gear ratio change - how does it affect the motor ? 8

[edison123]

My client, a car tire manufacturer, has a rubber mixing mill (called Banburry) driven by a 1500 HP/1000 RPM DC motor. The present rubber mill runs at 40 RPM with a 3-stage reduction gearing from the 1000 RPM motor. He tried to run the mill at 50 RPM (to increase productivity !!) but landed up in problems with the field weakened operation of the motor (vicious sparking, vibrations etc).

Now, he wants to change the gear boxes ratio from 1000 RPM/40 RPM (3 stages) to 800 RPM RPM/40 RPM (3 stages). With this arrangement, he figures he can increase the DC motor field current to reduce the motor speed to 800 RPM to get 40 RPM and then run it at the original 1000 RPM to get 50 RPM at mill end.

My query - if the gear box ratio is changed from 1000/40 to 800/40 RPM, what will the torque and horsepower implications to the DC motor ? More torque / More HP required at the reduced speed ?

 

[jbartos]

Suggestion: Assuming a DC motor with a separate excitation or self excitation (with approximately flat torque speed characteristics), the motor transition from 1000RPM to 800RPM will cause the motor torque to be approximately constant. However, the motor may deliver less HP at 800RPM since the mechanical load profile may have characteristics that have decreasing HP with decreasing RPM.

 

[jraef]

edison123,
Horsepower is the term used to describe a speed / torque relationship. The motor has a fixed MAXIMUM HP rating, so that stays essentially constant (other than overload capacity, temperature rises etc. etc.) As you lower the speed and the torque requirement stays constant, you are lowering the output HP at the same time. Once you go over base speed however, the max. HP can't go up so you begin to lose torque rapidly. At his 20% overspeed he may have lost as much as 50% of the motor's available torque (no I did not get out the calculator), so that may explain the strain on the motor!

When you lower the speed through gear reduction however, you AMPLIFY torque by the inverse ratio. The old speed reduction ratio was 1000:40 or 250:1, so the output shaft torque was 250 x the motor torque (ignore losses for sake of discussion). Now however he is changing the torque requrements via the gearboxes. At 800:40 or 200:1, the output shaft torque will only go up 200 x the motor torque, 20% less than before (200/250) although when he runs at 1000 rpm the output speed will be 20% faster (40/50). The problem then shifts to the work being performed. A Banbury mixer works by "exploding" the rubber slabs between high pressure rollers turning at slightly different speeds. If the Banbury can still shred with a 20% reduction in torque, no problem. Realistically though, it can probably do less effective shredding at faster speed: in other words probably no net gain in productivity, and maybe even a loss due to inefficiencies throughout the system! The proper people to consult on this would of course be Banbury, but if it were that simple, wouldn't they have already done it? Think about it, if they could have simply changed the gearbox ratio and offered increased throughput, why would they NOT? They probably already knew something different.

"Venditori de oleum-vipera non vigere excordis populi"

 

[GordS]

The load needs a specific torque at 40 rpm. It doesn't care what is driving it to get to 40 rpm. By reducing the gearbox ratio from 1000/40 to 800/40, the load torque reflected across the gearbox to the motor shaft should go up by a factor of 1.25.

The power required of the motor should be practically unchanged due to conservation of energy principle. The only difference would be due to relative gearbox inefficiencies.

 

[jraef]

I should add that the likely outcome of lower output shaft torque might be stalling the mixer alltogether! Those rubber blocks are mighty tough. Gear reductions and motor HP designs are a game played best by machine designers on something like that. One could argue that you might increase the motor HP to 1250 to compensate but then someone has to consider the torque inplications in the gearbox and support structure etc. etc. etc.

"Venditori de oleum-vipera non vigere excordis populi"

 

[AZEEMM1]

You might want to consider the implications on the gear drive end too.

Cranking up the torque by 20% (effective through the changing of gear ratios) may have advers effects on the pinions of the Mill and he may need to change more than just the Gearbox.

 

[Skogsgurra]

These kneading machines usually have a very speed dependent torque. It goes up at least proportional to the speed. So, if you plan to run the machine 50 RPM insteda of 40, it will probably mean that you need 20 percent more torque into the machine. Then again, if you change the gear ratio from 1000/40 to 800/40, the gearbox needs another 20 percent extra torque in addition to the former 20 percent. So you will need 44 percent (1.2 x 1.2 = 1.44) more torque from the motor. Not something you are likely to get from it. Not even if you overmagnetize it. The magnetization curve is rather flat above rated field.

In short: It will not work.

 

[Barry1961]

If you increase the work performed by the machine you increase the power required to drive the machine.

In a frictionless, linear world where it took 1500hp to run the machine at 40rpm it would take 1875hp to run the machine a 50rpm.

Another way to look at it would be that if it took 1200hp at 40rpm it would take 1500hp at 50rpm.

If you have some way of measuring how loaded the motor is now you might get an idea of how much more you can ask it to do.

I have no idea if the power required will increase in a linear fashion as the process speed increases.

