Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
MV Motor Starts
- Thread starter nhee
- Start date
Standard answer: It depends.
1) There is no "normal" for medium voltage motors. Some are capable of 1 or 2 starts-per-hour, others are capable of 1 or 2 starts-per-day. The only one who can tell you is the motor mfgs. If it's an old motor or a defunct mfg, post the nameplate data here, including model and/or serial number and someone may be capable of looking it up for you. There are a lot of really great motor people frequenting this forum.
2)Using a reduced voltage start does not increase the number of starts-per-hour/day of the motor. The energy to start from a dead stop is a finite curve defined by an amount of energy over a period of time from stat to full speed. All the RV start does it exchange peak height for acceleration time, the area under the curve is always identical.
3)The autotransformer is more likely to have a lower start capability than the motor. Notice I didn't say starts-per-hour, because some transformers may need more than 1 hour between starts depending on conditions.
Eng-Tips: Help for your job, not for your homework Read faq731-376![[pirate]](/data/assets/smilies/pirate.gif "[pirate] [pirate]")
1) There is no "normal" for medium voltage motors. Some are capable of 1 or 2 starts-per-hour, others are capable of 1 or 2 starts-per-day. The only one who can tell you is the motor mfgs. If it's an old motor or a defunct mfg, post the nameplate data here, including model and/or serial number and someone may be capable of looking it up for you. There are a lot of really great motor people frequenting this forum.
2)Using a reduced voltage start does not increase the number of starts-per-hour/day of the motor. The energy to start from a dead stop is a finite curve defined by an amount of energy over a period of time from stat to full speed. All the RV start does it exchange peak height for acceleration time, the area under the curve is always identical.
3)The autotransformer is more likely to have a lower start capability than the motor. Notice I didn't say starts-per-hour, because some transformers may need more than 1 hour between starts depending on conditions.
Eng-Tips: Help for your job, not for your homework Read faq731-376
- Moderator
- #3
Hi jraef,
I'm puzzled. I have no quarrel with your statement that "All the RV start does it exchange peak height for acceleration time, the area under the curve is always identical."
However, I was under the impression that part of the area under the curve was allowed, that is the heating eqivalent to normal full load running. I thought that the start heating that was harmful was the heating in excess of normal running heat.
Taken to the extreme if you use a VFD on current limit can you not make long slow starts with very little overheating of the motor?
Please don't misunderstand me. I'm not arguing, I'm asking.
Respectfully
I'm puzzled. I have no quarrel with your statement that "All the RV start does it exchange peak height for acceleration time, the area under the curve is always identical."
However, I was under the impression that part of the area under the curve was allowed, that is the heating eqivalent to normal full load running. I thought that the start heating that was harmful was the heating in excess of normal running heat.
Taken to the extreme if you use a VFD on current limit can you not make long slow starts with very little overheating of the motor?
Please don't misunderstand me. I'm not arguing, I'm asking.
Respectfully
electricpete
Electrical
The thermal energy added to the rotor of an unloaded motor during start is equal to the final kinetic energy of the rotor. Reduced voltage start doesn't change this. (vfd does... it's a whole different animal to change frequency than to change voltage).
I would go a step further than jraef and state that for a motor which has heavy torque load during start, the total energy added to the rotor during start is INCREASED by reducing the voltage (again talking reduced voltage start, not vfd).
It can be shown mathematically. I can post if anyone wants.
Also there is an intuitive argument. Let's say you have a rotor spinning at an initial speed. Than apply a stationary DC field. The rotor will come to a stop. How much energy was transfered to the source of that stationary magnetic field? Zero because power is the product of force times velocity (or torque times speed) and the stationary field isn't moving so no energy can be transferred to it through the field. Where did the energy go? Into the rotor thermal heating. By conservation of energy, the rotor thermal heating during this scenario must equal the initial kinetic energy.
Now compare this to a start. If you make your reference frame the frame of the rotating field, the experiment is the same. So the energy which will be dissipated in the rotor is the same.
This result also can be derived from the equivalent circuit. I can post but as I mentioned it is mathematical and tedious. (although easy for me to cut/paste because I already have it done).
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
I would go a step further than jraef and state that for a motor which has heavy torque load during start, the total energy added to the rotor during start is INCREASED by reducing the voltage (again talking reduced voltage start, not vfd).
