Question 4 of 7: Friction Hoist Duty Cycle, Motor Power, Production and Energy

Question 4: Friction Hoist Duty Cycle, Motor Power, Production and Energy (20 marks, one of three optional)

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.

4.1 — The hoist duty cycle

The duty cycle is the complete speed-versus-time (and, equivalently, position-versus-time) profile of one hoisting trip — acceleration from rest to full rope speed, travel at constant (full) velocity, controlled deceleration/retardation to a stop, and the non-hoisting dwell time at each end of the trip for loading and dumping (or "decking," where the conveyance is stopped and levelled at the shaft collar/loading pocket). The duty cycle defines, directly or indirectly, every major hoist selection parameter: maximum rope (line) speed, acceleration and deceleration rates (which set peak accelerating torque and hence motor/gearbox sizing), total cycle time (which, with skip/cage capacity, fixes the achievable production rate — the basis of Question 4.3), and the RMS (thermal) duty the motor must sustain averaged over the full cycle including its idle/dwell periods, which is what actually sizes the continuous motor rating rather than the instantaneous peak.

4.2 — Hoist motor power requirements: DC vs. AC

"Motor power requirements" means the power the hoist motor must deliver at every instant through the duty cycle — peak at the moment full-load acceleration begins (overcoming both the static hanging load and the inertia of the rotating/translating masses), falling to the steady, lower power needed to sustain constant velocity against the static load alone, and potentially negative (regenerative, i.e. the load drives the motor) during deceleration of a heavily loaded descending trip or an empty ascending one. DC systems (historically Ward-Leonard motor-generator sets, more recently SCR/thyristor DC drives) give smooth, continuous speed and torque control across the full range including true four-quadrant (motoring and regenerating, both directions) operation, long the standard for large mine hoists precisely because a hoist duty cycle demands exactly this kind of control. AC systems (modern variable-frequency drives, VFDs, feeding a squirrel-cage or synchronous motor) now achieve the same four-quadrant controllability through power electronics rather than a rotating motor-generator set, with lower maintenance (no commutator or brushes to service) and higher efficiency, and are the standard choice for new installations. In both cases the motor's continuous (RMS) rating is derived from the duty cycle's power-time profile — integrating the power demand (including the idle/dwell periods, which cool the motor without adding load) over one full cycle — while the peak/breakdown torque rating is set by the single highest instantaneous demand, at the start of full-load acceleration.

4.3 — Daily production

Given. Shift time 7.2 h, 3 shifts/day, skip capacity 11 t, cycle time 85 s/skip.

Find. Daily production (t/day).

Approach. Convert one shift to seconds, divide by the cycle time to get skips per shift, multiply by skip capacity and the number of shifts.

1. Skips per shift. $t_{shift} = 7.2\times3600 = 25{,}920\ \text{s}$, so $n_{skip} = \dfrac{25{,}920}{85} = 304.9\ \text{skips/shift}$ (continuous-duty idealisation, i.e. the hoist runs its full nominal cycle time back-to-back for the entire shift with no lost time).
2. Production per shift and per day. $\dot m_{shift} = 304.9\times11 = 3354\ \text{t/shift}$, so $\boxed{\dot m_{day} = 3\times3354 \approx 10{,}063\ \text{t/day}}$.

In practice a shift completes only whole skip cycles, not a fractional 304.9th skip — taking $\lfloor 304.9\rfloor = 304$ skips/shift gives $304\times11\times3 = 10{,}032\ \text{t/day}$, about 0.3% lower than the idealised continuous-duty figure above. The idealised value is the standard "nameplate" daily-production answer; the whole-cycle figure is the achievable-in-practice figure and is the more conservative number to carry into a mine-planning schedule.

4.4 — Approximate energy consumption per skip

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

Find. Electrical energy consumed per skip hoisted, and the useful mechanical energy actually delivered to the load.

Approach. The motor draws its average 730 kW only while the hoist is under power — the accelerating and constant-velocity portions of the cycle; multiply average power by that powered time to get the electrical energy consumed, then apply the hoist efficiency to find the useful mechanical output.

1. Powered time. $t_{pow} = t_{acc}+t_{const} = 6.5+63.5 = 70\ \text{s} = 0.01944\ \text{h}$ (the deceleration portion of the 85 s cycle is not separately powered at this average draw, consistent with only these two times being given).
2. Electrical energy consumed per skip. $E_{elec} = P_{avg}\times t_{pow} = 730\times0.01944 = \boxed{14.19\ \text{kWh/skip}}$.
3. Useful mechanical energy delivered. $E_{mech} = E_{elec}\times\eta = 14.19\times0.85 = \boxed{12.07\ \text{kWh/skip}}$, i.e. about 2.13 kWh/skip (15%) is lost to friction and heat in the drive train and is not available to do useful lifting work.