Delta induction motor: slip, rotor emf, turns ratio

Synchronous speed:

$$N_s = \frac{120f}{P} = \frac{120\times 50}{4} = 1500\ \text{rpm}$$

Slip

$$s = \frac{N_s - N}{N_s} = \frac{1500 - 1440}{1500} = 0.04 \ (4\%)$$

Frequency of rotor emf

$$f_r = sf = 0.04\times 50 = 2\ \text{Hz}$$

Rotor induced emf per phase at 1440 rpm

The rotor emf per phase is proportional to slip: $E_{2r} = sE_{2}$, where $E_2 = 130$ V is the standstill value.

$$E_{2r} = 0.04\times 130 = 5.2\ \text{V}$$

Stator to rotor turns ratio

The stator is delta connected, so the stator phase voltage equals the line voltage, 440 V. At standstill the induction motor behaves like a transformer, so

$$\frac{N_1}{N_2} = \frac{E_1}{E_2} = \frac{440}{130} = 3.385$$

The turns ratio is independent of speed, so it is the same at 1440 rpm.

Answer: slip $= 0.04$ ($4\%$); rotor frequency $= 2\ \text{Hz}$; rotor emf per phase $= 5.2\ \text{V}$; turns ratio $N_1/N_2 = 3.385$ (stator phase turns : rotor phase turns, about 3.38:1).