Motor Size

The question is a lot more complicated than just the weight. You have to say something about how the load is mounted, the moment of inertia of the load, whether you're fighting gravity, what kind of acceleration you need, and whether you're using any type of control system. There's probably more, but that's a start.

10tons of mass at 350 RPM? Do you want to rotate it? at what distance? It may consume huge power, or you want to move it horizontally?if yes in what time to what distance?

Its not a complete question.

For such low rpm applications motorised gear units are recommended.In this case if HP works out higher (More Than 60KW)then use a gear box which will ensure size of motor is kept to a minimum.Work out economics between a 6 and 8pole motor.To arrive at the rating you may have to give details as mentioned by other members.What is the distance to which u want to move?any reversals?

hello friend

i don't think with weight and rpm , we can design motor power ,so would please explain your application in detail which may help to design.

When Bodine Electric gets their web site up and running again, look for engineering info/handbook links. I've seen a good treatment of this topic here. It is from an old application brochure.

You really need to define this problem a LOT better though. Make friends with a mechanical engineer. You'll be able to tell who they are because their feet will be on the ground.

You are an electrical engineer?

From which college?

Mail order?

Your curriculum did not cover such basics as this?

Go to a real college and learn.

If you can't help, please let orthers do it, instead of trowing impolite comment.

Use whatever motor you want! Be advised though that the greater the horsepower the faster and farther you will be able to move your load for a given amount of time. The use of mechanical advantage will be needed in most cases. Was this specific enough for you?

I agree with Keywalker when he says use any motor you want. You will also need a big gear box or a multi-stage block & tackle betwen the motor and the load. Just make sure the motor is big enough to overcome the friction losses of the gear box.

Also make sure that you can stop the 10 tons once you get it moving.

Sounds like a job for a Hydraulics system....

There's just not enough information here to give a creditable answer. Please make up a sketch of the application, and answer the questions; horizontal load? vertical load, dragging across a surface with X coeficent of friction, how large is the object in contact with this surface, it's coeficent, is it supported along it's CG axis or off center? Startup load, stopping load and so on.

suppose you want to move 10 tons at 'x' distance.Therefore 10tons*x*9.8=total WORK you required.

formula for torque generated in motor is:

T=(E*I)/(2*PI*N)/60

WHERE, T=generated torque,

E=back e.m.f;

PI=3.141;

N=R.P.M of motor ;

now,E=back e.m.f=V-i.r;

where V=applied voltage to the motor ;

i=current drawn by armature

r=internal resistance of armature(coil) .

from above formulae you can find your REQUIRED WORK and GENEATED TORQUE

NOW,T*2*PI*N=POWER DEVELOPED.is MUST be > required work.

you have two formulae for finding E & I.after getting required work.

after finding E you have another formula for E which indicates & determines flux(let Ø hear) total no. of conductors (Z),no. of poles(P), & no.of parallel paths in winding(A).

is: E= (Ø*Z*N)(P/A);

WHERE, N is R.P.M,

A=no. of parallel paths (=2 for wave winding &

=P for lap winding)

this information may helpful you for designing motor.