Electromechanical Energy Conversion

conception Chapter 3 Electro robotlikely skillful nix Conversion Topics to covering 1. Introduction 3. vehemence and crookedness 5. skirmish 2. Electro-Motive trace (EMF) 4. double-Excited Actuators 6. Mechanical Comp binglents Introduction (Cont. ) For readiness conversion amidst electrical and robotlike forms, electromechanical whatsiss ar developed. In oecumenic, electromechanical nothing conversion impostures bath be dual-lane into three categories Transducers (for measurement and control), which transform signals of different forms. slips are microphones, pickups, and speakers Force producing devices ( elongate question devices), which produce deposits mostly for linear move drives, such(prenominal) as relays, solenoids (linear actuators), and electromagnets. free burning energy conversion equipment, which operate in rotating mode. A device would be known as a beginning if it convert mechanical energy into electrical energy, or as a motor if it does the o ther means around (from electrical to mechanical). Lorentz Force & EMF Lorentz embrace is the upshot on a point charge due to electro magnetised fields. It is given by the following equation in terms of the electric and magnetized fieldsF ? q(E? v? B) The induced electromotive force in a conductor of length l base with a speed v in a ordered magnetised field of fluxion density B burn down be determined by a e ? ?v? B? ? dl ? b In a ringlet of N turns, the induced emf can be reckon by e ? ? Concept constitute of electromechanical system modeling d? dt where ? is the flux linkage of the helix and the minus sign indicates that the induced current opposes the play of the field. It makes no conflict whether the variation of the flux linkage is a result of the field variation or cast movement. EMF EMF guinea pig EMF in a Linear Actuator Example resultantSketch L(x) and calculate the induced emf in the aggravation coil for a linear actuator shown below. Assuming infi nite permeableness for the magnetic core and ignore the fringing effect, we can express the self generalisation of the coil as L? x ? ? where Rg ? x? ? N2 ?o N 2 l ?d ? x? ? Rg ? x ? 2g L(x) L(0) 2g ?o ? d ? x? l O is the air gap reluctance. ? e? ? N 2l d? d ? Li ? di dL dx di ? ? L ? i =L? x ? ? i o v 2g dt dt dt dx dt dt EMF A champion Conductor in a Uniform Field e ? ? I dc If i=Imsin? t , e? Force and torsion Example Solution (Cont. ) If i=Idc , ?o N 2 l 2g ? Im ? Im ?o N 2l 2gFor a single conductor in a ordered magnetic field, we have v ? d ? x I m cos lettuce ? t ? vI m sin ? t ?o N l 2 2g ?o N 2 l 2g d Fm ? Il ? B ?o N 2 l In a rotating system, the contortion about an axis of rotation can be calculated by 2g d ? x cos? t ? v sin ? t ? ? T? r ? Fm v ? ? ? ? d ? x ? ? ? ? d ? x ? 2 ? 2 ? v 2 cos t ? arctan? ? where r is the radius vector from the axis towards the conductor. B Fm l I X Force and tortuousness A one after another Excited Actuator Consider a individ ually disturbed linear actuator. After a time interval dt, we notice that the speculator has moved for a distance dx under the action of the force F.The mechanical work done by the force playacting on the diver during this time interval is thus dWm ? Fdx Force and torque A individually Excited Actuator The amount of electrical energy that has been transferred into the magnetic field and born-again into the mechanical work during dt is dWe ? dWf ? dWm dWe ? eidt ? vidt? Ri2dt e ? d? dt ? v ? Ri Because dWf ? dW ? dW ? eidt ? Fdx ? id? ? Fdx e m we obtain From the total differential dW f ? ? , x ? ? ? W f , x ? i? Therefore, ? W f , x ? d? ? and ? W f , x ? ?x F dx ?W f ? ? , x ? ?x Force and Torque Force and Torque A Singly Excited Actuator (Cont. ) A Singly Excited Actuator (Cont. ) From the knowledge of electromagnetics, the energy stored in a magnetic field can be expressed as ? Wf ? ? , x? ? ? i? ? , x? d? In the diagram below, it is shown that the magnetic energy i s equivalent to the area above the magnetization or ? -i curve. Mathematically, if we dress the area underneath the magnetization curve as the coenergy (which does not survive somatogeneticly), i. e. 0 For a magnetically linear (with a constant permeability or a straight line magnetization curve such that the inductance of the coil is independent of the excitation current) system, the above formulation becomes 1 ? Wf ? ?, x? ? 