This document introduces AC machines and provides details about their classification and operation. It discusses synchronous generators and motors, as well as induction motors and generators. The key points covered are:
- AC machines are classified as synchronous or induction based on their operation. Synchronous machines include generators and motors while induction machines include motors and rarely generators.
- Generators convert mechanical energy to electrical energy, while motors convert electrical to mechanical. Their construction is similar but operation is reversed.
- Windings in AC machines can be concentrated or distributed. Distributed windings are used in induction machines' stators and rotors as well as in synchronous machine armatures.
- Double layer windings are more common than single layer above
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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ELECTRICAL MACHINES II
Lecturer: Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
Syllabus
I. Introduction to AC Machine
II. Synchronous Generators
III. Synchronous Motors
IV. Three-Phase Induction Machines
V. Three-Phase Induction Motors
VI. Induction Generators
VII. Induction Regulators
Recommended Textbook :
1) M.G.Say
Alternating Current Machines
Pitman Pub.
2) A.S. Langsdorf
Theory of AC Machinery
McGRAW-HILL Pub.
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IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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I. Introduction to AC Machines
Classification of AC Rotating Machines
•Synchronous Machines:
•Synchronous Generators: A primary source of electrical energy.
•Synchronous Motors: Used as motors as well as power factor
compensators (synchronous condensers).
•Asynchronous (Induction) Machines:
•Induction Motors: Most widely used electrical motors in both
domestic and industrial applications.
•Induction Generators: Due to lack of a separate field excitation, these
machines are rarely used as generators.
• Generators convert mechanical energy to electric energy.
Energy Conversion
• Motors convert electric energy to mechanical energy.
• The construction of motors and generators are similar.
• Every generator can operate as a motor and vice versa.
• The energy or power balance is :
– Generator: Mechanical power = electric power + losses
– Motor: Electric Power = Mechanical Power + losses.
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IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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AC winding design
The windings used in rotating electrical machines can be classified as
Concentrated Windings
• All the winding turns are wound together in series to form one multi-turn coil
• All the turns have the same magnetic axis
• Examples of concentrated winding are
– field windings for salient-pole synchronous machines
– D.C. machines
– Primary and secondary windings of a transformer
Distributed Windings
• All the winding turns are arranged in several full-pitch or fractional-pitch coils
• These coils are then housed in the slots spread around the air-gap periphery to
form phase or commutator winding
• Examples of distributed winding are
– Stator and rotor of induction machines
– The armatures of both synchronous and D.C. machines
Armature windings, in general, are classified under two main heads, namely,
Closed Windings
• There is a closed path in the sense that if one starts from any point on the
winding and traverses it, one again reaches the starting point from where
one had started
• Used only for D.C. machines and A.C. commutator machines
Open Windings
• Open windings terminate at suitable number of slip-rings or terminals
• Used only for A.C. machines, like synchronous machines, induction
machines, etc
Some of the terms common to armature windings are described below:
1. Conductor. A length of wire which takes active part in the energy-
conversion process is a called a conductor.
2. Turn. One turn consists of two conductors.
3. Coil. One coil may consist of any number of turns.
4. Coil –side. One coil with any number of turns has two coil-sides.
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IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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The number of conductors (C) in any coil-side is equal to the number of
turns (N) in that coil.
One-turn coil two-turn coil multi-turn coil
5. Single- layer and double layer windings.
Single- layer winding
• One coil-side occupies the total slot area
• Used only in small ac machines one coil-side per slot
Double- layer winding
• Slot contains even number (may be 2,4,6 etc.) of coil-sides in two layers
• Double-layer winding is more common above about 5kW machines
Two coil –sides per slot
4-coil-sides per slot
Coil-
sides
Coil-
sides
Coil -
sides
Overhang
Top layer
Bottom layer
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The advantages of double-layer winding over single layer winding are as follows:
a. Easier to manufacture and lower cost of the coils
b. Fractional-slot winding can be used
c. Chorded-winding is possible
d. Lower-leakage reactance and therefore , better performance of the machine
e. Better emf waveform in case of generators
6. Pole – pitch. A pole pitch is defined as the peripheral distance between
identical points on two adjacent poles. Pole pitch is always equal to 180o
7. Coil–span or coil-pitch. The distance between the two coil-sides of a coil is
called coil-span or coil-pitch. It is usually measured in terms of teeth, slots
or electrical degrees.
electrical.
8. Chorded-coil.
If the coil-span (or coil-pitch) is equal
in case the coil-pitch is
to the pole-pitch, then the coil is
termed a full-pitch coil.
less
if there are S slots and P poles, then pole pitch 𝑸𝑸 =
𝑺𝑺
𝑷𝑷
slots per pole
than pole-pitch, then it is called chorded,
short-pitch or fractional-pitch coil
if coil-pitch 𝒚𝒚 =
𝑺𝑺
𝑷𝑷
, it results in full-pitch winding
in case coil-pitch 𝒚𝒚 <
𝑺𝑺
𝑷𝑷
, it results in chorded, short-pitched or
fractional-pitch
Full-pitch coil Short-pitched or chorded coil
N S
Coil
span
Pole
pitch
N S
Coil
span
Pole
pitch
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IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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In AC armature windings, the separate coils may be connected in several different
manners, but the two most common methods are lap and wave
In polyphase windings it is essential that
The generated emfs of all the phases are of equal magnitude
The waveforms of the phase emfs are identical
The frequency of the phase emfs are equal
The phase emfs have mutual time-phase displacement of 𝜷𝜷 =
𝟐𝟐𝟐𝟐
𝒎𝒎
electrical
radians. Here m is the number of phases of the a.c. machine.
Phase spread
Where field winding on the rotor to produce 2 poles and the stator carries 12
conductors housed in 12 slots.
