Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations GregLocock on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Detecting Regenerated Voltages 11

Status
Not open for further replies.

buzzp

Electrical
Nov 21, 2001
2,032
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.
 
Replies continue below

Recommended for you

I should add another question: Does the location of the single phase condition affect the ability of the voltage protection device, I mentioned, to detect the loss of the a phase?
 
You are right that speed is decreasing, but the reason is not that you are "losing a pole". (actually every pole has windings from all 3 phases.... deenergizing one phase will distort and weaken the field but not change the number of poles).

So the syncronous (no-load speed) of the motor will not decrease with loss of one phase. But the flux is lower so the torque developed at any speed is lower. With a lower torque-speed curve, the intersection with load torque-speed curve will be lower... and speed drops further below syncrounous.

Someone has recently posted here an analysis of the increase in current on remaining phases due to open circuit. Based on the assumption that the change in speed is relatively small, the power will be constant. Neglecting change in power factor and efficiency, we can calculate the required increase in current on the two remaining phases to deliver the same power as the case with no open circuit. I don't remember the exact answer.

Generated voltage on the open phase. That's a good question too. And will it depend on motor load? Hmm.



 
After thinking about it for five minutes, my GUESS is that any voltage generated on the open phase will be far less than the line voltage. Here is my reasoning.

Look at 2-pole motor with pole-phase groups a, b', c, a', b, c'.

Assume c phase is open... what will be the voltage induced in c?

The c group overlaps 120 mechanical degrees with a' and b'. It also overlaps 60 degrees with a and b.

Since this is now a single phase system the flux from a is equal and opposite magnitude to flux from b. Contributions from 120 degrees of a' and b' cancel each other. Contributions from 60 degrees of a and b cancel each other. Voltage induced in C should be zero. Similar logic would show voltage induced in c' should be zero.

There may be some assymeteries and in particular the slight tangential component of the flux in the airgap which will cause unequal contributions from a and b. But this will be a small effect resulting in an induced voltage much lower than line.

That's just a guess.



 
Let me make a summary of your first post and tell me if I am right: The speed of the motor will drop, even at no-load, with the loss of a phase voltage. This speed will be reduced even further of a fully loaded motor.
Second post: I agree that the loss of any phase results in a single phase power supply. I think the current increases by a factor of 1.73 (if I remember the previous post you referred too).
I agree with all of your statements I just could explain why. When I was a newbie to motor protection design, the company I worked for touted protection against regenerated voltages. I was told this is because newer devices (not new now) monitored for voltage unbalance and this is why regenerated voltages were not a problem. I never pursued the real answer much, until now. If what I was told was true, then if the device was monitoring for low voltage, for example, then why wouldn't the average voltage protect against regenerated voltages? Maybe they only monitored one phase? Thank you.
 
I believe the no-load speed of motor will NOT drop upon loss of one phase. The number of poles remains the same, even with loss of one phase.

The slip will increase for a given load.

I can't provide any info on the subtleties of motor protection for loss of phase. Also bear in mind my response that there is no substantial generated voltage is a guess. I haven't read anything on the subject or done any measurements... I could be way off base.


 
The motor will keep the voltage on the open phase at a fairly high level. It is usually high enough that normal undervoltage relays cannot be used, because the required setting would be too high for normal operation with occasional voltage sags. Some 3-phase relays operated on an average of the three phase voltages, and this won't work either usually. If protection is to be provided by voltage detection, the best bet is a true negative sequence voltage relay, but even these may not detect the open phase if there is mostly motor load. The more static load on the bus, the more unbalanced the voltages will be for an open phase, and the more likely that a voltage relay will detect the condition. The best protection is to monitor current unbalance or negative sequence current. An open phase will increase the current unbalance significantly (zero current in the open phase and 1.73 x normal current in the remaining two phases) and can be quite reliably detected.
 
Thanks jwerthman. Your statement about regenerated voltage is similar to what busbar said on the other thread and I'm sure it's correct.

Can anyone figure out where I went wrong in my reasoning in analysizng the flux linked to a given pole-phase group (above). It seems pretty straightforward to me.
 
When you single phase a motor the big effect is a loss of torque. In general, I believe that the motor will only produce about 57% of the 3-phase torque.

Anyway, for no load operation the motor will run at speed. Performance loaded depends on the load torque requirements (duh!). This means that the motor will run at whatever speed the (available 57%) torque curve intersects the load torque. In some cases the motor will not turn at all and the starter will trip on the LRA of the active phases (overload trip). In some cases it will run at reduced speed due to load (not loss of poles) but during this situation the active phases will have a great slip and the starter will trip on overload. Finally, in some cases the motor will actually run at speed, but this is not common since most motors are not that lightly loaded.

