Question 6.10: A 1.2 m diameter pipeline must discharge 2.672 m³/s when flo......
A 1.2 m diameter pipeline must discharge 2.672 m³/s when flowing full. If the viscosity and pipes are as in Example 6.9, calculate the friction factor, λ, and the required hydraulic gradient.
v = 2.672 × 4 / (π × 1.2²) = 2.363 m /s
Re = VD/v = 2.363 × 1.2/1.005 × 10^{ – 6} = 2.821 \times 10^{ 6}
k/D = 0.00060/1.2 = 0.000500
The friction factor can either be found by solving equation (6.18) by trial and error or by solving equation (6.20) directly. Equation (6.20) is:
\frac{1}{\sqrt{\lambda } } = – 2 \log \left\lgroup\frac{k}{3.7 D} + \frac{2.51}{Re \sqrt{\lambda }} \right\rgroup (6.18)
\lambda = 0.0055 \left[1 + \left\lgroup20000 \frac{k}{D} + \frac{10^{6}}{Re} \right\rgroup^{{1}/{3}} \right]\lambda = 0.0055 \left[1 + \left\lgroup20000 \times 0.000500 + \frac{10^{6}}{2.821 \times 10^{6}} \right\rgroup^{{1}/{3}} \right]
\lambda = 0.0055\left[1 + \left(10.0 + 0.354\right)^{{1}/{3}} \right]
λ = 0.0175
The Darcy equation gives h_{F} = {\lambda LV^{2}}/{2gD} so S_{F} = {\lambda V^{2}}/{2gD} (since S_{F} = {h_{F}}/{L}).
S_{F} = 0.0175 × 2.363²/19.62 × 1.2
S_{F} = 0.0042 or 1 in 241
Solving equation (6.18) by trial and error gives λ = 0.0169, about a 4% difference. Note that the hydraulic gradient obtained above should have been 1 in 250 since this example is effectively Example 6.9 in reverse. Almost the same answer can be obtained from the Moody diagram – see the text above Example 6.9.