Numerical on EMF equation, Introduction to short-pitched winding

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
Lecture-5

2. 2
Learning Outcomes: - (Previous Lecture_04)
Students will be able to:
 Analyse the concept of harmonics and its sources in alternator.
 Determine the RMS value of complex waveforms.
 Know the disadvantages of harmonic components.

3. 3
Learning Outcomes: - (Today’s Lecture_05)
Students will be able to:
 Solve some numerical related to induced emf containing harmonics.
 Analyse and understand the techniques to eliminate the harmonics from the
generated emf.
 Analyse the effect of short-pitched and distributed winding on emf waveform.
 Solve some numerical to remove a particular harmonics from the emf waveform.

4. 4
Numerical on harmonics: -
1. A 3-phase, 50-Hz, 600 rpm, alternator has the flux distribution given by:
𝐵 = 𝑠𝑖𝑛𝜃 + 0.4𝑠𝑖𝑛3𝜃 + 0.2𝑠𝑖𝑛5𝜃.
The alternator has 30 slots with 30 coils each having 18 turns. If the armature has a diameter
of 1.2 m and length of 0.4 m, calculate
a. The RMS value of induced emf/phase, and
b. The instantaneous value of induced emf/phase.
Solution: -
a. The flux distribution has a fundamental, a 3rd harmonic and a 5th harmonic component.
Calculation of fundamental component: -
Peak value of the flux of fundamental component,
𝜑 𝑚1 = 𝐵 𝑚1 × 𝐴1 = 1 ×
𝜋𝐷𝐿
𝑃
=
𝜋×1.2××0.4
10
= 0.1508 Wb.
Average value of fundamental component of flux,
𝜑1 =
2𝜑 𝑚1
𝜋
=
2×0.1508
𝜋
= 0.096 Wb.
Number of turns/phase=
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑢𝑟𝑛𝑠
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝ℎ𝑎𝑠𝑒𝑠
=
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑢𝑟𝑛𝑠×𝑛𝑢𝑚𝑒𝑟 𝑜𝑓 𝑐𝑜𝑖𝑙𝑠/𝑡𝑢𝑟𝑛
3

5. 5
Contd…
Number of turns/phase=
30×18
3
= 180.
So, RMS vale of fundamental component of emf/phase,
𝐸1𝑝ℎ = 4.44𝑓𝜑𝑇𝑝ℎ = 4.44 × 50 × 0.096 × 180 = 3836.16 Volt.
Calculation of 3rd harmonic emf/phase: -
Peak value of the flux of 3rd harmonic component,
𝜑 𝑚3 = 𝐵 𝑚3 × 𝐴3 = 0.4 ×
𝜋𝐷𝐿
3𝑃
= 0.4 ×
𝜋×1.2××0.4
3×10
= 0.0201 Wb.
Average value of 3rd harmonic component of flux,
𝜑3 =
2𝜑 𝑚3
𝜋
=
2×0.0201
𝜋
= 0.0128 Wb.
So, RMS vale of 3rd harmonic component of emf/phase,
𝐸3𝑝ℎ = 4.44 × 3𝑓 𝜑3 𝑇𝑝ℎ = 4.44 × 3 × 50 × 0.0128 × 180 = 1534.46 Volt.
Calculation of 5th harmonic emf/phase: -
Peak value of the flux of 5th harmonic component,
𝜑 𝑚5 = 𝐵 𝑚5 × 𝐴5 = 0.2 ×
𝜋𝐷𝐿
5𝑃
= 0.2 ×
𝜋×1.2××0.4
5×10
= 0.006 Wb.

6. 6
Average value of 3rd harmonic component of flux,
𝜑5 =
2𝜑 𝑚5
𝜋
=
2×0..006
𝜋
= 0.00384 Wb.
So, RMS vale of 3rd harmonic component of emf/phase,
𝐸5𝑝ℎ = 4.44 × 5𝑓 𝜑3 𝑇𝑝ℎ = 4.44 × 5 × 50 × 0.00384 × 180 = 767.23 Volt.
So, effective RMS value of emf/phase,
b. Instantaneous value of induced emf/phase,
2 2 2 2 2 2
1 3 5 3836.16 1534.46 767.23 4202.3ph ph ph phE E E E Volt      
1 3ph 52 2 ×E 2
5425.15 2170.05 1085.03
ph phe E sin + sin3 E sin5
sin sin3 sin5
  
