A regular shaped cut and fill stope is shown. Pillars of size $4\ \text{m} \times 4\ \text{m}$ are left at an interval of $13\ \text{m}$ along the length of the stope. If the density of the mined ore is $2.5\ \text{tonne/m}^3$, and slices are extracted to the full stope width, the total tonnage of ore recovered from the 1st slice of the stope is $\times 10^3$. (rounded off to one decimal place)

Question

The primary purpose of mine ventilation is to:

Answer 1

Here is a second way to reach the same tonnage, working through the recovery percentage instead of subtracting a pillar volume directly.

First fix the pillar spacing. The pillars are $4\ \text{m} \times 4\ \text{m}$ and the drawing marks a $13\ \text{m}$ gap between the faces of neighbouring pillars, so pillar centres sit $4+13=17\ \text{m}$ apart. Over the $255\ \text{m}$ length of the stope this gives $255/17=15$ pillars.

Now work out what fraction of the plan area is lost to pillars. The full plan area of the slice is $255 \times 6.0 = 1530\ \text{m}^2$. The pillars take up $15 \times (4 \times 4) = 240\ \text{m}^2$ of that. So the fraction of area lost is $240/1530 = 0.1569$, meaning $15.69\%$ of the slice area is left as pillars and $84.31\%$ is actually mined out.

Next find the gross tonnage as if there were no pillars at all. Gross volume $= 255 \times 6.0 \times 4.8 = 7344\ \text{m}^3$, and at a density of $2.5\ \text{tonne/m}^3$ this is $7344 \times 2.5 = 18360$ tonnes.

Apply the recovery fraction to this gross tonnage: $18360 \times 0.8431 = 15479.7$ tonnes, which is the same $15.5 \times 10^3$ tonnes as before.

Let's summarize:

Pillar centre spacing is 17 m (4 m pillar plus 13 m gap), giving 15 pillars over 255 m.
Pillars remove about 15.7% of the slice's plan area.
Applying that loss to the gross slice tonnage gives the same answer as subtracting pillar volume directly.

So the 1st slice yields about $15.5 \times 10^3$ tonnes of ore.

Answer 2

Here is a second way to reach the same tonnage, working through the recovery percentage instead of subtracting a pillar volume directly.

First fix the pillar spacing. The pillars are $4\ \text{m} \times 4\ \text{m}$ and the drawing marks a $13\ \text{m}$ gap between the faces of neighbouring pillars, so pillar centres sit $4+13=17\ \text{m}$ apart. Over the $255\ \text{m}$ length of the stope this gives $255/17=15$ pillars.

Now work out what fraction of the plan area is lost to pillars. The full plan area of the slice is $255 \times 6.0 = 1530\ \text{m}^2$. The pillars take up $15 \times (4 \times 4) = 240\ \text{m}^2$ of that. So the fraction of area lost is $240/1530 = 0.1569$, meaning $15.69\%$ of the slice area is left as pillars and $84.31\%$ is actually mined out.

Next find the gross tonnage as if there were no pillars at all. Gross volume $= 255 \times 6.0 \times 4.8 = 7344\ \text{m}^3$, and at a density of $2.5\ \text{tonne/m}^3$ this is $7344 \times 2.5 = 18360$ tonnes.

Apply the recovery fraction to this gross tonnage: $18360 \times 0.8431 = 15479.7$ tonnes, which is the same $15.5 \times 10^3$ tonnes as before.

Let's summarize:

Pillar centre spacing is 17 m (4 m pillar plus 13 m gap), giving 15 pillars over 255 m.
Pillars remove about 15.7% of the slice's plan area.
Applying that loss to the gross slice tonnage gives the same answer as subtracting pillar volume directly.

So the 1st slice yields about $15.5 \times 10^3$ tonnes of ore.

Show Hint

Correct Answer: 15.5

Solution and Explanation

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