12.3: Parameter Estimation

We are now at the point of estimating the major parameters of the motors. Because we have a number of different motor geometries to consider, but because they share parameters in not too orderly a fashion, this section will have a number of sub-parts. First, we calculate flux linkage, then reactance.

Flux Linkage

Given a machine which may be considered to be uniform in the axial direction, flux linked by a single, full-pitched coil which spans an angle from zero to \(\ \pi / p\), is:

\(\ \phi=\int_{0}^{\frac{\pi}{p}} B_{r} R l d \phi\)

where \(\ B_{r}\) is the radial flux through the coil. And, if Br is sinusoidally distributed this will have a peak value of

\(\ \phi_{p}=\frac{2 R l B_{r}}{p}\)

Now, if the actual winding has \(\ N_{a}\) turns, and using the pitch and breadth factors derived in Appendix 1, the total flux linked is simply:

\[\ \lambda_{f}=\frac{2 R l B_{1} N_{a} k_{w}}{p}\label{30} \]

where

\(\ \begin{aligned}
k_{w} &=k_{p} k_{b} \\
k_{p} &=\sin \frac{\alpha}{2} \\
k_{b} &=\frac{\sin m \frac{\gamma}{2}}{m \sin \frac{\gamma}{2}}
\end{aligned}\)

The angle \(\ \alpha\) is the pitch angle,

\(\ \alpha=2 \pi p \frac{N_{p}}{N_{s}}\)

where \(\ N_{p}\) is the coil span (in slots) and \(\ N_{s}\) is the total number of slots in the stator. The angle \(\ \gamma\) is the slot electrical angle:

\(\ \gamma=\frac{2 \pi p}{N_{s}}\)

Now, what remains to be found is the space fundamental magnetic flux density \(\ B_{1}\). In the third appendix it is shown that, for magnets in a surface-mount geometry, the magnetic field at the surface of the magnetic gap is:

\[\ B_{1}=\mu_{0} M_{1} k_{g}\label{31} \]

where the space-fundamental magnetization is:

\(\ M_{1}=\frac{B_{r}}{\mu_{0}} \frac{4}{\pi} \sin \frac{p \theta_{m}}{2}\)

where \(\ B_{r}\) is remanent flux density of the permanent magnets and \(\ \theta_{m}\) is the magnet angle.

and where the factor that describes the geometry of the magnetic gap depends on the case. For magnets inside and \(\ p \neq 1\),

\(\ k_{g}=\frac{R_{s}^{p-1}}{R_{s}^{2 p}-R_{i}^{2 p}}\left(\frac{p}{p+1}\left(R_{2}^{p+1}-R_{1}^{p+1}\right)+\frac{p}{p-1} R_{i}^{2 p}\left(R_{1}^{1-p}-R_{2}^{1-p}\right)\right)\)

For magnets inside and \(\ p=1\),

\(\ k_{g}=\frac{1}{R_{s}^{2}-R_{i}^{2}}\left(\frac{1}{2}\left(R_{2}^{2}-R_{1}^{2}\right)+R_{i}^{2} \log \frac{R_{2}}{R_{1}}\right)\)

For the case of magnets outside and \(\ p \neq 1\):

\(\ k_{g}=\frac{R_{i}^{p-1}}{R_{s}^{2 p}-R_{i}^{2 p}}\left(\frac{p}{p+1}\left(R_{2}^{p+1}-R_{1}^{p+1}\right)+\frac{p}{p-1} R_{s}^{2 p}\left(R_{1}^{1-p}-R_{2}^{1-p}\right)\right)\)

and for magnets outside and \(\ p=1\),

\(\ k_{g}=\frac{1}{R_{s}^{2}-R_{i}^{2}}\left(\frac{1}{2}\left(R_{2}^{2}-R_{1}^{2}\right)+R_{s}^{2} \log \frac{R_{2}}{R_{1}}\right)\)

Where \(\ R_{s}\) and \(\ R_{i}\) are the outer and inner magnetic boundaries, respectively, and \(\ R_{2}\) and \(\ R_{1}\) are the outer and inner boundaries of the magnets.

Note that for the case of a small gap, in which both the physical gap g and the magnet thickness \(\ h_{m}\) are both much less than rotor radius, it is straightforward to show that all of the above expressions approach what one would calculate using a simple, one-dimensional model for the permanent magnet:

\(\ k_{g} \rightarrow \frac{h_{m}}{g+h_{m}}\)

This is the whole story for the winding-in-slot, narrow air-gap, surface magnet machine. For airgap armature windings, it is necessary to take into account the radial dependence of the magnetic field.

