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hammer mill - constant torque or variable torque? 4

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rhatcher

Electrical
Jan 17, 2001
636
I am considering a motor redesign to change the speed of a hammer mill that is used to shred steel. Does a hammer mill respond like a variable torque load or a constant torque load? If we change the speed, does the power change proportionally or as a square or cube?

Thanks
 
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Choosing a VFD has nothing to do with the torque-speed characteristic of the load. It has only to do with the desired short-term overload capacity.

If you only need 10% short-term overload torque, choose the Variable Torque rating. If you need 50%, choose the Constant Torque rating.

Due to the confusion these terms generate, some manufacturers are now labeling 10% rated drives as Normal Duty and 50% rated drives at Heavy Duty. Too bad they all don't switch to that terminology. It would be a lot less confusing.
 
Star for DickDV, I have made the mistake of underestimating the short term torque requirements of a similar load, only clever programming of the drive let me get away with it. But is was a sub-optimal solution.

cheers Niallnz
 
Thanks for the answer DickDv but that is not what I was looking for. There are no VFD's here. The hammer mill drive motor is 4000V, 4000hp, 600rpm. I am considering changing the speed of the drive motor by rewinding and am trying to characterize the load. So...

Does a hammer mill respond like a variable torque load or a constant torque load? If we change the speed, does the power change proportionally or as a square or cube?
 
That would be an "oops"! Sorry for misreading your post. Someone else needs to comment on the power characteristic of a hammer mill.
 
It would be a constant torque load characteristic as far as the physics of moving the hammers. No change in the work being done based solely on speed.


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Thanks jraef, a star for your answer. I was thinking the same thing about the hammer mill with no load. We are spinning a really heavy roll (100k lbs.) with hammers that will swing out under centrifugal force and with no resistance except for windage and bearings. I have amp readings for the same mill running with no load for two different motors at 450 and 600rpm. The results, calculating kva and estimating horsepower, indicate constant torque within 10%. That is close enough for me considering different design motors from different manufacturers and my "estimating."

What I am not sure about is what happens when you feed something into it. I only have amp readings under load for the 450 rpm motor. Call that P1. If this was a conveyor (constant torque), I would assume that P2, the power at 600 rpm, is P2 = P1 * (600/450).

If this were a fan or impeller pump (variable torque), I would assume that P2 = P1 * ((600/450)squared).

So...does anyone have any ideas about this? I am torn between the idea that it must be constant torque and the nagging feeling that perhaps it is variable torque. I do feel comfortable that it could be compared to cutting with a saw or turning with a lathe, but I do not know if those are CT or VT either. Can anyone help?
 
Those are both constant torque as well.

Consider this; as you feed material into the mill, does the rotor weight increase, i.e. does the material couple with the hammers? No. The impact force of the hammers on the material causes them to slow down, which increases the motor slip and causes it to pull more current to attempt to return to base speed, but the rotor inertia remains the same (forget the insignificant amount of crud build up on the hammers by the way). So the same holds true if you are starting at a different speed to begin with. The amount of torque required to return the rotor speed to "normal" is still based upon the rotor mass involved, which has not changed with speed.

Now the HP will change with speed, because HP is based on torque AND speed, so if you increase the speed and not the overall HP rating of the motor, you might find yourself in trouble, but that's a given.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376
 
rhatcher
My previous company had a great book on applications for VFD's. One of the special ones is "Drives with periodic load changes and surge loads" in the attachment. It goes into the science of sizing a drive for such applications. Very useful and has been one of my bibles for many years.

Note: the attachment is about 3.5MB
 
 http://files.engineering.com/getfile.aspx?folder=3a74a6b9-3caa-4110-8202-d19b889590f2&file=Applikationshandbuch_e.pdf
I assume you want to go up in speed to 720rpm???

Are you doing this change so you can feed more material? If so, then you'll need a bigger motor sized based on the new vs old feed ratio.

So, if you feed at the same rate, running at 720rpm vs 600rpm (20% speed increase) means the motor needs to be 4000hp * 1.2 = 4800hp. But then, you probably want to feed the material 20% faster so you now need 4800hp * 1.2 = 5760hp. So, to me, it looks like you'd want to have about a 6000hp motor.

Another issue I see is that the rotation energy of the mill is a squared function. It will take 1.44 times as much energy to accelerate to 720rpm compared to 600rpm. This will take more time and cause more heating of the motor stator and rotor. In these applications, it is often the acceleration of the huge inertia which is the most demanding part of the operation.
 
"rhatcher"
Welcome back Ray!

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Thank you all for your answers, especially jraef (another star for your 6/19 post), and thank you for the welcome back Pete. I've been away for a long time and it's good to see your familiar face.

I have concluded my work on this and (always with crossed fingers) submitted my design. However, if any of you are familiar with winding design theory please speak up. This is not a common subject and I am always looking for someone to trade ideas with.

Thanks again to everyone.
 
Sorry for forgetting you lionelhutz. Your calculations are correct (in my opinion) and much appreciated.

Thanks
 
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