I know motors are designed to start at +/- 10% of rated voltage. For instance, a 208-230V motor should be able to start at 187V. Does this mean that the motor can start under full load with the voltage at 187V before the motor is started? If so, the voltage drop will be even lower with the motor at full load. Is it reasonable for the motor to operate in this manner? Or, does this mean that with the motor at full load, it should see no less than 187V? Thank you for any help that can be provided.
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Starting Voltage for Motors 2
- Thread starter lanier
- Start date
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestions to lanier (Electrical) May 21, 2003 marked ///\\I know motors are designed to start at +/- 10% of rated voltage.
///For motor continuous steady state operations\\ For instance, a 208-230V motor should be able to start at 187V.
///Yes.\\ Does this mean that the motor can start under full load with the voltage at 187V before the motor is started?
///Yes.\\ If so, the voltage drop will be even lower with the motor at full load.
///Yes. However, the motor has another terminal voltage rating. It is called the motor starting minimum voltage. Some motors are built with 80% of motor terminal voltage, others at 75% of motor terminal voltage, etc.\\ Is it reasonable for the motor to operate in this manner?
///Yes.\\ Or, does this mean that with the motor at full load, it should see no less than 187V?
///The motor should not see less than motor rated terminal voltage less 10% for continuous operation at rated load. All statements above are for motors complying with NEMA MG-1 Standard.\\
///For motor continuous steady state operations\\ For instance, a 208-230V motor should be able to start at 187V.
///Yes.\\ Does this mean that the motor can start under full load with the voltage at 187V before the motor is started?
///Yes.\\ If so, the voltage drop will be even lower with the motor at full load.
///Yes. However, the motor has another terminal voltage rating. It is called the motor starting minimum voltage. Some motors are built with 80% of motor terminal voltage, others at 75% of motor terminal voltage, etc.\\ Is it reasonable for the motor to operate in this manner?
///Yes.\\ Or, does this mean that with the motor at full load, it should see no less than 187V?
///The motor should not see less than motor rated terminal voltage less 10% for continuous operation at rated load. All statements above are for motors complying with NEMA MG-1 Standard.\\
electricpete
Electrical
"Does this mean that the motor can start under full load with the voltage at 187V before the motor is started"
Accoring to NEMA, No - the voltage must not drop below 90% at any time, before, during and after start. If voltage is at 187% prior to start it will drop below 90% during start which does not meet NEMA requirement.
It is somewhat of an inconsistency in the NEMA requirements, which is what I think you are getting at. Many users specify (over and above NEMA requirements), that the motor must be able to start it's load at 80% terminal voltage.
Accoring to NEMA, No - the voltage must not drop below 90% at any time, before, during and after start. If voltage is at 187% prior to start it will drop below 90% during start which does not meet NEMA requirement.
It is somewhat of an inconsistency in the NEMA requirements, which is what I think you are getting at. Many users specify (over and above NEMA requirements), that the motor must be able to start it's load at 80% terminal voltage.
electricpete
Electrical
An obvious typographical correction:
'If voltage is at 90% prior to start it will drop below 90% during start...'
'If voltage is at 90% prior to start it will drop below 90% during start...'
-
1
- #5
Hi,
My explanation is as follows.
In fact the you should have indicated the HP(kW) and the motor name plate voltage rating and the type of the motor (Induction, Sync, DC etc) so that the discussion will be more realistic. Any how both IEC & NEMA allow +/- 10% of the motor rated voltage drop during steady state running and NEMA allows 15% drop at the motor terminals whereas IEC motors can have only down to 10% drop.
If I presume that the motor nameplate voltage to be 230 V, 3-phase ( a typical North American NEMA product) with D.O.L. starting.
Steady state : (+)10% of 230 V = 253 Volts & (-)10% of 230 V = 207 Volts.
If everything is fine and if this motor operates between 207 to 253 Volts, then it should deliver its nameplate power at the rated ambient temperature.
Starting: If we calculate the terminal voltage at the start, then we are allowed to go down to 85% of 230 V= 184 V. That means at the start, if the total inertia is within the stipulated limits, this motor should start without any trouble even at 184 V. But it should accelerate with in the given time.
