So I am working a design to rotate a large drum at 5 deg/min (very large inertia) sitting upright on a bearing. I am able to take care of the gear ratios and motor HP calcs but what I am stuck on is how to determine the required torque to move this drum. I know that T = I*alpha but I can input any range of alphas to get a range of torques. Alpha can be infinitely small in this equation which yields a low torque requirement but the motor will most certainly stall. Shouldn't alpha be dictated by the motor spec sheet then one could compare this to the stall torque? I missing something? Thanks in advance.
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 TugboatEng on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
AC Motor Selection 1
- Thread starter creighbm
- Start date
Your motor should be sized for peak, or max, torque seen by the prime mover shaft. Peak torque occurs (usually) when accelerating all the masses in the system from a zero start to max desired speed. So you have to determine how fast you want to get things moving, and that will determine max alpha (rotational acceleration) and subsequently peak torque. In some instances peak torque may be on the deceleration portion of the motion, but the arithmetic is similar.
Unlikely you could get a motor RPM to be exactly what you want, so you typically use a gearbox, selected as a standard from a catalog. That output RPM will drive your system through whatever gear train at the rotational speed you specify.
Peak torque at the driving shaft includes everything required of the motor to move: all rotational inertias and translational inertias, plus friction loads, gravity loads, wind loads, whatever. Account for everything.
After the final answer is calculated, then upsize it appropriately and select the matching standard unit from the catalog.
TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering
Unlikely you could get a motor RPM to be exactly what you want, so you typically use a gearbox, selected as a standard from a catalog. That output RPM will drive your system through whatever gear train at the rotational speed you specify.
Peak torque at the driving shaft includes everything required of the motor to move: all rotational inertias and translational inertias, plus friction loads, gravity loads, wind loads, whatever. Account for everything.
After the final answer is calculated, then upsize it appropriately and select the matching standard unit from the catalog.
TygerDawg
Blue Technik LLC
Virtuoso Robotics Engineering
I don't understand how you managed to calculate HP without torque.
Selection is iterative. Assuming you are starting the motor across the line, the motor will produce it's locked-rotor torque.
Feed that into your torque equation for the drum. Don't forget bearing friction. Solve for acceleration.
Calculate how long things will take to get up to speed at that acceleration. You'll need that to select the motor circuit breaker.
If the acceleration and time are acceptable, you're done. If not repeat with a different motor characteristic.
Selection is iterative. Assuming you are starting the motor across the line, the motor will produce it's locked-rotor torque.
Feed that into your torque equation for the drum. Don't forget bearing friction. Solve for acceleration.
Calculate how long things will take to get up to speed at that acceleration. You'll need that to select the motor circuit breaker.
If the acceleration and time are acceptable, you're done. If not repeat with a different motor characteristic.
-
1
- #4
electricpete
Electrical
Assuming you are using DOL start, I agree with MintJulep.
A shortcut to the analysis would be to consult NEMA MG-1 table 12-7 (attached).... you enter the table with speed and horsepower and it tells you the maximum inertia that the motor can start DOL. I believe it assumes that in addition to acceleration torque, the load demands additional torque proportional to speed^2 reaching full-load torque at full speed... that is probably a little bit conservative unless you have a lot of friction, but it doesn't hurt to throw some margin in there.
So basically the process suggested would be:
1 - Select a gear ratio and motor speed which will give 5 degrees/minute. (*)
2 - Determine effective inertia of your load based on the inverse square of speed ratio.
3 - Find the smallest HP motor whose associated load inertia in the table is above your effective inertia.
You may also want to think about breakaway torque associated with friction in your gear system and if you have any way to estimate that, verify it is not more than motor locked rotor torque.
If you suspect voltage droop at motor during start below nameplate, that's something to consider as well. I think the table assumes 90% of nominal voltage, but don't quote me on that.
If you have other requirements regarding start or stop or precise speed control (more precise than you can estimate slip) that you haven't told us, you'll have to check them.
(*) - There are only a finite number of speeds available for motors without vfd....3600,1800,1200, 900 RPM etc. You might compare what the system looks like ($) for a few different speeds: I think faster motors will tend to give lower motor cost and higher gear cost.
=====================================
(2B)+(2B)' ?
A shortcut to the analysis would be to consult NEMA MG-1 table 12-7 (attached).... you enter the table with speed and horsepower and it tells you the maximum inertia that the motor can start DOL. I believe it assumes that in addition to acceleration torque, the load demands additional torque proportional to speed^2 reaching full-load torque at full speed... that is probably a little bit conservative unless you have a lot of friction, but it doesn't hurt to throw some margin in there.