  
   
  

7. Magnetic Field (Flux Density Waveform) Induced EMF/phase Waveform

8. 8
Methods to eliminate the harmonic components from induced emf: -
 It has already been discussed in the previous lecture that even harmonic gets
cancels in the coil. So they do not cause any harm to the system.
 Polarity of induced emf in the conductors for fundamental component are opposite
but additive in nature.
 Polarity of induced emf in the conductors for 2nd harmonic component are same so
cancel out.
A
B C
D
+
-
-
+
Topconductor
Bottomconductor
e e
2e
volt +-
A
B C
D
+
- -
+
Topconductor
Bottomconductor
e e
0
volt +-
N
S




X
.
X
N
SS
N



 


X

9. 9
 So our aim should be eliminate odd harmonics.
 Amplitude of 3rd harmonic component is more as compared to that of 5th harmonic
and so on. So higher if the order of harmonic less is its amplitude.
 3rd harmonic and its multiples called triplen harmonics (odd harmonics multiple of
3 like 3rd , 9th, 15th, 21st ) can be removed from the line voltage by connecting the
3-phase armature winding in star (**Note: these harmonics will be there in phase
emf but eliminated from line emf).
ER
EYEB
-EY
ERY
For the fundamental component, line voltage,
So, in star connection, line voltage of the
fundamental component is 3 times the phase and
leads the phase voltage by 300.
   
0 0
0 0 0 0
0
0 120
cos(0 ) sin(0 ) cos( 120 ) sin( 120 )
(1.5 0.866) 3 30
RY R Y ph ph
ph
ph ph
E E - E E E
E j j
E j E
     
      
    
300

10. 10
For the 3rd harmonic component, line voltage
 So, in star connection, in line voltage 3rd harmonic component is absent.
 Not only the 3rd harmonic but also all its multiple will be absent in the line
voltage.
 So, our main focus should be to reduce 5th and the 7th order harmonics.
 The other methods to eliminate the harmonics and improve the emf waveform
are:
i. Using short pitched winding.
ii. Using distributed winding.
iii. Skewing the pole face/armature slots.
0 0
3 3 3 3 3
0 0
3 3
(3 0 ) (3 120 )
0 0 0
RY R Y ph ph
ph ph
E E E E E
E E
       
    

11. 11
Full pitched and Short pitched winding: -
 In a full pitch winding, the coil span (i.e. peripheral distance between two coil
sides of a coil) is always one pole pitch i.e. 1800 electrical, where as in a short-
pitched winding, the coil span is less than a pole pitch i.e. less than 1800
electrical.
N
S




X
.
Full Pitched Coil
Coil Span = Pole Pitch=
1800
electrical
N
S




X
.
Coil Span < Pole Pitch
< (1800
-α) electrical
α
Short Pitched Coil
Short
Pitched
Angle

12. 12
 If the induced emf/coil side is ‘E’, then the total induced in the coil side is twice
the induced emf/coil side, i.e. algebraic sum of induced emfs/coil side.
A
B C
D
+
-
-
+
Topcoilside
Bottomcoilside
E E
2E
volt +-
E E
2E
N
S N
S
E
E

Er

Full Pitched Coil
Short Pitched Coil
Resultant emf in short pitched coil
A
B C
D
+
-
-
+
Topcoilside
Bottomcoilside
E E
2E
volt
+-
N
S N
S


13. 13
 In full pitched coil as shown in Figure 1, coil
span = pole pitch = 9 slots = 1800 electrical.
 So, slot angle 𝛽 =
1800
9
= 200
electrical.
 In short pitched coil as shown in Figure 2,
coil span = 8 slots = 1600 electrical.
 So, coil of the Figure 2 is short pitched by 1
slot i.e. 𝛼 = 200
electrical.
 In short pitched coil as shown in Figure 3,
coil span = 7 slots = 1400 electrical.
 So, coil of the Figure 2 is short pitched by 2
slots i.e. 𝛼 = 400electrical.
Figure 1
Figure 2
Figure 3
1 2 3 4 5 6 7 8 9
Pole Pitch
Coil Span
Full Pitched Coil
1 2 3 4 5 6 7 8 9
Pole Pitch
Coil Span
Short Pitched Coil
1 2 3 4 5 6 7 8 9
Pole Pitch
Coil Span
Short Pitched Coil
10
10
10


