Guest Question.

400 V, 3-phase, 50 Hz, 6 pole induction motor is supplying a load of 20 kW, when the frequency of rotor currents is 2Hz. Estimate (i) Slip and speed at this load (ii) Rotor copper loss (iii) Speed of motor when supplying 30 kW load assuming torque-slip curve to be a straight line.

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Solution

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