Air-Gap Armature Windings

With no windings in slots, the conventional definition of winding factor becomes difficult to apply. If, however, each of the phase belts of the winding occupies an angular extent \(\ \theta_{w}\), then the equivalent to (31) is:

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

Next, assume that the “density” of conductors within each of the phase belts of the armature winding is uniform, so that the density of turns as a function of radius is:

\(\ N(r)=\frac{2 N_{a} r}{R_{w o}^{2}-R_{w i}^{2}}\)

This just expresses the fact that there is more azimuthal room at larger radii, so with uniform density the number of turns as a function of radius is linearly dependent on radius. Here, \(\ R_{w o}\) and \(\ R_{w i}\) are the outer and inner radii, respectively, of the winding.

Now it is possible to compute the flux linked due to a magnetic field distribution:

\(\ \lambda_{f}=\int_{R_{w i}}^{R_{w o}} \frac{2 l N_{a} k_{w} r}{p} \frac{2 r}{R_{w o}^{2}-R_{w i}^{2}} \mu_{0} H_{r}(r) d r\label{32}\)

Note the form of the magnetic field as a function of radius expressed in 80 and 81 of the second appendix. For the “winding outside” case it is:

\(\ H_{r}=A\left(r^{p-1}+R_{s}^{2 p} r^{-p-1}\right)\)

Then a winding with all its turns concentrated at the outer radius \(\ r=R_{w o}\) would link flux:

\(\ \lambda_{c}=\frac{2 l R_{w o} k_{w}}{p} \mu_{0} H_{r}\left(R_{w o}\right)=\frac{2 l R_{w o} k_{w}}{p} \mu_{0} A\left(R_{w o}^{p-1}+R_{s}^{2 p} R_{w o}^{-p-1}\right)\)

Carrying out (32), it is possible, then, to express the flux linked by a thick winding to the flux that would have been linked by a radially concentrated winding at its outer surface by:

\(\ k_{t}=\frac{\lambda_{f}}{\lambda_{c}}\)

where, for the winding outside, \(\ p \neq 2\) case:

\[\ k_{t}=\frac{2}{\left(1-x^{2}\right)\left(1+\xi^{2 p}\right)}\left(\frac{\left(1-x^{2+p}\right) \xi^{2 p}}{2+p}+\frac{1-x^{2-p}}{2-p}\right)\label{33} \]

where we have used the definitions \(\ \xi=R_{w o} / R_{s}\) and \(\ x=R_{w i} / R_{w o}\). In the case of winding outside, \(\ p=2\),

\[\ k_{t}=\frac{2}{\left(1-x^{2}\right)\left(1+\xi^{2 p}\right)}\left(\frac{\left(1-x^{4}\right) \xi^{4}}{4}-\log x\right)\label{34} \]

In a very similar way, we can define a winding factor for a thick winding in which the reference radius is at the inner surface. (Note: this is done because the inner surface of the inside winding is likely to be coincident with the inner ferromagnetic surface, as the outer surface of the outer winding ls likely to be coincident with the outer ferromagnetic surface). For \(\ p \neq 2\):

\[\ k_{t}=\frac{2 x^{-p}}{\left(1-x^{2}\right)\left(1+\eta^{2 p}\right)}\left(\frac{1-x^{2+p}}{2+p}+(\eta x)^{2 p} \frac{1-x^{2-p}}{2-p}\right)\label{35} \]

and for \(\ p=2\):

\[\ k_{t}=\frac{2 x^{-2}}{\left(1-x^{2}\right)\left(1+\eta^{2 p}\right)}\left(\frac{1-x^{4}}{4}-(\eta x)^{4} \log x\right)\label{36} \]

where \(\ \eta=R_{i} / R_{w i}\)

So, in summary, the flux linked by an air-gap armature is given by:

\[\ \lambda_{f}=\frac{2 R l B_{1} N_{a} k_{w} k_{t}}{p}\label{37} \]

where \(\ B_{1}\) is the flux density at the outer radius of the physical winding (for outside winding machines) or at the inner radius of the physical winding (for inside winding machines). Note that the additional factor \(\ k_{t}\) is a bit more than one (it approaches unity for thin windings), so that, for small pole numbers and windings that are not too thick, it is almost correct and in any case “conservative” to take it to be one.

