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Transformer sizing for motor starting 11
- Thread starter gentech
- Start date
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- #2
Gentech,
Suggest you identify the minimum voltage you can accept at the motor terminals for a successful start, then calculate volt drop in cables at locked rotor current, then calculate the maximum acceptable transformer impedance which will allow you to start. Calculate based on low supply voltage to the transformer.
I think the transformer size will be determined by impedance rather than rating.
Suggest you identify the minimum voltage you can accept at the motor terminals for a successful start, then calculate volt drop in cables at locked rotor current, then calculate the maximum acceptable transformer impedance which will allow you to start. Calculate based on low supply voltage to the transformer.
I think the transformer size will be determined by impedance rather than rating.
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- #3
Socttuk is correct.
If you do your calc correctly, you may find that a transformer rated 120% to 150% of kVA (HP) of the motor would provide acceptable voltage dip during the starting.
15 to 20% voltage dip during starting should normally be acceptable. Of course lesser is better.
If you do your calc correctly, you may find that a transformer rated 120% to 150% of kVA (HP) of the motor would provide acceptable voltage dip during the starting.
15 to 20% voltage dip during starting should normally be acceptable. Of course lesser is better.
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- #4
I somewhat disagree with scottyuk and rbulsars. I’m not sure about all the theoretical impedance stuff mentioned, however, if you were to size all of your system components based on locked-rotor current, you would GROSSLY overestimate. The following works from a practical standpoint:
For virtually all industrial applications that I am aware of, Locked-Rotor current is never used to size the equipment feeding the motor. The full-voltage starting current of a motor is approximately 6 times rated current (or approximately the locked rotor current). The only time that this may be a consideration is if the motor is “plugged” very frequently (start-stop-start-stop-start-stop-you-get-the-picture). Your application appears to be “normal duty” (assuming this is a fan, the motor will be running continuously for long periods of time), thus locked rotor amps would NOT be necessary to size the transformer (or starters, fuses, overloads, cabling, conduit, etc.) to feed the motor. The rated current on the nameplate is sufficient. In reality, the RMS (root-mean-squared) heating produced by the (very short duration) starting current will not harm the transformer (or the motor for that matter). By the way, concerning the fuses, you must use “dual-element, time-delay” fuses that are designed to “pass through” the starting current without melting.
For your specific application, a 3-phase, 460V, 75 hp motor requires approximately 95 amps (rated amps). Thus the kVA required to meet the motor demand is (460)(95)(sqrt(3))/1000 = 75.7 kVA. My recommendation would be to install a transformer rated a minimum of 100 kVA.
You indicate that this is the only motor to be fed. One other consideration is that it is extremely more expensive to have to engineer, design, purchase, install and checkout a transformer twice. If there is a good chance that this transformer will be used to supply future loads (like a second fan), I would take that into account.
For virtually all industrial applications that I am aware of, Locked-Rotor current is never used to size the equipment feeding the motor. The full-voltage starting current of a motor is approximately 6 times rated current (or approximately the locked rotor current). The only time that this may be a consideration is if the motor is “plugged” very frequently (start-stop-start-stop-start-stop-you-get-the-picture). Your application appears to be “normal duty” (assuming this is a fan, the motor will be running continuously for long periods of time), thus locked rotor amps would NOT be necessary to size the transformer (or starters, fuses, overloads, cabling, conduit, etc.) to feed the motor. The rated current on the nameplate is sufficient. In reality, the RMS (root-mean-squared) heating produced by the (very short duration) starting current will not harm the transformer (or the motor for that matter). By the way, concerning the fuses, you must use “dual-element, time-delay” fuses that are designed to “pass through” the starting current without melting.
For your specific application, a 3-phase, 460V, 75 hp motor requires approximately 95 amps (rated amps). Thus the kVA required to meet the motor demand is (460)(95)(sqrt(3))/1000 = 75.7 kVA. My recommendation would be to install a transformer rated a minimum of 100 kVA.
You indicate that this is the only motor to be fed. One other consideration is that it is extremely more expensive to have to engineer, design, purchase, install and checkout a transformer twice. If there is a good chance that this transformer will be used to supply future loads (like a second fan), I would take that into account.
