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Motor's Maximum Inertia 2
- Thread starter Kelley8
- Start date
RichLeimbach
Mechanical
The best way would be to model the rotating parts of the assembly (shaft, windings, etc.) in a 3D CAD package, set the correct material densities and then have the computer tell you what the combined interia is.
Alternatively, you could estimate it by checking the diameter of your rotor, weighing it and deviding by gravity to determine mass and then using the mass moment of interia calculation to give you a rough guess. The equation you would need would be Inertia about axis = 1/2* (mass) * ( radius of rotor)^2 .
This would probably get you within 15% or so, but for best results, use the CAD method.
- Rich
Alternatively, you could estimate it by checking the diameter of your rotor, weighing it and deviding by gravity to determine mass and then using the mass moment of interia calculation to give you a rough guess. The equation you would need would be Inertia about axis = 1/2* (mass) * ( radius of rotor)^2 .
This would probably get you within 15% or so, but for best results, use the CAD method.
- Rich
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2
- #3
electricpete
Electrical
Are you interested in estimating
1 - the inertia of the motor itself, or
2 - determining the max inertia which the rotor can safely accelerate.
For #2, this info is defined in NEMA MG-1 for NEMA-frame induction motors applied DOL. For larger motors there is no way other than to consult the manufacturer (or purchase specifications).
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1 - the inertia of the motor itself, or
2 - determining the max inertia which the rotor can safely accelerate.
For #2, this info is defined in NEMA MG-1 for NEMA-frame induction motors applied DOL. For larger motors there is no way other than to consult the manufacturer (or purchase specifications).
=====================================
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I assume you mean 'the maximum inertia that a motor is capable of handling'.
You need to think instead of the inertia ratio (IR)
IR = Jl/Jr
Jl = 'reflected' load inertia, what is the motor coupled to.
Jr = motor rotor inertia, found in any motor description
Here's my rule of thumb for a good solid crack at it:
Are you positioning then Jl < 10
Are you operating in velocity or torque mode then Jl < 50
Finding the load inertia can be immensely difficult by the way.
Good luck,
Robert Trask, PE
Los Angeles, CA USA
bob@mindspan.us
You need to think instead of the inertia ratio (IR)
IR = Jl/Jr
Jl = 'reflected' load inertia, what is the motor coupled to.
Jr = motor rotor inertia, found in any motor description
Here's my rule of thumb for a good solid crack at it:
Are you positioning then Jl < 10
Are you operating in velocity or torque mode then Jl < 50
Finding the load inertia can be immensely difficult by the way.
Good luck,
Robert Trask, PE
Los Angeles, CA USA
bob@mindspan.us
electricpete
Electrical
Another related formula:
J1 = J2*(N2/N1)^2
where
N1 = motor speed
N2 = load speed
J2 = load inertia
J1 = load inertis as seen by the motor.
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J1 = J2*(N2/N1)^2
where
N1 = motor speed
N2 = load speed
J2 = load inertia
J1 = load inertis as seen by the motor.
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- Thread starter
- #6
Thanks for the input.
Yes, I am trying to calculate the max. inertia that the motor can turn. I have some GE 15 Hp motor data sheets which gives you all the information you need, such as max. inertia, acceleration time w/ max inertia, cold and hot stall times, and full speed/torque curves, etc.
From Rich Leimbach's equation, I calculated ~0.6 lb-ft2 (using the density of Iron for rotor material). The GE motor data sheets that I have state 2.0 and 2.17 lb-ft2 each(for a reference). I suspect that my motor is in the ball park of these two machines.
Electricpete, the max. inertia that a 15 Hp motor can safely accelerate is 75 lb-ft2 (from NEMA, MG-1). Unfortunately, the motor in my application is starting a much larger load. I have almost no motor data other than nameplate.
From your formula, "J1 = J2*(N2/N1)^2", how would the load speed be different than the motor speed? Some kind of coupling?
Thanks.
Yes, I am trying to calculate the max. inertia that the motor can turn. I have some GE 15 Hp motor data sheets which gives you all the information you need, such as max. inertia, acceleration time w/ max inertia, cold and hot stall times, and full speed/torque curves, etc.
From Rich Leimbach's equation, I calculated ~0.6 lb-ft2 (using the density of Iron for rotor material). The GE motor data sheets that I have state 2.0 and 2.17 lb-ft2 each(for a reference). I suspect that my motor is in the ball park of these two machines.
Electricpete, the max. inertia that a 15 Hp motor can safely accelerate is 75 lb-ft2 (from NEMA, MG-1). Unfortunately, the motor in my application is starting a much larger load. I have almost no motor data other than nameplate.
From your formula, "J1 = J2*(N2/N1)^2", how would the load speed be different than the motor speed? Some kind of coupling?
Thanks.
electricpete
Electrical
Load speed would be different than motor speed in case of belt-driven or gear driven load. The terminology you used "referred to the motor shaft" implies that all inertia has been corrected to motor speed
I am confused by mindspan's response. It does not look similar to any criteria I have seen for motor inertia limits related to motor starting. Inertia of the motor itself is not of concern to the motor limits since the NEMA limits apply to external (load) inertia. It sounds to me like this terminology maybe arises from a control application of vfd? Is that the application of this post?
