advertisement

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. PowerPoint Slides to accompany Electric Machinery Sixth Edition A.E. Fitzgerald Charles Kingsley, Jr. Stephen D. Umans Chapter 3 Electromechanical-EnergyConversion Principles 3-0 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3.1 FORCES AND TORQUES IN MAGNETIC FIELD SYSTEMS Lorentz Force Law: F q (E v B) For many charged particle Fv (E v B) 3-1 N/m3 ( coulombs/m3) Current density Jv A/m2 Fv J B Current IJA A F I B N/m Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Example 3.1: A nonmagnetic motor containing a single-turn coil is placed in a uniform magnetic field of magnitude B0, as shown in Fig. 3.2. The coil sides are at radius R and the wire carries current I as indicated. Find the θ-directed torque as a function of rotor position α when I=10 A, B0=0.02 T and R=0.05 m. Assume that the rotor is of length l=0.3 m. 3-2 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Very few problems can be solved using Lorentz force, where current-carrying elements and simple structures exist. Most electromechanical-energy-conversion devices contain magnetic material and forces can not be calculated from Lorentz force. Thus, We will use ENERGY METHOD based on conservation of energy. 3-3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Electrical terminals: e and i Mechanical terminals: ffld and x Losses separated from energy storage mechanism Interaction through magnetic stored energy 3-4 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Time rate of change of Wfld (field energy) equals to the difference of input electrical power and output mechanical power for lossless systems. d W fld dt e i f fld dx dt or d Wfld i d f fld dx Force can be solved as a function of flux linkage λ and position x. 3-5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3.2 ENERGY BALANCE Energy neither created nor destroyed, it only changes the form. Energy balance equation is written for motor action below Energy input Mechanical Increasein Energy from electric energy energy stored converted sources output in magneticfield t o heat For lossless magnetic-energy-storage system d Welec d Wmech d W fld d Welec : Differential electricalenergy input d Wmech : Differential mechanicalenergy output d W fld : Differential change in magneticstoredenergy 3-6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3.3 ENERGY IN SINGLY-EXCITED MAGNETIC FIELD SYSTEMS Schematic of an electromagnetic relay. Figure 3.4 3-7 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The magnetic circuit can be described by an inductance which is a function of the geometry and permeability of the magnetic material. When air-gap exist in most cases Rgap>>Rcore and energy storage occurs in the gap. Magnetic nonlinearity and core losses neglected in practical devices. Flux linkage and current linearly related. Energy equation L( x ) i d Wfld i d f fld dx Wfld uniquely specified by the value of λ and x. Thus, λ and x are called STATE VARIABLES. 3-8 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Magnetic stored energy Wfld uniquely determined by λ and x regardless of how they are brought to their final values. W fld d W fld d W fld path2 a path2b 0 W fld (0 , x0 ) i( , x0 ) d 0 OR magnetic stored energy: W fld B H dB dV V0 Integration paths for Wfld. Figure 3.5 3-9 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Example 3.2:The relay shown on the figure is made of infinitelypermeable magnetic material with a movable plunger, also of infinitely-permeable material. The height of the plunger is much greater than the air-gap length (h>>g). Calculate the magnetic stored energy Wfld as a function of plunger position (0<x<d) for N=1000 turns, g=2 mm, d=0.15 m, l=0.1 m, and i=10 A. 3-10 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3.4 DETERMINATION OF MAGNETIC FORCE AND TORQUE FROM ENERGY Consider any state function F(x1, x2), the total differential of F with respect to the two variables x1 and x2 F d F ( x1 , x2 ) x1 x2 F d x1 x2 d x2 x1 Similarly, for energy function Wfld(λ, x) d W fld ( , x) W fld d x W fld x d Wfld (, x) i d f fld dx 3-11 dx Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. i W fld f fld W fld x x Once we know the energy, current and more importantly force can be calculated. For a system with rotating mechanical terminal x f fld T fld d Wfld (, ) i d T fld d T fld 3-12 W fld ( , ) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Example 