3-phase winding - phase spread is 120o
A
B
C E1
E2
E3
E4
E5
E6E7E8
E9
E10
E11
E12
1
2
3
4
5
6
7
8
9
10
11
12
N
S
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IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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Time phase angle is 120o
between EA, EB and EC
Maximum emf Em
Zero emf induced in conductor 4 (conductor 4 is cutting zero lines of flux)
induced in conductor 1�𝐸𝐸1 =
𝐸𝐸𝑚𝑚
√2
� R
the emf generated in conductor 7 is maximum (conductor 7 is cutting
maximum lines of flux from S pole)
the polarity of emf in conductor 7 will be opposite to that in conductor 1,
𝑬𝑬𝟕𝟕 =
𝑬𝑬 𝒎𝒎
√𝟐𝟐
, opposite to E1
similarly the emfs generated in conductors 2, 3, 5, 6 and in conductor 8 to 12
can be represented by phasors E2, E3 , E5 , E6 and E8 to E12
the slot angle pitch is given by 𝛾𝛾 =
180 𝑜𝑜
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
=
180 𝑜𝑜
6
= 30𝑜𝑜
if
𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝟏𝟏 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡𝑡𝑡 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝟐𝟐
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑒𝑒𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝟐𝟐 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡𝑡𝑡 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑒𝑒𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝟑𝟑
𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝟑𝟑 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡𝑡𝑡 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑒𝑒𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝟑𝟑
� 𝐸𝐸𝐴𝐴 = 𝐸𝐸1 + 𝐸𝐸2 + 𝐸𝐸3 + 𝐸𝐸4
Similarly, 𝐸𝐸𝐵𝐵 = 𝐸𝐸5 + 𝐸𝐸6 + 𝐸𝐸7 + 𝐸𝐸8 & 𝐸𝐸𝐶𝐶 = 𝐸𝐸9 + 𝐸𝐸10 + 𝐸𝐸11 + 𝐸𝐸12
the phase belt or phase band may be defined as the group of adjacent slots
belonging to one phase under one pole-pair
Conductors 1, 2, 3 and 4 constitute first phase group
Conductors 5, 6, 7 and 8 constitute second phase group
Conductors 9, 10, 11 and 12 constitute third phase group
the angle subtended by one phase group is called phase spread, symbol σ
𝜎𝜎 = 𝑞𝑞𝑞𝑞 = 4 × 30𝑜𝑜
where
𝑞𝑞 = 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝ℎ𝑠𝑠𝑠𝑠 =
𝑆𝑆
𝑃𝑃𝑃𝑃
EA
EB
EC
E1
E2
E3
E4
E12
E11
E10 E9
E5
E6
E7
E8
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IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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Sequence of phase-belts (groups)
Let
12-conductors can be used to obtain three-phase single – layer winding having a
phase spread of 60o
coil pitch or coil span y = pole pitch τ =
𝑆𝑆
𝑃𝑃
=
12
2
= 6
(𝜎𝜎 = 60𝑜𝑜
)
for 12 slots and 2 poles, slot angular pitch γ =30o
for 𝜎𝜎 = 60𝑜𝑜
, two adjacent slots must belong to the same phase
A
B
C
E1
E2
E3
E4
E5
E6E7E8
E9
E10
E11
E12
1
2
3
4
5
6
7
8
9
10
11
12
N
S
A′
B′
C′
3-phase winding, phase spread is 60o
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IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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(a)
(b)
Phase spread of 60o
(b) Time-phase diagram for the emfs generated in (a)
, 12 slots,2 pole winding arrangement
A
B
C
E1
E7
E2
-E8
E5
E6
E9
E10
-E11
-E12
-E4 -E3
120o
1 2 3 4 5 6 7 8 9 10 11 12
b
b
a
a
c
c
d
d
120o
120o
γ=30o
A A
′A C
′
C
′
B B C CA
′
B
′ B
′
B1A1 C1B2A2 C2
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IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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Double Layer Winding
synchronous machine armatures and induction –motor stators above a few
kW, are wound with double layer windings
if the number of slots per pole per phase 𝒒𝒒 =
𝑺𝑺
𝒎𝒎𝒎𝒎
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
winding, because 𝑞𝑞 =
36
3×4
= 3 𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖𝑖𝑖
a 3-phase winding with 30 slots and 4 poles is a fractional slot
winding, because 𝑞𝑞 =
30
3×4
=
5
2
𝑖𝑖𝑖𝑖 𝑛𝑛𝑛𝑛𝑛𝑛 𝑎𝑎𝑎𝑎 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖𝑖𝑖
the number of coils C is always equal to the number of slots S, C=S
1- Integral Slot Winding
Example: make a winding table for the armature of a 3-phase machine with
the following specifications:
Total number of slots = 24 Double – layer winding
Number of poles = 4 Phase spread=60
Coil-span = full-pitch
o
(a) Draw the detailed winding diagram for one phase only
(b) Show the star of coil-emfs. Draw phasor diagram for narrow-spread(σ=60o
)
connections of the 3-phase winding showing coil-emfs for phases A and B only.
Solution: slot angular pitch, 𝛾𝛾 =
4×180 𝑜𝑜
24
= 30𝑜𝑜
Phase spread, 𝜎𝜎 = 60𝑜𝑜
Number of slots per pole per phase, 𝑞𝑞 =
24
3×4
= 2
Coil span = full pitch =
24
4
= 6
11.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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(a)
Detailed double layer winding diagram for phase A for 3-phase armature
having 24 slots, 4 poles, phase spread 60o
12.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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(c) The star of coil emfs can be drawn similar to the star of slot emfs or star of
conductor emfs
Phasor diagram showing the phasor sum of coil-emfs to obtain phase voltages A
and B
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IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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2. integral slot chorded winding
Coil span (coil pitch) < pole pitch (y < τ)
The advantages of using chorded coils are:
To reduce the amount of copper required for the end-connections (or
over hang)
To reduce the magnitude of certain harmonics in the waveform of
phase emfs and mmfs
The coil span generally varies from 2/3 pole pitch to full pole pitch
Example. Let us consider a double-layer three-phase winding with q = 3,
p = 4, (S = pqm = 36 slots), chorded coils y/τ = 7/9
The star of slot emf phasors for a double-layer winding p = 4 poles,
q = 3 slots/pole/phase, m = 3, S = 36
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IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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Double-layer winding: p = 4 poles, q = 3, y/τ = 7/9, S = 36 slots.
