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

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.

Find. The pressure drop along the tunnel.

Approach. Model the rectangular tunnel with its hydraulic (equivalent) diameter, compute the Reynolds number and relative roughness, read the Darcy friction factor off the Moody diagram (or its Colebrook equation), then apply the Darcy–Weisbach head-loss formula converted to a pressure drop.

1. Hydraulic diameter. $$A=(2.6)(3.2)=8.32\ \text{m}^2\qquad P_w=2(2.6+3.2)=11.6\ \text{m}$$ $$D_e=\frac{4A}{P_w}=\boxed{2.869\ \text{m}}$$
2. Reynolds number and relative roughness. $$\nu_{air}=\frac{\mu}{\rho}=\frac{1.8\times10^{-5}}{1.21}=1.488\times10^{-5}\ \text{m}^2/\text{s}$$ $$Re=\frac{D_eV}{\nu}=\frac{(2.869)(8)}{1.488\times10^{-5}}=\boxed{1.543\times10^6}\qquad \frac{\epsilon}{D_e}=\frac{0.10}{2.869}=0.0349$$
3. Friction factor from the Moody diagram. At this Reynolds number and a large relative roughness of 0.035, the flow is fully rough (the curve is flat, independent of Re): $$\boxed{f\approx0.061}$$
4. Head loss and pressure drop. $$h_L=f\frac{L}{D_e}\frac{V^2}{2g}=(0.061)\left(\frac{1500}{2.869}\right)\frac{8^2}{2(9.81)}=103.9\ \text{m (of air)}$$ $$\Delta p=\rho_{air}gh_L=(1.21)(9.81)(103.9)=\boxed{1.23\ \text{kPa}}$$