SJK14,
I'm not suggesting to size the transformer or switchgear to be able to continuously supply locked rotor current. Normal design practise would be to use motor starting gear to suit the motor rating, unless there was something unusual about the load starting characteristics.
The permissible volt-drop through the transformer impedance depends on what starting torque the load requires: if the load requires a certain torque at standstill or during acceleration to full speed, then there will be a minimum voltage at which this torque can be developed. If the voltage available at the motor is too low to achieve the required torque because the volt-drop due to [motor current x xfmr impedance] is large, the motor will stall. The motor current remains high until it is almost at full speed, and consequenctly the volt-drop is large for much of the acceleration period. Thus the transformer, cable, and source impedances are important characteristics which should be considered when sizing the transformer.
In the specific case of a fan, the torque required is low and the motor will probably start with the transformer you suggest. But it is a quick and easy calculation to check that the motor will start, so isn't it better to be sure?
I'm not suggesting to size the transformer or switchgear to be able to continuously supply locked rotor current. Normal design practise would be to use motor starting gear to suit the motor rating, unless there was something unusual about the load starting characteristics.
The permissible volt-drop through the transformer impedance depends on what starting torque the load requires: if the load requires a certain torque at standstill or during acceleration to full speed, then there will be a minimum voltage at which this torque can be developed. If the voltage available at the motor is too low to achieve the required torque because the volt-drop due to [motor current x xfmr impedance] is large, the motor will stall. The motor current remains high until it is almost at full speed, and consequenctly the volt-drop is large for much of the acceleration period. Thus the transformer, cable, and source impedances are important characteristics which should be considered when sizing the transformer.
In the specific case of a fan, the torque required is low and the motor will probably start with the transformer you suggest. But it is a quick and easy calculation to check that the motor will start, so isn't it better to be sure?
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- #6
I usually check if the full load current of the motor is below transformer size serving it then I check the IEEE red book fig "Approximate Voltage Drop in a transformer Due to full load voltage starting of a motor" I will select a transformer from the curves that will allow only 85 percent voltage drop on the motor. In this case- it is a 100 kva transformer.
Discussion: The 85 percent voltage drop is used because the NEMA rating on coils of starters and relays is specified as a minimum of 85 percent of rated voltage before dropping out. coils will usually drop out at a lower voltage but that is a safe minimum.
Some motors will start as with as low as a 35 percent voltage dip (see REA Bulletin 160-3 Oct 1969 "Service to Induction Motors"
and still work, but the holding coils will fail long before the motors ever see this this drop.
The only place I could find this 85 percent requirement for motors is ANSI C50.41-1982. It indicates that motors are required to be able to start as long as the voltage is not less than 85 percent of rated. But this ANSI standard is for motors in power generating plants
Discussion: The 85 percent voltage drop is used because the NEMA rating on coils of starters and relays is specified as a minimum of 85 percent of rated voltage before dropping out. coils will usually drop out at a lower voltage but that is a safe minimum.
Some motors will start as with as low as a 35 percent voltage dip (see REA Bulletin 160-3 Oct 1969 "Service to Induction Motors"
and still work, but the holding coils will fail long before the motors ever see this this drop.The only place I could find this 85 percent requirement for motors is ANSI C50.41-1982. It indicates that motors are required to be able to start as long as the voltage is not less than 85 percent of rated. But this ANSI standard is for motors in power generating plants
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- Thread starter
- #7
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- #8
jbartos
Electrical
- Jan 15, 2000
- 6,440
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Suggestion: When it comes to larger motor sizes, computer software may be used for modeling. The transformer upstream will be included in the motor starting model showing the motor starting transients. Then, the actual design follows the results of the motor starting from the computer simulation.
Normally, industry standards are the design basis for the computer software.
Normally, industry standards are the design basis for the computer software.
