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the ux density:
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175 10 5 Wb = 0175 Wb/m2 = 2 = A w 00001 m2 and the magnetic eld intensity: B= H = 0175 Wb/m2 B B = = = 139 A t/m r 0 1,000 4 10 7 H/m
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Comments: This example has illustrated all the basic calculations that pertain to
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magnetic structures Remember that the assumptions stated in this example (and earlier in
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the chapter) simplify the problem and make its approximate numerical solution possible in a few simple steps In reality, ux leakage, fringing, and uneven distribution of ux across the structure would require the solution of three-dimensional equations using nite-element methods These methods are not discussed in this book, but are necessary for practical engineering designs The usefulness of these approximate methods is that you can, for example, quickly calculate the approximate magnitude of the current required to generate a given magnetic ux or ux density You shall soon see how these calculations can be used to determine electromagnetic energy and magnetic forces in practical structures The methodology described in this example is summarized in the following methodology box
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Magnetic Structures and Equivalent Magnetic Circuits Direct Problem: Given The structure geometry and the coil parameters (number of turns, current) Calculate The magnetic ux in the structure 1 Compute the mmf 2 Determine the length and cross section of the magnetic path for each continuous leg or section of the path 3 Calculate the equivalent reluctance of the leg 4 Generate the equivalent magnetic circuit diagram and calculate the total equivalent reluctance 5 Calculate the ux, ux density, and magnetic eld intensity, as needed Inverse Problem: Given The desired ux or ux density and structure geometry Calculate The necessary coil current and number of turns 1 Calculate the total equivalent reluctance of the structure from the desired ux 2 Generate the equivalent magnetic circuit diagram 3 Determine the mmf required to establish the required ux 4 Choose the coil current and number of turns required to establish the desired mmf
Consider the analysis of the same simple magnetic structure when an air gap is present Air gaps are very common in magnetic structures; in rotating machines, for example, air gaps are necessary to allow for free rotation of the inner core of the machine The magnetic circuit of Figure 1617(a) differs from
16
Principles of Electromechanics
i N l1 l2 l3
air gap
r 0
l5 (a)
the circuit analyzed in Example 162 simply because of the presence of an air gap; the effect of the gap is to break the continuity of the high-permeability path for the ux, adding a high-reluctance component to the equivalent circuit The situation is analogous to adding a very large series resistance to a series electrical circuit It should be evident from Figure 1617(a) that the basic concept of reluctance still applies, although now two different permeabilities must be taken into account The equivalent circuit for the structure of Figure 1617(a) may be drawn as shown in Figure 1617(b), where Rn is the reluctance of path ln , for n = 1, 2, , 5, and Rg is the reluctance of the air gap The reluctances can be expressed as follows, if we assume that the magnetic structure has a uniform cross-sectional area, A: R1 = l1 r 0 A R2 = l2 r 0 A R3 = l3 r 0 A
l4 R4 = r 0 A
l5 R5 = r 0 A
Rg = 0 Ag
(1631)
Figure 1617 (a) Magnetic circuit with air gap; (b) Equivalent representation of magnetic circuit with an air gap
Note that in computing Rg , the length of the gap is given by and the permeability is given by 0 , as expected, but Ag is different from the cross-sectional area, A, of the structure The reason is that the ux lines exhibit a phenomenon known as fringing as they cross an air gap The ux lines actually bow out of the gap de ned by the cross section, A, not being contained by the high-permeability material any longer Thus, it is customary to de ne an area Ag that is greater than A, to account for this phenomenon Example 163 describes in more detail the procedure for nding Ag and also discusses the phenomenon of fringing
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