Question 8 of 13: Head Loss in an Annular Duct

Question 8: Head Loss in an Annular Duct (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 head loss over the 5 m annulus.

Approach. Replace the true diameter with the hydraulic diameter D_h = 4A/P (which reduces to D_o−D_i for a concentric annulus), then apply the same Colebrook/Darcy–Weisbach procedure as Question 7.

1. Flow area and hydraulic diameter: $$A=\frac{\pi}{4}(D_o^2-D_i^2)=\frac{\pi}{4}(0.025^2-0.018^2)=2.364\times10^{-4}\ \text{m}^2, \qquad D_h=D_o-D_i=0.007\ \text{m}$$
2. Velocity and Reynolds number: $$V=\frac{Q}{A}=\frac{0.0005}{2.364\times10^{-4}}=\boxed{2.12\ \text{m/s}}, \qquad Re=\frac{VD_h}{\nu}=\frac{2.12(0.007)}{1.0\times10^{-6}}=1.48\times10^4$$
3. Relative roughness and friction factor (drawn tubing, ε = 0.0015 mm, very smooth): $$\frac{\epsilon}{D_h}=\frac{0.0000015}{0.007}=2.14\times10^{-4}, \qquad f\approx0.0283\ \text{(Colebrook at this Re, }\epsilon/D_h\text{)}$$
4. Head loss: $$h_L=f\frac{L}{D_h}\frac{V^2}{2g}=0.0283\left(\frac{5}{0.007}\right)\frac{2.12^2}{2(9.81)}=\boxed{4.61\ \text{m}}$$