It can be shown mathematically. I can post if anyone wants.
Also there is an intuitive argument. Let's say you have a rotor spinning at an initial speed. Than apply a stationary DC field. The rotor will come to a stop. How much energy was transfered to the source of that stationary magnetic field? Zero because power is the product of force times velocity (or torque times speed) and the stationary field isn't moving so no energy can be transferred to it through the field. Where did the energy go? Into the rotor thermal heating. By conservation of energy, the rotor thermal heating during this scenario must equal the initial kinetic energy.
Now compare this to a start. If you make your reference frame the frame of the rotating field, the experiment is the same. So the energy which will be dissipated in the rotor is the same.
This result also can be derived from the equivalent circuit. I can post but as I mentioned it is mathematical and tedious. (although easy for me to cut/paste because I already have it done).
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
- Moderator
- #5
Hello electricpete.
I apreciate the time and trouble you have taken.
I agree with the transfer of kinetic energy to the rotor.
Talking about starting an unloaded motor.
The thing that I am wondering about is;
1> When the motor is running at full load, there is heating of the stator and the rotor. When it is running at no load, there is still heating of the stator and the rotor, but much less.
I understand that the biggest issue with starting is heating.
2> I understand that the issues with heating during starting are heating above normal temperature.
If a motor is started across the line and the starting current is 600% Then 100% is normal, acceptable current and 500% is contributing to overheating.
If a motor is started with a reduced current scheme and the starting current is 300%, then 100% is normal and only 200% contributes to overheating. The starting time will probably be doubled but we have a saving in the overheating. The I^2 will probably make the saving in overheating considerable.
I apreciate the time and trouble you have taken.
I agree with the transfer of kinetic energy to the rotor.
Talking about starting an unloaded motor.
The thing that I am wondering about is;
1> When the motor is running at full load, there is heating of the stator and the rotor. When it is running at no load, there is still heating of the stator and the rotor, but much less.
I understand that the biggest issue with starting is heating.
2> I understand that the issues with heating during starting are heating above normal temperature.
If a motor is started across the line and the starting current is 600% Then 100% is normal, acceptable current and 500% is contributing to overheating.
If a motor is started with a reduced current scheme and the starting current is 300%, then 100% is normal and only 200% contributes to overheating. The starting time will probably be doubled but we have a saving in the overheating. The I^2 will probably make the saving in overheating considerable.
- Moderator
- #6
e-pete,
Seems to me we've had this very discussion before in this forum a few years ago. As I recall, I capitulated to your mathematical prowess, so I won't bother forcing you to prove it to me again. However, I also recall that your math was dependant upon the reduced voltage being applied for too long, where it ceases to be doing useful work in accelerating the load. I have no argument with this because that is the reality of most methods of RV starting that use 2 steps, such as RVAT, PW and Y-Delta, and even solid state using a step start. So you are right, I should have said that. Theoretically though, if the timing for coming out of reduced voltage were perfect, the energy could be identical regardless of load, no?
I have an IEEE paper somewhere (maybe this one? IEEE paper link ) on starting of large AC motors that explained the theoretical side of this very well, but now that I recall our earlier discussion, you were siding with the practical application side, and I agree.
Eng-Tips: Help for your job, not for your homework Read faq731-376
Seems to me we've had this very discussion before in this forum a few years ago. As I recall, I capitulated to your mathematical prowess, so I won't bother forcing you to prove it to me again. However, I also recall that your math was dependant upon the reduced voltage being applied for too long, where it ceases to be doing useful work in accelerating the load. I have no argument with this because that is the reality of most methods of RV starting that use 2 steps, such as RVAT, PW and Y-Delta, and even solid state using a step start. So you are right, I should have said that. Theoretically though, if the timing for coming out of reduced voltage were perfect, the energy could be identical regardless of load, no?
I have an IEEE paper somewhere (maybe this one? IEEE paper link ) on starting of large AC motors that explained the theoretical side of this very well, but now that I recall our earlier discussion, you were siding with the practical application side, and I agree.
