Question 5 of 7: Mine Ventilation Network Theory and a Four-Airway Parallel Circuit

Question 5: Mine Ventilation Network Theory and a Four-Airway Parallel Circuit (20 marks, one of three optional)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Kirchhoff's laws and Atkinson's equation in ventilation design

Atkinson's equation gives the frictional head loss of a single airway, $H = RQ^2$, where the airway resistance $R = \dfrac{kPL}{A^3}$: $k$ is the friction factor of the airway lining (an empirical drag coefficient depending on rock/support surface roughness), $P$ the rubbing-surface perimeter, $L$ the airway length, and $A$ the cross-sectional area — entering as $A^3$, so resistance is extremely sensitive to airway size (halving the area increases resistance eightfold). Atkinson's equation is the building block; a real mine ventilation circuit is a network of many such airways in series and parallel, and Kirchhoff's first (nodal/continuity) law — the air entering any junction equals the air leaving it — and Kirchhoff's second (mesh/loop) law — around any closed circuit, the algebraic sum of head losses (and any fan heads or natural ventilation pressure within the loop) is zero — are applied together across the whole network to solve for the unknown quantity distribution and pressure at every airway. In network design this is done by combining each airway's Atkinson resistance according to the network topology — resistances in series add directly, $R_{series}=\sum R_i$ (same $Q$ throughout, losses add), while resistances in parallel combine as $1/\sqrt{R_{eq}} = \sum 1/\sqrt{R_i}$ (same $H$ across every branch, per Kirchhoff's second law applied to the parallel loop, with the total $Q$ split between branches per Kirchhoff's first law) — and modern network-analysis software (VnetPC, Ventsim, ClimSim) applies both laws simultaneously through an iterative Hardy-Cross-type balancing routine to converge on the quantity and pressure distribution of a full branched circuit, exactly the calculation carried out by hand below for the simpler four-branch parallel case.

5.1–5.4 — Equivalent resistance, head loss, and branch flows

Given. $Q_{total}=47.19\ \text{m}^3/\text{s}$; $R_1=2.627$, $R_2=0.151$, $R_3=0.349$, $R_4=0.397\ \text{N}\cdot\text{s}^2/\text{m}^8$.

Find. $R_{eq}$, $H$, $Q_1$ through $Q_4$, and $\sum Q_i$.

Approach. All four branches connect the same two nodes, so each carries the same head loss $H$ (Kirchhoff's second law on the trivial single-loop network); combine resistances with the parallel-Atkinson rule, then back-solve $H$ from the given total flow, then each branch flow from $H$ and its own resistance.

1. Equivalent resistance. $\dfrac{1}{\sqrt{R_{eq}}} = \sum_{i=1}^{4}\dfrac{1}{\sqrt{R_i}} = \dfrac{1}{\sqrt{2.627}}+\dfrac{1}{\sqrt{0.151}}+\dfrac{1}{\sqrt{0.349}}+\dfrac{1}{\sqrt{0.397}} = 0.617+2.573+1.693+1.587 = 6.470$, so $\boxed{R_{eq} = 1/6.470^2 = 0.02389\ \text{N}\cdot\text{s}^2/\text{m}^8}$.
2. Head loss. Applying Atkinson's equation to the equivalent single resistance carrying the full flow: $\boxed{H = R_{eq}Q_{total}^2 = 0.02389\times47.19^2 = 53.19\ \text{Pa}}$.
3. Branch flows. Each branch sees the same $H$, so $Q_i = \sqrt{H/R_i}$: $Q_1=\sqrt{53.19/2.627}=\boxed{4.50\ \text{m}^3/\text{s}}$; $Q_2=\sqrt{53.19/0.151}=\boxed{18.77\ \text{m}^3/\text{s}}$; $Q_3=\sqrt{53.19/0.349}=\boxed{12.35\ \text{m}^3/\text{s}}$; $Q_4=\sqrt{53.19/0.397}=\boxed{11.58\ \text{m}^3/\text{s}}$.
4. Sum check. $\sum Q_i = 4.50+18.77+12.35+11.58 = \boxed{47.19\ \text{m}^3/\text{s}}$, matching the given total flow exactly — the internal consistency check that confirms $R_{eq}$ and $H$ were solved correctly (Kirchhoff's first law satisfied at both nodes).

The lowest-resistance airway (Airway 2, $R=0.151$) naturally carries by far the largest share of the flow (18.77 of 47.19 m³/s, about 40%) — because $Q\propto1/\sqrt{R}$, flow splits are far less sensitive to resistance than a naive inverse (not inverse-square-root) split would suggest, which is why a poorly proportioned parallel circuit can still leave a high-resistance airway (Airway 1 here) badly under-ventilated relative to the others even though every branch shares the identical head loss.