12.5: Appendix 1- Air-Gap Winding Inductance

In this appendix we use a simple two-dimensional model to estimate the magnetic fields and then inductances of an air-gap winding. The principal limiting assumption here is that the winding is uniform in the zˆ direction, which means it is long in comparison with its radii. This is generally not true, nevertheless the answers we will get are not too far from being correct. The style of analysis used here can be carried into a three-dimensional, or quasi-three dimensional domain to get much more precise answers, at the expense of a very substantial increase in complexity.

The coordinate system to be used is shown in Figure 10. To maintain generality we have four radii: \(\ R_{i}\) and \(\ R_{s}\) are ferromagnetic boundaries, and would of course correspond with the machine shaft and the stator core. The winding itself is carried between radii \(\ R_{1}\) and \(\ R_{2}\), which correspond with radii \(\ R_{w i}\) and \(\ R_{w o}\) in the body of the text. It is assumed that the armature is carrying a current in the z- direction, and that this current is uniform in the radial dimension of the armature. If a single phase of the armature is carrying current, that current will be:

\(\ J_{z 0}=\frac{N_{a} I_{a}}{\frac{\theta_{w e}}{2}\left(R_{2}^{2}-R_{1}^{2}\right)}\)

over the annular wedge occupied by the phase. The resulting distribution can be fourier analyzed, and the n-th harmonic component of this will be (assuming the coordinate system has been chosen appropriately):

\(\ J_{z n}=\frac{4}{n \pi} J_{z 0} \sin n \frac{\theta_{w e}}{2}=\frac{4}{\pi} \frac{N_{a} I_{a}}{R_{2}^{2}-R_{1}^{2}} k_{w n}\)

where the n-th harmonic winding factor is:

\(\ k_{w n}=\frac{\sin n \frac{\theta_{w e}}{2}}{n \frac{\theta_{w e}}{2}}\)

and note that \(\ \theta_{w e}\) is the electrical winding angle:

\(\ \theta_{w e}=p \theta_{w}\)

Now, it is easiest to approach this problem using a vector potential. Since the divergence of flux density is zero, it is possible to let the magnetic flux density be represented by the curl of a vector potential:

\(\ \bar{B}=\nabla \times \bar{A}\)

Taking the curl of that:

\(\ \nabla \times(\nabla \times \bar{A})=\mu_{0} \bar{J}=\nabla \nabla \cdot \bar{A}-\nabla^{2} \bar{A}\)

and using the coulomb gage

\(\ \nabla \cdot \bar{A}=0\)

we have a reasonable tractable partial differential equation in the vector potential:

\(\ \nabla^{2} \bar{A}=-\mu_{0} \bar{J}\)

Now, since in our assumption there is only a z- directed component of \(\ \bar{J}\), we can use that one component, and in circular cylindrical coordinates that is:

\(\ \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial A_{z}}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2}}{\partial \theta^{2}} A_{z}=-\mu_{0} J_{z}\)

For this problem, all variables will be varying sinusoidally with angle, so we will assume that angular dependence \(\ e^{j k \theta}\). Thus:

\[\ \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial A_{z}}{\partial r}-\frac{k^{2}}{r^{2}} A_{z}=-\mu_{0} J_{z}\label{40} \]

This is a three-region problem. Note the regions as:

\(\ \begin{array}{ll}
\mathrm{i} & R_{i}<r<R_{1} \\
\mathrm{w} & R_{1}<r<R_{2} \\
\text { o } & R_{2}<r<R_{s}
\end{array}\)

For i and o, the current density is zero and an appropriate solution to (40) is:

\(\ A_{z}=A_{+} r^{k}+A_{-} r^{-k}\)

In the region of the winding, w, a particular solution must be used in addition to the homogeneous solution, and

\(\ A_{z}=A_{+} r^{k}+A_{-} r^{-k}+A_{p}\)

where, for \(\ k \neq 2\),

\(\ A_{p}=-\frac{\mu_{0} J_{z} r^{2}}{4-k^{2}}\)

or, if \(\ k=2\),

\(\ A_{p}=-\frac{\mu_{0} J_{z} r^{2}}{4}\left(\log r-\frac{1}{4}\right)\)

And, of course, the two pertinent components of the magnetic flux density are:

\(\ \begin{aligned}
B_{r} &=\frac{1}{r} \frac{\partial A_{z}}{\partial \theta} \\
B_{\theta} &=-\frac{\partial A_{z}}{\partial r}
\end{aligned}\)

Next, it is necessary to match boundary conditions. There are six free variables and correspondingly there must be six of these boundary conditions. They are the following:

• At the inner and outer magnetic boundaries, \(\ r=R_{i}\) and \(\ r=R_{s}\), the azimuthal magnetic field must vanish.
• At the inner and outer radii of the winding itself, \(\ r=R_{1}\) and \(\ r=R_{2}\), both radial and azimuthal magnetic field must be continuous.

