Single Phasing effect on rotation of 3 Phase Motor

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mull982

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What happens to the rotation of a three phase 480V motor when a single phasing condition occurs? I'm assuming that if the motor is running then it will continue running in the same direction until the overloads trip of the motor burns up.

What about when a 3-phase motor is starting in a single phased condition? What happens with the fields within the stators and rotor, and will the motor start turning? Some have said that that the motor may jockey back and forth and then may even turn in the opposite direction? Is this true?

I'm trying to get a theoretical explanation to get my head around this concept.
 
Find a reference on how a normal single phase induction motor works, because that will explain it better than I can. Then consider what the energized coils are doing in a three phase motor with one terminal disconnected. You will quickly see the similarities.

In the single phase motor, you have an alternating magnetic field that is flipping back and forth at the supply frequency. Such a field has no 'rotational' qualities, but like a single piston tugging on a flywheel, it will maintain an already existing rotation. If you take a common three phase induction motor, and supply single phase power to 2 terminals, and then mechanically spin the shaft in one direction or the other, then it will continue to rotate in the 'started' direction.

The jogging back and forth situation may have to do with the mechanical load acting as a spring, or it may be related to motor slotting effects; I have no good explanation for that.

-Jon
 
A three phase motor will continue running if a phase is lost. Most three phase overload relays are not intended to provide true single phase protection, so the most likely scenario is the motor will eventually overheat and fail.

A three phase motor will not start, on its own, with only single phase. However, if the motor shaft is "manually" spun it may begin to turn and accelerate slightly. If the motor shaft is turned backwards then the motor would start in reverse.
 
jim dungar said:
A three phase motor will continue running if a phase is lost. Most three phase overload relays are not intended to provide true single phase protection, so the most likely scenario is the motor will eventually overheat and fail.
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Shouldn't the overloads take out the motor due to the fact that the current draw in a single phase condition will be 1.73 times that of the normal 3-phase operating condition?
 
mull982 said:
Shouldn't the overloads take out the motor due to the fact that the current draw in a single phase condition will be 1.73 times that of the normal 3-phase operating condition?
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Yes, theoretically. Most motors are underloaded and do not run at their FLC, so it may take a while to have the overload shut them off on single phasing.
 
mull982 said:
Shouldn't the overloads take out the motor due to the fact that the current draw in a single phase condition will be 1.73 times that of the normal 3-phase operating condition?
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First, if the motor is not fully loaded, then the currents will not reach the 1.73 x FLC level. Second, many people do not size their overload protection correctly. they are often sized 1.15x too large. Third, a normal (Class 20) overload relay is calibrated to trip after 6x FLC for 20secs. The curves I have seen show it would take at least 12 min to trip at about 1.4x.

Taking these three factors into account is why I said that most OL relays are not intended to provide single phase protection. This is one reason that the NEC was changed, in the 60's, to require an overload relay in each conductor.
 
In trying to get my head around this concept and have a complete theoretical understanding I have drawn a few picutes for myself. (Attached)

The first firgure on the top of the drawing (I) is a copy of a phasor diagram from an article I was reading online about single phase motor theory. I was having a little trouble understanding where these two different phasors came from with a single phase source through a single coil? What do the two different vecors represent?

The second figure in the middle (II) is the waveform representation of these vectors. I'm confused however how to represent this waveform and do not know if one sine wave is represented as in the first waveform or if these vectors represent two different sine waves 180deg apart. Which waveform is the correct representation of the vecots on top?

Last I was trying to draw which direction the magnetic field would appear across a coil as represented in the circuit with a coil hooked to a single phase source. The 5 circuits on the bottom match the 5 vector diagrams on the top of the page. Did I draw the direction of the mag fild correctly? Where would the rotor be in these circuits? I am trying to visualize how this one coil produces a pulsing magnetic field, and what is going on at both the terminals of this coil (source terminals) as a result of the source (vectos)

Lastly once I get my head around this I am understanding that for a single phase situation these vectors are 180deg apart. What about a three phase motor where the vectors are origonally 120deg apart? Are the two phases (vectors) still 120deg apart after a single phase condtion persists or do they become 180deg apart?
 
Single phase motor theory is different than a three phase motor being supplied with a single phase voltage.
 
Your protection using just OL is rather simplified, as many motors, especially the larger Hp sizes have additional protection that would trip either on loss of phase, current imbalance, or motor temperature. As the motor size increases usually the level of protection also increases, which means many addtional tripping options would protect the motor from phase loss.