Whether an ind/sync. motor is started at full load or no-load it DOESNOT affect the magnitude of the starting (locked rotor) current. It is always the same and it will be 6-8 times the motor name plate full load current (or given by the manufacturer or as per given in NEMA) The only change is in the accelerating time. When we start a pump with the discharge valve closed, then it will take a little shorter time than when it is started with open discharge valve. Finally, the motor will settle at no-load speed or the full load speed depending on the flow requirement. If we keep the discharge valve opening (number of turns) corresponding to the motor full load current and then start, after the starting inrush the motor current will settle down at full load reading. Typical example is a fire pump where we don’t want to manually throttle the discharge valve during a fire situation. Instead, the fire pump should deliver its full flow to the fire right from the start. But in a process plant what we do is, after the starting time, we manually (or by a DCS system ) throttle the discharge as per our flow requirement up to the full load of the motor. As a precaution we don’t keep the discharge valve tightly closed at the start because specially in large pumps it may lead to cavitation etc. So we used to “ crack open “ the discharge valve for about 50-60% through a “by-pass line”. Then the motor will start with 50-60% full load, but will not deliver fluid to the main system. After the starting we can throttle the main valve as per our flow requirement while slowly closing the by-pass valve.
Therefore there is no harm in starting a ind/sync. motor at full load, but we should make sure
1)during the prolonged starting time the over current protection should not unnecessary trip the motor.
2)minimum of 15% of the motor name plate voltage should be available at the motor
terminals during the start-up process.
3)during the steady state running, voltage available at the motor terminals should be between 207 & 253 Volts.
Regards!
My explanation is as follows.
In fact the you should have indicated the HP(kW) and the motor name plate voltage rating and the type of the motor (Induction, Sync, DC etc) so that the discussion will be more realistic. Any how both IEC & NEMA allow +/- 10% of the motor rated voltage drop during steady state running and NEMA allows 15% drop at the motor terminals whereas IEC motors can have only down to 10% drop.
If I presume that the motor nameplate voltage to be 230 V, 3-phase ( a typical North American NEMA product) with D.O.L. starting.
Steady state : (+)10% of 230 V = 253 Volts & (-)10% of 230 V = 207 Volts.
If everything is fine and if this motor operates between 207 to 253 Volts, then it should deliver its nameplate power at the rated ambient temperature.
Starting: If we calculate the terminal voltage at the start, then we are allowed to go down to 85% of 230 V= 184 V. That means at the start, if the total inertia is within the stipulated limits, this motor should start without any trouble even at 184 V. But it should accelerate with in the given time.
Whether an ind/sync. motor is started at full load or no-load it DOESNOT affect the magnitude of the starting (locked rotor) current. It is always the same and it will be 6-8 times the motor name plate full load current (or given by the manufacturer or as per given in NEMA) The only change is in the accelerating time. When we start a pump with the discharge valve closed, then it will take a little shorter time than when it is started with open discharge valve. Finally, the motor will settle at no-load speed or the full load speed depending on the flow requirement. If we keep the discharge valve opening (number of turns) corresponding to the motor full load current and then start, after the starting inrush the motor current will settle down at full load reading. Typical example is a fire pump where we don’t want to manually throttle the discharge valve during a fire situation. Instead, the fire pump should deliver its full flow to the fire right from the start. But in a process plant what we do is, after the starting time, we manually (or by a DCS system ) throttle the discharge as per our flow requirement up to the full load of the motor. As a precaution we don’t keep the discharge valve tightly closed at the start because specially in large pumps it may lead to cavitation etc. So we used to “ crack open “ the discharge valve for about 50-60% through a “by-pass line”. Then the motor will start with 50-60% full load, but will not deliver fluid to the main system. After the starting we can throttle the main valve as per our flow requirement while slowly closing the by-pass valve.
Therefore there is no harm in starting a ind/sync. motor at full load, but we should make sure
1)during the prolonged starting time the over current protection should not unnecessary trip the motor.
2)minimum of 15% of the motor name plate voltage should be available at the motor
terminals during the start-up process.
3)during the steady state running, voltage available at the motor terminals should be between 207 & 253 Volts.
Regards!
-
1
- #6
Hello Lanier
If I can perhaps throw a different slant on your question. The motor exhibits a speed torque curve that is peculiar to that particular motor design and can be very different for the curve of another motor. When the motor is stationary, (rotor not rotating), it will exhibit Locked rotor torque when full voltage is applied. The Locked Rotor Torque of motors can vary from less than 70% of Full Load Torque to greater than 250% of the rated torque of the motor. As the motor accelerates, this torque will reduce and then build back up, peaking at around 90% of rated speed.
