Residual Magnetism in an Induction Motor 4

[rockman7892]

I have had many discusses lately about when a motor was capable of regeneration. Obviously when the motor is driven past synchronous speed it generates a voltage but what about when the motor is shut off?

How long does it take the magnetic field in the motor to decay? Will there always be some sort of residual magnetism in the motor even after sitting for a while?

 

[davidbeach]

What type of motor? Induction motors need excitation current that they can't supply themselves to produce any real power current. Current drops to zero very rapidly (2 cycles would be about the outer limit). Residual voltage (open terminal) will last longer, depends on lots of things.

Other types of motors are different.

 

[rockman7892]

I'm sorry I should have specified. I'm focusing more on induction motors.

As an example If I had a motor that had been running for a while and then turned the motor off and let it sit for an hour whould there be any residual magnetism in the motor. If I were to spin the shaft of this motor and record voltage on the three leads would I be able to see a small amount of voltage assuming there is some residual magnetism that exists in the core (stator or rotor?)

 

[waross]

Probably yes but not much. The case is similar to an alternator bootstrapping on residual magnetism and the resulting voltage. I would expect less than 5% of rated voltage. The residual voltage MAY build to a higher value if capacitors are connected to the motor.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jraef]

I agree with Bill. It's a trick used by some people to use AC motors on small (<10HP 3 phase or 3HP 1 phase) island-mode wind or hydro generators, but there are a lot of variables having to do with the motor design, i.e. permeability of the steel in the laminations etc. etc. They connect caps in a sort of a tank circuit to build on the residual magnetism in order to create enough excitation, but it only works on small motors, and not always.

Here is a decent representation of voltage decay in larger 3 phase motors. I used this when designing DC injection brakes and we later had empirical test data that was right on track with what he shows here.

http://www.tpub.com/content/UFGScriteria/ufc_3-520-01/ufc_3-520-010159.htm

"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[rockman7892]

I guess one example that comes to mind is a fan motor which is sitting idle and is being spun in the oppisite direction due to a draft from somewhere in the process. Although this motor is off (Assume no caps) is there a possibility that this motor would be regenerating a voltage back to the open contactor? If this voltage was present back at the contactor and the contactor was closed, could this present a case of closing in out of phase voltages?

I'm also assuming that this residual magnetisim in the motor will have some effect on the inrush characteristic of the motor. If a residual magnetic field exists in the motor then a larger reactive inrush will be needed to first dissipate this field before establishing a new one?

 

[davidbeach]

Residual magnetism/residual voltage doesn't have much to do with it. A back spinning motor has an extremely high slip, and effectively a negative rotor resistance, therefore very high starting current.

 

[waross]

Yes that happens. One of the first digital ammeters on the market had a peak lock function that locked on to the highest part cycle peak.
When a motor starts, there is a transient current in the first 1/4 cycle due to the residual magnetism. This is much greater than the familiar starting surge of several seconds. That first transient was too fast to be seen by analog meters. The new digital meter would lock on to the transient which was several times the inrush, but varied over about a 2:1 range depending on the point on wave that the contactor closed and the orientation of the residual magnetism.
I have found that a greater problem with back spinning fans is the extended starting time. I worked on an installation where the fans were in parallel banks. If one tripped out, and we attempted to restart it, the time to bring it to a stop and then accelerate it in the correct direction was so long that the heaters in the overload relays would glow red hot before they radiated enough heat to the bi-metal strips and tripped out.
The I²R of the extended starting surge was probably hundreds if not thousands of times greater than the first 1/4 cycle transient resulting from residual magnetism and regenerated voltages.
Thanks for the graph jraef.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[rockman7892]

Davidbeach

That is interesting what you said about the slip and negative rotor resistance. Where can I find more information on this phenonemon, I am very curious to learn more?

We had instances where a backward fan trying to start would trip the MCP breaker and I did not understnad why since these breakers are set high only to respond to short circuits. I kept thinking that LRC in a motor was the highest current a motor would ever draw and therfore it should not have an effect on tripping a MCP but appearently my thought process was incorrect, and there are other factors to take into consideration.

Waross

So you are saying that it is possible to see 5% of the motor voltage occur at the leads when the shaft is spun by hand and there is residual magnetism in the core? So for a 5kV motor is it possible to see 250V when performing this test.

You also mentioned that a backwards spinning fan can regen a voltage with residual magnetism and there is a possibility that this can contribute to a high startig current. How does this coencide with what Davidbeach described

 

[Skogsgurra]

We have had this discussions a few times. One that I remember was if it possible to find what way a motor will rotate when connected to a three-phase grid. One way of doing it is to rotate the rotor and watch the phase sequence of the induced voltage. I think that I attached a set of wave-forms showing what happens. But can't find them any more. So here goes again.

This is a 400 V motor that has been "nudged" manually. Speed is not more than a few turns/minute. Voltage not more than a few hundred millivolts. Should be a lot more at 1500 RPM. Probably around 10 - 20 V.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[waross]

I didn't explain myself well.
First I suspect that the breaker on the back spinning fan is tripping on thermal because of the duration of the starting current rather than on the instantaneous magnetic trip.
The point that I was trying to make was that the first quarter transient may be so high that any additional current contribution due to residual voltage may be insignificant.
Further to David's post, you may have noticed the length of time to bring a back spinning fan to a stop is often enough to use up most if not all of the I²T of the protection devices.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[rockman7892]

Waross

The breakers used for most of the starters are MCP breakers and have only a magnetic element and therefore no thermal element to trip the breaker as you suggested.