So basically the process suggested would be:
1 - Select a gear ratio and motor speed which will give 5 degrees/minute. (*)
2 - Determine effective inertia of your load based on the inverse square of speed ratio.
3 - Find the smallest HP motor whose associated load inertia in the table is above your effective inertia.
You may also want to think about breakaway torque associated with friction in your gear system and if you have any way to estimate that, verify it is not more than motor locked rotor torque.
If you suspect voltage droop at motor during start below nameplate, that's something to consider as well. I think the table assumes 90% of nominal voltage, but don't quote me on that.
If you have other requirements regarding start or stop or precise speed control (more precise than you can estimate slip) that you haven't told us, you'll have to check them.
(*) - There are only a finite number of speeds available for motors without vfd....3600,1800,1200, 900 RPM etc. You might compare what the system looks like ($) for a few different speeds: I think faster motors will tend to give lower motor cost and higher gear cost.
=====================================
(2B)+(2B)' ?
- Thread starter
- #5
Thanks for all the responses. Just to clarify I haven't caclcuated the HP yet...just know that I can get there once I figure out the torque 
I guess the thing I have a hard time getting my head around is no matter how large the inertia the angular acceleration will always be a positive value (unless the friction high enough). Is there some value that NEMA assumes to come up with table 45?
I guess the thing I have a hard time getting my head around is no matter how large the inertia the angular acceleration will always be a positive value (unless the friction high enough). Is there some value that NEMA assumes to come up with table 45?
electricpete
Electrical
Table 45 represents capability of the motor, based on allowable short-term heating of the motor during DOL
start. It is primarily a function of total inertia (and influenced to lesser extent by load).
An interesting tidbit, the I^2*R heating deposited into the rotor during start of an induction motor under the idealized assumptions is equal to the final kinetic energy. KE = 0.5 * J * w^2 so it depends on speed w and inertia J..... and you don't have to know how fast it accelerated (alpha) to know that. The 2 simplifying assumptions were: 1 - equivalent circuit parameters do not vary with frequency (not 100% correct since deep bar effect changes R2 and X2) and 2 - load torque is zero (only accelerating torque is presen). The justification for considering load torque to be zero is that it typically doesn't increase the heating by much. If we retain assumption 1 and discard assumption 2, the total heating is equal to kinetic energy of rotating parts times a weighted average of Te/(Te-Tm) where Te is motor torque, Tm is load and friction torque, and the weighting is based on slip.
=====================================
(2B)+(2B)' ?
electricpete
Electrical
I went a little bit far afield in the last post but the key point is that from a simple standpoint, there is only a certain amount of inertia that a given motor can accelerate without overheating during the DOL start.
=====================================
(2B)+(2B)' ?
=====================================
(2B)+(2B)' ?
If the inertia is truly enormous, and there is very little actual torque load once up to final speed, the power required will depend totally on the rate of acceleration required to reach full final speed.
I would calculate the required motor torque during acceleration, and use that combined with the final motor rpm to calculate motor horsepower.
I would then fit a variable frequency drive, and set the ramp up/ramp down times to keep motor current within it's full load current rating.
How big and powerful the whole thing ends up needing to be then depends only on an the acceptable time to accelerate up to full speed.
I would calculate the required motor torque during acceleration, and use that combined with the final motor rpm to calculate motor horsepower.
I would then fit a variable frequency drive, and set the ramp up/ramp down times to keep motor current within it's full load current rating.
How big and powerful the whole thing ends up needing to be then depends only on an the acceptable time to accelerate up to full speed.
electricpete
Electrical
My comments assumed the application requires no limit on how long it takes to accelerate the load, other than that the motor must not damage the motor itself (or trip) during start.
=====================================
(2B)+(2B)' ?
=====================================
(2B)+(2B)' ?
electricpete
Electrical
... I should add that if you followed that chart the acceleration time will be less than 10 seconds.... probably just a few seconds. That can certainly be checked once you have narrowed down the motor if you have all data (motor and load inertia, load torque vs speed, motor torque vs speed)
=====================================
(2B)+(2B)' ?
=====================================
(2B)+(2B)' ?
