9.1: Relating Rating to Size

It is possible, even with the simple model we have developed so far, to establish a quantitative relationship between machine size and rating, depending (of course) on elements such as useful flux and surface current density. To start, note that the rating of a machine (motor or generator) is:

\(\ |P+j Q|=q V I\)

where \(\ q\) is the number of phases, \(\ V\) is the RMS voltage in each phase and \(\ I\) is the RMS current. To establish machine rating we must establish voltage and current, and we do these separately.

Voltage

Assume that our sinusoidal approximation for turns density is valid:

\(\ n_{a}(\theta)=\frac{N_{a}}{2 R} \cos p \theta\)

And suppose that working flux density is:

\(\ B_{r}(\theta)=B_{0} \sin p(\theta-\phi)\)

Now, to compute flux linked by the winding (and consequently to compute voltage), we first compute flux linked by an incremental coil:

\(\ \lambda_{i}(\theta)=\int_{\theta-\frac{\pi}{p}}^{\theta} \ell B_{r}\left(\theta^{\prime}\right) R d \theta^{\prime}\)

Then flux linked by the whole coil is:

\(\ \lambda_{a}=p \int_{-\frac{\pi}{2 p}}^{\frac{\pi}{2 p}} \lambda_{i}(\theta) n_{a}(\theta) R d \theta=\frac{\pi}{4} \frac{2 \ell R N_{a}}{p} B_{0} \cos p \phi\)

This is instantaneous flux linked when the rotor is at angle \(\ \phi\). If the machine is operating at some electrical frequency \(\ \omega\) with a phase angle so that \(\ p \phi=\omega t+\delta\), the RMS magnitude of terminal voltage is:

\(\ V_{a}=\frac{\omega}{p} \frac{\pi}{4} 2 \ell R N_{a} \frac{B_{0}}{\sqrt{2}}\)

Finally, note that the useful peak current density that can be used is limited by the fraction of machine periphery used for slots:

\(\ B_{0}=B_{s}\left(1-\lambda_{s}\right)\)

where \(\ B_{s}\) is the flux density in the teeth, limited by saturation of the magnetic material.

Current

The (RMS) magnitude of the current sheet produced by a current of (RMS) magnitude \(\ I\) is:

\(\ K_{z}=\frac{q}{2} \frac{N_{a} I}{2 R}\)

And then the current is, in terms of the current sheet magnitude:

\(\ I=2 R K_{z} \frac{2}{q N_{a}}\)

Note that the surface current density is, in terms of area current density \(\ J_{s}\), slot space factor \(\ \lambda_{s}\) and slot depth \(\ h_{s}\):

\(\ K_{z}=\lambda_{s} J_{s} h_{s}\)

This gives terminal current in terms of dimensions and useful current density:

\(\ I=\frac{4 R}{q N_{a}} \lambda_{s} h_{s} J_{s}\)

Rating

Assembling these expressions, machine rating becomes:

\(\ |P+j Q|=q V I=\frac{\omega}{p} 2 \pi R^{2} \ell \frac{B_{s}}{\sqrt{2}} \lambda_{s}\left(1-\lambda_{s}\right) h_{s} J_{s}\)

This expression is actually fairly easily interpreted. The product of slot factor times one minus slot factor optimizes rather quickly to \(\ 1 / 4\) (when \(\ \lambda_{s}=1\)). We could interpret this as:

\(\ |P=j Q|=A_{s} u_{s} \tau^{*}\)

where the interaction area is:

\(\ A_{s}=2 \pi R \ell\)

The surface velocity of interaction is:

\(\ u_{s}=\frac{\omega}{p} R=\Omega R\)

and the fragment of expression which “looks like” traction is:

\(\ \tau^{*}=h_{s} J_{s} \frac{B_{s}}{\sqrt{2}} \lambda_{s}\left(1-\lambda_{s}\right)\)

Note that this is not quite traction since the current and magnetic flux may not be ideally aligned, and this is why the expression incorporates reactive as well as real power.

