Question 2 of 6: Hoisting and Winding

Question 2: Hoisting and Winding (20 marks, optional — Section B)

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.

2.1 — Hoist designs

2.1.1 Single drum. One cylindrical drum winds a single rope to a single skip/cage; there is no counterbalance, so the motor must overcome the full rope-plus-conveyance weight on every up-trip and the drum must be sized to store the entire rope length. Simplest, cheapest, lowest headroom hoist, but the least power-efficient of the four (nothing offsets the dead weight being hoisted) and normally limited to shallow/medium shafts or sinking service.

2.1.2 Single drum, counterweighted. The single drum rope runs over a head-sheave to a counterweight that partially balances the empty-conveyance dead weight (though not the payload), so the motor need only overcome the payload plus rope-weight imbalance rather than the whole system weight. Reduces installed power and energy cost versus a plain single drum at modest extra headframe complexity; used on medium-depth single-conveyance shafts, especially man-hoists, where the smoother, more constant motor loading is also a safety benefit.

2.1.3 Double drum. Two independent drums on a common shaft, each carrying its own rope to its own skip, are clutched so the two skips can be run balanced (one ascending loaded while the other descends empty) OR de-clutched and run independently at different depths/speeds — the double drum's key advantage over Koepe. This flexibility (independent depths, one skip serviceable while the other hoists) makes it the standard choice for multi-level shafts and shaft sinking, at the cost of a heavier, wider headframe and drum than a single-rope system and a friction-type Koepe.

2.1.4 Blair multi-rope hoist. A single large-diameter drum is grooved for several (typically 4–6) parallel ropes attached to one large skip or cage, sharing the load across multiple smaller-diameter ropes instead of one large rope. This lets very large payloads be hoisted from very deep shafts without the prohibitively large single-rope diameter (and correspondingly huge drum/headframe) a one-rope system would need, since total breaking strength scales with rope COUNT as well as diameter; the trade-off is the complexity of keeping many ropes at exactly equal tension (rope-equalizing devices are required) and a taller headframe to clear multiple rope runs.

2.2 — Koepe (friction) mining hoists

2.2.1 Applications. Floor-mounted Koepe winders sit at the shaft collar with the friction sheave close to ground level and the headframe carrying only deflection sheaves, giving a lower headframe and easier maintenance access; headframe-mounted (tower-mounted) Koepe winders carry the drive and friction sheave at the TOP of the headframe directly above the shaft, the more common modern arrangement, giving the shortest possible rope run and the smallest sheave-to-conveyance offset, at the cost of moving the entire drive/maintenance platform to height.

2.2.2 Rope and guide types. Koepe winders use non-rotating, multi-rope, flattened-strand or round-strand ropes (never a single Lang-lay rope, since the friction-drive geometry needs balanced, non-rotating conveyances) together with tail ropes below the conveyances to maintain constant total rope weight on each side of the sheave through the hoisting cycle (see 2.2.6). Rigid steel guides (or, in some modern installations, rope guides) keep the skips/cages centred in the shaft against the lateral component of friction-drive rope tension.

2.2.3 Typical hoisting depths. Koepe/friction hoisting is best suited to deep-to-very-deep shafts, roughly 500–2,000+ m single-lift, because its light, small-diameter sheave (versus a drum that must store the full rope length) keeps headframe size and installed inertia manageable at depth where a drum hoist's ever-larger drum becomes impractical.

2.2.4 Major advantages. Lower moving inertia (a sheave, not a full rope-storing drum, so faster acceleration and less wasted energy); virtually unlimited depth capability (rope length is not limited by drum storage capacity); smaller, lighter headframe/drive components for a given depth; and multi-rope capacity giving high load capacity without a single oversized rope.

2.2.5 Disadvantages. The hoist relies on FRICTION between rope and sheave groove, so it is only usable where the loaded/empty conveyance weight ratio stays within the slip limit (Question 2.2.12, 2.3.4); it cannot easily run two conveyances independently to different depths (unlike a double drum); and rope/groove wear (lining wear, rope flattening) must be monitored closely since loss of friction risks slip rather than simply increased wear as on a drum.

2.2.6 Rope configuration, tail ropes and anti-fouling. Multiple head ropes run in parallel grooves over the friction sheave to each conveyance, fitted with rope-equalizing capels so all head ropes share load evenly; tail ropes (heavier per metre than the head ropes to balance the changing head-rope weight through the cycle) hang below each conveyance and loop in the shaft sump, keeping the total suspended weight on each side of the sheave constant regardless of conveyance position. Anti-fouling and damage control are managed by guide ropes or rigid shaft guides that keep all ropes laterally separated and centred, rope-turning/anti-spin devices at the tail-rope loop, and scheduled rope-condition inspection (magnetic flux/visual) rather than any single passive device.

