EE 306 Electromechanical Devices Module 2 Chapter One

  • Slides: 44
Download presentation
EE 306 Electromechanical Devices Module 2 (Chapter: One) Magnetic Circuits 1

EE 306 Electromechanical Devices Module 2 (Chapter: One) Magnetic Circuits 1

Lecture Contents ü Introduction ü Production of a Magnetic field ü B-H & i-H

Lecture Contents ü Introduction ü Production of a Magnetic field ü B-H & i-H Relations ü Magnetic Equivalent Circuits ü Magnetization Curve (Magnetic Behavior) ü Analysis of Series-Parallel Magnetic Circuits 2

Lecture Contents ü Introduction ü Production of a Magnetic field ü B-H & i-H

Lecture Contents ü Introduction ü Production of a Magnetic field ü B-H & i-H Relations ü Magnetic Equivalent Circuits ü Magnetization Curve (Magnetic Behavior) ü Analysis of Series-Parallel Magnetic Circuits 3

Introduction § Our course is concerned primarily with the study of energy conversion devices

Introduction § Our course is concerned primarily with the study of energy conversion devices that convert electrical energy into mechanical energy or the reverse. § Rotating electric machines, such as DC machines, Induction machines, and Synchronous machines, are the most important ones used to perform the energy conversion. § Almost all practical motors and generators convert energy from one form to another through the action of a magnetic field. § In all these devices, magnetic materials are used to shape and direct the magnetic fields that acts as a medium in the energy conversion process. 4

Advantage of using magnetic materials? § High flux density can be obtained in the

Advantage of using magnetic materials? § High flux density can be obtained in the machine which will result in large torque or large machine output per unit machine volume. § The size of the machine is greatly reduced by the use of magnetic materials. § Due to these major advantages, magnetic materials form a major part in the construction of electric machines. § Thus it is necessary to study magnetic circuits in order to understand the operation of any electric machine. 5

Action of magnetic field: ü Four principles describe the action of magnetic field in

Action of magnetic field: ü Four principles describe the action of magnetic field in these energy conversion devices: § A current carrying wire produces a magnetic field in the area around it. § A time changing magnetic field induces a voltage in a coil of wire if it passes through that coil. (Basis of transformer action. ) § A current-carrying wire in the presence of a magnetic field has a force induced on it. (Basis of motor action. ) § A moving wire in the presence of a magnetic field has a voltage induced in it. (Basis of generator action. ) 6

Lecture Contents ü Introduction ü Production of a Magnetic field ü B-H & i-H

Lecture Contents ü Introduction ü Production of a Magnetic field ü B-H & i-H Relations ü Magnetic Equivalent Circuits ü Magnetization Curve (Magnetic Behavior) ü Analysis of Series-Parallel Magnetic Circuits 7

Production of a Magnetic field ü Magnetic fields can be visualized as lines of

Production of a Magnetic field ü Magnetic fields can be visualized as lines of flux that form closed paths. Using a compass, we can determine the direction of the flux lines at any point. Note that the flux density vector B is tangent to the lines of flux. 8

Illustrations of the right-hand rule 9

Illustrations of the right-hand rule 9

Ampere’s Law Ø Ampere’s Law – the basic law governing the production of a

Ampere’s Law Ø Ampere’s Law – the basic law governing the production of a magnetic field by a current. Ø It relates the integrated magnetic field around a closed loop to the electric current passing through the loop. Ø Ampère’s law (generalization of Kirchhoff's law) states that the line integral of magnetic field intensity H around a closed path is equal to the sum of the currents flowing through the surface bounded by the path. 10

Ampere’s Law-Illustration § Consider a current carrying conductor is wrapped around a ferromagnetic core,

Ampere’s Law-Illustration § Consider a current carrying conductor is wrapped around a ferromagnetic core, § Applying Ampere’s Law, § Here the path of integration is the mean length of the core (lc) and the current enclosed is (Ni), 11

