Detecting Regenerated Voltages 11

[buzzp]

I would like to discuss the details of how a motor acts when a phase is lost and if votlage imbalance and single phase protection devices actually work. Here is what I know about the behavior of the motor:
If you lose a phase on a 3 phase motor, you are losing a pole. The speed of the motor drops, meanwhile the other pole is acting like a generator. What is the amplitude of the voltage generated? I think the voltage of the generated leg will be in phase as the 'lost'.
The voltage protection devices I am familiar with protect against single phase, low voltage, and voltage unbalance conditions. They assume that the generated voltage will not be the same amplitude as the line voltage. Is this a correct statement? What affect, if any, does multiple motors on the same line have on the ability of voltage protection device?
Thank you.

 

[electricpete]

I lied. I'm not gone yet. Still a few hours before my flight.

I should back off on my statement that "The rotor residual magnetism plays only a minor role in the steady-state excitation of the machine."

What I should say instead is: I don't think the induction generator can generate significant voltage without the caps. (analogous situation to open-circuited phase...no caps).

Here's what I think is going on in the standalone induction generator with caps attached.

Caps are selected to be in resonance with the motor magnetizing reactance. Rotor residual magnetism induces a small voltage in the magnetizing reactance. This small voltage induces a large circulating current within the loop formed by the magnetizing reactance and the cap (assuming single phase motor where both are connected line-to-ground...other connections must be considered separately). The voltage drop of that current flowing through either the magnetizing reactance or the cap (equal and opposite impedance) creates the high terminal voltage.

One thing that bothered me was the the busbar's link mentioned the induction generator maintained very close to 120v. That seems like quite a coincidence.

I'm curious what fraction the voltage would drop to if caps removed. I think it would be very low.

 

[gjones33]

To electricpete,

Pete, I just happened to have one of those single phase shaded pole motors they used to drive record players with handy and out of curiosity I hooked it up to my 'scope and spun the shaft by hand, I could easily obtain 1.6v p-p with it.

Now if a small motor spun by hand with no other input or connection to an external load other than an oscilloscope probe can produce an appreciable voltage then I'm sure you will be fully convinced that a larger motor hooked up to a supply can do it.

I dont think there is anymore I can say on this topic and I'll leave the last word with you.
Bon voyage.
Cheers,
G

 

[buzzp]

You guys have been busy! I feel lost now. But in reading through the posts I propose one thing: not knowing the internal connections of a motor that well let me propose a scenario and you guys tell me if I am missing something- say we have a single phase condition on a 30HP motor. The voltage protection is at the contactor (where monitored) and the single phase condition exists on the line side of the contactor. Wouldn't the internal connections of the motor produce a reading on the so called missing leg by means of the internal connections, ignoring regenerated voltages? Forgive me if this is too elementary, it is Monday morning though. Granted it wouldn't be the same magnitude as line voltage. Thank you.

 

[gjones33]

Hi buzzp,

In a word yes

Cheers,
G

 

[buzzp]

So what error does this produce when trying to measure regenerated voltages??

 

[gjones33]

Hi buzzp,

Can I ask a silly question but like you said it is Monday. Where are you going to measure your voltage between? The missing phase motor terminal and one of the other phases or between the missing phase terminal and neutral assuming a wye connected motor or, across the break in supply, assuming it's available?

In any case the voltage will depend on the load on the other two phases, the slip, etc. In other words it is not a reliable indication. Was this your point?

Cheers
G

 

[buzzp]

Yes, this was my point! I am glad someone caught it.

 

[jbartos]

Suggestion to gjones33 (Electrical) May 19, 2002 marked ///\\There will be a voltage close to the line voltage generated in the third phase whether it is wye or delta connected
///Please, could you prove this statement or refer to any literature?\\ and this is a function of the rotor magnetic field and the speed of the rotor not the current in the phases as you suggest.
///It can be proved that the degenerated delta to the two branches of windings have the current flowing from one of the remaining phase terminal to the other remaining phase terminal through the two stator motor windings; one winding is normal, e.g. having impedance Zab and the second winding is created by Zbc in series with Zca. Certainly, there will be some voltage generation due to the rotor BEMF; however, there will also be the voltage drop I x Zbc and I x Zca. Additionally, one can simplify the motor's loss of one phase by considering a delta connected transformer on its primary. The delta connected transformer on its primary will deliver more power in case of loss of one phase than the wye connected transformer on its primary. It needs to be recalled that the induction motor is somewhat a special case of the transformer. Whoever, can perform laboratory measurements will see the difference.\\\

 

[jbartos]

Suggestion to electricpete (Electrical) May 19, 2002 marked ///\\gjones
you say "the slip increases" while jbartos says that "we lose torque".
///Please, would you point to my posting where I stated "we lose torque."\\ I say you're both right.