2.2.7 Shaft sinking with a Koepe. Shaft sinking is possible with a Koepe winder but requires modification: a temporary, smaller-capacity sinking sheave/headgear (or a temporary drum hoist) is normally used during sinking itself because the sinking stage's rapidly changing depth and the need for independent, unbalanced single-rope operation do not suit the fixed, balanced-pair Koepe geometry well; the permanent Koepe installation is typically commissioned only once the shaft has reached final depth and permanent guides/loading pockets are in.

2.2.8 Capel inspection and skip-position adjustment. A capel is the conical/socketed end-fitting that terminates a hoist rope and anchors it to the conveyance or equalizing bar; because rope stretches (permanently and elastically) over its service life, the capel must periodically be removed, the rope end re-socketed (or the rope re-capped shorter), and the ropes re-equalized so that the skip loading and dumping positions in the shaft stay at the correct, calibrated depths despite the accumulated stretch.

2.2.9 Optimal hoisting depths, single vs. multi-rope. Single-rope Koepe winders suit shallower-to-medium depths (up to roughly 1,000 m) where one rope's capacity is sufficient; multi-rope Koepe winders extend the practical depth range well beyond that (1,500–2,000+ m) and to higher payloads, because sharing the load across several ropes avoids the single-rope diameter/weight that would otherwise become the limiting factor at extreme depth.

2.2.10 Skip type. Bottom-dump skips are preferred for Koepe installations: they discharge directly downward through a bottom door into a fixed receiving bin/chute at the collar, giving fast, low-spillage dumping without needing the skip itself to tilt, which keeps the loading/dumping cycle short and the rock stream contained; over-turning (tipping) skips, which rotate the whole skip body to dump, are more common on drum hoists where the extra headroom and slower dump cycle are less of a constraint. Spillage at load/dump is minimised with sealed, close-fitting chute/bin interfaces and dust/rock skirting at both loading pocket and dump point.

2.2.11 Personnel hoisting. Koepe winders are commonly adapted for personnel (man-cage) hoisting by fitting a purpose-built cage (with safety catches, over-speed/over-travel protection and a lower, controlled man-riding speed) in place of, or alongside, the ore skip; this is a very common application, particularly on production shafts that also serve as the mine's main personnel access, since the Koepe's smooth, low-inertia acceleration profile suits passenger comfort and safety standards well.

2.2.12 Slip ratio and hoist geometry. The maximum tension ratio a friction sheave can sustain before the rope slips in the groove is governed by the capstan (belt-friction) relation T1/T2 < e^μθ, where μ is the rope/lining friction coefficient and θ is the total wrap angle of rope on the sheave; slip occurs whenever the loaded-side tension exceeds e^μθ times the empty-side tension. Geometry can be modified to keep the design ratio below the limit by increasing the wrap angle θ (e.g. adding a deflection/bicycle sheave to wrap the rope further around the drive sheave), specifying a higher-friction sheave lining (raising μ, within lining-life limits), or — the approach actually used in Question 2.3 — adding counterbalancing tail-rope/equivalent-effective-weight mass so the STATIC T1/T2 ratio itself stays comfortably under the (commonly quoted) 1.6:1 practical slip limit for a headframe-mounted Koepe.

2.3 — Mine hoisting: cycle time, rope weight, tensions and slip check

Given. Friction (Koepe) hoist, headframe mounted; shaft depth 300 m; skip live load W_o = 4.5 mt (4,500 kg); skip dead load W_s = 5.4 mt (5,400 kg, dead/live ratio 1.2); four 25 mm flattened-strand 6×19 ropes at 2.68 kg/m each; sheave diameter 3.14 m; hoisting velocity V = 6.1 m/s; duty-cycle times t_a=10 s, t_v=39.75 s (≈40 s), t_r=8 s, t_d=10 s; hoist efficiency η=0.9; equivalent effective weight W_e = 16,818 kg read directly from Fig. 2.3.5 at a 3.14 m sheave diameter (Koepe curve); slip limit for a headframe-mounted Koepe = T1/T2 > 1.6.

Find. Hoist cycle time t_t (2.3.1); rope weight W_r (2.3.2); total load weight W_I (2.3.3); loaded/empty rope tensions T1, T2 and whether slippage occurs (2.3.4); the equivalent effective weight already read from Fig. 2.3.5 (2.3.5); and the total suspended load W (2.3.6).

Approach. Apply the "Additional Hints" formulas directly in the order given — sum the cycle segments, weigh the ropes, sum the loaded/empty sides, form the tension ratio, and combine with W_e for the total suspended load.

