Principles of Electromechanical Energy Conversion

Chapter One
Principle of Electromechanical energy conversion
By Yimam A.(MSc.)
March 17, 2022
By Yimam A.(MSc.) Chapter One March 17, 2022 1 / 60

Outline
1 Introduction
2 Salient Aspects of Conversion
3 Energy Balance
4 Determination of Magnetic force and torque from Energy
5 Forces and torques in systems with permanet magnet

Learning Objectives
At the end of this chapter the students should be able to:
ª Understand the concept of electromechanical energy conversion and analyze its
energy balance
ª Understand the components of energy conversion system
ª Understand the determination of force and torque from energy and coenergy
ª Understand how to analyze force and torque in systems with permanent magnet

Introduction
Introduction
Electrical energy is the most popular form of energy, because:
1 it can be transmitted easily for long distance, at high efficiency and reasonable
cost.
2 It can be converted easily to other forms of energy such as sound, light, heat or
mechanical energy.
A device which converts electrical energy into mechanical energy or mechanical
energy into electrical energy is known as electromechanical energy conversion
device.
Electromechanical energy conversions use a magnetic field as the medium of
energy conversion
There are Various electromechanical energy conversion devices.

Introduction
Cont....
Three categories of electromechanical energy conversion devices:
Transducers (for measurement and control)- small motion
Transform the signals of different forms. Examples: microphones, sensors and
speakers.
Force producing devices (translational force)- limited mechanical motion.
Produce forces mostly for linear motion drives, Example Actuators - relays,
solenoids and electromagnets.
Continuous energy conversion equipment.
Operate in rotating mode. Examples: motors and generators.

Introduction
Principle of conservation of energy
The principle of conservation of energy states that “the energy can neither be
create not destroyed. It can merely be converted from one form into another.
The total energy in a system is therefore constant.
In an electromechanical energy conversion device, the total input energy is equal
to the sum of following three components
Energy dissipated,
Energy stored, and
Useful output energy

Salient Aspects of Conversion
Electromechanical energy conversion takes place through the medium of magnetic
field. The following salient features are worth noting in this energy conversion :
i As with any energy conversion system, the principle of conservation of energy
holds good in case of electromechanical energy conversion
ii During electromechanical energy conversion, various losses occur in the system.

Cont....
The electrical energy loss (iR
) is due to current (i) flowing in the winding (having
resistance R) of the energy converter.
The field loss is the core loss due to changing magnetic field in the magnetic
core.
The mechanical loss is the friction and windage loss due to the motion of the
moving components.
iii Electromechanical energy conversion is a reversible process except for the losses
in the system.
iv Electromechanical conversion devices are built with air gaps in the magnetic
circuit to separate the fixed and moving parts. Most of the m.m.f. of the
windings is required to overcome the air gap reluctance so that most of the
energy is stored in the air gap and is returned to the electric source when the
field is reduced.

Cont....
v The electromechanical energy conversion system can be analysed by using princi-
ple of conservation of energy, laws of electric and magnetic field, electric circuits
and Newtonian mechanics.
vi The rotating electrical machines (motors and generators) continuously convert
electrical energy into mechanical energy or vice versa.

Energy Balance
Cont....
An electromechanical converter system has three essential parts:
1 An electrical system (electric circuits such as windings)
2 A magnetic system (magnetic field in the magnetic cores and air gaps)
3 A mechanical system (mechanically movable parts such as a rotor in an electrical
machine)

Energy Balance
Cont....
During this energy conversion, energy loss occurs due to three causes viz., (i) iR
loss in the winding of the energy converter (ii) core or field loss due to changing
magnetic field and (iii) mechanical loss is the friction and windage loss due to
the motion of moving parts.
If the energy losses in the electrical system, the coupling magnetic field and the
mechanical system are grouped with the corresponding terms.

Energy Balance
Cont....
Now consider a differential time interval dt during which an increment of elec-
trical energy dWe(excluding the iR
loss) flows to the system.
During this time dt, let dWf be the energy supplied to the field (either stored or
lost, or part stored and part lost) and dWm the energy converted to mechanical
form (in useful form or as loss, or part useful and part as loss). In differential
forms, it can be expressed as
dWe = dWm + dWf
Core losses are usually small, and if they are neglected, dWf will represent the
change in the stored field energy. Similarly, if friction and windage losses can be
neglected, then all of dWm will be available as useful mechanical energy output.

Energy Balance
Cont....
Consider the electromechanical system shown below. The movable part can be
held in static equilibrium by the spring.
Let us assume that the movable part is held stationary at some air gap and
the current is increased from zero to a value i. Flux will be established in the
magnetic system. Obviously,
dWm = 0
and
dWm = dWe

Energy Balance
Cont....