Hope this helps.

Barry1961

 

[edison123]

Thx guys. So, to recap :

As the motor speed is reduced (by increasing the field current and assuming there is no saturation), the motor output (in terms of armature amp-turns) is reduced to maintain the winding temp limits. The motor output torque stays the same at the lower speed (since torque = k*field flux density *armature amp-turns).

But the gear box output torque will now be reduced by 4/5 (=800/1000). If this lower torque is sufficient to run the rubber mill, the proposed system of 800/40 RPM gear box will work with a lower load on the motor.

But if the lower gear box torque (at 800 RPM) is insufficient (more likely given the rubber load on the mill), then to increase the torque, the motor amp-turns (or rather the motor amps) has to increase. This means a higher motor output requirement at a lower speed, which as stated above, is not possible due to temp rise issues. So, the proposed system will not work.

Correct me, if I am wrong.

 

[jbartos]

Comment: The motor may actually work well if the motor manufacturer built the higher HP motor than its nameplate rated HPs state.

 

[aolalde]

Edison123

My opinion is that the approach of the problem has been twisted for concepts as: “ he figures he can increase the DC motor field current to reduce the motor speed to 800 RPM to get 40 RPM and then run it at the original 1000 RPM to get 50 RPM at mill end.” Etc.


The simple fact is: the gear box ratio will be changed from 25/1 original to 20/1 (proposed).
If the actual motor power demand is close to 1500 HP at 1000 rpm, changing the Gear Box Ratio will result in overload of the motor.
However if the actual load of the motor is much lower than 1000 HP, it could work with the new gear box.
The load power demand will be increased because of the speed increase of 25% ( 50/40)
But the motor will continue running at 1000 rpm.
The increase of production is not free.

 

[jraef]

Wow, after reading all the other posts refering to 25:1 and 20:1, I realized I was off by a factor of 10!

Oh well, whats a zero between friends.

"Venditori de oleum-vipera non vigere excordis populi"

 

[macmil]

Disclosure; I am a mechanical guy.
You gave the motor nameplate data. What is the actual motor amps / horsepower in normal operation ?
On the assumption of constant torque in the mill, the horsepower requirement will increase by 50/40, 25%.
If, as someone suggested, the torque will increase, even linearly with speed, your motor horsepower requirement will increase by 1.25 x 1.25, or 56%. If the torque increases by say a square relationship, it will be even more. Will the motor be able to provide that ?
And, as someone else asked, will the gear box and other mechanical components take the increased torque ?
Unless the motor is currently running way below rated capacity, I would suggest a very cautious approach.

 

[edison123]

Thx guys for your valuable posts. I have advised the client to:

1. Check with the Banburry maker about the implications of 800/40 RPM and 1000/50 RPM operation (whether the production will really increase by increasing the speed, what is the mechnical stress on the banburry etc.).

2. Buy a new DC motor rated 1500 HP at 800 RPM since at present at 1000 RPM, the load is hitting a peak of 100% rated for a short duration (typically 12 to 15 secs).

As for the proposed new gear box, it is up to the gear manufacturer to take care of all the additional torque stresses because of lower speed.

 

[jbartos]

Suggestion: The application might potentially be fine tuned over a electromagnetic coupling, e.g.

http://www.magnadrive.com

 

[GOTWW]

Gearbox, reflects torque. Power reflects energy per unit time *efficiency. Gear boxes reduce efficiency.

 

[ScottyUK]

A useful electrical analogy to a mechanical gearbox is the transformer:

Transformer: Power = voltage x current
Prim. volts x prim. current = sec. volts x sec. current

Gearbox: Power = speed x torque
Input speed x input torque = output speed x output torque.

Obviously the gearbox is less efficient that the transformer, but then it was built by mechies, not electrical engineers .




-----------------------------------

Ask a silly question and you are laughed at for a few moments.

Don't ask the question and you might be laughed at for eternity.

 

[ICEMAN]

Your clients's motor is 1500HP based on rated speed of 1000 rpm. So from 0 rpm to 1000 rpm the power (HP) is directly proportional the speed; the speed is (almost) directly proportional to the applied armature voltage if we ignore the effects of armature reaction et al. If the gearing is such that the decrease is from 1000 rpm to 40 rpm then the power available will be very similar to 1500HP less losses.

The operating area in armature voltage control is based on a constant field flux. This is called the 'constant torque area' where the available torque is constant and subsequently the power varies with speed (voltage). If you reduce in speed you still have the same level of torque available and the power is within the speed torque envelope.

If you go into the 'field weakening area' where the armature voltage is kept constant and the speed is increased by reducing the field flux (speed is inversely proportion to flux) the power available will be fixed and the torque available with progressively reduce with increases in speed. In this condition the motor can reach instability if the nature of load is either constant torque or square load and the motor will stall; in doing so will inhibit the type of sparking you mentioned.

Tell your customer to stick to making tyres. Hopefully, he knows more about that subject.

ICEMAN

 

[jbartos]

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