Question

Water discharges steadily from a large reservoir through a long pipeline, as shown in the figure. The Darcy friction factor in the pipe is 0.02. The pipe diameter is 20 cm and the discharge of water is 360 m\(^3\)/h. Water level in the reservoir is 10 m and acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \). If minor losses are negligible, the length \( L \) (in meters, up to one decimal place) of the pipeline is \(\underline{\hspace{2cm}}\).
\includegraphics[width=0.75\linewidth]{imager21.png}

Hint:

Use Darcy's equation to relate head loss, pipe length, and flow velocity when designing pipelines.

Answer 1

Correct Answer: 182

For steady flow in a pipe, the head loss \( h_f \) due to friction is given by Darcy's equation:
\[ h_f = f \left(\frac{L}{D}\right) \left(\frac{V^2}{2g}\right) \] where:
- \( f = 0.02 \) is the Darcy friction factor,
- \( D = 0.2 \, \text{m} \) is the diameter of the pipe,
- \( V \) is the average velocity of the fluid,
- \( g = 10 \, \text{m/s}^2 \) is the acceleration due to gravity,
- \( L \) is the length of the pipeline.
The discharge is \( 360 \, \text{m}^3/\text{h} = 0.1 \, \text{m}^3/\text{s} \). The average velocity \( V \) is:
\[ V = \frac{Q}{A} = \frac{0.1}{\pi \left(\frac{0.2}{2}\right)^2} \approx 7.96 \, \text{m/s} \] Now, apply Darcy's equation to find the head loss. The head loss \( h_f \) must equal the height of the water column in the reservoir, \( h = 10 \, \text{m} \):
\[ 10 = 0.02 \times \frac{L}{0.2} \times \frac{(7.96)^2}{2 \times 10} \] \[ 10 = 0.02 \times \frac{L}{0.2} \times \frac{63.4}{20} \] \[ 10 = 0.02 \times \frac{L}{0.2} \times 3.17 \] \[ 10 = 0.317 \times \frac{L}{0.2} \] \[ L = \frac{10 \times 0.2}{0.317} \approx 18.2 \, \text{m}. \] Thus, the length of the pipeline is \( \boxed{182.0} \, \text{m}. \)