Motor Loads and Starting Currents

Can I recommend to you, to used a tong tester and clamp it to the motor lead and test run the motor with free load. From the reading of the tong tester you can see what will be your motor current without the load.

I do believed that you do have a tong tester with you as it is a very common equipment that every electrician should have or at least one.

Do you mean what will the starting current be without mechanical load, or do you mean what is the current required to run the motor without mechanical load?

No-load current will vary with different motors, but they don't do any work, so that information is not very useful.

Starting currents vary too, but if a load is applied to the motor with a clutch assembly, the motor current may spike to a value similar to the starting current with load attached, the only difference being that higher currents will occur at lower armature speeds.

The bottom line is that "there ain't no such thing as a free lunch." Energy will have to be expended to accelerate an inertial load to running speed, and that amount of energy doesn't change unless the inertial mass changes.

The no load current will be in the range of 25% to 40% of the full load current (FLA). So, in your case the no load current is in the range of 3.5 A to 6 A. As other person mentioned, you can easily measure it with a tong tester.

The starting current will be 6 to 8 times of the FLA. So, in your case the starting current is in the range of 87 A to 116 A (assumed DOL starting). If the motor is started with no load, the duration of this current will be less than 5 seconds.

- MS

The starting current of an induction motor is a function of the motor design and the starting voltage. The full voltage start current is independent of shaft load and is initially equal to the locked rotor current of the motor and gradually reduces as the motor accelerates, only dropping significantly when the motor reaches 80% speed.

For more information, see http://www.lmphotonics.com/m_control.htm

The inertia of the driven load influences the start time.

Best regards,
Mark.