A 600V line powers a three-phase induction motor with a nominal power of 100 hp (≈ 75 kW). Two wattmeters placed in the line wires indicate a total power of P_a = 70kW and an amperemeter indicates a line current of  I_a = 78A. Accurate measurements indicate a speed of rotation of  N = 1463 rpm. In addition, the following information is provided:
– The iron losses in the stator: P_fs = 2kW
– Ventilation and mechanical losses: P_mec = 1.2kw
– Resistance between two stator terminals : 2×R_s = 0.34Ω

We suppose STAR connexion.
1- Give the number of poles of machine stator and the slip s of the machine
2- Calculate the stator joule losses P_js
3- Calculate the electromechanical power delivered to the rotor P_em
4- Deduce the joule losses in the rotor P_jr
5- Calculate the useful torque T_u of the motor
6- Calculate the efficiency of the motor η

[bg_collapse view= »button-orange » color= »#4a4949″ expand_text= »Voir Correction » collapse_text= »Afficher Moins »]CORRECTION

1)N= 1463rpm →Ns = 1500rpm →4 poles(synchronism speed: 1500rpm).

S = (N_s – N)/N_s = (1500 -1463)/1500  = 2.4%

2) Stator Joule losses : P_js = 3R_sI_a² = 3×0.34/2 ×78²= 3.1kW

3) P_em = P_a – (P_js + P_fs) = 70 – 3.1 – 2 = 64.9kW

4) Rotor Joule losses: P_jr = s×P_em = 0.024×64.9 = 1.5kW

5) T_u = P_u/Ω_n = (64900 – 1300 -1200)/70000 = 407.3Nm

6) η = P_u/P_a = (64900 – 1300 -1200)/70000 = 89%

 

 

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