Laplace's equation of a scalar magnetic potential in an airgap is given in (A.1).

d2 Q d2Q

&2 ax2

Where Q = magnetic scalar potential

The equivalent current sheet and mmf distribution of the rotor is shown in Figure A.

Figure A. 1: mmf distribution of rotor

The mmf distribution can be broken down into sinusoidal harmonics of the form given in (A.2)

f 2wmx^

Where X

wavelength (m) harmonic number The general solution of (A.1) will therefore be of the form (A.3) Q( x, y) = Q( x)Q( y)

The boundary conditions for Q in the (y) direction are: Q(0)=Qm; mmf has a value at the rotor surface

Q(g)=0; the mmf is zero at the extremity of the airgap (when x=g) From [100] the solution of an equation of this form with exactly these boundary conditions is

From Maxwell's equation, (A. 5), the flux density in the y-direction can be found using (A.6).

This will be at a maximum at the surface of the rotor, y=0 and peak when x/A =1/4, as given in (A.7).

As g approaches infinity and the machine effectively becomes an ironless machine, the hyperbolic function tends to 1. Remembering that m is the harmonic number, it can be seen that for the first harmonic with an infinite airgap the flux density in the y direction is given by (A.8).

Assuming that the flux flows in straight lines and that the iron is infinitely permeable, all the mmf for the fundamental harmonic is dropped across the imaginary airgap.

ßoA A ßo ßo mmf - A1 -<Sg - <-Z-- By (A.9)

Where lg is the length of an effective airgap which gives the same reluctance.

Combining (A.8) and (A.9) shows that this effective airgap is (A.10).

From Figure A. 1 it can be seen that A=2(Wm+Ws), so the effective airgap becomes (A.11).

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