Energy Balance
Cont....
If core loss is neglected, all the incremental electrical energy input is stored as
incremental field energy. Now
e =
dλ
dt
dWe = eidt
From the above equations
dWf = idλ
When the flux linkage is increased from zero to λ, the energy stored in the field
is
Wf =
Z λ
0
idλ

Energy Balance
Cont...
(a) λ − i curve (b) B-H curve

Energy Balance
Cont....
Other useful expressions can also be derived for the field energy of the magnetic
system. Let
Hc is magnetic intensity in the core
Hg is magnetic intensity in the air gap
lc is length of the magnetic core material
lg is length of the air gap
Then
Ni = Hclc + Hglg
λ = Nϕ
= NAB

Energy Balance
Cont....
Where A is the cross-sectional area of the flux path
B is the flux density, assumed same throughout
From the above three equations
Wf =
Z
Hclc + Hglg
N
NAdB
For the air gap
Hg =
B
µo

Energy Balance
Cont....
From the above two equations
Wf =
Z
Hclc +
B
µo
lg
!
AdB
=
Z
HcdBAlc +
B
µo
dBlgA
!
=
Z
HcdB × volume of magnetic material +
B2
2µo
× volume of air gap
= wfc × Vc + wfg × Vg
= Wfc + Wfg

Energy Balance
Cont....
Where wfc is the energy density in the magnetic material
wfg is the energy density in the air gap
Vc is the volume of magnetic material
Vg is the volume of the air gap
Wfc is the energy stored in the magnetic material
Wfg is the energy stored in the air gap
In most cases Wfc can be neglected.
For a linear magnetic system,
Hc =
Bc
µc

Energy Balance
Cont....
Therefore
wfc =
Z
Bc
µc
dBc =
B2
c
2µc
and we know that for air gap
wfg =
B2
g
2µo

Energy Balance
Example
The dimensions of the relay system are shown in figure (a) below. The magnetic core
is made of cast steel whose B-H characteristic is shown in Figure (b) . The coil has
250 turns, and the coil resistance is 5 ohms. For a fixed air-gap length lg = 5mm, a
dc source is connected to the coil to produce a flux density of 1 Tesla in the air-gap.
Calculate
a) The required field current and the voltage of the dc source.
b) The energy stored in the air gap
c) The energy stored in the steel
d) The total field energy

Energy Balance
Cont...
(a) Dimension of relay system (b) B-H curve

Energy and Coenergy
Energy and Coenergy
The λ − i characteristics of an electromagnetic system depends on the air-gap
length and B-H characteristics of the magnetic material.
For a larger air-gap length the characteristic is essentially linear. The character-
istic becomes non linear as the air-gap length decreases.
The magnetic stored energy is a state function, determined uniquely by the
values of the independent state variables λ and x.
Coenergy: Here the force can be obtained directly as a function of the current.
The selection of energy or coenergy as the state function is purely a matter of
convenience.
For a magnetically-linear system, the energy and coenergy (densities) are nu-
merically equal:

Energy and Coenergy
Cont...
For a particular value of the air gap length, the energy stored in the field is
represented by the area A between the λ axis and the λ − i characteristic.
Wf =
Z λ
0
idλ
The area B between the i axis and the λ − i characteristic is known as the
coenergy and is defined as
W′
f =
Z i
0
λdi

Energy and Coenergy
Cont...
(a) Energy and Coenergy (b) λ−i characteristics for d/t air gap lengths

Energy and Coenergy
Cont...
From the figure of λ − i characteristic,
W′
f + Wf = λi
Note that
if the λ − i characteristic is non linear W′
f > Wf .
if the λ − i characteristic is linear W′
f = Wf .
The quantity of coenergy has no physical significance. However, it can be used
to derive expressions for force (torque) developed in an electromagnetic system

Energy and Coenergy
Cont...
For the lossless magnetic energy storage system in differential form,
dWe = dWm + dWf
dWe = idλ is differential change in electric energy input
dWm = fmdx is differential change in mechanical energy output
dWf = differential change in magnetic stored energy
The mechanical force fm is defined as acting from the relay upon the external
mechanical system and the differential mechanical energy output of the relay is
dWm = fmdx
Then, substitution dWe = idλ gives

Energy and Coenergy
Cont...
dWf = idλ − fmdx
Value of Wf is uniquely specified by
the values of λ and x, since the mag-
netic energy storage system is lossless.
dWf = idλ

Energy and Coenergy
Cont...
dWf is differential change in magnetic stored energy
The λ − i characteristics of an electromagnetic system depends on the air gap
length and B-H characteristics of the magnetic material.
For a larger air-gap length the characteristic is essentially linear.
The characteristic becomes non linear as the air-gap length decreases.

