝜑
=
6350.85
4.44×50×0.1016
= 282
So, the number of armature conductors/phase, 𝑍 𝑝ℎ = 2 × 𝑇𝑝ℎ = 2 × 282 = 564.
But number of slots/phase, 𝑆 𝑝ℎ =
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑙𝑜𝑡𝑠
3
=
90
3
= 30.
564 is not a multiple of 30. so we need to select the number of conductors which is close
to 564 but multiple of 30.
So, the number of conductors/phase are 570, with 19 conductors/slot.

21. 21
2. Find the no load line voltage of a star connected 4-pole, 50-Hz alternator from the
following data:
Flux per pole= 0.12 Wb, No. of slots = 48 with 4 conductor/slots.
Solution: -
Total number of armature conductors, 𝑍 𝑝ℎ =
𝑍 𝑇
3
=
48×4
3
= 64.
Number of turns/phase, 𝑇𝑝ℎ =
𝑍 𝑝ℎ
2
=
64
2
= 32.
So, induced emf/phase, 𝐸 𝑝ℎ = 4.44𝑓𝜑𝑇𝑝ℎ = 4.44 × 50 × 0.12 × 32 = 852.48 V.
Line value of no-load voltage, 𝐸 𝐿 = 3 × 𝐸 𝑝ℎ = 3 × 852.48 = 1476.54 V.

22. 22
Thank you