9.4: Continuous Approximation to Winding Patterns

Now let’s try to produce those surface current distributions with physical windings. In fact we can’t do exactly that yet, but we can approximate a physical winding with a turns distribution that would look like:

\(\ \begin{aligned}
n_{S} &=\frac{N_{S}}{2 R} \cos p \theta \\
n_{R} &=\frac{N_{R}}{2 R} \cos p(\theta-\phi)
\end{aligned}\)

Note that this implies that \(\ N_{S}\) and \(\ N_{R}\) are the total number of turns on the rotor and stator. i.e.:

\(\ p \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} n_{S} R d \theta=N_{S}\)

Then the surface current densities are as we assumed above, with:

\(\ K_{S}=\frac{N_{S} I_{S}}{2 R} \quad K_{R}=\frac{N_{R} I_{R}}{2 R}\)

So far nothing is different, but with an assumed number of turns we can proceed to computing inductances. It is important to remember what these assumed winding distributions mean: they are the density of wires along the surface of the rotor and stator. A positive value implies a wire with sense in the +z direction, a negative value implies a wire with sense in the -z direction. That is, if terminal current for a winding is positive, current is in the +z direction if n is positive, in the -z direction if n is negative. In fact, such a winding would be made of elementary coils with one half (the negatively going half) separated from the other half (the positively going half) by a physical angle of π/p. So the flux linked by that elemental coil would be:

\(\ \Phi_{i}(\theta)=\int_{\theta-\pi / p}^{\theta} \mu_{0} H_{r}\left(\theta^{\prime}\right) \ell R d \theta^{\prime}\)

So, if only the stator winding is excited, radial magnetic field is:

\(\ H_{r}=-\frac{N_{S} I_{S}}{2 g p} \sin p \theta\)

and thus the elementary coil flux is:

\(\ \Phi_{i}(\theta)=\frac{\mu_{0} N_{S} I_{S} \ell R}{p^{2} g} \cos p \theta\)

Now, this is flux linked by an elementary coil. To get flux linked by a whole winding we must ‘add up’ the flux linkages of all of the elementary coils. In our continuous approximation to the real coil this is the same as integrating over the coil distribution:

\(\ \lambda_{S}=p \int_{-\frac{\pi}{2 p}}^{\frac{\pi}{2 p}} \Phi_{i}(\theta) n_{S}(\theta) R d \theta\)

This evaluates fairly easily to:

\(\ \lambda_{S}=\mu_{0} \frac{\pi}{4} \frac{\ell R N_{S}^{2}}{g p^{2}} I_{s}\)

which implies a self-inductance for the stator winding of:

\(\ L_{S}=\mu_{0} \frac{\pi}{4} \frac{\ell R N_{S}^{2}}{g p^{2}}\)

The same process can be used to find self-inductance of the rotor winding (with appropriate changes of spatial variables), and the answer is:

\(\ L_{R}=\mu_{0} \frac{\pi}{4} \frac{\ell R N_{R}^{2}}{g p^{2}}\)

To find the mutual inductance between the two windings, excite one and compute flux linked by the other. All of the expressions here can be used, and the answer is:

\(\ M(\phi)=\mu_{0} \frac{\pi}{4} \frac{\ell R N_{S} N_{R}}{g p^{2}} \cos p \phi\)

Now it is fairly easy to compute torque using conventional methods. Assuming both windings are excited, magnetic coenergy is:

\(\ W_{m}^{\prime}=\frac{1}{2} L_{S} I_{S}^{2}+\frac{1}{2} L_{R} I_{R}^{2}+M(\phi) I_{S} I_{R}\)

and then torque is:

\(\ T=\frac{\partial W_{m}^{\prime}}{\partial \phi}=-\mu_{0} \frac{\pi}{4} \frac{\ell R N_{S} N_{R}}{g p} I_{S} I_{R} \sin p \phi\)

and then substituting for \(\ N_{S} I_{S}\) and \(\ N_{R} I_{R}\):

\(\ \begin{array}{l}
N_{S} I_{S}=2 R K_{S} \\
N_{R} I_{R}=2 R K_{R}
\end{array}\)

we get the same answer for torque as with the field approach:

\(\ T=2 \pi R^{2} \ell<\tau_{\theta}>=\frac{\mu_{0} \pi R^{3} \ell}{p g} K_{S} K_{R} \sin p \phi\)