Why and how exactly is electric motor torque limited?

Question

Inspired by this question and specifically this answer to it.

From my experience there's always some very specific limit to how much torque an electric motor can output. For example, an electric drill will often have a manually switched mechanical transmission - if one needs to drill some relatively weak materials (like wood) he will use the setting that outputs lets torque at higher RPM and if one want to drill steel or mix cement mortar he will use the settings for more torque at lower RPM.

The relation between RPM and torque is more or less clear if one imagines a set of two gears of different diameters and thinks that their radiuses are lever arms - three times more RPM automatically induces three times lower torque and vice versa.

But where does the limit to any given electric motor torque come from? Say I have some specific motor right now in front of me and it can output 40 Newton-meters at 500 RPM. Why exactly 40 and not more?

Answer 1

Let's imagine that we live in beautiful world of brush-less motors without friction.

Limitation must come from coil resistivity - the more power you pump into coils, the more losses you have due to their non-0 resistivity. At high RPM resistance also increases due to skin effect (reducing effective cross-section of the wire).

So if one want to have more force at higher RPM - he would need thicker wire in coils (copper, or even silver), and in case of high RPM - made of litz wire (this would be incredibly expensive).

But if you can make coils out of superconductor - problems solved, and your power is limited only by mechanics & coil commutation.

Answer 2

The small matter of the magnetic saturation of iron really ought to be mentioned in this discussion. In practise, the best iron cannot sustain a magnetic field of more than about 1 Tesla, or 1 volt-sec/m^2. We must then ask: how much current does it take to drive a field of that magnitude? That depends of course on the magnetic permeability of the iron and the length of the path, but even in the best case scenario of infinite permeability, the flux still has to cross the air gap between rotor and stator.

We can calculate the flux given the field by first dividing by 377 ohms (the impedance of free space) and then multiplying by the speed of light. You can verify that this comes to something on the order of 10^6 Amps/meter (the units check out nicely). If we take as a first guess that the air gap is one millimeter, it is evident that we need a current of at least 1000 amps (or ampere-turns if you like) to drive the field. In practise, this calls for a copper cross-section on the order of one square centimeter. Let's make it actually square so the width of the copper is one centimeter. Then let's make the width of the iron pole pieces the same.

Let's take a rotor circumference of one meter (radius approx. 6 inches) and alternate copper and iron so we get fifty pole pieces. Ideally, at each station we get the maximum field crossing the maximum amperage: (one Tesla) x (1000 amps) x 50 poles. You can verify that this comes to 50 000 Joules/m^2. That's not a torque, and that's not even a force yet, because we haven't accounted for the length of the rotor. Let's take a half meter: that gives a force of 25 kJoules/meter. It's a "force" because it's the force which is turning the motor; to convert it to a torque, we multiply by the radius of 0.15 meter. The torque is about 4 kNewton-meters.

How does this relate to a practical motor? We can get the horsepower by multiplying torque by rotational frequency. At 3600 rpm, the frequency in radians per second is of course 377. So we get a horsepower of 4000*377 = 1500 kW, or about 2000 hP.

This is off by a factor of about ten for a typical industrial motor of this size, which would likely be closer to 200 hP. But it puts you in the general ballpark. The only factor which is totally made up is the 1-millimeter spacing I assumed for the air gap between rotor and stator.

Answer 3

My Wikipedia suggestion for this problem is Faraday's law of induction. They sum it up in pretty much a single quote.

The induced electromotive force (EMF) in any closed circuit is equal to the time rate of change of the magnetic flux through the circuit.

There are lots of technicalities of motors and generators, but they're not necessary for this problem. The fundamental principle is that there is a wire spinning while in a magnetic field. The EMF, I'll denote $V$ for voltage, is quantified as follows with $r$ being the rotating radius of the coil assuming it's rectangular (as well as rotating in the right direction), $l$ is the other dimension of the rectangular loop, $B$ is the magnetic field, $\omega$ is the speed of rotation.

$$V = B r l \omega$$

If any single one of these factors had unlimited potential to increase then a motor could deliver infinite voltage. Of course they are all limited. The most obvious way to scale up power is to make a bigger machine.

There is one missing piece, which is that EMF refers to the voltage that can be either produced or converted into a mechanical action. That does not say anything of current, so taken at face value, such a simple coil rotating in a constant magnetic field would allow infinite power conversion if there were infinite current. Current in any wire or bundle of wires, is, of course limited by resistive heating limits. You can go find plenty of information about these limits but I will not cover them here. Yes, it is possible to use superconducting wires for both the primary coils as well as the magnetic field generating coils, but they also do not allow infinite energy conversion, and yes, there are companies that sell these.

http://www.amsc.com/products/motorsgenerators/index.html

I'm not familiar enough with the technology to say for sure, but I believe that the problem is still resistive heating. Superconductors generate much less heat, but each unit of heat they produce is much more expensive to remove if it's a low temperature superconductor. The 2nd law of thermodynamics gives a direct penalty on a heat flux out of a refrigerated system.