Interior Magnet Motors

For the flux concentrating machine, it is possible to estimate air-gap flux density using a simple reluctance model.

The air- gap permeance of one pole piece is:

\(\ \wp_{a g}=\mu_{0} l \frac{R \theta_{p}}{g}\)

where \(\ \theta_{p}\) is the angular width of the pole piece.

And the incremental permeance of a magnet is:

\(\ \wp_{m}=\mu_{0} \frac{h_{m} l}{w_{m}}\)

The magnet sees a unit permeance consisting of its own permeance in series with one half of each of two pole pieces (in series):

\(\ \wp_{u}=\frac{\wp_{a g}}{\wp_{m}}=\frac{R \theta_{p}}{4 g} \frac{w_{m}}{h_{m}}\)

Magnetic flux density in the magnet is:

\(\ B_{m}=B_{0} \frac{\wp_{u}}{1+\wp_{u}}\)

And then flux density in the air gap is:

\(\ B_{g}=\frac{2 h_{m}}{R \theta_{p}} B_{m}=B_{0} \frac{2 h_{m} w_{m}}{4 g h_{m}+R \theta_{p} w_{m}}\)

The space fundamental of that can be written as:

\(\ B_{1}=\frac{4}{\pi} \sin \frac{p \theta_{p}}{2} B_{0} \frac{w_{m}}{2 g} \gamma_{m}\)

where we have introduced the shorthand:

\(\ \gamma_{m}=\frac{1}{1+\frac{w_{m}}{g} \frac{\theta_{p}}{4} \frac{R}{h_{m}}}\)

The flux linkage is then computed as before:

\[\ \lambda_{f}=\frac{2 R l B_{1} N_{a} k_{w}}{p}\label{38} \]

Winding Inductances

The next important set of parameters to compute are the d- and q- axis inductances of the machine. We will consider three separate cases, the winding-in-slot, surface magnet case, which is magnetically “round”, or non-salient, the air-gap winding case, and the flux concentrating case which is salient, or has different direct- and quadrature- axis inductances.

Surface Magnets, Windings in Slots

In this configuration there is no saliency, so that \(\ L_{d}=L_{q}\). There are two principal parts to inductance, the air-gap inductance and slot leakage inductance. Other components, including end turn leakage, may be important in some configurations, and they would be computed in the same way as for an induction machine. As is shown in the first Appendix, the fundamental part of air-gap inductance is:

\[\ L_{d 1}=\frac{q}{2} \frac{4}{\pi} \frac{\mu_{0} N_{a}^{2} k_{w}^{2} l R_{s}}{p^{2}\left(g+h_{w}\right)}\label{39} \]

Here, \(\ g\) is the magnetic gap, including the physical rotational gap and any magnet retaining means that might be used. \(\ h_{m}\) is the magnet thickness.

Since the magnet thickness is included in the air-gap, the air-gap permeance may not be very large, so that slot leakage inductance may be important. To estimate this, assume that the slot shape is rectangular, characterized by the following dimensions:

\(\ h_{s}\) height of the main portion of the slot

\(\ w_{s}\) width of the top of the main portion of the slot

\(\ h_{d}\) height of the slot depression

\(\ w_{d}\) slot depression opening

Of course not all slots are rectangular: in fact in most machines the slots are trapezoidal in shape to maintain teeth cross-sections that are radially uniform. However, only a very small error (a few percent) is incurred in calculating slot permeance if the slot is assumed to be rectangular and the top width is used (that is the width closest to the air-gap). Then the slot permeance is, per unit length:

\(\ \mathcal{P}=\mu_{0}\left(\frac{1}{3} \frac{h_{s}}{w_{s}}+\frac{h_{d}}{w_{d}}\right)\)

Assume for the rest of this discussion a standard winding, with \(\ m\) slots in each phase belt (this assumes, then, that the total number of slots is \(\ N_{s}=2 p q m\)), and each slot holds two halfcoils. (A half-coil is one side of a coil which, of course, is wound in two slots). If each coil has \(\ N_{c}\) turns (meaning \(\ N_{a}=2 p m N_{c}\)), then the contribution to phase self-inductance of one slot is, if both half-coils are from the same phase, \(\ 4 l \mathcal{P} N_{c}^{2}\). If the half-coils are from different phases, then the contribution to self inductance is \(\ l \mathcal{P} N_{c}^{2}\) and the magnitude of the contribution to mutual inductance is \(\ l \mathcal{P} N_{c}^{2}\). (Some caution is required here. For three phase windings the mutual inductance is negative, so are the senses of the currents in the two other phases, so the impact of “mutual leakage” is to increase the reactance. This will be true for other numbers of phases as well, even if the algebraic sign of the mutual leakage inductance is positive, in which case so will be the sense of the other- phase current.)