2 L? x? and the force acting on the plunger is then F ?Wf ? ?, x? ?x 1 ? ? ? dL? x? 1 2 dL? x? ?i 2 ? L? x? ? dx 2 dx ? 2 we can obtain Wf ? i, x? ? i? ?Wf , x? ? Wf (? , x ) dW f ? i , x ? ? ? di ? id? ? dW f ? ? , x ? ? ? di ? Fdx Therefore, ? ?W f ? i , x ? ?i ? W f ? i , x ? ?i di ? and ?W f ? i , x ? ?x F? dx ? W f ? i , x ? ?x (? , i ) Wf ( i, x ) O i Force and Torque Force and Torque A Singly Excited Actuator (Cont. ) Example 1 Calculate the force acting on the plunger of a linear actuator as shown below. From the definition, the coenerg y can be calculated by iWf ? i , x? ? ? ? ? i , x? di ? 0 Wf ? i, x? ? ? (? , i ) Wf (? , x ) For a magnetically linear system, the above expression becomes Rg 1 L? x? i 2 2 Ni Wf ( i, x ) Rg and the force acting on the plunger is then F? ?Wf ? i , x ? ?x 1 dL? x ? ? i2 dx 2 O i (c) Force and Torque Force and Torque Singly Excited Rotating Actuator Solution to Example 1 Assume infinite permeability for the actuator core. The self inductance of the excitation idle words can be readily obtained as L? x? ? N 2 ? o N 2l? d ? x? ? 2Rg 2g Therefore, the force acting on the plunger is F? ? Rg Ni ?l 1 2 dL ? x ? 2 i ? ? o ? Ni ? 2 dx 4gThe minus sign of the force indicates that the direction of the force is to centralise the displacement so as to reduce the reluctance of the air gaps. Since this force is ca utilize by the variation of magnetic reluctance of the magnetic circuit, it is known as the reluctance force. Rg The singly unhinged linear actuator becomes a singly evoke rota ting actuator if the linearly movable plunger is replaced by a rotor. Through a derivation similar to that for a singly excited linear actuator, one can readily obtain that the torque acting on the rotor can be expressed as the negative partingial derived function of the energy stored in the agnetic field against the angular displacement or as the positive partial derivative of the coenergy against the angular displacement. Force and Torque Solution b) Voltage induced Example The magnetically-linear electro-mechanical circuit breaker as shown is singly-excited via a N-turn coil. Its magnetic reluctance varies with the angle ? as R ? Rm? ? R0 , where Rm and R0 are constant. pull a engineer the torque developed by the field from the system co-energy. When the device is excited with a direct current i=I, the angular displacement increases quadratically as ? ?t ? ? 1 ? t 2 ? ?t ? ? 0 , 2 where ? ? and ? 0 are constant. Find the voltage induced in the coil . Singly Excited Rotati ng Actuator Total turns, N = N1 + N2 Frame reluctance Rf ? rf 2 Gap reluctance Rg ? 2rg ? ? lf 2? 0 ? r wd 2lg ?0rd (2? ? ? ) , 2? ? 760 ? 1. 33 rad Rg(? ) Rcore ?g Rarmature Fm=Ni e(t ) ? ? N 2 IRm (? t ? ? ) R0 ? Rm 1 ? t 2 ? ?t ? ? 0 2 2 ? Singly Excited Rotating Actuator ? Singly Excited Rotating Actuator airgap length, lg = 0. 001 m airgap radius, r = 0. 0745 m airgap depth, d = 0. 0255 m frame length lf = 0. 496 m arm width w = 0. 024 m Singly Excited Rotating Actuator ? (? ) ? T? ? NI R f ? Rg (? ) lf Rf ? 2 ? ? r wd Magnetic flux at equilibrium ? NI ?0 ? ? ? R (? ) ? R f ?g ? ? ?0 NI lf 2 ? 0 ? r wd ? lg ?0rd? ? , Rg ? 2l g ?W f? ? ? ? N2 ? ?, ? R (? ) ? R ? f? ?g dRg dRg 2l g sign(? ) , where ? d? d? ?0 rd (2? ? ? )2 1 2 ? L(? ) 1 2 ? I ?I 2 2 122 ?1 IN 2 ?Rg (? ) ? R f ? 