Electromagnetic and Electromechanical Engineering Principles Notes 03 Basic Machines

Electromagnetic and Electromechanical Engineering Principles Notes 03 Basic Machines

Electromagnetic and Electromechanical Engineering Principles Notes 03 Basic Machines Marc T. Thompson, Ph.D. Thompson Consulting, Inc. 9 Jacob Gates Road Harvard, MA 01451 Phone: (978) 456-7722 Email: marctt@thompsonrd.com Website: http://www.thompsonrd.com © Marc Thompson, 2006-2008 Portions of these notes reprinted from Fitzgerald, Electric Motors, 6th edition Electromechanics Basic Machines 3-1 Course Overview --- Day 3 Electromechanics Basic Machines 3-2 Electrodynamic Levitation Demo • AC levitation with 60 Hz excitation from autotransformer Reference: Marc Thompson, “Electrodynamic Magnetic Suspension --- Models, Scaling Laws and Experimental Results,” IEEE Transactions on Education, vol. 43, no. 3, August 2000, pp. 336-342 Electromechanics Basic Machines 3-3 One Method to Calculate Lift Force • We can deduce force acting on the coil by making measurement of inductance of coil at different heights Electromechanics Basic Machines 3-4 Levitation Example Reference: Marc Thompson, “Electrodynamic Magnetic Suspension --- Models, Scaling Laws and Experimental Results,” IEEE Transactions on Education, vol. 43, no. 3, August 2000, pp. 336-342 Electromechanics Basic Machines 3-5 Circular Coil Above Conducting Aluminum Plate • Flux density plots at DC and 60 Hz via FEA • At 60 Hz, currents induced in plate via magnetic induction create lift force DC 60 Hz Electromechanics Basic Machines 3-6 Levitation Example • Coil inductance vs. coil height above aluminum plate at 60 Hz Reference: Marc Thompson, “Electrodynamic Magnetic Suspension --- Models, Scaling Laws and Experimental Results,” IEEE Transactions on Education, vol. 43, no. 3, August 2000, pp. 336-342 Electromechanics Basic Machines 3-7 Levitation Example • Curve fit for inductance: • Magnetic stored energy: • Levitation force: Reference: Marc Thompson, “Electrodynamic Magnetic Suspension --- Models, Scaling Laws and Experimental Results,” IEEE Transactions on Education, vol. 43, no. 3, August 2000, pp. 336-342 Electromechanics Basic Machines 3-8 Levitation Example Reference: Marc Thompson, “Electrodynamic Magnetic Suspension --- Models, Scaling Laws and Experimental Results,” IEEE Transactions on Education, vol. 43, no. 3, August 2000, pp. 336-342 Electromechanics Basic Machines 3-9 AC Magnetic Suspension Scaling Laws • Scaling law for zero speed EDS 2 P ∝ M 3 • Coil mass M = 0.35 kg; P = 257 W at h = 10 mm • Full scale suspension: For M = 1000 kg, P = 51 kW • Other tradeoffs: – Excitation frequency vs. guideway thickness Electromechanics Basic Machines 3-10 Example and Case Study: Solenoid Design • Follows Example 3.10 in Fitzgerald (pp. 153-155) • Note that dimension x is the distance from the top of the plunger to the top of the coil • As this dimension x varies, axial force on the plunger varies as well Electromechanics Basic Machines 3-11 Strategy • Use reluctance model • Find flux Φ • From flux, find flux linkage λ • From flux linkage, find inductance as a function of plunger position x • From L(x), find force Electromechanics Basic Machines 3-12 Reluctance Model • Reluctance of steel is very small if the steel doesn’t saturate Path length in direction of field ℜ = Areaμ() of fluxperpendicular path to field Electromechanics Basic Machines 3-13 Reluctance Model g ap: gUpper ℜ upper = μo()d π x g ap: gerLowℜlower = μo()d π a Electromechanics Basic Machines 3-14 Flux : Flux NI Φ = ℜupper +lower ℜ Electromechanics Basic Machines 3-15 Flux Linkage linFlukaxge: NI2 =λ N Φ = ℜupper +lower ℜ Electromechanics Basic Machines 3-16 as a Function of Plunger PositionInductance e: anctInduc λ N 2 L= = Iℜupper +lower ℜ g g ⎛ g ⎞ a+ x ⎜ ⎟⎛ ⎞ ℜupper + ℜlower = +⎜ = ⎟⎜ ⎟ μ πo dx μo da π⎝ o da μ⎠⎝ πx ⎠ ⎛μdaN π 2 ⎞⎛ x ⎞ ⎛ x ⎞ L() x ⎜ o ⎟ L = ⎜ ⎟⎜ ⎟ = o ⎜ ⎟ ⎝ g ⎠⎝a+ x⎠ ⎝a+ x⎠ Electromechanics Basic Machines 3-17 Find Force se icfore thngndifie, ctancudn the inowu koOnce y :easy I2 dL() x f = x 2 dx I 2 d ⎛ x ⎞ = Lo ⎜ ⎟ 2 dx ⎝a+ x⎠ I 2 a = L 2 o()a+ x2 Electromechanics Basic Machines 3-18 Solenoid Design Example Dimensions of solenoid design example Item Description Dimension a Height of solenoid backiron arm 12.5 mm h Height of solenoid center backiron arm 10 mm d Radius to backiron 10 mm g Airgap 0.5 mm rcoil Coil mean radius 7.75 mm x Variable offset of plunger top from top of 5 mm (variable) coil rplunger Plunger radius 9.5 mm Electromechanics Basic Machines 3-19 Design Example --- Analytic Result • For NI = 354 A-turns (DC) • Coil power dissipation ∼ 3.5 Watts Example solenoid design 4.5 4 3.5 3 2.5 2 Force, [N] Force, 1.5 1 0.5 0 024681012 x, [mm] Electromechanics Basic Machines 3-20 Simulation • Trust, but verify • 2D axisymmetric finite-element analysis program used • Finite Element Method Magnetics (FEMM) by Dave