Induction Motor Drives Questions and Answers

Answer: C

Correct answer: reduced voltage

With the rotor held still the machine is a short-circuited transformer, so rated voltage would drive five to seven times rated current and overheat the windings in seconds. The voltage is raised only until rated current flows, typically 15 to 25 per cent of rated, and the impedance measured then applies at full voltage because it is essentially linear.

Answer: D

Correct answer: reducing the flux to match the load

At part load the torque needed is small, but a machine held at full flux still carries the full magnetising current and the full core loss. Reducing the flux - lowering the voltage in step with the reduced torque demand - brings the iron loss and magnetising current down and can improve part-load efficiency appreciably. Raising the flux does the opposite, and a higher switching frequency merely increases the inverter loss.

Answer: A

Correct answer: operated with flux reduction or sized more closely

A machine running for long periods at a small fraction of its rating spends most of its energy on losses that do not depend on the load. The two remedies both address that: reduce the flux so the magnetising and core losses fall, or fit a smaller machine so the load is a larger fraction of its rating. Fitting a larger machine makes the problem worse, which is why oversizing for margin is a false economy.

Answer: B

Correct answer: no brushes to spark or wear

A cage induction motor has no brushes, no commutator and no sliding contacts, so there is nothing to spark in a flammable atmosphere and nothing to wear out or need replacing in a dirty one. A dc machine’s commutator sparks by its nature and its brushes must be inspected and changed. The efficiency advantage is real but modest; the maintenance and safety advantage is decisive.

Answer: C

Correct answer: Varying the armature resistance of the machine

An induction motor has no armature - that is a dc machine term, and the corresponding induction motor method is rotor resistance control, which works only on a wound rotor machine. Stator voltage variation, pole changing and frequency variation are all genuine methods, the last being the one that has displaced the others.

Answer: A

Both statements are true. In the rotor resistance control method, an external resistance is connected to the rotor through slip rings, and a squirrel cage rotor has no slip rings and its bars are short circuited, so the method cannot be used with it. A static Kramer drive takes the slip power from the rotor of a slip ring motor, rectifies it, and returns it as mechanical power to the shaft (through a DC motor), which is a slip power recovery scheme; the static Scherbius drive is the other example.

So true and true is right, option A.

Answer: A

The synchronous speed of the 12 pole motor on a 50 Hz supply is

\[ N_s = \frac{120 \times 50}{12} = 500\ \text{rpm} \]

At a slip of 0.045 the rotor speed is

\[ N = (1 - 0.045) \times 500 = 477.5\ \text{rpm} \]

\[ \begin{aligned} \omega = \frac{2\pi \times 477.5}{60} \\ &= 50\ \text{rad/s} \end{aligned} \]

The power rating and the rotor resistance are not needed to find the speed. The values of 100 and 200 rad/s are too high for a 12 pole motor, since even the synchronous speed is only 52.4 rad/s. The answer is 50 rad/s, option A.

Answer: C

The synchronous speed of the 12 pole motor on a 50 Hz supply is

\[ N_s = \frac{120 \times 50}{12} = 500\ \text{rpm} \]

\[ N = (1 - s)N_s = 0.955 \times 500 = 477.5\ \text{rpm} \]

\[ \begin{aligned} \omega = \frac{2\pi N}{60} \\ &= \frac{2\pi \times 477.5}{60} = 50\ \text{rad/s} \end{aligned} \]

The torque law of the fan (proportional to the speed squared) is not needed to find the speed at the stated slip. The values 45, 55 and 60 rad/s correspond to speeds of 430, 525 and 573 rpm, and the last two exceed the synchronous speed of 52.4 rad/s. The answer is 50 rad/s, option C.

Answer: A

The slip at which the induction motor develops its maximum torque is

\[ s_m = \frac{R_r'}{\sqrt{R_s^2 + (X_s + X_r')^2}} \]

With the frequency ratio \(\beta = 1\) (rated frequency) the reactances have their normal values.

\[ \begin{aligned} s_m \\ &= \frac{0.38}{\sqrt{0.66^2 + (1.14 + 1.71)^2}} \\ &= \frac{0.38}{\sqrt{0.4356 + 8.1225}} \\ &= \frac{0.38}{2.925} = 0.1299 \end{aligned} \]

The answer is 0.1299, option A. The values 0.562 and 0.661 would come from leaving out the stator reactance, and 0.235 has no basis in the data.