EMF EQUATION.ppt

EMF EQUATION

1. CONTENTS
EMF EQUATION OF AN ALTERNATOR
For full pitched concentric winding
For short pitched distributed winding

2. EMF EQUATION OF AN
ALTERNATOR
Considerthefollowing
=fluxperpoleinwb
p=Numberofpoles
N
s=Synchronousspeedinrpm
f=frequencyofinducedemfinHz
Z=totalnumberofstatorconductors
Z
ph=conductorsperphaseconnectedinseries
T
ph=Numberofturnsperphase
•Assumingconcentratedwinding,consideringoneconductor
placedinaslot

3. •AccordingtoFaradaysLawelectromagnetic
induction,
•Theaveragevalueofemfinducedperconductorin
onerevolutione
avg=d/dt
•e
avg=ChangeofFluxinonerevolution/Timetaken
foronerevolution
•ChangeofFluxinonerevolution=p×
•Timetakenforonerevolution=60/N
sseconds
•Hence e
avg= (p ×) / ( 60/N
s) = p ××N
s/ 60
•We know f = pN
s/120
•Hence pN
s/60 = 2f
•Hence e
avg= 2 f volts

4. •If there are Z conductors connected in series/phase,
•Hence average emf = 2 ×f Z volts
since Z = 2 T
ph
•Hence average emf per turn = 4 ×f volts
•i.e. If there are T
ph, number of turns per phase
connected in series,
•Then average emf induced in Tph turns is
E
ph,
avg= T
phx e
avg= 4 f T
phvolts
•Form factor = R.M.S. value/Average value = 1.11
………….. (if e.m.f. is assumed sinusoidal)
R.M.S. value = 1.11 ×Average value

5. •Hence RMS value of emf induced/phase
E = 1.11 ×E
ph,
avg
= 1.11 ×4 f T
ph volts
•This is the general emf equation for the machine
having concentrated and full pitched winding.
•In practice, alternators will have short pitched
winding and hence coil span will not be 180
o
, but
on or two slots short than the full pitch.

6. •PITCH FACTOR:
•Asshownintheabovefigure,considerthecoilshortpitchedby
anangle,calledchordingangle.
•Whenthecoilsarefullpitchedtheemfinducedineachcoilside
willbeequalinmagnitudeandinphasewitheachother.
•Hencetheresultantemfinducedinthecoilwillbesumofthe
emfinduced.

7. •HenceEc=E1+E2=2Eforfullpitchedcoils,
•Hencetotalemf=algebraicsumoftheemfs=vector
sumofemfsasshowninfigurebelow,
•Whenthecoilsareshotpitchedbyanangle,theemf
inducedineachcoilsidewillbeequalinmagnitude
butwillbeoutofphasebyanangleequalto
chordingangle.
•Hencetheresultantemfisequaltothevectorsumof
theemfsasshowninfigurebelow.

8. •HencetheresultantcoilemfisgivenbyE
c=2E
1cos
/2=2Ecos/2volts.
•Hencetheresultantemfintheshortpitchedcoilsis
dependantonchordingangle.
•Nowthefactorbywhichtheemfinducedinashort
pitchedcoilgetsreducediscalledPITCHFACTORand
DEFINITION:
•Itisdefinedastheratioofemfinducedinashort
pitchedcoiltoemfinducedinafullpitchedcoil.

9. •Pitch factor K
p= emf induced in a short pitched
coil/ emf induced in a full pitched coil
K
p = (2E cos /2 ) / 2E
•K
p= cos /2
•where is called chording angle.
Distribution Factor:
•Even though we assumed concentrated winding in
deriving emf equation, in practice an attempt is
made to distribute the winding in all the slots
coming under a pole.
•Such a winding is called distributed winding

10. •Inconcentratedwindingtheemfinducedinallthe
coilsideswillbesameinmagnitudeandinphase
witheachother.
•Incaseofdistributedwindingthemagnitudeofemf
willbesamebuttheemfsinducedineachcoilside
willnotbeinphasewitheachotherastheyare
distributedintheslotsunderapole.
•Thetotalemfwillnotbesameasthatinconcentrated
windingbutwillbeequaltothevectorsumofthe
emfsinduced.
•Henceitwillbelessthanthatintheconcentrated
winding
•Thefactorbywhichtheemfinducedinadistributed
windinggetsreducediscalleddistributionfactor

11. DEFINITION:
Itisdefinedasdefinedastheratioofemfinducedina
distributedwindingtoemfinducedinaconcentratedwinding.
DistributionfactorK
d=emfinducedinadistributedwinding/emf
inducedinaconcentratedwinding
=vectorsumoftheemf/arithmeticsumof
theemf
Let
E=emfinducedpercoilside
m=numberofslotsperpoleperphase,
n=numberofslotsperpole
=slotangle=180/n

12. •The emf induced in concentrated winding with m
slots per pole per phase = mE volts.
•Fig below shows the method of calculating the vector
sum of the voltages in a distributed winding having a
mutual phase difference of .

