Induction Motor Starting Current on "NO-Load" and "ON-Load"?

(1) Yes, the starting current can be significantly larger than the unloaded current. Why do you think many induction motors have a centrifugal switch connected to starter windings?

(2) I have not bothered to calculate any motor current conditions since I took a motor fundamentals class. There are many additional parameters (core losses, windage, bearing drag, squirrel cage topology and materials, poles, gaps, etc.) that must be known to calculate the motor operation. I prefer to use the motor manufacturer's specifications.

(3) Yes, there is an upper limit to the current that can be driven into any power grid driven device. That limit is usually established by the core saturation of an upstream transformer. The only time this needs to be known is in calculating the arc flash condition. (Same class) For safe usage of a motor, just follow the manufacturer's data in selecting proper circuit protection.

The induction motor start current does not depend upon load, but the starting duration does. For D.O.L. [direct-on-line] start, the current will be the same as locked rotor current[See NEC Table 430.7(B) Locked-Rotor Indicating Code Letters]

Net acceleration torque for any speed S is Tacc=TM-TL [this would be slight variable and we should state it is constant] where:

TM=motor torque TL=load torque

tstart=wk^2*(ωmax -0)/Tacc/308

TMrat = Php 5,252 / ωmax [rated torque-lbs*ft]

Php=Motor rated power [HP]

ωmax= Motor rated speed [rpm]

If it is NO-LOAD motor start TL (approx..)=0 then Tacc=TM then tstart will be very short.

See also:

Re: Induction motor, inrush current vs locked rotor current

If no load current is stamped on the nameplate we need not guess it. Similarly the maximum load(inertia)which a motor can start and the time taken to start should be stamped in the nameplate.

I've never seen no-load current on a motor nameplate.

It's not as low as you might expect (20-30% FLC springs to mind, but don't quote me) as the power factor is very low in those conditions.

Maybe we could help you more if you were to divulge why you are asking these questions.

If we should take Steinmetz Equivalent Circuit and we could establish the Xrotor and Rrotor for a slip=1[slip=(rpmsynchron-rpmactual)/rpmsynchron] for rpmactual=0 at start [see for instance: https://www.selinc.com/WorkArea/DownloadAsset.aspx?id=3690] neglecting Xm and Rfe we could get :

Istart=V/sqrt((Xs+X'r)^2+(Rs+R'r)^2)

As the load is represented here by the slip, as we can see at start the slip is not involved in Istart formula. That means Istart does not depend on load.

link not working

Try again:

https://www.selinc.com/WorkArea/DownloadAsset.aspx?id=3690

thanks.

Dear Mr. tigerdon,

The No Load Current will be 35 to 40 % of FULL LOAD CURRENT but Power Factor will be very LOW to the extent of 0.3 to 0.4.

For other details pl. Refer Standard Text Book of Elec.Engg - INDUCTION MOTOR - CIRCLE DIAGRAM, which will clear all your doubts.

DHAYANANDHAN.S