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3. Fractional Slot Windings
If the number of slots qof a winding is a fraction, the winding is called a fractional
slot winding.
Advantages of fractional slot windings when compared with integral slot windings
are:
1. a great freedom of choice with respect to the number of slot a possibility to
reach a suitable magnetic flux density
2. this winding allows more freedom in the choice of coil span
3. if the number of slots is predetermined, the fractional slot winding can be
applied to a wider range of numbers of poles than the integral slot winding
the segment structures of large machines are better controlled by using
fractional slot windings
4. this winding reduces the high-frequency harmonics in the emf and mmf
waveforms
Let us consider a small induction motor with p = 8 and q = 3/2, m = 3.
The total number of slots S = pqm = 8*3*3/2= 36 slots. The coil span y is
y = (S/p) = (36/8) = 4slot pitches
Fractionary q (q = 3/2, p = 8, m = 3,S = 36) winding- emf star,
16.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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The actual value of q for each phase under neighboring poles is 2 and 1,
respectively, to give an average of 3/2
Fractionary q (q = 3/2, p = 8, m = 3, S = 36) winding
slot/phase allocation & coils of phase A
Single – Layer Winding
One coil side occupies one slot completely, in view of this, number of coils
C is equal to half the number of slots S, 𝑪𝑪 =
𝟏𝟏
𝟐𝟐
𝑺𝑺
The 3-phase single –layer windings are of two types
1. Concentric windings
2. Mush windings
Concentric Windings
The coils under one pole pair are wound in such a manner as if these have
one center
the concentric winding can further be sub-divided into
1. half coil winding or unbifurcated winding
2. Whole coil winding or bifurcated winding
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IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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Half coil winding
For phase A only
The half coil winding arrangement with 2-slots per pole per phase and for
σ=60o
A coil group may be defined as the group of coils having the same center
The number of coils in each coil group = the number of coil sides in each
phase belt (phase group)
The carry current in the same direction in all the coil groups
whole coil winding
For phase A only
The whole coil winding arrangement with 2-slots per pole per phase
The number of coil sides in each phase belt (here 4) are double the number
of coils (here 2) in each coil group
There are P coil groups and the adjacent coil groups carry currents in
opposite directions
Example. Design and draw (a) half coil and (b) whole coil single layer
concentric windings for a 3-phase machine with 24-slots, 4-poles and 60o
phase
spread.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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(b) Whole-coil concentric winding
For slot pitch γ = 30o
& phase spread σ = 60o
The number of coils per phase belt = 2
,
The number of coils in each coil group = 1
The pole pitch=6
The coil pitch of 6 slot pitches does not result in proper arrangement of
the winding
In view of this, a coil pitch of 5 is chosen
Whole-coil winding arrangement of 24 slots, 4 poles, 60o
phase spread, single
layer concentric winding (three-plane overhang)
Mush Winding
The coil pitch is the same for all the coils
Each coil is first wound on a trapezoidal shaped former. Then
the short coil sides are first fitted in alternate slots and the long
coil sides are inserted in the remaining slots
The number of slots per pole per phase must be a whole number
The coil pitch is always odd
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IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
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For example, for 24 slots, 4 poles, single-layer mush winding, the pole pitch is 6
slots pitches. Since the coil pitch must be odd, it can be taken as 5 or 7. Choosing
here a coil pitch of 5 slot pitches.
Single – layer mush winding diagram for 24 slots, 4 poles and 60o
phase
spread
H.W: Design and draw
1. 3-phase, 24-slots, 2-poles single-layer winding (half coil winding)
2. a.c. winding: 3-phase, 4 -pole, 24- slots, double layer winding with full
pitch coils (phase B& phase C)
3. a.c. winding: 3-phase, 4 -pole, 24- slots, double layer winding with chorded
coils y/τ = 5/6
4. 10 -pole, 48- slots, fractional 3-phase double layer winding
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When balanced 3-phase currents flow in balanced 3-phase windings, a
rotating magnetic field is produced.
Rotating Magnetic Field
All 3-phase ac machines are associated with rotating magnetic fields in their
air-gaps.
For example, a 2-pole 3-phase stator winding
The three windings are displaced from
each other by 120o
along the air-gap
periphery.
Each phase is distributed or spread over
60o
(called phase-spread σ=60o
)
The 3-phase winding a, b, c is represented
by three full pitched coils, aa′ , bb′ , cc′
For instance, the concentrated full-pitched coil aa′ represents phase a winding
in all respects
A current in phase a winding establishes magnetic flux directed along the
magnetic axis of coil aa′
Positive currents are assumed to be flowing as indicated by crosses in coil-sides
a′ , b′ , c′
22.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
22
Magnetic flux plot
At the instant 1, the current in phase a is positive and maximum Im
At the instant 2, 𝒊𝒊𝒂𝒂 =
𝑰𝑰 𝒎𝒎
𝟐𝟐
, 𝒊𝒊𝒃𝒃 =
𝑰𝑰 𝒎𝒎
𝟐𝟐
and 𝒊𝒊𝒄𝒄 = −𝑰𝑰 𝒎𝒎
�𝒊𝒊𝒃𝒃 = 𝒊𝒊𝒄𝒄 = −
𝑰𝑰 𝒎𝒎
𝟐𝟐
�
At the instant 3, 𝒊𝒊𝒂𝒂 = −
𝑰𝑰 𝒎𝒎
𝟐𝟐
, 𝒊𝒊𝒃𝒃 = 𝑰𝑰 𝒎𝒎 and 𝒊𝒊𝒄𝒄 = −
𝑰𝑰 𝒎𝒎
𝟐𝟐
The 2 poles produced by the resultant flux are seen to have turned through
further 60o
The space angle traversed by rotating flux is equal to the time angle
traversed by currents
The rotating field speed, for a P-pole machine, is
𝟏𝟏
𝑷𝑷
𝟐𝟐�
revolution in one cycle
𝒇𝒇
𝑷𝑷
𝟐𝟐�
revolutions in f cycles
Production of rotating magnetic field illustrated by magnetic flux plot
23.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
23
𝒇𝒇
𝑷𝑷
𝟐𝟐�
revolutions in one second [because f cycles are completed in one
second]
Here f is the frequency of the phase currents. If ns
𝑛𝑛𝑠𝑠 =
𝑓𝑓
𝑃𝑃 2⁄
=
2𝑓𝑓
𝑃𝑃
denotes the rotating field speed
in revolutions per sec, then
Or
𝑁𝑁𝑠𝑠 =
120𝑓𝑓
𝑝𝑝
𝑟𝑟. 𝑝𝑝. 𝑚𝑚 [The speed at which rotating magnetic field revolves is
called the Synchronous speed]
Space phasor representation
When currents ia , ib , ic
Production of rotating magnetic field illustrated by space phasor m.m.fs.
