Handbook of Small Electric Motors MAZ Part 6 doc

CHAPTER 1 BASIC MAGNETICS Chapter Contributors Mark A. Juds Earl F. Richards William H. Yeadon 1.1 Electric motors convert electrical energy into mechanical energy by utilizing the properties of electromagnetic energy conversion. The different types of motors operate in different ways and have different methods of calculating the perfor- mance, but all utilize some arrangement of magnetic fields. Understanding the con- cepts of electromagnetics and the systems of units that are employed is essential to understanding electric motor operation.The first part of this chapter covers the con- cepts and units and shows how forces are developed. Nonlinearity of magnetic mate- rials and uses of magnetic materials are explained. Energy and coenergy concepts are used to explain forces, motion, and activation. Finally, this chapter explains how motor torque is developed using these concepts. 1.1 UNITS* Although the rationalized mks system of units [Système International (SI) units] is used in this discussion, there are also at least four other systems of units—cgs, esu, emu, and gaussian] that are used when describing electromagnetic phenomena. These systems are briefly described as follows. MKS. Meter, kilogram, second. CGS. Centimeter, gram, second. * Sections 1.1 to 1.12 contributed by Mark A. Juds, Eaton Corporation. ESU. CGS with e 0 = 1, based on Coulomb’s law for electric poles. EMU. CGS with µ 0 = 1, based on Coulomb’s law for magnetic poles. Gaussian. CGS with electric quantities in esu and magnetic quantities in emu. The factor 4π appears in Maxwell’s equation. Rationalized mks. µ 0 = 4π×10 −7 H/m, based on the force between two wires. Rationalized cgs. µ 0 = 1, based on Coulomb’s law for magnetic poles.The factor 4π appears in Coulomb’s law. The rationalized mks and the rationalized cgs systems of units are the most widely used.These systems are defined in more detail in the following subsections. 1.1.1 The MKS System of Units The rationalized mks system of units (also called SI units) uses the magnetic units tesla (T) and amps per meter (A/m), for flux density B and magnetizing force H, respectively. In this system, the flux density B is defined first (before H is defined), and is based on the force between two current-carrying wires.A distinction is made between B and H in empty (free) space, and the treatment of magnetization is based on amperian currents (equivalent surface currents). Total or normal flux density B, T B =µ 0 (H + M) (1.1) Intrinsic flux density B i ,T B i =µ 0 M (1.2) Permeability of free space µ 0 ,(T⋅m)/A µ 0 = 4π×10 −7 (1.3) Magnetization M, A/m M == (1.4) Magnetic moment M, J/T m = MV = pl (1.5) Magnetic pole strength p, A⋅m p = (1.6) 1.1.2 The CGS System of Units The rationalized cgs system of units uses the magnetic units of gauss (G) and oersted (Oe) for flux density B and magnetizing force H, respectively. In this system, the magnetizing force, or field intensity, H is defined first (before B is defined), and is m ᎏ l m ᎏ V B l ᎏ µ 0 1.2 CHAPTER ONE based on the force between two magnetic poles. No distinction is made between B and H in empty (free) space, and the treatment of magnetization is based on mag- netic poles. The unit emu is equivalent to an erg per oersted and is understood to mean the electromagnetic unit of magnetic moment. Total or normal flux density B, G B = H + 4π M (1.7) Intrinsic flux density B i ,G B i = 4πM (1.8) Permeability of free space µ 0 , G/Oe µ 0 = 1 (1.9) Magnetization M, emu/cm 3 M == (1.10) Magnetic moment m, emu m = MV = pl (1.11) Magnetic pole strength p, emu/cm p = (1.12) The magnetization or magnetic polarization M is sometimes represented by the sym- bols I or J, and the intrinsic flux density B i is then represented as 4πM, 4πI, or 4πJ. 1.1.3 Unit Conversions Magnetic Flux φ 1.0 Wb = 10 8 line = 10 8 maxwell = 1.0 V⋅s = 1.0 H⋅A = 1.0 T⋅m 2 Magnetic Flux Density B 1.0 T = 1.0 Wb/m 2 = 10 8 line/m 2 = 10 8 maxwell/m 2 = 10 4 G = 10 9 γ m ᎏ l m ᎏ v B i ᎏ 4π BASIC MAGNETICS 1.3 1.0 G = 1.0 line/cm 2 = 10 −4 T = 10 5 γ = 6.4516 line/in 2 Magnetomotive or Magnetizing Force NI 1.0 A⋅turn = 0.4 π Gb = 0.4 π Oe⋅cm Magnetic Field Intensity H 1.0 (A⋅turn)/m = 4π×10 −3 Oe = 4π×10 −3 Gb/cm = 0.0254 (A⋅turn)/in 1.0 Oe = 79.5775 (A⋅turn)/m = 1.0 Gb/cm = 2.02127 (A⋅turn)/in Permeability µ 1.0 (T⋅m)/(A⋅turn) = 10 7 /4π G/Oe = 1.0 Wb/(A⋅turn⋅m) = 1.0 H/m = 1.0 N/(amp⋅turn) 2 1.0 G/Oe = 4π×10 −7 H/m Inductance L 1.0 H = 1.0 (V⋅s⋅turn)/A = 1.0 (Wb⋅turn)/A = 10 8 (line⋅turn)/A Energy W 1.0 J = 1.0 W⋅s = 1.0 V⋅A⋅s = 1.0 Wb⋅A⋅turn = 1.0 N⋅m = 10 8 line⋅A⋅turn = 10 7 erg Energy Density w 1.0 MG⋅Oe = 7.958 kJ/m 3 = 7958 (T⋅A⋅turn)/m 1.0 (T⋅A⋅turn)/m 3 = 1.0 J/m 3 1.4 CHAPTER ONE Force F 1.0 N = 1.0 J/m = 0.2248 lb 1.0 lb = 4.448 N Magnetic Moment m 1.0 emu = 1.0 erg/Oe = 1.0 erg/G = 10.0 A⋅cm 2 = 10 −3 A⋅m 2 = 10 −3 J/T = 4π G⋅cm 3 = 4π×10 −10 Wb⋅m = 10 −7 (N⋅m)/Oe Magnetic Moment of the Electron Spin β=eh/4πm e 1.0 Bohr magneton =β=9.274 × 10 −24 J/T =β=9.274 × 10 −24 A⋅m 2 =β=9.274 × 10 −21 erg/G Constants Permeability of free space µ 0 = 1.0 G/Oe µ 0 = 4π×10 −7 (T⋅m)/(A⋅turn) µ 0 = 4π×10 −7 H/m Electron charge e = 1.602177 × 10 −19 C Electron mass m e = 9.109390 × 10 −31 kg Plank’s constant h = 6.6262 × 10 −34 J⋅s Velocity of light c = 2.997925 × 10 8 m/s Pi π=3.1415926536 Acceleration of gravity g = 9.807 m/s 2 = 32.174 ft/s 2 = 386.1 in/s 2 Miscellaneous Length l 1.0 m = 39.37 in = 1.094 yd 1.0 in = 25.4 mm Time t 1.0 day = 24 h 1.0 min = 60 s Velocity v 1.0 m/s = 3.6 km/h BASIC MAGNETICS 1.5 = 3.281 ft/s Acceleration a 1.0 m/s 2 = 3.281 ft/s 2 = 39.37 in/s 2 1.2 DEFINITION OF TERMS The Greeks discovered in 600 B.C . that certain metallic rocks found in the district of Magnesia in Thessaly would attract or repel similar rocks and would also attract iron.This material was called Magnes for the district of Magnesia, and is a naturally magnetic form of magnetite (Fe 3 O 4 ), more commonly known as lodestone. Lode- stone means “way stone,” in reference to its use in compasses for guiding sailors on their way. A bar-shaped permanent magnet suspended on a frictionless pivot (like a com- pass needle) will align with the earth’s magnetic field.The end of the bar magnet that points to the earth’s geographic north is designated as the north magnetic pole and the opposite end is designated as the south magnetic pole. If any tiny compass nee- dles are placed around the bar magnet, they will line up to reveal the magnetic field shape of the bar magnet. Connecting lines along the direction of the compass nee- dles show that the magnetic field lines emerge from one pole of the bar magnet and enter the opposite pole of the bar magnet.These magnetic field lines do not stop or end, but pass through the magnet to form closed curves or loops. By convention, the magnetic field lines emerge from the north magnetic pole and enter through the south magnetic pole.Two permanent magnets will attract or repel each other in an effort to minimize the length of the magnetic field lines, which is why like poles repel and opposite poles attract. Therefore, since the north magnetic pole of a bar magnet points to the earth’s geographic north, the earth’s geographic north pole has a south magnetic polarity. Hans Oersted discovered in 1820 that a compass needle is deflected by an elec- tric