Motor/Generator Analysis

Wouldn't that be quadrate subtraction, Jim?

Subtracting the resistance at 0 degrees from the impedance associated with the mismeasured inductance at an unknown phase angle...

That is,

Xt = 2 * pi * f * L (measured) Xl = sqrt( Xt^2 - R^2) L ( compensated ) = Xl / ( 2 * pi * f ) ?

Doug

Reply to
DGoncz
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I have found a candidate motor for use as a generator:

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I believe this motor has 8 poles, a synchronous speed of 900 rpm and full load slip of 12%. So it should be around 70-75% efficient, shouldn't it?

Would this be a suitable motor for further experiments with self-excited induction generators? It's used, but it's cheap, small, and seems efficient enough.

Doug

Reply to
DGoncz

The only ~$200 handheld LCR meter that doesn't make the assumption of lossless (pure) inductance or capacitance that I know of is the Extech model 380193 LCR Meter. I have one, and it works well. It measures L, C, or R at 120 Hz or 1000 Hz (but not R at DC), and for L and C also reports parasitic R.

The B+K model 875B LCR meter did not work for me because they only worked with very pure inductances. I have heard that the Wavetek meters also have the problem. The test is to take a relatively pure inductance, like one winding of a power transformer, and put a potentiometer in series. Does the reported inductance vary as the series resistance is increased? With the B+K, it just explodes, with the indicated inductance becoming a large factor bigger than the true value, so a 2-henry inductor was reported as 45 henries. Complete nonsense, rendering the meter useless.

Note that one-frequency and two-frequency LCR meters cannot detect self-capacitance in an inductor, or self-inductance in a capacitor. Only parasitic resistance can be detected.

Joe Gwinn

Reply to
Joseph Gwinn

My meter uses a test frequency of 100 Hz.

I measured the two windings and got R1 = 38.5 ohms, L1 = 0.149 H, R2 =

123.5 ohms, L2= 0.457 H.

So I got X1 = 94 ohm, X2 = 287 ohm.

And X1v = 85 ohm, X2v = 259 ohm, with sqrt(X1^2 - R1^2) etc... And L1v = 0.136 H, L2v = 0.413 H.

Doug

Reply to
DGoncz

Wouldn't that be quadrate subtraction, Jim?

Subtracting the resistance at 0 degrees from the impedance associated with the mismeasured inductance at an unknown phase angle...

That is,

Xt = 2 * pi * f * L (measured) Xl = sqrt( Xt^2 - R^2) L ( compensated ) = Xl / ( 2 * pi * f ) ?

*********************

sorry " quadrature addition" was just a careless reference to the fact that the impedances were in quadrature.

********************* Doug

My meter uses a test frequency of 100 Hz.

I measured the two windings and got R1 = 38.5 ohms, L1 = 0.149 H, R2 =

123.5 ohms, L2= 0.457 H.

So I got X1 = 94 ohm, X2 = 287 ohm.

And X1v = 85 ohm, X2v = 259 ohm, with sqrt(X1^2 - R1^2) etc... And L1v = 0.136 H, L2v = 0.413 H.

Doug

**************

This is an example of how easy it is to attribute spurious accuracy to inductance measurements made with an LCR meter. There is no way that the measurement is accurate to three significant figures and even the second figure is pretty dubious.

***************

I have found a candidate motor for use as a generator:

formatting link
I believe this motor has 8 poles, a synchronous speed of 900 rpm and full load slip of 12%. So it should be around 70-75% efficient, shouldn't it?

Would this be a suitable motor for further experiments with self-excited induction generators? It's used, but it's cheap, small, and seems efficient enough.

Doug

*****************

Certainly a more suitable than your existing 36 pole device but I'm still pretty doubtful. Small multipole induction motors suffer from low efficiency and comparatively high leakage inductance. Given a free choice I would go for a 2 or 4 pole motor with full load load slip of no more than 5%. Even with a motor as good as this there would still be a small chance that the rotor residual magnetism be too low to initiate self excitation.

However the advertised machine is a nice little motor at an attractive price Even if it doesn't work it's all useful experience and a handy motor to add to your stock.

