EEE336 Electromechanical Energy Conversion Dr. Bülent Dağ 1 Reference Books Textbook: S.J. Chapman, Electric Machinery Fundamentals, McGraw Hill Supplementary book: A.E. Fitzgerald, et.al., Electric Machinery, McGraw Hill, 6th ed. 2 Office Hours Tuesday: 10:30-12:30, Friday: 10:3012:30 Office #: 323 E-mail: bulentdag@gazi.edu.tr 3 Assesment Criteria Exam Number Weight Midterm 1 % 40 Final 1 % 60 4 Content of the Lecture • Electromagnetic circuits and energy conversion principles • Transformers • DC Machines – DC machine fundamentals – DC motors – DC generators • AC Machines – AC machine fundamentals – Three-phase Induction motors – Synchronous generators 5 Lecture 1 Introduction To Machinery Principles and Magnetic Circuits 6 Energy Conversion with Electrical Machines Motors Mechanical Energy Electrical Energy Generators Types of electrical machines: • DC Machines • AC Machines Induction machines Synchronous machines • Special machines (step motor, reluctance machine, BDCM) 7 Electrical Machines Definitions: • Shaft: The axis of rotation in the electric machines is called shaft. Direction of rotation can be either clock wise (CW) or counter clock wise (CCW). CCW will be assumed positive in this course. • Angular position (θ): The angle of the rotating object with respect to an arbitrary reference point. The unit is radians or degrees. Analogues to distance (x) in linear motion. • Angular velocity (ω): Angular velocity (speed) is the time rate at which the angular position (θ) changes. It is analogous to linear velocity (v). • Angular acceleration (α): Rate of change of angular velocity with respect to time. 8 Rotational Motion • m , Angular Velocity, radians/second (rad/sec) • fm, Angular Velocity, revolution/second (rev/sec) • nm, Angular Velocity, revolution/minute (rpm) • , Angular Acceleration, radians/second2 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 9 • Torque (T): Analogous to force (F) in linear motion 10 Newton’s Law of Motion Linear motion Rotational motion J: Moment of inertia of the rotating object, kg-m2 Work: Linear motion Rotational motion If F is constant, W = Fx, (Joules) Power: Linear motion If T is constant, W = Tθ, (Joules) (Joules/second = Watts) Rotational motion 11 Maxwell’s Equations Gauss’s Law Faraday’s Law Ampere’s Law 12 Electromagnetic Systems λ Faraday’s law Ampere’s law Lorentz’s laws 13 The Magnetic Field • Magnetic field intensity H (A-turns/m) is governed by Gauss and Ampere’s Law (Gauss Law), • (Ampere Law) Magnetic Flux density, B (Tesla or T) depends on the medium property and related to H as follows, Where, µ is magnetic permeability of medium µo is permeability of the free space = µr is the relative permeability of the material = 14 • Magnetic Flux density B (Tesla or T) accounts for the magnetic prosperities of the medium. • B vs H relationship is frequently expressed by a non-linear curve called B-H curve, as shown in figures below. • In most ferromagnetic materials, the curve starts at a very high slope which tends to be constant. This is the linear portion of the B-H curve. • At higher values of H, the flux density levels off and the material is said to be in saturation region. 15 A simple magnetic core 16 Magnetic Circuit (Weber, Wb) 17 Magnetic Circuits Flux equation can be rewritten as follows, Define magnetomotive force, F and reluctance, R (A-turns, A-t), (A-t/Wb), Then basic magnetic circuit equation in correlation with electric circuits, 18 Magnetic Circuits The reciprocal of the reluctance is called permeance, Λ For series connected parts; Polarity of mmf, F For parallel connected parts; Assumptions made in magnetic circuit analysis; 1. Mean core length and uniform flux density 2. No leakage flux 3. Constant permeability, µ 4. No flux fringing 19 Fringing and leakage flux leakage fringing Fringing: When there exists an air gap, flux may not follow the shortest path while jumping from one part to the other part, especially at the edges. As a result, effective cross sectional area of the air gap increases. Leakage: Some of the flux produced complete its turn around the winding through the air. 20 Example (1.1-Chapman) A ferromagnetic core is shown in the Figure. Three sides of this core are of uniform width while the fourth side is somewhat thinner. The depth of the core (into the page) is 10 cm. and the other dimensions are shown in the figure. There is a 200 turn coil wrapped around the left side of the core. Assumming relative permeability µr of 2500, how much flux will be produced by a 1 A input current? 21 Solution (1.1-Chapman) 22 Example (1.2-Chapman) A magnetic core with a relative permeability of 4000 is shown in the Figure. Assuming a fringing coefficient of 1.05 (or %5 increase in airgap cros sectional area) for the air gap, find a) total (equivalent) reluctance of the system b) the flux density in the air gap if i = 0.6 A. 23 24 The magnetic circuit is shown in figure (b). 