1. Dr. S &S.S GHANDHY GOVERNMENT ENGINEERING COLLEGE
Electrical Department
SUBJECT: Electromechanical Energy Conversion
SUBMITED BY :-
NAME
ENROMENT NO.
LETHWALA PAVAN
ASHISH MALKIYA
MARATHE POOJA
MEHTA KARNAV
GANDHI MITESH
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2. Basics - electromechanical energy conversion
The conversion of electrical energy into mechanical energy or mechanical energy into electrical energy is called Electromechanical energy conversion
An electrical generator is a machine which converts mechanical energy into electrical energy
An electrical motor is a machine which converts electrical energy into mechanical energy

3. Important Parts in electromechanical energy conversion
An electrical system
An mechanical system
A coupling field
a) electrical or
b) magnetic

4. Electromechanical Energy Conversion
Consider the block diagram depicted below.
Coupling
Field
Electric
System
Mechanic
System
WE = We WeL + WeS
Energy supplied by an electric source
Energy transferred to the coupling field by the electric system
Energy losses of the electric system. Basically, I2R
Energy stored in the electric o magnetic field

5. coupling medium
There are two types of coupling medium in electromechanical conversion.
Magnetic field as a coupling device
In Generator, On Application Of Mechanical Force On The Conductor, The Conductor Moves In The Opposite Direction Of The Magnetic Force Over Coming This Opposition The Energy .
The Magnetic Field Therefore Plays The Part Coupling Link.

6. In Motor, The Current Carrying Conductor Is Placed In The Magnetic Field. The Conductor Experience A Force That Tends To Move It.
Now Conductor Is Free To Move In The Direction Of This Force. Thus Magnetic Field, In This Case , Helps The Conversion Of Electrical Energy , Acting As Coupling Medium.

7. CONTD…. 2. Electric field as a coupling medium
In electronic microphones and in electrostatic voltmeters the electric field plays the part of coupling medium for energy conversion.
Taking a case of a capacitor having plates charged oppositely, separated by a dielectric. A force of attraction exists between the two plates. The plates tends to move towards each other. If one plate is allowed to move in, the direction of force. Electrical energy gets converted into mechanical energy.

8. CONTD….
If we compare these two coupling media, very small force is developed in electric field use but a large force in case of magnetic field media
Fmech > Fele

9. Electro mechanical energy conversion devices
Devices Which Carry Out Electrical Operations By Using Moving Parts Are Known As Electromechanical Devices.
Common Devices Are Generators And Motors
b) Generators
a) MOTOR
d) relay
c) Car Starter

10. Important points Worth noting in Electromechanical Energy Conversion
Electromechanical Energy Conversion can be analyses by using principle of energy conservation and electrical circuits and network.
Ignoring loss of energy in system is ideally a reversible process as show in fig.
In practical, during conversion some energy is lost forever due to copper loss, iron loss, mechanical loss.
In electro-mechanical devices some air-gap is to be maintained between static and moving parts.
Reluctances of air-gap is much more. Majority of the ampere turns of winding are required to overcome air gap reluctance. Most of the energy is store in the air gap, this is retuned to the electric source when the field is reduced.

11. Cont…
Taking into account electrical loss (I2 R). Field loss in core due to change in magnetic filed and loss due to rotation. i.e. friction and windage loss (mech loss) the modified electro-mechanical energy conversion system is represented as shown in fig.
V= Supply Voltage, I = Current, R = Resistance, loss input, output in watts

12. Derivation of Magnetic Energy Stored
V = Input applied Voltage; I = Current in the coil;
R = Resistance of the coil; e = Self or back developed in coil;
N = Number of turns of the coil; Ø = flux; Ψ = mmf i.e. Ψ = NØ
Supply voltage V has to send the current in the coil overcoming its resistance(r) and opposing emf ‘e’.
but……
… Ψ= NØ
(1)
This is voltage balance equation.

