Starting of Induction Motor

A1) It depends upon the inertia of whatever it is driving.

A2) Starting at about 5 times full load current dropping back to full load current or below in the time A1↑.

Current isn't consumed. It is drawn, taken or supplied, among other things.

Thank you can you give me the time current Characteristics for star delta starter for an example?

Not possible, as there is no information on what the motor is turning, which affects the gradient of the slope at any point on it.

The other factors that affect the slope are:

• The rating of the motor
• The voltage of the supply
• The torque/speed characteristic curve of the load
• The timer settings in the starter
• The rating of the supply
• The loads applied elsewhere on the supply
• Etc.

Please note that the magnitude of the starting current of an induction motors is decided by the motor parameters and is normally given by the motor manufacturer. The duration of the starting current is decided by the inertia (GD2) of the motor's rotor AND of the connected load. Obviously, the higher the total GD2, the more would be the starting time.

For example, if a motor with a rated full load current of 100A, draws, let us say, 6 times its rated current during starting (with a DOL starter), then even if the motor is not coupled to any load, it would still draw 600A; but, in such a case, this 600A starting current would last for a very small duration of, say, about 4 seconds. Now, if this motor is coupled to a centrifugal pump, then again this motor would draw the same 600A during starting, but now the starting current would last for about 10 seconds. If the same motor is coupled to a centrifugal blower, again there would be a starting current of 600A, but now, this might last for about 20 seconds.

With star delta starting, the acceleration time would increase for the same load, as the starting torque of an induction motor is proportional to the square of the applied voltage. And, with star delta starting, the voltage at the time of starting is reduced by root 3 times and thus the starting torque would reduce by 3 times and hence the longer acceleration time.

Hope it is clear.

As it was already said in the above posts, the starting time depends on the acceleration torque, motor rpm and inertia moment of the motor and the load. See-for instance:

http://www05.abb.com/global/scot/scot259.nsf/veritydisplay/bf88560fb1b0335cc2256fc6003e4e05/$File/Motor%20guide%20GB%2002_2005.pdf

section 4.3.3 Starting characteristics

ts=2*pi()*rpm/60*(JM+J'L)/Tacc ts =starting time [sec]

JM=motor inertia moment [kg*m^2 ] JL=load inertia moment [kg*m^2]

J'L=JL*(rpmLoad/rpmMotor)^2 if there is here a gearbox.

Tacc=Tmotor-Tload acceleration torque[Nm]

Tmotor=0.45*(Ts+Tmax) approx. Ts=start torque Tmax =maximum torque

Tload=kl*Tratedload Tratedload=Tnmot=rated motor torque [usually].

Kl=1/3 for centrifugal compressor or pump.