- Thread starter
- #9
What may further complicate things is that this motor powers the engine cooling fan in a 2000KW diesel generator. Under black start conditions the generator voltage will rise to normal, energize the transformer and then start the fan motor. My assumption is that the time it takes the motor starter to transfer, 25-50 mS, is enough time to ride out the transformer inrush and not melt the transformer primary fuses. I am considering placing a 0-5 second timer in the circuit. Any thoughts?
jb,
The transformer is protected by fuses. The transformer load is a trivial part of the output of a 2000kW gen-set, so the generator protection relay settings should be irrelevant to this problem. I agree that it would be relevant if discussing the unit auxiliary transformer of a large multi-MW set, which would be covered by a complex and overlapping set of protection functions, but that isn't the case here.
Gentech,
There seems little reason why you can't include such a delay timer, as the gen-set will have little or no cooling demand for the first few seconds of operation. I personally don't think the timer will be necessary but if you make allowance for the timer, you can always remove it without loss of face, whereas retrofitting one to cure a problem that your customer will think you should have foreseen will do you no favours. Spend the extra $20 and fit one.
The transformer is protected by fuses. The transformer load is a trivial part of the output of a 2000kW gen-set, so the generator protection relay settings should be irrelevant to this problem. I agree that it would be relevant if discussing the unit auxiliary transformer of a large multi-MW set, which would be covered by a complex and overlapping set of protection functions, but that isn't the case here.
Gentech,
There seems little reason why you can't include such a delay timer, as the gen-set will have little or no cooling demand for the first few seconds of operation. I personally don't think the timer will be necessary but if you make allowance for the timer, you can always remove it without loss of face, whereas retrofitting one to cure a problem that your customer will think you should have foreseen will do you no favours. Spend the extra $20 and fit one.
jbartos
Electrical
- Jan 15, 2000
- 6,440
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- 0
Suggestion to the previous posting marked ///\\ScottyUK (Electrical) Sep 14, 2003
jb,
The transformer is protected by fuses.
///It is good to know.\\ The transformer load is a trivial part of the output of a 2000kW gen-set,
///I did not address or comment on this issue.\\ so the generator protection relay settings should be irrelevant to this problem.
///I do not see anywhere that I mentioned it relevant to this problem.\\ I agree that it would be relevant if discussing the unit auxiliary transformer of a large multi-MW set, which would be covered by a complex and overlapping set of protection functions, but that isn't the case here.
///Agree.\\///I see that the main concern is in the protection of transformer and its sizing. Typically, the protection of the transformer (e.g. melted fuses) are covered under protective devices selection and coordination study. It appears the discussed time delay 0 - 5 second is being considered for the motor starter to have the motor starting delayed from 25 - 50 msec to (25 - 50 msec) + whatever will be set on the 0 - 5 second timer.\\
jb,
The transformer is protected by fuses.
///It is good to know.\\ The transformer load is a trivial part of the output of a 2000kW gen-set,
///I did not address or comment on this issue.\\ so the generator protection relay settings should be irrelevant to this problem.
///I do not see anywhere that I mentioned it relevant to this problem.\\ I agree that it would be relevant if discussing the unit auxiliary transformer of a large multi-MW set, which would be covered by a complex and overlapping set of protection functions, but that isn't the case here.
///Agree.\\///I see that the main concern is in the protection of transformer and its sizing. Typically, the protection of the transformer (e.g. melted fuses) are covered under protective devices selection and coordination study. It appears the discussed time delay 0 - 5 second is being considered for the motor starter to have the motor starting delayed from 25 - 50 msec to (25 - 50 msec) + whatever will be set on the 0 - 5 second timer.\\
gentech:
I think that transformer inrush is not your problem. As the generator comes to speed and developes the voltage, tranformer will be energized. If the voltage is building slowly , the inrush, if anything, will be less (much like reduced voltage motor starting).
It will be a good idea to have a few seconds (3-10 sec) of delay before the motor is started, so the full (or maximum)voltage is applied to the motor for starting. I would think this radiator fan motor is very small compared to the gen size so DOL starting should be acceptable,even preferred.
I think that transformer inrush is not your problem. As the generator comes to speed and developes the voltage, tranformer will be energized. If the voltage is building slowly , the inrush, if anything, will be less (much like reduced voltage motor starting).
It will be a good idea to have a few seconds (3-10 sec) of delay before the motor is started, so the full (or maximum)voltage is applied to the motor for starting. I would think this radiator fan motor is very small compared to the gen size so DOL starting should be acceptable,even preferred.