current, and for the first time showed that electricity and magnetism are related. The magnetic field around a current-carrying wire can be examined by placing many tiny compass needles on a plane perpendicular to the axis of the wire. This shows that the magnetic field lines around a wire can be envisioned as circles centered on the wire and lying in planes perpendicular to the wire. The direction of the magnetic field around a wire can be determined by using the right-hand rule, as follows (see Fig. 1.1). The thumb of your right hand is pointed in the direction of the current, where current is defined as the flow of positive charge from + to −.The fingers of your right hand curl around the wire to point in the direc- tion of the magnetic field. If the current is defined as the flow of negative charge from − to +, then the left-hand rule must be used. To summarize: 1. The north magnetic pole of a bar magnet will point to the earth’s geographic north. 2. Magnetic field lines emerge from the north magnetic pole of a permanent magnet. 3. Magnetic field lines encircle a current-carrying wire. 4. Magnetic field lines do not stop or end, but form closed curves or loops that always encircle a current-carrying wire and/or pass through a permanent magnet. 5. The right-hand rule is used with current flowing from positive to negative and with the magnetic field lines emerging from the north magnetic pole. 1.6 CHAPTER ONE 1.2.1 System Performance φ=magnetic flux NI = magnetomotive or magnetizing force ᏾ = reluctance ᏼ = permeance λ=flux linkage Figure 1.2 shows a magnetic circuit based on an electrical analogy. In general, a coil with N turns of wire and I amperes (amps) provides the magnetomotive force NI that pushes the magnetic flux φ through a region (or a material) with a cross sec- tional area a and a magnetic flux path length l. In the electrical analogy, a voltage V provides the electromotive force that pushes an electrical current I through a region. The amount of magnetomotive force required per unit of magnetic flux is called reluctance ᏾.The amount of voltage required per amp is called resistance R. ᏾ = (1.13) NI =φ᏾ (1.14) NI ᎏ φ BASIC MAGNETICS 1.7 FIGURE 1.1 Direction of flux. (Courtesy of Eaton Corporation.) FIGURE 1.2 Magnetic circuit with electrical analogy. (Courtesy of Eaton Corporation.) Electrical Analogy R = (1.15) V = IR (Ohm’s law) (1.16) ᏼ == (1.17) φ=NI ᏼ (1.18) λ=Nφ (1.19) V == (Faraday’s law) (1.20) Visualization of Flux Linkage l. Figure 1.3 shows 10 magnetic flux lines passing through 4 turns of wire in a coil.Each turn of the coil is linked to the 10 magnetic flux lines, like links in a chain. Therefore, the total flux linkage λ is obtained by multiply- ing the turns N by the magnetic flux φ, as defined in Eq. (1.19). In this case, the magnetic flux linkage λ=40 line turns, where the units of N are turns and the units of magnetic flux φ are lines. 1.2.2 Material Properties B = magnetic flux density H = magnetic field intensity µ=magnetic permeability The magnetic flux density B is defined as the magnetic flux per unit of cross-sectional area a. The magnetic field intensity H is defined as the mag- netomotive force per unit of magnetic flux path l. The magnetic permeability µ of the material is defined as the ratio between the magnetic flux density B and the magnetic field intensity. The permeabil- ity can also be obtained graphically from the magnetization curve shown in Fig. 1.4. In the electrical analogy, the current density J, the electric field intensity E, and the elec- trical conductivity σ are defined using ratios of similar physical parameters. B = (1.21) H = (1.22) µ= (1.23) B ᎏ H NI ᎏ l φ ᎏ a d(Nφ) ᎏ dt dλ ᎏ dt 1 ᎏ ᏾ φ ᎏ NI V ᎏ I 1.8 CHAPTER ONE FIGURE 1.3 Flux linkage visualization. (Cour- tesy of Eaton Corporation.) Electrical Analogy J = (1.24) E = (1.25) σ= (1.26) Permeability of free space, H/m or (T⋅m)/A: µ 0 = 4π×10 −7 (1.27) Relative permeability: µ r = (1.28) where µ is the permeability of a material at any given point. 