Electromagnet and Electromechanical Concepts

1. -Mr. M.N. Mestri
Department of Electrical Engineering,
ATS’S Sanjay Bhokare Group of Institutes,
Miraj.
mnmestri@gmail.com /
mestrimn@sbgimiraj.org
Electrical Machine-I

2. I. Single Phase Transformer. (7 Hours)
II. Three Phase Transformers. (8 Hours)
III. Electromechanical Energy Conversion Principles. (6 Hours)
IV. DC Generators. (9 Hours)
V. DC Motors. (9 Hours)
VI. Special Machines. (6 Hours)
Syllabus Contents
-Mr. M.N. Mestri 2

3. Energy in a magnetic systems,
Field energy and mechanical force,
Energy in singly and multiply excited magnetic
systems,
Determination of magnetic force and torque from
energy and co-energy,
Forces and torques in magnetic field systems,
Dynamic equations of electromechanical systems and
analytical techniques
Chapter 3rd: Lecture 1st
Electromechanical Energy Conversion Principles.
(6 Hours)
-Mr. M.N. Mestri 3

4. To convert any energy from one form to another form is known
as Conversion.
E.g.: Conversion of Electrical Energy into Chemical, Heat,
Magnetic or Mechanical Energy or Vice Versa.
In this chapter we are going to observe concepts of Conversion
from Electrical to Mechanical Energy, this conversion is known as
Electromechanical Conversion.
Background of Conversion (Extra)
-Mr. M.N. Mestri 4

5. What is Magnetism? (Extra)
5
 Magnetism is the force exerted by magnets when they attract
or repel each other.
-Mr. M.N. Mestri

6. Electric and Magnetic Circuit (Extra)
6
For Electric Circuit,
V= Supply Voltage
I= Current Flow
L= Load
Ri= Internal Resistance
For Magnetic Circuit,
V= Supply Voltage
I= Supply Current
Φ= Flux Flow
Si= Internal Reluctance
-Mr. M.N. Mestri

7. All Ferromagnetic Materials, like conducting materials
Iron, Steel, Silicon Steel, Copper, Aluminum etc.
In Insulation Material Magnetic Field can’t get created and can’t
get transferred.
Materials where Magnetic Field can get Created
-Mr. M.N. Mestri 7

8. In below figure we can observe Electromechanical System which
is operated with Magnetic System
Energy in a Magnetic Systems
-Mr. M.N. Mestri 8

9. From Previous Figure we can observe that, Copper Winding is Connected
on Magnetic Fixed Core to Produce Magnetic Flux in System when
Current flow is Started in a Coil and Iron Core.
As we know, when supply is started the current flow starts in a coil
and produce magnetic flux in a core as shown with dotted line in both
Moving and Fixed Core.
This magnetic flux creates magnetic polarities in a core such as North
“N” and South “S” as shown in figure.
As we know, alternate poles attract each other, due to which Moving
Core moves towards Fixed Core from Hinged Point and get attached.
This Momentum is known as Mechanical Energy and it is created due to
Electrical Input, Hence, it is known as Electromechanical Force.
Energy in a Magnetic Systems
-Mr. M.N. Mestri 9

10. Energy in a magnetic systems,
Field energy and mechanical force,
Energy in singly and multiply excited magnetic
systems,
Determination of magnetic force and torque from
energy and co-energy,
Forces and torques in magnetic field systems,
Dynamic equations of electromechanical systems and
analytical techniques
-Mr. M.N. Mestri 10
Chapter 3rd: Lecture 2nd
Electromechanical Energy Conversion Principles.
(6 Hours)

11. What is Magnetism? (Extra)
11
 Magnetism is the force exerted by magnets when they attract
or repel each other.
-Mr. M.N. Mestri

12. Earth Magnetism (Extra)
12
 Our Plant Earth is also a BIG Magnet
N
S
-Mr. M.N. Mestri

13. Magnet Experiment with Iron Filling (Extra)
13
Magnetic Lines
Magnet
-Mr. M.N. Mestri

14. Magnetic Lines (Extra)
14
Magnetic field lines are imaginary lines. Magnetic field lines are a
visual tool used to represent magnetic fields. They describe the
direction of the magnetic force on a north monopole at any given
position. The density of the lines indicates the magnitude of the
field.
Relation: Magnetic Lines α Magnetic Field (Directly Proportional)
Direction: North Polarity to South Polarity
-Mr. M.N. Mestri

