Numerical on induced emf containing harmonics

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-7

2. 2
Learning Outcomes: - (Previous Lecture_06)
 To solve some numerical related to induced emf containing harmonics.
 To analyse and understand the techniques to eliminate the harmonics from the
generated emf.
 To analyse the effect of distributed winding on emf waveform.

3. 3
Learning Outcomes: - (Today’s Lecture_07)
 To solve some numerical related to induced emf containing harmonics with short
pitched and distributed winding.

4. 4
Numerical on short pitched and distributed windings: -
1. A 3-phase, star connected, 16 pole alternator has 192 slots with 8 conductors per
slot, coil span = 160 electrical degrees, speed of alternator = 375 rpm, flux per pole =
55 mWb. Calculate the line and phase values of EMF generated.
Solution: -
Number of slots/pole/phase, m = 192/(16 x 3) = 3.
Turns/phase, Tph = Total armature conductors/(2 x 3) = (192 x 8)/6 = 256.
Frequency of generation, f = (PN)/120 = (16 x 375)/120 = 50 Hz.
Coil span is 1600 electrical, i.e. coil is short pitched by 1800-1600 = 200 electrical.
So, pitch factor, Kp = cos(α/2) = cos(200/2) = 0.9848.
Number of slots/poles/phase = 3.
So, slot angle, β = 600/3 = 200 electrical.

5. 5
Distribution factor, Kd = sin(mβ/2)/(m x sin(β/2)) = sin(300)/(3 x sin (200/2)) = 0.9598.
So, induced emf/phase = 4.44fφTphKpKd = 4.44 x 50 x 55 x 10-3 x 256 x 0.9848 x
0.9598 = 2954.503 Volts.
Line value of induced emf = 3 x induced emf/phase = 3 x 2954.503 = 5117.35 Volts.

6. 6
2. A 3-phase, 10-pole, 50-Hz, 600 rpm, star connected alternator has the flux
distribution given by:
𝑩 = 𝒔𝒊𝒏𝜽 + 𝟎. 𝟒𝒔𝒊𝒏𝟑𝜽 + 𝟎. 𝟐𝒔𝒊𝒏𝟓𝜽 + 𝟎. 𝟏𝒔𝒊𝒏𝟕𝜽 Wb/m2.
The alternator has 150 slots with 20 conductors/slot. The coil span is 12 slots. 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 expression of induced emf/phase.
c. RMS value of the line voltage.
d. Instantaneous expression of the line voltage.
Solution: -
The flux distribution has a fundamental component, a 3rd harmonic component, a
5th harmonic component and a 7th harmonic component.
Number of slots/pole/phase, m = 150/(10 x 3) = 5.
Since m is greater than ‘1’, the armature winding is distributed type.

7. 7
Slot angle, β = 3600/150 = 2.40 mechanical = (P/2) x 2.40 = (10/2) x 2.40
= 120 electrical.
Slot angle can also be calculated from the phase spread using the relation,
m β = 600 electrical, so, β = 600/5 = 120 electrical.
Pole pitch, Y = Number of slots/pole = 150/10 = 15 slots.
But, coil span is 12 slots. So the coil is short pitched by 15-12 = 3 slots.
Short pitched angle α = 3 x β = 3 x 120 = 360 electrical.
Number of conductors/phase, Zph = total armature conductors/3
= (150 x 20)/3 = 1000.
Number of turns/phase, Tph = Zph/2 = 1000/2 = 500.

8. 8
Calculation of RMS value of fundamental component
Eph1 = 4.44fφ1TphKp1Kd1
Where, φ1 is the average value fundamental component of flux/pole, Kp1 is the
pitch factor for the fundamental component and Kd1 is the distribution factor of the
fundamental component.
Peak value of fundamental flux, φm1 = Bm1 x A1 = 1 x (πDL)/P
= 1 x (π x 1.2 x 0.4)/10 = 0.1508 Wb.
Average value fundamental flux, φ1 = (2φm1)/ π = 0.096 Wb.
Kp1 = cos(α/2) = cos(360/2) = 0.9511.
Kd1 = sin(m β/2)/(m x sin(β/2)) = sin(5 x 120/2)/(5 x sin(120/2)) = 0.9567.