flow in their respective phase windings, then the
three stationary pulsation m.m.fs 𝐹𝐹𝑎𝑎
� ,𝐹𝐹𝑏𝑏
��� , 𝐹𝐹𝑐𝑐
� combine to give the resultant
m.m.f. 𝐹𝐹𝑅𝑅
��� which is rotating at synchronous speed.
At the instant 1,
𝑖𝑖𝑎𝑎 = 𝐼𝐼𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐹𝐹�𝑎𝑎 = 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚. 𝑚𝑚. 𝑓𝑓. 𝐹𝐹𝑚𝑚
𝑖𝑖𝑏𝑏 = 𝑖𝑖𝑐𝑐 = −
𝐼𝐼𝑚𝑚
2
𝑡𝑡ℎ𝑒𝑒 𝑚𝑚. 𝑚𝑚. 𝑓𝑓. 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐹𝐹�𝑏𝑏 = 𝐹𝐹�𝑐𝑐 =
𝐹𝐹𝑚𝑚
2
The resultant of m.m.fs. 𝑭𝑭�𝒂𝒂 , 𝑭𝑭�𝒃𝒃 , 𝑭𝑭�𝒄𝒄 is 𝑭𝑭� 𝑹𝑹 and its magnitude is given by
The vertical component of 𝑭𝑭�𝒃𝒃 & 𝑭𝑭�𝒄𝒄 cancel each other.
𝐹𝐹𝑅𝑅 = 𝐹𝐹𝑚𝑚 +
2𝐹𝐹𝑚𝑚
2
cos 60𝑜𝑜
=
3
2
𝐹𝐹𝑚𝑚
24.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
24
At the instant 2,
𝑖𝑖𝑎𝑎 = 𝑖𝑖𝑏𝑏 =
𝐼𝐼𝑚𝑚
2
& 𝑖𝑖𝑐𝑐 = −𝐼𝐼𝑚𝑚
𝑡𝑡ℎ𝑒𝑒 𝑚𝑚. 𝑚𝑚. 𝑓𝑓. 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐹𝐹�𝑎𝑎 = 𝐹𝐹�𝑏𝑏 =
𝐹𝐹𝑚𝑚
2
& 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐹𝐹�𝑐𝑐 = 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚. 𝑚𝑚. 𝑓𝑓. 𝐹𝐹𝑚𝑚
The resultant m.m.f. 𝐹𝐹𝑅𝑅 =
3
2
𝐹𝐹𝑚𝑚 [it rotate by a space angle of 60o
At the instant 3,
clockwise]
𝑖𝑖𝑎𝑎 = 𝑖𝑖𝑐𝑐 = −
𝐼𝐼𝑚𝑚
2
& 𝑖𝑖𝑏𝑏 = 𝐼𝐼𝑚𝑚
The resultant m.m.f. 𝐹𝐹𝑅𝑅 =
3
2
𝐹𝐹𝑚𝑚 [The resultant m.m.f. has turned through a
further space angle of 60o
Sinusoidal rotating mmf wave creates in phase sinusoidal rotating flux density
wave in the air gap; the peak value of B- wave is given by
from its position occupied at instant 2]
Where g is air-gap length
Example: Prove that a rotating magnetic field of constant amplitude is
produced when 3-phase balanced winding is excited by three-phase balanced
currents.
Solution: three – phase balanced currents given by
A constant-amplitude rotating m.m.f. or
rotating field is produced in the air-gap of a
three-phase machines at synchronous speed
------ (1)
25.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
25
The three mmfs Fa , Fb and Fc can be expressed mathematically as
Angle α is measured from the axis of phase a
The mmf of phase a can be expressed as
Similarly, for phases b & c,
The resultant mmf 𝐹𝐹𝑅𝑅(𝛼𝛼, 𝑡𝑡) can be obtained by adding the three mmfs given by
Eqs. (1), (2) and (3).
------ (2)
------ (3)
------ (4)
------ (5)
26.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
26
Eq.(5), therefore, reduces to
It can be shown that Eq.(6) represents a travelling mmf wave of constant amplitude
𝟑𝟑
𝟐𝟐
𝑭𝑭 𝒎𝒎
H.W: A three-phase, Y-connected winding is fed from 3-phase balanced supply,
with their neutrals connected together. If one of the three supply leads gets
disconnected, find what happens to the m.m.f. wave .
But
mmf
------ (6)
At
At
At
27.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
27
• A wire loop is rotated in a magnetic field.
Electromotive Force (EMF) Equation
– N is the number of turns in the loop
– L is the length of the loop
– D is the width of the loop
– B is the magnetic flux density
– n is the number of revolutions per seconds
• A wire loop is rotated in
a magnetic field.
