Guest Question.

4-pole, 50 Hz, 3-phase 400 V induction motor develops 20 KW including mechanical losses when running at 1440 rpm and the p.f is 0.8. Determine (i) rotor current frequency (ii) total input if the stator loss is 1000 W (iii) the line current.


The rotor current frequency corresponds to slip frequency sf where f is the normal frequency.


(i) Now, the synchronous speed of the motor with 4 pole at 50 Hz is


\(n_{s}=\frac{120\, f}{P}=\frac{120\times 150}{4}=1500\) r.p.m.


and rotor speed is 1440 r.p.m.


Hence %\(s=\frac{1500-1440}{1500}\times 100=4\)%


The rotor current frequency is \(sf=0.04\times 50=2\) Hz


(ii) Since power input to rotor \({I_{2}}^{2}R_{2e}\frac{1-s}{s}+{I_{2e}}^{2}R_{2}={I_{2}}^{2}R_{2efs}\)


The ratio \(=\frac{Mechanical\: power\: developed}{Power\: input\: to\: rotor}\)




Hence power input to rotor \(=\frac{Mechanical\: power\: developed}{1-s}\)


\(=\frac{20}{1-0.04}-\frac{20}{0.96}=20.83\) KW


Since stator loss is 1 KW, the total input to the stator is 21.83 KW


(iii) Hence the line current


\(\frac{21830}{\sqrt{3}\times 400\times 0.8}=39.4\) A