Alternator / Synchronous Generator

Alternator/
Synchronous Generator
EECE 205

DC Generator
Field/Stator
Rotor
Generates Field Flux
Carrying the conductor which
cuts flux & generates voltage
3

AC Generator
StatorRotor
Carrying the conductor
which cuts flux &
generates voltage
Generates Field Flux
4

Advantages of having
stationary armature
 The output current can be led directly from fixed
terminals on the stator (or armature windings) to
the load circuit, without having to pass it through
brush-contacts.
 It is easier to insulate stationary armature winding
for high a.c. voltages, which may have as high a
value as 30 kV or more. It is because they are
not subjected to centrifugal forces and also extra
space is available due to the stationary
arrangement of the armature.
5

Advantages of having
stationary armature
 The sliding contacts i.e. slip-rings are
transferred to the low-voltage, low-power
d.c. field circuit which can, therefore, be
easily insulated.
 The armature windings can be more easily
braced to prevent any deformation, which
could be produced by the mechanical
stresses set up as a result of short-circuit
current and the high centrifugal forces
brought into play.
6

Details of Construction
 Let’s see what is inside an alternator?
7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

 Stator
22

 Rotor
Salient (or projecting) Pole Type
Smooth Cylindrical Type/
Non-salient pole type
23

Speed and Frequency
Let
P = total number of magnetic poles
N = rotative speed of the rotor in r.p.m.
f = frequency of generated e.m.f. in Hz.
Since one cycle of e.m.f. is produced when a pair of
poles passes past a conductor, the number of cycles of
e.m.f. produced in one revolution of the rotor is equal
to the number of pair of poles.
26

Speed and Frequency
No. of cycles/revolution = P/2 and
No. of revolutions/second = N/60
N is known as the synchronous speed, because it is the speed at
which an alternator must run, in order to generate an e.m.f. of the
required frequency.
27

Ac machine windings
Lec Nasim & Capt Kazi Newaj
Faisal
28

Introduction
 The Armature winding of a machine is defined as an
arrangement of conductors' design to produce emfs by
relative motion in a magnetic field.
 Electrical machines employ groups of conductors distributed
in slots over the periphery of the armature.
 The groups of conductors are connected in various types of
series-parallel combination to form Armature winding.
 The conductors connected in series so as to increase the
voltage rating.
 They are connected in parallel to increase the current
rating.
 Some of the commonly used terms associated with
windings are as follow:
Faisal
29

Common Terminologies associated with ac windings
Conductor:
– The active length of a wire or strip in the slot.
Turn:
– A turn consists of two conductors separated from each other by a
pole pitch or nearly so, and connected in series as shown in fig.(a)
– The conductors forming a turn are kept a pole pitch apart in order
that the emf in two are additive to produce maximum resultant emf.
N S
Conductor
Conductor
a) Single turn coil
Pole-pitch
Faisal
30

 Coil: A coil may consist of a single turn or may consist of many
turns, placed in almost similar magnetic position, connected in series.
 Coil-Side: A coil consists of two coil sides, which are placed in two
different slots, which are almost a pole pitch apart.
 The group of conductors on one side of the coil form one coil side
while the conductors on the other side of the coil situated a pole pitch
(or approximately a pole pitch apart) forms the second coil side.
N S N S
Coil side
Conductor
a) Single turn coil b) 3 turns coil
Conductor
Faisal
31

 The connections joining the conductors form the end
connectors or in the mass, the overhang or end winding.
 When the coil sides forming a coil are spaced exactly one pole
pitch a part they are said to be of full-pitch.
 However, the coil span may be less than a pole pitch, in which
case the coil is described as short pitched or chorded.
Overhang
Single turn coil
Pole-pitch
Coil-sides
B D
C
Faisal
32

TYPES OF AC MACHINES WINDINGS
 They are two basic physical types for
the windings. They deal differently
with the mechanical problem for
arranging coils in sequence around
the armature.
 The two types are:
1. Single layer winding and
2. Double layer winding
Faisal
33

1. SINGLE LAYER WINDING
 Fig (a) below shows an
arrangement for a single layer
winding.
 In this type of winding
arrangement one coil side of a
coil occupies the whole of the
slot.
 Single layer winding are not
used for machine having
commutator.
 Single layer winding allow the
use of semi-closed and closed
types of slots.
Coil
side
Semi-closed
slot
Open slot
(a)
Faisal
34

