Mine Ventilation

Airway Resistance & Fan Laws

Atkinson's square law, series/parallel airway networks, and the fan laws — the backbone of every mine ventilation numerical.

PART 1

Topic Breakdown & Traps

The Engineering Principle

Air pushed through a mine airway behaves like a fluid in a rough duct: energy is lost to wall friction, and because the flow is fully turbulent the pressure drop rises with the square of the airflow (P ∝ Q²) — not linearly like Ohm's law. Atkinson's equation rolls the airway geometry (perimeter, length, cross-sectional area) and wall roughness (friction factor k) into a single number, the airway resistance R. Airways combine like resistors, but the square-law makes the parallel rule use reciprocal square-roots. A fan's duty (Q, P, power) scales with its speed through the fan laws, letting you predict a new operating point from a known one.

The Core Formula Matrix

Atkinson airway resistance

R = \frac{k O L}{A ^{3}}

where

k

= friction factor

[N\cdots^{2} / m^{4} \approx kg/m^{3}]

O

= airway perimeter

[m]

L

= length

[m]

A

= cross-sectional area

[m^{2}]

, and

R

= resistance

[N\cdots^{2} / m^{8}]

.

Atkinson square law (pressure drop)

P = R Q^{2}

P

= frictional pressure drop

[Pa = N/m^{2}]

Q

= airflow

[m^{3} / s]

.

Series airways (same Q through each):

R_{e q} = R_{1} + R_{2} + \dots

Parallel airways (same P across each):

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots

Fan laws (constant diameter, speed

N

\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}, \frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{2}, \frac{W ˙ _{2}}{W ˙ _{1}} = (\frac{N _{2}}{N _{1}})^{3}

Air power delivered:

\dot{W}_{ai r} = P Q [W]

Two airways (R₁, R₂) in parallel between junctions A–B, then airway R₃ in series to the upcast C.

The ‘IIT Traps’

⚠
Cube the area, not the square. Atkinson's R has $A^{3}$ in the denominator. Using $A^{2}$ (a very common slip) inflates R by a factor equal to the area — a classic distractor in MCQs.
⚠
Parallel ≠ Ohm's law. Because $P = R Q^{2}$ is non-linear, parallel airways combine as $1/ R_{e q} = \sum 1/ R_{i}$ , NOT $1/ R_{e q} = \sum 1/ R_{i}$ . Treating airways like electrical resistors gives a wrong (too-small) equivalent resistance.
⚠
Fan-law exponents differ. Quantity scales as $N^{1}$ , pressure as $N^{2}$ , and power as $N^{3}$ . Applying a single linear factor to all three is a frequent error.
⚠
You cannot add parallel pressures. Airflow splits between parallel paths so that each path carries the *same* pressure drop; the quantities add, the pressures do not.

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · MCQ

A straight mine airway has friction factor

k = 0.01 N\cdots^{2} / m^{4}

, length

L = 200 m

, perimeter

O = 8 m

and cross-sectional area

A = 4 m^{2}

. Using Atkinson's equation, the airway resistance

R

is:

Q2MEDIUM2 Marks · NAT

An airway has

k = 0.012 N\cdots^{2} / m^{4}

L = 300 m

, perimeter

O = 10 m

and area

A = 5 m^{2}

. If it carries an airflow of

Q = 20 m^{3} / s

, the frictional pressure drop across it is ______ Pa. (Round off to two decimal places.)

Q3HARD2 Marks · NAT

The plot shows a main fan whose characteristic is linearised as

P = a - b Q

with

a = 2000 Pa

and

b = 20 Pa\cdots/m^{3}

, operating against a mine of total resistance

R = 0.5 N\cdots^{2} / m^{8}

(mine curve

P = R Q^{2}

). The operating airflow

Q

at the intersection of the two curves is ______ m³/s. (Round off to two decimal places.)

Fan line P = a − bQ (a = 2000 Pa, b = 20) intersecting the mine resistance parabola P = 0.5·Q². The intersection is the operating point.

m³/s