We have a 75kw motor and recently replaced the all the contactors, following this when the motor ran up, it would trip the overload as soon as it switched to delta, If the motor had no load it switched over fine, we found by increasing the magnetic overload level the breaker will stay in under loaded conditions. I have since put a current clamp meter on all three legs and found two spike from 90A to 160A as the delta contactor comes in, the current level then settles and the motor runs fine, any suggestions as to what is happening here??
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Star Delta changeover fault 4
- Thread starter dave57
- Start date
dave57
Did you replace the contactors with vacuum contactors? There is an ongoing thread about this named "Vacuum Contactors" started by Moose2.
Have you tried to adjust the "transition" timer? If you have an open delta starter or closed transition starter the time for switching is critical. This timer is most likely your problem. Try making some minor changes and see if it clears the problem.
One thing I would caution you about is the actual transition. Instantaneous blinks in the power will cause some very strange "torque loading" for your equipment. This torque loading could cause you to break a shaft.
As a possible example in the industry that I work with, if a power company has an auto-reclosure device that resets before the contactor can drop out the motor tries to restart out of phase. This equates a broke shaft.
Hope this helps!!!
David
Did you replace the contactors with vacuum contactors? There is an ongoing thread about this named "Vacuum Contactors" started by Moose2.
Have you tried to adjust the "transition" timer? If you have an open delta starter or closed transition starter the time for switching is critical. This timer is most likely your problem. Try making some minor changes and see if it clears the problem.
One thing I would caution you about is the actual transition. Instantaneous blinks in the power will cause some very strange "torque loading" for your equipment. This torque loading could cause you to break a shaft.
As a possible example in the industry that I work with, if a power company has an auto-reclosure device that resets before the contactor can drop out the motor tries to restart out of phase. This equates a broke shaft.
Hope this helps!!!
David
- Thread starter
- #3
dave57:
I would recommend that you wait for an EE to answer this post. I would hate to cause a broke shaft. If you had a working system and just replaced the contacts it should work. It would help if you told the smart guy's if it is closed or open transition. I only pointed out the torque loading to help you advoid possible problems.
Note:
I would personally try reducing the transtion time.
David
I would recommend that you wait for an EE to answer this post. I would hate to cause a broke shaft. If you had a working system and just replaced the contacts it should work. It would help if you told the smart guy's if it is closed or open transition. I only pointed out the torque loading to help you advoid possible problems.
Note:
I would personally try reducing the transtion time.
David
- Thread starter
- #5
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3
- #6
Hello dave57
With a star delta starter, the motor is initially connected in star and in this configuration, the current is one third for the full voltage starting current and the torque is one third of the full voltage starting torque. After the timer operates, the star contactor openas and the delta contactor closes applying full voltage to the motor after an open transition.
There are two conditions here to consider.
1) the motor in delta will draw full voltage start current for the speed that it is rotating. It is very important that while connected in star, the motor is allowed to accelerate to full speed before switching to delta. If you switch to delta at less than 90% speed, you will have a very high start current until the motor reaches full speed. (typically 6 - 9 times the rated current of the motor.)
2) the open transition causes a transient current and resultant transient torque. This transient occurs because the motor in star will be spinning at part speed (hopefully close to full speed) with a rotating magnetic field in the stator inducing rotor current and that in turn creates a rotor field. When the star contactor is opened, we have a magetic field from the rotor, rotating inside an open circuit stator winding. This is a generator with the frequency of the generated voltage dependant on the rotor speed. The voltage across the delta contactor is the resultant of both the line voltage and the generated voltage, and depending on the phase relationship at the instant of switching, can result in a voltage that is less than line voltage or greater than line voltage. If the two voltages are equal in amplitued and exactly 180 degrees out of phase, we are effectively closing in on twice line voltage causing a very high current and torque transient. Lengthening the open time will allow the rotor field to reduce and will reduce the magnitude of the transients, but in many cases, allow the motor to slow resulting in a very high delta current after change over.
Your comments about the situation being fine under no load suggests to me that the motor is not up to full speed at switch over when loaded. This could be because of insufficient torque in star (nothing you can do) or insufficient time in star (lengthen the time in star will fix problem) The fast open transition switching transient can only be cured by using a closed transition or a transition free starter.
Hope this helps
Best regards, Mark Empson
With a star delta starter, the motor is initially connected in star and in this configuration, the current is one third for the full voltage starting current and the torque is one third of the full voltage starting torque. After the timer operates, the star contactor openas and the delta contactor closes applying full voltage to the motor after an open transition.
There are two conditions here to consider.