15. Magnetic Flux (Example) (Extra)
15
Definition: Magnetic flux is defined as the number of magnetic
field lines passing through a given closed surface. It provides the
measurement of the total magnetic field that passes through a
given surface area. Surface will be always Conducting Surface.
Examples:
1. Highway Road with High Traffic,
2. Village Road with Low Traffic.
Road = Conducting Material
Traffic = Magnetic Lines
Highway/ Village Road = Conducting Material Area
-Mr. M.N. Mestri

16. Magnetic Flux (Formula) (Extra)
16
Equation for Magnetic Flux (ΦB) = B x A x Cos(θ)………… (Weber)
Where:
B= Magnetic Lines/ Magnetic Flux Density,
A= Area of Conductor,
Cos(θ)= Angle of Conductor Plate
Rule: Cos(0)= 1, Cos(90)= 0
-Mr. M.N. Mestri

17. Magnetic Flux Density (Extra)
17
Definition: Magnetic flux density is magnetic flux passes through
unit area when conducting plate placed perpendicular.
Equation for Magnetic Flux Density (B)= ΦB/A …… (Weber/meter2)
-Mr. M.N. Mestri

18. Magnetic Field Strength (Extra)
18
Definition: Magnetic field strength, also called magnetic intensity
or magnetic field intensity, the part of the magnetic field in a
material that arises from an external current and is not intrinsic
to the material itself.
Formula: H=I/2πr ………… (Ampere-Turn/Meter)
Denoted: H
Unit: Ampere-Turns / Meter
-Mr. M.N. Mestri

19. Example for Reluctance and Permeability (Extra)
19
Relation Between Reluctance
and Permeability is they both
are Inversely Proportional to
each other
-Mr. M.N. Mestri

20. Reluctance (Extra)
20
Definition: The 'magnetic resistance' of a magnetic circuit to the
oppose flow of magnetic flux is called reluctance.
If Reluctance is more we can say than it reduce flow of flux with
losses.
Formula: Reluctance (Si)= l/μA ………… (Ampere-Turns / Weber)
Denoted: Si
Unit: Ampere-Turns / Weber
-Mr. M.N. Mestri

21. Permeability (Extra)
21
Definition: It is defined as the ability of Magnetic material to
permit Flow of Flux in a Material, it is dependent on Magnetic
Flux Density and Magnetic Field in Material.
It is also known as Reciprocal of Reluctivity.
If Permeability is more we can say than it will flow maximum flux
without losses.
Formula: Permeability (μ)= B/H ………… (Henry per Meter)
Denoted: Greek Letter Mu (μ)
Unit: Henry per Meter
-Mr. M.N. Mestri

22. Magneto Motive Force (Extra)
22
Definition: The magneto motive force (mmf) is a quantity
appearing in the equation for the magnetic flux in a magnetic
circuit, often called Ohm's law for magnetic circuits.
The MMF is also known as the magnetic potential. It is the
property of a material to give rise to the magnetic field. The
magneto motive force is the product of the magnetic flux and the
magnetic reluctance. The reluctance is the opposition offers by
the magnetic field to set up the magnetic flux on it.
-Mr. M.N. Mestri

23. Magneto Motive Force (Example) (Extra)
23
For Electric Circuit,
V= Supply Voltage
I= Current Flow
L= Load
Ri= Internal Resistance
For Magnetic Circuit,
V= Supply Voltage
I= Supply Current
Φ= Flux Flow
Si= Internal Reluctance
-Mr. M.N. Mestri

24. Magneto Motive Force (Extra)
24
Magnetic Motive Force is dependent on Number of Turns of
Conductor placed on a Magnetic Core and Current Flowing in a
Magnetic Core.
Formula: Magnetic Motive Force (Fm)= N*I ………… (Ampere-Turns)
Denoted: Fm
Unit: Ampere-Turns (Where, Ampere for Current and Turn for
Conductor Turns used in Winding n Core)
-Mr. M.N. Mestri

25. Energy in a magnetic systems,
Field energy and mechanical force,
Energy in singly and multiply excited magnetic
systems,
Determination of magnetic force and torque from
energy and co-energy,
Forces and torques in magnetic field systems,
Dynamic equations of electromechanical systems and
analytical techniques
-Mr. M.N. Mestri 25
Chapter 3rd: Lecture 3rd
Electromechanical Energy Conversion Principles.
(6 Hours)

