# Study-of-induction motor-3

EXERCISE : STUDY OF AN INDUCTION MOTOR

The descriptive plate (“la plaque signalétique”) of the induction motor of a milling machine shows the following indications:

3 ∼ 50Hz ;    Δ     220V     11A;        Υ       380V       6,4A ;              1455tr/min                cosφ = 0,80
1- This motor is supplied by an electric grid 220 V / 380 V; 50 Hz. How do we have to couple the stator windings of this machine? Justify your answer.
2- Give the number of poles of machine stator.
3- Calculate the nominal slip gn (in %).
4- A no-load test (“essai à vide”) was performed for this motor powered by the grid mentioned above. We obtained the following results:
– No-load current : I0 = 3.2A
– No-load absorbed power : Pa0 = 260W

We neglect the joule losses of the rotor at no-load: Pjr0 .  The mechanical losses Pmec are evaluated at 130 W.
The warm measurement of the resistance of a stator winding gives Rs = 0.65Ω. Deduce the iron losses in the stator at no-load : Pfs0.
In the following, we will consider that the iron losses in the stator is constant and equal to Pfs0.
5- We perform a test in the nominal conditions of the machine:
a) Calculate the stator joule losses Pjs                                        b) Calculate the joule losses in the rotor Pjr
c) Calculate the useful torque Cu of the motor                    d) Calculate the efficiency of the motor η

CORRECTION

1) Star coupling as this motor is supplied by an electric grid 220 V / 380 V: grid voltages are equal to those of the machine.
2) 4 poles (synchronism speed: 1500 rpm).

3) gn = (Ns – N)/Ns = (1500 – 1455)/1500 = 3%

4) First, we calculate:

– Stator Joule losses : Pjs0 = 3×Rs I02 = 3×0.65×3.22  =20W

We neglect the joule losses of the rotor at no-load: Pjr0
Based on power balance for no load test, we obtain: Pfs0 = Pa0 – (Pjs0 + Pjr0 + Pmec) = 260 – (20 + 0 + 130) = 110W

5) a) a) Stator joule losses: Pjs = 3×RsIn2 = 3×0.65×6.42 = 80W

b) Rotor Joule losses: Pjr = gnPem = gn [Pa – (Pjs + Pfs)] = 0.03×[3380 – (80 + 110)] = 95.6W; with  Pa = 3VnIncosφn = 3×220×6.4×0.8 = 3380W

c) Cu Pun = (3380 -80 -95.6 – 130)/1455×(π/30) = 19.46Nm

d) η = Pu/Pa = (3380 -80 -95.6 – 130)/3380  = 87.7%

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