2. DOUBLE LAYER WINDING
 The double layer winding have
identical coils with one coil side
of each coil lying in top half of the
slot and the other coil side in
bottom half of another slot exactly
or approximately one pole pitch.
Fig (a)
 Each layer may contain more than
one coil side in case large
numbers of coils are required (fig
c).
 Figure (c) shows the arrangement
where there are 8 coil sides per
slot. Open slots are frequently
used to house double layer
windings.
Top coil side
(top layer)
Bottom coil side
(Bottom layer)
Top
layer
Bottom
layer
Coil
sides
(a)
(c)(b)Lec Nasim & Capt Kazi Newaj
Faisal
35

NUMBER OF PHASES AND PHASES SPREAD
An ac winding, meant to be user for a 'm' phase
system, should produce emfs of equal
magnitude in all the phase.
These emfs should have identical waveforms
and equal frequency.
Their displacement in time should be y =2/m
electrical radians.
This is obtained by having similar pole phase
groups (a pole phase group is defined as a
group of coils of a phase under one pole) and
arranging the groups to have an effective
displacement of y =2/m electrical radians in
space.
Faisal
36

Short-pitch Winding : Pitch
factor/chording factor
 if the coil sides are placed in
slots 1 and 7, then it is full-
pitched.
 If the coil sides are placed in
slots 1 and 6, then it is short-
pitched or fractional-pitched
because coil span is equal to
5/6 of a pole-pitch. It falls
short by 1/6 pole-pitch or by
180°/6 = 30°
41

Advantages of Short-pitch
Winding
 They save copper of end connections.
 They improve the wave-form of the generated
e.m.f. i.e. the generated e.m.f. can be made to
approximate to a sine wave more easily and
the distorting harmonics can be reduced or
totally eliminated.
 Due to elimination of high frequency
harmonics, eddy current and hysteresis losses
are reduced thereby increasing the efficiency.
42

Disadvantages of Short-
pitch Winding
 The disadvantage of using short-pitched coils is
that the total voltage around the coils is
somewhat reduced.
 Because the voltages induced in the two sides
of the short-pitched coil are slightly out of
phase, their resultant vectorial sum is less than
their arithmetical sum.
43

Pitch factor or coil-span
factor
 The pitch factor or coil-span factor kp or kc is
defined as
 It is always less than unity.
44

factor
 Let ES be the induced e.m.f. in each side of the
coil. If the coil were full-pitched i.e. if its two
sides were one pole-pitch apart, then total
induced e.m.f. in the coil would have been =
2ES
45

factor
 If it is short-pitched by 30° (elect.) their resultant
is E which is the vector sum of two voltage 30°
(electrical) apart.
46

factor
47

factor
 In general, if the coil span falls short of full-pitch by an
angle (electrical)*, then
kc = cosα/2.
 Similarly, for a coil having a span of 2/3 pole-pitch,
kc = cos 60°/2 = cos 30° = 0.866.
 The value of will usually be given in the question, if not,
then assume kc = 1.
48

Example
 Calculate the pitch factor for the under-given
windings :
 (a) 36 stator slots,4-poles, coil-span, 1 to 8
 (b) 72 stator slots, 6 poles, coils span 1 to 10 and
 (c) 96 stator slots, 6 poles, coil span 1 to 12.
49

Example: Solution
 Calculate the pitch factor for the under-given windings :
 (a) 36 stator slots,4-poles, coil-span, 1 to 8
50
(a) Here, the coil span falls
short by (2/9) × 180° = 40°
Hence, α= 40°
kc = cos 40°/2 = cos 20° = 0.94

Distribution or Breadth Factor or
Winding Factor or Spread Factor
51

52
 Coils are not concentrated or bunched in one slot,
but are distributed in a number of slots to form
polar groups under each pole.
 These coils/phase are displaced from each other
by a certain angle. The result is that the e.m.fs.
induced in coil sides constituting a polar group
are not in phase with each other but differ by an
angle equal to angular displacement of the slots.

53
 3-phase single-layer
winding for a 4-pole
alternator.
 Total 36 slots
 9 slots/pole
 3 slots / phase / pole
 Angular displacement
between any two
adjacent slots = 180°/9
= 20° (elect.)

54
 If the three coils were
bunched in one slot, then
total e.m.f. induced in the
three sides of the coil would
be the arithmetic sum of the
three e.m.f.s. i.e. = 3 ES
 Since the coils are
distributed, the individual
e.m.fs. have a phase
difference of 20° with each
other.