You said you have a data sheet with max inertia. Doesn't that answer your question? Or are you looking at a different motor?
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I am confused by mindspan's response. It does not look similar to any criteria I have seen for motor inertia limits related to motor starting. Inertia of the motor itself is not of concern to the motor limits since the NEMA limits apply to external (load) inertia. It sounds to me like this terminology maybe arises from a control application of vfd? Is that the application of this post?
You said you have a data sheet with max inertia. Doesn't that answer your question? Or are you looking at a different motor?
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- Thread starter
- #9
Yes, I am looking at a different motor.
My motor - 15 Hp motor connected to a motor generator. 270 lb-ft2 flywheel connected to the shaft. Thermal overloads keep tripping on startup. Trying to evaluate bumping up the thermal overloads to 140% FLA as guided by the NEC.
Yes this motor is too small for the application but that's what we've got. If cost was no concern, we would swap the motor with one that is properly sized. Cheapest correction is to bump up the overloads to prevent tripping, provided that we can and sucessfully start the motor.
My motor - 15 Hp motor connected to a motor generator. 270 lb-ft2 flywheel connected to the shaft. Thermal overloads keep tripping on startup. Trying to evaluate bumping up the thermal overloads to 140% FLA as guided by the NEC.
Yes this motor is too small for the application but that's what we've got. If cost was no concern, we would swap the motor with one that is properly sized. Cheapest correction is to bump up the overloads to prevent tripping, provided that we can and sucessfully start the motor.
electricpete
Electrical
NEMA gives max inertia for this motor at 75 and you want to try to for 270 lb-ft^2.
Even if you are successful at getting it to work initially, over time (especially with frequent starts) you may develop rotor problems.
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Even if you are successful at getting it to work initially, over time (especially with frequent starts) you may develop rotor problems.
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jbartos
Electrical
- Jan 15, 2000
- 6,440
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Suggestion to Kelley8 (Electrical) Jan 20, 2004 marked ///\\I am looking for equations/methods to calculate a motor's maximum inertia.
///The maximum inertia the motor can turn is given by the motor torque-speed curve limiting boundary to load torque speed curve fitting below or touching the motor torque-speed curve. The form of each curve will determine maximum inertia the motor can turn without stalling. This can be solved analytically or graphically.
Often, the average motor shaft output torque T=Io x alfa is being considered for the acceleration time calculations with:
Io = Wk^2 /(2g) in lb*ft*s^2
Io ... Moment of inertia
Wk^2=(Wkmotor^2 + Wkload^2) .. Inertia of motor rotor and load rotor
The calculation of the maximum inertia will depend on the shape of the load torque-speed curve. This curve can be parametric and it can form a family of tangents to the motor torque-speed curve in a certain segment of the motor torque-speed curve.\\
Under technical information of a motor, they might state the maximum inertia referred to the motor shaft and the acceleration time with the maximum inertia load.
///The acceleration time is normally calculated for a specific motor HP, Wkmotor^2 and Wkrotor^2, which are considered maximum for the calculated acceleration time.
The acceleration time within certain speed intervals is:
Tr/min1 to Tr/min2 = (Wkmotor^2 + Wkload^2)(r/min1-r/min2)/(308xTn)
where:
Tn is the net average accelerating torque between rev/min1 and rev/min2.\\\
///The maximum inertia the motor can turn is given by the motor torque-speed curve limiting boundary to load torque speed curve fitting below or touching the motor torque-speed curve. The form of each curve will determine maximum inertia the motor can turn without stalling. This can be solved analytically or graphically.
Often, the average motor shaft output torque T=Io x alfa is being considered for the acceleration time calculations with:
Io = Wk^2 /(2g) in lb*ft*s^2
Io ... Moment of inertia
Wk^2=(Wkmotor^2 + Wkload^2) .. Inertia of motor rotor and load rotor
The calculation of the maximum inertia will depend on the shape of the load torque-speed curve. This curve can be parametric and it can form a family of tangents to the motor torque-speed curve in a certain segment of the motor torque-speed curve.\\
Under technical information of a motor, they might state the maximum inertia referred to the motor shaft and the acceleration time with the maximum inertia load.
///The acceleration time is normally calculated for a specific motor HP, Wkmotor^2 and Wkrotor^2, which are considered maximum for the calculated acceleration time.
The acceleration time within certain speed intervals is:
Tr/min1 to Tr/min2 = (Wkmotor^2 + Wkload^2)(r/min1-r/min2)/(308xTn)
where:
Tn is the net average accelerating torque between rev/min1 and rev/min2.\\\
RAMConsult
Electrical
Kelley8, You actually have enough information to solve your problem graphically. Take the motor's speed-torque curve and plot the speed-torque curve of the load on to it. The net difference between the two curves is the torque available for accelerating the load at any given speed. In order to continually accelerate the load there must always be a positive accelerating torque. A full graphical description of this method can be found on pages 192-193 in the 1964 edition of the Electrical Transmission and Distribution Reference Book by Central Station Engineers of the Westinghouse Electric Corporation.
