Question Bank • QB_SWCE_ADV_090 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Fluid Mechanics

Subtopic: Flow in open channels

Question Statement:

A hydraulic jump forms on a horizontal rectangular concrete channel apron of width $B = 3.0\text{ m}$. The initial supercritical flow depth before the jump is $y_1 = 0.40\text{ m}$ and the upstream velocity is $V_1 = 6.0\text{ m/s}$. Taking $g = 9.81\text{ m/s}^2$, the subcritical sequent depth of flow $y_2$ downstream of the jump in meters is ________ (round off to two decimal places). ``` Supercritical Stream Subcritical Flow y1 = 0.40 m .-~~~~~~~~~~~~~~~-._ y2 = ? ═══════════════' Roller Region `═══════════════════════ ──────────────────────────────────────────────────────────── Horizontal Apron Floor ```

Answer Key:

1.53

Step-by-Step Solution & Explanation

**Method 1: Bélanger Sequent Depth Formula via Froude Number** 1. Upstream Froude number $Fr_1$: $Fr_1 = \frac{V_1}{\sqrt{g y_1}} = \frac{6.0}{\sqrt{9.81 \times 0.40}} = \frac{6.0}{\sqrt{3.924}} = \frac{6.0}{1.9809} = 3.0289$ $Fr_1^2 = (3.0289)^2 = 9.1743$ 2. Bélanger equation for sequent depth ratio in rectangular channels: $\frac{y_2}{y_1} = \frac{1}{2} \left( \sqrt{1 + 8 Fr_1^2} - 1 \right)$ $\frac{y_2}{y_1} = \frac{1}{2} \left( \sqrt{1 + 8(9.1743)} - 1 \right) = \frac{1}{2} \left( \sqrt{74.394} - 1 \right) = \frac{1}{2} (8.6252 - 1) = 3.8126$ $y_2 = 0.40 \times 3.8126 = 1.525\text{ m} \approx 1.53\text{ m}$ **Method 2: Momentum / Specific Force Conservation** $M_1 = M_2 \implies \frac{q^2}{g y_1} + \frac{y_1^2}{2} = \frac{q^2}{g y_2} + \frac{y_2^2}{2}$ Discharge per unit width $q = V_1 y_1 = 6.0 \times 0.40 = 2.40\text{ m}^2/\text{s}$. $\frac{(2.40)^2}{9.81 \times 0.40} + \frac{(0.40)^2}{2} = \frac{5.76}{3.924} + 0.08 = 1.4679 + 0.08 = 1.5479\text{ m}^2$ Solving $\frac{5.76}{9.81 y_2} + \frac{y_2^2}{2} = 1.5479$ yields positive root $y_2 = 1.525\text{ m} \approx 1.53\text{ m}$.
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