55
 Their vector sum will be
 E = ES cos 20° + ES + ES cos 20°
= 2 ES cos 20° + ES
= 2 ES × 0.9397 + ES = 2.88 ES
 The distribution factor (kd) is
defined as

56
 In the present case

57
 General Case
 Let β be the value of angular
displacement between the
slots. Its value is

58
 General Case

Example 59
 Calculate the distribution factor for a 36-slots, 4-
pole, single-layer three-phase winding.

Practice 60
 Find the value of kd for an alternator with 9 slots per
pole for (i) One winding in all the slots

Equation of Induced E.M.F
 In one revolution of the rotor (i.e. in 60./N second)
each stator conductor is cut by a flux of ΦP webers.
62

Equation of Induced E.M.F
If there are Z conductors in series/phase, then Average
e.m.f./phase = 2f ΦZ volt = 4 f ΦT volt
R.M.S. value of e.m.f./phase = 1.11 × 4f ΦT = 4.44f ΦT
volt*.
This would have been the actual value of the induced
voltage if all the coils in a phase were (i) full-pitched
and (ii) concentrated or bunched in one slot (instead
of being distributed in several slots under poles). But
this not being so, the actually available voltage is
reduced in the ratio of these two factors.
63

Advantages of distributed
winding
 It also reduces harmonic emf and so wave form is
improved.
 It also diminishes armature reaction.
 Even distribution of conductors, helps for better
cooling.
 The core is fully utilized as the conductors are
distributed over the slots on the armature
periphery.
64

Disadvantages of
distributed winding
 The voltage generated in distributed winding is
less than the voltage generated in concentrated
winding.
65

Effect of Harmonics on Pitch
and Distribution Factors
 If the short-pitch angle is α degrees (electrical) for the
fundamental flux wave, then its values for different
harmonics are
 for 3rd harmonic = 3α ; for 5th harmonic = 5α and so on.
 pitch-factor, kc = cos α/2 —for fundamental
= cos 3α/2 —for 3rd harmonic
= cos 5α/2 —for 5th harmonic etc.
66

Effect of Harmonics on Pitch
and Distribution Factors
 Similarly, the distribution factor is also different for different
harmonics. Its value becomes
 Frequency is also changed. If fundamental frequency is 50 Hz
i.e. f1 = 50 Hz then other frequencies are :
 3rd harmonic,
f3 = 3 × 50 = 150 Hz, 5th harmonic, f5 = 5 × 50 = 250 Hz etc.
67

Example
 An alternator has 18 slots/pole and the first coil lies in slots 1 and
16. Calculate the pitch factor for (i) fundamental (ii) 3rd harmonic
(iii) 5th harmonic and (iv) 7th harmonic.
68

Example
 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 rpm and the
phase and line e.m.f. Assume full-pitched coil.
69

Practice
 A 3-phase, 32 pole alternator has a star-connected
winding with 288 slots. The flux per pole is 0.06 wb
sinusoidally distributed and the speed is 750 rpm.
Find the line emf. Assume full pitched coil.
70

Practice
 Find the no-load phase and line voltage of a
star-connected 3-phase, 6-pole alternator which
runs at 1200 rpm, having flux per pole of 0.1 Wb
sinusoidally distributed. Its stator has 54 slots
having double layer winding. Each coil has 8
turns and the coil is chorded by 1 slot.
71

Practice
 A 4-pole, 3-phase, 50-Hz, star-connected
alternator has 60 slots, with 4 conductors per slot.
Coils are short-pitched by 3 slots. If the phase
spread is 60º, find the line voltage induced for a
flux per pole of 0.943 Wb distributed sinusoidally
in space. All the turns per phase are in series.
72

Practice
 Calculate the R.M.S. value of the induced e.m.f.
per phase of a 10-pole, 3-phase, 50-Hz alternator
with 2 slots per pole per phase and 4 conductors
per slot in two layers. The coil span is 150°. The
flux per pole has a fundamental component of
0.12 Wb and a 20% third component.
73

Practice
 Calculate the R.M.S. value of the induced e.m.f. per phase of a
10-pole, 3-phase, 50-Hz alternator with 2 slots per pole per phase
and 4 conductors per slot in two layers. The coil span is 150°. The
flux per pole has a fundamental component of 0.12 Wb and a
20% third component.
74

