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24-MMP-A5 Surface Mining Methods and Design · December 2018

Question 21 of 27

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper
Paper: Surface Mining Methods and Design (09-MMP-A5), National Exam, December 2018 — 20 pages, compulsory Question 1 (40 marks, parts 1.1–1.8) plus THREE of five optional Questions 2–6 (20 marks each) normally constitute a complete paper. As a study resource, this solution answers Question 1 in full AND all five optional Questions 2–6.

Reference texts: Hustrulid, Kuchta & Martin, Open Pit Mine Planning and Design (3rd ed.) — truck-shovel match factor, dragline stripping geometry, capital cost indexes, open-pit scheduling; SME Mining Engineering Handbook (3rd ed.) — equipment costing, mine dewatering, cost-index escalation.

Question 4.2 capital cost indexes (6 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.

Given. Power-law cost model P=aXb for each machine class, with a, b and unit conventions as tabulated, and unit conversions 1 lb=0.4536 kg, 1 m³=1.308 yd³, 1 tonne=1.102 short ton.

Given data — power-law capital cost models (1997 $)
MachineabX unitMachine size given
Rotary drill4,0000.67pull-down force, lb55,000 kg
Cable shovel5,350,0000.74bucket capacity, yd³53 m³
Mechanical-drive truck20,0000.90capacity, short ton300 tonne

Find. The 1997 capital cost of each machine.

Approach. Convert each machine’s given size into the formula’s OWN native unit first, then apply P=aXb directly.

  1. 4.2.1 — drill. $$X = \dfrac{55{,}000\text{ kg}}{0.4536\text{ kg/lb}} = 121{,}252\text{ lb}$$ $$P = 4{,}000 \times 121{,}252^{0.67}$$ $$\boxed{P \approx \$10.19\text{ million}}$$
  2. 4.2.2 — shovel. $$X = 53\text{ m}^3 \times 1.308\text{ yd}^3/\text{m}^3 = 69.32\text{ yd}^3$$ $$P = 5{,}350{,}000 \times 69.32^{0.74}$$ $$\boxed{P \approx \$123.20\text{ million}}$$
  3. 4.2.3 — truck. $$X = 300\text{ tonne} \times 1.102\text{ ton/tonne} = 330.6\text{ short ton}$$ $$P = 20{,}000 \times 330.6^{0.90}$$ $$\boxed{P \approx \$3.70\text{ million}}$$
MachineConverted X1997 cost
55,000 kg rotary drill121,252 lb pull-down$10.19 million
53 m³ cable shovel69.32 yd³$123.20 million
300 tonne mechanical-drive truck330.6 short ton$3.70 million