2 2l Rr sign(? ) 1 ? ? I 2N 2 , where Rr ? g 2 2 ?0 rd 4 ?Rg (? ) ? R f ? (2? ? ? ) ?0 rd (2? ? ? ) Restoring Torque ?1, x ? 0 sign ( x ) ? ? 1, x ? 0 NI? 0d lf l ?g 2 ? r w r? Force and Torque Sin gly Excited Rotating Actuator Singly Excited Rotating Actuator (Cont. ) Torque Nm Flux mWb Flux, Torque for 2-pole motorEnergy In g eneral, 1. 5 Coenergy dW f ? id? ? Td ? dW f ? ? di ? Td ? ? i W f ? ? , ? ? ? ? i ? ? , ? ?d ? W f ? i , ? ? ? ? ? ?i , ? ?di ?W f ? ? , ? ? i? ?W f ? ? , ? ? T ?W f ? i , ? ? ?i ?W f ? i , ? ? T? 0 1. 0 mWb, Nm 0. 5 0 If the permeability is a constant, W f , ? ? ? 0. 0 0 5 10 15 20 25 30 35 40 45 rotor angle 50 55 60 65 70 75 80 1 ? 2 2 L ? 1 ? ? ? dL ? 1 2 dL ? ?i 2 ? L ? ? d ? 2 d? ? ? W f ? i , ? ? ? 2 T? T? 12 i L ? 2 1 2 dL ? i 2 d? Force and Torque Force and Torque Doubly Excited Rotating Actuator Doubly Excited Rotating Actuator (Cont. If a second winding is placed on the rotor, the singly excited actuator becomes a doubly excited actuator. The general principle for force and torque calculation discussed here is equally applicable to multi-excited systems. The differential energy and coenergy functions can be derived as dW f ? dWe ? dWm where dWe ? e1i1dt ? e2 i2 dt , e1 ? d? 1 dt , e2 ? d ? 2 dt , and dWm ? Td ? Hence, dW f 1 , ? 2 , ? ? ? i1d ? 1 ? i2 d ? 2 ? Td ? ? and ? W f 1 , ? 2 , ? ? ? W f 1 , ? 2 , ? ? ? W f 1 , ? 2 , ? ? d ? 1 ? d ? 2 ? d? 1 2 ? ? dW f ? i1 , i 2 , ? ? ? d i1 ? 1 ? i 2 ? 2 ? W f ? 1 , ? 2 , ? ? ? ?1 di1 ? ?2 di 2 ? T d ? ? ? W f ? i1 , i 2 , ? ? Therefore, T ? i1 di1 ? ? W f ? i1 , i 2 , ? ? ?Wf 1 , ? 2 , ? ? ? i2 or di 2 ? T? ? W f ? i1 , i 2 , ? ? Force and Torque Doubly Excited Rotating Actuator (Cont. ) Example 3 ? ? L? 1 For magnetically linear systems, ? ? 1 ? ? L11 ? ? ? L ? 2? ? 21 L 1 2 ? ? i1 ? L 22 ? ?i2 ? ? ? i1 ? ? ? 11 ?i ? ? ? ? ? 2? ? 21 or ? 1 2 ? ? ? 1 ? ? 22 ? ? ? 2 ? ? The magnetic energy and coenergy can then be expressed as W f ? ?1 , ? 2 , ? ? ? Therefore, d? ? W f ? i1 , i2 , ? ? Force and Torque and 1 1 2 ? 1 1 ? 12 ? 2 2 ? 2 ? ? 1 2 ? 1 ? 2 2 2 W f ? i 1 , i 2 , ? 1 1 L 1 1 i 12 ? L i 2 ? L 1 2 i1i 2 2 2 22 2 ? W f ? i 1 , i 2 , ? ? 1 2 d L 1 1 ? 1 2 d L 2 2 ? d L 1 2 ? ? i1 ? i2 ? i1i 2 2 2 T ?W f 1 , ? 2 , ? ? ? A magnetically-linear doubly-fed electromechanical actuator has deuce windings and a mechanical output with spatial rotary displacement ?. The self and mutual inductances of the windings are respectively L11 ? ? 5 ? cos(2? ) mH, L22 ? ? 50 ? 10 cos(2? ) mH, and L12 ? ? L21 ? ? 100 cos? mH. Brushless doubly-fed machine The first winding is supplied with i1 = 1. A while the second winding draws i2 = 20 mA. Determine a) The general electromagnetic torque of the actuator as a function of ? . b) The maximum torque that the actuator can develop. Solution to Example 3 (a) Solution to Example 3 (cont. ) Theenergystoredatthedoubly? fedactuatoris, 1 1 2 2 W f ? L11i1 ? L12 i1i2 ? L22 i2 2 2 1 1 ?3 2 ?3 2 ? (5 ? cos 2? ) ? 10 i1 ? (0. 1cos? )i1i2 ? (50 ? 10 cos 2? ) ? 10 i2 2 2 Theexpressionofelectromagnetictorqueisobtainedasfollows ? ?W f (i1 , i2 ,? ) T ? i1 ? 1. 5, i2 ? 0. 02 ? 2 2 1 ? (i 1 L11 ) ? (i1i2 L12 ) 1 ? (i2 L22 ) ? ? ? 2 2 1 1 (1. 5) 2 ( ? 2 sin 2? ) ? 10 ? 3 ? (1. 5)(0. 02)(? 0. 1sin ? ) ? (0. 02) 2 ( ? 20 sin 2? ) ? 10 ? 3 2 2 ?3 ? ? ( 2. 25 sin 2? ? 3 sin ? ) ? 10 Why Magnetic Field? Ratio of electrical and Magnetic Energy Densities in the air gap we ? 0 ? 0 E 2 1 ? ? wm B2 3. 6 ? 10 5 Saturation Flux absorption Bs = 2T in commonlyused magnetic materials Air breakdown voltage Ebd=1,000,000 V/m b) Atmaximumtorque, dT ?0 d? DifferentiatingTfrompart(a), 4. 5 cos 2? ? 3 cos ? ? 0 ? 1. 5 cos 2? ? cos ? ? 0 or 1. 5( 2 cos 2 ? ? 1) ? cos? ? 0 Solvingfor? bythequadraticformula, ?=55. 94and153. 25(extraneous)Substitutingthe appreciateof? intothetorqueexpressionyields, T(max) ? ?(2. 25 sin 2(55. 94) ? 3 sin(55. 94)) ? 10 ? 3 ? ?4. 57 ? 10 ? 