1) the motor in delta will draw full voltage start current for the speed that it is rotating. It is very important that while connected in star, the motor is allowed to accelerate to full speed before switching to delta. If you switch to delta at less than 90% speed, you will have a very high start current until the motor reaches full speed. (typically 6 - 9 times the rated current of the motor.)
2) the open transition causes a transient current and resultant transient torque. This transient occurs because the motor in star will be spinning at part speed (hopefully close to full speed) with a rotating magnetic field in the stator inducing rotor current and that in turn creates a rotor field. When the star contactor is opened, we have a magetic field from the rotor, rotating inside an open circuit stator winding. This is a generator with the frequency of the generated voltage dependant on the rotor speed. The voltage across the delta contactor is the resultant of both the line voltage and the generated voltage, and depending on the phase relationship at the instant of switching, can result in a voltage that is less than line voltage or greater than line voltage. If the two voltages are equal in amplitued and exactly 180 degrees out of phase, we are effectively closing in on twice line voltage causing a very high current and torque transient. Lengthening the open time will allow the rotor field to reduce and will reduce the magnitude of the transients, but in many cases, allow the motor to slow resulting in a very high delta current after change over.
Your comments about the situation being fine under no load suggests to me that the motor is not up to full speed at switch over when loaded. This could be because of insufficient torque in star (nothing you can do) or insufficient time in star (lengthen the time in star will fix problem) The fast open transition switching transient can only be cured by using a closed transition or a transition free starter.
Hope this helps
Best regards, Mark Empson
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1
- #7
I'd have to pretty much agree with Marke on this one; sounds like the motor is running too loaded or too slow when switched from wye to delta.
You may only be getting 2 phases to spike up on current because of the applied instantaneous voltage(it may be close to the zero crossover on the third phase at the time of transition).
How long was the duration of the spike?
Is the overload tripping upon transition, or is the short-circuit device (MCP) tripping? If it's actually tripping on a thermal overload, that may indicate that the initial wye starting current is high, also.
You may only be getting 2 phases to spike up on current because of the applied instantaneous voltage(it may be close to the zero crossover on the third phase at the time of transition).
How long was the duration of the spike?
Is the overload tripping upon transition, or is the short-circuit device (MCP) tripping? If it's actually tripping on a thermal overload, that may indicate that the initial wye starting current is high, also.
- Thread starter
- #8
Thanks to Marke and DanDel
To try and overcome the initial problem the run up time in star was reduced,surprisingly this actually helped the situation (avoided tripping) although we do not reach 90% speed prior to change over, Following taking the current readings this was one thing I was going to adjust to see if by increasing the run up time in star would improve the current curve, we are hitting our busiest production time now so for a few days I have been asked to leave it running, will get back next week and post the results
To try and overcome the initial problem the run up time in star was reduced,surprisingly this actually helped the situation (avoided tripping) although we do not reach 90% speed prior to change over, Following taking the current readings this was one thing I was going to adjust to see if by increasing the run up time in star would improve the current curve, we are hitting our busiest production time now so for a few days I have been asked to leave it running, will get back next week and post the results
cbarn24050
Industrial
Hi, you don't say what kind of load you have on the motor. If it's a fan, or certain types of pump then you will get tripping if the star time is too long. There is not much you can do about the current surges at changover, their normal for this kind of starter.
Hello Marke,
Top Class!!
One of the more authoritive and informative answers I have read.
Now I have a small query in your words
"----LENGTHENING the open time will allow the rotor field to reduce the magnitude of the transients......"
I doubt ,if it should be instead of LENGTHENING it should should be SHORTENING .Please reply.
Top Class!!
One of the more authoritive and informative answers I have read.
Now I have a small query in your words
"----LENGTHENING the open time will allow the rotor field to reduce the magnitude of the transients......"
I doubt ,if it should be instead of LENGTHENING it should should be SHORTENING .Please reply.
Hello kuwait
The rotor has a current flowing at the point where the star contactor is opened. If you look at the rotor circuit, it is essentially an R L parallel circuit. At the point where the star contactor is opened, the stator (driving) field is removed. The current continues to flow in the rotor and decays due to the time constant. The decay time of the generated voltage varies form machine to machine, but as the rotor field decays, so does the generated voltage. If the star to delta delay is long enough, the rotor field will decay to zero, but the problem is, that many machines are rapidly slowing during this time so it is not a practical option. Open star at 90% speed and close the delta 0.5 second later at 60% speed and you have almost locked rotor current flowing. In the case of a pump, it is not possible to provide an interlock delay between star and delta, however in the case of a high inertia load such as a centrifuge, it is a very real option and a one second delay will not cause any drop in speed, but will almost eleiminate the reclose transient.