The motor can start the driven load, provided that at all points on the speed torque curve, the torque developed by the motor exceeds the torque of the driven load. If the driven load has a high inertia, you need to have a healthy margin between the motor developed torque and the load torque in order to achieve an acceptable acceleration and prevent excess start times.
If the voltage is reduced from the rated voltage of the motor, the current into the motor will reduce, and the torque will reduce by the voltage reduction squared. If the voltage drops too low, then you will not be able to start your machine. Some motors that have a very high start torque, will be able to start the machine down to a low supply voltage, whereas others that have a low start torque will not be able to start the load with a small voltage drop.
To answer your question, you really need to look at the characteristics of the motor in question and see what the speed / torque curve looks like at the reduced voltage.
Best regards,
Mark Empson
If I can perhaps throw a different slant on your question. The motor exhibits a speed torque curve that is peculiar to that particular motor design and can be very different for the curve of another motor. When the motor is stationary, (rotor not rotating), it will exhibit Locked rotor torque when full voltage is applied. The Locked Rotor Torque of motors can vary from less than 70% of Full Load Torque to greater than 250% of the rated torque of the motor. As the motor accelerates, this torque will reduce and then build back up, peaking at around 90% of rated speed.
The motor can start the driven load, provided that at all points on the speed torque curve, the torque developed by the motor exceeds the torque of the driven load. If the driven load has a high inertia, you need to have a healthy margin between the motor developed torque and the load torque in order to achieve an acceptable acceleration and prevent excess start times.
If the voltage is reduced from the rated voltage of the motor, the current into the motor will reduce, and the torque will reduce by the voltage reduction squared. If the voltage drops too low, then you will not be able to start your machine. Some motors that have a very high start torque, will be able to start the machine down to a low supply voltage, whereas others that have a low start torque will not be able to start the load with a small voltage drop.
To answer your question, you really need to look at the characteristics of the motor in question and see what the speed / torque curve looks like at the reduced voltage.
Best regards,
Mark Empson
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: If the power distribution system has relatively high system impedance, then the motors may ordered custom-built or starting at 75% of the motor rated terminal voltage. These motors are more expensive. If the power distribution system has relatively small system impedance, i.e. the supply is stiff, then the motors may start at 85% or 90% of the motor rated terminal voltage. Some motors, e.g. ground water submersible pumps tend to have 90% motor starting voltage requirement.
Hello Lanier
I am with Marke on the importance of the speed/torque curve for the particular motor.If system voltage is as low as 187 volts,it will drop further when starting.
One cannot say with certainty,that under these conditions this drive will start under hheavy load and will be able to accelerate without tripping.
Load torque may very well call for higher Tq than what this motor will deliver at such a low voltage level.
GusD
I am with Marke on the importance of the speed/torque curve for the particular motor.If system voltage is as low as 187 volts,it will drop further when starting.
One cannot say with certainty,that under these conditions this drive will start under hheavy load and will be able to accelerate without tripping.
Load torque may very well call for higher Tq than what this motor will deliver at such a low voltage level.
GusD
Gentlemen,
This does not make sense to me. A large portion of installed motors are started via a reduced voltage starter of some sort and in those cases the 'terminal' voltage is well below the -10% level.
It would make a lot more sense to me if what was being refered to was the 'supply' voltage tolerance.
Could someone please explain.
Regards,
GGOSS
This does not make sense to me. A large portion of installed motors are started via a reduced voltage starter of some sort and in those cases the 'terminal' voltage is well below the -10% level.
It would make a lot more sense to me if what was being refered to was the 'supply' voltage tolerance.
Could someone please explain.
Regards,
GGOSS
electricpete
Electrical
My take on it is that there are two numbers of interest:
1 - At what voltage are you "guaranteed" by NEMA MG-1 that your motor will start the standard-inertia load without injurious heating. I know this is 90% for medium and large motors. I have never heard of the 85% number in NEMA standards, but maybe it is something unique to small motor?