These conditions may be summarized by:

\(\ \begin{aligned}
k A_{+}^{i} R_{i}^{k-1}-k A_{-}^{i} R_{i}^{-k-1} &=0 \\
k A_{+}^{o} R_{s}^{k-1}-k A_{-}^{o} R_{s}^{-k-1} &=0 \\
A_{+}^{w} R_{2}^{k-1}+A_{-}^{w} R_{2}^{-k-1}-\frac{\mu_{0} J_{z} R_{2}}{4-k^{2}} &=A_{+}^{o} R_{2}^{k-1}+A_{-}^{o} R_{2}^{-k-1} \\
-k A_{+}^{w} R_{2}^{k-1}+k A_{-}^{w} R_{2}^{-k-1}+\frac{2 \mu_{0} J_{z} R_{2}}{4-k^{2}} &=-k A_{+}^{o} R_{2}^{k-1}+k A_{-}^{o} R_{2}^{-k-1} \\
A_{+}^{w} R_{1}^{k-1}+A_{-}^{w} R_{1}^{-k-1}-\frac{\mu_{0} J_{z} R_{1}}{4-k^{2}} &=A_{+}^{i} R_{1}^{k-1}+A_{-}^{i} R_{1}^{-k-1} \\
-k A_{+}^{w} R_{1}^{k-1}+k A_{-}^{w} R_{1}^{-k-1}+\frac{2 \mu_{0} J_{z} R_{1}}{4-k^{2}} &=-k A_{+}^{i} R_{1}^{k-1}+k A_{-}^{i} R_{1}^{-k-1}
\end{aligned}\)

Note that we are carrying this out here only for the case of \(\ k \neq 2\). The \(\ k=2\) case may be obtained by substituting its particular solution in at the beginning or by using L’Hopital’s rule on the final solution. This set may be solved (it is a bit tedious but quite straightforward) to yield, for the winding region:

\(\ \begin{aligned}
A_{z}=& \frac{\mu_{0} J_{z}}{2 k}\left[\left(\frac{R_{s}^{2 k} R_{2}^{2-k}-R_{i}^{2 k} R_{1}^{2-k}}{(2-k)\left(R_{s}^{2 k}-R_{i}^{2 k}\right)}+\frac{R_{2}^{2+k}-R_{1}^{2+k}}{(2+k)\left(R_{s}^{2 k}-R_{i}^{2 k}\right)}\right) r^{k}\right.\\
&\left.+\left(\frac{R_{2}^{2-k}-R_{1}^{2-k}}{(2-k)\left(R_{i}^{-2 k}-R_{s}^{-2 k}\right)}+\frac{R_{s}^{-2 k} R_{2}^{2+k}-R_{i}^{-2 k} R_{1}^{2+k}}{(2+k)\left(R_{i}^{-2 k}-R_{s}^{-2 k}\right)}\right) r^{-k}-\frac{2 k}{4-k^{2}} r^{2}\right]
\end{aligned}\)

Now, the inductance linked by any single, full-pitched loop of wire located with one side at azimuthal position \(\ \theta\) and radius \(\ r\) is:

\(\ \lambda_{i}=2 l A_{z}(r, \theta)\)

To extend this to the whole winding, we integrate over the area of the winding the incremental flux linked by each element times the turns density. This is, for the n-th harmonic of flux linked:

\(\ \lambda_{n}=\frac{4 l k_{w n} N_{a}}{R_{2}^{2}-R_{1}^{2}} \int_{R_{1}}^{R_{2}} A_{z}(r) r d r\)

Making the appropriate substitutions for current into the expression for vector potential, this becomes:

\(\ \begin{aligned}
\lambda_{n}=& \frac{8}{\pi} \frac{\mu_{0} l k_{w n}^{2} N_{a}^{2} I_{a}}{k\left(R_{2}^{2}-R_{1}^{2}\right)^{2}}\left[\left(\frac{R_{s}^{2 k} R_{2}^{2-k}-R_{i}^{2 k} R_{1}^{2-k}}{(2-k)\left(R_{s}^{2 k}-R_{i}^{2 k}\right)}+\frac{R_{2}^{2+k}-R_{1}^{2+k}}{(2+k)\left(R_{2}^{2 k}-R_{i}^{2 k}\right)}\right) \frac{R_{2}^{k+2}-R_{1}^{k+2}}{k+2}\right.\\
&\left.+\left(\frac{R_{2}^{2-k}-R_{1}^{2-k}}{(2-k)\left(R_{i}^{-2 k}-R_{s}^{-2 k}\right)}+\frac{R_{s}^{-2 k} R_{2}^{2+k}-R_{i}^{-2 k} R_{1}^{2+k}}{(2+k)\left(R_{i}^{-2 k}-R_{s}^{-2 k}\right)}\right) \frac{R_{2}^{2-k}-R_{1}^{2-k}}{2-k}-\frac{2 k}{4-k^{2}} \frac{R_{2}^{4}-R_{1}^{4}}{4}\right]
\end{aligned}\)