Chapter 3 Electromechanical-Energy-Conversion
Principles
The electromechanical-energy-conversion process takes place through the medium of the
electric or magnetic field of the conversion device of which the structures depend on their
respective functions.
Transducers: microphone, pickup, sensor, loudspeaker
Force producing devices: solenoid, relay, electromagnet
Continuous energy conversion equipment: motor, generator
This chapter is devoted to the principles of electromechanical energy conversion and the
analysis of the devices accomplishing this function. Emphasis is placed on the analysis of
systems that use magnetic fields as the conversion medium.
The concepts and techniques can be applied to a wide range of engineering situations
involving electromechanical energy conversion.
Based on the energy method, we are to develop expressions for forces and torques in
magnetic-field-based electromechanical systems.
§3.1 Forces and Torques in Magnetic Field Systems
The Lorentz Force Law gives the force F on a particle of charge q in the presence of
electric and magnetic fields.
F = q (E + v × B )
(3.1)
F : newtons, q : coulombs, E : volts/meter, B : telsas, v : meters/second
In a pure electric-field system,
In pure magnetic-field systems,
Figure 3.1
F = qE
(3.2)
F = q(v × B )
(3.3)
Right-hand rule for F = q(v × B ) .
For situations where large numbers of charged particles are in motion,
Fv = ρ (E + v × B )
J = ρv
Fv = J × B
ρ (charge density): coulombs/m3, Fv (force density): newtons/m3,
J = ρ v (current density): amperes/m2.
1
(3.4)
(3.5)
(3.6)
Figure 3.2
Single-coil rotor for Example 3.1.
Unlike the case in Example 3.1, most electromechanical-energy-conversion devices contain
magnetic material.
Forces act directly on the magnetic material of these devices which are constructed of
rigid, nondeforming structures.
The performance of these devices is typically determined by the net force, or torque,
acting on the moving component. It is rarely necessary to calculate the details of the
internal force distribution.
Just as a compass needle tries to align with the earth’s magnetic field, the two sets of
fields associated with the rotor and the stator of rotating machinery attempt to align, and
torque is associated with their displacement from alignment.
In a motor, the stator magnetic field rotates ahead of that of the rotor, pulling on it
and performing work.
For a generator, the rotor does the work on the stator.
2
The Energy Method
Based on the principle of conservation of energy: energy is neither created nor destroyed;
it is merely changed in form.
Fig. 3.3(a): a magnetic-field-based electromechanical-energy-conversion device.
A lossless magnetic-energy-storage system with two terminals
The electric terminal has two terminal variables: e (voltage), i (current).
The mechanical terminal has two terminal variables: f fld (force), x (position)
The loss mechanism is separated from the energy-storage mechanism.
– Electrical losses: ohmic losses…
– Mechanical losses: friction, windage…
Fig. 3.3(b): a simple force-producing device with a single coil forming the electric
terminal, and a movable plunger serving as the mechanical terminal.
The interaction between the electric and mechanical terminals, i.e. the
electromechanical energy conversion, occurs through the medium of the magnetic
stored energy.
Figure 3.3
(a) Schematic magnetic-field electromechanical-energy-conversion device;
(b) simple force-producing device.
Wfld : the stored energy in the magnetic field
d Wfld
dx
= ei − f fld
dt
dt
dλ
e=
dt
d Wfld = idλ − f fld dx
(3.7)
(3.8)
(3.9)
Equation (3.9) permits us to solve for the force simply as a function of the flux λ
and the mechanical terminal position x .
Equations (3.7) and (3.9) form the basis for the energy method.
3
§3.2 Energy Balance
Consider the electromechanical systems whose predominant energy-storage mechanism is in
magnetic fields. For motor action, we can account for the energy transfer as
⎛ Energy input ⎞ ⎛ Mechanical ⎞ ⎛ Increase in energy ⎞ ⎛ Energy ⎞
⎜
⎟ ⎜
⎟ ⎜
⎟ ⎜
⎟
(3.10)
⎜ form electric ⎟ = ⎜ energy
⎟ + ⎜ stored in magnetic ⎟ + ⎜ converted ⎟
⎜ sources
⎟ ⎜ output
⎟ ⎜ field
⎟ ⎜ into heat ⎟
⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠
Note the generator action.
