ELE 2314 Principles of Machines and Power PDF

ELE 2314 Principles of Machines and Power Learning Outcome 2: MAGNETIC CIRCUITS 1Dr Azzeddine Ferrah  LO2: Magnetic Circuits: Objectives Apply basic physical concepts to explain the operation of, and solve problems related to, electrical machines....

ELE 2314 Principles of Machines and Power Learning Outcome 2: MAGNETIC CIRCUITS 1Dr Azzeddine Ferrah  LO2: Magnetic Circuits: Objectives Apply basic physical concepts to explain the operation of, and solve problems related to, electrical machines. 1. Describe the construction of simple magnetic circuits with and without air gap. 2. Explain the basic laws which govern the electrical machine operation such as Faraday's Law, Ampere - Biot - Savart's Law, and Lenz's Law 3. Apply Faraday's Law of Electromagnetic Induction, Ampere - Biot - Savart's Law, and Lenz's Law to solve for induced voltages and currents in case of simple magnetic circuits with movable parts. 4. Illustrate the principle of the electromechanical energy conversion in magnetic circuits with movable parts. 2  Review Magnetic Field A permanent magnet has a magnetic field surrounding it. Such a field consists of lines of force that radiate from the north pole to the south pole and back to the north pole through the magnetic material. Magnetic flux lines are invisible, but the effects can be visualized with iron filings sprinkled in a magnetic field 3  Review Flux and Flux Density The unit of flux is the weber. The unit of flux density is the weber/m2, which defines the unit tesla, (T), which is a very large unit. Flux density is given by the equation B=  A Fluxlines( where B = flux density (T) Area (m2)  = flux (Wb) A = area (m2)  TheTesla (T) is the SI unit for flux density, another unit called the gauss, from the CGS (cm-gram-second) system,is sometimes used (104 gauss = 1 T).  The instrument used to measure flux density is the Gaussmeter. 4  Review of Magnetism and Electromagnetism Flux and Flux Density Example: What is the flux density in a rectangular core that is 8 mm by 10 mm if the flux is 4 mWb?     A  8x10 3 x 10x10 3  80x10 6 m 2    4x10 3 6  50 Wb/m 2  50 T A 80x10 5  Electromagnetism  Is the production of a magnetic field by current in a conductor.  Many types of useful devices such as tape recorders, electric motors, speakers, solenoids, and relays are based on electromagnetism.  Magnetic flux lines surround a current carrying conductor.  The field lines are concentric circles.  As in the case of bar magnets, the effects of electrical current can be visualized with iron filings around the wire – the current must be large to see this effect. 6  Electromagnetism When the direction of current When current is left to right, the is right to left, the lines are in lines are in a counterclockwise a clockwise direction. direction. 7  Electromagnetism Right-Hand Rule An aid to remembering the direction of the lines of force is illustrated in the. Imagine that you are grasping the conductor with your right hand, with your thumb pointing in the direction of current. Your fingers indicate the direction of the magnetic lines of force. A Simple Electromagnet 8  Applications of electromagnets  Many objects around you contain electromagnets. They are found in electric motors and loudspeakers. Very large and powerful electromagnets are used as lifting magnets in scrap yards to pick up, then drop, old cars and other scrap iron and steel.  They are better than magnets because the magnetism can be turned off and on. 9  Magnetic Quantities Permeability () It defines the ease with which a magnetic field can be established in a given material. It is measured in units of the weber per ampere- turn meter. The permeability of a vacuum (0) is 4 x 10-7 weber per ampere-turn meter, which is used as a reference. Relative Permeability (r) It is the ratio of the absolute permeability to the permeability of a vacuum. r   0 10  Magnetic Quantities Reluctance (R) It is the opposition to the establishment of a magnetic field in a material. l R A R = reluctance inA-t/Wb l = length of the path  = permeability (Wb/A-t m). A = area in m2 Recall that magnetic flux lines surround a current-carrying wire. A coil reinforces and intensifiesthese flux lines. 