Question 4 of 13: Head Loss in a Reservoir-Fed Pipe

Question 4 — Head Loss in a Reservoir-Fed Pipe (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 total head loss between the reservoir surface and the pipe exit.

Approach. Write the general energy equation between the reservoir free surface (large area, V₁≈0, p=atm.) and the open pipe exit (p=atm., V₂=exit velocity), solving for h_L as the only unknown.

1. Exit velocity from continuity. $$A = \frac{{\pi}}{{4}}(0.050)^2 = 1.9635\times10^{{-3}}\text{{ m}}^2,\quad V_2=\frac{{Q}}{{A}}=\frac{{0.006}}{{1.9635\times10^{{-3}}}}=3.056\text{{ m/s}}$$
2. Energy equation, reservoir surface (1) to pipe exit (2). $$\frac{{p_1}}{{\rho g}}+z_1+\frac{{V_1^2}}{{2g}} = \frac{{p_2}}{{\rho g}}+z_2+\frac{{V_2^2}}{{2g}}+h_L$$ With p₁=p₂=0 (gauge), V₁≈0, z₁=9 m, z₂=4 m: $$h_L = (z_1-z_2)-\frac{{V_2^2}}{{2g}} = 5-\frac{{3.056^2}}{{2\times9.81}} = 5-0.476$$ $$h_L = \boxed{{4.52\text{{ m}}}}$$