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ENGIN2503 Surface Mining Operations and Equipment - Federation University

Question 1
You are required to do the blast design to facilitate the production request of stripping 2,000,000 m3 overburden annually

Solution:
150 mm holes
11 m benches
Density of compacted ANFO = 0.85 g/cm³
PF >0.65 kg/m³
Stripping 2000000 m³

Required: Blast design
Charge density = (Πr²)/4*ρ_comp

= Π/4*(150/1000)²*0.85*1000=15.021 kg/m

Rock factor for coal =6=A

Q=0.150*15.021*8

=18.025 kg

Stemming length T=12z/A (QS/100)^1/3

z=1.25 (for controlled)

T=(12*1.25)/6*((18.025*100)/100)^(1/3)

=6.555 m

Sub drill U=10*150=1500 mm=1.5 m

Blast hole length = length of benches*stand off distance

Assume a standoff distance of 1.5 m

Blast hole length =11+1.5=12.5 m

Charge length =12.5-6.555=5.945 m

Thus, one blast hole will have

charge density*charge length=15.021*5.945=89.300 kg

Volume of rock blasted =89.300/0.65=137.384 m^2

Number of holes =2000000/137.384=17558 holes annually

Thus, SB=137.384/11=12.489

Assuming S=1.15B,

1.15B*B=12.489
B=3.30 m
S=1.15*3.30=3.79 m

However, if face is inclined to 20° from vertical, then;

New length of blast holes will be 11/cos?20° +1.5=13.206 m

Charge length =13206-6.555=6.651 m

Explosive mass per hole =6.651*15.021=99.904 kg

Therefore, 1.15B²=99.904/(11*0.65)=13.973 m²

B=3.486 m

S=1.15*3.486=4.009 m

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Question 2

Illustrate with suitable cross-section sketches showing ore and waste components, how overall stripping ration can be;
a. Constant
b. Increasing
c. Decreasing over the life of an open-pit mine.

Solution:

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Question 3

Give advice to the mines concerning the practicality and utility of using a conveyor to either replace or supplement the existing systems with proper justification

Give advice to the mines as to what extra additional equipment they may need if they were to combine existing and new conveyor systems with proper justification

Solution:
Haul trucks are expensive, although they are known for perceived bias as a common and accepted way to move stone. Infrastructure building and maintenance is required. Trucks often develop mechanical problems which require drivers with comprehensive pay experience. Mining trucks have heavy engines and therefore use a lot of fuel and regular service that is costly.
In-pit crushing and conveying (IPCC) has outgrown other techniques in the recent past. The method allows the ore in the pit to be crushed and then removed using the conveyor system. This unit is portable, the crusher. Similar to buses, it helps to reduce prices. The system is also efficient, requiring little maintenance and requiring little operation. It also works in bad weather conditions.

Conveyors are robust and reliable tools used in automated storage and transport, as well as mining facilities. In tandem with computer-controlled pallet handling equipment, this allows more efficient distribution of mining materials. It is considered a labour-saving tool that enables large volumes to move quickly through a network, enabling miners to source and receive higher volumes of materials and lower labour costs.

Belt conveyors are the conveyors most commonly used because they are the most versatile and the least expensive. Mining materials and other necessary equipment are placed directly on the belt so that items of ordinary or irregular size, large or small, light and heavy, can be transported efficiently. Belt conveyors are also made of curved parts to move goods around a corner using tapered rollers and curved belting.

Belt conveyors can be used to transfer the material in a straight line or to adjust the altitude or direction. When bulk mining materials such as ore, coal, and gravel are moved, over gentle slopes or gentle curvatures, a drilled belt conveyor is used. The belt trough ensures that the material capable of flow within the edges of the belt is retained. The trough is achieved by holding the idler rollers in a horizontal angle at the sides of the idler frame.

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Question 4
A coal strip mining is designed to use dragline operation as illustrated below. The swell of the spoil is 1.25 to 1.30. You are required to design the effective radius Re and select a dragline that is suitable to the mine.

Solution:

Provided information:
Swelling Factor SF = 1.25 to 1.30. Take SF to be 1.30

H₀=16 m

H_c=4.5 m

W₀₁= W₀₂=40 m

Φ₀= Φ_c=70°

θ=40°

Effective radius Re= H₀/tan?Φ₀ + H_c/tanΦ_c + (W₀₂*H₀*SF)/(W₀₁*tan?θ )+ W_o1/4

Re=16/tan70° +4.5/tan70° +(40*16*1.3)/(40*tan?40°)+40/4

Re=42.250 m

Use truck-mounted dragline. This dragline will be suitable because it's not designed for overburden removal work.

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