Reclose quickly and you are reclosing on high generated voltage. The longer the open transition time, the lower the generated voltage and therefore the lower the transient.
dave57 Too much time in star can actually cause the motor to fail, but this requires a very long time relative to starting unless them motor is only at part speed. When the motor is connected in star, the start torque is reduced to one third of the full voltage starting torque, and often, this is just not enough torque to drive the motor through to full speed. In this case, if you extend the time in star, the motor will stop accelerating at part speed (when the start torque developed is equal to the load torque at that speed) and all you do is dissipate unnecesary heat in the motor. To set up the timer, set it long and time the time taken to either reach full speed, or to stop accelerating, and set the timer to that time. If it is less than full speed, you would be better off setting the timer to zero and going straigh to full voltage starting with out the star step.
Best regards, Mark Empson
The rotor has a current flowing at the point where the star contactor is opened. If you look at the rotor circuit, it is essentially an R L parallel circuit. At the point where the star contactor is opened, the stator (driving) field is removed. The current continues to flow in the rotor and decays due to the time constant. The decay time of the generated voltage varies form machine to machine, but as the rotor field decays, so does the generated voltage. If the star to delta delay is long enough, the rotor field will decay to zero, but the problem is, that many machines are rapidly slowing during this time so it is not a practical option. Open star at 90% speed and close the delta 0.5 second later at 60% speed and you have almost locked rotor current flowing. In the case of a pump, it is not possible to provide an interlock delay between star and delta, however in the case of a high inertia load such as a centrifuge, it is a very real option and a one second delay will not cause any drop in speed, but will almost eleiminate the reclose transient.
Reclose quickly and you are reclosing on high generated voltage. The longer the open transition time, the lower the generated voltage and therefore the lower the transient.
dave57 Too much time in star can actually cause the motor to fail, but this requires a very long time relative to starting unless them motor is only at part speed. When the motor is connected in star, the start torque is reduced to one third of the full voltage starting torque, and often, this is just not enough torque to drive the motor through to full speed. In this case, if you extend the time in star, the motor will stop accelerating at part speed (when the start torque developed is equal to the load torque at that speed) and all you do is dissipate unnecesary heat in the motor. To set up the timer, set it long and time the time taken to either reach full speed, or to stop accelerating, and set the timer to that time. If it is less than full speed, you would be better off setting the timer to zero and going straigh to full voltage starting with out the star step.
Best regards, Mark Empson
purplepete
Electrical
Guys,
you have a classic sitiuation of an incorrectly-connected transition, even though 99 % of the time they work. Without going into the technicalities (Sprecher and Schuh explain it fully in their A4-sized technical book on motor starting) the easy way is to check that the other end of the motor winding which is connected to Red phase ENDS UP connected to the blue phase when it goes into delta.
I have seen a number of motor starter manuals which show this INCORRECTLY. This applies for red-white blue phase sequence. If this is done, the spike on transition (even with your 150 to 200 millisec delay in changeover) will be limited to about 3 or 4 times motor FLC.
If it ends up connected to the White phase in delta, the spike will be anything from 10 to 20 times motor FLC. It usually trips the magnetic (instantaneous) trip of the upstream circuit breaker. I've not seen it take out a thermal overload before.
you have a classic sitiuation of an incorrectly-connected transition, even though 99 % of the time they work. Without going into the technicalities (Sprecher and Schuh explain it fully in their A4-sized technical book on motor starting) the easy way is to check that the other end of the motor winding which is connected to Red phase ENDS UP connected to the blue phase when it goes into delta.
I have seen a number of motor starter manuals which show this INCORRECTLY. This applies for red-white blue phase sequence. If this is done, the spike on transition (even with your 150 to 200 millisec delay in changeover) will be limited to about 3 or 4 times motor FLC.
If it ends up connected to the White phase in delta, the spike will be anything from 10 to 20 times motor FLC. It usually trips the magnetic (instantaneous) trip of the upstream circuit breaker. I've not seen it take out a thermal overload before.
Hello purplepete
I was waiting for this "classic" argument.
This statement appears every now and then, particularly from supporteres / manufacturers of star/delta starters and it is true when applied to laboratory test conditions or very lightly loaded applications with a reasonable inertia. The theory is that the motor will be operating at synchronous speed, and the voltage at reclose will lag the supply by a small amount. If you connect the winding between the correct phases, the effective reclose voltage will be lower. In practice however, the motor is never at synchronous speed and the phase angle between the generated voltage and the supply will be variable and so in practice, the results are more random.
Best regards, Mark Empson
I was waiting for this "classic" argument.
