Environment, Ventilation & Hazards

Mine Inundation (Water Hazards)

Hydrostatic pressure behind a water body, the time to pump out a flooded volume, and the pumping power required.

PART 1

Topic Breakdown & Traps

The Engineering Principle

Inundation is the sudden inrush of water into workings — from old flooded workings, aquifers or surface water. The static hydrostatic pressure at depth

h

below a water surface is

ρ g h

. Recovering a flooded mine means pumping out the stored volume at the available pump rate (time = volume/rate), and the pumping power to lift water against the head is

ρ g Q H / η

The Core Formula Matrix

Hydrostatic pressure:

p = ρ g h

.

Pump-out time:

t = \frac{V}{Q}

(

V

= flooded volume,

Q

= pump rate).

Pumping power:

P = \frac{ρ g Q H}{η}

(

Q

in m³/s,

H

= head).

The ‘IIT Traps’

⚠
**Convert $Q$ to m³/s** for power (divide m³/h by 3600).
⚠
**Hydrostatic pressure depends only on depth $h$ **, not the water body's lateral extent.
⚠
**Pump power divides by efficiency $η$ ** — input power exceeds the hydraulic power.

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · NAT

The hydrostatic pressure at the base of a

50 m

column of water (

ρ = 1000 kg/m^{3}

g = 9.81

) is ______ kPa. (Round off to two decimal places.)

kPa

Q2MEDIUM2 Marks · NAT

A flooded section holds

18, 000 m^{3}

of water; a pump delivers

50 m^{3} / h

. The time to pump it dry is ______ hours. (Round off to the nearest whole number.)

hours

Q3HARD2 Marks · NAT

Pumping

50 m^{3} / h

against a

50 m

head at

70%

efficiency requires a power of ______ kW. (Round off to two decimal places.)