13. •WhenmislargecurveACENwillformthearcofa
circleofradiusr.
•FromthefigurebelowAC=2×r×sin/2
•Hencearithmeticsum=m×2rsin/2
•NowthevectorsumoftheemfsisANasshownin
figurebelow=2×r×sinm/2
•ThedistributionfactorK
d=vectorsumoftheemf/
arithmeticsumoftheemf
=(2rsinm/2)/(m×2rsin/2)
K
d=(sinm/2)/(msin/2)
•Inpracticalmachinesthewindingswillbegenerally
shortpitchedanddistributedovertheperipheryofthe
machine

14. •Hence in deducing the emf equation both pitch factor
and distribution factor has to be considered.
•Hence the general emf equation including pitch factor
and distribution factor can be given as
•EMF induced per phase = 4.44 f T
ph×K
pK
dvolts
•E
ph= 4.44 K
pK
df T
phvolts
•Hence the line Voltage E
L= ×phase voltage
= E
ph.

15. Effect of harmonics on Pitch and Distribution
factor
(a)Ifshort–pitchangleorchordingangleisforthe
fundamentalfluxwavethenitsvaluesfordifferent
harmonicsare
•For 3
rd
harmonics = 3
•For 5
th
harmonics = 5 and so on
•Pitch factor K
p= cos /2 ………for fundamental
= cos 3 /2 ………for 3
rd
harmonics
= cos 5 /2 ………..for 5
th
harmonics
•Distribution factor K
c= (sin nm /2) /(m sin n /2)
Where n = order of the harmonics
For n = 1 K
c= (sin m /2) /(m sin /2)
….. For fundamental

16. •For n = 3 K
c= (sin 3m /2) /(msin3 /2)
…………For 3
rd
harmonics
•For n = 5 K
c= (sin 5m /2) /(msin5 /2)
……......For 5
th
harmonics
•(b) Frequency also changes if the fundamental
frequency is 50Hz i.e. f
1= 50Hz
–For 3
rd
harmonics f
3= 3 ×50 = 150Hz
–For 5
th
harmonics f
5= 5 ×50 = 250Hz

17. SOLVED PROBLEMS
1.A3,50Hz,starconnectedsalientpole
alternatorhas216slotswith5conductorsper
slot.Alltheconductorsofeachphaseare
connectedinseries;thewindingisdistributed
andfullpitched.Thefluxperpoleis30mW
andthealternatorrunsat250rpm.Determinbe
thephaseandlinevoltagesofemfinduced.

18. Given Data:
N
s= 250 rpm, f = 50 Hz,
m = 3, Ss = 216, Zs = 5, = 30 mWb
To Find:
E
ph,E
line
Solution:
Step 1:
P = 120 ×f/N
s= 120 ×50/250 = 24 poles
= 180
o
/ number of slots/pole = 180
o
/ (216/24)
= 20
Step 2:
K
d= ( sin m /2) / (m sin /2)
= ( sin 3 ×20 / 2) / (3 sin 20/2)
= 0.9597
Pitch factor Kp = 1 for full pitched winding

19. Step 3:
T
ph= Z
ph/2 ; Z
ph= Z/m = Z/3
Z = conductor/ slot x number of slots
T
ph= Z/6 = 216 x 5 /6 = 180
Step 4:
E
ph= 4.44 K
p K
df T
phvlolts
= 4.44 ×1 ×0.9597 ×50 ×30 ×10
-3
×180
= 1150.488 volts
Step 5:
Line Voltage E
Line= ×phase voltage = E
ph
= ×1150.488
= 1992.65 volts

20. SOLVED PROBLEM
2.A3,16pole,starconnectedsalientpole
alternatorhas144slotswith10conductorsper
slot.Thealternatorisrunat375rpm.The
terminalvoltageofthegeneratorfoundtobe
2.657kV.Determinethefrequencyofthe
inducedemfandthefluxperpole.

21. GIVEN DATA:
Ns = 375 rpm, p =16,E
L= 2.657 kV, Z
ss= 10, S
s= 144, phase = 3
TO FIND:
1. f, 2. Φ
FORMULA USED:
a) E
ph = 4.44 K
pK
dfΦT
phvlolts
b) f = P N
s /120
SOLUTION:
Step : 1
f = 16 ×375/120 = 50 Hz
Step : 2
Assuming full pitched winding K
p= 1
Number of slots per pole per phase m = (S
s)/ (p ×phase) = 144/(16 x 3) = 3
Step :3
Slot angle = 180
0
/ number of slots/pole = 180
0
/9 = 20
0
Step :4
Distribution factor Kd = ( sin m /2) / (m sin /2)
= ( sin 3 x 20 / 2) / (3 sin 20/2)
= 0.9597

22. Step:5
Turns per phase T
ph= 144 ×10/ 6 = 240
Step:6
E
ph = E
L/3 = 2.657/3 = 1.534 kV
Step:7
E
ph = 4.44 K
pK
df ΦT
phvlolts
1534.0 = 4.44 ×1 ×0.9597 ×50 ×Φ×240
Φ= 0.03 wb= 30 mwb

23. Unsolved problems
1.A4pole,3phase,50Hz,starconnected
alternatorhas60slotswith4conductorsper
slot.Thecoilsareshortpitchedby3slots.If
thephasespreadis60
0
,findthelinevoltage
inducedforafluxperpoleof0.943wb.

24. Solution hint’s
•Slot angle = phase spread/ number of slots per pole/phase
= 60/5 = 12
•Distribution factor K
d= (sin m /2) / (m sin /2)
•Pitch factor = cos/2
•Coils are short chorded by 3 slots
•Slot angle = 180/number of slots/pole
= 180/15 = 12
•Therefore coil is short pitched by = 3 x slot angle
= 3 x 12 = 36
0