 

[busbar]

buzzp— I think it's generally recognized that any (upstream) loss at motor terminals has to be reflected at the voltage-unbalance relay terminals. If they are not the same, all bets are off.

The advertisements for a number of voltage-sensing/phase-loss devices claim responding to about ±15% voltage imbalance. I may be misinterpreting the advertising claims but, that seems to be an ineffective level of protection if allowed to remain beyond a very short interval.

In many cases, it seems that a persistent voltage imbalance of less than 15% could yield unacceptable motor heating and consequent coil-to-ground and/or turn-to-turn insulation damage.

 

[buzzp]

Busbar- most unbalance relays I have seen usually have an adjustable unbalance setting from 5-15% or sometimes lower. The ones with fixed unbalance trip levels are usually set to 6% (the ones I am familiar with). They usually have a trip time of 2 seconds for unbalance. Also, unbalance percent does not have a sign attached to it.

 

[electricpete]

gjones -

I am fully convinced that if you spin your motor at full speed you may get a high voltage approaching full line voltage.

I am not at all convinced that this provides an explanation for the voltage generated on the open phase of a motor. (although as I've said many times I cannot provide any alternative explanation myself for this regenerated voltage that apparently does exist).

Your residual magnetism motor amounts to a syncronous permanent magnet generator. The voltage you generate will correspond to the speed of the rotor. [As an unrelated side discussion...it would be interesting to see if you could sustain your voltage under load. An experiement would be to short the stator windings and spin it.... then stop spinning and remove the short, spin again and see if voltage remains].

If you want to consider the rotor during the one-phase open case as a permanent magnet, then you would come to the conclusion that the generated voltage is "syncronous" to the rotor speed. But it cannot possibly be syncronous to the rotor speed (under load conditions) and also syncronous to the power frequency, since these will differ depending on slip. There is no way that we could get a third induced phase 120 degrees apart from the 2 applied power phase because the speed/frequency would be different... phase is undefined.

The solution to the above contradiction is that we cannot view the rotor as a simple permanent magnet.... the rotor field is not stationary with respect to the rotor... it moves at slip speed with respect to the rotor. More importantly the heart of the matter is that the rotor field is NOT the only field we need to consider. It is a FACT that in a normal-condition induction motor with leakage reactance neglected for simplicity, the air-gap flux is independent of the rotor flux. The reason is that a load-component of the stator current flows to create a field which is equal and opposite to the rotor field (as I have mentioned a few times ;-). Whether this cancellation holds when only 2 of 3 phases are energized (open-phased why motor) is open to debate.

Likewise there is no direct comparison between the open-phased motor and an induction generator. The induction generator draws it's excitation from either external supply or external capacitors. The open-phase motor cannot draw any excitation on the open phase.

One more distinction to draw...Your residual magnetism generator is also different than an induction generator.... one is induction and one is syncronous. I believe residual rotor magnetism plays an important role only in establishing the initial field of the standalone induction generator with caps. I don't believe residual rotor magnetism can possibly play a role during steady state under-load operation of an induction generator because the frequency of any static residual field would not match the terminal frequency.

 

[jbartos]

Suggestion: Reference:
M.G. Say "Alternating Current Machines," John Wiley & Sons, Inc., 1978 Paragraph Single-Phasing on page 331
It stresses, if single-phasing occurs, a loaded motor may stall because of disconnection of the stator terminal introduces a negative phase sequence torque. Figure 8.58 Normal and single-phasing performances show current versus load curves for three-phase and single-phase conditions and torque versus speed relationships for 3-phase and 1-phase conditions. Clearly, the 1-phase condition torque is about (1/sqrt3) x 3-phase torque and the single-phase rotor or stator current is approximately 2 times the rated 3-phase condition stator or rotor current at rated load 1.0 per unit. However, there are the motor stator winding connections not mentioned.

 

[jbartos]

Small encore: Above mentioned Reference Figure 8.58 shows curves from no load to 1.25 pu load and from 0 speed to 1.0 p.u. speed. This is somewhat useful to this thread since some postings deal with light shaft load conditions. E.g. 1-phase condition torque at zero speed is equal to zero and the three phase condition torque at zero speed is equal to 1.0 pu.
Three phase stator current at no load is about 0.3 pu and one phase condition stator current at no load is about 0.45 pu.