26. Field Energy and Mechanical Force
26
 We know,
 V= Input Voltage= I*R+E
 I= Current in the Coil
 R= Resistance of the Coil
 E= Self or Back emf developed in Coil= N*(dφ/dt)
 N= Number of Turns of the Coil
 φ= Flux (Developed due to Current Flow in Coil)
 Ψ= mmf= N*I= N*φ …………(As I Directly Proportional to φ)
 Ψ= mmf= N*I= S*φ …………(Where S is Reluctance)
 Reluctance, S= l/μ*A …………(Standard Equation for Reluctance)
 Where, μ= Permeability, A= Area of Conductor and l=Length of Conductor
-Mr. M.N. Mestri

27. Field Energy and Mechanical Force
27
Relation of Energy Stored:
When, we supply voltage to a coil, current flow takes place in a coil,
coil resistance opposes flow of current and due to flow of current some
self emf is generated in a coil.
Due to Which Voltage will be, V= I*R + emf
Therefore, V=I*R+E => E= V-I*R
But we know, emf (E)= N*(dφ/dt)
Therefore, V= I*R+N*(dφ/dt)
We can rewrite equation as, V= I*R+(d(N*φ)/dt)
Therefore, V= I*R+(dψ/dt) …………(As mmf (ψ)= N*φ) (Also E= dψ/dt)
-Mr. M.N. Mestri

28. Field Energy and Mechanical Force
28
 Relation of Energy Stored:
 Therefore, V= I*R+(dψ/dt)
 Thus, this equation is Voltage Balanced Equation
 To Determine Power and Energy Balance Equation we must Multiple Current (I)
with Above Equation.
 Power Balanced Equation, V*I= I2*R+I*(dψ/dt)
 By multiplying above equation with dt we will get Energy Balanced Equation, as
Energy=Power*Time
 Therefore, Energy Balanced Equation, V*I*dt= I2*R*dt + I*dψ
 Rearranging Equation, (V-I*R)*I*dt= I*dψ
 We know, E= V-I*R, Therefore, E*I*dt= I*dψ
 Therefore, E= I*dψ/I*dt => dψ/dt
-Mr. M.N. Mestri

29. Field Energy and Mechanical Force
29
To Determine Total Field Energy with Co-Energy Factor:
We know, E*I*dt= I*dψ
From Above Equation, E*I*dt is Electrical Energy Output it is a
Energy which Converts Electrical Energy into Magnetic Energy
This Energy is also Called as “Co-Energy”
Co-Energy= E*I*dt (Which Electrical to Magnetic Energy)
Also we can say that, I*dψ is Magnetic Energy
Therefore, Electrical Output (Co-Energy)= Magnetic Energy
Therefore, E*I*dt= I*dψ
-Mr. M.N. Mestri

30. Field Energy and Mechanical Force
30
To Determine Total Field Energy with Co-Energy Factor:
From Equation, E*I*dt= I*dψ we can determine Total Field
Energy of Whole Surface by Integrating this Equation.
-Mr. M.N. Mestri

31. Field Energy and Mechanical Force
31
To Determine Total Field Energy with Co-Energy Factor:
After Solving Whole Integration we will get Final Equation of
Total Field Energy
-Mr. M.N. Mestri

32. Field Energy and Mechanical Force
32
To Determine Mechanical Force:
We know,
We know Basic Rule of Mechanical Work Done.
Therefore, Mechanical Work Done (Fm)= Force*Displacement
To Determine Mechanical Force= Fm/ Displacement
Mechanical Force= (d/dx)*(WField(ψ,x))
Therefore, Mechanical Force = (d(WField(ψ,x)/dx)
We know, WField= (1/2)*φ2*S
-Mr. M.N. Mestri

33. Field Energy and Mechanical Force
33
To Determine Mechanical Force:
We know,
We know, from mmf, φ= mmf/S
Therefore, φ2= mmf2/S2
-Mr. M.N. Mestri

34. Field Energy and Mechanical Force
34
To Determine Mechanical Force:
Therefore,
Therefore, Mechanical Force = (d(WField(ψ,x)/dx)
Substituting Value of Field Energy in Above Equation
-Mr. M.N. Mestri

35. Field Energy and Mechanical Force
35
To Determine Mechanical Force:
Therefore,
We replace L with Reluctance S as L is only Imaginary Quantity
Substitute value of Reluctance in equation.
-Mr. M.N. Mestri

