Section 4: Soil and Water Conservation Engineering
Fluid Mechanics
Subtopic: Laminar and turbulent flow in pipes, Darcy–Weisbach and Hazen–Williams equations
Question Statement:
A smooth irrigation pipe of internal diameter $D = 0.20\text{ m}$ and length $L = 500.0\text{ m}$ conveys water at a steady discharge of $Q = 0.050\text{ m}^3/\text{s}$. The Darcy-Weisbach friction factor is $f = 0.024$. Taking acceleration due to gravity $g = 9.81\text{ m/s}^2$ and $\pi = 3.1416$, the head loss due to pipe friction in meters is ________ (round off to two decimal places).
Answer Key:
7.75
Step-by-Step Solution & Explanation
**Method 1: Darcy-Weisbach with Mean Flow Velocity**
1. Cross-sectional area of pipe:
$A = \frac{\pi}{4} D^2 = \frac{3.1416}{4} \times (0.20)^2 = 0.031416\text{ m}^2$
2. Mean flow velocity:
$V = \frac{Q}{A} = \frac{0.050}{0.031416} = 1.59155\text{ m/s}$
3. Velocity head:
$\frac{V^2}{2 g} = \frac{(1.59155)^2}{2 \times 9.81} = \frac{2.53303}{19.62} = 0.12910\text{ m}$
4. Head loss due to friction:
$h_f = \frac{f L V^2}{2 g D} = \frac{0.024 \times 500.0}{0.20} \times 0.12910 = 60.0 \times 0.12910 = 7.746\text{ m} \approx 7.75\text{ m}$
**Method 2: Direct Discharge Formulation**
$h_f = \frac{8 f L Q^2}{\pi^2 g D^5} = \frac{8 \times 0.024 \times 500.0 \times (0.050)^2}{(3.1416)^2 \times 9.81 \times (0.20)^5} = \frac{0.240}{96.820 \times 0.00032} = \frac{0.240}{0.030982} = 7.746\text{ m} \approx 7.75\text{ m}$