Question

An 4-pole 50 Hz induction motor with rotor resistance of 0.2 Ω produces a maximum torque of 10 N-m at 1200 rpm. At maximum torque, what will be rotor reactance and the difference between synchronous and asynchronous speeds (slip) respectively?
1. 2 Ω and 0.1 p u
2. 1 Ω and 0.1 p u
3. 2 Ω and 0.2 p u
4. 1 Ω and 0.2 p u

Answer 1

Correct Answer - Option 4 : 1 Ω and 0.2 p u

Concept:

The slip at maximum torque is given by  \({s_m} = \dfrac{{{R_m}}}{{{X_m}}}\)

Where Rm = Motor resistance per phase

Xm = Motor reactance per phase

n a three-phase induction motor, synchronous speed is

\({N_s} = \dfrac{{120f}}{P}\)

Where,

Ns = Synchronous Speed in rpm

f = supply frequency in Hz

P = number of poles

Rotor speed Nr = Ns (1 – s)

Per Unit rotor speed(asynchronous) is given by

\(N_{r\ (pu)}=\dfrac{N_r}{N_s}=(1-s)\)

Per Unit Synchronous speed N_s(pu) = N_s/N_s = 1 PU

Application:

Given:

Number of Poles (P) = 4

f = 50 HZ implies \({N_s} = \dfrac{{120f}}{P}=1500 \ rpm\)

Rotor resistance at Maximum Torque = 0.2 Ω

N_r = 1200 rpm at Maximum Torque

Slip at maximum Torque is given by

\({s_m} = \dfrac{N_s-N_r}{N_s}=\dfrac{1500-1200}{1500}=0.2=\dfrac{R_m}{X_m}\)

Therefore,

Rotor reactance at Maximum Torque is

\(Xm = \dfrac{0.2}{0.2}\)

= 1 Ω

The difference between synchronous and asynchronous speeds is

N_s(pu) - N_r(pu) = 1 - (1 - s) = s

= 0.2 PU