Question

A coal seam of \(2.4\ \text{m}\) thickness is mined with a DERD shearer at a cutting speed of \(0.6\ \text{km/h}\). The web depth of the cut is \(0.5\ \text{m}\). The average cross-sectional area of coal on the AFC during transportation is \(0.24\ \text{m}^2\). For the evacuation of cut coal from the face with 10% spillage, the required minimum velocity of the AFC, in \(m\ s^{-1}\), is . (rounded off to two decimal places)

Hint:

Find the volume of coal the shearer cuts each second from the seam thickness, web depth and cutting speed, then think about how much of that actually has to be carried away once some of it spills off the face.

Answer 1

Correct Answer: 0.75

Step 1: Understand where the coal volume comes from.
As a shearer advances along the face, it cuts a slice of coal whose cross-section is the seam thickness times the web depth of the cut. Moving at the cutting speed, this gives a steady volume of coal produced per unit time, which the AFC (armoured face conveyor) must carry away.

Step 2: Convert the cutting speed and find the cutting rate.
\(0.6\ \text{km/h} = 600\ \text{m/h} = 600/3600 = 0.1667\ \text{m/s}\).
Volumetric rate of coal cut \(= \text{thickness} \times \text{web depth} \times \text{cutting speed} = 2.4 \times 0.5 \times 0.1667 = 0.2\ \text{m}^3/\text{s}\).

Step 3: Account for the 10% spillage.
Spillage means 10% of the coal that is cut never actually loads onto the conveyor, it falls off the face before reaching the AFC. So the AFC only has to carry the remaining 90% of the cut volume.
Volume the AFC must actually handle \(=0.9 \times 0.2=0.18\ \text{m}^3/\text{s}\).

Step 4: Find the minimum belt velocity.
The volumetric capacity of a conveyor is its cross-sectional area times belt velocity, so \(0.18=0.24 \times v\).
\(v=0.18/0.24=0.75\ \text{m/s}\).

Final Answer:
The AFC needs a minimum velocity of about \(0.75\ \text{m/s}\) to clear the cut coal. \[ \boxed{0.75} \]