Plugging as a braking method consists of
Answer: A
Reversing the applied voltage or the phase sequence makes the machine develop torque against its rotation, which stops it very rapidly. Both the supply energy and the stored kinetic energy end up as heat in the machine, so it is hard on the windings.
Please log in or register to report this question.
Of the three electrical braking methods, the one that wastes the most energy is
Answer: B
Plugging dissipates the stored kinetic energy and an equal quantity drawn from the supply, all inside the machine. Dynamic braking wastes only the stored energy, and in an external resistance, while regeneration wastes almost none.
Please log in or register to report this question.
A flywheel is added to a drive with a fluctuating load in order to
Answer: C
Correct answer: equalise the load on the motor
A flywheel stores energy during the light part of the cycle and gives it back during the heavy part, so the motor sees something close to the mean load instead of the peaks. The motor can then be rated for the average rather than the maximum, which is a large saving on a press, a shear or a rolling mill. It works only if the speed is allowed to fall as energy is drawn out, which is why it requires a machine with a drooping characteristic.
Please log in or register to report this question.
Load equalisation with a flywheel requires the motor to have
Answer: D
The flywheel can only give up energy if the speed falls, and only take it up if the speed rises, so some droop is essential. A synchronous motor holding an exactly constant speed cannot be used for load equalisation at all.
Please log in or register to report this question.
A flywheel is unsuitable for load equalisation in a drive that must
Answer: A
Every acceleration and deceleration must now carry the flywheel's inertia as well, which slows the drive and wastes energy. Flywheels suit steady-speed drives with fluctuating torque, such as presses and rolling mills.
Please log in or register to report this question.
The energy stored in a flywheel of inertia \( J \) at angular speed \( \omega \) is
Answer: B
The square dependence means most of the useful energy lies in the upper part of the speed range, so a small percentage drop in speed releases a great deal. It is also why flywheels are run as fast as their material will allow.
Please log in or register to report this question.
A drive has an inertia of 10 kg·m², a motor torque of 60 N·m and a load torque of 20 N·m. Its angular acceleration is
Answer: C
Correct answer: 4 rad/s²
From the torque equation,
\[ \frac{d\omega}{dt} = \frac{T - T_l}{J} = \frac{60 - 20}{10} = 4 \ \mathrm{rad/s^2} \]The 6 rad/s squared answer uses the motor torque alone and forgets that the load torque must be overcome before anything is left to accelerate the inertia.
Please log in or register to report this question.
That drive reaches 100 rad/s from rest in
Answer: D
Correct answer: 25 s
With constant accelerating torque the acceleration is constant, so
\[ t = \frac{\omega}{d\omega/dt} = \frac{100}{4} = 25 \ \mathrm{s} \]equivalently \( J\omega/(T - T_l) = 10 \times 100/40 \). Twenty-five seconds is a long run-up, and it reflects a large inertia against a modest torque margin - doubling the motor torque would cut it to ten seconds.
Please log in or register to report this question.
A motor of inertia 0.5 kg·m² drives through a 5:1 reduction a load of inertia 20 kg·m². The total inertia referred to the motor shaft is
Answer: A
Correct answer: 1.3 kg·m²
The load inertia refers through the square of the ratio and is then added to the motor’s own:
\[ J_{total} = 0.5 + \frac{20}{5^2} = 0.5 + 0.8 = 1.3 \ \mathrm{kg\,m^2} \]The 20.5 answer adds the load inertia unreferred, which would make the drive sixteen times harder to accelerate than it is. The square law is what makes gearing so effective at taming a large slow-moving inertia.
Please log in or register to report this question.
A load torque of 500 N·m acts on a shaft driven through a 10:1 reduction of 90 per cent efficiency. The torque referred to the motor is
Answer: B
Correct answer: 55.6 N·m
Torque refers with the first power of the ratio, and the gearbox loss must be supplied by the motor, so the efficiency divides:
\[ T_{referred} = \frac{T_l}{i\eta} = \frac{500}{10 \times 0.9} = 55.6 \ \mathrm{N\,m} \]Answering 50.0 forgets the efficiency, and multiplying by it instead of dividing gives 45.0 - which would have the gearbox delivering more than it received. During braking the power flows the other way and the efficiency multiplies instead.
Please log in or register to report this question.
