Question 5 of 6: Hoist Duty Cycle, Motor Power and Production

Question 5: Hoist Duty Cycle, Motor Power and Production (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.

5.1 — Hoist duty cycle

The hoist duty cycle is the complete, repeating time-and-velocity profile a skip or cage follows over one round trip — typically a trapezoidal profile of acceleration, constant (rope) velocity travel, deceleration, and a stationary dwell for loading/dumping (decking) — together with the payload and distance the hoist must move on every such cycle. It is the single design specification from which every other hoist-selection parameter is derived, because it fixes how hard, how often and how fast the machine must work over its service life, not just what it must be able to do once.

From the duty cycle, the following hoist selection parameters follow directly, in order: the required hoisting depth (shaft depth, from the given rope travel); the required maximum rope/skip velocity (set by the constant-velocity portion of the profile needed to close the cycle time within the target); the required acceleration/deceleration rate (from the accel/decel time and the velocity to be reached, which in turn sizes the motor's peak torque, not just its average power); the resulting cycle time and hence the achievable trips/hour and daily/annual production (Part 5.3 below is exactly this calculation); the rope and drum/sheave sizing (from the static and dynamic loads implied by payload, skip weight and acceleration, per Question 1.2); and finally the motor power rating and duty class (from the peak, RMS and average power the profile demands over a full shift — Question 5.2's subject). In short, the duty cycle is the input every other hoist design decision is derived from, not an output of the design.

5.2 — Hoist motor power requirements: DC vs. AC

Hoist motor power requirements means the electrical power (and, more precisely, the torque-versus-time profile) the hoist motor must be able to deliver at every instant of the duty cycle described in 5.1 — not simply a single average figure, since a hoist motor must supply a large torque during acceleration, a lower steady torque at constant velocity, and (for a regenerative system) may need to ABSORB power during deceleration of a descending loaded skip or an ascending empty one.

DC systems (historically Ward-Leonard motor-generator sets, now more often thyristor/SCR-controlled DC drives) obtain their power requirement from direct armature-current and field-current control: torque is directly proportional to armature current, so the required power at each instant of the duty cycle is read straight off the torque-time profile via $P = T\omega$, with the controller regulating armature voltage/current to track the required acceleration and velocity profile precisely — DC's simple, direct torque-current relationship is what historically made it the preferred technology for the fine, smooth speed control a friction hoist's duty cycle demands.

AC systems (modern variable-frequency-drive induction or synchronous motors) instead obtain the equivalent power/torque profile by varying the SUPPLY FREQUENCY (and voltage, held in a fixed ratio to frequency below base speed) delivered to the motor, which shifts the induction motor's torque-speed curve so that the operating point tracks the required duty-cycle torque at each instantaneous speed; a modern VFD digitally computes and outputs the frequency/voltage profile needed to reproduce the same acceleration/constant-velocity/deceleration duty cycle a DC drive would follow, and can equally regenerate power back to the supply (or a braking resistor) during a decelerating, overhauling load. In both cases the underlying REQUIREMENT — the instantaneous power the duty cycle demands — is identical; DC and AC differ only in the electrical mechanism (current control vs. frequency control) used to deliver it.

5.3 — Daily production

Given. Shift time 7.2 hours; 3 shifts/day; skip capacity 11 t; cycle time 85 s/skip (one complete round trip).

Find. Daily production (tonnes/day).

Approach. Convert the shift time to seconds, divide by the cycle time to get skips per shift, multiply by shifts/day for skips/day, then multiply by skip capacity.

1. Skips per shift. $$\text{skips/shift} = \dfrac{7.2\times3600\ \text{s}}{85\ \text{s/skip}} = \dfrac{25{,}920}{85} = 304.94\ \text{skips/shift}$$
2. Skips per day and daily production. $$\text{skips/day} = 304.94\times3 = 914.82\ \text{skips/day}$$ $$\boxed{\text{Daily production} = 914.82\times11\ \text{t} = 10{,}063\ \text{t/day}}$$ (rounding to a whole skip per shift, 304 skips $\times$ 3 shifts $\times$ 11 t $=$ 10,032 t/day is the practically achievable figure; the continuous-cycle value above is the theoretical hoist capacity used for design/comparison purposes.)

5.4 — Energy consumption per skip hoisted

Given. Average power consumed 730 kW; hoist efficiency 85%; acceleration time 6.5 s; constant-velocity time 63.5 s.

Find. Approximate energy consumption per skip.

Approach. Multiply the average power by the powered duration to get the electrical energy consumed, then apply the hoist efficiency to find the useful mechanical energy actually delivered to the skip.

1. Powered duration and energy consumed. $$t_{\text{powered}} = 6.5+63.5 = 70\ \text{s}$$ $$E_{\text{consumed}} = P_{\text{avg}}\times t_{\text{powered}} = 730\ \text{kW}\times70\ \text{s} = 51{,}100\ \text{kJ} = \boxed{51.1\ \text{MJ}\ (14.2\ \text{kWh})\ \text{per skip}}$$
2. Useful energy delivered to the skip. $$E_{\text{useful}} = E_{\text{consumed}}\times\eta_{\text{hoist}} = 51.1\ \text{MJ}\times0.85 = \boxed{43.4\ \text{MJ}\ (12.1\ \text{kWh})\ \text{per skip}}$$ The difference between the two figures, roughly 7.7 MJ per skip, is the energy lost to motor, gearing, sheave-bearing and rope-bending friction — the shortfall the 85% hoist efficiency accounts for.