Saturday, January 12, 2013

Again with the rotor bars

I had a big thing going last year about the rotor bars in an induction motor, and I still can't figure it out. But once in a while someone comes along and posts a comment which purports to set everything straight, as one of my readers did yesterday on this post. John wants me to analyze the motor as magnet-on-magnet action instead of rotor bars in a magnetic field.

It's a good theory but I don't buy it. John doesn't deal with the essential problem: how do you calculate  the torque? The problem is that ILxB actually gives you pretty much the right answer, but fails to answer the question I asked with this picture:

how do you get torque if the field lines curve away from the rotor bars, avoiding the copper almost entirely?

I was tempted to answer this question by suggesting that the field lines exert a torque not on the copper bars, but on the atomic-level circulating currents in the magnetic domains of the iron. We know that iron can amplify a magnetic field up to a thousand times; this can only mean that the atomic-level circulating currents on the surface of the iron are a thousand times greater than the macroscopic copper-wound currents that we can see. Does this explain the torque of a motor?

I don't think so. My problem with this explanation is all of a sudden, there's way too much torque. Because the standard macroscopic analysis gets me in the right ballpark. I don't want a correction that makes the torque a thousand times bigger. The problem with the standard analysis isn't the's that nagging question: don't the magnetic field lines curve away from the copper to stay inside the iron?

I want to show what I mean when I say the standard calculation gives the right answer (assuming the field passes through the copper bars, so I put some dimensions on my picture. With a 3-cm radius, the rotor shown here would look OK in an ordinary 1/2HP motor. It's not hard to see that there are about 10 square centimeters of rotor bar cross section, which would typically sustain a current of around 2000 Amps. A good strong magnetic field achievable with laminated iron would be around 1 Tesla (that's volt-seconds-per-meter-squared). Is this enough to calculate the torque? Just about...I still need to give you the length of the rotor bars. Let's call it 12 centimeters:

 (2000 Amps)(0.12 meters)(1 volt-sec/square meter) = 240 Newtons

To get the torque, we multiply by the radius (3 cm) which gives a torque of about 7 N-m. Now horsepower is torque times frequency (in radians per sec.) Remember that 3600 RPM is 377 radians per second...except the typical motor is four-pole, so you only get half of that...and also, the utilization factor of the copper bars is certainly under 50% with the sinusoidally varying currents and the phase shifts. Lumping it all together I get:

 (7.2 Newton-meters)(1/2 * 377 radians/sec)(1/2) = 650 Watts

which is one horsepower...just about where we want it.

Except for one problem: the magnetic field inside the copper bars is very low. Look at the picture again. The field goes into the iron. I could try and fix this by analyzing the atomic-level currents in the magnetic domanis (1000x greater than the copper bar current) that would really mess up my calculation. So I still really don't know what to do.

The irony is that I think I'm going to explain the great paradoxes in quantum mechanics but I still can't seem to figure out how a motor works.


John said...

The current does not need to interact with a surrounding magnetic field to generate a force.

A moving charge in a magnetic field does not generate a force. I know , I know we are told F=q(uxB), but what is qu? A current. Remember:
J = delxH.
Thus we can see that:
Were H is the field generated by the moving charge and B is the surrounding field density.

So in the rotor bars it isn't the electrons interacting with the external magnetic field, but the magnetic field they are generating that is interacting with the external field.

The magnetic field generated by a wire carrying a current is:
where r is the distance from the wire. So the force per unit length of wire is going to be:
F = ((delx(uI/2PIr))xB)dl .
Thus integrating over l you get:
F= L*uI*B/(2PIr).
This is why the formula: F = ILxB gives the right answer even though physically the electronics don't pass through field.

Marty Green said...

This is a very promising argument but I am unable to personally verify the integration. I agree there ought to be a way of analyzing it as a field-on-field interaction, but I still find it very odd that you should not get the same result by alternatively analyzing by ILxB. Shouldn't they both give the same answer?