• The magnetic flux through
the loop changes by the position
• Position 1 all flux links with
the loop
• Position 2 the flux linkage
reduced
• The change of flux linkage
induces a voltage in the loop
• The induced voltage is an ac voltage
• The voltage is sinusoidal
• The rms value of the induced voltage loop is:
The r.m.s value of the generated emf in a full pitched coil is
𝐸𝐸 =
𝐸𝐸𝑚𝑚𝑚𝑚𝑚𝑚
√2
, where 𝐸𝐸𝑚𝑚𝑚𝑚𝑚𝑚 = 𝜔𝜔𝑟𝑟 𝑁𝑁∅ = 2𝜋𝜋𝜋𝜋𝜋𝜋∅ [∅ = 𝐵𝐵𝐵𝐵𝐵𝐵]
∴ 𝐸𝐸 =
𝐸𝐸𝑚𝑚𝑚𝑚𝑚𝑚
√2
= √2 𝜋𝜋 𝑓𝑓𝑓𝑓∅ = 4.44𝑓𝑓𝑓𝑓∅
( ) ( )tLDBt ωcos=Φ
nπω 2=
( ) ( ) ( )[ ] ( )tLDBN
dt
td
LDBN
dt
td
NtV ωω
ω
sin
cos
==
Φ
=
2
ωLDBN
Vrms =
E
E
28.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
28
Winding Factor (Coil Pitch and Distributed Windings)
Pitch Factor or Coil Pitch
The ratio of phasor (vector) sum of induced emfs per coil to the arithmetic sum
of induced emfs per coil is known as pitch factor (Kp) or coil span factor (Kc)
which is always less than unity.
Let the coil have a pitch short by angle θ electrical space degrees from full pitch
and induced emf in each coil side be E,
• If the coil would have been full pitched, then total induced emf in the coil
would have been 2E.
• when the coil is short pitched by θ electrical space degrees the resultant
induced emf, ER
𝐸𝐸𝑅𝑅 = 2𝐸𝐸 cos
𝜃𝜃
2
in the coil is phasor sum of two voltages, θ apart
Pitch factor, 𝑲𝑲𝒑𝒑 =
𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷 𝒔𝒔𝒔𝒔𝒔𝒔 𝒐𝒐𝒐𝒐 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 𝒔𝒔𝒔𝒔𝒔𝒔 𝒐𝒐𝒐𝒐 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆
=
𝟐𝟐𝟐𝟐 𝐜𝐜𝐜𝐜𝐜𝐜
𝜽𝜽
𝟐𝟐
𝟐𝟐𝟐𝟐
= 𝐜𝐜𝐜𝐜𝐜𝐜
𝜽𝜽
𝟐𝟐
Example. The coil span for the stator winding of an alternator is 120o
. Find
the chording factor of the winding.
Solution: Chording angle, 𝜃𝜃 = 180𝑜𝑜
− 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 180𝑜𝑜
− 120𝑜𝑜
= 60𝑜𝑜
Chording factor, 𝐾𝐾𝑝𝑝 = cos
𝜃𝜃
2
= cos
60 𝑜𝑜
2
= 0.866
E
E
E
𝜽𝜽
𝟐𝟐
𝜽𝜽
𝟐𝟐
𝜽𝜽
29.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
29
The ratio of the phasor sum of the emfs induced in all the coils distributed in a
number of slots under one pole to the arithmetic sum of the emfs induced(or to the
resultant of emfs induced in all coils concentrated in one slot under one pole) is
known as breadth factor (K
Distribution Factor
b) or distribution factor (Kd
𝐾𝐾𝑑𝑑 =
𝐸𝐸𝐸𝐸𝐸𝐸 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝐸𝐸𝐸𝐸𝐸𝐸 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑒𝑒 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
)
=
𝑃𝑃ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 ℎ𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
The distribution factor is always less than unity.
Let no. of slots per pole = Q and no. of slots per pole per phase = q
Induced emf in each coil side = E
Angular displacement between the slots, 𝛾𝛾 =
180 𝑜𝑜
𝑄𝑄
c
The emf induced in different coils of one phase under one pole are
represented by side AC, CD, DE, EF… Which are equal in magnitude (say
each equal Ec ) and differ in phase (say by γo
) from each other.
γ
γ
γ/2
γ/2
γ/2
qγ
A
B
C
D
E F
E
E
E E
E
O
30.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
30
If bisectors are drawn on AC, CD, DE, EF… they would meet at common point
(O). The point O would be the circum center of the circle having AC, CD, DE,
EF…as the chords and representing the emfs induced in the coils in different slots.
EMF induced in each coil side, 𝐸𝐸𝑐𝑐 = 𝐴𝐴𝐴𝐴 = 2𝑂𝑂𝑂𝑂 sin
𝛾𝛾
2
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴ℎ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑞𝑞 × 2 × 𝑂𝑂𝑂𝑂 sin
𝛾𝛾
2
∴ The resultant emf, 𝐸𝐸𝑅𝑅 = 𝐴𝐴𝐴𝐴 = 2 × 𝑂𝑂𝑂𝑂 sin
𝐴𝐴𝐴𝐴𝐴𝐴
2
= 2 × 𝑂𝑂𝑂𝑂 sin
𝑞𝑞𝑞𝑞
2
& distribution factor, 𝑘𝑘𝑑𝑑 =
𝑃𝑃ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 ℎ𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
=
2 × 𝑂𝑂𝑂𝑂 sin
𝑞𝑞𝑞𝑞
2
𝑞𝑞 × 2 × 𝑂𝑂𝑂𝑂 sin
𝛾𝛾
2
=
𝐬𝐬𝐬𝐬 𝐬𝐬
𝒒𝒒𝜸𝜸
𝟐𝟐
𝒒𝒒 𝐬𝐬𝐬𝐬 𝐬𝐬
𝜸𝜸
𝟐𝟐
Example. Calculate the distribution factor for a 36-slots, 4-pole, single layer 3-
phase winding.