36. Field Energy and Mechanical Force
36
To Determine Mechanical Force:
Therefore,
From Equation we can say that, l is air gap length in Reluctance
Equation
Substitute, value of μ in above Equation as per Reluctance
Equation
And we know Flux Density (B) is given by φ/A
-Mr. M.N. Mestri

37. Field Energy and Mechanical Force
37
To Determine Mechanical Force:
By Substituting Permeability and Flux Density Value we get,
Thus, we can say that, to operate Mechanical Force we need to
have Electromechanical Conversion and the Mechanical Force will
be Dependent on Flux Density in Material and Air Gap and Area
of Conductor on which Mechanical Work to be Conducted.
-Mr. M.N. Mestri

38. Energy in a magnetic systems,
Field energy and mechanical force,
Energy in singly and multiply excited magnetic
systems,
Determination of magnetic force and torque from
energy and co-energy,
Forces and torques in magnetic field systems,
Dynamic equations of electromechanical systems and
analytical techniques
-Mr. M.N. Mestri 38
Chapter 3rd: Lecture 4th
Electromechanical Energy Conversion Principles.
(6 Hours)

39. Energy in Singly and Multiply Excited
Magnetic Systems
39
Singly Excited:
In this method coil is used which is supported on fixed part and
necessary excitation is produced which magnetizes fixed as well
as movable part.
E.g. Magnetic Relay and Single Phase AC Machine
Doubly Excited:
In this method two coils are magnetized to obtain motion.
E.g. DC Motors
-Mr. M.N. Mestri

40. Energy in Singly and Multiply Excited
Magnetic Systems
40
Multiple Single Excited:
In this method Three Windings are excited at Stator side with 3
Phase supply and Rotor receives power for motion or operation.
E.g. Three Phase AC Machine
Multiple Doubly Excited:
In this method Three Windings are excited at Stator side with 3
Phase supply and Rotor is excited with DC Excitation Supply for
motion or operation.
E.g. Three Phase Synchronous Motor
-Mr. M.N. Mestri

41. Energy in Singly and Multiply Excited
Magnetic Systems
41
 Energy in Singly Excited Magnetic Systems:
 In Figure we have Static Core and Free Moving
Rotating Core.
 Stator is wound with Electrical Excitation Coil and
Rotor has no coil.
 Initial Position of Rotor is 1 as shown in figure,
after supply started, magnetic flex will link with
rotor and Creates N and S poles as shown in
figure in Stator and Rotor Part.
 As poles are opposite they will attract and Rotor
will move to position 2, as it has low reluctance
path.
 We need to determine this torque.
 From equation of Voltage we can determine Energy
in Singly Excited Magnetic System.
-Mr. M.N. Mestri

42. 42
 We know,
 V= Input Voltage= I*R+E
 I= Current in the Coil
 R= Resistance of the Coil
 E= Self or Back emf developed in Coil= N*(dφ/dt)
 N= Number of Turns of the Coil
 φ= Flux (Developed due to Current Flow in Coil)
 Ψ= mmf= N*I= N*φ …………(As I Directly Proportional to φ)
 Ψ= mmf= N*I= S*φ …………(Where S is Reluctance)
 Reluctance, S= l/μ*A …………(Standard Equation for Reluctance)
 Where, μ= Permeability, A= Area of Conductor and l=Length of Conductor
-Mr. M.N. Mestri
Energy in Singly and Multiply Excited
Magnetic Systems

43. 43
 When, we supply voltage to a coil, current flow takes place in a coil, coil
resistance opposes flow of current and due to flow of current some self emf is
generated in a coil.
 Due to Which Voltage will be, V= I*R + emf
 Therefore, V=I*R+E => E= V-I*R
 But we know, emf (E)= N*(dφ/dt)
 Therefore, V= I*R+N*(dφ/dt)
 We can rewrite equation as, V= I*R+(d(N*φ)/dt)
 Therefore, V= I*R+(dψ/dt) …………(As mmf (ψ)= N*φ) (Also E= dψ/dt)
-Mr. M.N. Mestri
Energy in Singly and Multiply Excited
Magnetic Systems

44. 44
 Therefore, V= I*R+(dψ/dt)
 Thus, this equation is Voltage Balanced Equation
 To Determine Power and Energy Balance Equation we must Multiple Current (I)
with Above Equation.
 Power Balanced Equation, V*I= I2*R+I*(dψ/dt)
 By multiplying above equation with dt we will get Energy Balanced Equation, as
Energy=Power*Time. Therefore,
 Energy Balanced Equation, V*I*dt= I2*R*dt + I*dψ
-Mr. M.N. Mestri
Energy in Singly and Multiply Excited
Magnetic Systems

