Motor/Generator Analysis

Reactance is at a right angle to resistance so it is incorrect to add them arithmetically. This is easy to accomodate in MathCAD since it can do arithmetic with complex variables. Just multiply your X terms by i which MathCAD understands to be the complex operator sqrt(-1). You can direct MathCAD to use j instead of i since j is more commonly used in electrical engineering. Using the j convention, an R in series with an X has impedance R + j*X. In polar notation, this would be a Z with an angle theta. Z is the magnitude, volts/amps and theta is the phase angle between voltage and current.

Sample MCad sheet available if you'd like. Email me -- and tell me which version of MCad you're running.

Doug, you don't simply add reactance and resistance, since they are orthogonal quanities (at right angles to one another)..

To compute Impedance, Z, simply use the same formula as you would use for computing the hypotenuse of a right triangle (The Pythagorean Thorem). In this case:

Z = SQRT(Resistance Squared + Reactance Squared).

A simple empirical method of determining the impedance is simply to connect a variable resistor (potentiometer) in series with the generator, apply a 60-Hz AC voltage (say 24-Volts), then adjust the variable resistor until the AC voltage across it is equal to the AC voltage across the generator windings. You can then measure the resistance of the adjustable resistor with a simple ohm meter and its value will be numerically equal to the impedance of the generator, since the voltage drop across each will be equal.

E = IZ = IR, where the AC current passing though both the generator and the variable resistor (when connected in series) are of course equal.

Harry C.

Doug, as a follow on, it is important to note that the current draw of a motor is typicallly stated for its maximum horsepower load, hence tells you almost nothing about its impedance/reactance.

Harry C.

Thanks, Harry, that was what I meant when I wrote

sqrt( R1^2 + X1^2)

You've confirmed that these are vector quantities.

I have a calibrator output on my 'scope that is a 60 Hz *square wave* but I'd need a sine wave to use the potentiometer method, right?

Doug

Doug, since reactance varies with frequency you really need a 60-hz sinewave source (something like a doorbell transformer or toy train transformer should do nicely).

A pulse or square wave contains many higher frequency harmonics which would confuse the measurement.

Hope this helps, Harry C.

I have ordered two 10-1134 motors and eleven 5 mfd capacitors from Surplus Center. I have ordered an 11.1111 mfd 50 VDC cap sub box from Electronic Parts. I intend to mate the hanger threads on the end bells of the motors with a close nipple and install a threaded rod and cap nuts to join the shafts. One motor will be the prime mover, and the other will be the test generator.

I have written to Terry Given to see if he will recruit me into the IEEE.

I have ordered an LCR meter and a less expensive cap sub box that should be rated 200 VDC instead of 50 VDC.

Doug

The LCR meter is on the way from a private ebay seller overseas by air mail. The cap sub box has not been shipped yet. The motors are on their way by Parcel Post.

Doug

snipped-for-privacy@aol.com wrote in sci.electronics.design and rec.crafts.metalworking:

The LCR meter has arrived from Hong Kong.

The inductance of the motor coils in parallel is 112 microhenries.

The resonant capacitance is (confirm?) 62 microfarads.

There are 36 poles on this split-capacitor motor.

The synchronous speed would be (confirm?) 400 rpm.

The motor runs at 225 (rated) rpm.

The motor runs with 175/400 = 44% slip.

I need to run the motor at 400+44% = 576 rpm.

I have to do that because of the impedance protection, right?

So the efficiency will indeed be low.

At 90 pedal rpm, with my existing cog, I will need a

x * 90 / 8 = 576; x = 576 * 8 / 90 = 51 tooth cog, which is just what I have.

However, if I splice two motors together at 225 rpm, I will have to recompute.

I recall that damping reduces the apparent frequency of an impulse driven resonant system, and wonder if the substantial resistance of this impedance protected motor will reduce the continuously driven resonant frequency, or whether my recollection only applies to impulse driven resonant systems.

Yours,

Doug Goncz Replikon Research Seven Corners, VA 22044-0394

(see previous thread in Google; it is likely expired in your reader.)

Doug

I've picked up this thread late so I've probably missed important bits. However the following comments may be useful.

If I've understood the post correctly you are aiming to use an impedance protected 36 pole motor as a self excited induction generator.

Self excited induction generators rely on the tiny residual pattern of magnetisation of the rotor being reinforced by the current flowing in the near resonant stator winding circuit. It has to be operating close to resonance for the current build up to be large enough to reinforce the rotor field pattern. It has to be on the capacitative side of resonance to permit the phase angle of the stator current to reinforce the rotor field pattern.

It is a positive feedback regenerative system and on a large efficient motor the output can build up to far beyond its rated motor power until limited by magnetic saturation. This effect is sometimes used for regenerative braking of single and three phase motors and can result in a spectacularly short stopping time.

With a care and control of speed, self excited induction generator systems are possible but they're pretty touchy devices. If you're unlucky with the the rotor iron they may not retain enough initial magnetism to enable the output to build up (manufacturers strive to reduce this because it degrades the efficiency when used as a motor) Also it must use a reasonably efficient motor for the magnetic feedback to exceed the system losses.

Efficiency is your major problem. An impedance protected motor means a motor with deliberately large leakage inductance so that the impedance of this inductance limits the current that flows when the motor is stalled or overloaded. With limited stalled current the starting torque (already poor because it is a capacitor run machine) has to be boosted by the use of a high resistance rotor and this results in your observed very high slip speed. Even if there were no other losses of any kind the motor efficiency could not be any better than the % synchronous speed - 56%. With other losses taken into account the motor efficiency is probably no better than 40%.