Solution: No. of slots per pole, 𝑄𝑄 =
36
4
= 9
No. of slots per pole per phase, 𝑞𝑞 =
𝑄𝑄
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
=
9
3
= 3
Angular displacement between the slots, 𝛾𝛾 =
180 𝑜𝑜
𝑄𝑄
=
180 𝑜𝑜
9
= 20𝑜𝑜
Distribution factor, 𝐾𝐾𝑑𝑑 =
sin
𝑞𝑞𝑞𝑞
2
𝑞𝑞 sin
𝛾𝛾
2
=
sin
3×20 𝑜𝑜
2
3 sin
20 𝑜𝑜
2
=
1
3
sin 30 𝑜𝑜
sin 10 𝑜𝑜 = 0.96
31.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
31
Example1. A 3-phase, 8-pole, 750 r.p.m. star-connected alternator has 72 slots on
the armature. Each slot has 12 conductors and winding is short chorded by 2 slots.
Find the induced emf between lines, given the flux per pole is 0.06 Wb.
Solution:
Flux per pole, ∅ = 0.06 𝑊𝑊𝑊𝑊
𝑓𝑓 =
𝑝𝑝𝑝𝑝
60
=
4×750
60
= 50 𝐻𝐻𝐻𝐻
Number of conductors connected in series per phase,
𝑍𝑍𝑠𝑠 =
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑝𝑝𝑝𝑝𝑝𝑝 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ×𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑁𝑁𝑁𝑁𝑁𝑁 𝑏𝑏𝑏𝑏𝑏𝑏 𝑜𝑜𝑜𝑜 𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
=
12×72
3
= 288
Number of turns per phase, 𝑇𝑇 =
𝑍𝑍𝑠𝑠
2
=
288
2
= 144
Number of slots per pole, 𝑄𝑄 =
72
8
= 9
Number of slots per pole per phase, 𝑞𝑞 =
𝑄𝑄
3
=
9
3
= 3
Angular displacement between the slots, 𝛾𝛾 =
180 𝑜𝑜
𝑄𝑄
=
180 𝑜𝑜
9
= 20𝑜𝑜
Distribution factor, 𝐾𝐾𝑑𝑑 =
sin
𝑞𝑞𝑞𝑞
2
𝑞𝑞 sin
𝛾𝛾
2
=
sin
3×20 𝑜𝑜
2
3 sin
20 𝑜𝑜
2
=
1
3
sin 30 𝑜𝑜
sin 10 𝑜𝑜 = 0.96
Chording angle, 𝜃𝜃 = 180𝑜𝑜
×
2
9
= 40𝑜𝑜
Pitch factor, 𝐾𝐾𝑝𝑝 = cos
𝜃𝜃
2
= cos
40 𝑜𝑜
2
= cos 20𝑜𝑜
= 0.94
Induced emf between lines, 𝐸𝐸𝐿𝐿 = √3 × 4.44 × 𝐾𝐾𝑑𝑑 × 𝐾𝐾𝑝𝑝 × ∅ × 𝑓𝑓 × 𝑇𝑇
= �3 × 4.44 × 0.96 × 0.94 × 0.06 × 50 × 144 = 2998 𝑉𝑉
32.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
32
the variation of magnetic potential difference along the air –gap periphery is
of rectangular waveform and of magnitude
1
2
𝑁𝑁𝑁𝑁
Magnetomotive Force (mmf) of AC Windings
M.m.f. of a coil
The amplitude of mmf wave varies with time, but not with space
The air –gap mmf wave is time-variant but space invariant
The air –gap mmf wave at any instant is rectangular
Mmf distribution along air-gap periphery
The fundamental component of rectangular wave is found to be
𝐹𝐹𝑎𝑎1 =
4
𝜋𝜋
∙
𝑁𝑁𝑁𝑁
2
cos 𝛼𝛼 = 𝐹𝐹1𝑝𝑝 cos 𝛼𝛼
Where
α = electrical space angle measured from the magnetic axis of the stator coil
33.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
33
Here F1p , the peak value of the sine mmf wave for a 2-pole machine is given by
𝐹𝐹1𝑝𝑝 =
4
𝜋𝜋
∙
𝑁𝑁𝑁𝑁
2
𝐴𝐴𝐴𝐴 𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
When i=0 F1p =0
i=Imax
The mmf distribution along the air gap periphery depends on the nature of
slots, winding and the exciting current
=√𝟐𝟐𝑰𝑰
For 2-pole machine 𝐹𝐹1𝑝𝑝𝑝𝑝 =
4
𝜋𝜋
∙
𝑁𝑁√2𝐼𝐼
2
𝐴𝐴𝐴𝐴 𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
For p-pole machine 𝐹𝐹1𝑝𝑝𝑝𝑝 =
4
𝜋𝜋
∙
𝑁𝑁√2𝐼𝐼
𝑃𝑃
𝐴𝐴𝐴𝐴 𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
M.m.f of distributed windings
The effect of winding distribution has changed the shape of the mmf wave,
from rectangular to stepped
Developed diagram and mmf wave of the machine
(each coil has Nc turns and each turn carries i amperes)
34.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
34
Example: a 3-phase, 2-pole stator has double-layer full pitched winding with 5
slots per pole per phase. If each coil has Nc turns and i is the conductor
current, then sketch the mmf wave form produced by phase A alone.
A 3-phase, 2-pole stator with double-layer winding having 5 slots per pole per phase
For any closed path around slot 1, the total current enclosed is 2Nci ampere
Magnetic potential difference across each gap is
𝟏𝟏
𝟐𝟐
[𝟐𝟐𝑵𝑵𝒄𝒄 𝒊𝒊] = 𝑵𝑵𝒄𝒄 𝒊𝒊
The mmf variation from −𝑵𝑵𝒄𝒄 𝒊𝒊 to +𝑵𝑵𝒄𝒄 𝒊𝒊 at the middle of slot 1
The mmf variation for slot 1′
is from +𝑵𝑵𝒄𝒄 𝒊𝒊 to −𝑵𝑵𝒄𝒄 𝒊𝒊
The mmf variation for coil 11′
is of rectangular waveform with amplitude
±𝑁𝑁𝑐𝑐 𝑖𝑖 . similarly, the rectangular mmf waveforms of amplitude ±𝑁𝑁𝑐𝑐 𝑖𝑖 are
sketched for the coils 22′
, …, 55′
The combined mmf produced by 5 coils is obtained by adding the ordinates
of the individual coil mmfs.
