3 phase Induction motor under low load conditions

It is not efficient. Whoever told you this doesn't know what they are speaking of.

Running a motor designed for delta in star mode (assuming it is a NEMA configured 12 lead motor) will reduce the torque to 33% of normal, and reduce the current to 33% as well. But f the load is still at 50% of normal, the motor will overload.

Even if the motor were only loaded to 33% of capacity, the potential savings is minimal at best. You will have a very slight drop in magnetization current as apercentage of total current, so the efficiency will look ever so slightly better.

Thank you very much for your reply.

Suppose that the load in less than 33% of rated and the torque is sufficient for the use, then is there any other consequences to the lifetime of the motor if it is continued to be used in star mode?

is there any other -?-

NONE

If the load is less than 33%, why do you have such a large motor? If the answer is that sometimes you need more power, then you should not ever connect it in star. The first time you need more power, you will overload the motor.

I should have noted: if you are using dual voltage IEC motors and using the higher voltage rating, you motor will be in star anyway!

Tried to add this in Edit mode but typed too slow...

AC motor power consumption is composed of 3 main elements:

1. Load power; the amount of power it takes for the motor to rotate the load in it's attempt to pull back into the slip speed. This represents approximately 70% of the power consumed. If the load is less, the power is less. You don't need anything fancy to save this energy. The motor only uses what it needs.
2. Winding losses; the amount of power lost due to magnetization of the iron core and the resistance in the copper windings. This represents about 25% of the power consumed overall. The copper losses are fixed as long as all of the windings are used, so there is no savings potential there. The iron losses are as a percentage of the applied voltage and the overall power consumption, so the component of those losses which is based on the applied voltage can be reduced by reducing the effective voltage, which is in effect what you would do. At best, this represents only about 10% of the total power consumed, but don't forget, it is dependent on the load power in the first place, so if your load power (from above) is already below 33%, the savings percentage drops as well.
3. Windage and friction losses; the power lost to moving air across the motor for cooling and the bearing friction encountered in the motor, load and power transmission components. Again, these are fixed losses that account for around 5% of the total power consumed.

So in a nutshell, you can save a reduced amount of a reduced amount by reducing the effective motor voltage, but it's hard to predict how much the savings would be in advance and comes with the risks mentioned above.

Thanks indeed for your explanations!

As you have correctly mentioned, these motors are sometimes used for higher powerapplications. But at the moment since the loads are aroung 30% at an average, therefore it is being used in star mode. The torque and speed found to be okey for their applications. My worry was whether there will be any negative impacts to the motor that is continuously running in delta.

After your explanation I feel it is okey to run in star under low load conditions, becase the current is going to be less.Isn't it?

It should be OK.

The important thing is to have the correct overload installed / set when running in star.

Hi JRaef : Can you clarify something for me ? I thought that the current would be reduced to 1/(sq.rt. 3) in star mode and of course the torque would be reduced by the square of the current or ( 1/( sq.rt. 3) )squared = 1/3 ??