Magnetic Field Intensity (H) § It is the measure of the effort that a

Magnetic Field Intensity (H) § It is the measure of the effort that a current is putting into the establishment of magnetic field. § In this sense, H (Ampere turns per metre) is known as the effort required to induce a magnetic field. § The strength of the magnetic field flux produced in the core also depends on the material of the core. § Thus, ü B = magnetic flux density (webers per square meter, Tesla (T)) ü μ= magnetic permeability of material (Henrys per meter) ü H = magnetic field intensity (ampere-turns per meter) 12

Core Permeability (µ) § A property of a magnetic material which indicates the ability

Core Permeability (µ) § A property of a magnetic material which indicates the ability of magnetic circuit to carry electromagnetic flux. § The constant µ may be further expanded to include relative permeability which can be defined as below: § µr is the relative permeability (No Units) § µ 0 is the permeability of free space (air) (Henry per metre) § The value of relative permeability is dependent upon the type of material used. The higher the amount permeability, the higher the amount of flux induced in the core. Relative permeability is a convenient way to compare the magnetizability of materials. 13

Magnetic Flux(ø) § Now, to measure the total flux flowing in the ferromagnetic core,

Magnetic Flux(ø) § Now, to measure the total flux flowing in the ferromagnetic core, consideration has to be made in terms of its cross sectional area (CSA). § Therefore, § Assuming that the flux density in the ferromagnetic core is constant throughout hence the equation simplifies to be: 14

Magneto motive force (MMF): (F) § Force which drives or tends to drive the

Magneto motive force (MMF): (F) § Force which drives or tends to drive the magnetic flux through a magnetic circuit. § Flux is due to the existence of mmf. § Unit is AT (ampere-turn). Reluctance (R) § It is the opposition of a magnetic circuit to settling up of a magnetic flux in it. § Reluctance of a magnetic circuit is analogous to the resistance of an electric circuit. 15

Lecture Contents ü Introduction ü Production of a Magnetic field ü B-H & i-H

Lecture Contents ü Introduction ü Production of a Magnetic field ü B-H & i-H Relations ü Magnetic Equivalent Circuits ü Magnetization Curve (Magnetic Behavior) ü Analysis of Series-Parallel Magnetic Circuits 16

Magnetic Equivalent Circuits § The complete closed path followed by any group of magnetic

Magnetic Equivalent Circuits § The complete closed path followed by any group of magnetic lines of flux is referred to as a magnetic circuit. Magnetic circuit Equivalent electrical circuit 17

Analogy with electric circuits § The flow of magnetic flux induced in the ferromagnetic

Analogy with electric circuits § The flow of magnetic flux induced in the ferromagnetic core can be made analogous to an electrical circuit. § The magnetic circuit equivalent equation is: 18

Analogy with electric circuits 19

Analogy with electric circuits 19

Determining the polarity of mmf: To easily determine the direction of flux, the ‘right

Determining the polarity of mmf: To easily determine the direction of flux, the ‘right hand curl’ rule is utilized: a) The direction of the curled fingers determines the current flow. b) The resulting thumb direction will show the magnetic flux flow. 20

Reluctance (R) & Permeance (P): § The element of R in the magnetic circuit

Reluctance (R) & Permeance (P): § The element of R in the magnetic circuit analogy is similar in concept to the electrical resistance. It is basically the measure of material resistance to the flow of magnetic flux. § We Know, And Therefore: Permeance (P) Units: AT/weber 21

Reluctances in series & parallel: § Reluctance in this analogy obeys the rule of

Reluctances in series & parallel: § Reluctance in this analogy obeys the rule of electrical resistance (Series and Parallel Rules). § For Series connected reluctances: § For Parallel connected reluctances: 22

Lecture Contents ü Introduction ü Production of a Magnetic field ü B-H & i-H

Lecture Contents ü Introduction ü Production of a Magnetic field ü B-H & i-H Relations ü Magnetic Equivalent Circuits ü Magnetization Curve (Magnetic Behavior) ü Analysis of Series-Parallel Magnetic Circuits 23

