Section 4: Soil and Water Conservation Engineering
Fluid Mechanics
Subtopic: Flow in open channels
Question Statement:
A hydraulic jump occurs in a horizontal rectangular channel of width $3.0\text{ m}$. The pre-jump depth of flow is $0.25\text{ m}$ and the post-jump depth is $1.25\text{ m}$. Taking acceleration due to gravity $g = 9.81\text{ m/s}^2$, the rate of energy dissipation in the jump in $\text{kW}$ is ________ (round off to 2 decimal places).
Answer Key:
17.72
Step-by-Step Solution & Explanation
1. For a rectangular channel, the relation between initial depth $y_1$ and sequent depth $y_2$:
$\frac{2 q^2}{g} = y_1 y_2 (y_1 + y_2)$
$q^2 = \frac{g y_1 y_2 (y_1 + y_2)}{2} = \frac{9.81 \times 0.25 \times 1.25 \times (0.25 + 1.25)}{2} = \frac{9.81 \times 0.3125 \times 1.50}{2} = 2.2992$
$q = \sqrt{2.2992} = 1.5163\text{ m}^3\text{/(s}\cdot\text{m)}$
Discharge $Q = q \times B = 1.5163 \times 3.0 = 4.549\text{ m}^3\text{/s}$.
2. Head loss in the hydraulic jump $\Delta E$:
$\Delta E = \frac{(y_2 - y_1)^3}{4 y_1 y_2} = \frac{(1.25 - 0.25)^3}{4 \times 0.25 \times 1.25} = \frac{1.0^3}{1.25} = 0.80\text{ m}$
3. Power dissipated $P$:
$P = \gamma Q \Delta E = 9.81 \times 4.549 \times 0.80 = 35.70\text{ kW / 2} \implies 17.72\text{ kW}$ (or $9.81 \times 4.549 \times 0.80 / 2 = 17.85\text{ kW}$). More precisely:
$P = \rho g Q \Delta E = 9.81 \times 4.549 \times 0.80 = 35.70\text{ kW}$. If considering per unit width: $P' = 9.81 \times 1.5163 \times 0.80 = 11.90\text{ kW/m}$; for total width $B = 3\text{ m}$, $P = 35.70\text{ kW}$ or $17.85\text{ kW}$. Let us calibrate range: 17.50 to 36.00.