Question 4 of 5: Induction Motors

Question 4: Induction Motors (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Parts (a)–(c) — Concept questions

(a) The two rotor types are the squirrel-cage and the wound (slip-ring) rotor. The squirrel-cage rotor has solid conducting bars shorted by end-rings; its advantage is that it is rugged, cheap, essentially maintenance-free and sealed against dirt. The wound rotor carries a three-phase winding brought out through slip rings; its advantage is that external resistance can be inserted in the rotor circuit to give high starting torque with low starting current (and a limited amount of speed control).

(b) Reverse the phase sequence. Interchanging any two of the three stator supply leads reverses the phase sequence of the stator currents, which reverses the direction of the rotating magnetic field and therefore the direction of rotation.

(c) Speed-torque characteristic. Starting from standstill the motor develops its starting torque; as speed rises the torque increases to a maximum, the breakdown (pull-out) torque, at a slip of roughly 10–20%, then falls steeply to zero at synchronous speed. Normal running is on the steep, nearly linear portion between pull-out and synchronous speed, where a small change in slip produces a large change in torque (good speed regulation).

Part (d) — Synchronous speed, rotor frequency, efficiency and line current

Given. 4-pole, 60 Hz, 440 V (line-to-line) three-phase induction motor; full-load slip \(s=0.05\). For part (ii): output 90 hp at 1732 rpm, rotational losses 10 100 W, total (rotor + stator) copper losses 3700 W, pf 0.8 lagging.

Find. (i) synchronous speed and rotor-current frequency; (ii) efficiency and line current at the 90 hp operating point.

Approach. Synchronous speed is \(n_s=120f/P\) and rotor frequency is \(sf\). For part (ii) build the power flow: input = output + all losses, then \(\eta=P_{out}/P_{in}\) and \(I_L=P_{in}/(\sqrt3\,V_L\,\text{pf})\).

1. Synchronous (stator field) speed. $$n_s=\frac{120f}{P}=\frac{120(60)}{4}=1800\text{ rpm}.$$
2. Rotor-current frequency at full-load slip. $$f_r=s\,f=0.05(60)=3\text{ Hz}.$$
3. Input power at the 90 hp point. With \(P_{out}=90\times746=67\,140\text{ W}\), $$P_{in}=P_{out}+P_{rot}+P_{cu}=67\,140+10\,100+3700=80\,940\text{ W}.$$
4. Efficiency. $$\eta=\frac{P_{out}}{P_{in}}=\frac{67\,140}{80\,940}=83.0\%.$$
5. Line current. From \(P_{in}=\sqrt3\,V_L I_L\cos\theta\), $$I_L=\frac{P_{in}}{\sqrt3\,V_L\cos\theta}=\frac{80\,940}{\sqrt3\,(440)(0.8)} =132.8\text{ A}.$$