Download safsafsfsdfsfsfsdfsfsfsfsdfs and more Exercises Electric Machines in PDF only on Docsity!
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,
F = qE (3.2)
In pure magnetic-field systems,
F = q ( v × B ) (3.3)
Errore.
Figure 3.1 Right-hand rule for F = q ( v × B ).
For situations where large numbers of charged particles are in motion,
Fv =ρ ( E + v × B ) (3.4)
J = ρ v (3.5)
Fv = J × B (3.6)
ρ (charge density): coulombs/m
3 , Fv (force density): newtons/m
3 ,
J = ρ v (current density): amperes/m
2 .
Errore.
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.
§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
=
into heat
converted
Energy
field
storedinmagnetic
Increaseinenergy
output
energy
Mechanical
sources
formelectric
Energyinput
(3.10)
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
dW elec (^) = dW mech+ dW fld (3.11)
where
dW elec = id λ= differential electric energy input
dW mech (^) = f fld dx = differential mechanical energy output
dW fld = 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.
dW (^) elec = 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
dW elec (^) = eidt = dW mech+ dW fld (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.
Errore.
Figure 3.4 Schematic of an electromagnetic relay.
λ = L ( ) xi (3.14)
dW (^) mech = f fld dx (3.15)
dW fld = id λ − f fld dx (3.16)
W fld 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. W fld is
the same regardless of how λ and x are brought to their final values. See Fig. 3.
where tow separate paths are shown.
Figure 3.5 Integration paths for W fld.
= +
path 2b
fld path2a
W fld λ 0 , x 0 dW fld dW (3.17)
On path 2a, d λ = 0 and f fld = 0. Thus, dW fld = 0 on path 2a.
On path 2b, dx = 0.
Therefore, (3.17) reduces to the integral of id λ over path 2b.
λ W x i x d
=
0 fld 0 0 0 0 , , (3.18)
For a linear system in which λ is proportional to i , (3.18) gives
′=
′ = ′ ′=
0
2
fld (^02)
1 , , Lx
d Lx
W x i xd (3.19)
V : the volume of the magnetic field
fld ( 0 )
B
V
W = H ⋅ dB ′ dV
(3.20)
If B = μ H ,
2
fld V 2
B W dV
= (^)
(3.21)
§3.4 Determination of Magnetic Force and Torque form Energy
The magnetic stored energy W fld is a state function, determined uniquely by the values of the
independent state variables λ and x.
dW fld ( λ , x ) = id λ− f fld dx (3.22)
2 1
1 2 1 2 1 2
,
x x
F F dF x x dx dx x x
∂ ∂ = + ∂ ∂
(3.23)
( ) dx
x
W d
W dW x
x λ
∂
∂
∂
∂
fld fld fld ,^ (3.24)
Comparing (3.22) with (3.24) gives (3.25) and (3.26):
x
W x i
∂
∂
fld , (3.25)
λ
x
W x f ∂
∂ = −
fld , fld (3.26)
Once we know W fld 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
dx
dLx
x Lx Lx
f 2
2 2
fld 2 2
λ
=
∂
∂ = − (3.27)
dx
i dLx f 2
2
fld =^ (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 T fld.
dW fld (λ ,θ )= id λ− T fld d θ (3.29)
∂
∂ = −
fld , fld
W T (3.30)
For linear magnetic systems for which λ = L ( ) θ i :
λ λ θ L
W
2
fld 2
1 , = (3.31)
λ
d
dL
L L
T 2
2 2
fld 2
1
2
1
∂
∂ = − (3.32)
(3.33)
d
i dL T 2
2
fld =^ (3.34)
Figure 3.9 Magnetic circuit for Example 3.4.
By analogy to (3.18) in §3.3, the coenergy can be found as (3.41)
λ W x i x d
=
0 fld 0 0 0 0 , , (3.18)
i W i x i x di fld (^0)
For linear magnetic systems for which λ = L ( x ) i ,
2 fld 2
1 W ′^ i , x = Lxi (3.42)
dx
i dLx f 2
2
fld =^ (3.43)
(3.43) is identical to the expression given by (3.28).
For a system with a rotating mechanical displacement,
W ( i ) ( i ) di
i ′ (^) = ′ ′
fld 0
i
W i T
∂
∂ ′
fld , fld (3.45)
If the system is magnetically linear,
2 fld 2
1
W ′ i , θ = L θ i (3.46)
d
i dL T 2
2
fld =^ (3.47)
(3.47) is identical to the expression given by (3.33).
In field-theory terms, for soft magnetic materials
′ (^) = ⋅ V
H W B dH dV
0 fld (^0) (3.48)
dV
H W
v
′ = 2
2
fld
(3.49)
For permanent-magnet (hard) materials
′ = ⋅ V
H
H
W B dH dV c
0 fld (3.50)
For a magnetically-linear system, the energy and coenergy (densities) are numerically equal:
2 2
2
1 / 2
1
λ L = Li ,
2 2
2
1 / 2
1
B μ = μ H. For a nonlinear system in which λ and i or B and
H are not linearly proportional, the two functions are not even numerically equal.
