Question 1 of 7: Feasibility Costing, Ventilation, Hoisting, Mining-Method Selection and Backfill

Question 1: Feasibility Costing, Ventilation, Hoisting, Mining-Method Selection and Backfill (40 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.

1.1 — Estimating capital and operating costs at feasibility stage

A feasibility-stage cost estimate is built to roughly the AACE Class 3–2 level of accuracy (typically ±15–30%), which is far more defined than the order-of-magnitude Class 5 scoping estimate but not yet the Class 1 "definitive" estimate used for a construction go-ahead. The starting point is always the deposit itself: its geometry (tabular vein vs. massive/disseminated, dip, plunge, depth, continuity) and the surrounding rock mass conditions (RMR/Q-system rating, in-situ stress, hydrogeology) together dictate the feasible mining method and the access strategy — shaft, decline or adit — before a single cost line is estimated. Once the method and access are fixed, capital cost is built up from quantities: development metres (lateral and vertical) at a unit cost per metre that reflects the ground support class the rock conditions demand, a fixed-plant estimate (headframe/hoist, ventilation, backfill plant, processing plant) scaled from equipment vendor quotes or factored from a similar recently built mine of comparable tonnage, and a mobile-equipment fleet sized from the planned production rate and cycle times. Operating cost is estimated bottom-up per unit process (drilling, blasting, mucking, haulage, ground support, ventilation, backfill placement, dewatering) using labour, consumable and power unit costs, then rolled up to a $/tonne-mined figure and cross-checked against benchmark $/tonne data from operating mines with a similar method, depth and rock mass. Both capital and operating estimates should be run at more than one point (a base case plus sensitivities on grade, dilution, exchange rate and metal price) because the mining method itself is rarely fixed until the trade-off study between cut-and-fill/backfilled methods (higher unit cost, higher recovery, lower dilution) and bulk methods such as longhole or caving (lower unit cost, more dilution) is closed out.

1.2 — The basic law of mine ventilation and system terminology

Airflow through a fixed network of mine airways obeys the same friction relationship as pipe flow: pressure loss (head) rises with the square of the volumetric flow (the Atkinson equation), not linearly. Written for a single airway and summed (in series or parallel, as the network dictates) over the whole mine,

where H is the head (Pa, or in H₂O) the fan must overcome, Q is the air quantity (m³/s or cfm), R is the airway (or mine) resistance, k is the friction factor, L is airway length, O is perimeter and A is cross-sectional area. Because R is fixed by the physical dimensions of the airways, doubling the quantity moved quadruples the head the fan must supply — this quadratic law is what makes the mine's own H–Q curve a parabola through the origin.

• Static head of a system, Hs — the portion of total head consumed overcoming friction and shock losses in the airways (skin friction on airway walls, turns, splits, doors, obstructions); it excludes the velocity (kinetic) head needed just to accelerate the air, and is the term normally plotted as the "mine characteristic."
• Measured system head for an actual air flow, Hm — the pressure differential actually read across the fan (or the mine, via a pressure survey) at the air quantity currently passing, i.e. the real operating point taken from gauges/manometers rather than calculated from R and Q; it is the point where the fan's own H–Q curve and the mine's characteristic curve intersect.
• Mine characteristic curve — the plot of total system head Hs (or Ht = Hs + velocity head) against air quantity Q, H = RQ², that represents the combined resistance of every airway in the ventilation circuit. Superimposing the fan's own performance curve on the same axes and reading off their intersection gives the operating point the installation will actually deliver; Question 2 works this exact construction.

1.3 — Dual-skip hoisting cycle time

Given.

Find. The total cycle time, in seconds, to load, hoist and dump one skip.

Approach. Build the one-way hoist as a trapezoidal velocity profile — accelerate from rest to V, travel at constant V, then decelerate back to rest — and add the fixed loading/dumping/creep time; the return trip of the counterbalancing skip runs simultaneously in a dual-skip system, so it does not add to the cycle.

1. Convert the hoisting speed to SI. $$V = \dfrac{400\ \text{m/min}}{60} = 6.667\ \text{m/s}$$
2. Time and distance to reach full speed. With constant acceleration $a = 1\ \text{m/s}^2$: $$t_{acc} = \dfrac{V}{a} = \dfrac{6.667}{1} = 6.667\ \text{s}, \qquad d_{acc} = \dfrac{V^{2}}{2a} = \dfrac{6.667^{2}}{2} = 22.22\ \text{m}$$ Deceleration is symmetric, so $t_{dec} = t_{acc} = 6.667\ \text{s}$ and $d_{dec} = d_{acc} = 22.22\ \text{m}$.
3. Distance and time at constant speed. The accel/decel ramps together cover $2 \times 22.22 = 44.44\ \text{m}$ of the 500 m hoist, leaving $$d_{const} = 500 - 44.44 = 455.56\ \text{m}, \qquad t_{const} = \dfrac{455.56}{6.667} = 68.33\ \text{s}$$
4. Total travel (hoist) time. $$t_{travel} = t_{acc} + t_{const} + t_{dec} = 6.667 + 68.33 + 6.667 = 81.67\ \text{s}$$
5. Total cycle time. Adding the fixed load/dump/creep allowance: $$\boxed{t_{cycle} = t_{fixed} + t_{travel} = 60 + 81.67 = 141.7\ \text{s} \approx 2\ \text{min}\ 22\ \text{s}}$$

1.4 — Cut-and-fill and longhole mining methods

Cut and fill. The stope is mined upward (or, less commonly, downward) in horizontal lifts a few metres high; each lift is mucked out, ground-supported (bolts and mesh in the freshly exposed back), and then backfilled — typically with cemented rock fill or hydraulic/paste tailings — before the next lift is opened directly above. The fill both supports the walls of the adjoining, already-mined stope and becomes the working floor for the lift above. Selective mining is possible because each lift can follow narrow, irregular ore boundaries, giving low dilution and high recovery at the cost of a low mining rate and a high unit cost driven by the fill cycle and the frequent ground-support work.

