Braking of Electrical Drives Questions and Answers

Answer: A

Correct answer: the kinetic energy of the rotating parts

In dynamic braking the machine is disconnected from the supply and its armature connected across a resistor, so the only source of energy is the kinetic energy stored in the rotating parts:

All of it ends up as heat, mostly in the external resistor. Plugging is quite different: there the supply continues to deliver energy throughout, so the total dissipated is several times the stored kinetic energy.

Answer: B

The supply continues to deliver energy while the machine also absorbs the stored kinetic energy, and both end up as heat inside the machine. That doubling is why plugging is so hard on the windings and is reserved for occasional use.

Answer: C

Regeneration wastes only the conversion losses, dynamic braking wastes the stored energy in a resistor, and plugging wastes that and an equal amount from the supply. Cost, complexity and availability run the other way.

Answer: D

Correct answer: cannot accept power in the reverse direction

Regeneration requires something to receive the returned energy. A diode rectifier front end conducts in one direction only, so the energy has nowhere to go and the dc link voltage simply rises until protection operates. An isolated supply, or a weak one with no other load, presents the same difficulty. The remedies are an active front end, a dual converter, or a braking chopper and resistor that dissipates the energy instead.

Answer: A

Regenerated power needs another train accelerating nearby or a substation able to invert, and without either the braking train must dump the energy in resistors. Wayside storage and reversible substations are the two modern answers.

Answer: B

When the dc link voltage rises above a threshold, the chopper connects the resistor and the surplus becomes heat there rather than tripping the drive. It is much cheaper than an active front end where the braking energy is modest.

Answer: C

Correct answer: the energy per stop and the frequency of stopping

Two things must be checked. The energy of a single stop, \( \tfrac{1}{2}J\omega^2 \), sets the peak temperature rise and hence the thermal mass the resistor needs. The number of stops per hour sets the average power to be dissipated and hence the continuous rating and the cooling. A resistor adequate for one stop may fail on a duty of six an hour, which is exactly what the following calculation in this topic shows.

Answer: D

Correct answer: \( J\omega/(T_b + T_l) \) for a passive load torque

The equation of motion during braking is

for a passive load, whose torque always opposes motion and so assists the braking. Integrating from \( \omega \) to zero gives

An active load such as a suspended weight is the case with the minus sign, since gravity continues to drive the machine and the braking torque must overcome it as well.

Answer: A

Correct answer: 30 s

With no load torque the braking torque acts alone:

Half a minute is a long stop, and it reflects a large inertia braked by a modest torque. Doubling the braking torque would halve the time; the energy to be absorbed, however, would be unchanged, since that depends only on the kinetic energy stored.

Answer: B

Correct answer: 90 kJ

The energy is the stored kinetic energy of the rotating parts:

Note that this does not depend on the braking torque or on how long the stop takes - a harder brake removes the same energy in less time, which raises the peak power but not the total.