rating of induction motor

Is it Three Phase or Single Phase ?

both

If I understand, the ampere rating is given on the name plate.

Try these: kW = I x E x 1.73 x PF / 1000 (Must be three phase since you use 1.73)

HP = I x E x 1.73 x Eff x PF / 746 (If you need HP)

The amperage on the name plate is at full load.

Good luck,

James

Hastings,

Don't you need to use the efficiency for the KW version as well? I had a motor recently that was:

6.3 KV, 257 Amp, 0.85 PF, 0.945 eff, which resulted in a power of

6.3*257.0.85*0.945*1.732= 2252KW.

Hi Sandeep,

I think that, you don't need recalculate the ampere rating of induction motor.

Just use the ratted current of Motor manufacturer, and try to use the famous manufacturer e.g. Siemens, Fuji, Mitsubishi, AEG,.....

Best regard.

Hi ,

Few months ago , i also got the same problem while caluculating the amp rating of induction motor. what i concluded is we must take the factor of efficiency .If u get the correct answer plz let me know at pavan5ee@gmail.com . plz.

have a nice day .

pavan.

Hi friends, this is d induction motor rating provided on nameplate= 5 kw , current 9.2 a , v= 400+-10 , eff. 86 percent , p.f =0.92 , rpm=2880. now we use d formula den p = 1.732#400#9.2#0.92. d value of p is not comes to 5 kw. plz prove it.

Yes. looks good.

5857 using only PF, 5037 with eff factored.

James

OK james, it means i have to consider eff. can u tell me wat r d other equipments in which we need to consider dis eff. like in case d.g sets we didt consider eff. to calculate current. in their rating kva , voltage, current r required eff. is not y it is ? plz clerify

Yes It is approximate.

P_motor=√3*U*I*cosφ(PF)*µ(Eff)=1.732*400*9.2*0.92*0.86=5043W=5.04KW.

Therefore value in nameplate is 5KW.

And can you try again: Siemens motor with 400V three phases, 8.2A, 4KW, Cosφ=0.83, µ=85%, 1440 rpm.

And please don't recalculate something, that were standardized. Three phases Induction Motor is one of them.