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Unformatted text preview: Unit 09 Emma Ettaé EE 3410 ElectroMechanical Energy Conversion Conversion of mechanical to electrical energy is possible
due to mechanical force exerted on electric charge that moves in a
magnetic ﬁeld, as well the force exerted on a charge by electric
ﬁeld. This dynamic interaction between electric charge and
magnetic ﬁeld is described by the Lorentz Law dﬁ=dq(i+;x1§).
Suppose, that a conducting bar moves along rails a and b in a magnetic ﬁeld of the ﬂux density [3 with velocity v , as shown in
Fig. l. '5' ‘6 M
Dal Fig. l A Lorentz’s force is exerted on electric charge q in the moving bar
of the value F'L = q (17 X F?) .
An observer of the force on charge q inside of the bar can conclude,
that an electric ﬁeld of intensity 2, such that I? L = (131. 2
has occurred in the bar. This ﬁeld intensity is
EL = t7 x [3.
Due to the Lorentz’s force, the electric charge is displaced as shown
in Fi . l.
g A difference of electric potential e between rails a and b
occurs due to the charge displacement. The voltage between two conductors is equal to mechanical work needed for transportation of
the unit charge between these two conductors, namely a a IF d 7 J‘qg d7
Wab = b = b
q q ‘1
Since the charge is displaced againsLthe force exerted by the electric
ﬁeld, this work has to be done by an external agent that transports
the charge, thus the voltage e is positive. Since the angle between vectors {3 and f? is 90deg, while the angle between F? and d7 is 0, then, assuming that the ﬂux density 3 between conductors a and b
is constant, “ab=e= =ajEdi=aj(vxi2)di.
b b a a
uab =e= I(§X§)di=vB Idl=Blv.
b b Thus, if a conductor is moved in a magnetic ﬁeld in such a way that
direction of this conductor is perpendiculm to its displacement and it
moves in a plane perpendicular to the vector of the ﬁeld density,
then a voltage proportional to the magnetic ﬂux density, the bar
velocity and its length, occurs on the conductor. When rails a and b are open, then the electric ﬁeld that
occurs due to the bar movement is entirely balanced by the electric
ﬁeld created by displaced electric charge and consequently, the
resultant electric ﬁeld in the bar is equal to zero. No electric force is
exerted on the electric charge in the bar, thus it is not displaced any
more. Let the rails be connected with a resistor as shown in
Fig.2. Due to the voltage between rails, a current ﬂows through the
resistor causing energy dissipation as a heat. The resistor active
power is P = e 1‘.
Any reduction of the electric charge in the rail, due to the current
ﬂow, would result in a reduction of the electric ﬁeld caused by this
charge in the moving bar and consequently, the Lorentz’s force
exerted on the charge in the bar would prevail. This results in the
current ﬂow in the bar, the current, which is equal to the current in
the resistor. Fig. 2 The current in the bar ﬂaws in the magnetic ﬁeld. Thus, a Lorentz’s
force is exerted on each element of the bar dz; = mi x a ,
and oriented against the direction of the bar displacement. If the bar
moves in the plane perpendicular to the magnetic ﬂux density vector
and its displacement is perpendicular to direction of the current, then
assuming that the flux density between rails does not change, the
total force exerted on the bar is a a a
FL: deL=jidix§=iBjd1=BiL
b b b
Such a force, if not balanced by another force would decelerate the
bar. It moves at a constant velocity on the condition that a
mechanical force is exerted on the bar. An external mechanical
agent, referred as a primary mover, is needed to provide such a mechanical force.
The mechanical work that has to be done by the primary
mover for the bar displacement by dx is dWm =Fmdz=Fmdx, and the primary mover power Since the mechanical force has to balance the Lorentz’s force, the
mechanical power of the primary mover is Pm=Fmv=FLv=(BiI)v=(Blv)i=ei=P Thus, the mechanical power of the primary mover is equal to the
active power P of the resistor. It means that mechanical energy
delivered to the system by the primary mover is transported as
electrical energy to the resistor and dissipated as a heat. At a ﬁxed
velocity of the bar, the generated voltage e is constant, thus the
resistor power is 2 P=5R—=Pm, thus, the mechanical power of the primary mover depends only on
the resistance R. It illustrates how the load power affects the power of the primary mover.
Let another movable conducting bar is placed on the rails in a magnetic ﬁeld as shown in Fig. 3. The current in this bar causes
that a Lorentz’s force is exerted on the bar FML=BMil. The bar velocity is kept constant on the condition that the Lorentz’s
force is balanced by the mechanical force exerted by a mechanical
load, i.e., FML = FM“. The bar displacement in the magnetic ﬁeld
causes that a voltage
eM = BM i VM
is generated. At the assumption that the rails resistance is equal to
zero, voltages generated on the generator and the motor bars have to
be equal,
eM = ec.
Thus,
BM VM = 86 VG.