45. 45
 We can say that, We= Wloss + Wf
 We know in initial condition Rotor is Steady, Therefore we can say that,
Mechanical Work (Wm) done will be zero (Wm=0)
 In below equation L is imaginary value we can consider it as L=N.
-Mr. M.N. Mestri
Energy in Singly and Multiply Excited
Magnetic Systems

46. Energy in Singly and Multiply Excited
Magnetic Systems
46
Energy in Multiply Excited Magnetic Systems:
In Figure we have Static Core and Free
Moving Rotating Core.
Stator and Rotor both are wound with
Electrical Excitation Coil.
Resistance, Voltage and Current are shown in
Figure, R1 and R2 are Resistance, V1 and V2
are Voltages and I1 and I2 are Current of
Stator and Rotor Respectively.
In the operation Self and Mutual Inductance
takes place L and M respectively.
φ1 and φ2 are two coil Fluxes.
-Mr. M.N. Mestri

47. Energy in Singly and Multiply Excited
Magnetic Systems
47
Mmf equations for both coils,
Therefore Voltage Equation will be,
-Mr. M.N. Mestri

48. Energy in Singly and Multiply Excited
Magnetic Systems
48
Substitute value of mmf in previous equation, we get,
To get Power Equation Multiple Both sides with Current we get,
-Mr. M.N. Mestri

49. Energy in Singly and Multiply Excited
Magnetic Systems
49
To obtain Energy Equation we need to Integrate Stator and Rotor
Equations with respect to time and Adding them we get,
-Mr. M.N. Mestri
A= Useful Electrical Energy Input
B= Field Energy Stored in a Electrical System
C= Electrical to Mechanical Energy Transfer

50. Energy in Singly and Multiply Excited
Magnetic Systems
50
From equation,
A= Useful Electrical Energy Input
B= Field Energy Stored in a Electrical System
C= Electrical to Mechanical Energy Transfer
Therefore to determine Magnetic Energy in Doubly Excited
System we need to consider Mechanical Output is Zero
Hence, we need to Neglect dL1, dL2 and dM values
Thus, Total Magnetic Energy Stored will be,
-Mr. M.N. Mestri

51. Energy in Singly and Multiply Excited
Magnetic Systems
51
Total Magnetic Energy Stored will be,
-Mr. M.N. Mestri

52. Energy in a magnetic systems,
Field energy and mechanical force,
Energy in singly and multiply excited magnetic
systems,
Determination of magnetic force and torque from
energy and co-energy,
Forces and torques in magnetic field systems,
Dynamic equations of electromechanical systems and
analytical techniques
-Mr. M.N. Mestri 52
Chapter 3rd: Lecture 5th
Electromechanical Energy Conversion Principles.
(6 Hours)

53. Determination of Magnetic Force and Torque
from Energy and Co-Energy
53
Torque for Doubly Excited System:
We know, Total Magnetic Energy Stored for Multiply Excited
Magnetic System will be,
In all Previous Equation we have Neglected Mechanical Work Done
as we have taken only Initial Conditions
Now, we are determining Torque of Singly and Doubly Excited
Magnetic System.
We know Torque is Mechanical Quantity, Hence we need to
consider Mechanical Work Done.
-Mr. M.N. Mestri

54. Determination of Magnetic Force and Torque
from Energy and Co-Energy
54
We know, Total Magnetic Energy Stored for Multiply Excited
Magnetic System will be,
Thus, to determine Mechanical Work Done we need to make
derivate of Above Stored Energy Equation with respect to Time.
-Mr. M.N. Mestri

55. Determination of Magnetic Force and Torque
from Energy and Co-Energy
55
To get Total Value we need to Integrate Whole Equation with
respect to Time.
-Mr. M.N. Mestri

56. Determination of Magnetic Force and Torque
from Energy and Co-Energy
56
Simplifying Above Equation and Making Split of Mechanical Energy
Stored and Mechanical Work Done.
Here, Wf= Wstored+Wm is shown, Where dL and dM quantities are
of Mechanical Work Done.
-Mr. M.N. Mestri

