A 400 V, 15 kW, 4 pole, 50 Hz, Y-Connected induction motor has full load slip of 4%. The output torque of the machine at full load is

Concept:

In an induction motor

Full load torque (T_FL) is given by

​\({T_{FL}} = \frac{{{P_{out}\times60}}}{{2\pi {N_r}}}\)

No-load torque is given by

\({T_{NL}} = \frac{{{P_{in}\times60}}}{{2\pi {N_s}}}\)

Synchronous speed (N_S) is given by

\({N_s} = \frac{{120\;f}}{P}\)

Slip is given by

\(s = \frac{{{N_s} - {N_r}}}{{{N_s}}}\)

Where,

P_out = Output power

P_in = Input power

N_r = Rotor speed in rpm

s = slip

f = supply frequency in Hz

P = Number of poles

Calculation:

Given –

P = 4, f = 50 Hz,

 s = 4%, P­_out = 15 kW

\({N_s} = \frac{{120 \times 50}}{4} = 1500\;rpm\)

\(0.04 = \frac{{1500 - {N_r}}}{{1500}}\)

N_r = 1440 rpm

Now the output torque of the machine at full load can be calculated as

\({T_{FL}} = \frac{{15 \times {{10}^3\times 60}}}{{2\pi \times 1440}}\)

T_FL = 99.4718 Nm