torque and power calculation

[Nachooli]

Dear friends..
i was trying to figure out the required torque(NM)and power of motor(HP)to rotate a cylinder, placed vertically on a turn table.
cylider OD=2 meter , ID=1.5 meter. weight of the cylinder is 7500 kg. required constant RPM = 0.1 . any help please..

 

[dicer]

Your missing information required for the calculations.
Like is it on frictionless bearings, is in just sitting on the table (not a turn table)(I miss read), then there is the coefficient of friction etc etc.
Turn table? Same deal. Is there a gear reduction unit? How much power is required to turn it.

 

[jraef]

You will likely need to know your desired acceleration rate as well.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
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[Nachooli]

thanks dicer and jraef for valued time..the cylinder is placed on a rotating platform,which is fitted with a smooth bearing.so i think coefficient of friction and the power to rotate the platform can be considered as negligible.only the job mass matters...thanks..

 

[jraef]

Maybe, but the amount of torque needed to accelerate in 1 second is very different than the amount of torque required to acceleraqte in 1 minute.

t = WK² x rpm
308 x T av.

T = WK² x rpm
308 x t

WK2 = inertia in lb.ft.2
t = accelerating time in sec.
T = Av. accelerating torque lb.ft..

Inertia reflected to motor = Load Inertia * (Motor rpm/Load rpm)²






"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[7anoter4]

I'll try to translate in SI what jraef said in Imperial units.
First of all, if the friction force of the cylinder on the table is negligible then the induction motor will reach the no-load velocity [close to synchronous speed].
As jraef said, the start time does matter here.
We have to calculate the inertia moment of the cylinder.
J=mass*dia^2/8
mass=weight/g [kg] g=the gravity acceleration=9.807 m/sec^2.
mass=7500/9.807= 764.76 kg
So J=764.76*2^2/8= 382.38 Kg.m^2
Let's say the synchronous speed of the motor [for 50 Hz] will be 750 rpm.
The inertia moment "seen" from the motor [the reduction ratio 750/0.1=7500] will be
382.38/7500^2= 6.8/10^8 kg.m^2 [of course, negligible].
The friction will be negligible also. See:

http://www.skf.com/portal/skf/home/products?maincatalogue=1&lang=en&newlink=1_0_31

Let's take a SKF thrust ball bearing of for 16000 N load and 100 rpm. The losses power will be 0.0117 w [negligible].If there is not other force acting on the cylinder as a load you have a motor working on no-load[could be any motor].

 

[jraef]

Hmmm, my initial post was not that clean because of underline and alignment issues, someone cleaned that up. Thanks to whomever that was!

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[7anoter4]

What you said, jraef, it is obviously correct. But, someone said once upon the time:
"The God stays in the details".