Question

A bord and pillar panel has 24 developed pillars of size \(40\ \text{m} \times 40\ \text{m}\) (centre to centre) under extraction. The following information is given:
Gallery width = \(4\ \text{m}\)
Extraction height = \(3\ \text{m}\)
Extraction ratio during depillaring = 80%
Specific gravity of coal = 1.4
Incubation period = 6 months
Average working days per month = 25
In order to extract the panel within the incubation period, the minimum rate of production, in tonne per day, is . (rounded off to one decimal place)

Hint:

Work out the true size of each pillar once the gallery width is taken out of the centre to centre spacing, then think about how much coal that leaves to recover within the given incubation period.

Answer 1

Correct Answer: 696.7

Step 1: Find the actual size of a developed pillar.
The 40 m figure given is the centre to centre spacing between pillars, which includes the gallery driven around each pillar during development. The true solid pillar width is this spacing minus the gallery width: \(40-4=36\ \text{m}\). So each pillar measures \(36\ \text{m} \times 36\ \text{m}\) in plan.

Step 2: Find the total coal standing in the pillars.
Volume of coal in all 24 pillars \(=24 \times 36 \times 36 \times 3=93312\ \text{m}^3\), using the given extraction height of 3 m.

Step 3: Apply the extraction ratio for depillaring.
Only 80% of the coal standing in the pillars is actually won during depillaring, the rest is lost as stooks or left for safety.
Recoverable volume \(=0.8 \times 93312=74649.6\ \text{m}^3\).

Step 4: Convert to tonnage and divide by the time available.
Specific gravity 1.4 means density \(=1.4\ \text{tonne/m}^3\).
Mass to be extracted \(=74649.6 \times 1.4=104509.44\ \text{tonnes}\).
Time available \(=6\ \text{months} \times 25\ \text{days/month}=150\ \text{days}\).
Minimum daily production rate \(=104509.44/150=696.7296\ \text{tonne/day}\), which rounds to \(696.7\ \text{tonne/day}\).

Final Answer:
The panel must be worked at a minimum of about \(696.7\ \text{tonne/day}\) to clear the incubation period. \[ \boxed{696.7} \]