Applied Mathematics
Essay by Ernest Rase Sacedor • June 30, 2017 • Course Note • 78,760 Words (316 Pages) • 1,597 Views
MINING ENGINEERING 1
APPLIED MATHEMATICS
- 1 OPEN PIT MINING PROBLEM:
August, 1981. A new mining project has a blocked positive reserve of 2,000,000 WMT of iron ore. Specific gravity of the ore is 4 and moisture content is 7.5%. With a waste to ore volumetric ratio of 3:1 it was decided to mine the ore by open pit. The Mining Engineer recommended to the management that before the ore extraction starts, 100,000 metric tons of ore be exposed first. The occurrence is such that no matter how the orebody is approached, the waste to ore ratio remains constant.
- Compute volume of overburden to be stripped to expose 100,000 dry metric tons of ore.
- At a waste stripping rate of 1,000 cubic meters per day, how long will it take to do the initial stripping before mining starts.
- After the initial stripping, it is imperative that the waste to ore ratio is maintained with the same waste stripping rate of 1,000 cubic meters per day. Compute ore extraction rate in dry metric tons per day.
- Allowing 10% ore dilution, what is the expected life of the mine assuming 330 working days a year.
Solution:
Volume of waste (overburden)
a. ------------------------------------- = 3
Volume of ore
Volume of ore = 100,000 DMT / 0.925 = 108,108 WMT x 1 cu.m/4 MT
= 27,027 cu.m.
Volume of overburden = 3 x volume of ore
Volume of overburden = 3 x 27,027 cu.m.
Volume of overburden = 81,081 cu.m.
- Let : N = no. of days to do initial stripping
81,081 cu.m.
N = ------------------
1,000 cu.m./day
N = 81 days
Waste Extraction rate
- ---------------------------- = 3
Ore Extraction rate
Waste Extraction rate
Ore Extraction rate = --------------------------
3
1,000 cu.m./day
Ore Extraction rate = --------------------
3
Ore Extraction rate = 333.33 cu.m./day x 4 WMT/cu.m.
Ore Extraction rate = 1,333.33 WMT/day x (1-Moisture) DMT/WMT
Ore Extraction rate = 1,333.33 WMT/day x (1-0.075) DMT/WMT
Ore Extraction rate = 1,233.33 DMT/day
Total Ore Tonnage
- Life of Mine = -------------------------
Ore Extraction rate
2,000,000 WMT x 1.1
Life of Mine = ------------------------------------------
1,333.33 WMT/day x 330 days/year
Life of Mine = 5 years
A.2 OPEN PIT MINING PROBLEM:
August, 1984. An exploration project has resulted in the delineation of a mineable ore reserve of 2,000,000 WMT of low grade gold ore. Specific gravity of both the ore and waste is 2.5 with a moisture content of 8%. Mine management decided to mine the ore by open pit with a waste to ore ratio of 3:1 and involving a pre-stripping of 120,000 m3 waste. Assuming that the waste to ore ratio will remain constant throughout the life of the mine.
- Compute the total tonnage in DMT of ore to be exposed after the preliminary stripping activities.
- At a waste stripping capacity of equipment at 2500 WMTPD, how long will it take to do the initial stripping before mining starts?
- With the same stripping rate, what will be the ore extraction rate in dry metric tons per day?
- Allowing 10% for ore dilution, what is the projected life of the mine assuming 300 working days a year?
Solution:
a. Volume of waste
---------------------- = 3
Volume of ore
120,000 cu.m.
--------------------- = 3
Volume of ore
Volume of ore = 40,000 cu.m.
Tonnage of ore = 40,000 cu.m. x 2.5 WMT/cu.m
Tonnage of ore = 100,000 WMT x 0.92 DMT/WMT
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