1. 2.3.1 — Hoist cycle time. Sum the four duty-cycle segments (accelerate, full-speed run, retard, decking/rest): $$t_t = t_a + t_v + t_r + t_d = 10 + 39.75 + 8 + 10 = \boxed{67.75\ \text{s}\ (\approx 68\ \text{s})}$$
2. 2.3.2 — Rope weight. Total unit weight of the four parallel ropes times the shaft depth (with h_h≈0 per the check note): $$W_r = (4 \times 2.68\ \text{kg/m}) \times 300\ \text{m} = 10.72 \times 300 = \boxed{3{,}216\ \text{kg}\ (3.216\ \text{mt})}$$
3. 2.3.3 — Total load weight. Sum rope, dead and live loads on the loaded side: $$W_I = W_r + W_s + W_o = 3{,}216 + 5{,}400 + 4{,}500 = \boxed{13{,}116\ \text{kg}\ (13.116\ \text{mt})}$$
4. 2.3.4 — T1, T2 and slippage. The loaded side carries the full load weight; the empty side carries only its dead load plus rope weight: $$T1 = W_I = 13{,}116\ \text{kg}, \qquad T2 = W_s + W_r = 5{,}400 + 3{,}216 = 8{,}616\ \text{kg}$$ $$\frac{T1}{T2} = \frac{13{,}116}{8{,}616} = \boxed{1.52}$$ Since 1.52 < 1.6 (the stated slip threshold for a headframe-mounted Koepe sheave), no slippage occurs under these operating conditions; the rope/lining friction is adequate to hold the load with margin. If a future duty point pushed the ratio above 1.6, the standard remedies are the ones set out in 2.2.12 — increase the wrap angle θ, use a higher-friction lining, or add counterbalancing tail-rope weight to reduce the STATIC imbalance.
5. 2.3.5 — Equivalent effective weight. Read directly off Fig. 2.3.5 (Koepe curve) at the given 3.14 m sheave diameter, as stated in the question: $$W_e = \boxed{16{,}818\ \text{kg}\ (37{,}000\ \text{lb})}$$ (this term represents the effective inertial mass of the rotating sheave/drum reflected back to the rope line, additional to the physically suspended weights, and is obtained empirically from the manufacturer's diameter-vs-effective-weight chart rather than computed from first principles.)
6. 2.3.6 — Total suspended load. Combine the equivalent effective weight with the live load and BOTH sides' dead loads and rope weights (the total system inertia the motor must accelerate/decelerate): $$W = W_e + W_o + 2W_s + 2W_r = 16{,}818 + 4{,}500 + 2(5{,}400) + 2(3{,}216) = \boxed{38{,}550\ \text{kg}\ (38.55\ \text{mt})}$$

2.4 — Duty cycle: power at each phase

Given. The duty cycle is the plot of hoist power against time over one complete hoisting cycle: it consists of an acceleration phase t_a in which power ramps from zero to a peak, a constant full-speed running phase t_v, and a retardation (deceleration) phase t_r in which power falls through zero to a negative (regenerative) value as the motor is used to brake the load, followed by a "rest"/decking interval. Creep is the short, very-low-speed final approach to the loading/dumping position at the end of travel — run deliberately slowly so the conveyance can be positioned precisely at the pocket without overshoot. Decking is the stationary dwell (t_d = 10 s here) once the conveyance is at rest at the loading/dumping station, during which the skip is loaded or dumped and no hoisting power is drawn at all. From 2.3.6, the total suspended load W = 38,550 kg; the live load W_o = 4,500 kg; V = 6.1 m/s; η = 0.9; t_a=10 s, t_r=8 s.

Find. The six named duty-cycle powers P1–P6 and the four composite powers Pa–Pe (2.4.1–2.4.11), using the hint formulas with the metric-HP conversion 745.5 = 76×9.81 (W in kg, V in m/s) for the acceleration/regeneration terms and 76 for the steady-running terms.