The resultant mmf waveform consists of a series of steps each of height
𝟐𝟐𝑵𝑵𝒄𝒄 𝒊𝒊 = (conductors per slot) (conductor current)
The amplitude of the resultant mmf wave is 𝟓𝟓𝑵𝑵𝒄𝒄 𝒊𝒊 .
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
36
Harmonic Effect
The flux distribution along the air gaps of alternators usually is non-
sinusoidal so that the emf in the individual armature conductor likewise is
non-sinusoidal
The sources of harmonics in the output voltage waveform are the non-
sinusoidal waveform of the field flux.
Fourier showed that any periodic wave may be expressed as the sum of a d-c
component (zero frequency) and sine (or cosine) waves having fundamental
and multiple or higher frequencies, the higher frequencies being called
harmonics.
The emf of a phase due to the fundamental component of the flux per pole is:
𝐸𝐸𝑝𝑝ℎ1 = 4.44𝑓𝑓𝐾𝐾𝑤𝑤1 𝑇𝑇𝑝𝑝ℎ ∅1
Where 𝐾𝐾𝑤𝑤1 = 𝑘𝑘𝑑𝑑1. 𝐾𝐾𝑝𝑝1 is the winding factor. For the nth harmonic
𝐸𝐸𝑝𝑝ℎ𝑛𝑛 = 4.44𝑛𝑛𝑛𝑛𝑛𝑛𝑤𝑤𝑤𝑤 𝑇𝑇𝑝𝑝ℎ ∅𝑛𝑛
The nth harmonic and fundamental emf components are related by
𝐸𝐸𝑝𝑝ℎ 𝑛𝑛
𝐸𝐸𝑝𝑝ℎ1
=
𝐵𝐵𝑛𝑛 𝐾𝐾𝑤𝑤𝑤𝑤
𝐵𝐵1 𝐾𝐾𝑤𝑤1
The r.m.s. phase emf is:
𝐸𝐸𝑝𝑝ℎ = ��𝐸𝐸𝑝𝑝ℎ1
2
+ 𝐸𝐸𝑝𝑝ℎ3
2
+∙∙∙ +𝐸𝐸𝑝𝑝ℎ𝑛𝑛
2
�
All the odd harmonics (third, fifth, seventh, ninth, etc.) are present in the
phase voltage to some extent and need to be dealt with in the design of ac
machines.
Because the resulting voltage waveform is symmetric about the center of
the rotor flux, no even harmonics are present in the phase voltage.
In Y- connected, the third-harmonic voltage between any two terminals will
be zero. This result applies not only to third-harmonic components but also to
any multiple of a third-harmonic component (such as the ninth harmonic).
Such special harmonic frequencies are called triplen harmonics.
37.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
37
The pitch factor of the coil at the harmonic frequency can be expressed as
𝐾𝐾𝑝𝑝𝑝𝑝 = cos
𝑛𝑛𝑛𝑛
2
where n is the number of the harmonic
Elimination or Suppressed of Harmonics
Field flux waveform can be made as much sinusoidal as possible by the
following methods:
1. Small air gap at the pole centre and large air gap towards the pole ends
2. Skewing: skew the pole faces if possible
3. Distribution: distribution of the armature winding along the air-gap
periphery
4. Chording: with coil-span less than pole pitch
5. Fractional slot winding
6. Alternator connections: star or delta connections of alternators suppress
triplen harmonics from appearing across the lines
For example, for a coil-span of two-thirds �
2
3
𝑟𝑟𝑟𝑟� of a pole pitch
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠, ∝=
2
3
× 180𝑜𝑜
= 120𝑜𝑜 (𝑖𝑖𝑖𝑖 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑)
𝐶𝐶ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, 𝜃𝜃 = 180𝑜𝑜
− 𝛼𝛼 = 180𝑜𝑜
− 120𝑜𝑜
= 60𝑜𝑜
𝐾𝐾𝑝𝑝1 = cos
𝑛𝑛𝑛𝑛
2
= cos
60𝑜𝑜
2
= cos 30𝑜𝑜
= 0.866
For the 3rd
harmonic: 𝐾𝐾𝑝𝑝3 = cos
3×60𝑜𝑜
2
= cos 90𝑜𝑜
= 0;
Thus all 3rd
(and triplen) harmonics are eliminated from the coil and
phase emf . The triplen harmonics in a 3-phase machine are normally eliminated
by the phase connection.
Example: An 8-pole, 3-phase, 60o
spread, double layer winding has 72 coils in 72
slots. The coils are short-pitched by two slots. Calculate the winding factor for the
fundamental and third harmonic.