The ability to identify a lossless-energy-storage system is the essence of the energy method.
This is done mathematically as part of the modeling process.
For the lossless magnetic-energy-storage system of Fig. 3.3(a), rearranging (3.9) in form
of (3.10) gives
dWelec = dWmech + dWfld
(3.11)
where
dWelec = id λ = differential electric energy input
dWmech = f fld dx = differential mechanical energy output
dWfld = differential change in magnetic stored energy
Here e is the voltage induced in the electric terminals by the changing magnetic stored
energy. It is through this reaction voltage that the external electric circuit supplies power
to the coupling magnetic field and hence to the mechanical output terminals.
dWelec = ei dt
(3.12)
The basic energy-conversion process is one involving the coupling field and its action and
reaction on the electric and mechanical systems.
Combining (3.11) and (3.12) results in
dWelec = ei dt = dWmech + dWfld
(3.13)
§3.3 Energy in Singly-Excited Magnetic Field Systems
We are to deal energy-conversion systems: the magnetic circuits have air gaps between the
stationary and moving members in which considerable energy is stored in the magnetic field.
This field acts as the energy-conversion medium, and its energy is the reservoir between
the electric and mechanical system.
Fig. 3.4 shows an electromagnetic relay schematically. The predominant energy storage
occurs in the air gap, and the properties of the magnetic circuit are determined by the
dimensions of the air gap.
Figure 3.4
Schematic of an electromagnetic relay.
4
λ = L( x )i
dWmech = f fld dx
(3.14)
(3.15)
dWfld = idλ − f fld dx
(3.16)
Wfld is uniquely specified by the values of λ and x . Therefore, λ and x are
referred to as state variables.
Since the magnetic energy storage system is lossless, it is a conservative system. Wfld is
the same regardless of how λ and x are brought to their final values. See Fig. 3.5
where tow separate paths are shown.
Figure 3.5
Integration paths for Wfld .
Wfld (λ0 , x0 ) =
∫
∫
dWfld +
path 2a
dWfld
(3.17)
path 2b
On path 2a, d λ = 0 and f fld = 0 . Thus, dWfld = 0 on path 2a.
On path 2b, dx = 0 .
Therefore, (3.17) reduces to the integral of id λ over path 2b.
λ0
Wfld (λ0 , x 0 ) = ∫ i (λ , x0 ) dλ
0
For a linear system in which λ is proportional to i , (3.18) gives
λ
λ λ′
1 λ2
′
′
′
Wfld (λ , x ) = ∫ i (λ , x ) dλ = ∫
dλ =
0
0 L( x )
2 L( x )
V : the volume of the magnetic field
Wfld = ∫
V
(∫
B
0
H ⋅ dB′
)
dV
(3.18)
(3.19)
(3.20)
If B = µ H ,
⎛ B2 ⎞
Wfld = ∫ ⎜
⎟ dV
V 2µ
⎝
⎠
5
(3.21)
Figure 3.6 (a) Relay with movable plunger for Example 3.2.
(b) Detail showing air-gap configuration with the plunger partially removed.
6
§3.4 Determination of Magnetic Force and Torque form Energy
The magnetic stored energy Wfld is a state function, determined uniquely by the values of the
independent state variables λ and x .
dWfld (λ , x ) = idλ − f fld dx
dF ( x1 , x2 ) =
dWfld (λ , x ) =
∂F
∂x1
∂Wfld
∂λ
∂F
∂x2
x1
∂Wfld
∂x
λ
dx1 +
x2
dλ +
x
(3.22)
dx2
(3.23)
dx
(3.24)
Comparing (3.22) with (3.24) gives (3.25) and (3.26):
i=
∂Wfld (λ , x )
∂λ
f fld = −
(3.25)
x
∂Wfld (λ , x )
∂x
(3.26)
λ
Once we know Wfld as a function of λ and x , (3.25) can be used to solve for i (λ , x) .