This is not quite yet the whole story. The limit on \(\ B_{s}\) is easily understood to be caused by saturation of magnetic material. The other important element on shear stress density, \(\ h_{s} J_{s}\) is a little more involved.

We will do a more complete derivation of winding reactances shortly. Here, start by noting that the per-unit, or normalized synchronous reactance is:

\(\ x_{d}=X_{d} \frac{I}{V}=\frac{\mu_{0} R}{p g} \frac{\lambda_{s}}{1-\lambda_{s}} \sqrt{2} \frac{h_{s} J_{s}}{B_{s}}\)

While this may be somewhat interesting by itself, it becomes useful if we solve it for \(\ h_{s} J_{a}\):

\(\ h_{s} J_{a}=x_{d} g \frac{p\left(1-\lambda_{s}\right) B_{s}}{\mu_{0} R \lambda_{s} \sqrt{2}}\)

That is, if \(\ x_{d}\) is fixed, \(\ h_{s} J_{a}\) (and so power) are directly related to air- gap \(\ g\). Now, to get a limit on \(\ g\), we must answer the question of how far the field winding can “throw” effective air- gap flux? To understand this question, we must calculate the field current to produce rated voltage, no- load, and then the excess of field current required to accommodate load current.

Under rated operation, per- unit field voltage is:

\(\ e_{a f}^{2}=v^{2}+\left(x_{d} i\right)^{2}+2 x_{d} i \sin \psi\)

Or, if at rated conditions \(\ v\) and \(\ i\) are both unity (one per- unit), then

\(\ e_{a f}=\sqrt{1+x_{d}^{2}+2 x_{d} \sin \psi}\)

Thus, given a value for \(\ x_{d}\) and \(\ \psi\), per- unit internal voltage \(\ e_{a f}\) is also fixed. Then field current required can be calculated by first estimating field winding current for “no-load operation”.

\(\ B_{r}=\frac{\mu_{0} N_{f} I_{f n l}}{2 g p}\)

and rated field current is:

\(\ I_{f}=I_{f n l} e_{a f}\)

or, required rated field current is:

\(\ N_{f} I_{f}=\frac{2 g p\left(1-\lambda_{p}\right) B_{s}}{\mu_{0}} e_{a f}\)

Next, \(\ I_{f}\) can be related to a field current density:

\(\ N_{f} I_{f}=\frac{N_{R S}}{2} A_{R S} J_{f}\)

where \(\ N_{R S}\) is the number of rotor slots and the rotor slot area \(\ A_{R S}\) is

\(\ A_{R S}=w_{R} h_{R}\)

where \(\ h_{R}\) is rotor slot height and \(\ w_{R}\) is rotor slot width:

\(\ w_{R}=\frac{2 \pi R}{N_{R S}} \lambda_{R}\)

Then:

\(\ N_{f} I_{f}=\pi R \lambda_{R} h_{R} J_{f}\)

Now we have a value for air- gap \(\ g\):

\(\ g=\frac{2 \mu_{0} k_{f} R \lambda_{R} h_{R} J_{f}}{p\left(1-\lambda_{s}\right) B_{s} e_{a f}}\)

This then gives us useful armature surface current density:

\(\ h_{s} J_{s}=\sqrt{2} \frac{x_{d}}{e_{a f}} \frac{\lambda_{R}}{\lambda_{s}} h_{R} J_{f}\)

We will not have a lot more to say about this. Note that the ratio of \(\ x_{d} / e_{a f}\) can be quite small (if the per-unit reactance is small), will never be a very large number for any practical machine, and is generally less than one. As a practical matter it is unusual for the per-unit synchronous reatance of a machine to be larger than about 2 or 2.25 per-unit. What this tells us should be obvious: either the rotor or the stator of a machine can produce the dominant limitation on shear stress density (and so on rating). The best designs are “balanced”, with both limits being reached at the same time.