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I am trying to understand the efficiency of the following permanent magnet synchronous motor Motor Datasheet (note we are using the medium voltage model), and its relationship with the input voltages and currents. The application is an electric racecar.

I made the following graph from the data given in the sheet. The blue line is directly taken from the datasheet graph, and the red line is generated from the torque constant on the first page of the datasheet. I understand that the difference between these lines is due to magnetic saturation of the motor.

Current and torque curves

The graph on the right is what I understood to be the efficiency (the graph data / the ideal torque). At 120Nm of torque (~300A) the efficiency appears to be 75%. However, when looking at the datasheet's efficiency map, the efficiency at 120Nm could be as high as 94%.

Given the following equations, that does not make sense, unless the voltage vs rotational frequency graph curves in the opposite direction, enough to compensate.

Pin = Vin * Iin
Pout = Wm(Vin) * Tq(Iin)
Efficiency = Pout / Pin = torqueEfficiency * speedEfficiency

Additionally, I was wondering how the efficiency could be mathematically defined as a function of the input harmonics (we are wondering how to determine the ideal switching frequency of our inverter).

Am I fundamentally misunderstanding electric machines?

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  • \$\begingroup\$ consider the similarity between saturation and field weakening. \$\endgroup\$
    – user16324
    Oct 19, 2019 at 20:07
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    \$\begingroup\$ "Efficiency[%] 1.05" - that's either ridiculously low, or over-unity! The data in your link is more believable. \$\endgroup\$ Oct 20, 2019 at 10:50
  • \$\begingroup\$ That's my bad. The efficiency on that graph should not be measured in %. Yes, mathematically my equation results in an efficiency greater than 1, implying that there are other inefficiencies in the system. \$\endgroup\$ Oct 20, 2019 at 22:42

1 Answer 1

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Efficiency vs. Current:

Define an ideal motor as one that is 100% efficient. Then a real motor is an ideal motor wrapped in problems, like armature resistance, core losses, windage, friction, and other mechanical losses.

If \$k_m(i) = \frac{T_{out}}{i}\$ is the ideal motor's torque "constant" at any given current (I put "constant" in quotes, because here it depends on current), then by conservation of energy, it simply must be the reciprocal of the ideal motor's speed constant: \$\omega = \frac{V_a}{k_m(i)}\$. You can even, after an adventure in SI units, show that \$ \mathrm{\frac{N \cdot m}{A}} \$ has the same units as \$\mathrm{\frac{V \cdot s}{radians}}\$.

So it's certainly mathematically possible for the efficiency to be high at high currents, as long as the copper losses are low enough. If that's really true, I'd expect that a motor excited at a constant voltage would actually increase in speed at high torque. I find this highly counter-intuitive, but -- that's what drops out of the data sheet's claims.

Losses vs. Harmonics

I did not see anything in the data sheet that goes into this, although I may have overlooked this.

Yes, harmonics could be a problem, because some types of core losses increase with increasing frequency. You may need to just experiment with this -- if you have a high-enough power square wave generator, you could just hit the thing with the harmonics you expect and measure the power consumed (or measure the current and calculate the power).

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  • \$\begingroup\$ Efficiency and Current: Is that generally the case? Or is there an issue with the datasheet? \$\endgroup\$ Oct 20, 2019 at 22:52
  • \$\begingroup\$ How are you getting your torque vs. current, graph, when there's just a constant given in the data sheet? Or am I missing something? They just give torque & power vs. RPM; they don't say that they're holding current constant. \$\endgroup\$
    – TimWescott
    Oct 21, 2019 at 0:31
  • \$\begingroup\$ page 3 of the datasheet shows the torque vs. current graph \$\endgroup\$ Oct 21, 2019 at 21:54
  • \$\begingroup\$ D'oh! Either the data sheet lies, or when the motor is driven with a constant voltage, it speeds up under load. I want one. \$\endgroup\$
    – TimWescott
    Oct 21, 2019 at 22:08

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