1.2.3 System Properties ᏾ = reluctance ᏼ = permeance L = inductance The reluctance ᏾, as defined in equation (1.17), can be written in a form based on the material properties and the geometry (µ, a, and l), as shown in Eq. (1.29). µ ᎏ µ 0 J ᎏ E V ᎏ l I ᎏ a BASIC MAGNETICS 1.9 FIGURE 1.4 Magnetization B-H curve showing permeability µ. (Courtesy of Eaton Corporation.) ᏾ === (1.29) The same can be done for the permeance ᏼ, as shown in equation [1-30]. ᏼ == = (1.30) The inductance L is defined as the magnetic flux linkage λ per amp I, as shown in Eq. (1.31).The inductance can be written in a form based on the material properties and the geometry (µ, a, and L), also shown in Eq. (1.31). The B–H curve can be eas- ily changed into a λ–I curve, as shown in Fig. 1.5, and the inductance can then be obtained graphically. L == =N 2 = N 2 ᏼ (1.31) 1.2.4 Energy W e = input electric energy W f = stored magnetic field energy W m = output mechanical energy W co = magnetic field coenergy Energy Balance. All systems obey the first law of thermodynamics, which states that energy is conserved. This means that energy is neither created nor destroyed. Therefore, an energy balance can be written for a general system stating that the change in energy input to the system ∆W e equals the change in energy stored in the system ∆W f plus the change in energy output from the system ∆W m .This energy bal- ance is illustrated in the following equation. ∆W e =∆W f +∆W m (1.32) φ ᎏ NI Nφ ᎏ I λ ᎏ I µa ᎏ l φ ᎏ NI 1 ᎏ ᏾ l ᎏ µa Hl ᎏ Ba NI ᎏ φ 1.10 CHAPTER ONE FIGURE 1.5 Magnetization λ-I curve showing inductance L. (Courtesy of Eaton Corporation.) [...]... 0 A comparison of the results from Eqs (1.59) and (1 .63 ) shows that the change in mechanical energy is equal to the change in magnetic coenergy ∆Wm = Wco (1 .65 ) The mechanical force can be calculated as follows, by substituting Eq (1.38) into Eq (1 .65 ) F ∆X = ∆Wco ∆Wco F= ᎏ ∆X dWco F= ᎏ dX (1 .66 ) (1 .67 ) in the limit as ∆X→0 (1 .68 ) The mechanical torque can be calculated using Eq (1 .66 ) and the radius... dr11 11 dᏼ11 = µ0 ᎏ = µ0 ᎏ 2 l11 π2r2 11 (1. 161 ) 1 ᏼ11 = µ0 ᎏ π ͵ g/2 + h g/2 dr11 ᏼ11 = 0.32µ0h (1. 162 ) (1. 163 ) Total Permeance The total permeance of the air gap is equal to the sum of the individual parallel flux paths, as follows: ᏼtotal = ᏼ1 + ᏼ2 + ᏼ3 + ᏼ4 + ᏼ5 + 2( 6 + ᏼ7) + 4(ᏼ8 + ᏼ9 + ᏼ10 + ᏼ11) 1.3.1 (1. 164 ) Summary of Flux Path Permeance Equations All of the flux path permeances are based on... length Ferrite and rare-earth permanent magnets have effective poles at 95 percent of the magnet length; therefore, no magnetic flux is generated in this path and the permeance is zero 1 d1 = ᎏ 3 (1. 166 ) 1 r1 = ᎏ 2 (1. 167 ) 1.31 BASIC MAGNETICS 1 1 ᎏ . charge e = 1 .60 2177 × 10 −19 C Electron mass m e = 9.109390 × 10 −31 kg Plank’s constant h = 6. 6 262 × 10 −34 J⋅s Velocity of light c = 2.997925 × 10 8 m/s Pi π=3.14159 265 36 Acceleration of gravity. + F (1 .60 ) W co2 = ͵ I 2 0 λ dI = D + C + G (1 .61 ) W co = (D + C − F) − (G + F) (1 .62 ) ∆W co = D + C − F (1 .63 ) ∆W m = W co (1 .64 ) A comparison of the results from Eqs. (1.59) and (1 .63 ) shows. Richards William H. Yeadon 1.1 Electric motors convert electrical energy into mechanical energy by utilizing the properties of electromagnetic energy conversion. The different types of motors operate in different