Section 4: Soil and Water Conservation Engineering
Specific Energy, Critical Flow & Hydraulic Jump
Subtopic: Specific energy and critical flow
Question Statement:
In the specific energy diagram for a triangular flume having side slopes $1:1$ (horizontal:vertical), the discharge is $Q = 1.5\text{ m}^3/\text{s}$. Compute the critical depth $y_c$ in meters for $g = 9.81\text{ m/s}^2$ (round off to two decimal places).
Answer Key:
0.9
Step-by-Step Solution & Explanation
For triangular channel with side slope $m = 1$:
$\frac{Q^2}{g} = \frac{A_c^3}{T_c} = \frac{(m y_c^2)^3}{2 m y_c} = \frac{m^2 y_c^5}{2} = \frac{y_c^5}{2}$.
$y_c^5 = \frac{2 Q^2}{g} = \frac{2 \times (1.5)^2}{9.81} = \frac{4.5}{9.81} = 0.4587$.
$y_c = (0.4587)^{1/5} \approx 0.855\text{ m}$.