Electrical Machines by Dr. P.S. Bimbhra

1. Mechanical Work DoneMechanical Work Done
in a Singly excited systemin a Singly excited system
by Dr. P.S. Bimbhra

2. Mechanical Work DoneMechanical Work Done
 Consider the simple magnetic relay again, the movable end ofConsider the simple magnetic relay again, the movable end of
armature is assumed to be held in open position.armature is assumed to be held in open position.
 When the switch ‘S’ is closed, the exciting current increasesWhen the switch ‘S’ is closed, the exciting current increases
from Zero(0) to Ifrom Zero(0) to I11= V= Vtt / r and flux linkage rises from Zero to/ r and flux linkage rises from Zero to ΨΨ1.1.
HereHere ‘r’ is coil resistance.‘r’ is coil resistance.

3.  Consider the fig. shown:-Consider the fig. shown:-
 The moveable end of the Armature is assumed in closedThe moveable end of the Armature is assumed in closed
Position.Position.
 When switch ‘S’ is closed the exciting current rises from Zero toWhen switch ‘S’ is closed the exciting current rises from Zero to
I1 = Vt / r.I1 = Vt / r.
 Whereas flux linkage rises from zero toWhereas flux linkage rises from zero to ΨΨ2 [2 [ΨΨ2 >2 > ΨΨ1].1].

4. ► The energy stored in the magnetic field is now given by theThe energy stored in the magnetic field is now given by the
cross-hatched area OCDO.cross-hatched area OCDO.
► During the armature movement from its open position, theDuring the armature movement from its open position, the
reluctance of the magnetic path decreases, thereby increasingreluctance of the magnetic path decreases, thereby increasing
the flux linkage from its initial value ofthe flux linkage from its initial value of ΨΨ1.1.

5. ► These increments in flux linkage induces a counter Emf in theThese increments in flux linkage induces a counter Emf in the
coil, which opposes the flow of exciting current I,coil, which opposes the flow of exciting current I,
ie.ie.
The that magnitude of counter Emf induced in the exciting coilThe that magnitude of counter Emf induced in the exciting coil
depends on how fast the armature moves.depends on how fast the armature moves.

6. The two extreme cases of armature motion are consideredThe two extreme cases of armature motion are considered
below:-below:-
► Slow Movement: -Slow Movement: -
 With the armature in open positionWith the armature in open position
the exciting current is Ithe exciting current is I1;1; the fluxthe flux
linkage arelinkage are ΨΨ1 and operating point1 and operating point
is ‘A’.is ‘A’.
 In closed position the flux linkageIn closed position the flux linkage
isis ΨΨ2 , current is ‘2 , current is ‘II11’ and operating’ and operating
point is ‘C’.point is ‘C’.
 The slow movement of armature causes almost negligibleThe slow movement of armature causes almost negligible
counter Emfcounter Emf
asas Time
12 Ψ−Ψ
=

7. ► As a result , exciting current ‘IAs a result , exciting current ‘I11’ remains almost constant during’ remains almost constant during
armature movement and operatingarmature movement and operating
point is ‘A’ reaches new operatingpoint is ‘A’ reaches new operating
point ‘C’ along vertical line ACpoint ‘C’ along vertical line AC
as shown:--as shown:--
Now magnetic energy storedNow magnetic energy stored
Changed during the process—Changed during the process—
WWfldfld =(Energy stored in closed position)-(Energy stored in open)=(Energy stored in closed position)-(Energy stored in open)
WWfldfld = Area OA’CDFO – Area OAA’FO= Area OA’CDFO – Area OAA’FO

8. ► Electrical energy input during this change:-Electrical energy input during this change:-
Area ACDFA’AArea ACDFA’A
ButBut
WWelecelec =W=Wmechmech + W+ Wfldfld
Wmech = Area OACDFO – Area OA’CDFOWmech = Area OACDFO – Area OA’CDFO
= Area OACA’O= Area OACA’O
=Ψ−Ψ=Ψ= ∫
Ψ
Ψ
)(. 121
2
1
1 IdIWelec

9. Instantaneous MovementInstantaneous Movement
 Here the armature is assumed to move from open to closedHere the armature is assumed to move from open to closed
position instantaneously.position instantaneously.
 According to constant flux linkage theorem:-According to constant flux linkage theorem:-
““ The flux linkage with an inductive circuit can’tThe flux linkage with an inductive circuit can’t
change suddenly."change suddenly."
 Here also during the fastHere also during the fast
movements of armature themovements of armature the
flux linkage does not change &flux linkage does not change &
remain constant atremain constant at ΨΨ1.1.
 The operating point thereforeThe operating point therefore
travels horizontally fromtravels horizontally from
A to A’.A to A’.