With respect to the generated voltage in the open phase, I need to think about that one....I am not certain right now what the magnitude would be and it is not clear to me where you would connect the relay to detect this voltage since the circuit could be open anywhere between the incoming power supply and the motor or the open could be in the motor winding itself.

I believe that the most reliable way to detect a single phase condition would be a current loss relay. This would actuate whether the problem was an open circuit anywhere downstream of the starter or a voltage loss upstream.


 
I think we are all in agreement that current monitoring will detect a single phase condition.
I have heard a lot of stories about voltage monitors being effective and not effective at detecting regenerated voltages. No one has yet been able to explain one way or the other. Again, does anyone know where experimental data can be obtained? I have been looking for this for quite sometime. Given the arguments both ways, I am not content to believe either way just because someone says it is true. I do know, that even large motor protection companies make this claim and I tend to agree with them at this point. Maybe one of you guys can do an experiment and see what happens, just kidding. Single phase conditions are generally caused by a blown fuse or utility issues. If a phase in the motor opens then there should be no voltage on the phase to ground (or phase to phase for that matter) outside of the motor, depending on where the open winding is located.
My concern is not with a no-load motor operation because this is not a real world scenario. Lightly loaded maybe. Any real data would be greatly appreciated. Thanks to all for your opinions.
 
buzzp, apologies for the slow followup. See:


Based in info in the Siemens paper, a {probably longer-term} 5% voltage unbalance calls for a ~¼ reduction in load to prevent premature failure. Estimate a level of protection—is 5% voltage unbalance {highest level in the Siemens graph} an OK starting point?

Some manufacturer’s literature contains statements like:
• “The relay releases when one phase-neutral voltage drops below 70% of the other phase-neutral voltages or when the phase sequence is wrong.”
• “Phase Loss: 18% Low Voltage in one phase”
• “Phase Loss: <75% of set point”
Maybe I’m misinterpreting their statements, but it seems like those numbers translate to several multiples of the chosen 5% level [and also reflected in IEEE guides/standards.]

But, some literature contains unabashed statements like, “Phase Monitor Relays protect against unbalanced voltages or single phasing regardless of any regenerative voltages” in claims for some socketed or din-rail modules. That is simply not always the case. Significant temperature drift and very low transient immunity are problems with a fair number of socketed/din-rail so-called protective devices. A cheesy ¾-turn ‘full-range’ adjustment does not give me any additional confidence. I am not saying that this type of relay is completely useless, but the “specs” {and lack of same} should be carefully read and understood. Be careful in looking for a device that promises to end to all your motor problems. The specifier needs to review claims and compare them to published numbers, to see that they mesh. I would consider reliance on voltage-based protection during motor running conditions as giving a false sense of security.

I believe that voltage-based protection can be effectively used, within defined limitations, to inhibit starting on one or a group of motors, but that a current-based device is best applied to individual motors to terminate running. The need should be based on the historic, statistical likelihood of phase loss/imbalance, the cost of new or repaired motors and probably most important, the cost of production outages. A $120 device should not be tasked as expected to furnish faithful burnout protection.

It the end, the answer must be, “what’s going to cost the least amount of money,” and be based on the long run; not just initial cost. Electrical protection is often described as a philosophy, so economic operation has many opinions. The objectionable degree of unbalance and the time it is allowed to persist should be carefully reviewed.


Minor related discussion in thread237-20355 thread238-20643
 
Electripete is correct on the speed and Rhatcher does a nice job explaining the reasons. Rhatcher gets a star!
 
Suggestions to the original posting &quot;buzzp (Electrical)May 8, 2002&quot; marked ///\\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.
///Or more poles or more exactly pole-pairs, depending how many pole-pairs the motor has.\\ The speed of the motor drops,
///The slip increases at the induction motor, the synchronous motor keeps its synchronous speed.\\ meanwhile the other pole is acting like a generator.
///Not necessarily, the delta connection will have the winding somewhat energized. The Wye winding will be disconnected at one star corner, therefore, there will be no noticeable current flowing outside of the motor terminal with loss of phase.\\ What is the amplitude of the voltage generated?
///At the wye connection, it will the wye leg Erms x sqrt2.\\ I think the voltage of the generated leg will be in phase as the 'lost'.
///On the Wye connected leg, the rms generated voltage will be about the same as the lost rms voltage across that leg.\\ 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.
///There are various principles on which the voltage protective devices operation is based on. Some of them use sophisticated computations of the voltage negative sequence.\\ Is this a correct statement? What affect, if any, does multiple motors on the same line have on the ability of voltage protection device?
///It depends on a protection device type, a principle of operation, and what kind of motors are on the same line.\\
 
When you operate a motor off a three pahse supply and then remove one phase, the torque potential is dramatically reduced, but provided that the motor does not stall, the non driven phase will act as a generator generating a voltage that is correctly phased. If the shaft load is very low, the gnerated voltage will be comparative in magnitude and phase to the missing phase. This renders voltage monitoring devices next to useless unless the sensitivity is high and then there are nuisance trips due to supply variations. As the shaft load increases, the generated voltage reduces and voltage monitoring devices can work.

