Question 6.1: A three-phase, two-pole, 60-Hz induction motor is observed t......

The power dissipated in the stator winding is given by

P_{stator} = 3I_1^2R_1= 3(22.6)^2 0.2 = 306  W

Hence the air-gap power is

P_{gap} = P_{input}  –  P_{stator}= 15.7  –  0.3 = 15.4  kW

The synchronous speed of this machine can be found from Eq. 4.41

n_s=\left( \frac{120}{poles}\right)f_e   r/min \quad \quad (4.41)

n_s=\left( \frac{120}{poles}\right)f_e= \left( \frac{120}{2}\right) 60 =3600  r/min

and hence from Eq. 6.1,

s=\frac{n_s -n}{n_s} \quad \quad (6.1)

the slip is s = (3600 – 3502)/3600 = 0.0272. Thus, from Eq. 6.23,

P_{rotor} = s P_{gap} = 0.0272 × 15.4  kW = 419  W