I am trying to understand the magnitude of current associaeted with what Davidbeach explained, since it must be a current magnitude only (not time related) that is tripping these MCP breakers.

I agree that overloads would trip based upon what you described with starting time, but do not see this happening on an MCP breaker.

 

[davidbeach]

Find a good electric machines text and look at the circuit model of an induction motor. The rotor resistance is R₂*((1-s)/s), where s is slip and defined as ([&omega;]_sync-[&omega;]_m)/[&omega;]_sync. [&omega;]_sync is synchronous speed and [&omega;]_m is the mechanical speed. Negative speed ensures that slip will be greater than 1. When slip is greater than 1, (1-s)/s is negative and the rotor resistance component of the circuit model becomes negative.

 

[rockman7892]

davidbeach

I see what you are saying now and how the equation that you have posted would make the rotor resistance negative.

It is my understanding however that the impedance of the rotor is made up of:

Z_rotor=R_rotor((1-s)/s) + jX_rotor

The current in the rotor is then determined by:

I_rotor= E_rotor / Z_rotor

I would think that becuase the Z_rotor impedance is is a function of squaring both R_rotor and jX_rotor terms and then taking the square root of thier sums the the overall rotor impedance will always be positive. With this being said, I dont see how this negative rotor resistance would lead to a large amount of current.

I am trying to see my way through this theoryetically however I may be missing something where equations do not apply.

 

[davidbeach]

Thought I had a plot handy that showed motor current vs. slip for slip greater than 1, but I don't seem to and aren't going to derive one. All the plots I do have show rotor, and therefore stator, current increasing with increasing slip. Torque on the other hand continues to decrease as slip increases and I have many plots at hand that show torque vs. slip for slip up to 2. Higher current and lower torque means you won't catch that back spinning fan unless you are specifically set up to do so.

 

[waross]

My field experience is in agreement with your theoretical explanation, David. We tripped out a lot of times and destroyed a couple of overload relays trying to restart back spinning fans. I don't remember that we were ever successful. If a fan tripped out we would have to shut down the bank and then restart all three together.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Going back to the equivalent circuit, see figure 2b here:

http://www.reliance.com/prodserv/motgen/b7097_2.htm

It shows the s dependence shows up simply in R2/s. As s continues to increase beyond 1, current continues to decrease although as you observe relatively slowly because the impedance is primarily inductive as we pass down to zero speed and reverse speed.

For sake of addressing the negative resistance element, I'll mention that we can separate R2/s into two components.,... one that "delivers power" to the shaft (under normal forward rotation) and one that corresponds to rotor losses. This is accomplished by subracting R2 (resistive loss portion) from R2/s, which results in R2/s – R2 = R2*(1-s)/s (this is the component associated with power delivery to the shaft). If we are in steady state with speed between 0 and sync (slip between 1 and 0), then the term is positive corresponding to "motor action" in the sense that real power flows from electrical system to mechanical system. If we go below 0 speed (above slip=1), then you can imagine that to maintain this state of affairs in steady state we have to be applying force to the rotor in direction opposite motor torque. This corresponds to generator action since real power would be flowing from mechanical to electrical system. That is the physical singificance of the he term R2*(1-2)/s becoming negative.

I may be out of line, but it occurs to me it may be productive if the original poster can take a step back and describe what he is seeing from beginning to end.

I gather a motor is spinning backwards and trips during start on instantaneous for MCP breaker?

Trips every time? Trips intermittently? Would you care to share the motor data and the setpoint? Maybe a high-efficiency motor? Have you monitored the current? If so what kind of clamp-on ? How fast is it spinning backwards? How long does it take to start onj a successful start? Ever tripped when not spinning backwards?

Tripping of MCP breakers is a subject near and dear to my heart and I have shared some of my pain and lessons ont eh forums. I agree backward rotation could in theory increase the initial current but it is certainly not the only possible cause of an instantaneous trip.

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[electricpete]

Correction to the first line:

As s continues to increase beyond 1, current continues to decrease ...

should have been

As s continues to increase beyond 1, current continues to increase ...

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

 

[rockman7892]

electricpete

Thanks for your explanation. There was really no particular that I am troubleshooting involving this matter but yet came about from a discussion I was having with an electrician. We were talking about fans in the plant (480V about 100hp)and he had mentioned that when these fans are spinning backwards they trip the MCP when trying to start in the correct direction. He was trying to say that this was becasue the the fan was regenerating voltage however I argued that this was not the case and confirmed this with what was said above.

I was stating to him that theoretically during this case we should not trip a MCP breaker. My reasoning for this was that a motor had a LRC which was a current that was never surpassed, and was a maximum motor current. If the breaker did not trip during any other starts of this motor in the forward direction, then the breaker setting must be high enough where it is above this LRC. With that being said I stated that no other condition in the motor should cause the motor to exceed a LRC current draw, and therefore should not trip the breaker since it was able to stay closed when it saw LRC during starting. I explained that I could understand the motor overloads tripping since these were time dependent but I just couldn't see this tripping the MCP if it was set above LRC.

I can understand what you and others are saying above about the negative rotor impedance causing higher current (I understand plot I have but not the equation as I posted earlier) but I am wondering if this resulting current can be higher than the motors LRC. Can this resulting current be higher than the LRC? If not then I wouldn't this would trip the breaker.

I also understand that residual magnetism in the core can cause a larger inrush current as described above but I would think this would be negligable.

 

[davidbeach]

If locked rotor current is the motor zero speed current, slip = 1; then the motor at negative speed, slip > 1, would tend to draw more than locked rotor current.