Question

Mining Engineering

At a open pit mine the bench height is 15m, the vertical blasthole diameter is 250mm and the rock has a specific gravity of 2.9. If ANFO with a density of 0.8g/cc is used with a square pattern where S = B, determine: 1. Burden 2. Spacing 3. Subdrill 4. Stemming length 5. Blasthole length 6. Charged length 7. Rock broken/blasthole 8. Powder factor

Answer 1

Ans) Given,

Bench height = 15 m

Blast hole diameter (D) = 250 mm or 0.25 m

Specific gravity = 2.9

ANFO density = 0.80 g/cc or 800 kg/m³

1) Burden = 40 D

= 40 x 0.25

= 10 m

2) Spacing = 1.15 Burden

= 1.15 x 10

= 11.5 m

3) Sub drill = (3 to 15) D so let sub drill be 10 D

= 10 x 0.25 = 2.5 m

4) Stemming length = 20 D or 0.95 x Burden ,whichever is greater

= 20(0.25) = 5 m

or 0.95(10) = 9.5 m > 5 m

=> Stemming = 9.5 m

5) Blast hole length = Bench height + Sub drill

= 15 + 2.5

= 17.5 m

6) Charged length = Blast hole length - Stemming length

= 17.5 - 9.5

= 8 m

7) Rock Broken/Blast hole = Burden x Spacing x Bench height x Density of rock

= 10 x 11.5 x 15 x 2.9 x 1000

= 5002.5 ton/ blast hole

8) Powder factor(PF) = Total explosive in the blast / Volume of rock blasted

Volume of blast hole (V_b) = L D² / 4

= (8)(0.25)² / 4

= 0.393 m³    

Total explosive mass = ANFO density x Volume of blast hole

= 800 x 0.393

= 314.4 kg

  Volume of rock blasted = Amount / density = 5002.5 x 1000 kg / 2900 kg/m³  

= 1725 m³

=> PF = 314.4 kg / 1725 m³

= 0.182 kg/m³