Mining Methods & Machinery

Underground Metal Mining: Stoping & Dilution

Dilution as the waste fraction of the muck pile, how it lowers the milled grade, and the tonnage that actually reaches the mill.

PART 1

Topic Breakdown & Traps

The Engineering Principle

In metal stoping, some barren wall rock is unavoidably mined with the ore — this is dilution, the waste fraction of the total material hauled. Dilution lowers the mill feed grade below the in-situ ore grade and increases the tonnage sent to the mill. Controlling dilution is central to stope design economics.

The Core Formula Matrix

Dilution:

D = \frac{waste}{ore + waste} \times 100%

.

Diluted (mill) grade:

g_{mi l l} = g_{or e} (1 - D) + g_{w a s t e} D

.

Tonnage to mill:

T_{mi l l} = \frac{T _{or e}}{1 - D}

The ‘IIT Traps’

⚠
Dilution is waste ÷ *total* (ore+waste), not waste ÷ ore.
⚠
Waste grade is often ~0, so diluted grade ≈ $g_{or e} (1 - D)$ .
⚠
More dilution ⇒ more tonnes milled at a lower grade — both hurt economics.

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · MCQ

A muck pile contains

90 t

ore and

10 t

waste. The dilution is:

Q2MEDIUM2 Marks · NAT

Ore grading

8 g/t

Au is diluted

20%

with barren (

0 g/t

) waste. The mill feed grade is ______ g/t. (Round off to two decimal places.)

g/t

Q3HARD2 Marks · NAT

1000 t

of ore is mined at

20%

dilution, the total tonnage hauled to the mill is ______ t. (Round off to the nearest whole number.)