Suppose two uniform bars meet at an abrupt plane and there is a magnetic field induced by an extended core and coil structure off stage as in:

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Lucas Wilson
5 years ago
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1 Class Notes Week of 1/9 /3/017 ENGN1931F The magnetic structures that are central to the design of generators, transformers, motors, inductors, solenoids, etc. can be complex. We need a way to make approximate calculations of H, B, and from the current in a coil or coils embedded in the structures. One way to approach this is break the structure up into simple segments for which a relation between flux and magnetomotive force can be expressed in terms of a reluctance. The analogy is to Ohms law, current and voltage in that flux is thought of as a sort of magnetic current, magnetomotive force as similar to electromotive force and reluctance as analogous to resistance. We applied this to a simple uniform cross section structure on lm Friday. The reluctance for a linear magnetic material of high permeability to be where l m is AC the length of the magnetic path, and the denominator is the product of permeability and cross sectional area. Permeability plays the same role as conductivity in the case of linear resistive materials. Suppose two uniform bars meet at an abrupt plane and there is a magnetic field induced by an extended core and coil structure off stage as in: The dashed lines are paths that follow the direction of B. Think of it as the path of someone very small carrying a compass and walking in the direction the compass points. These lines are actually the result of solving the equation for B 0 subject to the boundary conditions that B vanishes outside the structure (not enough H there to sustain B in the low permeability of air) and B is parallel to the axis of the system at the end faces. Applying flux conservation to the joint, we draw a bounding box around the joint with two faces just back from the junction and perpendicular to the central axis. The outer surfaces are outside the magnetic material and so have no flux through them. To keep the problem simple assume that B is uniform across the material so that the flux entering from the left is 1 AC 1B1 and on the right the flux leaving the volume is AC B. By conservation of flux, these two fluxes must be

2 ACB ACH equal so. Suppose we have a closed structure consisting of one segment each of these AC1B1 AC1H1 two materials and that there is a magnetomotive force of magnitude Ni from a coil wrapped through l1 l this object. By Ampere s law, Ni H11 l Hl 1 where l 1 and l 1 are AC1 AC the lengths of the two segments. This is the equivalent of two resistances in series in an application of Kirchhoff s voltage law. It is left as a reading and thinking assignment to prove that parallel reluctances behave as expected. Air gaps are a common part of magnetic structures, introduced to allow motors to rotate and inductors to store energy more efficiently, that is, more energy per unit volume of space. For a first approximation, we ignore any fringing of the B field (equivalent to the assumption that the gap is small compared to the crosswise linear dimensions). Then by the same argument as for a change in cross section, the B field in the gap is the same as the B field in the magnetic material. From this it follows that lg g. You might assume that this reluctance is small compared to that of the rest of a total 0 Ag structure if the gap length is short compared to the size of the rest of the structure. However the lm relevant comparison is lg to. For all the common soft magnetic materials, the relative permeability r is very high. For the common silicon electrical steels it is usually around 4000 and that is large enough that the air gap is often the controlling factor in how much magnetomotive force is needed to get a particular B field. /1/017 A voltage is required to generate the current in the coil that generates the H field and that is related to the current through Faraday s Law. The Maxwell Heaviside equations that encapsulate this law are: B E Edl BdS t t t C If we take the ends of the coil and hold them close together, the voltage between the ends is by definition v Edl so: C M v N and if L then v L t t dt dt S N di N di m Usually the argument that the flux in the coil is N times the flux in the core is left to the reader s intuition. However, in class I showed you a Scotch TM and wire model for the surface of integration for the flux in a loosely wound helix. It showed that the approximation is likely very good if the flux is forced m