3 Nm electric automobile Machines Electric motor converts electrical energy into mechanical dubiousness. The reverse task, that of converting mechanical motion into electrical energy, is accomplished by a generat or or dynamo. In many cases the two devices differ only in their exercise and minor construction details, and some applications use a single device to fill both roles. For example, traction motors used on locomotive often perform both tasks if the locomotive is equipped with ynamic brakes. Introduction Electric Motors Electric Machine Insulation Class DC Motors Universal (DC/AC) AC Motors instalment Synchronous Stepping Motors Brushless DC Motors Coreless DC Motors Linear Motors MEMS Nano Motors A critical chemical element in the reduced life of electrical equipment is heat. The type of insulation used in a motor depends on the operating temperature that the motor impart experience. Average insulation life decreases rapidly with increases in motor midland operating temperatures. Electric motor converts electrical energy into mechanical motion Lorentz force on any wire when it is onducting electricity while contained in spite of appearance a magnetic field Rotor rotatin g part Stator stationary part Armature part of the motor across which the voltage is supplied MaglevMagnetic Levitation Three material body AC induction motors rated 1 Hp (750 W) and 25 W with pure motors from CD player, toy and CD/DVD drive reader head traverse DC Generators / Dynamos AC Generators / Alternators As the first electrical generator capable of delivering power for industry, the dynamo uses electromagnetic principles to convert mechanical rotation into a pulsing direct electric current through the use of a commutator.Without a commutator, the dynamo is an example of an alternator, which is a synchronous singly-fed generator. With an electromechanical commutator, the dynamo is a classical music direct current (DC) generator. The DC generator can operate at any speed within mechanical limits but always outputs a direct current waveform. Mechanical energy is used to rotate the coil (N turns, area A) at uniform angular speed ? in the magnetic field B, it will produce a sinusoidal emf in the coil Permanent Magnet DC Generators d? d ? ? ( NBA cos ? ) dt dt ? NBA? sin ? t e(t ) ? ? http//micro. magnet. fsu. edu/electromag/ java/generator/dc. tml Automotive alternator Rotor emf and current are induced by rotating magnetic field http//micro. magnet. fsu. edu/electromag/java/generator/ac. html Mechanical Components Mechanical Components Mass and Inertia The mechanical component which stores kinetic energy is a caboodle in a translational system, and a irregular of inertia in a rotational system. Mass and Inertia (Cont. ) The kinetic energy stored by a mass moving at a velocity v, or a moment of inertia rotating at an angular speed ?. can be calculated by ? x M T F J Wk ? 1 Mv2 2 d? d 2? T? J 2 ? J dt dt dv d 2x F? M ?M dt 2 dt 1 J? 2 2 (translational system) rotational system) comparability with the relationships of voltage, current, and magnetic energy in an inductor V? L By the Newtons second law, we have Wk ? or di dt and WL ? 1 Li2 2 we may c ompliments a mass or a moment of inertia as an inductor which stores magnetic energy, if we let J? L M? L or Mechanical Components Mechanical Components Springs An nonsuch stand out is a device with negligible mass and mechanical losses, whose contortion is a single-valued function of the utilize force or torque. A linear ideal forge has deformation proportional to force or ? 1 torque. Springs (Cont. ) For a given contortion of x and ? the potential energy stored in a spring is 1 1 W p ? ? Td ? ? K ? 2 W p ? ? Fdx ? Kx 2 T x1 F ? K ? x 1 ? x o ? ? Kx (linear spring) (torsional spring) Comparing with the relationships of electric charge, voltage and electric energy in a capacitor Q V? C F 2 2 WC ? and 1 1 Q2 VQ ? 