Equation (3.26) can be used to solve for the mechanical force f fld (λ , x) . The partial
derivative is taken while holding the flux linkages λ constant.
For linear magnetic systems for which λ = L( x)i , the force can be found as
f fld = −
∂ ⎛ 1 λ2 ⎞
⎜
⎟
∂x ⎜⎝ 2 L( x ) ⎟⎠
f fld =
=
λ
λ2
2 L(x )
i dL( x )
2 dx
2
dL( x )
dx
(3.27)
2
7
(3.28)
Figure 3.7 Example 3.3. (a) Polynomial curve fit of inductance.
(b) Force as a function of position x for i = 0.75 A.
For a system with a rotating mechanical terminal, the mechanical terminal variables become
the angular displacement θ and the torque Tfld .
dWfld (λ ,θ ) = idλ − Tfld dθ
(3.29)
Tfld = −
∂Wfld (λ , θ )
∂θ
(3.30)
λ
For linear magnetic systems for which λ = L(θ )i :
Wfld (λ , θ ) =
Tfld = −
∂
∂θ
1 λ2
2 L(θ )
⎛ 1 λ2 ⎞
⎜⎜
⎟⎟
⎝ 2 L(θ ) ⎠
Tfld =
=
λ
1 λ2 dL(θ )
2 L(θ )2 dθ
i dL(θ )
2 dθ
(3.31)
(3.32)
(3.33)
2
Figure 3.9 Magnetic circuit for Example 3.4.
8
(3.34)
§3.5 Determination of Magnetic Force and Torque from Coenergy
Recall that in §3.4, the magnetic stored energy Wfld is a state function, determined uniquely
by the values of the independent state variables λ and x .
dWfld (λ , x ) = idλ − f fld dx
(3.22)
dWfld (λ , x ) =
∂Wfld
∂λ
dλ +
x
∂Wfld (λ , x )
i=
∂λ
f fld
∂Wfld
∂x
(3.24)
dx
λ
(3.25)
x
∂W (λ , x )
= − fld
∂x
(3.26)
λ
Coenergy: from which 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.
′ (i, x) is defined as a function of i and x such that
The coenergy W fld
Wfld′ (i , x ) = iλ − Wfld (λ , x )
d (iλ ) = idλ + λdi
dWfld′ (i , x ) = d (iλ ) − dW fld (λ , x)
(3.34)
(3.35)
(3.36)
dWfld′ (i , x ) = λ di + f fld dx
′ (i, x) can be seen to be a state function of the two
From (3.37), the coenergy W fld
(3.37)
independent variables i and x .
dWfld′ (i , x ) =
λ=
′
∂W fld
∂i
di +
∂x
x
∂Wfld′ (i , x )
∂i
f fld =
′
∂W fld
∂Wfld′ (i , x )
∂x
dx
i
(3.39)
x
(3.40)
i
For any given system, (3.26) and (3.40) will give the same result.
9
(3.38)
By analogy to (3.18) in §3.3, the coenergy can be found as (3.41)
λ0
Wfld (λ0 , x 0 ) = ∫ i (λ , x0 ) dλ
0
Wfld′ (i , x ) = ∫ λ (i ′ , x ) di ′
i
0
For linear magnetic systems for which λ = L( x)i ,
1
Wfld′ (i , x ) = L( x ) i 2
2
2
i dL( x )
f fld =
2 dx
(3.43) is identical to the expression given by (3.28).
For a system with a rotating mechanical displacement,
Wfld′ (i ,θ ) = ∫ λ (i ′ , θ ) di ′
i
0
Tfld =
∂Wfld′ (i , θ )
∂θ
(3.18)
(3.41)
(3.42)
(3.43)
(3.44)
(3.45)
i
If the system is magnetically linear,
Wfld′ (i , θ ) =
1
L (θ )i 2
2
i 2 dL(θ )
Tfld =
2 dθ
(3.47) is identical to the expression given by (3.33).