57. Determination of Magnetic Force and Torque
from Energy and Co-Energy
57
While observing Mechanical Work in Motor we can Observe that
the Torque developed will be always Angular Torque with Respect
to Angle θ.
As it is Dependent on Angle θ, it is changing value as Angle
Changes.
Therefore Mechanical Work Done Will be,
Thus this is a Final Torque Equation for Doubly Excited Magnetic
System
-Mr. M.N. Mestri

58. Determination of Magnetic Force and Torque
from Energy and Co-Energy
58
Torque for Singly Excited System:
This is a Final Torque Equation for Doubly Excited Magnetic
System
We know in Singly Excited Magnetic System we are not supplying
Current at Secondary or Rotor Side.
Thus from Doubly Excited Magnetic System Equation we can say
that Secondary or Rotor Current (I2) will be Equal to Zero.
-Mr. M.N. Mestri

59. Determination of Magnetic Force and Torque
from Energy and Co-Energy
59
This is a Final Torque Equation for Doubly Excited Magnetic
System
In Above Equation Substituting I2=0.
-Mr. M.N. Mestri

60. Energy in a magnetic systems,
Field energy and mechanical force,
Energy in singly and multiply excited magnetic
systems,
Determination of magnetic force and torque from
energy and co-energy,
Forces and torques in magnetic field systems,
Dynamic equations of electromechanical systems and
analytical techniques
-Mr. M.N. Mestri 60
Chapter 3rd: Lecture 6th
Electromechanical Energy Conversion Principles.
(6 Hours)

61. Dynamic Equations of Electromechanical
Systems and Analytical Techniques
61
Consider one Electromechanical Relay who has One Singly Excited
Electrical Port and One Mechanical Load Port as Shown in Figure.
K= Spring
B= Damper
M= Mass
F= Force (Load)
-Mr. M.N. Mestri

62. Dynamic Equations of Electromechanical
Systems and Analytical Techniques
62
Representing Relay System in Electromechanical Block Manner
L= Inductance of Device= (μ0*A*N2)/(2(l0-x)) …… Standard Equ.n
v= Voltage
i= Current
R= Resistance
e= emf
ψ= mmf
-Mr. M.N. Mestri

63. Dynamic Equations of Electromechanical
Systems and Analytical Techniques
63
To Determine Dynamic Equation:
With Reference to Voltage Balanced Equation,
-Mr. M.N. Mestri

64. Dynamic Equations of Electromechanical
Systems and Analytical Techniques
64
We know Co-Energy in Singly Excited System,
To Find Field Force Created by Magnetic Flux in Relay System,
-Mr. M.N. Mestri

65. Dynamic Equations of Electromechanical
Systems and Analytical Techniques
65
 Basic Rules to Convert Mechanical Loads
into Mathematical Equations
 Force is Direct Value so we can Consider
Force Load directly in Equation as “F”.
 Spring is also Direct Value but Dependent
on Spring Size, so we need to Consider
Spring always with Size Factor as “K*r”.
 Damper has its own Displacement as per
Change in Time in its Cavity or Cylinder,
Thus we must consider Damper with respect
to Time “B*(dx/dt)”.
 Similarly, the Mass is Stable Quantity in
any System, but as no body is rigid body,
it may get vary with respect to Time, but
Time Required will be in Derivate of
Damper Value. Thus Mass must be
Considered as “M*(dx2/dt2)”.
-Mr. M.N. Mestri

66. Dynamic Equations of Electromechanical
Systems and Analytical Techniques
66
Depending on Previous Rules of
Load, Force Equation will be,
Substituting Value of Force Ff,
-Mr. M.N. Mestri

67. Dynamic Equations of Electromechanical
Systems and Analytical Techniques
67
We know, Voltage Balanced
Equation,
Substituting Value of Force Ff,
-Mr. M.N. Mestri

68. Dynamic Equations of Electromechanical
Systems and Analytical Techniques
68
We know Voltage Balanced Equation,
Substitute Value of L(x) in Voltage Balanced Equation,
Also, to Determine Final Force for Dynamic Operation, we must
Substitute Value of L(x) in Force Equation with Loads.
-Mr. M.N. Mestri

69. Dynamic Equations of Electromechanical
Systems and Analytical Techniques
69
We know Force Equation with Loads,
Substitute Value of L(x) in Force Equation,
Thus, We can say that, Dynamic Equation with Voltage Balanced
and with Force with Load, both Equations are Non-Linear
Equations in Electromechanical System.
-Mr. M.N. Mestri

70. -Mr. M.N. Mestri 70
Thank You…