0.0012 0.0012 0.0012 25 Example (1.3-Chapman) Figure (a) shows a simplified rotor and stator for a dc motor. The mean path length of the stator is 50 cm, and its crosssectional area is 12 cm2. The mean path length of the rotor is 5 cm, and its crosssectional area also may be assumed to be 12 cm2. Each air gap between the rotor and the stator is 0.05 cm wide, and the cross-sectional area of each air gap (including fringing) is 14 cm2. The iron of the core has a relative permeability of 2000, and there are 200 turns of wire on the core. If the current in the wire is adjusted to be 1 A, what will the resulting flux density in the air gaps be? 26 Solution (1.3-Chapman) 27 Magnetic Behavior of Ferromagnetic Materials • • • Constant permeability assumption is not always valid Ferromagnetic cores have non-linear B-H characteristics To obtain actual B-H plot, apply Vac at rated value and plot integral of Vac vs. i on an oscilloscope in x-y mode. This plot is called magnetization curve. Vac Remember that, and where, 28 Magnetic Behavior of Ferromagnetic Materials • Therefore, reluctance of the core is not constant, actually. • To find reluctance, we need permeability, • To find , first we need the operating point on B-H curve. • If permeability is not constant then there are two solution approaches according to given parameters; 1. If B is given and current is asked: From B-H curve determine operating H for the core. Either by using directly Ampere’s law or by determining permeability, µ at the operating point and using magnetic circuits, calculate the current. 2. If current is given and B is asked: The only solution way is to obtain a second relation between B and H of the core using Ampere’s law for the given magnetic system. Then cross-section of B-H curve of the core and B-H relation from the system is the operating point. 29 Example (1.4-Chapman) Find the relative permeability of the typical ferromagnetic material whose magnetization curve is shown in the figure below for, (a) H = 50. (b) H = 100. (c) H = 500. and (d) H = 1000 A-turns/m Solution: a) H = 50 A-t/m d) H = 1000 A-t/m B = 0.25 T B = 1.51 T 30 Example (1.5-Chapman) A square magnetic core has a mean path length of 55 cm and a cross-sectional area of 150 cm2. A 200 turn coil of wire is wrapped around one leg of the core. The core is made of a material having the same magnetization curve in previous example. (a) How much current is required to produce 0.012 Wb of flux in the core? (b) What is the core's relative permeability at that current level? (c) What is its reluctance? Solution: The required magnetic field density, B in the core is, , from magnetizing curve the required intensity, A-t/m a) b) c) 31 Example: T magnetic core has a B-H curve as shown below. What is the air gap flux density if the applied current is 0.86 A? Neglect fringing and leakage. 32 Solution: From Ampere’s Law, 33 Faraday’s Law: Induced voltage from a time varying magnetic field Faraday’s law is the basis of transformer operation. It states that if a flux passes through a turn of a coil of wire, a voltage will be induced in the turn of wire that is directly proportional to the rate of change in the flux with respect to time. In equation form, If a coil has N turns and if the same flux passes through all of them, then the voltage induced across the whole coil is given by, 34 Definition: Flux linkage, λ Definition: Inductance, L (Henry (H) = Wb-t/A) Example (Fitzgerald-1.3) The magnetic circuit of Fig (a) consists of an N-turn winding on a magnetic core of infinite permeability with two parallel air gaps of lengths g1 and g1 and areas A1 and A2, respectively. Find, (a) the inductance of the winding and (b) the flux density B1 in Gap 1 when the winding is carrying a current i. Neglect fringing effects at the air gap. 35 Solution (Fitzgerald-1.3) a) b) 36 Two winding construction Note that i1 and i2 are to produce flux in the same direction in this example. Flux linkage of coil 1 Flux linkage of coil 2 Self inductance Mutual inductance 37 Note that the resolution of the resultant flux linkages into the components produced by i1 and i2 is based on superposition of the individual effects and therefore implies a linear flux-mmf relationship (characteristic of materials of constant permeability). For a magnetic circuit with a single winding; a) Then, for static systems with constant air gap and permeability (L is constant); b) For electromechanical systems where exist moving parts (L is not constant: air gap changes. This, in turn changes reluctance); For systems with multiple windings; total flux linkage of each coil must be included. Then, for a system with two coils, for example; 38 Power at the terminals of a winding [W, J/s] Thus, the change in magnetic stored energy ΔW in the magnetic circuit in the time interval t1 to t1 is, [Joules] For a single-winding system of constant inductance, the change in magnetic stored energy as the flux level is changed from λ1 to λ1 can be written as; The total magnetic stored energy at any given value of λ can be found from setting λ1 equal to zero; 39 Example (Fitzgerald): The magnetic circuit shown in the figure has dimensions Ac = Ag = 9 cm2, g = 0.05 cm, lc = 30 cm, and N = 500 turns. Assume µr = 70000 for core material. Find (a) the inductance L, (b) the magnetic stored energy W for Be = 1 T, and (c) the induced voltage e for a 60-Hz time-varying core flux of the form Be = 1 sin ωt T where ω = 2π60 = 377. Solution; a) c) b) 40 Properties of magnetic materials randomly oriented dipoles • • • • • external field All materials are composed of magnetic dipoles. The flexibility of magnetic dipole to rotate around is a measure of permeability, µ In case of an AC excitation, the dipole orientations does not follow the same path. This is called hysteresis. When an external field is applied to the randomly oriented dipoles and removed, some flux remains in the material in the direction of applied field. This is called residual flux. The required opposite mmf to make the net flux zero again, is called magnetic coercivity. Permanent magnets (hard magnetic materials) have very high residual flux and coercivity compared to ferromagnetic materials (soft magnetic materials). 41 AC excitation The flux is created by a current, iφ, called magnetizing current. Since, ϕ – i curve can be obtained from B-H curve and iφ, can be obtained from ϕ – i curve as in Fig. (a) and (b) above. 42 Magnetic losses As can be seen from Fig. a, phase of the fundamental component of the magnetizing current is less than 90o . This means that the energy input associated with magnetizing current has reactive and real parts. Real part of this energy is dissipated as losses and results in heating of the core. Two loss mechanisms are associated with time-varying fluxes in magnetic materials; Eddy current and Hysteresis loss. The reactive power is associated with energy storage in the magnetic field. This reactive power is not dissipated in the core; it is cyclically supplied and absorbed by the excitation source. Eddy current losses: Time varying magnetic fields generate electrical fields and corresponding circulating currents inside the material, as well. These currents cause ohmic I2R losses. In order to reduce eddy current losses, thin sheets are laminated and put together to cut eddy current paths. Hysteresis losses: Coercive mmf causes hysteresis losses, given by, For one cycle of variation the energy input, Hysteresis power loss, W/m3 , where 1.5 < n < 2.5 43 Permanent magnets Permanent magnets are characterized by high residual flux values and high coercive force values. They are operated in 2nd quadrant of their hysteresis loop. 2nd quadrant of Alnico-5 • • • • 2nd quadrant of M-5 steel Remanent magnetization, Br corresponds to flux density which would remain in a closed magnetic structure without any applied mmf. The coercivity Hc is the value of magnetif field intensity (proportional to mmf) required to reduce magnetic flux density to zero. Due to high remanent magnetization, in permanent magnets there is no need for current to produce flux. Permanent magnets are widely used in applications such as refrigerator door holders, loudspeakers and DC motors. 44 Example (Fitzgerald): As shown in the figure, a magnetic circuit consists of a core of high permeability (µ→∞), an air gap of length g = 0.2 cm, and a section of magnetic material of length lm = 1.0 cm. The cross-sectional area of the core and gap is equal to Am = Ag = 4 cm2. Calculate the flux density Bg in the air gap if the magnetic material is (a) Alnico 5 and (b) M-5 electrical steel. Solution: Since the core permeability is assumed infinite, H in the core is negligible. Then, Load line equation a) Intersection point of the load line and B-H curve of Alnico-5 is shown in Fig. a, b) Intersection point of the load line and B-H curve of M-5 steel is shown in Fig. c, 45 Equivalent model of permanent magnets If B-H curve of a magnet in 2nd quadrant has wide linear characteristics as in the figure, then the magnet can be equivalently represented by an mmf and a series reluctance as in the equations below, where, lm is mean length, and Am is effective crossectional area of the magnet. µp is the slope of the B-H curve as shown in the figure. Example: Repeat the two previous example for a magnet with B-H characteristics as in the figure above with Hc = 250 kA/m and Br = 1 T. Solution: The equivalent magnetic circuit of the structure is shown below, Rm Fm Rg 46 Magnetization curves for common permanent-magnet materials 47