In field-theory terms, for soft magnetic materials
H0
Wfld′ = ∫ ⎛⎜ ∫ B ⋅ dH ⎞⎟ dV
V⎝ 0
⎠
2
µH
Wfld′ = ∫
dV
v
2
For permanent-magnet (hard) materials
H0
Wfld′ = ∫ ⎛⎜ ∫ B ⋅ dH ⎞⎟ dV
V ⎝ Hc
⎠
10
(3.46)
(3.47)
(3.48)
(3.49)
(3.50)
For a magnetically-linear system, the energy and coenergy (densities) are numerically equal:
1
1
1 2
1
λ / L = Li 2 , B 2 / µ = µH 2 . For a nonlinear system in which λ and i or B and
2
2
2
2
H are not linearly proportional, the two functions are not even numerically equal.
Wfld + Wfld′ = λi
(3.51)
Figure 3.10 Graphical interpretation of energy and coenergy in a singly-excited system.
Consider the relay in Fig. 3.4. Assume the relay armature is at position x so that the
device operating at point a in Fig. 3.11. Note that
f fld = −
∂Wfld (λ , x )
∂x
− ∆Wfld
∆x →0
∆x
≅ lim
λ
and
λ
f fld =
∂Wfld′ (i , x )
∂x
∆Wfld′
∆x → 0 ∆x
≅ lim
i
i
Figure 3.11 Effect of ∆x on the energy and coenergy of a singly-excited device:
(a) change of energy with λ held constant; (b) change of coenergy with i held constant.
11
The force acts in a direction to decrease the magnetic field stored energy at constant flux
or to increase the coenergy at constant current.
In a singly-excited device, the force acts to increase the inductance by pulling on
members so as to reduce the reluctance of the magnetic path linking the winding.
Figure 3.12 Magnetic system of Example 3.6.
12
§3.6 Multiply-Excited Magnetic Field Systems
Many electromechanical devices have multiple electrical terminals.
Measurement systems: torque proportional to two electric signals; power as the product of
voltage and current.
Energy conversion devices: multiply-excited magnetic field system.
A simple system with two electrical terminals and one mechanical terminal: Fig. 3.13.
Three independent variables: {θ , λ1 , λ 2 } , {θ , i1 , i2 } , {θ , λ1 , i2 } , or {θ , i1 , λ 2 } .
dWfld (λ 1 , λ 2 , θ ) = i 1dλ 1 + i 2 dλ 2 − Tfld dθ
(3.52)
Figure 3.13 Multiply-excited magnetic energy storage system.
i1 =
i2 =
∂Wfld (λ 1 , λ 2 ,θ )
∂λ 1
∂Wfld (λ 1 , λ 2 ,θ )
Tfld = −
∂λ 2
(3.53)
λ 2 ,θ
(3.54)
λ 1 ,θ
∂Wfld (λ 1 , λ 2 ,θ )
∂θ
(3.55)
λ 1 ,λ 2
To find Wfld , use the path of integration in Fig. 3.14.
(
)
Wfld λ 10 , λ 20 ,θ 0 = ∫
λ 20
0
i 2 (λ 1 = 0 , λ 2 , θ = θ 0 )dλ 2 + ∫
λ 10
0
(
)
i 1 λ 1 , λ 2 = λ 20 ,θ = θ 0 dλ 1 (3.56)
(
)
Figure 3.14 Integration path to obtain Wfld λ 10 , λ 20 ,θ 0 .
13
In a magnetically-linear system,
λ 1 = L 11i 1 + L 12 i 2
λ 2 = L 21i 1 + L 22 i 2
(3.57)
L 12 = L 21
(3.59)
(3.58)
Note that Lij = Lij (θ ) .