It means that at velocity VG ﬁxed by the primary mover, the load bar
velocity is ﬁxed as well. The mechanical power of the primary
mover is equal to PGm =FomVG =FGLVG =(BGi0vG =(BGIVG)i=eGi=PGe:
and it is equal to the mechanical power of the load, because PGe=PMe =eMi = (BMIVM)i = (BMiDvM = Fva =FvaM =PMm. Fig. 3 This result is obtained on the assumption that there are no
mechanical power losses on friction in the generator and the load
and no active power losses on resistances of the generator, rails and
the motor. At such assumptions electrical and mechanical powers of
the generator and the motor are mutually equal, i.e., PGm=PGe=P=PMe=PMm. All powers are speciﬁed, alternatively, by products of voltages and
currents or by products of forces and velocities, namely P=ui=Fv. The velocity and voltage in such products are enforced by the
primary mover. The velocity of the motor bar could be changed only
by a change in ﬂux density BM/BG ratio (or by change of the length
of the motor conductor). The force and consequently the current are
imposed by the mechanical load force. In such a way, both the
primary mover and the load contribute to the power value. The ﬁrst
contributes through the velocity and the voltage; the second
contributes through the force and the current. For example, if the weigh of mass m in Fig. 3 is removed, then mechanical power of the
primary mover declines to zero, because no current is needed to
move the unloaded motor bar. On the other side, an inﬁnite force is
needed to stop this bar. Consequently, the current and the mecha
nical power of the primary mover approach an infinity. Practically, however, mechanical power of the primary
mover is limited. Thus, when the mechanical power approaches this
limit, the primary mover slows down and the generated voltage
declines. At the same time, generator, transmission line and motor
have resistances that limit the current. If r denotes the total
resistance of the generator, transmission line and the motor, then the
current cannot be higher than z m
r Fundamentals of electromechanical energy conversion
were discussed above under the assumption that the whole system is
ideal, it means, the mechanical components are ﬁictionless, and the
electrical components are resistance—less. Moreover, In real system
only a part of energy of a primary source, like natural fossil, heat,
water or wind could be harnessed by a primary mover. A consi
derable part of it is lost. Generators have friction, need ventilation,
energy is dissipated in windings. Only a part of energy delivered to a
transmission line is recovered at its end. Motor windings are heated,
rotating parts dissipate energy because of ﬁiction. Thus, only a part
of the harnessed energy is used by a customer. A substantial part is
lost. A diagram of energy ﬂow and dissipation is shown in Fig. 4. [max Not convened energy TRANSMISSlON Resistance Friction Resistance
Friction Reelstance GENERATOR MOTOR Fig. 4 Figure 3 explains the very essence and mechanism of
electromechanical energy conversion, puriﬁed however from all
technical factors. Some of them are crucial for technical conversion.
First of all, generators and motors are not linear but rotational
devices, though linear motors are needed sometimes for special
purposes, for example, in transportation. Moreover, movement of a
conductor could be superseded by a movement of the magnetic ﬁeld,
as shown in Fig. 5, in this case, the ﬁeld of a permanent magnet. ’ Fig. 5 It solves the issue of the current transfer from the conductor where
the voltage is generated to stationary terminals of the generator.
Such a generator cannot, of course, operate continuously. A single
conductor is not sufﬁcient as well for generating voltages on a level
required in electrical power systems. Moreover, the voltage genera
ted in situation considered is unidirectional. Transformer cannot be
used for changing this voltage to other level. To generate bi
directional periodic voltage, the magnetic ﬂux linked with the volt age generating loop has to change its direction periodically. It is Thus, device shown in Fig. 7 operates as threephase generator. It is
possible in a rotational generator shown in Fig. 6. referred to as a synchronous generator. Fig. 6 It is built of a cylindrical stator, made of laminated ferromagnetic
material with a coil of the shape shown in Fig.6b, placed in the
stator slots. The magnetic field is created by a rotor made of a
permanent magnet, driven by the primary mover. I If the rotor rotates with the angular velocity of a) radians
per second, a) = 279’, where f denotes the number of rotations per
second, then the magnetic flux linked with the coil changes as (D = (15mm cosa = (15m cos a)! At the ﬂux and the induced voltage orientation as shown in Fig. 6,
i.e., according to the righthand screw rule, e=—N‘2—?=N¢mw sinwt = JEEsmw: where E =iN¢mw =J§an4>m J5
Thus, such a device operates as a singlephase generator of an AC
voltage. If two other coils, displaced mutually by 120°, are placed in
the stator as shown in Fig. 7 Fig.7 Particular coils are speciﬁed by indices R, S and T. The magnetic
ﬂux linked with these coils, due to their 120° displacement is (DR = (15mm, cosa = 45m“ cos a)! ,
(.05 = rpm cos(a —120°) = om cos (art 120°), :15, = rpm cos(a — 240°) = rpm cos (wt— 240°) . Hence the voltage induced in these coils is equal to eR =N dz“ =N<Dmar sine)! =J§Esmm,
do
es =N dis =J§Esin(wr—120°),
eT =.Ndqu =J§Esin(an—240°). , d1 ...
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This note was uploaded on 02/06/2012 for the course EE 3410 taught by Professor Staff during the Fall '08 term at LSU.
 Fall '08
 Staff

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