If the motor torque is always greater than the load torque graphically at all speeds, then time becomes your enemy since you will be drawing locked rotor amps from zero through at least 40-80% of rated speed, causing the time-overcurrent elements (overloads) to trip to keep the motor from severely overheating; i.e., you are exceeding the rated accelerating time. The integral cooling fan is only moving (%ratedspeed/100)**3 of the normal air flow at rated speed so overheating is problematic.
You also need to ensure that at some point around half-rated speed the available motor torque does not approach the required load torque since certain induction motor designs may exhibit a momentary dip in torque that will prevent the motor from accelerating past that point. Of course if the load torque exceeds the motor torque at any speed the motor will stall.
If the motor torque is always greater than the load torque graphically at all speeds, then time becomes your enemy since you will be drawing locked rotor amps from zero through at least 40-80% of rated speed, causing the time-overcurrent elements (overloads) to trip to keep the motor from severely overheating; i.e., you are exceeding the rated accelerating time. The integral cooling fan is only moving (%ratedspeed/100)**3 of the normal air flow at rated speed so overheating is problematic.
You also need to ensure that at some point around half-rated speed the available motor torque does not approach the required load torque since certain induction motor designs may exhibit a momentary dip in torque that will prevent the motor from accelerating past that point. Of course if the load torque exceeds the motor torque at any speed the motor will stall.
Hello Kelly8
The maximum load inertia that can be started by a motor is dependent on the thermal inertia of the rotor winding. This is related to the mass of the material used in the rotor winding. Nema give a minimum specification and you should be safe operating to this level, however in some cases, the actual figures could be a lot higher than specified in Nema.
In the non Nema world, is is common for manufacturers to specify the maximum load inertia as seen at the motor shaft, or to specify the maximum locked rotor time which also determines the maximum load inertia that can be started. Looking at data sheets, there are tremendous differences between similar motors. I have found that the maximum ful voltage start times for average motors can vary from 10 seconds to 45 seconds and for special motors, considerably more. I am not aware of an easy way to calculate this from the information that would normally be available if it is not specified by the manufacturer.
For a high inertia load, it is important to establish:
1) is ther enough torque across the full speed range to accellerate the load to full speed, and 2) is the motor capable of sustaining the start condition for that period of time.
Best regards,
Mark Empson
The maximum load inertia that can be started by a motor is dependent on the thermal inertia of the rotor winding. This is related to the mass of the material used in the rotor winding. Nema give a minimum specification and you should be safe operating to this level, however in some cases, the actual figures could be a lot higher than specified in Nema.
In the non Nema world, is is common for manufacturers to specify the maximum load inertia as seen at the motor shaft, or to specify the maximum locked rotor time which also determines the maximum load inertia that can be started. Looking at data sheets, there are tremendous differences between similar motors. I have found that the maximum ful voltage start times for average motors can vary from 10 seconds to 45 seconds and for special motors, considerably more. I am not aware of an easy way to calculate this from the information that would normally be available if it is not specified by the manufacturer.
For a high inertia load, it is important to establish:
1) is ther enough torque across the full speed range to accellerate the load to full speed, and 2) is the motor capable of sustaining the start condition for that period of time.
Best regards,
Mark Empson
- Thread starter
- #14
Thanks for the input,
this motor has been in operation for 37 years. It is run continuously and started only after power interuption or maintenence ~ 2 to 3 times/ year. Even though its load is much higher than NEMA standards, it accomplishes its task by being bump started by operators. It is turned on for 3-4 secs and off for 20-30 secs. On the third jog, it is left in run while it reaches full speed (1770 rpm). This starting procedure is to prevent the thermal overloads from tripping. We have been doing this for over 20 years.
I have no data on the motor except for name-plate. I do not have the coveted speed-torque curves that I need. I am assuming a typical Design B curve shape. I have used the accelerating time equation that jbartos mentioned, while assuming the avg. accelerating torque at twice the 100% rated torque. My calculations show that the acceleration time is very close to the overload trip times.
this motor has been in operation for 37 years. It is run continuously and started only after power interuption or maintenence ~ 2 to 3 times/ year. Even though its load is much higher than NEMA standards, it accomplishes its task by being bump started by operators. It is turned on for 3-4 secs and off for 20-30 secs. On the third jog, it is left in run while it reaches full speed (1770 rpm). This starting procedure is to prevent the thermal overloads from tripping. We have been doing this for over 20 years.
I have no data on the motor except for name-plate. I do not have the coveted speed-torque curves that I need. I am assuming a typical Design B curve shape. I have used the accelerating time equation that jbartos mentioned, while assuming the avg. accelerating torque at twice the 100% rated torque. My calculations show that the acceleration time is very close to the overload trip times.
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