Question 2 of 6: Mine Ventilation — Kirchhoff's Laws, Fan Heating, Parallel Airways and the Mine Characteristic Curve

Question 2: Mine Ventilation — Kirchhoff's Laws, Fan Heating, Parallel Airways and the Mine Characteristic Curve (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.

2.1.1 — Kirchhoff's laws and Atkinson's equation

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 — the same conservation-of-mass statement used to split the known total 47.19 m³/s among the four parallel airways in Part 2.1.3 below. Kirchhoff's second (mesh/loop) 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 law used to balance a branched, multi-path network with more than one route between two points — and it is exactly what forces every one of the four parallel airways in Part 2.1.3 to see the identical head loss H, since all four connect the same two junctions. Atkinson's equation, $H = RQ^2$, gives the frictional head loss of a single airway from its resistance $R = kPL/A^3$ (friction factor k, rubbing perimeter P, length L and cross-sectional area A) and the quantity Q flowing through it.

In mine ventilation design these three relationships are used together, iteratively: the mine plan is idealised as a network of nodes (junctions/airway ends) and branches (airways, each assigned a resistance from Atkinson's equation using its planned dimensions); Kirchhoff's laws are then solved simultaneously across the whole network (by hand for a simple network, or by a Hardy-Cross-type iterative solver such as VnetPC or Ventsim for a real branched mine) to find the airflow distribution and total system resistance the main fan(s) must overcome, which in turn sets the fan duty (head and quantity) needed to deliver the required minimum air quantity to every working place.

2.1.2 — Large mine-wide ventilation fans and northern air heating

Large mine-wide (main) fans are axial-flow or centrifugal machines, typically several metres in diameter and driven by motors from a few hundred kW up to several MW, installed on surface at the collar of an intake or exhaust shaft/raise (or, less commonly, underground on a booster duty). Their performance characteristics are described by a fan curve — head versus quantity at a given fan speed and blade pitch, falling from a maximum static head at zero flow to zero head at free-delivery flow — together with efficiency and power curves; the mine's own system (characteristic) curve, which rises with the square of quantity per Atkinson's equation, is plotted on the same axes, and their intersection is the fan's actual operating point (Part 2.2 develops this system curve directly). Axial fans typically offer higher efficiency at lower head and the ability to reverse airflow (important for emergency/fire ventilation reversal); centrifugal fans handle higher heads more readily. Many large main fans use variable pitch (axial) or variable inlet vanes/variable speed drives to shift the fan curve and match a changing mine system resistance as workings develop, without needing a full impeller change.

In a typical northern Canadian operation, outside air drawn in through the main intake fan in winter can be well below 0°C, which would freeze water lines, shotcrete and shaft/collar structures and create serious icing hazards at the collar and in the upper workings if admitted untreated. The standard solution is a direct-fired or indirect (glycol-loop) air heater installed at (or immediately after) the main intake fan, burning propane, natural gas or diesel (direct-fired) or circulating a heated glycol loop through a coil in the airstream (indirect, avoiding combustion products in the intake air), sized to raise the full fan quantity from the coldest design outside temperature to a target of a few degrees above 0°C before the air enters the shaft/decline and mine workings.

2.1.3 — Four parallel mine airways

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

Find. The equivalent resistance R_eq (2.1.3.1), the common head loss H (2.1.3.2), the individual branch quantities Q1–Q4 (2.1.3.3), and their sum (2.1.3.4).

Approach. Because all four airways share the same two end nodes, each carries the same head loss H, 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 2.1.3.1 — 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 square-root (not reciprocal-sum) combination law for airways in parallel: $$\dfrac{1}{\sqrt{R_{eq}}} = \sum_{i=1}^{4}\dfrac{1}{\sqrt{R_i}} = \dfrac{1}{\sqrt{2.627}}+\dfrac{1}{\sqrt{0.151}}+\dfrac{1}{\sqrt{0.349}}+\dfrac{1}{\sqrt{0.397}} = 0.6171+2.5740+1.6931+1.5872 = 6.4714$$ $$\boxed{R_{eq} = \dfrac{1}{6.4714^2} = 0.02389\ \text{N}\cdot\text{s}^2/\text{m}^8}$$
2. Part 2.1.3.2 — 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 2.1.3.3 — 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 2.1.3.4 — 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 quantity exactly, confirming the equivalent-resistance and head-loss calculations of Steps 1–2 are internally consistent.

2.2 — Mine characteristic curve

Given. At $Q_1 = 190\ \text{m}^3/\text{s}$: static head $H_{s1} = 500\ \text{Pa}$, total head $H_{t1} = 750\ \text{Pa}$.

Find. $H_s$ and $H_t$ at $Q = 375\ \text{m}^3/\text{s}$ (2.2.1), the plotted characteristic curve (2.2.2), and $H_s$/$H_t$ at $Q = 285\ \text{m}^3/\text{s}$ read from that curve (2.2.3).

Approach. The mine's airway network is a fixed, unchanged system between one flow condition and the next, so its resistance $R$ is constant and both heads scale with the mine characteristic law established in Part 2.1.1, $H = RQ^2$ — i.e. $H_2 = H_1 (Q_2/Q_1)^2$ — which is plotted as a parabola through the given point for 2.2.2 and read back off the same curve for 2.2.3.

1. Calibrate the resistance ratio from the given point. Since $R$ is unchanged, $H_2/H_1 = (Q_2/Q_1)^2$; this ratio is used directly rather than solving for R and Hs/Ht separately.
2. 2.2.1 — heads at 375 m³/s. $(Q_2/Q_1)^2 = (375/190)^2 = 3.895$. $$H_{s} = 500 \times 3.895 = 1947.7\ \text{Pa}, \qquad \boxed{H_{t} = 750 \times 3.895 = 2921.6\ \text{Pa}}$$
3. 2.2.2 — the curve. Plotting $H = RQ^2$ through the calibration point over $0 \le Q \le 400\ \text{m}^3/\text{s}$ and $0 \le H \le 3000\ \text{Pa}$ gives the two parabolas in Fig. 2.2.2 (static head lower, total head upper by the constant velocity-head increment).
4. 2.2.3 — heads at 285 m³/s read from the curve. $(285/190)^2 = 2.25$. $$H_{s} = 500 \times 2.25 = 1125.0\ \text{Pa}, \qquad \boxed{H_{t} = 750 \times 2.25 = 1687.5\ \text{Pa}}$$ — consistent with the graph, since both the plotted curve and the direct calculation come from the same $H=RQ^2$ law.