13. Cont…
Similarly to find power and energy balance equations. Multiplying equ. By I then
This is power balance equations. Now multiply by time dt .
This is energy balance equ.
Re- arranging the terms we get,
(V-ir) i.dt = idΨ as V-ir =e
In equ e.idt is electrical energy output or electrical energy is converted into magnetic energy. This is called as “co-energy” ”
Co-energy = e.idt
and I dΨ = Magnetic energy or stored energy

14. Cont… We can find total energy stored (wfield) by integration=
Total energy stored =

15. Graphical representaion of Co-energy and field energy (in open state)
Wco=co-energy in area OAC=
Wfield =field energy in area OAB=

16. Graphical representaion of Co-energy and field energy (in closed steady state)
In this case , there is no air-gap and flux path is through only iron core which has magnetic saturation limit.

17. Graphical representaion of Co-energy and field energy (in closed steady state) cont…
As current increases the flux-linkage decreases. so magnetization is no more a linear but a curve.
In (1) is open steady state energy and (2) is close steady state energy
In store energy
close steady state energy > open steady state energy
Comparision of open and close energy

18. Classification of Excitation system
Singly excited: in this a single coil is used which is supported on fixed part and necessary excitation is produced which is magnetizes fixed as well as movable part.
Ex. Magnetic relays , induction machine. These are used for motion through limited distances or through small angle. .
Magnetic relays
induction machine

19. Classification of Excitation system(contd..)
2. Doubly excited: have two exciting coils ,dc motors , commutator motors and synchronous motors are doubly excited type machine.
D.C motor
synchronous motor

20. Classification of Excitation system(contd..)
3. Multiple Singly excited: stator part excited by 3 winding with 3-phase supply and rotor receives power by transformer action i.e. 3-phase induction motor.
3-phase induction motor.

21. Classification of Excitation system(contd..)
4. Multiple doubly excited: 3-phase synchronous motor’s 3-phase windings are excited by 3-phase AC supply at the same time rotor single windings is excited by a separate d.c. voltages
3-phase synchronous motor

22. Singly Excited Magnetic Field System
Single coil is used in singly excited magnetic system.
It is supported on fixed part.
Necessary excitation is produced which magnetizes fixed and moveable part. Ex Magnetic relays, Induction machine.
Induction machine

23. Voltage Equation of Singly Excited Magnetic System
Take the basic construction of a static core and a
rotable part rotor.
Stator is wound with an exciting coil whereas
rotor has no coil.
It is simply a magnetic material part supported on shaft and rotate.
Initial position is 1.
Putting on the voltage V the coil carries the
current I and a magnetic flux ø is produced. Stator
is magnetized.
The same flux passes through rotor body and it
gets with opposite polarities by induction.

24. Once it comes to position 2 rotor stops rotating
Being opposite stator and rotor poles, rotor is attracted by a force and takes position 2.
This is because 2 position has less reluctance than 1 position of the rotor.
Once it comes to position 2 rotor stops rotating
V= Supply Voltage, I = Current, R = Resistance , e = Self induce emf of coil
Therefore,
V = I*R + e
e = N* dø/dt
Where,
N = number of turns of coil
Ø = magnetic flux

25. Hence,
V = I*R + d (N. Ø)/dt
V = I*R + d ψ/dt
Multiplying above equation with I on both sides,
V*I = I*I*R + I* d ψ/dt
Multiplying above equation with time dt,
V*I dt = I*I*R dt + I* dψ/dt

26. Integrating above equation, We = Wloss + Wf
Total electrical input = Electrical energy loss + Electrical energy
converted into
mechanical energy
Now,
Wf = We(energy stored Wme(energy spent in doing
in magnetic field) mechanical work)
From above relation we can write static equation when rotor is
Steady and Wme = 0
Energy stored is Wf = integration of I dψ = (L*I*I)/2
Where ,
ψ = L*I

27. Doubly Excited Magnetic Field System
The system is excited by two different coils.
One coil is wound on the static part ‘stator’ and the rotor also has coil wound on it as shown in figure.
For example DC motors , Commutator motor and synchronous motor.
Commutator motor

28. Energy Equation For Doubly Excited Magnetic System
Stator coil has resistance R1 and rotor coil has resistance R2.
Respective exciting voltages are V1 and V2 and currents in coils are I1 and I2.
Ø1 and Ø2 are respective fluxes by coil 1 and coil 2.
Ψ1 and Ψ2 are total flux linkages of coil 1 and coil 2.
L1 ,L2 and M are self and mutual inductances of coil 1 and coil 2.

