A rectangular development heading of dimension 3 m × 2.8 m is to be blasted with holes of 2.4 m in length. If the pull factor is 0.95 and swell factor is 1.20, the volume of blaste... A rectangular development heading of dimension 3 m × 2.8 m is to be blasted with holes of 2.4 m in length. If the pull factor is 0.95 and swell factor is 1.20, the volume of blasted rock per round, in m³, is _____ (round off up to 2 decimals). Read more

Steps to Solve

1. Calculate the volume of the hole created by blasting

The volume of the hole is given by the formula for the volume of a cuboid, which is:

$$ \text{Volume}_{hole} = \text{length} \times \text{height} \times \text{width} $$

Given the dimensions:

• Length = 2.4 m (length of the hole)
• Height = 3 m (height of the development heading)
• Width = 2.8 m (width of the development heading)

So, the volume of the hole is:

$$ \text{Volume}_{hole} = 2.4 \times 3 \times 2.8 $$

Calculating this gives:

$$ \text{Volume}_{hole} = 2.4 \times 3 = 7.2 , \text{m}^2 $$

Now multiply by the width:

$$ \text{Volume}_{hole} = 7.2 \times 2.8 = 20.16 , \text{m}^3 $$

2. Adjust for the pull factor

Next, we adjust the volume of the hole using the pull factor (0.95):

$$ \text{Adjusted Volume} = \text{Volume}_{hole} \times \text{pull factor} $$

Substituting the values:

$$ \text{Adjusted Volume} = 20.16 \times 0.95 $$

Calculating this gives:

$$ \text{Adjusted Volume} = 19.152 , \text{m}^3 $$

3. Adjust for the swell factor

Finally, we adjust for the swell factor (1.20):

$$ \text{Final Volume} = \text{Adjusted Volume} \times \text{swell factor} $$

Substituting the values:

$$ \text{Final Volume} = 19.152 \times 1.20 $$

Calculating this gives:

$$ \text{Final Volume} = 22.9824 , \text{m}^3 $$

4. Round the result

Rounding the final volume to two decimal places gives:

$$ \text{Final Volume} \approx 22.98 , \text{m}^3 $$