3 along the axis of the winding through all turns by a core structure but always underestimates some flux. Without a core the approximation is quite poor. Any inductor with a non zero current has energy stored within it. To see this, consider starting with zero current and integrating the power delivered by a power supply to the coil as it forces a coil current, i. t t i di 1 U vidt Li dt L i di Li dt The equivalent formula for stored electrostatic energy is U Cv. This energy had to exist somewhere within the coil and magnetic structure and it does so in the B field itself. I gave you a handout and discussed the derivation slightly in class for the energy density in space. The result is that at any point is space with a magnetostatic field, the energy stored per unit volume or per cubic meter is given by 1 H B B for linearly permeable materials and HdB in the general case. The stored energy is the volume integral of these densities over the entire structure of the device. /3/017 With the energy relations we can now calculate the force exerted on the solenoid that I used to show what the course is about on the first day of class. To keep the calculation simple, assume: 0 Current is a constant DC current (avoids time dependent forces and current changing with inductance) Neglect all resistance in the coil Linearly permeable core Uniform cross section (really built that way) B is uniform across the cross section Simplified structure with same air gap By a magnetic circuit calculation: 0Ni B lm x r To find the force on the bar, we calculate the work done by the power supply if I force the bar to move a small increment of x. We equate that work to the sum of the change in stored field energy plus the du dw work done on the bar or W U F x and in the limit that x 0, F dx dx

4 Notice that the total energy taken from the supply for a small displacement is twice the work done on the bar. The other half of the power goes into an increase in the stored energy of the fields. I did a quick very rough estimate of the parameters of the example solenoid. The cross section is about 1 cm of 1e 4 sq. m. The relative permeability of silicon core iron is about Doremeyer gave a specification for this device of 46 oz pull at ¼ opening. At that opening there is no inductance that is significant at 60 Hz relative to the measured resistance of 60 ohms. I assumed 30 AWG copper wire which is a common coil gauge when significant inrush currents can be expected. The current at 10 volts AC will be A RMS so I assumed I = A. I used a ruler and estimate an average turn as being 3. With 30 AWG having 109 ohms per thousand feet, this means N = 00 turns which I rounded down to 000. With NI = 4000 Amp turns and the path length in the core as 10 cm, the predicted force is 1.9 Nt or 40 oz. This is probably a somewhat fortuitous agreement! In the calculations so far I have assumed linear permeability but this is really not realistic. We need a short introduction to the properties of magnetic materials for several reasons, among which are: A physical picture of the basic mechanisms helps understand many aspects of the design of machines Losses in cores as a function of construction (and capital investment) and operating frequency depend on the true relation of B to H and on the variable electrical conductivity of these materials Selection of materials and their shapes is influenced by their properties

5 While quantum mechanics tells us of the ultimate nature of magnetic properties, it does not do a good job of predicting properties and so some knowledge of what metallurgists have found and how others have made use of those findings is useful to us. In classical electricity and magnetism a B field is the result of a current, that is, it arises from the flow of charge and there is no magnetic equivalent to charge itself. Indeed, in special relativity the B field is a consequence of observing a charge from a reference frame that is moving relative to the rest frame of the charge. However, the observed fact is that individual atoms often exhibit a magnetic field implying classically that there must be an internal, probably circulating current. Classically this does not make sense since such a moving charge would be continuously accelerated and therefore would have to radiate electromagnetic energy. Atoms do not do this. Instead quantum mechanics says that the electrons appear to have angular momentum as if they were spinning (hence the term spin ) and may also have orbital angular momentum as if they rotated about the nucleus. If these momenta have certain magical values, the motion may be stable in the same manner as Bohr postulated would be the case for the hydrogen atom. Bohr s picture is about as deep as we will go but, if you are interested, consider reading Richard Feynman s book, QED, for real insight into the depth of the mystery. The first step to taking a view of atoms as containing a small recirculating current into an understanding of macroscopic phenomena is to consider the classical view of what this type of current will appear as at our human scale. This will be the subject of our next class.

The initial magnetization curve shows the magnetic flux density that would result when an increasing magnetic field is applied to an initially

The initial magnetization curve shows the magnetic flux density that would result when an increasing magnetic field is applied to an initially MAGNETIC CIRCUITS The study of magnetic circuits is important in the study of energy systems since the operation of key components such as transformers and rotating machines (DC machines, induction machines,