11  Magnetic Quantities Magneto-Motive Force (MMF) The cause of the magnetic flux is called magnetomotive force which is related to the current and number of turns of the coil. Fm = NI Fm = magnetomotive force (A-t) N = number of turns of wire in a coil I = current (A) 12  Magnetic Quantities V Fm Ι Ohm’s law for magnetic circuits is:  R R   I Fm  V R  R Exercise 1 How much flux is established in the magnetic path of the Figure shown if the reluctance of the material is 0.28 X 105 At/WB? 53 Exercise 2  5  5.36 106Wb 0.28  10 What flux is in a core that is wrapped with a 300 turn coil with a current of 100 mA if the reluctance of the core is 1.5 x 107 A-t/Wb ? 300 100   2 103Wb 15  10 6 13  Magnetic Quantities The magnetomotive force (mmf) is not a true force in the physics sense, but can be thought of as a cause of flux in a core or other material. Iron core Current in the coil causes flux in the iron core. What is the mmf if a 250 turn coil has 3 A ofcurrent? Fm = 750 A-t 14  Magnetic Quantities Magnetic field intensity It is the magnetomotive force per unit length of a magnetic path. H=Fm or H = NI l l H= Magnetic field intensity (A-t/ m) Fm = magnetomotive force (A-t) l = average length of the path (m) N = number of turns I = current (A) Magnetic field intensity represents the effort that a given current must put into establishing a certain flux density in a material. 15  Magnetic Quantities If a material is permeable, then a greater flux density will occur for a given magnetic field intensity. The relation between B (flux density) and H (the effort to establish the field) is B = H  = permeability (Wb/A-t m). H= Magnetic field intensity (Wb/A-t m) This relation between B and H is valid up to saturation, when further increase in H has no affect on B. 16  Magnetic Quantities As the graph shows, the flux density depends on both the material and the magnetic field intensity. Magnetic material Saturation begins Flux density, B, (Wb//m2) Non-magnetic material Magnetic Field Intensity, H, (At/m) 17  Magnetic Quantities As H is varied, the magnetic hysteresis curve is developed. B B B Saturation Bsat BR Hsat H H H 0 0 Hsat H=0 H 0 Bsat Saturation B (a) (b) (c ) (d) B B Bsat HC H = 0 H H H 0 0 Hsat H BR HC HC B B (e) (f) (g) 18  Magnetization Curve A B-H curve is referred to as a magnetization curve for the case where the material is initially unmagnetized. 2.0 Annealed iron Flux Density, B, (T) 1.5 1.0 0.5 0 0 200 400 600 800 1000 1200 1400 1600 1800 2000 Magnetic Field Intensity, H,(A-t/m) 19  Magnetization Curve A B-H curve can be read to determine the flux density in a given core. The next slide shows how to read the graph to determine the flux density in an annealed iron core. 2.0 Annealed iron Flux Density, B, (T) 1.5 1.0 0.5 0 0 200 400 600 800 1000 1200 1400 1600 1800 2000 Magnetic Field Intensity, H,(A-t/m) 20  Magnetization Curve Annealed iron core I= 380 turns 0.9A What is B for the core? NI 380 t0.9 A H  1487 A-t/m l 0.23 m Reading 2.0 Annealed iron the graph, Flux Density, B, (T) 1.5 B = 1.63 T 1.0 0.5 0 0 200 400 600 800 1000 1200 1400 1600 1800 2000 Magnetic Field Intensity, H,(A-t/m) 21  B-H Curve 22  Relative Motion S When a wire is moved across a magnetic field, there is a relative motion between the wire and the magnetic field. N When a magnetic field is moved past a stationary wire, there is also relative motion. S In either case, the relative motion results in an induced voltage in the wire. N  Induced Voltage The induced voltage due to the relative motion between the conductor and the magnetic field when the motion is perpendicular to the field is given by vind = B l v B = flux density in T l = length of the conductor in the magnetic field in m v = relative velocity in m/s (motion is perpendicular) 24  Faraday’s Law 25  Faraday’s Law Faraday also experimented generating current by relative motion between a magnet and a coil of wire. The amount of voltage induced across a coil is determined by two factors: 1. The rate of change of the S magnetic flux with respect to the coil. 2. The number of turns of wire in the coil. - V+ Voltage is indicated only when magnet is moving. 