Question 5.13.1: Develop a two-dimensional model for the field in the gap reg...

We reduce the three-dimensional motor geometry to an approximate two-dimensional geometry [21]. This is done by introducing a cylindrical cutting plane at the mean radius of the magnets. This cylinder is then imagined to be unrolled into a two-dimensional surface of infinite extent into and out of the page, as illustrated in Fig. 5.40a. This two-dimensional geometry can be reduced further by exploiting the symmetry that results from the repeating magnetic structure. Specifically, the field is symmetric about the vertical center lines of each pole. The field solution for the reduced geometry is similar to that of the periodic geometry of Example 3.6.2. Therefore, we follow closely the solution procedure presented there. As depicted in Fig. 5.40b, there are three regions to consider. In this figure, l is the arc length from the center of one pole to the next at the mean radius, and x is the distance along the ‘‘straightened’’ arc. Therefore, if R_{1} and R_{2} are the inner and outer radii of the magnet, respectively, and N_{pole} = the number of poles, then

l = \frac{π(R_{1} + R_{2})}{N_{pole}}.

The variable x is related to the angular measure Φ along the arc as follows:

Φ = \frac{x}{l} \frac{2π}{N_{pole}} .                (5.263)

The magnetization for the three regions is

M_{s} = \left\{\begin{matrix} M_{s}\hat{y} \qquad (Region   1)\\ -M_{s}\hat{y}  \qquad (Region   2) \qquad (5.264)\\ 0 \qquad (Region   3). \end{matrix} \right.

Because M_{s} is constant in each region, the equation for the magnetic scalar

∇^{2} φ_{m} = 0                    (5.265)

for all three regions. The three regions under consideration are collectively bounded by four lines that form a rectangular area. The vertical sides of the rectangle, which are defined by the y-axis and the line x = l, represent lines of symmetry for the field, along which the normal component H_{n} is zero. Moreover, the boundaries that form the top and bottom of the rectangle are bordered by soft-magnetic materials (flux return plates). For the sake of simplicity, we assume that the flux plates have infinite permeability (μ = ∞). Consequently, the tangential component of the field vanishes along these boundaries. When we impose these boundary conditions we find that the general solutions for the magnet and gap regions are potential reduces to

φ_{mag}(x,y) = \sum\limits_{n=1}^{∞} C_{mag,n} sinh (\frac{nπ(y-h)}{l}) cos(\frac{nπx}{l})             (5.266)

and

φ_{gap}(x,y) = \sum\limits_{n=1}^{∞} C_{gap,n} sinh (\frac{nπy}{l}) cos(\frac{nπx}{l}) ,            (5.267)

respectively, where C_{mag,n} and C_{gap,n} are constants that need to be determined. At the interface between the magnet and the gap, the tangential component of H must be continuous:

H_{mag,t}(x,g) = H_{gap,t}(x,g)               (0 < x < l).               (5.268)

This condition, along with the fact that the functions sin(nπx/l) are orthogonal on (0, l), implies that

C_{mag,n} = C_{gap,n} \frac{sinh(nπg/l)}{sinh(nπ(g-h)/l)}.            (5.269)

In addition, the normal component of B must also be continuous

B_{mag,n}(x,g) = B_{gap,n}(x,g)               (0 < x < l).               (5.270)

As

B_{mag,n} =  \left\{\begin{matrix}μ H_{mag,n} + μ_{0} M_{s} \qquad (0  <  x  <  l/2)\\ μ H_{mag,n} – μ_{0} M_{s} \qquad (l/2 <  x  <  l), \end{matrix} \right.                (5.271)

then Eq. (5.270) reduces to

  \sum\limits_{n=1}^{∞} \frac{nπ}{l} C_{gap,n} K (n, h, g, l, μ) cos (\frac{nπx}{l}) = \left\{\begin{matrix} \frac{μ_{0}}{μ} M_{s} \qquad (0  <  x  <  l/2)\\ – \frac{μ_{0}}{μ} M_{s} \qquad (l/2 <  x  <  l)\end{matrix} \right.

where

K (n, h, g, l, μ) = -\frac{μ_{0}}{μ} cosh(\frac{nπg}{l}) + sinh(\frac{nπg}{l}) coth(\frac{nπ}{l} (g  –  h)).              (5.272)

Again, by exploiting the orthogonality of the functions cos(nπx/l) on (0, l), it follows that

C_{gap,n} =  \frac{4lμ_{0}M_{s}}{μ} \frac{(-1)^{(n-1)/2}}{(nπ)^{2} K (n, h, g, l, μ)}             (n = 1, 3, 5, . . .).              (5.273)

Substituting Eq. (5.273) into Eq. (5.267) gives the complete solution for the gap region

φ_{gap}(x,y) = \frac{4lμ_{0}M_{s}}{μ}\sum\limits_{n = 1, 3, 5, . . .}^{∞} \frac{(-1)^{(n-1)/2}}{(nπ)^{2} K (n, h, g, l, μ)}  sinh(\frac{nπy}{l}) cos (\frac{nπx}{l}).

Therefore, the vertical or axial component of the field in this region is

B_{gap,y}(x,y) = \frac{4μ_{0}^{2}M_{s}}{μ}\sum\limits_{n = 1, 3, 5, . . .}^{∞} \frac{(-1)^{(n-1)/2}}{(nπ) K (n, h, g, l, μ)}

× cosh(\frac{nπy}{l}) cos (\frac{nπx}{l}),              (5.274)

where x equals the horizontal distance along a circumferential arc at the mean radius of the magnet, y equals the vertical distance above the lower flux plate, g is the gap height, t_{m} is the thickness of the magnet, h = g + t_{m}, l = π (R_{1} + R_{2})/N_{pole}, and R_{1} and R_{2} are the inner and outer radii of the magnet.

Calculations: We apply Eq. (5.274) to a motor with the following parameters:

M_{s}  = 4.0 × 10^{5}   A/m
g = 6.0 mm
t_{m} = 4.0 mm
R_{1} = \frac{40}{π} mm
R_{2} = \frac{120}{π} mm
l = 20 mm
N_{pole} = 8.                (5.275)

Field values are computed with μ = 1.0 μ_{0}, 1.5 μ_{0}, and 2.0 μ_{0}, and with the vertical position in the gap set to y = 2 mm. Ten values are computed for each value of μ. Specifically, field values are computed at Φ = 0, 5, 10, . . . , 45° (i.e., from the center of one pole to that of its neighbor). These data are shown in Fig. 5.41. Notice that the field decreases with increasing μ.