i1 =
i2 =
L 22 λ 1 − L 12 λ 2
(3.60)
D
− L 21λ 1 + L 11λ 2
D
D = L 11 L 22 − L 12 L 21
The energy for this linear system is
λ 2 0 L 11 (θ 0 )λ 2
λ 10 L 22 (θ 0 )λ 1 − L 12 (θ 0 )λ 2
0
dλ 1
dλ 2 + ∫
Wfld λ 10 , λ 20 , θ 0 = ∫
0
0
D(θ 0 )
D(θ 0 )
(
(
)
dWfld′ (i 1 , i 2 ,θ ) = λ 1 di 1 + λ 2 di 2 + Tfld dθ
λ2 =
Tfld =
Wfld′ (i 1 , i 2 ,θ 0 ) = ∫
i 20
0
∂Wfld (i 1 , i 2 , θ )
∂i 2
∂Wfld′ (i 1 , i 2 , θ )
(3.64)
(3.66)
i 2 ,θ
(3.67)
i 1 ,θ
∂θ
λ 2 (i 1 = 0 , i 2 ,θ = θ 0 )di 2 + ∫
(3.68)
i 1 ,i 2
λ 10
0
λ 1 (i 1 , i 2 = i 2 ,θ = θ 0 )di 1
0
For the linear system described as (3.57) to (3.59)
1
1
Wfld′ (i 1 , i 2 ,θ 0 ) = L 11 (θ )i 12 + L 22 (θ )i 22 + L 12 (θ )i 1i 2
2
2
2
′
∂Wfld (i 1 , i 2 , θ 0 )
i 1 dL11 (θ ) i 22 dL22 (θ )
dL (θ )
=
+
+ i 1i 2 12
Tfld =
∂θ
2 dθ
2 dθ
dθ
i ,i
1
(3.63)
(3.65)
∂Wfld (i 1 , i 2 , θ )
∂i 1
(3.62)
)
L 12 (θ 0 )
1
1
λ1 λ 2
L 11 (θ 0 )λ 220 +
L 22 (θ 0 )λ 120 −
=
2 D(θ 0 )
2 D(θ 0 )
D(θ 0 ) 0 0
Coenergy function for a system with two windings can be defined as (3.46)
Wfld′ (i 1 , i 2 ,θ ) = λ 1i 1 + λ 2 i 2 − Wfld
λ1 =
(3.61)
(3.69)
(3.70)
(3.71)
2
Note that (3.70) is simpler than (3.63). That is, the coenergy function is a relatively
simple function of displacement.
The use of a coenergy function of the terminal currents simplifies the determination of
torque or force.
Systems with more than two electrical terminals are handled in analogous fashion.
14
Figure 3.15 Multiply-excited magnetic system for Example 3.7.
Figure 3.16 Plot of torque components for the multiply-excited system of Example 3.7.
15
Practice Problem 3.7
Find an expression for the torque of a symmetrical two-winding system whose
inductances vary as
L11 = L22 = 0.8 + 0.27 cos 4θ
L12 = 0.65 cos 2θ
for the condition that i1 = −i2 = 0.37 A .
Solution: Tfld = −0.296 sin 4θ + 0.178 sin 2θ
___________________________________________________________________
System with linear displacement:
(
)
Wfld λ 10 , λ 20 , x0 = ∫
(
λ 20
0
)
Wfld′ i 10 , i 20 , x 0 = ∫
i 2 (λ 1 = 0 , λ 2 , x = x0 )dλ 2 + ∫
λ 10
0
λ 20
0
λ 2 (i 1 = 0 , i 2 , x = x0 )di 2 + ∫
f fld = −
∂Wfld (λ 1 , λ 2 , x )
f fld = −
∂x
∂Wfld′ (i 1 , i 2 , x )
∂x
(
)
, x = x )di
i 1 λ 1 , λ 2 = λ 20 , x = x0 dλ 1
λ 10
0
λ 1 (i 1 , i 2 = i 2
0
0
1
(3.72)
(3.73)
(3.74)
λ 1 ,λ 2
(3.75)
i 1 ,i 2
For a magnetically-linear system,
Wfld′ (i 1 , i 2 , x ) =
f fld
1
1
L 11 ( x )i 12 + L 22 ( x )i 22 + L 12 ( x )i 1i 2
2
2
2
2
i 1 dL11 (x ) i 2 dL22 ( x )
dL ( x )
=
+
+ i 1i 2 12
2 dx
2 dx
dx
16
(3.76)
(3.77)