Solution: No. of slots per pole, 𝑄𝑄 =
72
8
= 9
No. of slots per pole per phase, 𝑞𝑞 =
𝑄𝑄
𝑚𝑚
=
9
3
= 3
38.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
38
Angular displacement between the slots, 𝛾𝛾 =
180 𝑜𝑜
𝑄𝑄
=
180 𝑜𝑜
9
= 20𝑜𝑜
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠, ∝=
180𝑜𝑜
× 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑖𝑖𝑖𝑖 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑁𝑁𝑁𝑁. 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑝𝑝𝑝𝑝𝑝𝑝 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
=
180𝑜𝑜(9 − 2)
9
= 140𝑜𝑜
𝐶𝐶ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, 𝜃𝜃 = 180𝑜𝑜
− 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 180𝑜𝑜
− 140𝑜𝑜
= 40𝑜𝑜
For the fundamental component
𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝐾𝐾𝑑𝑑 =
sin 𝑞𝑞𝑞𝑞 2⁄
𝑞𝑞 sin 𝛾𝛾 2⁄
=
sin 3 ×
20𝑜𝑜
2
3 sin
20𝑜𝑜
2
= 0.96
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃ℎ 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝐾𝐾𝑝𝑝 = cos
𝜃𝜃
2
= cos
40𝑜𝑜
2
= 0.94
𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝐾𝐾𝑤𝑤 = 𝐾𝐾𝑑𝑑 × 𝐾𝐾𝑝𝑝 = 0.96 × 0.94 = 0.9
For the third harmonic component (n=3)
𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝐾𝐾𝑑𝑑3 =
sin 𝑛𝑛𝑛𝑛𝑛𝑛 2⁄
𝑞𝑞 sin 𝑛𝑛𝑛𝑛 2⁄
=
sin
3 × 3 × 20𝑜𝑜
2
3 sin
3 × 20𝑜𝑜
2
= 0.666
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃ℎ 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝐾𝐾𝑝𝑝3 = cos
3𝜃𝜃
2
= cos
3 × 40𝑜𝑜
2
= 0.5
𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝐾𝐾𝑤𝑤3 = 𝐾𝐾𝑑𝑑3 × 𝐾𝐾𝑝𝑝3 = 0.666 × 0.5 = 0.333
Example3: Calculate the r.m.s. value of the induced e.m.f. per phase of a 10-pole,
3-phase, 50Hz alternator with 2 slots per pole per phase and 4 conductors per slot
in two layers. The coil span is 150o
.the flux per pole has a fundamental component
of 0.12Wb and a third harmonic component.
Solution: No. of slots/pole/phase, 𝑞𝑞 = 2
No. of slots/pole, 𝑄𝑄 = 𝑞𝑞𝑞𝑞 = 2 × 3 = 6
No. of slots/phase =2𝑝𝑝𝑝𝑝 = 10 × 2 = 20
No. of conductors connected in series, 𝑍𝑍𝑠𝑠 = 20 × 4 = 80
No. of series turns/phase, 𝑇𝑇 =
𝑍𝑍𝑠𝑠
2
=
80
2
= 40
39.
IInnttrroodduuccttiioonn ttoo AACCMMaacchhiinneess Dr. SSuuaadd IIbbrraahhiimm SShhaahhll
39
Angular displacement between adjacent slots, 𝛾𝛾 =
180 𝑜𝑜
𝑄𝑄
=
180 𝑜𝑜
6
= 30𝑜𝑜
𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝐾𝐾𝑑𝑑 =
sin 𝑞𝑞𝑞𝑞 2⁄
𝑞𝑞 sin 𝛾𝛾 2⁄
=
sin
2 × 30𝑜𝑜
2
2 sin
30𝑜𝑜
2
= 0.966
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝐾𝐾𝑝𝑝 = cos
𝜃𝜃
2
= cos
(180𝑜𝑜
− 150𝑜𝑜)
2
= cos 15𝑜𝑜
= 0.966
Induced emf per phase (fundamental component),
𝐸𝐸𝑝𝑝ℎ1 = 4.44 𝐾𝐾𝑑𝑑 𝐾𝐾𝑝𝑝∅𝑓𝑓𝑓𝑓
= 4.44 × 0.966 × 0.966 × 0.12 × 50 × 40 = 994.4 𝑉𝑉
For third harmonic component of flux
𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝐾𝐾𝑑𝑑3 =
sin 𝑞𝑞𝑞𝑞𝑞𝑞 2⁄
𝑞𝑞 sin 𝑛𝑛𝑛𝑛 2⁄
=
sin
2 × 3 × 30𝑜𝑜
2
3 sin
3 × 30𝑜𝑜
2
= 0.707
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝐾𝐾𝑝𝑝3 = cos 3
(180𝑜𝑜
− 150𝑜𝑜)
2
= cos 45𝑜𝑜
= 0.707
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹, 𝑓𝑓3 = 3 × 𝑓𝑓 = 3 × 50 = 150
Flux per pole, ∅3 =
1
3
× 0.12 ×
20
100
= 0.008 𝑊𝑊𝑊𝑊
Induced emf per phase (third harmonic component)
𝐸𝐸𝑝𝑝ℎ3 = 4.44 𝐾𝐾𝑑𝑑3 𝐾𝐾𝑝𝑝3∅3 𝑓𝑓3 𝑇𝑇
= 4.44 × 0.707 × 0.707 × 0.008 × 150 × 40 = 106.56 𝑉𝑉
Induced emf per phase,
𝐸𝐸𝑝𝑝ℎ = �𝐸𝐸𝑝𝑝ℎ1
2
+ 𝐸𝐸𝑝𝑝ℎ3
2
= �(994.4)2 + (106.56)2 = 1000 𝑉𝑉
40.
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40
H.W
1. Three-phase voltages are applied to the three windings of an electrical machine. If any two
supply terminals are interchanged, show that the direction of rotating mmf wave is
reversed,through its amplitude remains unaltered.
2. A 3-phase 4-pole alternator has a winding with 8 conductors per slot. The armature has
a total of 36 slots. Calculate the distribution factor. What is the induced voltage per phase
when the alternator is driven at 1800 RPM, with flux of 0.041 Wb in each pole?
(Answer. 0.96, 503.197 Volts/phase)
3. A 10 MVA ,11 KV,50 Hz ,3-phase star-connected alternator s driven at
300 RPM. The winding is housed in 360 slots and has 6 conductors per
slot, the coils spanning (5/6) of a pole pitch. Calculate the sinusoidally distributed
flux per pole required to give a line voltage of 11 kV on open circuit, and the full-
load current per conductor. (Answer. 0.086 weber , 524.864 Amps)
4. A three phase four pole winding is excited by balanced three phase 50 Hz
currents. Although the winding distribution has been designed to minimize the
harmonics, there remains some third and fifth spatial harmonics. Thus the phase
A mmf can be written as
Similar expressions can be written for phase B (replacing θ by θ -120o
and ωt by ωt-
120o
) and phase C (replacing θ by θ +120o
and ωt by ωt+120o
Derive the expression for the total three phase mmf, and show that the fundamental
and the 5
).
th
harmonic components are rotating