- Thread starter
- #12
So I updated the design such that a motor/gear box drives an internal ring gear. I ran through the calcuations again and say if I set the acceleration time to be 1 second the required torque is less than the output torque of my existing motor/gear box combo. Does this mean I need to vary the acceleartion time until the 2 values match? I found this equation that calculates the acceleration time:
t = (WK^2*(Speed Difference))/(308*Acceleration Torque)
This works fine if you are simply rotating a flywheel but I am trying to see how this works with a pinion driving a ring gear. Any advice? Also this only deals with inertia...not the other external forces like bearing friction.
t = (WK^2*(Speed Difference))/(308*Acceleration Torque)
This works fine if you are simply rotating a flywheel but I am trying to see how this works with a pinion driving a ring gear. Any advice? Also this only deals with inertia...not the other external forces like bearing friction.
Not sure you are clear, so I'll try to make it clearer
From your 5 deg/min output speed using a standard 1700 Rpm motor, I get a gear ratio of about 19,000:1
You first need to determine the overall inertia referenced to the motor shaft and it is usually the sum of the motor inertia and the second shaft inertia of all the gears in the train reflected or
Iref= I motor + ID2/(D2/D1)^2
where
ID2= inertia of gear D2
since ID2/(D2/D1)^2 is probably >> ID3/(n31^2 and Idrum/(19000)^2
n31 gear ratio between D3 and D1
Now from the dynamics,
t= V/a and V = 1700 rpm=1700*2PI/60=178 rad/sec. Let's say t= 5 seconds, then
a=178/5=35.6 rad/sec^2
The driving torque required would be
T=35.6* Iref
The actual motor torque needed would be this plus the friction torque.
Strictly speaking you get the friction at each shaft and reference it back to the motor like
Tf=sum of Tfx/nx1
where Tfx is the friction torque at shaft x and nx1 is the gear ratio from that shaft to the motor. It is more usual to get a lumped gearbox efficiency to take care of torque, windage, viscosity, etc. Using E=70 to 80 percent
Then the required motor torque would be greater or equal to
35.6*Iref/E
Now look for the smallest motor to give you this.This may seem a bit tricky since Iref contains the motor inertia ,but it should not be a problem.
From your 5 deg/min output speed using a standard 1700 Rpm motor, I get a gear ratio of about 19,000:1
You first need to determine the overall inertia referenced to the motor shaft and it is usually the sum of the motor inertia and the second shaft inertia of all the gears in the train reflected or
Iref= I motor + ID2/(D2/D1)^2
where
ID2= inertia of gear D2
since ID2/(D2/D1)^2 is probably >> ID3/(n31^2 and Idrum/(19000)^2
n31 gear ratio between D3 and D1
Now from the dynamics,
t= V/a and V = 1700 rpm=1700*2PI/60=178 rad/sec. Let's say t= 5 seconds, then
a=178/5=35.6 rad/sec^2
The driving torque required would be
T=35.6* Iref
The actual motor torque needed would be this plus the friction torque.
Strictly speaking you get the friction at each shaft and reference it back to the motor like
Tf=sum of Tfx/nx1
where Tfx is the friction torque at shaft x and nx1 is the gear ratio from that shaft to the motor. It is more usual to get a lumped gearbox efficiency to take care of torque, windage, viscosity, etc. Using E=70 to 80 percent
Then the required motor torque would be greater or equal to
35.6*Iref/E
Now look for the smallest motor to give you this.This may seem a bit tricky since Iref contains the motor inertia ,but it should not be a problem.
- Thread starter
- #14
Thanks zekeman. Dumb question but if I choose a motor with a higher torque does this simply mean the acceleration time will decrease? The is the issue I am running into is this motor is driving an internal gear and I am checking the loads on the teeth.
Not necessarily, the acceleration is
T/Iref, but as T goes up so does Iref (Iref=Imotor+Igr), snce Iref contains the motor inertia and it increases with increasing motor torque. Also, using a motor with excess torque could result in damage to the gearbox.
There is also a crucial caveat in the motor selection process not previously mentioned for a low speed high inertia output mass and that is, you have to consider the gearbox and try to get the smallest one to do the job. It would start and coincide with the largest time to come to speed (i.e. the smallest acceleration ) at the load and that would dictate the torque that the output gear must handle.
That time is NOT arbitrary.The difference in 5 seconds to 1 minute is in the ratio of 12:1 for the output torque required and correspondingly a much smaller gearbox.
Finally, for your problem your gear ratio using a 1700RPM motor is 19,000 which is a bad idea for a gearbox. You might be better off using 2 belt driven ones like a household clothes dryer or a low speed motor ( like a stepping motor) with correspondingly lower ratio gearbox.
T/Iref, but as T goes up so does Iref (Iref=Imotor+Igr), snce Iref contains the motor inertia and it increases with increasing motor torque. Also, using a motor with excess torque could result in damage to the gearbox.