2 - How low a voltage can the actual motor (possibly with margin to withstand duty beyond MG-1 requirements) start the actual load (often less than standard inertia, also possibly undersized with respect to horsepower/torque). There is often plenty of margin as indicated in all my parenthetic statements, but maybe not readily quantified. Using voltage-square law to compare torque speed curves would be a good first step in most cases. For a detailed evaluation that would applied to large motor this would not be sufficient... merely meeting the torque speed curve at all points does NOT assure a satisfactory start... would require a more detailed analysis that can only be done by the motor vendor. Nevertheless there are many thumbrules that can be applied with guidance from experienced folks like Mark in cases where wide margin is available. Also with small inexpensive motors the need for rigorous analysis is not as great.
1 - At what voltage are you "guaranteed" by NEMA MG-1 that your motor will start the standard-inertia load without injurious heating. I know this is 90% for medium and large motors. I have never heard of the 85% number in NEMA standards, but maybe it is something unique to small motor?
2 - How low a voltage can the actual motor (possibly with margin to withstand duty beyond MG-1 requirements) start the actual load (often less than standard inertia, also possibly undersized with respect to horsepower/torque). There is often plenty of margin as indicated in all my parenthetic statements, but maybe not readily quantified. Using voltage-square law to compare torque speed curves would be a good first step in most cases. For a detailed evaluation that would applied to large motor this would not be sufficient... merely meeting the torque speed curve at all points does NOT assure a satisfactory start... would require a more detailed analysis that can only be done by the motor vendor. Nevertheless there are many thumbrules that can be applied with guidance from experienced folks like Mark in cases where wide margin is available. Also with small inexpensive motors the need for rigorous analysis is not as great.
Hi,
1) For medium and small motors NEMA MG-1 products allows a drop of (-)10%.
2)Clause 20.14.2.2 of NEMA-MG1 always specifies a (-)15% voltage drop at start for large motors if the inertia and starting torque are within the limits.
3) IEC-34 specifies a factor of 1.3 between the motor and load torque curves upto the breakdown torque,taking into cosideration a voltage drop of 10% at the start.
Always it is the available voltage at the motor terminals and not the supply voltage. Due to the drop they are not equal.Bus voltage may be within limits. But due to the inrush current, the impedance drop is making a lesser voltage available at the motor terminals. A very good example is DOL starting of a 15000 HP motor through a captive (dedicated) transformer.
Regards!
1) For medium and small motors NEMA MG-1 products allows a drop of (-)10%.
2)Clause 20.14.2.2 of NEMA-MG1 always specifies a (-)15% voltage drop at start for large motors if the inertia and starting torque are within the limits.
3) IEC-34 specifies a factor of 1.3 between the motor and load torque curves upto the breakdown torque,taking into cosideration a voltage drop of 10% at the start.
Always it is the available voltage at the motor terminals and not the supply voltage. Due to the drop they are not equal.Bus voltage may be within limits. But due to the inrush current, the impedance drop is making a lesser voltage available at the motor terminals. A very good example is DOL starting of a 15000 HP motor through a captive (dedicated) transformer.
Regards!
electricpete
Electrical
Thanks Kirbanda, I was not aware of that clause. But I see the title is "Low voltage option". Per paragraph 20.14.2 the standard required starting condition is 90%. (assuming standard inertia, load torque varying with speed squared and reaching motor rated power at rated speed).
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: Visit
(for: What is the minimum recommended starting voltage for AC blower motors?)
etc. for more info
(for: What is the minimum recommended starting voltage for AC blower motors?)
etc. for more info
Hello Lanier.
GGoss's question regarding reduced voltage starting being used extensively in the industry.Reduced voltage starting
also means reduced torque starting.Some load types or different power systems,may require that you start at reduced torque(mechanical concerns)and low starting currents.The laws of motor speed/torque curves for the particular drive are still valid.Motor has to be able to start and accelerate to full speed.Motor torque curve cannot fall below load torque curve,or it will stall.Terminal voltage has to be high enough to maintain pak torque at near Sync speed,withoutexceeding motor rated slip.
GusD
GGoss's question regarding reduced voltage starting being used extensively in the industry.Reduced voltage starting
also means reduced torque starting.Some load types or different power systems,may require that you start at reduced torque(mechanical concerns)and low starting currents.The laws of motor speed/torque curves for the particular drive are still valid.Motor has to be able to start and accelerate to full speed.Motor torque curve cannot fall below load torque curve,or it will stall.Terminal voltage has to be high enough to maintain pak torque at near Sync speed,withoutexceeding motor rated slip.
GusD
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