1. 2.4.1 P1 — excess acceleration power. Extra power above steady state needed to accelerate the full suspended load W to V in time t_a (impulse force F=WV/(g t_a), power = F×V, with g folded into the 745.5 HP constant): $$P1 = \frac{W V^2}{745.5\, t_a} = \frac{38{,}550 \times 6.1^2}{745.5 \times 10} = \boxed{192.5\ \text{HP}}$$
2. 2.4.2 P2 — regenerated deceleration power. Same relation using the retardation time t_r in place of t_a, taken negative since the motor absorbs (regenerates) rather than supplies this power: $$P2 = -\frac{W V^2}{745.5\, t_r} = -\frac{38{,}550 \times 6.1^2}{745.5 \times 8} = \boxed{-240.6\ \text{HP}}$$
3. 2.4.3 P3 — steady running power. Power to lift only the live payload W_o at the constant hoisting velocity (76 already includes g, per the check note in 1.3): $$P3 = \frac{W_o V}{76} = \frac{4{,}500 \times 6.1}{76} = \boxed{361.2\ \text{HP}}$$
4. 2.4.4 P4 — motor-efficiency running term. Additional power drawn because of motor inefficiency at the end of acceleration, scaled off P3 by (1−η)/η: $$P4 = \frac{1-\eta}{\eta} \times P3 = \frac{0.1}{0.9} \times 361.2 = \boxed{40.1\ \text{HP}}$$
5. 2.4.5 Pa — acceleration peak. Sum of the positive power components at the instant acceleration ends: $$Pa = P1 + P3 + P4 = 192.5 + 361.2 + 40.1 = \boxed{593.8\ \text{HP}}$$
6. 2.4.6 Pb — full-speed running power. Steady-state power drawn throughout t_v (no acceleration term): $$Pb = P3 + P4 = 361.2 + 40.1 = \boxed{401.3\ \text{HP}}$$
7. 2.4.7 Pc — total retardation power. Steady running terms combined with the (negative) regenerated power: $$Pc = P2 + P3 + P4 = -240.6 + 361.2 + 40.1 = \boxed{160.7\ \text{HP}}$$
8. 2.4.8 P5 — rotor-inertia power, acceleration. Motor efficiency applied directly to the acceleration peak Pa (per the check note): $$P5 = \eta \times Pa = 0.9 \times 593.8 = \boxed{534.4\ \text{HP}}$$
9. 2.4.9 P6 — rotor-inertia power, retardation. Same magnitude, stated negative (power required to retard, not drive, the motor rotor): $$P6 = -\eta \times Pa = \boxed{-534.4\ \text{HP}}$$
10. 2.4.10 Pd — total accelerating power (hoist + motor rotor). $$Pd = Pa + P5 = 593.8 + 534.4 = \boxed{1{,}128.2\ \text{HP}}$$
11. 2.4.11 Pe — total retarding power (hoist + motor rotor). Subtracting the (negative) P6 adds its magnitude: $$Pe = Pc - P6 = 160.7 - (-534.4) = \boxed{695.1\ \text{HP}}$$

2.5 — RMS power

Given. Pd = 1,128.2 HP, Pb = 401.3 HP, Pe = 695.1 HP (Question 2.4); t_a=10 s, t_v=39.75 s, t_r=8 s.

Find. The RMS motor power over the full cycle, in HP and kW.

Approach. Apply the given weighted-RMS formula directly, using the source's own weighting denominator (which allots the acceleration interval double credit — half at the start of the window and, via the trailing 0.25t_a term, a further quarter — reflecting the thermal loading of the repeated accel/decel transients on the motor).

1. Weighting denominator. $$0.5t_a + t_v + 0.5t_r + 0.25t_a = 0.5(10) + 39.75 + 0.5(8) + 0.25(10) = 5 + 39.75 + 4 + 2.5 = \boxed{51.25\ \text{s}}$$
2. RMS power. $$P_{rms} = \sqrt{\frac{Pd^2 t_a + Pb^2 t_v + Pe^2 t_r}{51.25}} = \sqrt{\frac{1{,}128.2^2(10) + 401.3^2(39.75) + 695.1^2(8)}{51.25}}$$ $$P_{rms} = \sqrt{\frac{12{,}728{,}400 + 6{,}401{,}700 + 3{,}865{,}500}{51.25}} = \sqrt{448{,}700} = \boxed{669.8\ \text{HP}}$$
3. Convert to kW. $$P_{rms} = 669.8 \times 0.746 = \boxed{499.6\ \text{kW}}$$

2.6 — Energy consumption per trip

Given. Pb = 401.3 HP (full-speed running power); t_a=10 s, t_v=39.75 s; η=0.9; 0.746 kW/HP.

Find. Energy consumed per hoisting trip, in kWh.

Approach. The running (non-regenerative) power Pb is drawn for the combined accelerate-plus-full-speed duration (t_a+t_v); converting to kW and dividing by the motor efficiency gives the electrical energy drawn from supply, and by 3,600 s/hr converts the power×time product to kWh.

1. Energy per trip. $$E = \frac{0.746 \times Pb \times (t_a + t_v)}{3{,}600 \times \eta} = \frac{0.746 \times 401.3 \times (10 + 39.75)}{3{,}600 \times 0.9}$$ $$E = \frac{0.746 \times 401.3 \times 49.75}{3{,}240} = \frac{14{,}895}{3{,}240} = \boxed{4.60\ \text{kWh/trip}}$$