There is also a crucial caveat in the motor selection process not previously mentioned for a low speed high inertia output mass and that is, you have to consider the gearbox and try to get the smallest one to do the job. It would start and coincide with the largest time to come to speed (i.e. the smallest acceleration ) at the load and that would dictate the torque that the output gear must handle.
That time is NOT arbitrary.The difference in 5 seconds to 1 minute is in the ratio of 12:1 for the output torque required and correspondingly a much smaller gearbox.
Finally, for your problem your gear ratio using a 1700RPM motor is 19,000 which is a bad idea for a gearbox. You might be better off using 2 belt driven ones like a household clothes dryer or a low speed motor ( like a stepping motor) with correspondingly lower ratio gearbox.
- Thread starter
- #16
The motor is actually driving a bearing with an internal gear so the ratio is not exactly 19000:1. In any case what I did was solve for the torque inertia assuming some angular acceleration based on some delta t. Then I compared the resulting torque (including bearing friction) to the torque a specific motor/geabox combo would produce and varied the angular acceleration until the required torque matched the output shaft torque. Is this a correct approach? I tried reflecting inertias using the equation I2 = I1*(N1/N2)^2 but I can't seem to figure out the inertia of the motor. Now when computing the reflected inertia does it matter that the motor is at some radius away from where the load inertia is reference about? For example the intertia is the the roll inertia about the centerline of the bearing. The motor is being driven at 42" from the centlerline...is I1 in the equation I2 = I1*(N1/N2)^2 still the load inertia?
It is patently incorrect to match a fictitious time with an arbitrary gearbox. Why can't you state the minimum time to get to speed and maybe we can give further assistance.
And since you are having so much trouble grasping this, I suggest that you state the problem in its entirety and show us your design configuration.
And where did you get this "I2 = I1*(N1/N2)^2 ". I hope I didn't send it. And you DON'T figure out how to get the inertia of the motor,the manufacturer gives it to you.
And since you are having so much trouble grasping this, I suggest that you state the problem in its entirety and show us your design configuration.
And where did you get this "I2 = I1*(N1/N2)^2 ". I hope I didn't send it. And you DON'T figure out how to get the inertia of the motor,the manufacturer gives it to you.
- Thread starter
- #18
I should back up a couple steps. The minimum time to get up to speed is not a requirement...the motor just has to have enough power to rotate the drum. Now being that the motor interfaces with a 50" bearing with a 42" internal gear I am trying to do 2 things:
1. Verify the motor has enough power
1. Compute the forces that acts on the bearing teeth.
Perhaps I am over thinking this (wouldn't be the first time!) and should use the locked-rotor torque for this second calculation.
As for the equation I2 = I1*(N1/N2)^2 it came from Also when I said "can't figure out the inertia of the motor" I meant I couldn't find it on the data sheets...need to contact the vendor.
1. Verify the motor has enough power
1. Compute the forces that acts on the bearing teeth.
Perhaps I am over thinking this (wouldn't be the first time!) and should use the locked-rotor torque for this second calculation.
As for the equation I2 = I1*(N1/N2)^2 it came from Also when I said "can't figure out the inertia of the motor" I meant I couldn't find it on the data sheets...need to contact the vendor.
So all you need to do is to overcome friction and forget about the inertia of the motor.
Looks like the first stage could be a dryer type drive, i.e. a belt drive from the motor shaft to the 50" drum that could give you a 100 to 200 gear ratio and then add the next gear stage between the internal gear to get you to 19000,
GR2=19000/200=85
or something like it .Now to check the time it would take, use
Ibearig/(50/d)^2 for the reflected inertia added to the motor inertia = I and use the formula
t=V/a=V*I/T
V= angular speed of motor , radians/sec
T= motor torque
I= equivalent (reflected) inertia= Imotor+Ibearing/(50/d)^2
d= shaft diameter of motor
Looks like the first stage could be a dryer type drive, i.e. a belt drive from the motor shaft to the 50" drum that could give you a 100 to 200 gear ratio and then add the next gear stage between the internal gear to get you to 19000,
GR2=19000/200=85
or something like it .Now to check the time it would take, use
Ibearig/(50/d)^2 for the reflected inertia added to the motor inertia = I and use the formula
t=V/a=V*I/T
V= angular speed of motor , radians/sec
T= motor torque
I= equivalent (reflected) inertia= Imotor+Ibearing/(50/d)^2
d= shaft diameter of motor
- Status
Similar threads
- Question
- Locked
- Question
- Locked
- Question