2



14. 14
Induced emf in Short pitched winding: -
A
B C
D
+
-
-
+
Topcoilside
Bottomcoilside
E E
2E
volt +-
E E
2E
N
S N
S
Full Pitched Coil
Short Pitched Coil
Resultant emf in short pitched coil
E
E

Er

2 cos
2
rE E
 
    
 
Pitch factor, 𝐾 𝑝 =
𝑉𝑒𝑐𝑡𝑜𝑟 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑒𝑚𝑓 𝑖𝑛 𝑠ℎ𝑜𝑟𝑡 𝑝𝑖𝑡𝑐ℎ𝑒𝑑 𝑐𝑜𝑖𝑙
𝐴𝑙𝑔𝑒𝑏𝑟𝑖𝑐 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑒𝑚𝑓 𝑖𝑛 𝑓𝑢𝑙𝑙 𝑝𝑖𝑡𝑐ℎ𝑒𝑑 𝑐𝑜𝑖𝑙
=
2𝐸𝑐𝑜𝑠
𝛼
2
2𝐸
= 𝑐𝑜𝑠
𝛼
2
So, the induced emf in a short pitched is reduced by a factor, ‘𝐾 𝑝’
A
B C
D
+
-
-
+
Topcoilside
Bottomcoilside
E E
2E
volt
+-
N
S N
S


15.  Pitch factor,
 RMS value of induced emf/phase with a 5th harmonic component is
 So, to eliminate the 5th harmonic, we chose the short pitch angle ‘𝛼’ so that it will make
the ‘Kp5’ term zero.
cos
2
3
cos 3
2
5
cos 5
2
7
cos 7 ,
2
rd
p
th
th
for fundamental component of emf
for harmoniccomponent of emf
K
for harmoniccomponent of emf
for harmoniccomponent of emf and soon




  
  
 
  
  
  
 
 
   

 
 
 
1 1 1
5 5 5
4.44
4.44 (5 )
ph ph p
ph ph p
E f T K
E f T K



 
1 5
5
, cos , cos
2 2
p pWhere K and K
    
    
   
0 0
5
5 5
cos 0 90 36
2 2
pK
 

 
      
 

16. 16
 After elimination of the 5th harmonic component, waveform of the induced emf with 7th
harmonic component is shown in the Figure shown below.
 Mean square error in the complex waveform with 5th and 7th harmonics = 21.7469.
 Mean square error in the complex waveform with 7th harmonic = 7.3469.
 Effect of rest of harmonic components can be reduced by using distributed winding.

17. 17
Advantages and disadvantages of short pitch winding: -
Advantages:
i. Eliminates harmonic components in the emf waveform.
ii. Elimination of harmonics reduces the iron losses there by improving the machine
efficiency.
iii. Short Pitching reduces the amount of copper needed for end connection.
iv. Reduction of copper reduces the copper loss and cost of the machine.
v. Reduction in length of coil reduces the leakage inductance which in turn reduces the
internal voltage drop in the coil and improves the voltage regulation.
Disadvantages:
i. Reduces the generated emf by a factor (called pitch factor ).' 'pK

18. 18
Concentrated and Distributed winding: -
 In concentrated winding, all the turns of the coil belong to one phase will have the
same magnetic axis, e.g. field winding of DC machine, Transformer windings, field
winding of salient pole alternator etc.
 For rotating electrical machines, if all the conductors belonging to one phase under one
pole are kept in one slot, then the armature winding is said to be concentrated.
DC Machine main field
and interpole winding
Trnsformer winding

19. 19
 So, for concentrated winding:
i. A single phase, 2-pole machine should have 2 slots, 4-pole machine should have
4 slots, and so on, i.e. in a single phase machine number of armature slots must
be equal to number of poles.
ii. A three phase, 2-pole should have 6 slots, 4-pole machine should have 12 slots
and so on, i.e. in a three phase machine the number slots must be equal to three
time the number of poles.
N
S




X
.
N
S




X
.
X
. ..
X
.
X
Two pole machine with
concentrated winding
Four pole machine with
concentrated winding