Energy and Coenergy
Cont...
For a particular value of air-gap
length, the field energy is represented
by the area between λ axis and λ − i
characteristic.
The area between i axis and λ−i char-
acteristic is known as the coenergy

Determination of Magnetic force and torque from Energy
The magnetic stored energy Wf is a state function, determined uniquely by the
independent state variables λ and x. This is shown explicitly by
dWf (λ, x) = idλ − fmdx
For any function of two independent variables F(x1, x2), the total differential
equation of F with respect to the two state variables x1 and x2 can be written
dF(x1, x2) =
∂F(x1, x2)
∂x1
dx1 +
∂F(x1, x2)
∂x2
dx2
Therefore, for the total differential of Wf
dWf (λ, x) =
∂Wf (λ, x)
∂λ
dλ +
∂Wf (λ, x)
∂x
dx

Cont...
And we know that
By matching both equations, the current:
i =
∂Wf (λ, x)
∂λ
Where the partial derivative is taken while holding x constant and the mechanical
force:
fm = −
∂Wf (λ, x)
∂x
Where the partial derivative is taken while holding λ constant.

Cont...
For a linear magnetic system for which λ = L(x)i
Wf (λ, x) =
Z λ
0
i(λ, x)dλ =
Z λ
0
λ
L(x)
dλ =
1
2
λ2
L(x)
and the force, fm can be found directly:
fm = −
∂Wf (λ, x)
∂x
= −
∂
∂x
1
2
λ2
L(x)
!
=
λ2
2L(x)2
dL(x)
dx

Determination of Force from Coenergy
The coenergy W′
f is defined as
W′
f = iλ − Wf (λ, x)
and the differential coenergy dW′
f :
dW′
f (i, x) = d(iλ) − dWf (λ, x)
We know previously that
By expanding d(iλ):
d(iλ) = idλ + λdi

Cont...
So, the differential coenergy dW′
f :
dW′
f (i, x) = d(iλ) − dWf (λ, x)
= idλ + λdi − (idλ − fmdx)
= λdi + fmdx
By expanding dW′
f (i, x):
dW′
f (i, x) =
∂W′
f (i, x)
∂i
di +
∂W′
f (i, x)
∂x
dx

Cont...
From the previous result:
dW′
f (i, x) = λdi + fmdx
By matching both equations, λ
λ =
∂W′
f (i, x)
∂i
where the partial derivative is taken while holding x constant and the mechanical
force:
fm =
∂W′
f (i, x)
∂x
where the partial derivative is taken while holding i constant.

Cont...
For a linear magnetic system for which λ = L(x)i:
W′
f (i, x) =
Z i
0
λ(i, x)di =
Z λ
0
L(x)idi = L(x)
i2
2
and the force, fm can be found directly:
dW′
f (i, x) =
∂W′
f (i, x)
∂x
=
∂
∂x
L(x)
i2
2
!
=
i2
2
dL(x)
dx

Determination of Torque from Energy
For a system with a rotating mechanical terminal, the mechanical terminal vari-
ables become the angular displacement θ and the torque T. Therefore, equation
for the torque:
T = −
∂Wf (λ, θ)
∂θ
where the partial derivative is taken while holding λ constant.

Cont...
For a system with a rotating mechanical terminal, the mechanical terminal vari-
ables become the angular displacement θ and the torque T.
Therefore, equation for the torque:
T = −
∂W′
f (i, θ)
∂θ
where the partial derivative is taken while holding λ constant.

Determination of Force Using Energy or Coenergy?
The selection of energy or coenergy as the function to find the force is purely a
matter of convenience.
They both give the same result, but one or the other may be simpler analytically,
depending on the desired result and characteristics of the system being analyzed.

Direction of Force Developed
By using energy function:
fm = −
∂Wf (λ, x)
∂x
The negative sign shows that the force acts in a direction to decrease the
magnetic field stored energy at constant flux.
By using coenergy function
fm = +
∂W′
f (i, x)
∂x
The positive sign emphasizes that the force acts in a direction to increase the
coenergy at constant current.

Cont...
By using inductance function:
fm =
i2
2
dL(x)
dx
The positive sign emphasizes that the force acts in a direction to increase the
inductance at constant current.

B-H Curve and Energy Density
In a magnetic circuit having a substantial air gap g, and high permeability of
the iron core, nearly all the stored energy resides in the gap.
Therefore, in most of the cases we just need to consider the energy stored in the
gap. The magnetic stored energy,
Wf =
Z λ
0
idλ
in which i = Hglg
N
and dλ = d(Nϕ) = d(NAB) = NAdB
Therefore
Wf =
Z B
0
Hglg
N
NAdB = Alg
Z B
0
HdB

Cont...
However Alg is volume of the air gap.
dividing both sides of the above equation by the volume Alg results in
wf =
Wf
Alg
=
Z B
0
HdB
where wf =
R B
0 HdB is energy per unit volume , is known as energy density.
The area between the B-H curve and B axis represents the energy density in the
air gap.
In the same manner w′
f =
R H
0 BdH is coenergy per unit volume
The area between the B-H curve and H axis represents the coenergy density in
the air gap.

