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

Let's work this out on a per-minute basis instead of straight into seconds, which keeps the numbers a little rounder before the final conversion.

The cutting speed is $0.6\ \text{km/h}$. Converting to metres per minute: $0.6\ \text{km/h} = 600\ \text{m/h} = 10\ \text{m/min}$.

The shearer removes a slice of coal whose cross-section is the seam thickness times the web depth of cut, so the volume cut per minute is $2.4 \times 0.5 \times 10 = 12\ \text{m}^3/\text{min}$.

Now bring in the spillage. Only 90% of what is cut ever makes it onto the AFC, the rest falls off the face before it is picked up, so the conveyor only needs to move $0.9 \times 12 = 10.8\ \text{m}^3/\text{min}$.

Convert this to a per second basis to match the units asked for: $10.8\ \text{m}^3/\text{min} = 10.8/60 = 0.18\ \text{m}^3/\text{s}$.

The conveyor's carrying capacity is its cross-sectional area times its belt speed, so with an area of $0.24\ \text{m}^2$: $v = 0.18/0.24 = 0.75\ \text{m/s}$.

Let's summarize:

• The shearer cuts coal at a steady volumetric rate set by seam thickness, web depth and cutting speed.
• Spillage of 10% means the AFC only has to move 90% of that cut volume.
• Dividing the required volumetric flow by the AFC's cross-section gives the minimum belt speed.

So the AFC must run at a minimum of $0.75\ \text{m/s}$.