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Hydraulic jump definition and downstream Froude number

Short answer

A hydraulic jump is the turbulent transition from supercritical to subcritical flow. Energy principle fails because dissipation in the roller is unknown; momentum is used instead. For Fr1=3.5 and y1=1.2 m, y2=5.37 m and Fr2=0.37.

What is hydraulic jump? Why is energy principle not applied for the analysis of the jump? Water flows in a 5 m wide rectangular channel at Froude number 3.5; the depth of flow is 1.2 m. If water undergoes a hydraulic jump, what is the Froude number downstream of jump?

Hydraulic jump

A hydraulic jump is the sudden, turbulent rise of the water surface that occurs when a supercritical flow (Fr>1Fr > 1) changes into subcritical flow (Fr<1Fr < 1) in an open channel. It has a violent roller, strong mixing and large energy loss.

Why the energy principle is not used

The jump contains intense turbulence and eddies, and the energy dissipated (converted to heat) is unknown and cannot be written as a simple term. Hence the energy equation cannot be solved for the unknown depth. Instead the momentum equation is used, because the forces on the short control volume are known (the hydrostatic forces at the ends), while friction on the short, horizontal bed is negligible.

Numerical part

Given: b=5b = 5 m, Fr1=3.5Fr_1 = 3.5, y1=1.2y_1 = 1.2 m.

Sequent depth

y2=y12(1+8Fr12−1)=0.6(1+8(12.25)−1)=5.37 my_2 = \frac{y_1}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right) = 0.6\left(\sqrt{1 + 8(12.25)} - 1\right) = 5.37\ \text{m}

Velocities

V1=Fr1g y1=3.59.81×1.2=12.01 m/sV_1 = Fr_1\sqrt{g\,y_1} = 3.5\sqrt{9.81 \times 1.2} = 12.01\ \text{m/s} V2=V1y1y2=12.01×1.25.37=2.68 m/sV_2 = V_1\frac{y_1}{y_2} = 12.01 \times \frac{1.2}{5.37} = 2.68\ \text{m/s}

Froude number after the jump

Fr2=V2g y2=2.689.81×5.37=0.37Fr_2 = \frac{V_2}{\sqrt{g\,y_2}} = \frac{2.68}{\sqrt{9.81 \times 5.37}} = 0.37

Answer: Fr2≈0.37Fr_2 \approx 0.37 (subcritical; y2=5.37y_2 = 5.37 m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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