Meeker at Foster-Miller Electromechanics Basic Machines 3-21 Simulation --- 2D Axisymmetric FEA Electromechanics Basic Machines 3-22 Simulation --- 2D Axisymmetric FEA • Shown for x = 5 mm • FEA shows fx = 1.6 Newtons, Pdiss = 3.5 Watts Electromechanics Basic Machines 3-23 Comparison of Analytic vs. FEA • Note that FEA shows lower force than that predicted analytically • Probably due to saturating iron reducing B and hence force Example solenoid design 4.5 4 3.5 3 2.5 2 Force, [N] Force, 1.5 1 0.5 0 024681012 x, [mm] Electromechanics Basic Machines 3-24 Introduction to Rotating Machines • Elementary concepts and terminology • AC machines overview • DC machines overview • MMF of distributed windings • Magnetic fields in rotating machines • Generated voltage • Torque • Linear machines Electromechanics Basic Machines 3-25 Motor Terminology • Armature winding --- windings that carry AC currents • Stator --- stationary part of the motor • Rotor --- rotating part of the motor • DC machine --- armature winding found on the rotor • AC machine --- armature winding typically on the stator • Field winding --- windings that carry DC current and produce main operating flux Electromechanics Basic Machines 3-26 Types of Machines • AC machines • Synchronous motor • Induction motor • DC machines Electromechanics Basic Machines 3-27 Simple 2-Pole Synchronous Generator • This is a “salient pole” machine. • Field winding excited by brushes contacting slip rings • It advantageous to have the high power armature winding mounted to the stator for cooling purposes Electromechanics Basic Machines 3-28 Flux Density and Generated Voltage • Idealized analysis assuming sinusoidal magnetic flux density in airgap varying with angle θa and flux produced by field winding only • As the rotor rotates, there is a dλ/dt which results in a sinusoidal armature voltage Electromechanics Basic Machines 3-29 Flux Density and Generated Voltage • What is rotation speed needed to supply 60 Hz in a 2-pole machine? • Frequency in Hz = rotation speed … hence “synchronous” generation occurs. 2-pole synchronous machine rotates at 3600 RPM to produce 60 Hz f606060⋅ ⋅ ω == =3600 RPM s p/2 2/2 Electromechanics Basic Machines 3-30 Four-Pole Single Phase Synchronous Generator • This is also a “salient pole” machine • Generated frequency is twice that of the two-pole machine • 4-pole synchronous machine rotates at 1800 RPM to produce 60 Hz ⎛ poles ⎞ θ ae = ⎜ ⎟θ a ⎝ 2 ⎠ θae = electrical angle θa = mechanical angle Electromechanics Basic Machines 3-31 Distribution of Airgap Flux in 4-Pole Gen. • Note that electrical angle is twice the mechanical angle, since there are 2 N-S electrical pole pairs per rotor rotation Electromechanics Basic Machines 3-32 PM Machine • 4 pole machine; field provided by NdFeB magnets Electromechanics Basic Machines 3-33 PM Machine • 4 pole machine; field provided by NdFeB magnets Electromechanics Basic Machines 3-34 Nonsalient Pole Machine • 2-poles; windings are distributed in slots in the rotor • Suitable for high-speed generator Electromechanics Basic Machines 3-35 Three-Phase Generators 2-pole 4-pole Y-connection of windings Electromechanics Basic Machines 3-36 Induction Machines • Stator winding is excited with AC currents, as in a synchronous machine • AC flows in the rotor windings; rotor currents are produced by magnetic induction (i.e. transformer action) into short- circuited rotor windings • We can think of an induction machine as a type of transformer where power is transformed between rotor and stator, with a change of frequency and flow of mechanical power • Induction motor is the most common motor Electromechanics Basic Machines 3-37 Typical Induction Motor Torque-Speed Curve • The rotor of an induction machine does not rotate synchronously with the stator field; it “slips” ω −ω s = s r ω = synchronous speed ω s s ωr = rotor speed • Note that at full “synchronous speed” the motor generates zero torque Electromechanics Basic Machines 3-38 DC Machines • Armature winding of a DC generator is on the rotor with current conducted from it by brushes • Field winding is on the stator and is excited by DC current Electromechanics Basic Machines 3-39 Elementary DC Machine with Commutator Electromechanics Basic Machines 3-40 DC Machine Characteristics • Brushes cause a rectification Electromechanics Basic Machines 3-41 MMF of Concentrated Full-Pitch Winding • This is an extreme example • MMF is “magnetomotive force” • Remember, by Ampere’s law a current forces H to flow • A current in a wire causes a change in the MMF • Note first harmonic of airgap MMF (Fag1) Electromechanics Basic Machines 3-42 MMF of Concentrated Full-Pitch Winding Fundamental component of airgap MMF ⎛ 4 ⎞⎛ NI ⎞ Fag1 = ⎜ ⎟⎜ ⎟cosθ a ⎝ π ⎠⎝ 2 ⎠ Peak value of airgap MMF ⎛ 4 ⎞⎛ NI ⎞ ()Fag1 = ⎜ ⎟⎜ ⎟ peak ⎝ π ⎠⎝ 2 ⎠ Electromechanics

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