5) A 480 V, 60 Hz, three-phase, four-pole induction motor has the following parameters:

R₁ = 0.30; R₂ = 0.20; Xeq = 2.00

At full load, the motor speed is 1760 rpm. Calculate the following:

a) Slip of the motor: (n: 5 points; S: 5 points)

b) Developed torque of the motor at full load (5 points)

c) Developed power: (w: 5 points; Pa: 5 points)

d) Rotor current as seen from the stator winding (5 points)

e) Copper losses of the motor (5 points)

1. The synchronous speed of the motor can be computed by using Equation 12.9:

n. = 120=120=1800 rpm

w.n,= 1800 = 188.5 rad/s

Using Equation 12.15, we can compute the slip of the motor:

S="

1800 1760

-0.0222

1800

2. Equation 12.49 or Equation 12.50 can be used to compute the load torque:

T₁ =

3V²R₂

¹()'0²

Su. (R.)+x2] 0.0222-188.5 [(0.3+)4]

3. The angular speed of the motor is

W =

==1760=184.3 rad/s

The developed power of the motor is

P₁=Tw=121.56 x 184.3 = 22.4kW

4. The rotor current can be computed using Equation 12.46:

= 121.56 Nm

29.1 A

√(012) 3

5. The copper losses of the motor are in the stator and rotor windings, that is,

Peu=Poul + Peu2=3(1) Req=3x (29.1)²(0.3+0.2)=1.27 kW