Question 3 of 6: Design of Ventilation Facilities

Question 3: Design of Ventilation Facilities (20 marks)

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 mine ventilation

Kirchhoff's first (nodal/continuity) law states that at any junction of a ventilation network the air flowing in must exactly equal the air flowing out — a direct statement of mass conservation for the (effectively incompressible) mine air, and it is what splits the known total 47.19 m³/s among the four parallel airways solved below. Kirchhoff's second (mesh/loop) law states that around any closed circuit in the network, the algebraic sum of head (pressure) losses — including any fan heads or natural ventilation pressure acting within that loop — is zero; applied to a set of parallel airways connecting the same two junctions, it forces every one of those airways to see the identical head loss H, which is exactly the condition used in Part 3.1 below. Atkinson's equation, $H = RQ^2$, gives the frictional head loss of a single airway in terms of its resistance $R = k\,P\,L/A^3$ (friction factor k, rubbing perimeter P, airway length L and cross-sectional area A) and the quantity Q flowing through it — the ventilation-network analogue of a pipe-friction head-loss equation, with resistance playing the role of a (quadratic, not linear) hydraulic resistance.

In practice these three relationships are used together, iteratively, to design a mine ventilation system: the planned mine layout is idealised as a network of nodes (junctions, shaft/raise/drift ends) and branches (airways, each assigned a resistance from Atkinson's equation using its planned dimensions and expected friction factor); Kirchhoff's laws are then solved simultaneously across the whole network — by hand for a simple branched system such as the one below, or by a Hardy-Cross-type iterative solver (VnetPC, Ventsim) for a real multi-level mine — to find the airflow distribution and the total system resistance the main fan(s) must overcome. That total system head and required total quantity together set the main fan's required duty point, which in turn drives the fan selection and, ultimately, the ventilation shaft/raise sizing decisions that are made at the design stage.

3.1 – 3.4 — Four parallel mine airways

Given. Four airways connect the same two junctions in parallel, carrying a combined total quantity $Q_{tot} = 47.19\ \text{m}^3/\text{s}$.

Find. The equivalent resistance R_eq (3.1), the common head loss H_l (3.2), the individual branch quantities Q₁–Q₄ (3.3), and their sum (3.4).

Approach. Because all four airways share the same two end junctions, Kirchhoff's second law makes each carry the same head loss H; Atkinson's equation then gives $Q_i=\sqrt{H/R_i}$ for every branch. Summing the branches by Kirchhoff's first law and equating to the known total gives the square-root combination rule for parallel airways, $1/\sqrt{R_{eq}} = \sum 1/\sqrt{R_i}$, from which R_eq, then H, then each Q_i follow in turn.

1. Part 3.1 — equivalent resistance. Since $Q_{tot}=\sqrt{H}\sum(1/\sqrt{R_i})$ and also $Q_{tot}=\sqrt{H/R_{eq}}$, equating the two gives the parallel-airway combination law: $$\dfrac{1}{\sqrt{R_{eq}}} = \sum_{i=1}^{4}\dfrac{1}{\sqrt{R_i}} = \dfrac{1}{\sqrt{2.627}}+\dfrac{1}{\sqrt{0.151}}+\dfrac{1}{\sqrt{0.349}}+\dfrac{1}{\sqrt{0.397}} = 0.6171+2.5740+1.6931+1.5872 = 6.4714$$ $$\boxed{R_{eq} = \dfrac{1}{6.4714^2} = 0.02389\ \text{N}\cdot\text{s}^2/\text{m}^8}$$
2. Part 3.2 — head loss of the parallel combination. Applying Atkinson's equation to the equivalent resistance found in Step 1, at the known total quantity: $$H_l = R_{eq}\,Q_{tot}^2 = (0.02389)(47.19)^2 = \boxed{53.19\ \text{Pa}}$$
3. Part 3.3 — quantity in each airway. Every airway sees the same H_l from Step 2, so $Q_i=\sqrt{H_l/R_i}$ for each:

   Individual airway flows

   Airway	R (N·s2/m8)	Qi = √(Hl/Ri) (m3/s)
   1	2.627	4.50
   2	0.151	18.77
   3	0.349	12.35
   4	0.397	11.58

   $$\boxed{Q_1=4.50,\ Q_2=18.77,\ Q_3=12.35,\ Q_4=11.58\ \ \text{m}^3/\text{s}}$$ The lowest-resistance airway (2) carries by far the largest share — flow divides in inverse proportion to $\sqrt{R_i}$, not to $R_i$ directly, so a modest resistance difference produces a large flow-split difference.
4. Part 3.4 — sum of the branch flows. $$\sum Q_i = 4.50+18.77+12.35+11.58 = \boxed{47.19\ \text{m}^3/\text{s}}$$ This reproduces the given total exactly, confirming the equivalent-resistance and head-loss calculations of Steps 1–2 are internally consistent.