Mine Ventilation Characteristic Curve: 24-MMP-A2 Underground Mining Methods and Design, May 2013, Q2

Question 2: Mine Ventilation Characteristic Curve (15 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.

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 $Q = 375\ \text{m}^3/\text{s}$ (2.1) and at $Q = 285\ \text{m}^3/\text{s}$ read from the plotted curve (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 $H = RQ^2$ established in 1.2 — i.e. $H_2 = H_1 (Q_2/Q_1)^2$ — which is then plotted as a parabola through the origin for 2.2 and read back off the same curve for 2.3.

Fig. 2.2 — Mine characteristic curves, $H = RQ^2$, calibrated through the given point (190, 500/750); axes span 0–400 m³/s and 0–3000 Pa as specified. Both required reading points (2.1 at 375 m³/s, 2.3 at 285 m³/s) are marked.

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.
2.1 — heads at 375 m³/s. $(Q_2/Q_1)^2 = (375/190)^2 = 3.895$. $$H_{s} = 500 \times 3.895 = 1947.7\ \text{Pa}, \qquad \boxed{H_{t} = 750 \times 3.895 = 2921.6\ \text{Pa}}$$
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 (static head lower, total head upper by the constant velocity-head increment).
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.

Question 2 — final results
Air quantity, Q (m³/s)	Static head, Hs (Pa)	Total head, Ht (Pa)
190 (given)	500	750
375 (2.1)	1947.7	2921.6
285 (2.3)	1125.0	1687.5

Topic. The mine (system) characteristic curve and its quadratic scaling law.

Key relations. $H = RQ^2$ with $R$ constant for a fixed airway network; hence $H_2 = H_1(Q_2/Q_1)^2$ for any two flow states of the same system.

Why this works. Airway resistance depends only on the airways' geometry and friction factor, none of which changes when the flow rate changes, so every point on the mine's own H–Q curve is fixed by that one geometric relationship; only the fan's operating point (which airway resistance the fan is asked to overcome) moves. This is what lets a single calibration point (190 m³/s, 500/750 Pa) determine the entire curve.

Common pitfall. Assuming head scales linearly with quantity (as if $R$ were a simple linear "restrictor"), which underestimates the head needed at higher flow by a wide margin because of the square-law relationship; also confusing the mine's characteristic (rises with Q) with the fan's performance curve (falls with Q) — their intersection, not either curve alone, is the real operating point.

Source. Hartman & Mutmansky, SME Mining Engineering Handbook, 3rd ed., mine ventilation chapter (Atkinson equation, system/fan characteristic curves).

Question 2 of 7: Mine Ventilation Characteristic Curve

Question 2: Mine Ventilation Characteristic Curve (15 marks)