29. Flux linkages with coil 1 and coil2,
Ψ1 = L1*I1 + M*I2
Ψ2 = L2*I2 + M*I1
Voltage equations,
V1 = I1*R1 + d Ψ1/dt
V2 = I2*R2 + d Ψ2/dt
Putting the value of Ψ1 in voltage equation,
V1 = I1*R1 + d (L1I1 + MI2)/dt
V1 = I1*R1 + L1dI1 /dt + I1 dL1 /dt + MdI2 /dt + I2 dM /dt
Multiplying with I1 on both sides,
V1*I1 = I1* I1*R1 + L1* I1 dI1 /dt + I1 * I1 dL1 /dt + M *
I1 dI2 /dt + I1 * I2 dM /dt

30. Similar equation for rotor side,
V2*I2 = I2* I2*R2 + L2* I2 dI2 /dt + I2 * I2 dL2 /dt + M * I2 dI1 /dt +
I1 * I2 dM /dt
Adding both equations and multiplying with dt,
(V1*I1 + V2*I2)dt = (I1* I1*R1 + I2* I2*R2 )dt + (L1* I1 dI1 +
L2* I2 dI2 + M * I1 dI2+ M * I2 dI1 +
2* I1 * I2 dM + I1 * I1 dL1 + I2 * I2 dL2 )dt
(V1*I1 + V2*I2- I1* I1*R1 - I2* I2*R2 )dt = (L1* I1 dI1 + L2* I2 dI2 +
I1 * I1 dL1 + I2 * I2 dL2 )dt
+ (M * I1 dI2+ M * I2 dI1 +
2* I1 * I2 dM)dt
Useful electrical energy input = Field energy stored in the system
+ Electrical to Mechanical energy
transfer

31. Wfe = ½*L1*I1*I1 + ½*L2*I2*I2 +M*I1*I2
From above derived relations we can find the stored magnetic
energy in doubly excited system considering steady state where
mechanical output is treated as zero the entire electrical energy is stored in the system in magnetic energy form.
Steady terms in equation dL1 , dL2 , dM is zero and after
integrating the equation,
Wfe = ½*L1*I1*I1 + ½*L2*I2*I2 +M*I1*I2

32. Electromagnetic Torque Equation For Doubly Excited System
Stored energy is used to mechanical work.
Mechanical work done = Rate of change of stored energy
= dWfe/dt
dWfe/dt = d (½*L1*I1*I1 + ½*L2*I2*I2 +M*I1*I2 )/dt
= ½*L1*d(I1*I1)/dt +½*I1*I1*dL1/dt + ½*L2*d(I2*I2)/dt
+½*I2*I2*dL2/dt +M*I2*dI1/dt+M*I1*dI2/dt+I1*I2dM/dt
= L1*I1d(I1)/dt + ½*I1*I1*dL1/dt + L2*I2(dI2/dt)
+ ½*I2*I2*dL2/dt+M*I2*dI1/dt+M*I1*dI2/dt
+I1*I2dM/dt

33. Integrating above equation w.r.t dt,
Wfe = (L1*I1dI1 + L2*I2dI2 +M*I1*dI1+M*I1*dI2)+( ½*I1*I1*dL1
+ ½*I2*I2*dL2+I1*I2*dM)
Wfe = Wstored + Wme
Wme = ½*I1*I1*dL1 + ½*I2*I2*dL2+I1*I2*dM
In rotation while mechanical energy is produced the terms dL1 ,
dL2, dM are not constant but vary with θm.
Differentiating above equation w.r.t θm ,
Torque equation = dWme/dθm =½*I1*I1*dL1/dθm+½*I2*I2*dL2/dθm I1*I2*dM/dθm
Putting I2 = 0, we get torque equation for singly excited system,
Torque equation = ½*I1*I1*dL1/dθm

34. Comparison between Singly and Doubly Excited Systems
Singly Excited System
One coil takes active part in energy conversion process.
It has winding on stationary part only.
It works on induction or asynchronous principle.
Used in constant speed application.
Produce useful startup torque.
E.g. Magnetic relays, D.C. generator, Induction machine
Doubly Excited System
Two coils take active part in energy conversion process.
It has winding on stationary and rotating part.
It works on synchronous principle.
Used in variable speed application.
Do not produce useful startup torque.
E.g. DC motor , Commutator and Synchronous motor

35. THANK YOU