1. Dr. S &S.S GHANDHY GOVERNMENT ENGINEERING COLLEGE Electrical Department SUBJECT: Electromechanical Energy Conversion NAMEENROMENT NO. LETHWALA PAVAN130230109024 ASHISH MALKIYA130230109025 MARATHE POOJA130230109026 MEHTA KARNAV130230109027 GANDHI MITESH130230109028 SUBMITED BY :- GUIDED BY: M.R.VASAVDA

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 1)An electrical system 2)An mechanical system 3)A coupling field a) electrical or b) magnetic

4. Electromechanical Energy Conversion Consider the block diagram depicted below. 4 Electric System Coupling Field Mechanic System W E = W e + W eL + W eS Energy supplied by an electric source Energy transferred to the coupling field by the electric system Energy losses of the electric system. Basically, I 2 R Energy stored in the electric o magnetic field

5. coupling medium There are two types of coupling medium in electromechanical conversion. 1.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 F mech > F ele

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 (I 2 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 (w field ) by integration Total energy stored (w field ) = = Total energy stored =

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

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 1.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.. induction machine Magnetic relays

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. 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. Classification of Excitation system(contd..) 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. 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, W e = W loss + W f Total electrical input = Electrical energy loss + Electrical energy converted into mechanical energy Now, W f = W e(energy stored + W me(energy spent in doing in magnetic field) mechanical work) From above relation we can write static equation when rotor is Steady and W me = 0 Energy stored is W f = 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 R 1 and rotor coil has resistance R 2. Respective exciting voltages are V 1 and V 2 and currents in coils are I 1 and I 2. Ø 1 and Ø 2 are respective fluxes by coil 1 and coil 2. Ψ 1 and Ψ 2 are total flux linkages of coil 1 and coil 2. L 1,L 2 and M are self and mutual inductances of coil 1 and coil 2.

29. Flux linkages with coil 1 and coil2, Ψ 1 = L 1 *I 1 + M*I 2 Ψ 2 = L 2 *I 2 + M*I 1 Voltage equations, V 1 = I 1 *R 1 + d Ψ 1 /dt V 2 = I 2 *R 2 + d Ψ 2 /dt Putting the value of Ψ 1 in voltage equation, V 1 = I 1 *R 1 + d (L 1 I 1 + MI 2 )/dt V 1 = I 1 *R 1 + L 1 dI 1 /dt + I 1 dL 1 /dt + MdI 2 /dt + I 2 dM /dt Multiplying with I 1 on both sides, V 1 *I 1 = I 1 * I 1 *R 1 + L 1 * I 1 dI 1 /dt + I 1 * I 1 dL 1 /dt + M * I 1 dI 2 /dt + I 1 * I 2 dM /dt

30. Similar equation for rotor side, V 2 *I 2 = I 2 * I 2 *R 2 + L 2 * I 2 dI 2 /dt + I 2 * I 2 dL 2 /dt + M * I 2 dI 1 /dt + I 1 * I 2 dM /dt Adding both equations and multiplying with dt, (V 1 *I 1 + V 2 *I 2 )dt = (I 1 * I 1 *R 1 + I 2 * I 2 *R 2 )dt + (L 1 * I 1 dI 1 + L 2 * I 2 dI 2 + M * I 1 dI 2 + M * I 2 dI 1 + 2* I 1 * I 2 dM + I 1 * I 1 dL 1 + I 2 * I 2 dL 2 )dt (V 1 *I 1 + V 2 *I 2 - I 1 * I 1 *R 1 - I 2 * I 2 *R 2 )dt = (L 1 * I 1 dI 1 + L 2 * I 2 dI 2 + I 1 * I 1 dL 1 + I 2 * I 2 dL 2 )dt + (M * I 1 dI 2 + M * I 2 dI 1 + 2* I 1 * I 2 dM)dt Useful electrical energy input = Field energy stored in the system + Electrical to Mechanical energy transfer

31. 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 dL 1, dL 2, dM is zero and after integrating the equation, W fe = ½*L 1 *I 1 *I 1 + ½*L 2 *I 2 *I 2 +M*I 1 *I 2

32. Electromagnetic Torque Equation For Doubly Excited System Stored energy is used to mechanical work. Mechanical work done = Rate of change of stored energy = dW fe /dt dW fe /dt = d (½*L 1 *I 1 *I 1 + ½*L 2 *I 2 *I 2 +M*I 1 *I 2 )/dt = ½*L 1 *d(I 1 *I 1 )/dt +½*I 1 *I 1 *dL 1 /dt + ½*L 2 *d(I 2 *I 2 )/dt +½*I 2 *I 2 *dL 2 /dt +M*I 2 *dI 1 /dt+M*I 1 *dI 2 /dt+I1*I2dM/dt = L 1 *I 1 d(I 1 )/dt + ½*I 1 *I 1 *dL 1 /dt + L 2 *I 2( dI 2 /dt) + ½*I 2 *I 2 *dL 2 /dt+M*I 2 *dI 1 /dt+M*I 1 *dI 2 /dt +I 1 *I 2 dM/dt

33. Integrating above equation w.r.t dt,  W fe = (L 1 *I 1 dI 1 + L 2 *I 2 dI 2 +M*I 1 *dI 1 +M*I 1 *dI 2 )+( ½*I 1 *I 1 *dL 1 + ½*I 2 *I 2 *dL 2 +I 1 *I 2 *dM)  W fe = W stored + W me  W me = ½*I 1 *I 1 *dL 1 + ½*I 2 *I 2 *dL 2 +I 1 *I 2 *dM In rotation while mechanical energy is produced the terms dL 1, dL 2, dM are not constant but vary with θm. Differentiating above equation w.r.t θ m, Torque equation = dW me /dθ m =½*I 1 *I 1 *dL 1 /dθ m +½*I 2 *I 2 *dL 2 /dθ m +I 1 *I 2 *dM/dθ m Putting I 2 = 0, we get torque equation for singly excited system, Torque equation = ½*I 1 *I 1 *dL 1 /dθ m

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