Motor starting current

[gar]

220325-0123 EDT

The purpose of this question is to see how different persons respond.

You can assume the internal impedance of the power source to the motor is zero, and therefore that input voltage to the motor is a constant.

Assume the motor is a single phase induction motor with a start winding that drops out when motor reaches near full speed.

The question is how does the motor current vary from time t = 0 to when the motor reaches full speed for variations in the inertia load on the motor mechanical output?

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[Julius Right]

Single phase induction motor starts by assistance of a capacitor ,an inductance or a resistance in auxiliary winding.
If the schematic diagram you meant is as attached, then the centrifugal switch it is open only if the starting is finished that means the slip is constant already [ the speed in constant].

 

[Julius Right]

A mathematical correct calculation is difficult since there are two rotor current frequency : one for forward speed[s*f] and another for backward speed[(2-s)*f]
The first stage of inrush current depending only on the time constant [ ratio of total equivalent induction per total resistance] it could be neglected.
Usually, in the same mode calculation as in three phase case.
See-for instance-ABB The Motor Guide
tst=(JM+J'L)*K1/Tacc
where
the average motor torque [TM] it will be approximate 0.45*(Tm+Ts)
Tm maximum torque ; Ts start torque [Nm]
tst start time in seconds
JM and J’L the motor and the load inertia moment
K1 depends on pole numbers
Tacc=TM-KL*TL where TM it is motor average torque TL load torque
KL it is a constant depending on load type:

	Lift motion	Fan	Piston pump	Flywheel
KL	1​	1/3​	1/2​	1​

No poles	2​	4​	6​	8​	10​
K1	314​	157​	104​	78​	62​