10. ► After the armature has closed the operating point travels from A’ to C alongAfter the armature has closed the operating point travels from A’ to C along
the closed position magnetization curve since the final operating point has tothe closed position magnetization curve since the final operating point has to
be ‘C’.be ‘C’.
► Change in magnetic stored energy:Change in magnetic stored energy:
WWfldfld = Area OA’FO – Area OAA’FO= Area OA’FO – Area OAA’FO
ButBut
WWelecelec = W= Wfldfld + W+ Wmechmech
=Area OA’FO - Area OAA’FO +W=Area OA’FO - Area OAA’FO +Wmechmech
Therefore—Therefore—
WWmechmech = Area OAA’O= Area OAA’O
∫
Ψ
Ψ
=Ψ=
1
1
0.dIWelec

11. ► From the above Eqns we can easily make out the followingFrom the above Eqns we can easily make out the following
deductions:-deductions:-
1)1) During the fast movement of the arm there is no electricalDuring the fast movement of the arm there is no electrical
energy input.energy input.
2)2) Mechanical energy output = reduction in magnetic energyMechanical energy output = reduction in magnetic energy
storedstored

12. Lecture 4Lecture 4
Doubly-Excited Magnetic SystemDoubly-Excited Magnetic System

13. Doubly-Excited Magnetic SystemDoubly-Excited Magnetic System
► In this article we again will find the expressionIn this article we again will find the expression for :-for :-
1)1) Electrical Energy Input.Electrical Energy Input.
2)2) Magnetic Field Energy Stored.Magnetic Field Energy Stored.
3)3) Mechanical Work Done.Mechanical Work Done.

14. Doubly-Excited Magnetic SystemDoubly-Excited Magnetic System
► Consider the Fig Shown below:-Consider the Fig Shown below:-
► The model shown consist of a stator with NThe model shown consist of a stator with Nss turns andturns and
energized from source 1 andenergized from source 1 and
the rotor with Nthe rotor with Nrr turns and isturns and is
excited from source 2.excited from source 2.
 We know that for a singlyWe know that for a singly
excited system the differentialexcited system the differential
electrical energy input is givenelectrical energy input is given
by:- dWby:- dWelec =elec =I.dI.dΨΨ
ThereforeTherefore
for two energy sources 1&2for two energy sources 1&2
dWdWelecelec = I= Is.s.ddΨΨss +I+Irr.d.dΨΨr Eqn(1)r Eqn(1)

15. ► HereHere ΨΨs &s & ΨΨr are the instantaneous flux linkage of stator andr are the instantaneous flux linkage of stator and
rotor windings and can be expressed in terms of self androtor windings and can be expressed in terms of self and
mutual inductance respectively as:-mutual inductance respectively as:-
ΨΨs = Ls = Lss IIss + M+ Msrsr IIr andr and ΨΨr = Lr = Lrr IIrr + M+ Msrsr IIss
WhereWhere
LLr =r = self–inductance of stator windingself–inductance of stator winding
LLs =s = self-inductance of rotor windingself-inductance of rotor winding
MMsrsr,M,Mrs =rs = mutual inductance btwn rotor and stator windingsmutual inductance btwn rotor and stator windings
 Initially the space angle between rotor and stator axes isInitially the space angle between rotor and stator axes is θθr andr and
both the current Is and Ir are assumed zeros.both the current Is and Ir are assumed zeros.
 When the stator and rotor coils are switched on to thereWhen the stator and rotor coils are switched on to there
respective supplies current rises from zero to Is and Irrespective supplies current rises from zero to Is and Ir
respectively.respectively.