Cont...

Cont...
For a linear magnetic circuit, B = µH or H = B
µ
, energy density:
wf =
Z B
0
HdB =
Z B
0
B
µ
dB =
B2
2µ
and coenergy density:
w′
f =
Z H
0
BdH =
Z H
0
µHdH =
µH2
2
In this case, it is obvious that w′
f = wf .

Example
The λ − i relationship for an electromagnetic system is given by
i =
λg
0.09
!2
which is valid for the limits 0 i 4A and 3 g 10 cm. For current i = 3A
and air gap length g = 5cm, find the mechanical force on the moving part using
(a) Energy of the field.
(b) Coenergy of the field

Rotating Machines
Most of the energy converters, particularly the higher-power ones, produce ro-
tational motion.
The essential part of a rotating electromagnetic system is shown in the figure.
The fixed part is called the stator,the moving part is called the rotor.
The rotor is mounted on a shaft and is free to rotate between the poles of the
stator
Let consider general case where both stator rotor have windings carrying
current ( is and ir )

Cont...

Cont...
Assume general case, both stator and rotor have winding carrying currents (non-
uniform air gap – salient pole rotor)
The system stored field energy, Wf can be evaluated by establishing the stator
current is and rotor current ir and let system static, i.e. no mechanical output
Stator and rotor flux linkage λ is expressed in terms of inductances L (which
depends on position rotor angle θ)
Considering the system to be static,that is mechanical energy output wm is
zero,the stored field energy dWf is equal to the electrical energy input dWe and
given by

Cont...
The stored field energy is
dWe = dWf = esisdt + erirdt
= isdλs + irdλr
λs = Lssis + Lsrir
λr = Lrsis + Lrrir
λs
λr
=
Lss Lsr
Lrs Lrr
=
is
ir

Cont...
Stored field energy
dWf = isd(Lssis + Lsrir) + ird(Lsris + Lrrir)
= Lssisdis + Lsrisdir + Lsrirdis + Lrrirdir
= Lssisdis + Lrrirdir + Lsr(isdir + irdis)
= Lssisdis + Lrrirdir + Lsrd(isir)
The magnetic field Stored energy is given by
Wf = Lss
Z is
0
isdis + Lrr
Z ir
0
irdir + Lsr
Z isir
0
d(isir)
=
1
2
Lssi2
s +
1
2
Lrri2
r + Lsrisir

Cont...
In a linear magnetic system, energy and coenergy are the same, that is, W′
f = Wf
T =
∂W′
f (i, θ)
∂θ
=
1
2
i2
s
dLss
dθ
+
1
2
i2
r
dLrr
dθ
+ isir
dLsr
dθ

Cont...
First two terms represents torques produced in the machine because of variation
of self inductance with rotor position.
This component of torque is called the reluctance torque and do not depend on
the direction of currents in stator or rotor windings.
The third term represents torque produced by the variation of the mutual in-
ductance between the stator and rotor windings. This component of torque is
called the alignment torque or Electromagnetic torque or interaction torque.
This torque is developed by the interaction of stator and rortor magnetic fields
and depends on the direction of the currents in the stator and rotor.

Example
In a electromagnetic system, the rotor has no winding (i.e. we have a reluctance
motor) and the inductance of the stator as a function of the rotor position θ is
Lss = Lθ + L2 cos 2θ The stator current is is = Ism sin ωt
1 Obtain an expression for the torque acting on the rotor
2 Let θ = ωmt + δ where ωm is the angular velocity of the rotor and δ is the rotor
position at t = 0 Find the condition for the non-zero average torque and obtain
the expression for the average torque. t

Forces and torques in systems with permanet magnet
Forces and torques in Systems with permanent magnet
2

Cont...

Cont...
The dc magnetizing curve of the permanent magnet is drawn in Figure (a) which
upon linear extrapolation at the lower B-end.
W′
f (x) =
µRAH′
cd2
2 [d + 2(µR/µO)X]
The force on the armature is given by
Ff =
AB2
r
µo
h
1 + 2 (µr/µo) (x/d)2
i
You can read the whole concept
https://www.eeeguide.com/forces-in-system-with-permanent-magnets/ or
Electrical Machines by Kothari in section 4.5

Thank You!

Principles of Electromechanical Energy Conversion

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