Given:
Pole (P) = 8
Frequency (f) = 50 Hz
Rotor resistance (r2) = 0.08 Ω
Rotor speed at maximum torque = 650 RPM
Let's consider the slip at maximum torque as sm,
Slip at maximum torque is given by
\(s_m = \dfrac {r_2}{x_2}\)
Consider additional rotor resistance to be added = r2'
Rotor reactance = x2
Synchronous speed Ns = 120f/P = 120 × 50 / 8 = 750 RPM
As the speed of the motor at maximum toque condition is 650 RPM
Slip at maximum torque is given by
\(s_m = \dfrac {N_s - N_r}{N_s} =\dfrac {750-650}{750}= 0.1333\)
At maximum torque
\(s_m =\dfrac {r_2}{x_2} ⇒ 0.1333 = \dfrac {0.08}{x_2}\)
x2 = 0.6 Ω
To get the maximum torque at starting the slip at maximum torque should be equal to slip at starting (s=1).
⇒ \(1 =\dfrac {{r_2+r_2'}}{ x_2}\)
x2 = r2 + r2'
0.6 = 0.08 + r2'
r2' = 0.52 Ω
∴ The value of additional rotor resistance to be added (r2') at the time of starting is 0.52 Ω.