3-phase induction machine at slip 0.05

Given: $V_L=400$ V, output $=80$ kW, $f=50$ Hz, $P=4$ poles, $s=0.05$, stator : rotor turns $=1.5$. The stator is taken as star-connected.

a) Synchronous speed:

$$N_s=\frac{120f}{P}=\frac{120\times50}{4}=1500\ \text{rpm}$$

b) Motor speed:

$$N=N_s(1-s)=1500\times0.95=1425\ \text{rpm}$$

c) Rotor current frequency:

$$f_r=sf=0.05\times50=2.5\ \text{Hz}$$

d) Rotor induced voltage at running speed. Stator phase voltage:

$$V_1=\frac{400}{\sqrt3}=230.9\ \text{V}$$

Standstill rotor emf per phase $E_2=\frac{V_1}{1.5}=153.9$ V. At slip $s$ the rotor emf is $sE_2$:

$$E_{2r}=0.05\times153.9=7.70\ \text{V per phase}$$

(Line value $= \sqrt3\times7.70 = 13.33$ V.)

e) Rated torque (shaft output 80 kW at 1425 rpm):

$$\omega=\frac{2\pi\times1425}{60}=149.23\ \text{rad/s}, \qquad T=\frac{P_{out}}{\omega}=\frac{80000}{149.23}=536.1\ \text{N·m}$$

Answer: a) 1500 rpm, b) 1425 rpm, c) 2.5 Hz, d) 7.70 V per phase, e) 536.1 N·m.