Hydraulic Jump: 22-Agric-A4 Fluid Flow, December 2013, Q1

Question 1a: Hydraulic Jump (choose 1a or 1b — equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A hydraulic jump in a wide rectangular channel raises the depth from $y_1 = 0.40\ \text{m}$ upstream to $y_2 = 1.40\ \text{m}$ downstream.

Quantity	Value
Upstream depth, $y_1$	0.40 m
Downstream depth, $y_2$	1.40 m
Channel	wide (unit width, $q = Vy$)

Find. The upstream velocity $V_1$, the downstream velocity $V_2$, the critical depth $y_c$, and the percent of the upstream specific energy dissipated in the jump.

Figure 1a — Hydraulic jump in a wide channel: supercritical approach flow ($y_1$, $V_1$) jumps to a subcritical, deeper downstream flow ($y_2$, $V_2$) with energy dissipated in the turbulent roller.

Approach. Use the wide-channel hydraulic-jump (conjugate-depth) relation to back out the upstream Froude number from the given depth ratio, get $V_1$ from $Fr_1$, get $V_2$ from continuity ($q=V_1y_1=V_2y_2$), then evaluate the critical depth and the specific-energy loss across the jump.

Upstream Froude number from the conjugate-depth relation. For a rectangular jump, $$\frac{y_2}{y_1} = \tfrac12\!\left(\sqrt{1+8Fr_1^2}-1\right).$$ With $y_2/y_1 = 1.40/0.40 = 3.50$, solving for $Fr_1$: $$\sqrt{1+8Fr_1^2} = 2(3.50)+1 = 8.00 \;\Rightarrow\; Fr_1^2 = \frac{8.00^2-1}{8} = 7.875 \;\Rightarrow\; \boxed{Fr_1 = 2.81}.$$
Upstream velocity $V_1$. $$V_1 = Fr_1\sqrt{g y_1} = 2.81\sqrt{(9.81)(0.40)} = 2.81(1.981) = \boxed{V_1 \approx 5.56\ \text{m/s}}.$$
Downstream velocity $V_2$ from continuity. Unit discharge $q = V_1y_1 = (5.56)(0.40) = 2.22\ \text{m}^2/\text{s}$, so $$V_2 = \frac{q}{y_2} = \frac{2.22}{1.40} = \boxed{V_2 \approx 1.59\ \text{m/s}}.$$
Critical depth. For a wide channel, $y_c = (q^2/g)^{1/3}$: $$y_c = \left(\frac{2.22^2}{9.81}\right)^{1/3} = (0.5028)^{1/3} = \boxed{y_c \approx 0.80\ \text{m}}.$$ Since $y_1 < y_c < y_2$, the jump indeed carries the flow from supercritical to subcritical, as required.
Percent dissipation. Specific energies $E = y + V^2/2g$ give $$E_1 = 0.40+\frac{5.56^2}{19.62}=1.975\ \text{m}, \qquad E_2 = 1.40+\frac{1.59^2}{19.62}=1.529\ \text{m},$$ $$\Delta E = E_1-E_2 = 0.446\ \text{m} \;\Rightarrow\; \boxed{\%\ \text{dissipated} = \frac{\Delta E}{E_1}\times100 \approx 22.6\%}.$$

Quantity	Result
(a) $V_1$	5.56 m/s
(b) $V_2$	1.59 m/s
(c) Critical depth $y_c$	0.80 m
(d) Energy dissipated	0.446 m (≈ 22.6% of $E_1$)

Concept & Theory

Topic. The hydraulic jump — a standing, dissipative transition from supercritical to subcritical open-channel flow.

Key relations. Conjugate-depth (Bélanger) equation $y_2/y_1=\tfrac12(\sqrt{1+8Fr_1^2}-1)$; Froude number $Fr=V/\sqrt{gy}$; continuity $q=Vy$; critical depth $y_c=(q^2/g)^{1/3}$; specific energy $E=y+V^2/2g$.

Why this works. A hydraulic jump conserves momentum (not energy) across the short, violently turbulent roller, which is exactly what the conjugate-depth relation encodes. Because the two depths are both given here, the momentum relation alone pins $Fr_1$ without needing the discharge in advance; continuity then supplies $V_2$ and the unit discharge needed for $y_c$.

Common pitfall. Applying the energy equation ($y_1+V_1^2/2g=y_2+V_2^2/2g$) across the jump instead of momentum — a jump always loses energy, so the energy equation is exactly the tool used afterward to find how much was lost, not to relate $y_1$ and $y_2$ in the first place.

Source. White, Fluid Mechanics, Ch. 10 (Open-Channel Flow) — Hydraulic Jump section.

Question 1 of 6: Hydraulic Jump

Question 1a: Hydraulic Jump (choose 1a or 1b — equal value)