Magnetization Curve (Magnetic Behavior) 24

Magnetization Curve (Magnetic Behavior) 24

Lecture Contents ü Introduction ü Production of a Magnetic field ü B-H & i-H

Lecture Contents ü Introduction ü Production of a Magnetic field ü B-H & i-H Relations ü Magnetic Equivalent Circuits ü Magnetization Curve (Magnetic Behavior) ü Analysis of Series-Parallel Magnetic Circuits 25

Procedure to obtain Equivalent Magnetic Circuits m. m. f Source: 1. Locate the source

Procedure to obtain Equivalent Magnetic Circuits m. m. f Source: 1. Locate the source of m. m. f. 2. Identify the polarity of the m. m. f source (right hand thumb or curl rule). 3. Determine the magnitude of the m. m. f Reluctances: 1. Determine the number of reluctances required to model the physical circuit (based of mean length of flux, area of cross section & permeability of materials). Connection: 1. Connect the various reluctance components either in series or parallel so as to represent the given physical circuit. 26

Magnetic Circuits-Exercise The magnetic circuit for the toroidal coil can be analyzed to obtain

Magnetic Circuits-Exercise The magnetic circuit for the toroidal coil can be analyzed to obtain an expression for flux. Magnetomotive force is Where the reluctance is so and the magnetic flux is 27

Leakage flux Ø 28

Leakage flux Ø 28

Magnetic Circuits Example-1: Magnetic circuit below relative permeability of the core material is 6000

Magnetic Circuits Example-1: Magnetic circuit below relative permeability of the core material is 6000 its rectangular cross section is 2 cm by 3 cm. The coil has 500 turns. Find the current needed to establish a flux density in the gap of Bgap=0. 25 T. 29

Medium length of the magnetic path in the core is lcore=4*6 -0. 5=23. 5

Medium length of the magnetic path in the core is lcore=4*6 -0. 5=23. 5 cm, and the cross section area is Acore= 2 cm*3 cm = 6*10 -4 m 2 the core permeability is 30

The core reluctance is the gap area is computed by adding the gap length

The core reluctance is the gap area is computed by adding the gap length to each dimension of crosssection: thus the gap reluctance is: 31

Total reluctance is based on the given flux density B in the gap, the

Total reluctance is based on the given flux density B in the gap, the flux is thus magnetomotive force is thus the coil current must be 32

Example-2 In the magnetic circuit shown below, the relative permeability of the ferromagnetic material

Example-2 In the magnetic circuit shown below, the relative permeability of the ferromagnetic material is 1200. All dimensions are in centimeters, and the magnetic material has a square cross-sectional area. Assuming no magnetic leakage and fringing determine a. The air gap flux b. The air gap flux density c. The magnetic field intensity in the air gap. Considering the fringing effect, the effective area of the air gap is 8% larger than its physical size, repeat the above calculations 33

Solution: 34

Solution: 34

35

35

36

36

37

37

Example-3: The core of the magnetic circuit shown below is composed of cast steel

Example-3: The core of the magnetic circuit shown below is composed of cast steel and of cast iron (the related characteristics are shown in the curve figure below). Each material has a mean length of 1 m. The cross section area of the core is 0. 1 m 2. a) What would be the total mmf in order to have a flux density of 0. 6 Tesla (Wb/m 2) in the core? b) Find the core flux. c) What will be the current needed for the magnetization coil of 400 turns? d) What is the power delivered by the source of 350 V (DC)? 38

39

39

40

40

41

41

42

42

Example-4: Two coils are wound a toroidal core as shown in figure below. The

Example-4: Two coils are wound a toroidal core as shown in figure below. The core is made of silicon steel and has a square cross section. The coil currents are i 1=0. 28 A and i 2=0. 56 A. (a) Determine the flux density at the mean radius of the core. (b) Assuming a constant flux density over the cross section of the core, determine the flux in the core. (c) Determine the relative permeability of the core. 43

44

44