This statement appears every now and then, particularly from supporteres / manufacturers of star/delta starters and it is true when applied to laboratory test conditions or very lightly loaded applications with a reasonable inertia. The theory is that the motor will be operating at synchronous speed, and the voltage at reclose will lag the supply by a small amount. If you connect the winding between the correct phases, the effective reclose voltage will be lower. In practice however, the motor is never at synchronous speed and the phase angle between the generated voltage and the supply will be variable and so in practice, the results are more random.
Best regards, Mark Empson
- Thread starter
- #15
Gents
Many thanks for your support and comments, makes for Interesting reading, due to production demands we have not been able to shut this machine down yet, we plan to shut it down on 4th Feb when we will be able to investigate further
will keep you posted
Many thanks for your support and comments, makes for Interesting reading, due to production demands we have not been able to shut this machine down yet, we plan to shut it down on 4th Feb when we will be able to investigate further
will keep you posted
Marke,
For your edification, I submit the following link. I am not in a position to dispute his conclusions. In fact, your point about this discussion being foisted by those in a position to benefit from selling Wye-Delta starters made me look at it again and conclude that the writer appears to be trying to sell Closed Transition starters, so you have me wondering. Nevertheless it is a well written explaination. Let me know what you think.
It is posted by a manufacturer of Fire Pump controllers who I used to sell for, and they sell an inexpensive device called a Leading Phase Monitor for Wye-Delat starters that can be hooked up to the motor to indicate a connection pattern that will result in nuisance tripping. Their claim is that by simply swapping leads or "rolling" the connection one can reduce the transition spike. Quando Omni Flunkus Moritati
For your edification, I submit the following link. I am not in a position to dispute his conclusions. In fact, your point about this discussion being foisted by those in a position to benefit from selling Wye-Delta starters made me look at it again and conclude that the writer appears to be trying to sell Closed Transition starters, so you have me wondering. Nevertheless it is a well written explaination. Let me know what you think.
It is posted by a manufacturer of Fire Pump controllers who I used to sell for, and they sell an inexpensive device called a Leading Phase Monitor for Wye-Delat starters that can be hooked up to the motor to indicate a connection pattern that will result in nuisance tripping. Their claim is that by simply swapping leads or "rolling" the connection one can reduce the transition spike. Quando Omni Flunkus Moritati
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: The star-delta started motor may become a solution when the motor is somewhat overrated. It is true that the efficiency of the motor will be lower; power factor will be somewhat higher. However, there is also the efficiency of the load on the motor shaft to be taken into consideration. If the motor efficiency decreases from lower 90s % to lower 80s % and the motor load is less efficient, than such star-delta starting solution can be found reasonable. Apparently, this is how the star-delta started motors may have been applied in the more remote past. If one exaggerates this reasoning, there may be 100HP motor powering ~33.3HP load. However, this is what the star connection of 100HP motor delivers on the shaft. Then, such motor-load combination does not even need to be switched to the delta connection. Now, supposing that the motor will start ~66.6HP on its shaft via the star-delta start. Then, the star connection will deliver about ~33.3HP to the shaft and the delta connection will provide the full power to the shaft load that is ~66.6HP. Now, the overrated motors that have 66.6% of the rated HP are found in some applications for some reasons or others.
Marke, I'm interested in your comment about a one second delay for the star-delta transition. I've worked with centrifuges the past 10+ years and every machine I commissioned had the high transient when switching to delta, even when at full speed. I don't have the drawings in front of me, but I believe the delta contact coils were energized by the contacts on the transition timer, so the switchover was as fast as the mechanical interlock would allow. Is the one second delay an additional timer added to the output of the transition timer, or is this done by the mechanical interlock? Mike Bensema
Hello mbensema
Yes, the normal configuration of a star delta starter is as you describe which minimises time. Some argue that this minimises the phase drift between the supply voltage and the generated voltage. To add a delay to allow the rotor field to decay, you would need to add an extra timer. Just a delay ON timer in series with the delta contactor coil.
This is OK for high inertia loads, but not for low inertia loads, as the delay would allow th motor to slow below 85% speed and you would be better off just going for a full voltage starter.
Best regards, Mark Empson
Yes, the normal configuration of a star delta starter is as you describe which minimises time. Some argue that this minimises the phase drift between the supply voltage and the generated voltage. To add a delay to allow the rotor field to decay, you would need to add an extra timer. Just a delay ON timer in series with the delta contactor coil.
This is OK for high inertia loads, but not for low inertia loads, as the delay would allow th motor to slow below 85% speed and you would be better off just going for a full voltage starter.
Best regards, Mark Empson
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: Visit
etc. for more info on star-delta and other soft starts.
etc. for more info on star-delta and other soft starts.
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