16. ► If now the rotor is not allowed to move, then dWIf now the rotor is not allowed to move, then dWmechmech is zerois zero
and we have:and we have:
dWdWelecelec = 0 + dW= 0 + dWfldfld
dWdWelecelec = dW= dWfldfld = Is. d= Is. dΨΨs + Ir. ds + Ir. dΨΨrr
=Is.d(Ls.Is + Msr.Ir) + Ir.d(Lr.Ir + Mrs.Ir) -----Eqn(1a)=Is.d(Ls.Is + Msr.Ir) + Ir.d(Lr.Ir + Mrs.Ir) -----Eqn(1a)
 Since the rotor is not allowed to move, reluctances & thereforeSince the rotor is not allowed to move, reluctances & therefore
the inductances are constant. Hence dLs, dLr, dMsr are allthe inductances are constant. Hence dLs, dLr, dMsr are all
zeros.zeros.
ThereforeTherefore
dWdWfldfld =IsLs dIs + IsMsr dIr +IrLr dIr + IrMsr dIs=IsLs dIs + IsMsr dIr +IrLr dIr + IrMsr dIs
dWdWfldfld = IsLs dIs + IrLr dIr + Msr d(Is.Ir)= IsLs dIs + IrLr dIr + Msr d(Is.Ir)

17. ► The magnetic field energy stored in establishing currents fromThe magnetic field energy stored in establishing currents from
zero to Is and Ir, is given by:-zero to Is and Ir, is given by:-
--------Eqn(2)--------Eqn(2)
For Torque assume the rotor to move through a virtualFor Torque assume the rotor to move through a virtual
displacement in the direction of ‘Tdisplacement in the direction of ‘Tee’, therefore the inductances’, therefore the inductances
Ls, Lr, Msr must varyLs, Lr, Msr must vary
∫ ∫∫ ++=
Is IsIr
srsr
Ir
rrrsssfld IIdMdIILdIILW
0 00
)(
rssrrrss IIMLILI ++= 22
2
1
2
1

18. ► Therefore the differential electrical energy input dWelec is:-Therefore the differential electrical energy input dWelec is:-
► Differential magnetic energy stored during virtual displacementDifferential magnetic energy stored during virtual displacement
ddθθ is :-is :-
rrrrrsrrsrsrsssssselec dLIdILIdMIIdIMIdLIdILIdW 22
+++++=
srrsssrr dMIIdIMI ++
srsrrssrrrrrrsssssfld dIIMdIIMdIILdLIdIILdLIdW +++++= 22
2
1
2
1
srsr dMII+

19. ► NowNow
dWdWmechmech =T=Tee ddθθrr
Therefore by-Therefore by-
dWdWelecelec = dW= dWmechmech +dW+dWfldfld
The Underline term in the above eqn will get cancelled and we have:-The Underline term in the above eqn will get cancelled and we have:-
-------Eqn(3)-------Eqn(3)

20. ► From the above eqn(3) we can make out thatFrom the above eqn(3) we can make out that
Torque depends on :Torque depends on :
► Instantaneous values of currents Is ans Ir.Instantaneous values of currents Is ans Ir.
► The angular rate of change of inductance.The angular rate of change of inductance.
------Eqn(4)------Eqn(4)
Now integrating Eqn(1a) with constant current gives:-Now integrating Eqn(1a) with constant current gives:-
------Eqn(5)------Eqn(5)
Comparing Eqn’s(2)(3)(5) reveals that if electrical energy inputComparing Eqn’s(2)(3)(5) reveals that if electrical energy input
takes place at constant current then half of it is converted totakes place at constant current then half of it is converted to
mechanical energy and rest half is stored in magnetic field.mechanical energy and rest half is stored in magnetic field.

21. Generated EMF in a MachineGenerated EMF in a Machine
► Consider the figure shown:-Consider the figure shown:-
► The B-wave moves towards leftThe B-wave moves towards left
With a speed ofWith a speed of ωω elect. Rads/secelect. Rads/sec
OrOr ωωm mech. Rads/sec.m mech. Rads/sec.
 At the origin the coil sides areAt the origin the coil sides are
Located in the interpolar regionLocated in the interpolar region
Where the full pole flux links theWhere the full pole flux links the
Coil. At any time ‘t’ the coil hasCoil. At any time ‘t’ the coil has
Relatively moved by-Relatively moved by-
elect radselect radstωα =

