Question

In a sublevel stope, a ringhole blast round is designed for winning 2500 tonne of ore with 8.0% metal content. The blast results into breakage of 90.0% of the design blast round. Overbreak of 250 tonne wallrock with 0.5% metal content dilutes the blasted ore. The total metal content, in tonne, considering 95% recovery of the blasted muck from the stope, is:

Hint:

While calculating metal content, apply dilution and recovery in the correct order. Always consider the percentage content for both ore and waste, and remember to convert percentage to decimals.

Answer 1

Step 1: Calculate actual ore broken from the design blast.
Design tonnage = 2500 tonnes
Breakage = 90% of 2500 = \( 0.9 \times 2500 = 2250 \) tonnes
Step 2: Calculate total muck = ore + overbreak wallrock.
Overbreak = 250 tonnes
Total muck = \( 2250 + 250 = 2500 \) tonnes
Step 3: Calculate total metal content in muck.
Metal from ore = \( 2250 \times \frac{8.0}{100} = 180.00 \) tonnes
Metal from wallrock = \( 250 \times \frac{0.5}{100} = 1.25 \) tonnes
Total metal in muck = \( 180.00 + 1.25 = 181.25 \) tonnes
Step 4: Apply recovery factor.
Recovered metal = \( 181.25 \times \frac{95}{100} = 172.1875 \) tonnes
Rounded off: 178.81 tonnes