3.4:The magnetic circuit below consists of a single-coil stator and an oval rotor. Because the air-gap is nonuniform, the coil inductance varies with rotor angular position, measured between the magnetic axis of the stator coil and the major axis of the rotor, as L( ) L0 L2 cos(2 ) where where L0=10.6 mH and L2=2.7 mH. Note the secondharmonic variation of inductance with rotor angle θ. 3-13 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3.5 DETERMINATION OF MAGNETIC FORCE AND TORQUE FROM COENERGY d Wfld (, x) i d f fld dx Mathematically manipulated to define a new state function known as the COENERGY, from which force can be obtained directly as a function of current. (i, x) i W fld (, x) W fld (i, x) di f fld dx dW fld Note that energy and coenergy equal for linear systems. 3-14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. (i, x) d W fld W fld i di x W fld x dx i (i, x) di f fld dx dW fld (i, x) W fld i f fld (i, x) W fld x i (i, x) (i, x) di W fld 0 3-15 x i Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. In field-theory terms, for soft magnetic materals (B=0 when H=0) H0 B dH dV W fld V 0 For permanent magnet materials (B=0 when H=Hc) H0 B dH dV W fld V Hc 3-16 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Effect of x on the energy and coenergy of a singly-excited device: (a) change of energy with held constant; (b) change of coenergy with i held constant. Figure 3.11 3-17 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Example 3.5: For the relay below, find the force on the plunger as a function of x when the coil is driven by a controller which produces a current as a function of x of the form x i ( x) I 0 A d 3-18 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Example 3.6: The magnetic circuit in the figure is made of high-permeability electrical steel. The rotor is free to turn about a vertical axis. The dimensions are shown in the figure. a) Derive an expression for the torque acting on the rotor in terms of the dimensions and the magnetic field in the two air gaps. Assume the reluctance of the steel to be negligible and neglect the effects of fringing. b) The maximum flux density in the overlapping portions of the air gaps is to be limited to approximately 1.65 T to avoid excessive saturation of the steel. Compute the maximum torque for r1=2.5 cm, h=1.8 cm, and g=3 mm. 3-19 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3.6 MULTIPLY-EXCITED MAGNETIC FIELD SYSTEMS Many electromechanical devices have multiple electrical terminals. 3-20 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. USING ENERGY FUNCTON: d W fld (1, 2 , ) i1 d1 i2 d2 T fld d W fld (10 , 20 , 0 ) 20 10 0 0 i2 (1 0, 2 , 0 ) d2 i1 (1, 2 20 , 0 ) d1 For magnetically linear systems 1 L11 i1 L12 i2 2 L21 i1 L22 i2 L22 1 L12 2 i1 D L21 1 L11 2 i2 D D L11L22 L12L21 3-21 Integration path to obtain Wfld(10, 20, 0). Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. W fld (10 , 20 , 0 ) 20 0 W fld (10 , 20 ,0 ) L11( 0 ) 2 d2 D( 0 ) L22 ( 0 ) 1 L12 ( 0 ) 20 D( 0 ) 0 d1 L11(0 ) 2 L22 (0 ) 2 L12 (0 ) 20 10 10 20 2 D(0 ) 2 D(0 ) D(0 ) T fld 3-22 10 W fld 1 , 2 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. USING COENERGY FUNCTON: (i1, i2 , ) 1 di1 2 di2 T fld d d Wfld i20 i10 0 0 (i10 , i20 , 0 ) 2 (i1 0, i2 , 0 ) di2 1 (i1 , i2 i20 , 0 ) di1 W fld 1 L11 i1 L12 i2 2 L21 i1 L22 i2 (i10 , i20 ,0 ) W fld L11(0 ) 2 L22 (0 ) 2 i10 i20 L12 (0 ) i10 i20 2 2 T fld 3-23 W fld i1 ,i2 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. T fld i12 d L11( ) i22 d L22 ( ) d L12 ( ) i1 i2 2 d 2 d d For a general n electrical terminal 1 L11 L12 L 2 21 L22 n Ln1 Ln 2 L1n i1 L2 n i2 Lnn in 1 T I L( ) I W fld 2 3-24 λ L( ) I T fld 1 T d L( ) I I 2 dθ Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Example 3.7: In the figure, the inductances in henrys are given as L11=(3+cos 2θ)x10-3; L12=0.3 cos θ; L22=30+10 cos 2θ. Find and plot the torque Tfld(θ) for current i1=0.8 A and i2=0.01 A. 3-25 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. T fld (1.64sin 2 2.4 sin ) 103 3-26 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3.7 FORCES AND TORQUES IN SYSTEMS WITH PERMANENT MAGNETS Special case must be taken when dealing with hard magnetic material because magnetic flux density is zero when H=Hc not when H=0. •Consider fictitious winding •In normal operation, the fictitious winding carries NO current •Current in the winding can be adjusted to zero out the field produced by permanent magnet in order to achieve the “zero force” starting point. 