Here, \(f = 50 Hz; P = 6; f_{r} = 2Hz; Output = 20 kW\)

Synchronous speed, \(N_{s} = \frac{120f}{P} = \frac{120 \times 50}{6} = 1000 r.p.m.\)

Slip, \(S = \frac{f_{r}}{f} = \frac{2}{50} = 0.04 or, \mathbf{4\%} \)

Rotor speed, \(N = N_{s} – S N_{s} \hspace{30 pt} \hspace{30 pt} \left\lgroup since S = \frac{N_{s} – N}{N_{s}} \right\rgroup \\[0.5cm] \hspace{30 pt} \qquad \quad \quad = 1000 – 0.04 \times 1000 = \mathbf{960 rpm} \)

\(Rotor copper loss = \frac{S}{1 – S} \times mech. power developed \\[0.5cm] \hspace{30 pt} \hspace{30 pt} \qquad \, = \frac{0.04}{1 – 0.04} \times 20 = 0.833 kW = \mathbf{833 W} \)

 

It is given that torque-slip curve to be straight line.

\(\therefore T \propto S as well as T \propto output \)

 

Hence, \(S \propto output\)

\(\therefore S_{1} \propto output_{1} and S_{2} \propto output_{2} \)

 

or \(\frac{S_{2}}{S_{1}} = \frac{30 kW }{20 kW} \)

or \( S_{2} = \frac{30}{50} \times 0.04\)

or \( S_{2} = 0·06 \)

∴ Speed, \( N = N_{s} \left(1 – S\right) = 1000 \left(1 – 0.06 \right) = \mathbf{940 rpm} \)