With the uH to mH correction your sums are OK but this level of efficiency is too low for a succesful induction generator.

Jim

I find the rotor has 48 conductive bars. I could have this wrong.

By substituting the rotor, could I increase the efficiency of this motor? That might be much easier than rewinding. I have removed and replaced rotors from shafts before. It seems any rotor with the right number of bars and less than or equal to the linear size, with the right or smaller bore, would do to make *some* improvement. I'm guessing I'd want a rotor with fewer bars, but I don't know the physics.

Can any reader show me the math relating 60Hz, 36 poles, 400 rpm, 48 bars, and 225 rpm?

I don't get it. I take 7200 / 32 = 225. To get it, I need to go to Falls Church's library, the Mary Riley Stiles library. There are not one but two editions of Audel's there.

Doug

I find the following encouraging note at:

"As seen in the Table above, smaller motors and lower-speed motors typically have higher relative slip. However, high-slip large motors and low-slip small motors are also available."

I need a low-slip small motor.

Doug

The precise number of bars in the rotor makes little difference to motor efficiency and bears no particular relation to motor rated or synchronous speed. The only thing that matters is the inductance to resistance ratio of the effective shorted turn that surrounds the rotor iron circuit i.e. the L/R time constant.

If there a lot of iron but not much copper/aluminium, the resistive component is high and this is generally called a high resistance rotor. This gives good starting torque but lousy efficiency because of the extra power dissipated in the rotor. This results in full load speed well below synchronous speed and it is this speed difference that is responsible for the extra rotor dissipation.

Unless starting torque is a big problem, motors are designed with the maximum possible amout of copper/aluminium in the rotor. Only the minimum necesary amount iron is retained to enable it to handle the stator induced flux density. This results in very low slip frequency and high efficiency.

While fitting a low resistance rotor to your machine may give some efficiency improvement it is unlikely to be enough to get it to operate as useful induction generator. The main problem which will still remain is the extremely high leakage inductance which is inherent in the geometry of a small 36 pole stator.

Jim

I understand that leakage inductance is a defect in an otherwise perfect transformer, and it is clear why toroidal transformers would have low leakage inductance.

But "where" is the leakage inductance in a motor? Restated, in what way is a motor like a transformer?

I've looked all over the web and haven't found this information. I did find many documents referring to leakage inductance in motors.

I've only put a few hundred dollars into this generator. Perhaps I should let go of it. I have a DC generator that has produced loads of power with little drag.

Doug

In many ways.

You take a primary winding (stator) excite it with voltage to create magnetic fields. THe magnetic fields then (across the air gap) induce currents in the rotor (secondary). You have an air gapped transformer, with a shorted secondary.

Now, a 3 ph motor running as a rotary phase converter, is acting even more like a transformer. THe rotor currents (once the rotor is turning) create a field, that interacts (again across the airgap) with the other two stator windings, phase displacing the voltage you already have applied across them.

jk

An induction motor IS a transformer. The primary is the stator winding(s) and the secondary is the rotor bars with the short circuit acting as a zero resistance load.

The induced current in these rotor bars produces the magnetic field pattern which generates the rotary torque. Exactly similar forces occur in a heavily loaded or short circuited convential transformer but these are not normally noticed because the primary and secondary windings are firmly anchored in their correct positions.

Leakage inductance is best thought of in terms of flux which is generated by one winding and which closes in on itself without intersecting the field pattern of the associated secondary.

The leakage inductance of a motor winding is higher than that of a similarly rated convential transformer because of the much larger physical separation between primary and secondary. An example of leakage flux is the flux that passes from stator tooth tip to tooth tip without intersecting the rotor iron circuit.

The above ignores the effect of rotor rotation which changes both the frequency and the amplitude of the rotor currents. Nevertheless it's still a pretty good working guide to what's happening. A fuller explanation is of course possible but not really appropriate for a metalworking group.

Jim

To properly use an LCR meter, I believe the following would be appropriate.

1) Realize the meter measures total impedance and computes inductance using the impedance figure in the place of reactance. (I read this on a web page) 2) Measure the inductance and resistance, or capacitance and resistance. 3) Find the impedance associated with the inductance, using the meter's test frequency. 4) Recombine the impedance and resistance to find the reactance using Pythagoras. 5) Once again using the meter's test frequency, find the new inductance or capacitance associated with the reactance.

Right? Assuming the real component is strictly L-R or R-C.

Doug

Most LCR meters are pretty crude devices and simply indicate the scalar impedance of the test device. However the dial calibration is based on the assumption that the test device is a pure lossless L or C.

IF the loss component of an inductance is pure series R loss AND you know the value at the test frequency (which may be considerably higher than the DC value), as you have assumed, the true value of L can be arrived at by quadrature addition. Based on AC and DC LCR measurements on an air cored coil this method can give reasonably accurate results.

However, if the test piece is an iron cored component the measurement does not take into account the shunt losses (iron eddy currents, hysteresis etc.) and permeability variation both with frequency and flux density. These are all second order effects but mean that the "true" inductance of an iron cored device is a pretty variable quantity unless the measurement conditions are closely defined.

Fortunately, with power frequency electric motors, the low working frequency and the significantly air gapped iron circuit reduces the effect of these second order components. However measurement accuracies are likely to be limited to one, or at best, two significant figures,

Jim