Induction Motor Drives Questions and Answers

Answer: A

Correct answer: a little over 90 per cent overall

A modern motor of a few kilowatts is around 90 to 93 per cent efficient and the inverter around 96 to 98, so the product is a little over 90 per cent at rated load. The figure falls at light load, which is why the flux reduction discussed elsewhere in this topic is worth doing, and no drive is anywhere near 100 per cent since both the copper losses and the switching losses are unavoidable.

Answer: B

Correct answer: fan power falls with the cube of the speed

For a fan on a fixed system the flow varies with speed, the pressure with the square of the speed, and the power with the cube: \[ \frac{P_2}{P_1} = \left(\frac{N_2}{N_1}\right)^{3}. \] Running at 80 per cent speed therefore takes about half the power, and at 50 per cent about an eighth. That cube law is the whole economic case for variable-speed fan and pump drives.

Answer: C

Correct answer: dissipates the pressure it drops

A damper does not reduce the work the fan does; it throttles the flow by adding resistance, and the pressure dropped across it is converted straight to heat and turbulence. The fan still runs at full speed against a higher pressure and absorbs almost as much power as before. A drive reduces the speed instead, so the cube law applies and the energy is never generated at all.

Answer: D

Rated current at low frequency gives rated copper loss with almost no air flow, so the machine overheats unless it is cooled independently or a larger frame is used. It is one of the most common errors in variable-speed retrofits.

Answer: A

Correct answer: a short-circuited transformer

At standstill the slip is unity, the rotor is a shorted secondary at full supply frequency, and the only thing limiting the current is the leakage impedance. Five to seven times rated current is the usual result. As the machine accelerates the effective rotor resistance \( R_2/s \) rises, the current falls, and the machine settles at its running point - which is why the high current lasts only for the run-up.

Answer: B

Correct answer: a converter in the rotor circuit rated for the slip power

That is the whole point of the arrangement. With the stator connected directly to the grid and only the slip power passing through the rotor converter, a machine operating within about plus or minus 30 per cent of synchronous speed needs a converter rated at roughly 30 per cent of the machine output. The saving in converter cost and losses is what made the doubly fed generator the dominant wind turbine topology for many years.

Answer: C

Correct answer: flow into the rotor from the supply

The rotor electrical power is \( P_r \approx -sP_{ag} \), so above synchronous speed the slip is negative and \( P_r \) takes the same sign as the air gap power. For a doubly fed machine motoring above synchronous speed that means power must be injected into the rotor from the supply, while below synchronous speed the slip power flows the other way and is recovered to the supply rather than burnt in a resistance. A doubly fed generator reverses both directions, since its air gap power flows outward.

Answer: D

The machine has no field of its own, so the reactive current establishing the flux must come from somewhere, and on a network it comes from the network. Standing alone, it needs capacitors and the arrangement is difficult to control.

Answer: A

Correct answer: the largest load peak it can ride through

The breakdown or pull-out torque is the maximum on the torque-slip curve. A load demanding more than this stalls the machine, because beyond the peak the torque falls as the slip rises and there is no stable operating point. It is typically two to three times full-load torque, and that margin is what lets a motor survive a momentary overload or a supply dip.

Answer: B

Raising the rotor resistance moves the torque peak towards standstill and increases the full-load slip, which suits punch presses and hoists needing a soft characteristic. The penalty is greater rotor loss at rated load.