Question 7 of 13: Head Loss in an Outhouse Vent Pipe

Question 7: Head Loss in an Outhouse Vent 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. Total head loss in the pipe, expressed as a differential pressure (Pa).

Approach. Compute the mean velocity and Reynolds number, obtain the Moody friction factor for this hydraulically smooth pipe, sum the major (pipe-friction) and minor (entrance/elbow/exit) losses in metres of the flowing fluid (air), then convert to a pressure using the air density.

1. Velocity and Reynolds number. $$A = \frac{\pi(0.1)^2}{4} = 7.854\times10^{-3}\ \text{m}^2, \qquad V = \frac{Q}{A} = \frac{0.03056}{7.854\times10^{-3}} = 3.89\ \text{m/s}$$ $$Re = \frac{\rho V D}{\mu} = \frac{1.19(3.89)(0.1)}{1.8\times10^{-5}} = \boxed{25{,}720}\ \text{(turbulent)}$$
2. Friction factor (Moody chart / smooth-pipe correlation, ε/D ≈ 0). Using the Blasius smooth-pipe fit (equivalent to reading the ε/D=0 curve of the Moody diagram at this Re): $$f = \frac{0.316}{Re^{0.25}} = \frac{0.316}{25{,}720^{0.25}} = \boxed{0.0250}$$
3. Major (friction) head loss. $$h_{f} = f\frac{L}{D}\frac{V^2}{2g} = 0.0250\left(\frac{3.5}{0.1}\right)\frac{3.89^2}{2(9.81)} = 0.675\ \text{m (air)}$$
4. Minor losses (entrance + elbow + submerged exit). $$\Sigma K = 0.5+0.9+1.0 = 2.4, \qquad h_{minor} = 2.4\frac{V^2}{2g} = 1.852\ \text{m (air)}$$
5. Total head loss and equivalent differential pressure. $$h_L = h_f + h_{minor} = 0.675+1.852 = 2.53\ \text{m (of air)}$$ $$\Delta p = \rho_{air}\,g\,h_L = 1.19(9.81)(2.53) = \boxed{29.5\ \text{Pa}}$$