Question 4 of 6: Kirchhoff's Laws, Atkinson's Equation and Four Parallel Mine Airways

Question 4: Kirchhoff's Laws, Atkinson's Equation and Four Parallel Mine Airways (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 first (nodal/continuity) law states that at any junction of a ventilation network the air flowing in must equal the air flowing out — applied directly below to split the known total 47.19 m³/s among the four parallel airways. Kirchhoff's second (loop/mesh) law states that around any closed circuit in the network the algebraic sum of head losses (including any fan heads or natural ventilation pressure within the loop) is zero; it is the tool used to balance a branched series–parallel network with more than one path between two points, and is not needed for this purely-parallel four-airway problem, which is closed with the nodal law alone. Atkinson's equation, $H = RQ^2$, gives the frictional head loss of a single airway from its resistance $R = kPL/A^3$ (friction factor, rubbing perimeter, length and cross-sectional area) and quantity Q; because all four airways here connect the SAME two points (a single junction to a single junction), Kirchhoff's second law forces every airway to carry the identical head loss H, which is exactly the physical fact the parallel combination rule below is built on. Ventilation engineers use these three relationships together, iteratively (in hand calculation, or via a Hardy-Cross-type network solver such as VnetPC/Ventsim for a real branched mine circuit), to size fan duty and confirm that every working place receives its required minimum air quantity.

Given.

Total quantity through the parallel network: $Q_{tot} = 47.19\ \text{m}^3/\text{s}$ (100,000 cfm).

Find. The equivalent resistance R_eq, the common head loss H, the individual branch quantities Q1–Q4, and their sum.

Approach. Because all four airways share the same two end nodes, each carries the same head loss H (Kirchhoff's second law), so $Q_i = \sqrt{H/R_i}$ for every branch; summing the branches and equating to the known total gives the parallel combination rule $1/\sqrt{R_{eq}} = \sum 1/\sqrt{R_i}$, from which R_eq, then H, then each Q_i follow in turn.

1. Part 4.1 — 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 square-root (not reciprocal-sum) combination law for airways in parallel: $$\dfrac{1}{\sqrt{R_{eq}}} = \sum_{i=1}^{4}\dfrac{1}{\sqrt{R_i}} = \dfrac{1}{\sqrt{2.627}}+\dfrac{1}{\sqrt{0.151}}+\dfrac{1}{\sqrt{0.349}}+\dfrac{1}{\sqrt{0.397}} = 0.6170+2.5734+1.6927+1.5871 = 6.4702$$ $$\boxed{R_{eq} = \dfrac{1}{6.4714^2} = 0.02389\ \text{N}\cdot\text{s}^2/\text{m}^8}$$
2. Part 4.2 — head loss of the parallel combination. Applying Atkinson's equation to the equivalent single-airway resistance found in Step 1 at the known total quantity: $$H = R_{eq}Q_{tot}^2 = (0.02389)(47.19)^2 = \boxed{53.19\ \text{Pa}}$$
3. Part 4.3 — quantity in each airway. Every airway shares the same H found in Step 2, so $Q_i=\sqrt{H/R_i}$ for each:

   Individual airway flows

   Airway	R (N·s2/m8)	Qi=√(H/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, as expected — quantity divides in inverse proportion to $\sqrt{R_i}$, not to $R_i$ directly.
4. Part 4.4 — 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 quantity exactly, confirming the equivalent-resistance calculation of Step 1 and the head-loss calculation of Step 2 are internally consistent.