Power Notes 8

Power Notes 8 - Unit 09 Emma Ettaé EE 3410...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Unit 09 Emma Ettaé EE 3410 Electro-Mechanical 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 fiiction-less, 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 electro-mechanical 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 three-phase 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 right-hand screw rule, e=—N‘2—?=N¢mw sinwt = JEEsmw: where E =iN¢mw =J§an4>m J5 Thus, such a device operates as a single-phase 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 ...
View Full Document

This note was uploaded on 02/06/2012 for the course EE 3410 taught by Professor Staff during the Fall '08 term at LSU.

Page1 / 3

Power Notes 8 - Unit 09 Emma Ettaé EE 3410...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online