Airflow 0 m³/sFan pressure 0 PaFan shaft power 0 kWMax CH₄ at a face 0 %Sim time 0:00FPS 0

Drag to orbit, wheel or pinch to zoom, tap an airway to inspect it. Particles show the air at 4× real speed; their colour turns from blue through yellow to red as methane rises (1 % is the action level used here). The drawing is a schematic section: each level drift stands for a much longer run of workings.

Mine Ventilation Network

Atkinson's law H = R·Q², fan curves, Hardy Cross network solution and methane dilution in a cut-away mine.

Tube colour shows

Main fan

How it works. The fan adds pressure H_f(Q) that falls as flow rises. The airways resist with H = R·Q². The network settles where the two balance, so more resistance means less air, not more pressure. Speed scales the whole curve by the fan laws (Q ∝ N, H ∝ N², power ∝ N³).

Doors, regulators, booster

How it works. A regulator is a sliding plate that leaves an opening φ·A, adding R_reg = ρ / (2 C_d² (φA)²). A closed door still leaks through a small gap. An open door on the short Level 1 route lets air short-circuit past the working levels. A booster fan adds pressure inside one airway. A roof fall multiplies one airway's resistance.

Methane and time

How it works. Each airway is a well-mixed volume. Methane enters at the faces and is carried to the return, c ≈ q / (Q + q). Halving the airflow roughly doubles the concentration. Gas moves with the air, so a change takes minutes of simulated time to arrive; use the time warp slider to watch it.

Selected airway

Airway (or tap it in the 3D view)

Live readouts

Fan curve and network curve

Airflow in every airway (m³/s)

Methane at the faces (% CH₄)

Hardy Cross solver

How it works. Start with any flows that satisfy continuity at every junction. For each independent loop, compute the pressure imbalance ΣR·Q|Q| − H_fan and correct all its airways by ΔQ = −Σh / Σ(dh/dQ). Repeat until the corrections vanish. Drag the slider to watch the 3D flows move from the first guess to the converged answer.

Experiments

Each button solves the network many times with the other controls as set and lists the result.

Series and parallel lab

How it works. In series the resistances add, R = R₁+R₂+R₃, and every airway carries the same flow. In parallel the square roots of the conductances add, 1/√R = Σ 1/√Rᵢ, and each airway takes a share Qᵢ ∝ 1/√Rᵢ. The numbers below come from the Hardy Cross solver; the closed forms are shown beside them.

Independent numerical checks

Controls and missions

Drag or one finger: orbit; wheel or pinch: zoom; the four view buttons jump to a level; tap an airway to read its flow, speed and pressure drop.
Missions: (1) keep every face below 1 % CH₄ with the Level 3 regulator at 10 %; (2) find the smallest fan speed that holds all faces under 1 %; (3) show how much air the open Level 1 door wastes; (4) restore safe levels after the roof fall in Stope 2A using only regulators.

The science

Atkinson's law. Turbulent flow in an airway loses pressure H = R·Q² (Pa), with R = k·P·L / A³ (N s² m⁻⁸), where k is the friction factor (kg m⁻³, typically 0.004 to 0.02 for mine airways [typical values, check a mine-ventilation text]), P the perimeter, L the length and A the area; shock losses at bends and junctions are added as a fixed R. Series: R = ΣRᵢ. Parallel: 1/√R = Σ 1/√Rᵢ.

Fan curve. A quadratic H_f = a₀ − a₁Q − a₂Q² on the stable side of the characteristic, scaled by the fan laws to speed ratio n: H_f(Q,n) = n²a₀ − n·a₁Q − a₂Q². The operating point is where H_f equals the network curve. Air power is H·Q; shaft power divides by a simple parabolic efficiency curve. The stall region left of the pressure peak is not modelled. The fans are illustrative, not manufacturer data.

Network solution. Kirchhoff's laws: flows balance at each junction, pressure drops around each loop sum to zero (fan heads count as negative drops). The solver is Hardy Cross, with loops taken from a spanning tree. An independent nodal Newton solver (unknowns are junction pressures) is used for the cross-check.

Regulators and doors. An orifice of free area φA gives R = ρ / (2 C_d² (φA)²) with C_d = 0.8, ρ = 1.2 kg m⁻³. A closed door leaks through 0.03 m².

Methane. Dilution: c ≈ q / (Q + q), so the airflow needed for a limit c is Q = q(1−c)/c. Each airway is a well-mixed cell, V dc/dt = Q·c_in + q − (Q+q)c, mixed at junctions by flow. Alert 0.5 % and action 1 % are illustrative levels (real limits depend on the jurisdiction and the activity); the lower explosive limit of methane in air is about 5 %. The model shows dilution only: it does not predict ignition or explosions, and it has no buoyancy layering, desorption dynamics or gas drainage.

Simplifications. Steady incompressible flow (flows re-balance instantly; gas transport is dynamic), constant air density, no natural ventilation pressure, no leakage between airways other than the door, rectangular airways, one fan curve per fan, illustrative friction and emission values. The drawing is a schematic section: the model lengths of the level drifts (400 to 1500 m) are far longer than their drawn lengths, so the resistances resemble a mine of kilometre scale. The winding cage and cart are decoration. Assets built with Blender; scene by Three.js.