26  Faraday’s Law Faraday also experimented generating current by relative motion between a magnet and a coil of wire. The amount of voltage induced across a coil is determined by two factors: 1. The rate of change of the S magnetic flux with respect to the coil. 2. The number of turns of wire in the coil. - V+ Voltage is indicated only when magnet is moving. 27  Magnetic Field around a Coil Just as a moving magnetic field induces a voltage, current in a coil causes a magnetic field. The coil acts as an electromagnet, with a north and south pole as in the case of a permanent magnet. South North 28 Machines Basic Requirements  Presence of a magnetic field can be produced by:  Use of permanent magnets  Use of electromagnets (Mostly used)  Then one of the following methods is needed:  Motion to produce electric current (generator)  Electric current to produce motion (motor) 27 Ampere’s Law  Any current carrying conductor will produce magnetic field around itself. Magnetic field around a wire:  Thumb indicates direction of current flow  Finger curl indicates the direction of field 30 Lorentz Force The Lorentz force is the force on a point charge due to electromagnetic fields. It is given by the following equation in terms of the electric and magnetic fields. F  qE  v  B The first term in the Lorentz Force Equation represents the electric force Fe acting on a charge q within an electric field is given by. Fe  qE The electric force is in the direction of the electric field. 31 Lorentz Force The second term in the Lorentz Force Equation represents magnetic force Fm(N) on a moving charge q(C) is given by Fm  qv  B where the velocity of the charge is v (m/sec) within a field of magnetic flux density B (Wb/m2). Check the units. The magnetic force is at right angles to the magnetic field. The magnetic force requires that the charged particle be in motion. 32 Lorentz Force A segment of a current- carrying wire in a magnetic field B.The magnetic force exerted on each charge making up the current is q v x B and the net force on the segment of length L is I L x B. F  I L B 33 Lorentz Force thumb Direction thrust (force) Use Fleming’s left-hand first finger (or motor) rule to predict the direction of the field force. Second finger current If the thumb and first two fingers of the left hand are placed comfortably at right angles to each other, With the first finger pointing in the direction of the field and the second finger pointing in the direction of the current flow Then the thumb points in the direction of the thrust (force). 34  Points to remember: 1. The direction of the field is from the N pole to the S pole. 2. The direction of the current is from the positive (+) terminal of the power pack to the negative (-) terminal, i.e. conventional current direction. 3. The rule applies only where the current and field directions are at right angles. Demonstration A current carrying wire in a magnetic N field. N Field Field Force Force Conductor Conductor carrying current S S carrying out of page current into page 35 Magnetic Equivalent Circuit  i lc + N F  - i Analogy between magnetic circuit and electric circuit E R 36 Magnetic Equivalent Circuit l c  Air Gap i + c N lg F - g lc lg c  ; g  c Ac 0 Ag Ni  Ni  H c l c  H g l g C   g Flux density c g Bc  ; Bg  Ac Ag 37 Parallel Magnetic Circuit T  1  2 RT  R3  R1 // R2 38 Electric vs Magnetic Circuit 39 Leakage Flux Part of the flux generated by a current-carrying coil wrapped around a leg of a magnetic core stays outside the core. This flux is called x. leakage flux Useful flux 40 Fringing Effect  The effective area provided for the flow of lines of magnetic force (flux) in an air gap is larger than the cross-sectional area of the core. This is due to a phenomenon known as fringing effect. Air gap – to avoid flux saturation when too much current flows - To increase reluctance 41 Example 1 Refer to Figure below, calculate:- 1) Flux 2) Flux density 3) Magnetic intensity Given r = 1,000; no of turn, N = 500; current, i = 0.1 A. cross sectional area, A = 0.0001m2 , and means length core lC = 0.36 m. lc i N 1. 