Practice
 A 4-pole, 50-Hz, 3-phase, Y-connected
alternator has a single-layer, full-pitch winding
with 21 slots per pole and two conductors per
slot. The fundamental flux is 0.6 Wb and air-gap
flux contains a third harmonic of 5% amplitude.
Find the r.m.s. values of the phase e.m.f. due to
the fundamental and the 3rd harmonic flux and
the total induced e.m.f.
 Ans: 3,550 V; 119.5 V; 3,553 V
75

Factors Affecting
Alternator Size
 The efficiency of an alternator always increases as its
power increases.
 Power output per kilogram increases as the alternator
power increases.
 However, as alternator size increases, cooling problem
becomes more serious. cooling system becomes ever
more elaborate as the power increases. Low-speed
alternators are always bigger than high speed alternators
of the same power.
76

Alternator on Load
 As the load on an alternator is varied, its terminal
voltage is also found to vary as in d.c. generators.
 This variation in terminal voltage V is due to the
following reasons:
 1. voltage drop due to armature resistance Ra
 2. voltage drop due to armature leakage
reactance XL
 3. voltage drop due to armature reaction
80

Alternator on Load
 (a) Armature Resistance
 The armature resistance/phase Ra causes a
voltage drop/phase of IRa which is in phase with
the armature current I. However, this voltage drop
is practically negligible.
81

Alternator on Load
 (b) Armature Leakage Reactance
 When current flows through the armature
conductors, fluxes are set up which do not cross
the air-gap, but take different paths. Such fluxes
are known as leakage fluxes.
 The leakage flux is practically independent of
saturation, but is dependent on I and its phase
angle with terminal voltage V. This leakage flux
sets up an e.m.f. of self-inductance which is
known as reactance e.m.f. and which is ahead of
I by 90°.
82

Alternator on Load
 (b) Armature Leakage Reactance
 Hence, armature winding is assumed to possess
leakage reactance XL such that voltage drop due
to this equals IXL. A part of the generated e.m.f. is
used up in overcoming this reactance e.m.f.
83

Alternator on Load
 (c) Armature Reaction
 As in d.c. generators, armature reaction is the
effect of armature flux on the main field flux. In
the case of alternators, the power factor of the
load has a considerable effect on the armature
reaction.
 We will consider three cases :
 (i) when load of p.f. is unity
 (ii) when p.f. is zero lagging and
 (iii) when p.f. is zero leading.
84

Alternator on Load
 in a 3-phase machine the combined ampere-turn
wave (or m.m.f. wave) is sinusoidal which moves
synchronously.
 This amp-turn or m.m.f. wave is fixed relative to
the poles, its amplitude is proportional to the load
current, but its position depends on the p.f. of the
load.
85

Alternator on Load
 Consider a 3-phase, 2-pole alternator having a
single-layer winding, as shown in Fig. 37.24 (a).
 For the sake of simplicity, assume that winding of
each phase is concentrated and that the number
of turns per phase is N. Further suppose that the
alternator is loaded with a resistive load of unity
power factor, so that phase currents Ia, Ib and Ic
are in phase with their respective phase voltages.
86

Alternator on Load
 Consider a 3-phase, 2-pole
alternator having a single-layer
winding. For the sake of
simplicity, assume that winding
of each phase is concentrated
and that the number of turns per
phase is N. Further suppose that
the alternator is loaded with a
resistive load of unity power
factor, so that phase currents Ia,
Ib and Ic are in phase with their
respective phase voltages.
87

Alternator on Load
 When Ia has a maximum value, Ib
and Ic have one-half their
maximum values.
 The m.m.f. (= NIm) produced by
phase a-a’ is horizontal, whereas
that produced by other two
phases is (Im/2) N each at 60° to
the horizontal. The total armature
m.m.f. is equal to the vector sum
of these three m.m.fs.
 Armature m.m.f.
= NIm + 2.(1/2 NIm) cos 60°
= 1.5 NIm
88

Alternator on Load
 When Ia has a maximum value, Ib and Ic have
one-half their maximum values.
 The m.m.f. (= NIm) produced by phase a-a’ is
horizontal, whereas that produced by other two
phases is (Im/2) N each at 60° to the horizontal.
The total armature m.m.f. is equal to the vector
sum of these three m.m.fs.
 Armature m.m.f. = NIm + 2.(1/2 NIm) cos 60°
= 1.5 NIm
89

Alternator on Load
 At this instant t1, the m.m.f. of the main field is
upwards and the armature m.m.f. is behind it by
90 electrical degrees.
90