22. ► The B-wave can be expressed as:-The B-wave can be expressed as:-
B=BB=Bpp SinSinθθ
== BBpp Sin(P/2)Sin(P/2)θθm [Bp= peak flux linkage]m [Bp= peak flux linkage]
The fluxThe flux ΦΦ linking the coil can belinking the coil can be
Completed by integrating over theCompleted by integrating over the
Mechanical angle. Thus:-Mechanical angle. Thus:-
== (2/P) 2 B(2/P) 2 Bpp llrr CosCos
= (2/P) 2 B= (2/P) 2 Bpp llrr CosCos ωωtt
== ΦΦ CosCos ωωt [t [ΦΦ=(4B=(4Bppllrr)/P])/P]
Then flux linkage of the coil at any time ‘t’ is:-Then flux linkage of the coil at any time ‘t’ is:-
λλ=N=Nφφ= N= N ΦΦ CosCos ωωt [N= Number of turns of the coil]t [N= Number of turns of the coil]
∫
+Π
=
α
α
θθφ dSinlB
P
rp
2
∫
+Π
=
α
α
θθφ dSinlB
P
rp
2
α

23. ► Hence the coil induced emf is:-Hence the coil induced emf is:-
► The Rms value of the emf induced is:-The Rms value of the emf induced is:-
tSinN
dt
d
e ωω
λ
Φ=−=
Φ=ΦΠ= fNfNE 44.42

24. ► Rotating Magnetic field in a Three phase SystemRotating Magnetic field in a Three phase System
► Consider that three phase of an AC wdgs are carrying balancedConsider that three phase of an AC wdgs are carrying balanced
alternating currents:-alternating currents:-
► IIaa = I= Imm CosCos ωωtt
► IIbb = I= Imm Cos (Cos (ωωt-120)t-120)
► IIcc = I= Imm Cos (Cos (ωωt-240)t-240)

25. ► These currents will now setup three pulsating mmf waves whichThese currents will now setup three pulsating mmf waves which
have a time phase difference of 120 from each other.have a time phase difference of 120 from each other.
► Since the magnetic axes are located 120 apart in space fromSince the magnetic axes are located 120 apart in space from
each other, the three mmf’s can be expressed as:-each other, the three mmf’s can be expressed as:-
FFaa = F= Fmm CosCos ωωt Cost Cos θθ ------------------------Eqn(1)------------------------Eqn(1)
FFbb = F= Fmm Cos (Cos (ωωt-120) Cos (t-120) Cos (θθ-120 )---------Eqn(2)-120 )---------Eqn(2)
FFcc = F= Fmm Cos (Cos (ωωt-240) Cos (t-240) Cos (θθ-240)----------Eqn(3)-240)----------Eqn(3)
Now,Now,
The resultant of the above 3 mmf’s is:-The resultant of the above 3 mmf’s is:-
F= FF= Faa + F+ Fbb + F+ Fcc
Ie.Ie. FF= F= Fmm [[CosCos ωωt Cost Cos θθ + Cos (+ Cos (ωωt-120) Cos (t-120) Cos (θθ-120 )+-120 )+
Cos (Cos (ωωt-240) Cos (t-240) Cos (θθ-240)]-------Eqn(4)-240)]-------Eqn(4)

26. ► Simplifying the above Eqn(4) trignometricallySimplifying the above Eqn(4) trignometrically:-:-
► Since:-Since:-
Cos(Cos(θθ++ωωt-480t-480) = Cos() = Cos(θθ++ωωt-120t-120))
Putting this in above Eqn we have—Putting this in above Eqn we have—
)]480()240()([
2
1
)]([
2
3
−++−+++
+−=
tCostCostCosF
tCosFF
m
m
ωθωθωθ
ωθ
)]240(
)120()([
2
1
)(
2
3
−++
−++++−=
tCos
tCostCosFtCosFF mm
ωθ
ωθωθωθ

27. ► Now the peak value of the mmf wave:-Now the peak value of the mmf wave:-
► Cos (Cos (θθ--ωωtt)=1=Cos 0)=1=Cos 0
((θθ--ωωtt) =0) =0
((θθ==ωωtt))
So,So,
Peak EmfPeak Emf
FFpeakpeak =(3/2)F=(3/2)Fm --------m --------Eqn(5)Eqn(5)