Calculation for proper power (in kW)

In summary, a three phase 220V/50Hz motor must be able to rotate a shaft with a mass of 1500 kg, diameter of 0.4m, and length of 3m. It needs to achieve a speed of 600 RPM in 5 seconds and have enough strength to maintain that speed. The motor is connected to the shaft through transmission belts in a 3:1 ratio. To calculate the power needed for this motor, torque is determined by multiplying the moment of inertia (30 kg-m^2) by the angular acceleration (12.56 r/s/s). This results in a torque of 376 kg-m^2/s^2, or 62.7 n-m at the motor.
  • #1
mcupwr
3
0
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,
 
Physics news on Phys.org
  • #2


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


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


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


Big Thanks.
 

FAQ: Calculation for proper power (in kW)

What is the purpose of calculating proper power in kW?

Calculating proper power in kilowatts (kW) is important for understanding the amount of energy needed to perform a certain task and for properly sizing and selecting equipment such as generators or motors.

How do I calculate proper power in kW?

To calculate proper power in kW, you will need to know the voltage and current of the system. The formula for calculating kW is kW = (voltage x current) / 1000. Make sure to use the appropriate units for voltage and current (e.g. volts and amps).

What factors can affect the proper power calculation in kW?

Several factors can affect the proper power calculation in kW, including the power factor (the ratio of real power to apparent power), the efficiency of the equipment, and any losses in the system (such as resistance in wires or other components).

Why is it important to consider the power factor in the calculation for proper power?

The power factor is important because it reflects the efficiency of the system. A low power factor means that the system is not using energy efficiently and may result in higher energy costs. It is important to consider the power factor in the calculation for proper power to ensure accurate results.

Can proper power in kW be calculated for both AC and DC systems?

Yes, proper power can be calculated for both AC (alternating current) and DC (direct current) systems. However, the formula for calculating kW may differ slightly depending on the type of system. For AC systems, the power factor must also be taken into account.

Back
Top