Mine Ventilation

Mechanical Ventilation & Mine Fans

Air power, fan efficiency and the operating point where a fan's pressure–quantity curve meets the mine's resistance characteristic.

PART 1

Topic Breakdown & Traps

The Engineering Principle

A mine fan supplies the pressure that drives air through the mine's resistance (

P = R Q^{2}

). The useful air power delivered is the product of that pressure and the airflow,

P \cdot Q

; the shaft (motor) power is larger by the reciprocal of the fan's efficiency. The mine actually runs where the fan characteristic (pressure falling with quantity) intersects the rising mine characteristic

P = R Q^{2}

— the operating point.

The Core Formula Matrix

Mine characteristic:

P = R Q^{2}

.

Air power (useful output):

P_{a} = P Q

[W]

(pressure in Pa,

Q

in m³/s).

Fan efficiency / shaft power:

η = \frac{P _{a}}{P _{s ha f t}} \Rightarrow P_{s ha f t} = \frac{P Q}{η}

.

Operating point: solve

R Q^{2} = P_{f an} (Q)

for

Q

(fan curve meets mine curve).

Fan curve P = 3000 − 20Q meeting the mine resistance R = 1.0 at Q ≈ 45.7 m³/s.

The ‘IIT Traps’

⚠
**Air power is $P \cdot Q$ , not $P / Q$ .** Pressure (Pa) × quantity (m³/s) gives watts directly.
⚠
Shaft power divides by efficiency. The motor must supply more than the air power: $P_{s ha f t} = P_{a} Q / η$ ... i.e. $P Q / η$ .
⚠
Operating point balances both curves. Reading the fan's free-delivery or shut-off point instead of the intersection gives the wrong duty.

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · MCQ

A fan develops

2000 Pa

while passing

50 m^{3} / s

. The air power delivered is:

Q2MEDIUM2 Marks · NAT

The fan above (air power

100 kW

) has an efficiency of

0.75

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

Q3HARD2 Marks · NAT

A fan with characteristic

P = 3000 - 20 Q

(Pa) serves a mine of resistance

R = 1.0 N\cdots^{2} / m^{8}

. The operating airflow is ______ m³/s. (Round off to two decimal places.)

m³/s