W fld + W fld′ = λ i (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
λ λ
x
W
x
W x f x (^) Δ
−Δ ≅ ∂
∂ = − Δ→
fld
0
fld fld lim
, and
i
x i x
W
x
W i x f Δ
Δ ′ ≅ ∂
∂ ′
Δ→
fld
0
fld fld lim
,
Figure 3.11 Effect of Δ 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.
§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 , λ 1 , λ 2 }, { θ , i 1 , i 2 }, { θ , λ 1 , i 2 }, or { θ , i 1 , λ 2 }.
dW (^) fld (λ 1 , λ 2 ,θ)= i 1 d λ 1 + i 2 d λ 2 − T fld d θ (3.52)
Figure 3.13 Multiply-excited magnetic energy storage system.
( )
λθ
(^1) ,
fld 1 2 1
2
, ,
∂
∂
W i (3.53)
( )
λθ
(^2) ,
fld 1 2 2
1
, ,
∂
∂
W i (3.54)
( )
1 , 2
fld 1 2 fld
, ,
λ λ
∂
∂ = −
W T (3.55)
To find W fld , use the path of integration in Fig. 3.14.
0
1 2 0 2 1 0
fld 1 ,^2 , 0 2 0 , , , 0 ,
(^2010)
0 0
λ λ W = i = = d + i = = d
(3.56)
Figure 3.14 Integration path to obtain fld ( 1 , 2 , 0 )
0 0
W λ λ θ.
In a magnetically-linear system,
λ 1 = L 11 i 1 + L 12 i 2 (3.57)
λ 2 = L 21 i 1 + L 22 i 2 (3.58)
L 12 (^) = L 21 (3.59)
Note that Lij = Lij ( θ).
D
L L i
22 1 12 2 1
= (3.60)
D
L L i
21 1 11 2 2
= (3.61)
D = L 11 L 22 − L 12 L 21 (3.62)
The energy for this linear system is
0 0 0 0
(^2010 )
0 0
1 2 0
2 12 0 22 0 1 0
2 11 0 2 0
0 1 0
22 0 1 12 0 2
0 2 0
11 0 2 fld 1 2 0
2
1
2
1
, ,
λ λ
D
L L D
L D
d D
L L d D
L W
= + −
− = +
(3.63)
Coenergy function for a system with two windings can be defined as (3.46)
W fld ′ ( i 1 , i 2 ,θ ) =λ 1 i 1 +λ 2 i 2 − W fld (3.64)
dW (^) fld′^ ( i 1 , i 2 , θ )= λ 1 di 1 + λ 2 di 2 + T fld d θ (3.65)
( )
θ
(^1) ,
fld 1 2 1
2
, ,
i
i
W i i
∂
∂ = (3.66)
( )
θ
(^2) ,
fld 1 2 2
1
, ,
i
i
W i i
∂
∂ = (3.67)
( )
1 , 2
fld 1 2 fld
, ,
i i
W i i T
∂
∂ ′ = (3.68)
fld 1 2 0 0 2 1 2 0 2 0 1 , , 0 , , , , 0
(^2010) W i i i i di i i i di
i
λ ′ = = = + = =
(3.69)
For the linear system described as (3.57) to (3.59)
( ) ( ) ( ) 12 ( ) (^12)
2 22 2
2 fld 1 2 0 11 1 2
1
2
1
W ′ i , i ,θ = L θ i + L θ i + L θ ii (3.70)
( ) (^) ( ) ( ) ( )
θ
θ
θ
θ
θ
θ
θ
θ
d
dL ii d
i dL
d
W i i i dL T
ii
12 12
22
2 11 2
2 1
,
fld 1 2 0 fld 2 2
, ,
1 2
= + + ∂
∂ ′ = (3.71)
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.
Practice Problem 3.
Find an expression for the torque of a symmetrical two-winding system whose
inductances vary as
L 11 = L 22 = 0. 8 + 0. 27 cos 4 θ
L 12 = 0. 65 cos 2 θ
for the condition that i 1 (^) = − i 2 = 0. 37 A.
Solution : T fld =− 0. 296 sin 4 θ + 0. 178 sin 2 θ
System with linear displacement:
fld 1 2 0 0 2 1 2 0 2 0 1 , , 0 , , , , 0
(^2010)
0 0
λ λ λ λ λ λ λ λ λ
λ λ W x = i = x = x d + i = x = x d
(3.72)
0
1 2 0 2 1 0
fld 1 ,^2 , 0 2 0 , , , 0 ,
(^2010)
0 0
W ′^ i i x = i = i x = x di + i i = i x = x di
λ λ λ λ (3.73)
( )
1 , 2
fld 1 2 fld
, ,
λ λ
λ λ
x
W x f ∂
∂ = − (3.74)
( )
1 , 2
fld 1 2 fld
, ,
i i
x
W i i x f ∂
∂ ′ = − (3.75)
For a magnetically-linear system,
( ) ( ) ( ) 12 ( ) (^12)
2 22 2
2 fld 1 2 11 1 2
1
2
1 W ′^ i , i , x = L xi + L xi + L xii (3.76)
dx
dL x ii dx
i dL x
dx
i dL x f
12 12
22
2 11 2
2 1 fld 2 2
= + + (3.77)
