Section 4: Soil and Water Conservation Engineering
Specific Energy, Critical Flow & Hydraulic Jump
Subtopic: Specific energy and critical flow
Question Statement:
In the rectangular channel specific energy diagram, for discharge intensity $q = 1.5\text{ m}^3/(\text{s}\cdot\text{m})$, flow occurs at depth $y = 0.5\text{ m}$. Calculate the specific energy $E$ at this section in meters (round off to two decimal places).
Answer Key:
0.96
Step-by-Step Solution & Explanation
$v = \frac{q}{y} = \frac{1.5}{0.5} = 3.0\text{ m/s}$.
Velocity head $\frac{v^2}{2g} = \frac{3.0^2}{2 \times 9.81} = \frac{9}{19.62} = 0.4587\text{ m}$.
Specific energy $E = y + \frac{v^2}{2g} = 0.5 + 0.4587 = 0.9587\text{ m} \approx 0.96\text{ m}$.