Constant Torque Motor

[desertfox]

Hi

Hope somebody can sort my confusion out:-

I have a very old assembly drawing (1985) which shows an old DC motor on it and the only information I have on it are these figures 5.5kW at 1500 rpm and 2.75kW at 750rpm.
Clearly the torque at these two speeds is the same and I understand that basically it's the same torque up to 1500 rpm.
Now my problem, I have a screw jack that is driven through this motor,in fact I have two screw jacks and the jacks come with a standard ratio of 8:1. I need a power of 2.75Kw to each jack according to the screw jack catalogue, however to raise the jack in the time required I only need an input speed of about 480rpm and not 1500rpm which is needed according to the formula

P=2*3.142*N*T/60
So I was tempted to place a speed reducer between the motor and jack but I have been advised that this May not be necessary because I have sufficient torque at 480rpm , therefore reduce the motor speed to 480rpm and forget the power.
Now I am having a problem understanding this because I believe that unless I achieve the correct power for the jack it cannot perform the required lift in ten seconds, Am I wrong?

Thanks in advance

Desertfox

 

[itsmoked]

Lots of possibilities here for a fail. If both these screw jacks are going to be lifting the same load you could need substantially more force than you think. Is there any way you can actually measure the torque you need? Say, with the jacks installed using a torque wrench?

Keith Cress
kcress -

http://www.flaminsystems.com

 

[desertfox]

Hi itsmoked

It's all theoretical I'm afraid however I don't have an issue with the jacks, I know the torque required to lift the load based on the jack manufacturers catalogue and I also know that the motor gives a constant torque of 35Nm, which when stepped up through the jacks is 140Nm on each jack, this includes efficiency factors.
All I really want to know is whether I should put another speed reducer after the motor.

Thanks

Desertfox

 

[waross]

TRIUMF is Canada's national laboratory for particle and nuclear physics. Its headquarters are located on the south campus of the University of British Columbia in Vancouver, British Columbia. TRIUMF houses the world's largest cyclotron,[1] a source of 500 MeV protons, which was named an IEEE Milestone in 2010.[2] TRIUMF's activities involve particle physics, nuclear physics, nuclear medicine, and materials science.
The particle accelerator is housed inside a large shell. Screw jacks are used to lift the shell when access is reqired.
I heard from several sources that when the accelerator was first constructed, the screw jacks would not perform as expected.
The torque values used for the selection of the motors and reduction gear sets were based on sliding friction, not static friction.

The moral of this anecdote is that it may be wise to consider both motor peak torque in relation to static friction and motor running torque in relation to sliding friction.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[desertfox]

Hi warts

Thanks for your post but again the jacks have been sized from the manufacturers catalogue, what I am failing to understand is the relationship between power and torque in this application.
Because it's a constant torque motor I know I have sufficient torque to do the lift even at 480 rpm, I thought because the jack catalogue sized the jack on power before deriving the torque, I was under the impression the power was important, however I was advised that the power isn't important provided I have sufficient torque which I have and I am within the speed band of the motor i.e. 0-1500rpm which I am, is this correct?

 

[waross]

Torque and Power.
Torque is the twisting force of the motor.
Power is the torque times the speed.
If you don't have enough torque the jacks won't lift.
The load will use only as much torque as it needs.
The more power you have, the faster you can deliver the required torque and the faster you may lift the load.
If you use a gear reduction, even though you may not need it, the motor will run faster but will be called upon to deliver less torque. The current will be less and so the motor will run cooler. The shaft mounted fan will be turning faster and cooling will be better.
There is no free lunch. You will need more voltage to run the motor faster. Very roughly, if you use a 2:1 reducer, the motor will run twice as fast to turn the jack screws at the same speed and need twice as much voltage but only half as much current. The work done will be the same. The same load lifted in the same amount of time, so the power consumption will be approximately the same.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[BrianE22]

I should stay away from numbers because I'm getting too old. I think you need a greater than 2:1 speed reducer (torque multiplier). Your motor torque rating is 310 in-lbs (= 63025*HP/RPM). Each jack requires 484 in-lbs.

I think you are assuming constant power from your motor as the speed of the motor gets below 750 RPM. In reality, below a certain speed, motors are in the constant (limited) torque realm.

 

[Compositepro]

It would be more correct to call it a torque limit rather than a constant torque. 2.75 kW and 750 rpm is a point on a line.
If you exceed the torque limit the motor will slow and stall (and eventually burn-up without separate fan cooling). If you have no mechanical load on the motor then it will speed-up, but not much more than 750 rpm, and the torque output of the motor will be zero.

 

[rhatcher]

Although a DC motor is capable of producing constant torque from zero speed to rated speed, your motor may not be designed or rated for that use.

A standard motor will be cooled by internal fans mounted to te motor shaft. At lower speeds, the internal fan will not provide enough cooling and the motor will overheat. A 2:1 constant torque speed range is typical for a self cooled motor. However, if the motor is equipped with a blower, it can operate down to zero speed since the flow of cooling ir is constant regardless of speed.

 

[desertfox]

Hi All

Thanks for all the responses however I was advised:-

"Provided you have the torque and speed that you require then there is no need to worry about the power" I wanted to know if that was correct or not?

 

[Compositepro]

That statement is pretty accurate. Power is just torque times speed. Speed is limited by mechanical design of the motor (bearings, centrifugal force on windings. Torque is limited by electrical current heating the motor vs. the heat dissipation.

 

[waross]

Actually the speed of most DC motors is limited when the back EMF equals the applied EMF. At full load the speed is often close to the speed limit.
An exception is the series motor. There are not a lot of applications for a series motor other than engine starters.
Torque is limited by iron saturation at low speeds. As the speed increases, the back EMF increases and the current drops. When the current drops enough that the core is no longer saturated, the current determines the torque.
This is the limit of starting or maximum torque.
You are talking about maximum continuous torque.
For varying loads the torque may at times exceed the maximum continuous torque. Then we must consider the RMS torque, which should not exceed the maximum allowable continuous torque.
The current is determined by the difference between the applied EMF and the back EMF.
When the load is increased on a shunt, compound or PM field motor, the motor slows down. As a result the back EMF drops and the difference between back EMF and applied EMF increases. This causes the armature current to increase, and with increased current the torque increases. The EMF difference divided by the effective resistance of the armature circuit gives the armature current.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[sensij1]

Referring back to the OP, if the jack screw catalog is specifying 2.75 kW at 1500 RPM, and you are planning to operate more slowly, then the power requirement will scale down with speed, with maybe a little non-linearity due to speed based friction/lubrication changes in the jack. So yes, as long as motor can deliver the required torque, you will be fine. As other posters have suggested, in the long run, the motor might be the weak link if it is continuously running hotter than it would if the reducer were installed.

 

[mikekilroy]

Of course it is correct, as all replies have alluded.

Some actuator catalogs show motor rating required in HP for easy figuring by non technical folks. But as you find, it is easy to misread their catalogs and not see that HP rating is ONLY for a given MAX speed and given load.

Your quote above tells all, and HP or Kw required is simply not required as shown in previous posts. Actuators should have a speed vs load curve for each motor for each actuator. An example is attached for reference.

As you increase the load, you increase the required HP linearly. As you increase the speed, you increase the required HP linearly. So if you increase both at same time, the required HP increases by both: ie., if you double the load and triple the speed you increase the HP by 6x. Since most motors are generally constant torque devices, going slower than this max speed generally is ok down to the motor/drives min capable speed.

www.KilroyWasHere<dot>com