If the loss of supply is well upstream from the motor / starter installation, the generated voltage can supply other electrical equipment on that supply. This can render current monitoring protection less effective also. If the electrical load is high, then the generated voltage can be expected to drop making the voltage sensor effective.
If you monitor the current vectors, you can discriminate between current drawn from the supply and current generated by the motor.

The two phase operation of a three phase motor is a known means of regenerating the absent third phase. Mark Empson
 
I am not concerned with line-ground readings because the motor could care less about this. Any device being used to protect a motor that monitors line-ground is a waste of money, unless it is a single phase motor. Everyone keeps talking about lightly loaded motors in this thread, my concern is with almost fully loaded or fully loaded motors. Also, I wanted to talk about a real single phase condition (total loss of phase) not a brown out condition, sag, dip, or anything along those lines as the voltage unbalance detection is suitable to handle this phenomenon with no problems.

Well thanks for all the posts, unfortunately I have not learned what I wanted to from this post. I will keep searching for some real data and will post it here when and if I find it. Thanks to all.
 
Hi Buzzp,

From my experience when a three phase motor running on full load looses a phase it will run with greatly increased current and slight loss of speed. The voltage generated in the third phase will be very close to the voltage in the live phases and will be in the correct phase, i.e 120 out of phase with the live phases. In this condition a great deal of heat is generated and the motor is running for a rewind. I appreciate this is an re-iteration of most of what has been said previously. I just thought I might simplify it it a bit.

Regards,
gjones33

 
If your concern is for motors that have a high loading, then a) the motor will begin to stall and draw a high current on the two remaining phases and b) the regenerated voltage on the open phase will be very low and a voltage imbalance relay will trip. Mark Empson
 
ok guys, I have to admit that I don't get it.....would marke and/or gjones33 please explain the how an induction motor that is operating in a single phase condition can generate voltage on the missing phase that is similar in magnitude to the missing phase and is 120 degrees out of phase with each of the active phases.

Also, for marke...please further the explanation by detailing the 'known means' that an induction motor operating in a single phase condition can be used to generate the third phase and drive three phase loads. If I am reading you correctly you could, for example, boost the 220V supply from your house to 460V, attach a 3-phase motor (single phased of course) to the transformer output, and then use that motor as a power supply for 460V 3-phase equipment.

PS: the only analytical approach to answering this question has come from electricpete (stars to you buddy on all posts in this thread). Unfortunately, his analysis is that it won't happen. I am hoping that you guys can clarify what he and I don't see.
 
Hello rhatcher and others.
First, let us consider the induction motor and how it works. With a three phase induction motor, there are effectively three sets of windings that are displaced by 120 degrees that are energised by the applied three phase. These windings are split into subwindings to create the extra poles for multipole machines. When the stator winding is energised with three phases, there is a rotating magnetic field generated spinning in one direction and at a speed determined by the supply frequency and number of poles. This rotating magnetic field is coupled to the short circuited winding in the rotor and induces a current to flow. The frequency of the current flowing in the rotor is equal to the difference between the stator magnetic field frequency and the rotor speed. The rotor current in turn creates a magnetic field that is equal in speed to the stator field speed. This is because the rotor field is spinning relative to the rotor and the rotor current frequency. Relative to the stator, it is the rotor current frequency plus the rotor speed.

If we consider a situation of a single phase supply (or two phase two wire) then we have two counter rotating magnetic fields in the stator. Both have equal magnitude and frequency but spin in opposite directions. A stationary rotor will develop equal torque in both directions so will not rotate. A spinning rotor will however synchronise onto the torque field spinning in the same direction. This is why we need to have a start winding on a single phase motor. The start winding creates a second magnetic component that is displaced and effectively enhances the rotating field in one direction.

If the motor is spinning at full speed, there is a rotating magnetic field in the rotor and that induces a voltage in the open circuit winding. The physical positioning of the winding establishes the 120 degree offset. If the slip is low, the voltage will be of the same order of magnitude as the missing phase.

In terms of the rotating two phase to three phase converter, yes, once ou have th machine up to full speed, you can actually draw three phase from a three phase motor driven by two phases, but you do need to get to full speed first. This can be done by using a single phase motor in conjunction wih the three phase motor. Obviously, the impedance is relatively high and the motor must be unloaded.
Mark Empson
 
Actually, an unloaded or lightly loaded three phase motor can be used as a crude 1-phase to 3-phase converter. This has been done on farms for decades. The voltages become more unbalanced as more 3-phase load is placed on the system, but it will work. If the motor is loaded very heavily, it will overheat from the unbalanced current, hence the need for unloaded or lightly loaded.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor