Airway Resistance & Networks
Air loses pressure as it flows through rough airways. Learn Atkinson's equation, friction factors, airway resistance (p = RQ²), shock losses, series and parallel combinations, natural splitting of airflow, regulators, and ventilation network analysis with software.
A ventilation system is a network of airways, each resisting flow. Fans must supply enough pressure to overcome the total resistance at the required flow. Understanding resistance lets engineers size airways, choose fans and balance flows.
An analogy
Airways are like electrical resistors and fans like batteries — but for air, pressure is proportional to flow squared, not flow.
Friction factor
k depends on wall roughness: ≈ 0.004 for smooth concrete-lined shafts, 0.008–0.012 for typical blasted rock drives, 0.015–0.02 for irregular or obstructed airways. Area has a huge effect (A³): slightly larger drives greatly cut pressure loss.
Series and parallel
Airways in series add resistances. In parallel, air splits so that pressure drop is equal in each branch; the combined resistance is lower than any branch. Doubling identical parallel airways cuts resistance to one quarter.
Regulation
Natural splitting sends most air down low-resistance paths. Regulators (adjustable openings) add resistance to branches that get too much air, ensuring each working place gets its share — at the cost of extra fan pressure.
Network analysis
Kirchhoff's laws for ventilation (Σ Q = 0 at junctions, Σ p = 0 around loops) are solved by Hardy Cross iteration in software such as Ventsim, VnetPC and VUMA, which also simulate heat, gases, fires and contaminant spread.
Formulas
- Atkinson's equation:
p = k C L Q² / A³ - Square law:
p = R Q², R = k C L / A³ - Series:
R_s = R₁ + R₂ + … - Parallel:
1/√R_p = 1/√R₁ + 1/√R₂ + … - Air power:
P_air = p Q
Worked example: Atkinson pressure drop
Given: Drift 5 × 5 m (A 25, C 20), L 1000 m; k 0.012, Q 100 m³/s
Find: p and R.
- p = 0.012 × 20 × 1000 × 100²/25³ = 153.6 Pa
- R = 153.6/100² = 0.0154 N·s²/m⁸
Answer: 154 Pa, R = 0.0154
Worked example: Parallel split
Given: R₁ = 0.0154, R₂ = 0.030; Total 150 m³/s
Find: R_p, p and split.
- 1/√R_p = 8.07 + 5.77 = 13.84 → R_p = 0.00522
- p = 0.00522 × 150² = 117 Pa
- Q₁ = √(117/0.0154) = 87 m³/s; Q₂ = √(117/0.03) = 63 m³/s
Answer: 117 Pa; 87 / 63 m³/s split
Common mistakes
- Wrong: Assuming pressure ∝ Q. — Right: Pressure ∝ Q² — doubling flow needs 4× pressure and 8× power.
- Wrong: Adding parallel resistances like series. — Right: Use 1/√R_p = Σ 1/√R_i.
Applications
- Airway sizing: Trade-off of excavation cost vs fan power.
- Flow balancing: Regulators for each level and stope.
- Emergency planning: Simulating fires and smoke spread.
Key takeaways
- p = kCLQ²/A³ = RQ².
- Series add R; parallel add 1/√R.
- Bigger airways save large amounts of power.
- Use network software for real mines.
Going further
- Shock losses: Bends, junctions and obstructions add equivalent length.
- Hardy Cross method: Iterative network balancing.
- Fire simulation: Transient smoke and fume modelling.