2 2C we may regard a spring as an electric capacitor which stores electric potential energy, if we let T ? K 1 ? ?o ? ? K ? K? 1 C Friction Friction deterrent exampleling Friction force that opposes the sex act motion or tendency of such motion of two surfaces in contact. Friction between the two purposes converts kinetic energy into heat.Coefficient of clangoring (Frictional coefficient) dimensionless scalar value which describes the ratio of the force of encounter between two bodies and the force atmospheric pressure them together, needs not be less than 1 under wakeless conditions, a expel on concrete may have a coefficient of friction of 1. 7. Static friction (stiction) occurs when the two objects are not moving relative to each other Rolling friction occuring when one object rolls on another (like a cars wheels on the ground), is stiction as the patch of the pall in contact with the ground, at any point while the tire spins, is stationary relative to the ground.Kinetic (or dynamic) friction occurs when two objects are moving relative to each other and rub together Sliding friction is when two objects are rubbing against each other. Fluid friction is the friction between a solid object as it moves through a suave or a gas. The drag of air on an airplane or of water on a swimmer are two examples of legato friction. Lu-Gre Model (1995) ? 0 , ? 1 bristles stiffness and damping coefficient ?2 viscous friction FC , F S Coulomb and Stribeck friction ? F f ? ? 0 z ? ? 1z ? ? 2v ? z? v? v z g (v ) 2 1 g (v ) ? FC ? ( F S ? FC ) e ? v / v S ) ?0 Mechanical Components Mechanical Components moist The mechanical wet is analogous to electrical resistor in that it dissipates energy as heat. An ideal damper is a device that exhibits no mass or spring effect and exerts a force that is a function of the relative velocity between its two parts. A linear ideal damper has a force proportional to the relative velocity. In all cases a damper produces a force that opposes the relative motion of the two parts. Mechanical friction occurs in a variety of situations under many different physical conditions.Sometimes friction is unwanted but must be tolerated and accounted for analytically, as, for example, in bearings, skid electrical contacts, and the aerodynamic drag on a moving body. In other cases friction is desired and is designed into equipment. Examples are vibration dampers and surprise absorbers. d ? x2 ? x1 ? dt dx ?B dt F? B ? B? R d 2 ? ?1 ? dt d? ?B dt T? B Damper (Cont. ) Mechanical Components Mechanical Components Damper (Cont. ) The damping due to Coulomb friction, as shown by the characteristic, can be regarded as a nonlinear resistor, which can declare the voltage across it to be constant.The Coulomb friction force can be expressed as Damper (Cont. ) There is another charitable of damping caused by the drag of a viscous fluid in pissed off flow. 2 F ? ? Bs d x2 ? x1 dt F ? ?d Fn ? ? d Fn d ? x2 ? x1 ? dt ? ? ? Bs dx dt d ? x2 ? x1 ? dt ? 2 ? R ? B s dx dt dx dt ? dx dt or T ? ? Bs d 2 ? ?1 ? dt Comparing with V=RI, we may conclude that ?F R? d n dx dt ? ? ? Bs d? dt ? ? 2 2 ? R ? B s d ? dt MR Dampers as a semi-active device MR Damper New Models Non-symmetrical Model (200 7) ? F ( x) ? c0 x ? ko ( x ? x0 ) ? ?z ? ? ? z ? (? ? ( ? ? ? sign( zx) z ) x n hysteresis variable, ? , ? , ? , ? , n, c0 , k0 model parameters Bouc-Wen Model ? F ( x ) ? c0 x ? k o ( x ? x0 ) ? ? z ? ? ? ? z ? ? ? z x z n ? 1 ? ? x z n ? ? x z hysteresis variable , ? , ? , ? , ? , n , c 0 , k 0 model parameters Static Hysteresis Model (2006) ? F ( x) ? cx ? kx ? ?z ? f 0 ? z ? tanh( ? x ? sign( x)) z hysteresis variable, ? , ? , f 0 , c, k model parameters Minimally-Parameterised Model (2007) ? F G ( x ) ? D ( x ), F ( x) ? ? 1 ? F2 G ( x ) ? D ( x ), b G ( x) ? a ? ? 1 ? exp ( cx ) ? D ( x ) ? rexp? ( x / 2? ) 2 ? 0 x ? 0, x