1.75x10-5 Wb 2. 0.175 Wb/m2 3. 139 AT/m 43 Example 2 lg  Movable i part N Data- 1T – 700 at/m The Figure represents the magnetic circuit of a relay. The coil has 500 turns and the mean core path is lc = 400 mm. When the air-gap lengths are 2 mm each, a flux density of 1.0 Tesla is required to actuate the relay. The core is cast steel. a.Find the current in the coil. (6.93 A) b.Compute the values of permeability and relative permeability of the core. (1.4 x 10-3 , 1114) c. If the air-gap is zero, find the current in the coil for the same flux density (1 T) in the core. ( 0.6 A) 44  Example 3 Find the value of I required to establish a magnetic flux of  = 0.75  10-4 Wb in the series magnetic circuit as shown in Figure Ex3. Calculate the force exerted on the armature (moving part) when the flux is established. The relative permeability for the steel is r = 1424. Fig Ex3 44  Solution (Example 3) The above device can be analysed by its magnetic circuit equivalent and its electric circuit analogy as shown in figure below. (a) Magnetic circuit equivalent and (b) electric circuit analogy 45  Solution (Example 3) From the Gauss law (analogy to KCL in electric circuit), the flux density for each section is  0.7510-4 Wb B   0.5 T. A 1.510-4 m2 The magnetic field intensity of steel is B B 0.5 T Hc     279 At/m. c  r 0 1424 4 10-7 Hm-1 The magnetic field intensity of air gap is B B 0.5 T Hg     3.98105 At/m. g 0 1 4 10 Hm -7 -1 The magnetomotive force drops are H c lc  279 At/m10010-3 m  28 At. 46 H g l g  3.98105 At/m2 10-3 m  796 At.  Applying Ampère circuital law (analogy to KVL in electric circuit), NI  H c l c  H g l g  28 At  796 At 200 t I  824 At I  4.12 A. Energy stored in air gap, W = Volume of air gap, Vg × magnetic energy density, W0 Bg2  Ag lg . 20 W = mechanical energy to close the air gap = Force, P × length of the air gap, lg B g2 Ag l g   Pl g 2 0 Ag B g2 P 2 0 1.5  10  4 m 2  0.5 T  2  2  4  10 -7 Hm -1 47  14.92 N  Example 4 Determine the value of I required to establish a magnetic flux of 2 = 1.510-4 Wb in the section of the core indicated in Figure Ex4. The relative permeability for the steel at region bcde, be, and efab are 2 = 4972, 1 = 4821, and T = 2426, respectively. Fig Ex4 48  Solution (Example 4) The above device can be analysed by its magnetic circuit equivalent and its electric circuit analogy as shown in figure below. (a) Magnetic circuit equivalent and (b) electric circuit analogy 49  Solution (Example 4) lbcde 0.2 Rbcde    53.35103 At/Wb 20 A 4972 0  6 10 m -4 2 lbe 0.05 Rbe    13.76103 At/Wb 10 A 4821 0  6 10 m -4 2 lefab 0.2 Refab    109.34103 At/Wb T 0 A 2426 0  6 10 m-4 2 Since 1 Rbe   2 Rbcde  R 1  2 bcde  5.816 10  4 Wb Rbe From Gauss Law,  T  1   2  (1.5  5.816) 10 4 Wb  7.316 10 4 Wb 50  Solution (Example 4) Applying Ampère circuital law for loop 1, NI   T Refab  1 Rbe  0  T Refab  1 Rbe 80  8 I   1.76 A N 50 51  Example 5 The core of Figure Ex5 is made of cast steel. Calculate the current I that is needed to establish a flux of g = 6  10-3 Wb at the air gap if fringing field is neglected. [Hint: Additional information can be obtained from the B-H curve on Slide 55] Fig Ex5 52  Solution (Example 5) (a) Magnetic circuit equivalent and (b) electric circuit analogy 53  Consider each section in turn. For air gap, Bg = g/Ag = 6  10-3 Wb/2  10-2 m2 = 0.3 T. Hg = Bg/0 = 0.3 T/(4  10-7 H/m) = 2.388  105 At/m. For section ab and cd, Bab = Bcd = Bg = 0.3 T. Hab = Hcd = 250 At/m. (refer to B-H curve in figure 4.26 in lec note). Apply Ampère circuital law (KVL) at loop 2, Since the flux in leg da flows in the opposite direction of the flux in leg ab, leg cd, and air gap, the corresponding term of Hdalda will be subtractive. Also, NI = 0 for loop 2. Thus,  loop2 NI  loop2 Hl 0  H abl ab  H g l g  H cd lcd  H da l da  2500.25  2.388  105 0.25  10 3   2500.25  0.2 H da  62.5  59.7  62.5  0.2H da 54  184.7  0.2 H da  B-H curve (Notes) 55  Solution (Example 5) Thus, Hda = 923.5 At/m. Bda = 1.12 T (from B-H curve). 