Mining Methods & Machinery

Pumps & Comminution

Dewatering-pump power from flow and head, and the grinding energy predicted by Bond's law of comminution.

PART 1

Topic Breakdown & Traps

The Engineering Principle

A mine dewatering pump raises water at a flow rate

Q

against a total head

H

; the hydraulic power is

ρ g Q H

, and the shaft (motor) power is that divided by the pump efficiency. Size reduction — comminution — is governed empirically by Bond's law, which says the work per tonne is proportional to the difference of the reciprocal square-roots of the product and feed sizes, scaled by the material's work index

W_{i}

The Core Formula Matrix

Pump (hydraulic) power:

P_{h} = ρ g Q H

; shaft power:

P = \frac{ρ g Q H}{η}

ρ

= fluid density,

Q

= flow

[m^{3} / s]

H

= total head

[m]

η

= efficiency.

Bond's law (work index

W_{i}

in kWh/t, sizes in µm):

W = 10 W_{i} (\frac{1}{P _{80}} - \frac{1}{F _{80}}) kWh/t

F_{80}

P_{80}

= 80%-passing feed and product sizes.

The ‘IIT Traps’

⚠
Divide hydraulic power by efficiency for shaft power. $ρ g Q H$ is the useful output; the motor draws more.
⚠
**Bond uses $1/ P_{80} - 1/ F_{80}$ ** (product minus feed reciprocals). Reversing the order flips the sign.
⚠
**Keep $Q$ in m³/s and $H$ in m** so $ρ g Q H$ comes out in watts; mixing L/s or m³/h needs conversion first.

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · MCQ

A pump delivers

0.05 m^{3} / s

of water against a total head of

100 m

. The hydraulic power (ideal) is:

Q2MEDIUM2 Marks · NAT

If that pump (

P_{h} = 49.05 kW

) has an efficiency of

0.70

, the shaft power required is ______ kW. (Round off to two decimal places.)

Q3HARD2 Marks · NAT

A mill reduces ore from

F_{80} = 10000 μ m

P_{80} = 100 μ m

. With a Bond work index

W_{i} = 15 kWh/t

, the specific grinding energy is ______ kWh/t. (Round off to two decimal places.)

kWh/t