Longhole (sublevel open stoping). Large-diameter, long (10–40 m) blastholes are drilled in fan or parallel patterns from development on the top and/or bottom of the stope, then blasted in a sequence of rings; the broken ore falls into an open, unsupported void and is mucked from drawpoints at the base with LHD loaders, with no fill placed between the stope walls until the whole panel is complete (if at all). Because drilling, blasting and mucking are decoupled and each employs bulk, high-capacity equipment, longhole achieves a much higher mining rate and lower unit cost than cut-and-fill, but requires a competent, self-supporting orebody and hangingwall (open spans are large) and carries more dilution risk from wall sloughing as the void grows.

1.5 — Shrinkage and sub-level caving mining methods

Shrinkage. The stope is mined upward in horizontal slices, but — unlike cut-and-fill — the broken ore itself (which bulks roughly 30–40% in volume when blasted) is left in the stope and used as the working platform and temporary wall support; only enough muck is drawn off after each slice (about a third) to keep a working gap beneath the back for the next round. Once the top of the stope is reached the remaining ore (roughly two-thirds of the total) is drawn in a final, largely unsupported and hazardous full-scale draw. It needs a steeply dipping, regular orebody and reasonably competent walls (they stand unsupported for the whole stope life), gives a low mining rate because most of the broken ore sits idle as platform (tying up working capital), but needs little standing development and no backfill plant.

Sub-level caving. The orebody is developed on a series of closely spaced horizontal sublevels; each sublevel is ring-drilled and blasted from the top down, and the broken ore is drawn through drawpoints while the overlying caved waste rock progressively fills the space left behind — the method deliberately induces surface subsidence and requires an orebody and cap rock that will cave predictably. It is a high-tonnage, low unit-cost, highly mechanized bulk method well suited to large, steeply dipping deposits, but dilution from waste mixing into the drawn ore rises steadily through each draw and ultimate ore recovery is correspondingly lower than a selective method such as cut-and-fill.

1.6 — Pump power requirement delivered to the motor

Given.

Find. The power that must be delivered to the motor, in kW and horsepower.

Approach. Compute the hydraulic (water) power the pump must impart to the flow from $P = \rho g Q H$, then divide by the pump and motor efficiencies in series — each stage loses energy on the way from electrical input to hydraulic output — to recover the power that must be delivered at the motor.

1. Hydraulic power. With clear water $\rho = 1000\ \text{kg/m}^3$, $g = 9.81\ \text{m/s}^2$, $Q = 0.0126\ \text{m}^3/\text{s}$: $$P_{hyd} = \rho g Q H = 1000 \times 9.81 \times 0.0126 \times 183 = 22{,}620\ \text{W} = 22.62\ \text{kW}$$
2. Power delivered to the motor. The pump and motor each dissipate part of the input; dividing by both efficiencies gives the power the motor must actually deliver: $$P_{motor} = \dfrac{P_{hyd}}{\eta_{pump}\,\eta_{motor}} = \dfrac{22.62}{0.70 \times 0.85} = \dfrac{22.62}{0.595} = 38.0\ \text{kW}$$
3. Convert to horsepower. $$\boxed{P_{motor} = 38.0\ \text{kW} \times 1.341\ \text{hp/kW} = 51.0\ \text{hp}}$$

1.7 — Backfilling material types

Hydraulic fill is de-slimed mill tailings (typically the coarser fraction after cyclone classification, minus fines and clay-sized particles that would drain poorly) delivered underground as a slurry through a borehole/pipe range and allowed to settle and drain behind a permeable barricade at the stope entrance; it is inexpensive because it uses a waste product already generated by the mill, but it needs good drainage (the barricade and any embedded drainage layer must handle the return water) and gives comparatively low, variable strength unless cemented.

Paste fill is whole (unclassified) tailings thickened to a high solids content (typically 70–85% by weight, non-segregating, with the consistency of wet concrete) and usually blended with a cement or binder before being pumped underground through a pipeline; because it carries almost no free water it needs little or no drainage at the barricade, achieves higher and more consistent engineered strength than hydraulic fill for a given binder addition, and can be placed in stopes with limited drainage capacity, at the cost of a more complex, higher-capital surface paste plant.

Rock fill is waste rock (development muck, quarried rock, or crushed aggregate) hauled or tipped underground and end-dumped or conveyed into the void; it is cheap where waste rock is already being generated, provides good drainage and immediate bearing capacity, but is bulky to place uniformly and, uncemented, offers only frictional support (cemented rock fill is used where higher, engineered strength is required, e.g. as a rib pillar substitute).

Flocculated fill is tailings or hydraulic fill dosed with a flocculant (a polymer that agglomerates fine particles) before or during placement to accelerate settling and improve drainage of an otherwise slow-draining, fines-rich slurry; it lets a mine backfill with a tailings stream that would otherwise be unsuitable for straight hydraulic fill (too fine, too slow to drain) without building a full paste/thickening plant, trading a modest reagent cost for usable placement rates.