Jim

Reply to
pentagrid

I do understand that L1v is accurate to about two digits. I can show the tolerance buildup if anyone cares to see it. For example 2 is an exact number, pi can be brought in to any precision, but L is, judging from what Jim has written, accurate to maybe 10%. So with a tolerance on each number, the tolerance or precision on the result can be had.

And L is of course, suspect as well.

However, I have computed the resonant capacitor for the 1 winding as

  1. blah blah blah microfarads, and understand that I can pick out 50 ufd of capacitor and even measure the cap and still be off by 10%.

But I did include an extra digit. So sorry. Mea culpa.

Correct me if I am wrong, but it seems, looking at a Bode plot, that generator operation should be arranged at the nose of the curve, the inflection point between positive and negative curvature, where bandwidth is measured. Yes? Behavior is linear there.

Doug

Reply to
DGoncz

If you want to call RF/AF bridges crude. Just not uP I guess. Some are. Owner and long time user of low tech to high tech versions.

Martin

Reply to
lionslair at consolidated dot

snip

My comments referred only to LCR "METERS". A good RF/AF Bridge could well handle the highlighted problems.

Jim

Reply to
pentagrid

OK, I went to my hoarding/clutter group tonght and put a deadline on the motor/generator project.

Basically, what I am hearing is:

Slow motors don't generate well. Small motors don't generate well. Both, because the poles are small.

Well, this is a small, slow motor, so I am going to try to see if the Q at the nose of the curve is greater than one with math and measurement. And I will try to make it work, but at some point I am going to have to let this one go. Pity.

Doug

Reply to
DGoncz

Well, I received two motors and eleven capacitors from Surplus Center.

I used a short 1/8 pipe nipple to join the motors and added a handle made from another nipple and a bit of 1/4-20 threade rod between the mounting brackets. I ran another rod through the hollow shafts and added finishing washers on the ends to keep the rod concentric. I topped the rod with cap nuts. The motors turn together with a pronounced squeal.

So now I am sitting here with a $143 cap sub box I don't dare use because another is coming UPS from ebay today. If that box tests out, I'll return the expensive one ordered in haste. I plan to run the motors together and use various capacitances on the driven motor with the oscilloscope to look for generator action, one value at a time.

Doug

Reply to
DGoncz

I got the capacitance sub box from ebay! Lenny, our UPS driver, brought it by at 5:37 PM.

The sub box checked out with a few slightly unreliable switches.

With the sub box in place the motor/generator pair produced an odd looking wave on the scope. With more capacitance, the waveform adopted a more nearly sine wave form. There was a very small change in amplitude.

I need to add more of the capacitors. I think I have to wire up a 10,

20, 20 uf sub box.

Doug

Reply to
DGoncz

The waveform produced by the motor/generator spans 15 milliseconds.

That's 67 Hz.

I was expecting a slower wave, a lower frequency. But of course! The motor isn't running under a full load, so the speed is closer to the synchronous speed. Still, the output should be a lower frequency than the input.

Doug

Reply to
DGoncz

OK, the waveform might span 16, 17, or 18 microseconds. That would be about right.

I have found something quite heartening, an easy DOS program to analyze RCL/LCR parallel tanks:

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"Introduction C is in parallel with L and R in series. Z is the impedance measured across C. This deceptively simple circuit is found in narrow-band tuned amplifiers, wideband video amplifiers, impedance-matching L-networks and in filters, etc. When R is less than 1.554*Sqrt(L/C) there is a hump, Zmax, in the Z vs frequency response. As R decreases, Zmax increases and moves to higher frequencies. When R > Sqrt(L/C) the angle of Z is always -ve and unity power, commonly defined as the resonant condition, does not occur at any frequency. For smaller values of R the angle of Z is +ve at low frequencies and passes through zero (unity power factor) on the LF side of the frequency of Zmax which is itself lower than the LC series resonant frequency. At much smaller values of R, giving high Q, all three frequencies converge on a common value."

When I try my values for R, L, and C, the program says, in effect, "No resonance".

