Question

A 25 hp, 400 V, 50 Hz, 4-pole, star connected induction motor has the following impedances per phase in ohms referred to the stator side: Rs= 0.641 Ω, R! r = 0.322 Ω : Xs=1.106 Ω, X! r = 0.464 Ω and Xm = 26.30 Ω. Rotational losses are assumed constant and are 1.1 kW and the core losses are assumed negligible. If the slip is 2.2% at rated voltage and frequency, find i) speed ii) stator current iii) power factor iv) output and input power and v) efficiency of motor​

Answer 1

Answer:The equivalent circuit of the motor is shown

 

(i) The synchronous speed for 4 pole and 50 Hz supply is 1500 rpm and since the slip is 2.5% the motor speed is 1500 (1 – 0.025) = 1462.51500(1–0.025)=1462.5 r.p.m.

 

(ii) The total impedance as seen from stator terminals

 

Z=(0.5+j1.2)+\frac{j25(14+j1.2)}{14+j25+j1.2}Z=(0.5+j1.2)+  

14+j25+j1.2

j25(14+j1.2)

​

 

 

=10.42+j7.64=12.92\angle 36.3^{\circ}\Omega=10.42+j7.64=12.92∠36.3  

∘

Ω

 

Therefore, the stator current is

 

I_{s}=\frac{400}{\sqrt{3}(12.92\angle 36.3)}=17.87\angle -36.3I  

s

​

=  

3

​

(12.92∠36.3)

400

​

=17.87∠−36.3

 

Stator copper loss =3\times 17.87^{2}\times 0.5=479=3×17.87  

2

×0.5=479 Watt.

 

(iii) Hence the p.f. angle is 36.3 lag and the p.f. is

 

\cos 36.3=0.806cos36.3=0.806 lag

 

To find out efficiency we should find out the mechanical power developed, for which we require rotor current.

 

The rotor current from the diagram

 

I_{2}=I_{1}\cdot \frac{j25}{j25+14+j1.2}=I_{1}\cdot \frac{j25}{14+j26.2}I  

2

​

=I  

1

​

⋅  

j25+14+j1.2

j25

​

=I  

1

​

⋅  

14+j26.2

j25

​

 

 

=17.87\angle -36.3\cdot \frac{j25}{14+j26.2}=15\angle -8.2^{\circ}=17.87∠−36.3⋅  

14+j26.2

j25

​

=15∠−8.2  

∘

 

 

Therefore, the rotor copper loss is

 

3\times 15^{2}\times 0.35=2363×15  

2

×0.35=236 Watts

 

The power input to motor

 

\sqrt{3}VI\cos \theta =\sqrt{3}\times 400\times 17.87\times 0.806  

3

​

VIcosθ=  

3

​

×400×17.87×0.806

 

= 9978 Watts

 

The power output is given as

 

Power input – stator and rotor core and frictional loss

 

– stator copper loss – rotor copper loss

 

9978 – 800 – 479 – 2369978–800–479–236

 

= 8463 Watts

 

Hence efficiency of the motor

 

\frac{8463}{9978}\times 100=84.8  

9978

8463

​

×100=84.8%

Explanation: