A mine airway plan solved by the Hardy Cross method. Wider, warmer branches carry more air. Dots show flow direction.
Each airway b has an Atkinson resistance Rb (N s²/m⁸) so the frictional pressure drop is h = R Q|Q|. Airflows obey Kirchhoff's laws: the flow into every junction equals the flow out (ΣQ = 0) and the pressure drops around any closed loop balance the fan pressure (Σh = Pfan).
The Hardy Cross method solves the loop equations iteratively: Δq = -Σh / (2ΣR|Q|) per loop. This page uses a Newton iteration on the three independent loop flows, which is the same idea with the full Jacobian. The fan follows P = P0 - K Q|Q| with the fan laws Q∝n, P∝n² for speed n. The operating point is where the fan curve meets the system curve P = RsysQ².
Methane dilution: a gas make qg (m³/s) entering an airway carrying Q gives a concentration c = qg/(Q + qg); 1 % is a typical warning level, 2 % a typical withdrawal level (limits vary by jurisdiction).
Simplifications: incompressible air at constant density, constant resistances, no natural ventilation pressure, no leakage between airways, no fan stall region. It is a teaching model, not a design tool for real mines.