20. 20
 So, for concentrated winding, number of slots/pole/phase is always ‘1’, i.e. all the coil sides
of one phase under one pole are kept in one slot.
 In distributed winding all the turns of the winding belonging to one phase will not have the
common magnetic axis.
 The coil sides of one phase under one pole are kept in different slots, i.e. number of
slots/pole/phase is greater than 1.
** Note: To know the winding whether concentrated of distributed, we need to calculate the
number of slots/pole/phase. If it is ‘1’, then the winding is concentrated and if it is greater than
‘1’ the winding is distributed.
 For example if there are 360 conductors in a 3-phase, 4-pole alternator,
i. For concentrated winding, they are to be arranged in 3x4 =12 slots.
So, number of conductors in one slot = 360/12 = 30.
ii. For distributed winding with 2 solts/pole/phase, these 30 conductors are to be arranged in
two different slots. So number of slots will 24 and conductors/slot will be 15.
iii. For distributed winding with 3 solts/pole/phase, these 30 conductors are to be arranged
in three different slots. So number of slots will 36 and conductors/slot will be 10.

21. N
S




X
.
X
. ..
X
.
X
Two pole machine with
concentrated winding
(1 slot/pole/phase)
One slot for one
phase under one
Pole, i.e. 30 conductors
in one slot for
R-phase under N-pole N
S




X
.
X
. ..
X
.
X
X
.
.
X
X
.
Two pole machine with
Distributed winding
(2 slots/pole/phase)
Two slots for one
phase under one
Pole, i.e. 15 conductors
in one slot.
Slot angle 𝜷 =
𝟑𝟔𝟎 𝟎
𝟔
= 𝟔𝟎 𝟎
Slot angle 𝜷 =
𝟑𝟔𝟎 𝟎
𝟏𝟐
= 𝟑𝟎 𝟎
Number of slots/pole/phase, m = 1 Number of slots/pole/phase, m = 1
 Phase spread or phase belt is defined as the angle subtended by the conductors of one
phase under one pole.

22. 22
 In the 1st figure,
Number of slots/pole/phase, m = 1.
A lot angle β = 600 electrical.
So, phase spread = mβ = 600 electrical.
 In the 2nd figure,
Number of slots/pole/phase, m = 2.
A lot angle β = 300 electrical.
So, phase spread = mβ = 2x300 = 600 electrical.
N
S




X
.
X
. ..
X
.
X
N
S




X
.
X
. ..
X
.
X
X
.
.
X
X
.
 Suppose we have a 2-pole and a 4-pole machine with 48 slots.
For the 2-pole machine,
Number of slots/pole/phase, m = 48/(2x3) = 8
Slot angle, β = 3600/48 = 7.50 mechanical = 7.50 electrical (as it a 2-pole machine)
So, phase spread, m β = 8x7.50 = 600 electrical.
For the 4-pole machine,
Number of slots/pole/phase, m = 48/(4x3) = 4
Slot angle, β = 3600/48 = 7.50 mechanical = (P/2) x 7.50 = (4/2) x 7.50 =150 electrical
(as electrical angle is (P/2) x mechanical angle).
So, phase spread, m β = 4 x 150 = 600 electrical.

23. 23
Induced EMF in Distributed Winding: -
 In a concentrated winding, as all the conductors of one phase under one pole
are kept in one slot (i.e., m = 1), the total induced emf is algebraic sum of
induced emf/conductor.
 In distributed winding, conductors of the same phase under pole are
distributed in more than one slot (i.e. m > 1). So the induced emfs will have
a phase difference of β (= slot angle).
 So the resultant induced emf will be the vector sum of induced emfs in ‘m’
number of slots.
 As vector sum is always less than the algebraic sum, distributed winding
gives rise to a reduced emf.
 So the induced emf in a distributed winding is reduced by a factor called
distribution factor ‘𝐾 𝑑’.