2 = BdaA = 1.12  0.02 = 2.24  10-2 Wb. 1 = 2 + 3 = 2.84  10-2 Wb. Bdea = 1/A = (2.84  10-2)/0.02 = 1.42 T. Hdea = 2100 At/m (from B-H curve). Apply Ampère circuital law at loop 1, NI = Hdealdea + Hdalda = (2100)(0.35) + 184.7 = 919.7 At I = 919.7/200 ≈ 4.6 A 56 Electromagnetic Induction  An emf can be induced in a coil if the magnetic flux through the coil is changed. This phenomenon is known as electromagnetic induction. d  The induced emf is given by e  N dt  Faraday’s law: The induced emf is proportional to the rate of change of the magnetic flux.  This law is a basic law of electromagnetism relating to the operating principles of transformers, inductors, and many types of electrical motors and generators. 45 Electromagnetic Induction  Faraday's law is a single equation describing two different phenomena: The motional emf generated by a magnetic force on a moving wire, and the transformer emf generated by an electric force due to a changing magnetic field.  The negative sign in Faraday's law comes from the fact that the emf induced in the coil acts to oppose any change in the magnetic flux.  Lenz's law: The induced emf generates a current that sets up a magnetic field which acts to oppose the change in magnetic flux. 46 Self-Inductance d d  From Faraday’s law eN  dt dt  Where  is the flux linkage of the winding is defined as   N  For a magnetic circuit composed of constant magnetic permeability, the relationship between  and i will be linear and we can define the inductance L as L  i  It can be shown later that N2 L  eq 47 Magnetic Stored Energy We know that for a magnetic circuit with a single winding   N  Li d d d and e  (N)  (Li) dt dt dt For a static magnetic circuit the inductance L is fixed di eL dt For a electromechanical energy device, L is time varying di dL eL i dt dt 60 Magnetic Stored Energy The power p is d p  ie  i dt Thus the change in magnetic stored energy 2 2  t2 W   pdt   id   L d  1 2L  2 2  12  t1 1 1 The total stored energy at any  is given by setting 1 = 0: 1 W  2  L i2 2L 2 61  Magnetic units It is useful to review the key magnetic units from this chapter: Quantity SI Unit Symbol Magnetic flux density Tesla B Flux Weber  Permeability Weber/ampere-turn-meter  Reluctance Ampere-turn/Weber R Magnetomotive force Ampere-turn Fm Magnetizing force Ampere-turn/meter H 62  Quiz 1. A unit of flux density that is the same as a Wb/m2 is the a. ampere-turn b. ampere-turn/weber c. ampere-turn/meter d. tesla 63  Quiz 2. If one magnetic circuit has a larger flux than a second magnetic circuit, then the first circuit has a. a higher flux density b. the same flux density c. a lower flux density d. answer depends on the particular circuit. 64  Quiz 3. The cause of magnetic flux is a. magnetomotive force b. induced voltage c. induced current d. hysteresis 65  Quiz 4. The measurement unit for permeability is a. weber/ampere-turn b. ampere-turn/weber c. weber/ampere-turn-meter d. dimensionless 66  Quiz 5. The measurement unit for relative permeability is a. weber/ampere-turn b. ampere-turn/weber c. weber/ampere-turn meter d. dimensionless 67  Quiz 6.The property of a magnetic material to behave as if it had a memory is called a. remembrance b. hysteresis c. reluctance d. permittivity 68  Quiz 7. Ohm’s law for a magnetic circuit is a. Fm = NI b. B = H F c.   m R d. R  l A 69  Quiz 8. The control voltage for a relay is applied to the a. normally-open contacts b. normally-closed contacts c. coil d. armature 70  Quiz 9. A partial hysteresis curve is shown. At the point indicated, magnetic flux B a. is zero b. exists with no magnetizing force BR c. is maximum H d. is proportional to the current 71  Quiz 10.When the current through a coil changes, the induced voltage across the coil will a. oppose the change in the current that caused it b. add to the change in the current that caused it c. be zero d. be equal to the source voltage 72  Quiz Answers: 1. 6. 2. 7. 3. 8. 4. 9. 5. 10. 73  END OF LEARNING OUTCOME 2 74