2-Pole vs. 4-Pole Motor Using VFD & Torque Problem

No way to determine from here what the LOAD requires by way of torque. But you do know that the motor you have now provides enough, and it is running at 80% load (or so you say). Do the math.

Calculate the torque from each motor, take the torque from motor A x 80%.

If that value is greater than B, you can use B.

Probably not though. Quick rough check:

11kW motor x 80% = 8.8kW needed

13.2kW motor x 100% x (1450/2940) = 6.5kW deliverable

The only way it will work is if, by chance, your "80% load" on motor A is still more than the load actually needs. If you are calculating "80% load" by simply looking at current, that's not really a true indicator of load kW. There is a slight chance you are off by enough of a margin because of power factor and efficiency at load speed. If you determined "80% load" by using a kW meter so it really is kW, then it isn't going to work.

Oh damn, I hope I didn't just do someone's homework!

"If by reducing the RPM Torque of this motor increases. If so it can easily Drive the Belt

because in that case motor have approximately double the torque than previous one i-e"

When you reduce speed with a gearbox you are correct, the torque increases inversely proportional to the speed decrease. But when you reduce speed with a VFD, the torque remains constant. So running a 2940RPM motor at 1450RPM by turning down the speed with a VFD will not give you more torque, it will still be the same, 42.83 N.m.

Is there some element in the drive train such as a pulley or sprocket that can be easily changed so that the faster motor can be run more to the high range of speed?

For that matter, would it hurt if the conveyor belt ran faster? It would then carry less weight load.

"For that matter, would it hurt if the conveyor belt ran faster? It would then carry less weight load."

Solution from this side it looks difficult.Actually this belt extract material from a Silo. This Silo (Clinker Doom) has two discharge gates. These discharge gates have either Fully open or Fully Close. One is big gate (280 T/h) and other is small (130 t/h). When both gates are open and belt is running at 55 RPM (Pully) we have approximately 400 t/h of material to two mills of 180 t/h.

We need this arrangement 13.2 KW, 2940 RPM in emergency situation.

Even if 2940 rpm motor had same full-load torque as 1450 rpm, it is only going half speed. Cooling fan will deliver less than half normal flow - motor would overheat and fail quickly. Operating continuously above rated torque means stator and rotor currents are above rated and overheating will occur. Losses increase with square of current, so pushing much over rated soon gives trouble.

This Video contain some of information here required

http://www.youtube.com/watch?v=KJPMYb8BygM

You're right. With that set-up, speeding up the belt won't help. (I was hoping that an easy suggestion would work.)

I may not understand properly what you can do with your process. You have two discharge gates, does the problem motor drive both?? I was thinking that if 1450 rpm motor was 80% loaded, then 280T/h is 70 % of 400 T/h and 0.8 x 0.7 is 0.56 which means you have more chance of the 2940 rpm motor surviving [ to give you 70% production] -- but do not forget the cooling problem which I described in post #6!!

Based on all of the calculations done by many members on this board. You need to have a spare motor on site. As all of my thoughts have already been posted by others. This is your only real option.

If you can advise the current motor current under existing load conditions OR the actual measured motor speed we can estimate the actual load to see if the v2 pole motor has any possibility to drive this.

Also, consider starting torque, what's your starting situation?

Doesn't your VFD give a load indication anyway (torque, motor amps etc?)

VFD is ACS 800

It has many option on display unit like

RPM

Frequency

Power

Torque

etc

So what are the running and starting loads look like in terms of torque and motor current?