24. 24
Distribution factor: -
 In distributed winding, the induced
emfs have a phase difference of β,
whereas in concentrated winding
there is no phase difference.
N
S




X
.
X
. ..
X
.
X
30 Conductors
kept in one slot
N
S




X
.
X
. ..
X
.
X
X
.
.
X
X
.
30 Conductors
kept in two slots

Distribution factor,
In the diagram shown in the figure,
Vector sum = AB,
Algebraic sum = m x AC, where AC
is the total induced emf in one slot.
vectorsumof inducedemf in distributed winding
algebraicsumof inducedemf in concentrated winding
dK 
G
G
H

25. 25
Vector sum = AB = 2 x AG,
Algebraic sum = m x AC = m x (2 x AH).
G
H
Vector sum = AB = 2 x AG,
Algebraic sum = m x AC = m x (2 x AH).
In triangle OAH,
AH = OA sin(β/2), and AC = 2 x AH
So, algebraic sum of emf = m x 2OA sin(β/2)
In triangle OAB,
AG = OA x sin(mβ/2)
So, vector of emf = AB = 2 x AG
= 2 x OA sin(mβ/2).
 So, distribution factor,
vectorsum 2OAsin(mβ/2) sin(mβ/2)
=
Algebriacsum m×2OAsin(β/2) msin(β/2)
dK  

26. 26
Number of
slots/pols/phas
e (m)
Phase difference
between the induced
emf (in electrical
degree)
Distribution factor
(Kd) for
fundamental
component
Distribution factor
(Kd5) for 5th
Harmonic
component
Distribution factor
(Kd7) for 7th
Harmonic
component
1 60 1.0 1.0 1.0
2 30 0.9659 0.2588 -0.2588
3 20 0.9598 0.2176 -0.1774
4 15 0.9577 0.2053 -0.1576
5 12 0.9567 0.2000 -0.1494
6 10 0.9561 0.1972 -0.1453
7 8.5714 0.9558 0.1955 -0.1429
8 7.5 0.9556 0.1944 -0.1413
9 6.6667 0.9555 0.1937 -0.1403
10 6 0.9554 01932 -0.1395
Distribution factor with different slots/pole/phase (m): -

27. 27
 Distribution factor for different harmonic components are :
sin( / 2)
sin( / 2)
sin(3 / 2)
3
sin(3 / 2)
sin(5 / 2)
5
sin(5 / 2)
sin(7 / 2)
7 ,
sin(7 / 2)
rd
d
th
th
m
for fundamental component of emf
m
m
for harmoniccomponent of emf
m
K
m
for harmoniccomponent of emf
m
m
for harmoniccomponent of emf
m















and soon













28. 28
Induced emf with 5th and 7th harmonic
Mean square error = 30.162e-3
Induced emf with 7th harmonic
Mean square error = 10.19e-3

29. 29
 Effect of 7th harmonic component can be
reduced by using distributed winding.
 For the machine shown in the figure let the
rms value of induced emf/turn is 1 V.
 So, the induced emf in the concentrated
winding = 30 V (as all the turns have the
same magnetic axis, the induced emf is
simply the algebraic sum i.e. 30 turns x 1
V = 30V
N
S




X
.
X
. ..
X
.
X
30 Conductors
kept in one slot
N
S




X
.
X
. ..
X
.
X
X
.
.
X
X
.
30 Conductors
kept in two slots

 Induced emf in the distributed winding, with m = 2, will not be
the algebraic sum. This is because the 30 turns are now divided
into two groups 15 turn in each group with a phase displacement
of ‘β’ (3600/12 = 300 electrical).
 So the resultant emf will the vector sum of two 15 V sources
with a phase difference of 300 electrical i.e.
15 V
15 V
Ev

2 15 cos(30 / 2) 28.9778vE V   

30. 30
 So, distribution factor
 Distribution factor can also be calculated as:
vectorsumof emf 28.9778
= 0.9659
Algebriacsumof emf 30
dK  
sin(mβ/2) sin(2 30 / 2)
0.9659
msin(β/2) 2 sin(30 / 2)
dK

  

Mean Square Error = 1.2319
with m = 2
Phase difference = 60
Phase difference = 300

31. 31
Mean Square Error = 0.3571
with m = 10

32. 32
Advantages and disadvantages of distributed winding: -
Advantages: -
 Reduces harmonic component in emf and so, voltage wave form is improved.
 Even distribution of conductors, in the stator periphery helps for better cooling.
 The core is fully utilized as the conductors are distributed over the slots on the
armature periphery.
 Disadvantages: -
 Reduces the RMS value of the induced emf.

33. 33
Thank you