Still with 30 ufd across the main winding, 1 VAC is present, and a mighty clean sine wave at that.

I have ordered a 1 pole, 12 position switch and will add 10 ufd per position. It's a shorting switch. I may build a second switch up with

50 ufd per position. We will see....

Doug

Reply to
DGoncz

Long range diagnosis is pretty dangerous but the following comments may help.

The fact that there is any output means the rotor is retaining significant residual magnetism. The amount depends on its previous history. It could be maximised by temporarily DC energising the stator with the rotor stationary. Use about twice the rated full load current.

With output generated by residual magnetism the output should rise roughly directly proportional to speed and be at synchronous frequency - it's a "permanent" magnet alternator.

Regenerative generation is signalled by a much more rapid rise in output with speed as stator resonance is approached.

Significant regenerative generation can only occur if the stator resonance is of high enough Q. The necesary minimum Q depends mainly on motor efficiency. As a pure guess I would expect Q>2 to be essential and it may need to be as high as 5.

Jim

Reply to
pentagrid

Thanks, Jim, that's enormously helpful. I am in search of Q, be it with this motor or another.

I have computed with

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using input values 136000 microhenries, 51000000 picofarads, 23 ohms, and 0.00006 MHz,

that there is "no impedance hump".

Can you or any reader verify this by running the program?

Can anyone explain why, since R < 1.554 * sqrt (L/C), there is no impedance hump? I must be missing the fine print....

Doug

Reply to
DGoncz
[ ... ]

Hmm ... consider the following:

1) One four-pole 12 position switch

2) One 10 uF capacitor.

3) Two 20 uF capacitors.

4) one 40 uF capacitor

Wire switch so in the following positions you have the following caps connected:

0 None

1 Single 10 uF

2 Single 20 uf

3 One 20 uF and one 10 uF in parallel (30 uf total).

4) Two 20 uF in parallel (40uF total)

5 Two 20 uF and one 10uF in parallel (50 uF total) or One 40 uf and one 10uf capacitor in parallel (50 uF total)

6 One 40 uF and one 20 uF in parallel (60 uF total)

7 One 40 uF and one 20 uF and one 10 uf in parallel (70 uF total)

8 One 40 uf and two 20 uf in parallel (80 uF total)

9 One 40 uF, two 20 uF and one 10 uF in parallel (90 uf total

So -- you have 0-90 uF in 10 uF steps from four capacitors. This is pretty much how capacitor decade boxes are made.

If you want to extend the range a bit, add another 20 uf (and another deck) and you can get up to 110 uF from 12 steps.

Essentially, each capacitor has its own deck, and which are connected in is determined by the wiring to the terminals of the switch.

I would suggest avoiding the switching while it is powered, as it will burn the contacts each time you switch.

An alternative way would be to install one toggle switch for each capacitor, and simply sum the values of the switches which are on.

Enjoy, DoN.

Reply to
DoN. Nichols

I do understand that halving the resistance requires halving the number of turns and this could change

R < 1.554 * sqrt(L/C)

Doug

Reply to
DGoncz

You're being seduced by illusory accuracy again! With iron cored inductors you are working with only reasonable approximations to the behaviour of a complex component and this is compounded by pretty dubious measurement accuracy.

Your DC value of R assumes that this is the only loss mechanism and totally ignores the shunt losses arising from eddy currents and the iron circuit losses.

In the same way as Q=(rootL/C)/R for series losses Q=r/(rootL/C) is for shunt losses where r is the effective value of the shunt loss component. The combined effect of both types of losses must be taken into account.

With the setup you have there is no method of a accurately measuring the separate or combined losses and, since you are only taking part of the series loss into account for your Q calculation, your result is necessarily pretty optimistic.

Jim

Reply to
pentagrid

Even if I run the resistance down near zero, the LCR.EXE program still says "No impedance hump."

Is the program broken? I doubt it.

Am I doing something wrong? I don't know.

L = 136000 uH, C = 51000000 pf, R = 23 ohms, f = 0.00006 MHz.

Doug

Reply to
DGoncz

That's a good design, Don.

Doug

Reply to
DGoncz

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