We will make two other assumptions here. The standard one is that the winding “coil throw”, or span between sides of a coil, is \(\ \frac{N_{s}}{2 p}-N_{s p}\). \(\ N_{s p}\) is the coil “short pitch”. The other is that each phase belt will overlap with, at most two other phases: the ones on either side in sequence. This last assumption is immediately true for three- phase windings (because there are only two other phases. It is also likely to be true for any reasonable number of phases.

Noting that each phase occupies \(\ 2 p\left(m-N_{s p}\right)\) slots with both coil halves in the same slot and \(\ 4 p N_{s p}\) slots in which one coil half shares a slot with a different phase, we can write down the two components of slot leakage inductance, self- and mutual:

\(\ \begin{aligned}
L_{a s} &=2 p l\left[\left(m-N_{s p}\right)\left(2 N_{c}\right)^{2}+2 N_{s p} N_{c}^{2}\right] \\
L_{a m} &=2 p l N_{s p} N_{c}^{2}
\end{aligned}\)

For a three- phase machine, then, the total slot leakage inductance is:

\(\ L_{a}=L_{a s}+L_{a m}=2 p l \mathcal{P} N_{c}^{2}\left(4 m-N_{s p}\right)\)

For a uniform, symmetric winding with an odd number of phases, it is possible to show that the effective slot leakage inductance is:

\(\ L_{a}=L_{a s}-2 L_{a m} \cos \frac{2 \pi}{q}\)

Total synchronous inductance is the sum of air-gap and leakage components: so far this is:

\(\ L_{d}=L_{d 1}+L_{a}\)

Air-Gap Armature Windings

It is shown in Appendix 2 that the inductance of a single-phase of an air-gap winding is:

\(\ L_{a}=\sum_{n} L_{n p}\)

where the harmonic components are:

\(\ \begin{aligned}
L_{k}=& \frac{8}{\pi} \frac{\mu_{0} l k_{w n}^{2} N_{a}^{2}}{k\left(1-x^{2}\right)^{2}}\left[\frac{\left(1-x^{2-k} \gamma^{2 k}\right)\left(1-x^{2+k}\right)}{\left(4-k^{2}\right)\left(1-\gamma^{2 k}\right)}\right.\\
&+\frac{\xi^{2 k}\left(1-x^{k+2}\right)^{2}}{(2+k)^{2}\left(1-\gamma^{2 k}\right)}+\frac{\xi^{-2 k}\left(1-x^{2-k}\right)^{2}}{(2-k)^{2}\left(\gamma^{-2 k}-1\right)} \\
&\left.+\frac{\left(1-\gamma^{-2 k} x^{2+k}\right)\left(1-x^{2-k}\right)}{\left(4-k^{2}\right)\left(\gamma^{-2 k}-1\right)}-\frac{k}{4-k^{2}} \frac{1-x^{2}}{2}\right]
\end{aligned}\)

where we have used the following shorthand coefficients:

\(\ \begin{aligned}
x &=\frac{R_{w i}}{R_{w o}} \\
\gamma &=\frac{R_{i}}{R_{s}} \\
\xi &=\frac{R_{w o}}{R_{s}}
\end{aligned}\)

This fits into the conventional inductance framework:

\(\ L_{n}=\frac{4}{\pi} \frac{\mu_{0} N_{a}^{2} R_{s} L k_{w n}^{2}}{N^{2} p^{2} g} k_{a}\)

if we assign the “thick armature” coefficient to be:

\(\ \begin{aligned}
k_{a}=& \frac{2 g k}{R_{w o}} \frac{1}{\left(1-x^{2}\right)^{2}}\left[\frac{\left(1-x^{2-k} \gamma^{2 k}\right)\left(1-x^{2+k}\right)}{\left(4-k^{2}\right)\left(1-\gamma^{2 k}\right)}\right.\\
&+\frac{\xi^{2 k}\left(1-x^{k+2}\right)^{2}}{(2+k)^{2}\left(1-\gamma^{2 k}\right)}+\frac{\xi^{-2 k}\left(1-x^{2-k}\right)^{2}}{(2-k)^{2}\left(\gamma^{-2 k}-1\right)} \\
&\left.+\frac{\left(1-\gamma^{-2 k} x^{2+k}\right)\left(1-x^{2-k}\right)}{\left(4-k^{2}\right)\left(\gamma^{-2 k}-1\right)}-\frac{k}{4-k^{2}} \frac{1-x^{2}}{2}\right]
\end{aligned}\)

and \(\ k=n p\) and \(\ g=R_{s}-R_{i}\) is the conventionally defined “air gap”. If the aspect ratio \(\ R_{i} / R_{s}\) is not too far from unity, neither is \(\ k_{a}\). In the case of \(\ p=2\), the fundamental component of \(\ k_{a}\) is:

\(\ k_{a}=\frac{2 g k}{R_{w o}} \frac{1}{\left(1-x^{2}\right)^{2}}\left[\frac{1-x^{4}}{8}-\frac{2 \gamma^{4}+x^{4}\left(1-\gamma^{4}\right)}{4\left(1-\gamma^{4}\right)} \log x+\frac{\gamma^{4}}{\xi^{4}\left(1-\gamma^{4}\right)}(\log x)^{2}+\frac{\xi^{4}\left(1-x^{4}\right)^{2}}{16\left(1-\gamma^{4}\right)}\right]\)

For a q-phase winding, a good approximation to the inductance is given by just the first space harmonic term, or:

\(\ L_{d}=\frac{q}{2} \frac{4}{\pi} \frac{\mu_{0} N_{a}^{2} R_{s} L k_{w n}^{2}}{n^{2} p^{2} g} k_{a}\)

Internal Magnet Motor

The permanent magnets will have an effect on reactance because the magnets are in the main flux path of the armature. Further, they affect direct and quadrature reactances differently, so that the machine will be salient. Actually, the effect on the direct axis will likely be greater, so that this type of machine will exhibit “negative” saliency: the quadrature axis reactance will be larger than the direct- axis reactance.

A full- pitch coil aligned with the direct axis of the machine would produce flux density:

\(\ B_{r}=\frac{\mu_{0} N_{a} I}{2 g\left(1+\frac{R \theta_{p}}{4 g} \frac{w_{m}}{h_{m}}\right)}\)

Note that only the pole area is carrying useful flux, so that the space fundamental of radial flux density is:

\(\ B_{1}=\frac{\mu_{0} N_{a} I}{2 g} \frac{4}{\pi} \frac{\sin \frac{p \theta_{m}}{2}}{1+\frac{w_{m}}{h_{m}} \frac{R \theta_{p}}{4 g}}\)

Then, since the flux linked by the winding is:

\(\ \lambda_{a}=\frac{2 R l N_{a} k_{w} B_{1}}{p}\)

The d- axis inductance, including mutual phase coupling, is (for a q- phase machine):

\(\ L_{d}=\frac{q}{2} \frac{4}{\pi} \frac{\mu_{0} N_{a}^{2} R l k_{w}^{2}}{p^{2} g} \gamma_{m} \sin \frac{p \theta_{p}}{2}\)

The quadrature axis is quite different. On that axis, the armature does not tend to push flux through the magnets, so they have only a minor effect. What effect they do have is due to the fact that the magnets produce a space in the active air- gap. Thus, while a full- pitch coil aligned with the quadrature axis will produce an air- gap flux density:

\(\ B_{r}=\frac{\mu_{0} N I}{g}\)

the space fundamental of that will be:

\(\ B_{1}=\frac{\mu_{0} N I}{g} \frac{4}{\pi}\left(1-\sin \frac{p \theta_{t}}{2}\right)\)

where \(\ \theta_{t}\) is the angular width taken out of the pole by the magnets.

So that the expression for quadrature axis inductance is:

\(\ L_{q}=\frac{q}{2} \frac{4}{\pi} \frac{\mu_{0} N_{a}^{2} R l k_{w}^{2}}{p^{2} g}\left(1-\sin \frac{p \theta_{t}}{2}\right)\)