# Calculation for proper power (in kW)

1. Nov 1, 2008

### mcupwr

Hi all,

I'm new here and not so good in physic.
I need help with calculation for proper power (in kW) of three phase 220V/50Hz motor. This motor must be able to rotate shaft which have mass of 1500 kg, diameter 0.4m, and length 3m. Here is important that it must achieve speed of 600 RPM in 5 sec, and it must be strong enough to "control" shaft that it keep constant speed later.
Motor is not connected directly to the shaft it is connected with transmission belts in 3:1 ratio.
Maybe it is not the best proper ratio so here is welcome any advice.

Thanks for any help in advance,

2. Nov 1, 2008

### Staff: Mentor

Re: Torque

With the information given, that's a relatively simple calculation if you assume there are no other complications. However if this is a real-world situation, you're talking about a large motor and some serious danger if anything is missed (like friction, low rpm performance of the motor, resistance of anything connected to it).

You need to hire a real, live engineer to work on this problem for you.

3. Nov 1, 2008

### mcupwr

Re: Torque

Thanks for replay,

I just need rough calculation of motor power, without real world problems for now. Can you, or someone else please show how to calculate power for given data above?

Thanks again,

4. Nov 1, 2008

### Staff: Mentor

Re: Torque

Well, ok....

Torque is moment of inertia times angular acceleration.

Your angular acceleration is:
600 rpm / 60 / 5 * 2*3.14 = 12.56 r/s/s
Your moment of inertia is:
I=mr^2/2 = 1500*.2^2/2=30 kg-m^2

So, torque = 30*12.56=376 kg-m^2/s^2
Now using f=ma to convert the units gives 188 n-m (at the shaft. The motor torque will be 188/3=62.7 n-m)

Power is torque times angular speed, so your power is 188*600/ 60*2*3.14= 23613 W or 23.6 kW

5. Nov 2, 2008

Re: Torque

Big Thanks.