Alternator on Load
 At this instant, the poles are in the
horizontal position. Also Ia = 0, but Ib
and Ic are each equal to 0.866 of their
maximum values. Since Ic has not
changed in direction during the
interval t1 to t2, the direction of its
m.m.f. vector remains unchanged. But
Ib has changed direction, hence, its
m.m.f. vector will now be in the
position shown. Total armature m.m.f.
is again the vector sum of these two
m.m.fs. Armature m.m.f. = 2 × (0.866
NIm) × cos 30° = 1.5 NIm.
91

Alternator on Load
 Armature m.m.f. remains constant with time
 it is 90 space degrees behind the main field
m.m.f., so that it is only distortional in nature.
 it rotates synchronously round the armature
i.e. stator.
92

Alternator on Load
 For a lagging load of zero power factor, all
currents would be delayed in time 90° and
armature m.m.f. would be shifted 90° with respect
to the poles. Obviously, armature m.m.f. would
demagnetise the poles and cause a reduction in
the induced e.m.f. and hence the terminal
voltage.
 For leading loads of zero power factor, the
armature m.m.f. is advanced 90° with respect to
the position. The armature m.m.f. strengthens the
main m.m.f. In this case, armature reaction is
wholly magnetising and causes an increase in the
terminal voltage.
93

Alternator on Load
 Unity Power Factor
 The armature flux is cross-magnetising. The result is
that the flux at the leading tips of the poles is reduced
while it is increased at the trailing tips. However, these
two effects nearly offset each other leaving the
average field strength constant. In other words,
armature reaction for unity p.f. is distortional.For
leading loads of zero power factor, the armature
m.m.f. is advanced 90° with respect to the position.
The armature m.m.f. strengthens the main m.m.f. In
this case, armature reaction is wholly magnetising
and causes an increase in the terminal voltage.
94

Alternator on Load
 Zero P.F. lagging
 The armature flux whose wave has moved
backward by 90°) is in direct opposition to the
main flux. Hence, the main flux is decreased.
Therefore, it is found that armature reaction, in
this case, is wholly demagnetising, with the
result, that due to weakening of the main flux,
less e.m.f. is generated. To keep the value of
generated e.m.f. the same, field excitation will
have to be increased to compensate for this
weakening.
95

Alternator on Load
 Zero P.F. leading
 Armature flux wave has moved forward by 90°
so that it is in phase with the main flux wave.
This results in added main flux. Hence, in this
case, armature reaction is wholly magnetising,
which results in greater induced e.m.f. To
keep the value of generated e.m.f. the same,
field excitation will have to be reduced
somewhat.
96

Synchronous Reactance
 it is clear that for the same field excitation, terminal
voltage is decreased from its no-load value E0 to V
(for a lagging power factor). This is because of
 1. drop due to armature resistance, IRa
 2. drop due to leakage reactance, IXL
 3. drop due to armature reaction.
 The drop in voltage due to armature reaction may
be accounted for by assumiung the presence of a
fictitious reactance Xa in the armature winding. The
value of Xa is such that IXa represents the voltage
drop due to armature reaction.
97

Synchronous Reactance
 The leakage reactance XL (or XP) and the armature reactance Xa
may be combined to give synchronous reactance XS.
Hence, XS =XL + Xa
 Therefore, total voltage drop in an alternator under load is
= IRa + jIXS = I(Ra + jXS) = IZS
 where ZS is known as synchronous impedance of the armature
98

Vector Diagrams of a
Loaded Alternator
 unity p.f.
99

Loaded Alternator
 Lagging p.f.
10
0

Loaded Alternator
 Leading p.f.
10
1

Voltage Regulation
 It is clear that with change in load, there is a
change in terminal voltage of an alternator. The
magnitude of this change depends not only on the
load but also on the load power factor.
 The voltage regulation of an alternator is defined
as “the rise in voltage when full-load is removed
(field excitation and speed remaining the same)
divided by the rated terminal voltage
102

Voltage Regulation
 E0 − V is the arithmetical difference and not the
vectorial one.
 In the case of leading load p.f., terminal voltage
will fall on removing the full-load. Hence,
regulation is negative in that case.
 The rise in voltage when full-load is thrown off is
not the same as the fall in voltage when full-load is
applied.
103

Example 105
 A 3-phase, star-connected alternator supplies a load of 10
MW at p.f. 0.85 lagging and at 11 kV (terminal voltage). Its
resistance is 0.1 ohm per phase and synchronous reactance
0.66 ohm per phase. Calculate the line value of e.m.f.
generated.

Alternator / Synchronous Generator

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