Question 4 of 6: Open-Pit Fragmentation Design – Iron Ore

Question 4: Open-Pit Fragmentation Design – Iron Ore (15 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.

(a) Explosive selection and square pattern for X50=30 cm

Given. Square pattern (S=B), ρ_rock=3.0 g/cm³, UCS=300 MPa, E=110 GPa, V_p=6 km/s, De=311 mm, H=14 m, target X₀₁&sub2;= 30 cm. ANFO ρ=0.8 g/cm³ up to emulsion ρ=1.4 g/cm³.

Find. Explosive choice (justified) and the burden/spacing/subdrill/ stemming that deliver a 30 cm average fragment.

Approach. (1) Use the rock's own UCS/E via Lilly's Blastability Index to get a data-driven Kuznetsov rock factor A (rather than an unsupported guess). (2) Use the rock/explosive acoustic impedance match (Z=ρ·VOD, available directly from the given E and V_p) to choose where in the ANFO–emulsion range to explosive to sit – better impedance matching couples more shock energy into this hard, high-modulus ore instead of reflecting it back into the borehole. (3) Bisection-solve the burden on the Kuznetsov equation to hit the 30 cm target.

1. Lilly's Blastability Index (data-driven rock factor A). Massive rock: RMD=50; no joint/RQD data given so assume widely-spaced/massive, RDI=50; UCS>50 MPa so the hardness factor uses Young's modulus: $$HF=\frac{E}{3}=\frac{110}{3}=36.7$$ $$A=0.06(RMD+RDI+HF)=0.06(50+50+36.7)=\boxed{8.2}$$
2. Impedance matching to select the explosive. Rock impedance $$Z_{rock}=\rho_{rock}V_p=3000(6000)=18.0\ \text{MPa}\cdot\text{s/m}$$ Candidate explosives: $$Z_{ANFO}=850(4500)=3.83,\quad Z_{emulsion}=1400(5500)=7.70,\quad Z_{blend}=1200(5000)=6.00\ (\text{all MPa}\cdot\text{s/m})$$ $$\frac{Z_{ANFO}}{Z_{rock}}=0.21,\quad \frac{Z_{blend}}{Z_{rock}}=0.33,\quad \frac{Z_{emulsion}}{Z_{rock}}=0.43$$ Straight ANFO couples poorly into this hard, stiff ore (impedance ratio only 0.21); straight emulsion couples best but at the highest unit cost. A 70/30 emulsion/ANFO blend (ρ≈1.2 g/cm³, VOD≈5000 m/s) is selected – a materially better impedance match than plain ANFO (ratio 0.33 vs 0.21) at a lower cost than straight emulsion.
3. Bisection-solve the square pattern for X₀₁&sub2;=30 cm. Using S=B, J=0.3B, T=0.7B and the blend (ρ=1200 kg/m³): $$X_{50}=A\left(\frac{V}{Q}\right)^{0.8}Q^{1/6},\quad V=B^2H,\quad Q=(H+J-T)\cdot\text{Area}\cdot\rho$$ solving numerically gives $$\boxed{B=S=9.06\ \text{m},\ J=2.72\ \text{m},\ T=6.34\ \text{m}}$$
4. Resulting charge and powder factor. Hole depth = H+J = 16.72 m; charge length = 16.72−6.34 = 10.38 m; Area(311 mm) = 0.0760 m². $$Q=10.38\times0.0760\times1200=\boxed{946\ \text{kg/hole}},\qquad K=\frac{Q}{B^2H}=\frac{946}{9.06^2\times14}=\boxed{0.82\ \text{kg/m}^3}$$

(b) Fragmentation resulting from the design (Kuz–Ram)

Given. Design from part (a): B=S=9.06 m, single-charge column (no decks), charge length 10.38 m, bench height H=14 m.

Find. The full Rosin–Rammler size distribution the design produces (not just X₀₁&sub2;).

Approach. Compute Cunningham's uniformity index n from the pattern geometry, then combine with the X₀₁&sub2; already found in part (a) to get the Rosin–Rammler characteristic size X_c and read off the distribution.

1. Uniformity index. With drilling deviation W≈0.05B (no drilling- accuracy data given – check) and a single, un-decked charge (BCL=CCL): $$n=\left(2.2-14\frac{B}{De_{mm}}\right)\sqrt{\frac{1+S/B}{2}}\left(1-\frac{W}{B}\right) \left(\frac{|BCL-CCL|}{L}+0.1\right)^{0.1}\frac{L}{H}$$ $$n=(2.2-14\times0.0291)(1)(0.95)(0.1)^{0.1}(0.741)=\boxed{1.00}$$
2. Characteristic size and the distribution. From X₀₁&sub2;=X_c(ln2)^1/n: $$X_c=\frac{30}{(0.693)^{1/1.00}}=\boxed{43.2\ \text{cm}}$$ $$X_{20}=9.7\ \text{cm},\quad X_{50}=30.0\ \text{cm}\ (\text{by construction}),\quad X_{80}=69.5\ \text{cm},\quad X_{90}=99.4\ \text{cm}$$

n≈1.0 is a moderately uniform result – typical for a large-diameter, single-deck production pattern – and the X₁₀≈1 m tail indicates the shovel/crusher operation should expect and plan for occasional oversize requiring secondary breakage, even though the design average (X₀₁&sub2;) meets the 30 cm target exactly.

(c) Sequence and timing for a five-row blast, one free face, rope-shovel loading

With one initial vertical free face (not the horizontal top surface), the round must be initiated and sequenced to progressively relieve away from that single face – an echelon (V1/diagonal) firing pattern is recommended: Row 1 (nearest the free face) fires first, breaking directly into it; Row 2 fires next into the void Row 1 has just created; and so on through Row 5, with a delay interval of roughly 25 ms per metre of burden (a standard rule of thumb; here ≈25×9.06 ≈ 225 ms between rows) to give each row's muck time to move and its void to open before the next row's detonation needs it. Within each row, adjacent holes are staggered by a short (a few ms) delay in a diagonal (echelon) sequence rather than all firing simultaneously, so that muck is actively directed and heaved obliquely toward the free face/existing muckpile rather than merely being loosened in place. This heave/displacement is specifically important because loading is by rope shovel: rope shovels dig best against a well- defined, consolidated toe with the muckpile cast forward and piled with a consistent profile (unlike a hydraulic excavator, which tolerates a tighter, less-displaced pile) – an echelon-delayed, adequately-relieved five-row sequence is what produces that shovel-friendly muckpile shape, and also keeps the maximum instantaneous charge at the single-hole value (946 kg) throughout the round, since no two holes anywhere in the pattern share a delay number.