Starting current and torque in star cicuit

Look here.

When the windings of a 3-phase motor are connected in STAR:

• the voltage applied to each winding is reduced to only (1 /._'/^_'3) [1 divided by root three] of the voltage applied to the winding when it is connected directly across two incoming power service line phases in DELTA.
  
• the current per winding is reduced to only (1 /._'/^_'3) [1 divided by root three] of the normal running current taken when it is connected in DELTA.
  
  
• so, because of the Power Law V [in volts] x I [in amps] = P [in watts],
  
  the total output power when the motor is connected in STAR is:
  
  P_{S =} [V_L x (1/._'/^_'3)] x [I_D x (1/._'/^_'3 )] = P_D x (1/3) [one third of the power in DELTA]
  
  where:
  V_L is the line-to-line voltage of the incoming 3-phase power service
  I_D is the line current drawn in DELTA
  P_S is the total power the motor can produce when running in STAR
  P_D is the total power it can produce when running in DELTA.
  
  
• when the motor is connected in STAR is that the total output torque is only 1/3 of the total torque it can produce when running in DELTA.

It is already explained.

Here is some additional:

Starting current: If the motor is started with star winding, the voltage applied to the winding is 1/√3 of the supply voltage and hence, the current is also 1/√3 of the Delta winding starting current.

Starting Torque: Torque developed is directly proportional to square of voltage and so the star winding starting torque is (1/√3)² = 1/3 of the Delta winding starting torque.

Please see the thread http://cr4.globalspec.com/thread/54100 for avoiding the confusion about star/delta configuration of motor.

- MS

Thanks to all.

It means that we can not run such a motor in star where a high starting torque require such as conveyor belts air compressors as in star config, the torque reduces 1/3 of motor full load torque. Am I right?

2nd question, what is the max starting torque can produce an AC motor?

Best Regards

Compressor load demand also is low when the RPM is low. That hence may not be a problem.

What usually is done is that you get the load characteristics curve - speed/torque characteristics (or speed- power and convert into the speed torque)

get the same for motor. Then overlap one above other, the intersection will give you the operating point.

If you see the load torque is more than the motor, it will not start.

Conveyor is another problem, since if it is fully loaded, the torque in the start is highest (static friction) then it goes down. So either

Take the motor of higher capacity.

Start the conveyor without load and then start loading it.

Slip Ring Induction motors ( will have better starting torque) also more loss. There are certain double cage motors acting with high resistance cage at start (almost like SRIM) and then as the RPM increases it works as normal SQIM.

DC Motors - best starting torque.

Variable Frequency Drives - that may provide you with high starting torque (even there are VFDs that can give > FL Torque at start)

All this decision will be based on your load characteristic curve. Y must take care of the starting torque, in conveyor and similar the requirement is taken to be 20% higher - and although VFD will deliver it, whether motor shaft can withstand it or not is questionable.

BTW: Smaller Pumps, compressors etc at our place are usually started with star-delta without any problem.