9. 9
Eph1 = 4.44fφ1TphKp1Kd1 = 4.44 x 50 x 0.096 x 500 x 0.9511 x 0.9567
= 9696.08 Volt.
Calculation of RMS value of 3rd harmonic component
Peak value of 3rd harmonic flux, φm3 = Bm3 x A3 = 0.4 x (πDL)/(3P)
= 0.4 x (π x 1.2 x 0.4)/(3x10) = 0.0201 Wb.
Average value 3rd harmonic flux, φ3 = (2φm3)/ π = 0.0128 Wb.
Kp3 = cos(3α/2) = cos(3 x 360/2) = 0.5878.
Kd3 = sin(3m β/2)/(m x sin(3β/2)) = sin(3 x 5 x 120/2)/(5 x sin(3 x 120/2)) = 0.6472 .
Eph3 = 4.44(3f)φ1TphKp1Kd1 = 4.44 x (3 x 50) x 0.0128 x 500 x 0.5878 x 0.6472
= 1621.52 Volt.

10. 10
Calculation of RMS value of 5th harmonic component
Peak value of 5th harmonic flux, φm5 = Bm5 x A5 = 0.2 x (πDL)/(5P)
= 0.2 x (π x 1.2 x 0.4)/(5x10) = 0.00603 Wb.
Average value 5th harmonic flux, φ5 = (2φm5)/ π = 0.00384 Wb.
Kp5 = cos(5α/2) = cos(5 x 360/2) = 0.
Kd5 = sin(5m β/2)/(m x sin(5β/2)) = sin(5 x 5 x 120/2)/(5 x sin(5 x 120/2)) = 0.2 .
Eph5 = 4.44(5f)φ5TphKp5Kd5 = 4.44 x (5 x 50) x 0.00384 x 500 x 0 x 0.2 = 0.
Calculation of RMS value of 7th harmonic component
Peak value of 7th harmonic flux, φm7 = Bm7 x A7 = 0.1 x (πDL)/(7P)
= 0.1 x (π x 1.2 x 0.4)/(10x10) = 0.00151 Wb.
Average value 7th harmonic flux, φ7= (2φm7)/ π = 0.0009613 Wb.

11. 11
Kp7 = cos(7α/2) = cos(7 x 360/2) = -0.5878.
Kd7 = sin(7m β/2)/(m x sin(7β/2)) = sin(7 x 5 x 120/2)/( 5 x sin(7 x 120/2)) = -0.1494
Eph7 = 4.44(7f)φ7TphKp7Kd7 = 4.44 x (7 x 50) x 0.0009613 x 500 x (-0.5878) x
(-0.1494) = 65.5934 Volt.
a. RMS value of induced emf/phase,
2 2 2 2 2 2 2
1 3 5 7 9696.08 1621.52 65.5934 9830.95ph ph ph ph phE E E E E V       

12. 12
b. Instantaneous expression of induced emf/phase,
c. RMS value of line voltage
d. Instantaneous expression of line voltage
Note:
1. In the line voltage there is no 3rd harmonic component as the armature
winding is star connected.
2. 5th harmonic component is totally eliminated as short pitched angle is 360
electrical.
2(9696.08sin 1621.52sin3 65.5934sin 7 )
13712.33sin 2293.18sin3 92.763sin 7
e   
  
  
  
   
2 2
3 9696.08 3 65.5934 16794.49LE Volt    
 0 0
0
2 3 9696.08sin( 30 ) 3 65.5934sin(7( 30 )
23750.45sin( 30 ) 160.67sin(7 210)
e  
 
      
   

13. 13
Waveform of phase voltages

14. 14Waveform of Line voltages

15. 15
Thank you