Pressure Drop in a Sloping Mine Ventilation Tunnel: 04-BS-7, December 2014, Q7

Question 7 — Pressure Drop in a Sloping Mine Ventilation Tunnel (5 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.

Quantity	Value
Tunnel width × height	2.6 m × 3.2 m
Length, L	1.5 km
Roughness (irregular profile), ε	0.10 m
Mean velocity, V	8 m/s
Air at 15°C: ρ	1.21 kg/m³, μ = 1.8×10⁻⁵ Ns/m²

Find. The pressure drop Δp over the tunnel length.

Fig. Q7 — rectangular unlined tunnel cross-section (dashed = nominal profile, solid wavy = actual irregular rock surface).

Approach. Replace the rectangular duct with its hydraulic diameter; compute Reynolds number and relative roughness; solve the Colebrook–White equation (the same relation the attached Moody chart plots graphically) for the friction factor; apply the Darcy–Weisbach relation for head loss and convert to a pressure drop using air density.

Hydraulic diameter. $$D_e = \frac{4A}{P} = \frac{4(2.6\times3.2)}{2(2.6+3.2)} = \frac{33.28}{11.6} = \boxed{2.869\text{ m}}$$
Reynolds number and relative roughness. ν_air = μ/ρ = 1.8×10⁻⁵/1.21 = 1.488×10⁻⁵ m²/s: $$Re = \frac{D_e V}{\nu} = \frac{2.869\times8}{1.488\times10^{-5}} = \boxed{1.543\times10^{6}}, \qquad \frac{e}{D}=\frac{0.10}{2.869}=0.0349$$
Friction factor (Colebrook–White / Moody chart). At this Re and e/D, the fully-rough regime dominates: $$\frac{1}{\sqrt f} = -2\log_{10}\!\left(\frac{e/D}{3.7}+\frac{2.51}{Re\sqrt f}\right) \;\Rightarrow\; \boxed{f = 0.0609}$$
Head loss and pressure drop. $$h_L = f\left(\frac{L}{D_e}\right)\!\left(\frac{V^2}{2g}\right) = 0.0609\times\frac{1500}{2.869}\times\frac{8^2}{19.62} = 103.9\text{ m (of air column)}$$ $$\Delta p = \rho_{air}\, g\, h_L = 1.21\times 9.81\times 103.9 = \boxed{1.23\text{ kPa}}$$

Quantity	Result
Hydraulic diameter, D_e	2.87 m
Reynolds number, Re	1.54×10⁶
Friction factor, f	0.0609
Pressure drop, Δp	1.23 kPa

Topic: Frictional pressure loss in a non-circular duct, via the Moody chart.

Key relations: D_e = 4A/P; Darcy–Weisbach h_L = f(L/D)(V²/2g); Colebrook–White implicit friction-factor equation (what the Moody chart plots).

Why this works: substituting the hydraulic diameter for the true (non-circular) shape lets the standard circular-pipe correlations apply, accurately for turbulent flow; at e/D≈0.035 this duct sits deep in the fully-rough zone, where f becomes essentially independent of Re — a useful sanity check against the flattened right-hand region of the chart.

Common pitfall: using the tunnel's width or height directly as "D" instead of the hydraulic diameter, or forgetting the roughness here comes from the STATED irregular-profile deviation (10 cm), not a typical pipe-wall texture value.

Source: White, Fluid Mechanics, 8th ed., §6.4–6.5 (Moody chart, Colebrook equation, non-circular ducts).

Question 7 of 13: Pressure Drop in a Sloping Mine Ventilation Tunnel

Question 7 — Pressure Drop in a Sloping Mine Ventilation Tunnel (5 marks)