Answer to Question #237176 in Molecular Physics | Thermodynamics for Unknown346307

Question

A 10 pole, 50 Hz, Y connection 3-phase induction motor having a rating of 60 kW and 415V.

The slip of the motor is 5% at 0.6 power factor lagging. If the full load efficiency is 90%,

calculate:

(i) Input power

(ii) Line current and phase current

(iii) Speed of the rotor (rpm)

(iv) Frequency of the rotor

(v) Torque developed by the motor (if friction and windage losses is 0)

Answer 1

(i)

$η = \frac{O/P \;power}{Input \;power} \\ Input \; power = \frac{O/P \;power}{η} \\ = \frac{60 \times 1000}{0.9} = 66.66 \;kW$

(ii)

$P = \sqrt{3}V_2I_2cosθ \\ I_2 = \frac{P}{\sqrt{3} V_2 cosθ} \\ = \frac{66.66 \times 1000}{\sqrt{3} \times 415 \times 0.6} \\ = 154.75 \;A \\ I_2 = \sqrt{3}I_{ph} \\ I_{ph} = \frac{I_2}{\sqrt{3}} = 88.24 \;A$

(iii)

$N= N_s(1-s) \\ N_s = \frac{120f}{P} = \frac{120 \times 50}{10} = 600 \;rpm \\ = 600(1-0.05) \\ = 570 \;rpm$

(iv)

$f_r = sf = 0.05 \times 50 = 2.5 \;Hz$

(v)

$P =τω \\ τ = \frac{P}{ω} = \frac{60 \times 1000 \times 60}{2 \pi \times 570} = 1005.65 \;Nm$