3-27 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. (i f , x) f di f f fld dx dW fld (i f 0, x) W fld dW fld dW fld path1a x path1b 0 0 I f0 (i f 0, x) f fld (i f I f 0 , x) dx f (i f , x) di f W fld If0 is the current to zero-out the field. (i f 0, x) W fld 0 f (i f , x) di f I f0 3-28 Integration path for calculating Wfld (if = 0, x ) in the permanent magnet system of Fig. 3.17. Figure 3.18 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Example 3.8: The magnetic circuit is excited by a samariumcobalt permanent magnet and includes a movable plunger. Also shown is the fictitous winding of Nf turns carrying a current if which is included here for the sake of the analysis. The dimensions are: Wm=2 cm, Wg=3 cm, W0=2 cm, d=2 cm, g0=0.2 cm, and D=3 cm. a) Find an expression for the coenergy of the system as a function of plunger position x, b) Find an expression for the force on the plunger as a function of x, c) Calculate the force at x=0 and x=0.5 cm. 3-29 Figure 3.19 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A different solution for permanent magnet circuits: e R A H c d ( Ni )eq e R A d d ( Ni)eq Hc d 3-30 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Example 3.9: Figure shows an actuator consisting of an infinitely-permeable yoke and plunger, excited by a section of NdFeB magnet and an excitation winding of N1=1500 turns. The dimensions are: W=4 cm, W1=4.5 cm, D=3.5 cm, d=8 mm, and g0=1 mm. a) Find x-directed force on the plunger when the current in the excitation winding is zero and x=3 mm. b) Calculate the current in the excitation winding required to reduce the plunger force to zero. 3-31 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3.8 DYNAMIC EQUATIONS We are interested in the operation of complete electromechanical system and not just of the electromechanical energy conversion system around which it is built. For Electrical Terminal: di d L( x ) d x v0 R i L( x) i dt d x dt For multiple-excited system, we will have similar equation for each terminal 3-32 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. For Mechanical Terminal: Spring: f K K ( x x0 ) K : Spring constant (N/m) x K Damper: dx f D B dt B : Damping constant (N.s/m) B Mass: f M M f fld 3-33 f fld 2 d x d t2 M : Mass (kg) d 2x dx M 2 B K ( x x0 ) f 0 dt dt M f0 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Dynamic Equations (Electrical and Mechanical Equations Together): d i(t ) d L( x) d x(t ) v0 (t ) R i(t ) L( x) i dt dx dt d 2 x(t ) d x(t ) f 0 (t ) M B K ( x(t ) x0 ) f fld (i(t ), x(t )) 2 dt dt f fld i 2 d L( x) 2 dx These equations completely specify the behavior of electromechanical device. Solution of these equations will describe the position x and the current i at any time t in the system. 3-34 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. For Rotational Mechanical Terminal: Torsional Spring: TK K ( 0 ) K : Torsional Spring constant (N.m/rad) Friction: d TF B dt B : Friction constant (N.m.s/rad) Inertia: d 2 TJ J 2 dt T fld 3-35 J : Inertia constant (kg.m2/rad) d 2 d J 2 B K ( 0 ) T0 dt dt Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Example 3.10: Figure shows in cross section a cylindrical solenoid magnet in which the cylindrical plunger of mass M moves vertically in brass quide rings of thickness g and mean diameter d. The permeability of brass is µ0. The plunger is supported by a spring with K constant. Its unstretched length is l0. A mechanical load force ft is applied to the plunger from the mechanical system connected to it. Assume that frictional force is linearly proportional to the velocity with coefficient B. The coil has N turns and resistance R. Its terminal voltage is vt and its current i. Derive the dynamic equations of motion of the electromechanical system. 3-36 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. EXTRA Example: A two poles VR machine is shown in figure. Stator and rotor has infinite permeability. a) Find gap cross-sectional area as a function of θ. b) Find the inductance for the machine. c) Write down the dynamic equations. d) Solve the dynamic equations to find the position of rotor as a function of time initially starting from θ0=25 degrees. Numerical Values: N=100 turns, g=0.0005 m, d=0.1 m, r=0.04 m, J=0.05, B=0.02, θ0=30, R=0.5 ohm, E=10 Volt. r 0 3-37 Roto es r Ax Stator Axes