Eng-Tips: Help for your job, not for your homework Read faq731-376
- Thread starter
- #8
- Moderator
- #9
There are curves, formulas and rules of thumb for the reduced life expectancy of a distribution transformer that is operated in excess of it's temperature rating. It has to do with the accelerated aging of the insulation at elevated temperatures. I would suspect that the motor and the auto transformer will suffer similar but not identical effects. In the case of the motor, by the time you can see it or can measure it it may be too late. On the other hand, if your maintenance tests show that the insulation is degrading, you can schedule a rewind and avoid the possibility of arc damage to the stator in the event of a burnout.
We designed a control scheme for a compressor motor that was limited to 3 starts per hour. When the compressor was in backup mode it could be called upon to start at any time the plant air usage exceeded the capacity of the primary compressor. Our control scheme compelled the motor to run for at least 20 minutes after each start. As I remember we tied the timer to the unloader circuit so that the motor actually ran for at least 20 minutes after every demand for air. That avoided short cycling if the plant demand called for extra air only every 10 or 15 minutes.
We designed a control scheme for a compressor motor that was limited to 3 starts per hour. When the compressor was in backup mode it could be called upon to start at any time the plant air usage exceeded the capacity of the primary compressor. Our control scheme compelled the motor to run for at least 20 minutes after each start. As I remember we tied the timer to the unloader circuit so that the motor actually ran for at least 20 minutes after every demand for air. That avoided short cycling if the plant demand called for extra air only every 10 or 15 minutes.
LionelHutz
Electrical
jraef is right in that the starts/period of time can vary widely between different MV motors.
I've found that a properly applied reduced voltage starter will usually increase the heating in the motor, but it's a fairly insignificant amount.
This is how I generally try to explain it in a simplified manner.
The torque the motor produces is divided into a "maintaining" torque and an accelerating torque. The maintaining torque is the torque required by the load just to continue spinning at whatever part speed it's presently spinning. Since this torque is required by the load it is a fixed requirement. The accelerating torque is the excessive torque over the maintaining torque that the motor produces and is used to accelerate the load.
As you reduce the voltage you lower the accelerating torque while the maintaining torque remains the same. If you reduce the voltage enough you end up with no accelerating torque and the motor stalls. At this point, all the current applied to the motor is just holding it at the part speed and heating the motor. The motor will run indefinately at this part speed until the overload trips or it burns out.
Now, if you increase the voltage just a little bit more the motor would be accelating but at an extremely slow rate. Since the accelerate rate is so small the motor is still operating in basically the same mode as when it's stalled and it will still reach it's thermal limit before reaching full speed.
So, instead, you increase the voltage a fair bit and you get a decent amount of accelating torque and the motor accelerates in a reasonable amount of time.
I think explaing the motor heating as the motor stalls or approaches a stall really shows how the heating increases. At times with decent accelerating torque it's not quite so easy to show.
Waross, to somewhat address your one post. If you hold the motor rotor stopped and apply 100% current the motor will overheat. So, saying that 100% current is "normal" current during the motor acceleration is not really true.
I've found that a properly applied reduced voltage starter will usually increase the heating in the motor, but it's a fairly insignificant amount.
This is how I generally try to explain it in a simplified manner.
The torque the motor produces is divided into a "maintaining" torque and an accelerating torque. The maintaining torque is the torque required by the load just to continue spinning at whatever part speed it's presently spinning. Since this torque is required by the load it is a fixed requirement. The accelerating torque is the excessive torque over the maintaining torque that the motor produces and is used to accelerate the load.
As you reduce the voltage you lower the accelerating torque while the maintaining torque remains the same. If you reduce the voltage enough you end up with no accelerating torque and the motor stalls. At this point, all the current applied to the motor is just holding it at the part speed and heating the motor. The motor will run indefinately at this part speed until the overload trips or it burns out.
Now, if you increase the voltage just a little bit more the motor would be accelating but at an extremely slow rate. Since the accelerate rate is so small the motor is still operating in basically the same mode as when it's stalled and it will still reach it's thermal limit before reaching full speed.
So, instead, you increase the voltage a fair bit and you get a decent amount of accelating torque and the motor accelerates in a reasonable amount of time.
I think explaing the motor heating as the motor stalls or approaches a stall really shows how the heating increases. At times with decent accelerating torque it's not quite so easy to show.
Waross, to somewhat address your one post. If you hold the motor rotor stopped and apply 100% current the motor will overheat. So, saying that 100% current is "normal" current during the motor acceleration is not really true.
- Status
Similar threads
- Question
- Question
- Question