- Thread starter
- #16
All,
I found this excel application at GE's website. Click the link and then "Motor Voltage Drop".
I found this excel application at GE's website. Click the link and then "Motor Voltage Drop".
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- #17
Wow, what a lengthy thread!
I think that some of the respondents are not recognizing that a fan load is the easiest of all loads to start (NEARLY zero starting torque).
Contrary to some posts, I'd have to say that the transformer could be sized based on the continous rating and that the transformer impedance is desirable to reduce the starting current of the motor.
Does anybody want to venture a guess on a worst-case fan, bearing (and belt?) starting torque; maybe 10 percent?
I wouldn't think that this would take a lot of calculation to be sure of a safe start.
I think that some of the respondents are not recognizing that a fan load is the easiest of all loads to start (NEARLY zero starting torque).
Contrary to some posts, I'd have to say that the transformer could be sized based on the continous rating and that the transformer impedance is desirable to reduce the starting current of the motor.
Does anybody want to venture a guess on a worst-case fan, bearing (and belt?) starting torque; maybe 10 percent?
I wouldn't think that this would take a lot of calculation to be sure of a safe start.
jbartos
Electrical
- Jan 15, 2000
- 6,440
- 0
- 0
Suggestion: Visit
for Transient Motor Starting Simulation software
for Transient Motor Starting Simulation software
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- #19
Let remind ourselves the basics, since we went a bit off track and some suggestions are not correct:
(By the way fully agree with scottyUK and rbulsara)
The commonly held misconception about motor starting is that:
The start current of a motor is dependant of a motor driven load. (artym)
The fact is:
The start current of a motor is dependant on motor design, rotor speed and stator voltage from zero speed until full speed is reached.
The load will influence the time taken for the motor to reach full speed.
The current/speed curve of the motor is independent of all external influences other then stator voltage
When a motor is atationary, the motor current is limited by the effective series impedance of the rotor and stator.
At very low speeds, the dominant impedance is the rotor.
At high speeds, the stator impedance can become influential.
Therefore:
The rotor determines the starting characteristics of the motor while the stator has an influence on the full speed characteristics.
Motors, including large motors, are design for DOL full voltage starting(except most Slip Rings as they have external rotor resistance).
It is the allowable voltage drop on the plant/power system that may require other starting methods. Hence, limitation of power source must always be considered when applying motors, especially large motors, all up stream imedances must be known and starting volts drop calculated.
The motor in the question is small in kW and the original question was not re. volt drop but the transformer size; the question was correctly answered in the first replay by scottyUK.
Aquarius is certainly correct about the starting current drawn by a motor being independent of the load.
ScottyUK did answer it in his first line of the very first post:
"Suggest you identify the minimum voltage you can accept at the motor terminals for a successful start, then calculate volt drop in cables at locked rotor current, then calculate the maximum acceptable transformer impedance which will allow you to start. Calculate based on low supply voltage to the transformer."
The next question is: "the minumum voltage you can accept at the motor terminals for a successful start".
What I hope to add is: that the minimum acceptable voltage at the motor terminals for reliable starting IS load-dependent and very low for a fan load.
Aquarius also correctly pointed out that the main consideration of the questioner was transformer sizing.
I'd have to differ with ScottyUK regarding the sizing being likely driven by impedance rather than rating (in this special case of a fan load).
ScottyUK did answer it in his first line of the very first post:
"Suggest you identify the minimum voltage you can accept at the motor terminals for a successful start, then calculate volt drop in cables at locked rotor current, then calculate the maximum acceptable transformer impedance which will allow you to start. Calculate based on low supply voltage to the transformer."
The next question is: "the minumum voltage you can accept at the motor terminals for a successful start".
What I hope to add is: that the minimum acceptable voltage at the motor terminals for reliable starting IS load-dependent and very low for a fan load.
Aquarius also correctly pointed out that the main consideration of the questioner was transformer sizing.
I'd have to differ with ScottyUK regarding the sizing being likely driven by impedance rather than rating (in this special case of a fan load).
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