Phase loss is the ultimate in voltage unbalance, and although from a voltage perspective the motor could continue to run (with damage to insulation), the heating effect due increased current results in a current unbalance that is 6 to 10 times the voltage unbalance. A properly size OL would disconnect the motor fairly quickly. If running it would probably stall under load, and would probably not start. if it did start, the current drawn would be near locked rotor, and this would again allow the protection to open the circuit. insulation under this condition can fail within less than 90 secs.
 
I've also seen that a single phased motor will only be able to supply 58% or 1.73 times less power. I'm assuming that this is because the 1.73 factor for 3 phase power disapears. If this factor then disaperars then what causes the increased 1.73x current?
 
I once stopped a fractional horsepower motor by hand with a glove to prove that it was the motor and not a gearbox which I could turn by hand. It was single phasing as I thought.
 
jim dungar said:
Single phase motor theory is different than a three phase motor being supplied with a single phase voltage.
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Can you briefly explain this in terms of starting, and starting currents?
 
mull982 said:
I've also seen that a single phased motor will only be able to supply 58% or 1.73 times less power. I'm assuming that this is because the 1.73 factor for 3 phase power disapears. If this factor then disaperars then what causes the increased 1.73x current?
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It seems you are thinking of a single phase load across the open side of an open delta. One would think the two transformers would deliver 2/3 the power of a balanced, full delta. However, as a result of the PFs of the two transformers, the power is 2*VI*cos(30) which is 58 of the power rating of a full delta.
 
rattus said:
It seems you are thinking of a single phase load across the open side of an open delta. One would think the two transformers would deliver 2/3 the power of a balanced, full delta. However, as a result of the PFs of the two transformers, the power is 2*VI*cos(30) which is 58 of the power rating of a full delta.
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I have read that when a three phase motor single phases it may keep turning until it overheats, or the motor may stall because there is less power and torque delivered to the motor. I read that this reduced power and torque comes in the form of 1.73x less then what the motor was operating at 3-phase.

Revisiting my post with my attached sketches, I think I almost have my head around singe phase motor operation. The one thing that I just can seem to visualize however is how a single phase L-L voltage is represented vectorally. For instance in the first figure there are two different vectors. What are these vectors, are they the voltage at the source terminals? Since there is only a single phase then I have figured that there is only one sine wave answering my own question regarding waveforms from that post.

Can anyone help me visualize vectorally what is going on at the terminals of a L-L single phase source? I understand that these terminals are 180 deg apart but since theres only one sine wave I cant quite picture this vectorally.
 
Can anyone help me visualize vectorally what is going on at the terminals of a L-L single phase source? I understand that these terminals are 180 deg apart but since theres only one sine wave I cant quite picture this vectorally.
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If all you have is 2 conductors (i.e. the terminals of the motor) you can not have any phase difference between them. Would you say the two ends of a straight line are 180? apart?
 
480sparky said:
You can also try here.
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I like that article. A lot of the ones I have read don't do justice to the Negative Sequence currents issue causing additional heat. That is why you can smoke a motor on single phasing even if it is not fully loaded and the overload relay doesn't trip.
 
Ok I've kind of shifted topic here because I am now confused about how to conceptually view a single phase 2-wire source. I have seen preveious threads about this topic and still am not able to completely grasp what is going on.

In one of the threads I saw the following comment regarding the two legs of a single phase source:

"They both hit zero at the same time, and they both hit peak values at the same time. But one is hitting a positive peak at the same time the other is hitting a negative peak. In other words, when they hit zero together, one is about to go positive and the other is about to go negative. That is what is meant by saying they are in opposition".

Trying to understand this I have put together another drawing to help conceptually visualize this. Now I understand that since it is one wire it cannot be 180deg out of phase but why does it appear that way? How is the above comment explained in reference to terminals A and B on my drawing?

The only way I can figure to explain this comment is if we had simultaniously flowing sine wave voltage flowing in one direction for A and the other direction for B as shown in the figure in the middle of the page. This would then be represented by the figures I had posted before showing the two vectors. I am confused however because these diagrams show two vectors when in reality we should only have two phase vectors. Are these two vectors just representing what is going on when looking at terminals A and B? Why are they different if there is only one source, and no tapped nuetral point on the coil?
 
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