Question 1 of 6: Question 1 (Part A, Question A1): Hydraulic Jump on a Spillway and its Froude-Scaled Model

Question 1 (Part A, Question A1): Hydraulic Jump on a Spillway and its Froude-Scaled Model (20 marks)

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.

Find. The upstream Froude number, the model discharge that reproduces it, the sequent (post-jump) depth in both channels, and the downstream Froude number in both.

Approach. Free-surface flow with a jump is governed by gravity and inertia, so the model must be a Froude model; get the prototype depth from continuity, form the Froude number, impose Froude equality to size the model, and close each channel with the Belanger sequent-depth relation.

1. Part (a) — recover the approach depth from continuity. For a rectangular channel the discharge is Q = b y V, so $$y_1 = \frac{Q_p}{b_p V_{1p}} = \frac{3548}{(100)(10)} = 3.548\ \text{m}$$ The spillway therefore carries a sheet of water about 3.55 m deep moving at 10 m/s.
2. Part (a) — form the upstream Froude number. With $Fr = V/\sqrt{g y}$ the depth just computed gives $$Fr_1 = \frac{V_{1p}}{\sqrt{g\,y_1}} = \frac{10}{\sqrt{(9.81)(3.548)}} = \frac{10}{5.900} = \boxed{1.695}$$ The value exceeds unity, which is exactly the condition for a jump to be possible: the approach flow is supercritical, and it is only weakly so, which places this jump in the undular to weak class.
3. Part (b) — impose Froude similarity to fix the model depth. Dynamic similarity of a gravity-driven free surface requires $Fr_m = Fr_p$. Because the pump fixes $V_{1m} = 1$ m/s, the model depth follows from $$y_{1m} = \frac{V_{1m}^{2}}{g\,Fr_1^{2}} = \frac{1^{2}}{(9.81)(2.8731)} = 0.03548\ \text{m}$$ that is 35.5 mm. Comparing this with the prototype depth gives the length scale $\lambda = y_{1m}/y_{1p} = 0.01000 \approx 1/100$, which is precisely the width ratio 1 m : 100 m already built into the apparatus. The rig is therefore a geometrically consistent 1:100 Froude model, and the speed ratio $V_m/V_p = \sqrt{\lambda} = 0.1$ is automatically satisfied by the pump.
4. Part (b) — the laboratory discharge. Applying continuity to the model channel, $$Q_m = b_m\,y_{1m}\,V_{1m} = (1)(0.03548)(1) = \boxed{0.03548\ \text{m}^{3}\text{/s}}$$ or about 35.5 L/s — a very manageable laboratory flow. The same number follows from the scaling law $Q_m = Q_p\lambda^{5/2} = 3548 \times 10^{-5}$, which is a useful independent check on the arithmetic.
5. Part (c) — sequent depth from the momentum equation. Applying the momentum equation across the jump (pressure forces plus momentum flux, with bed friction over the short jump length neglected) yields the Belanger equation $$\frac{y_2}{y_1} = \frac{1}{2}\left(\sqrt{1 + 8\,Fr_1^{2}} - 1\right) = \frac{1}{2}\left(\sqrt{1 + 8(2.8731)} - 1\right) = 1.9487$$ The depth ratio is a function of $Fr_1$ alone, so it is identical in model and prototype — which is the whole point of building a Froude model.
6. Part (c) — apply the ratio in both channels. Multiplying each approach depth by 1.9487 gives $$\begin{aligned} y_{2p} &= (3.548)(1.9487) = \boxed{6.914\ \text{m}} \\ y_{2m} &= (0.03548)(1.9487) = \boxed{0.06914\ \text{m}} \end{aligned}$$ The spillway water level rises from 3.55 m to about 6.91 m, and the laboratory level from 35.5 mm to 69.1 mm. The model depth is again exactly one hundredth of the prototype, confirming the scaling.
7. Part (d) — downstream Froude number. Continuity through the jump gives $V_2 = V_1 y_1 / y_2$, and substituting this into the definition of the Froude number produces a relation that involves only the depth ratio: $$Fr_2 = \frac{V_2}{\sqrt{g y_2}} = Fr_1\left(\frac{y_1}{y_2}\right)^{3/2} = (1.695)\left(\frac{1}{1.9487}\right)^{3/2} = \boxed{0.623}$$ Because the expression contains no length or velocity scale, the answer is the same in the laboratory and on the spillway: $Fr_{2m} = Fr_{2p} = 0.623$. The corresponding speeds are $V_{2p} = 5.132$ m/s and $V_{2m} = 0.5132$ m/s, again in the ratio $\sqrt{\lambda} = 0.1$. The value is below unity, confirming that the flow leaves the jump subcritical as it must.
8. Closing check — how much energy the jump destroys. The specific-energy loss across a jump is $\Delta E = (y_2 - y_1)^{3}/(4 y_1 y_2)$, which for the spillway gives $\Delta E = 0.389$ m of head, or $\rho g Q \Delta E = 13.5$ MW dissipated in the stilling basin. That is the quantity the model is ultimately built to measure, and it scales as $\lambda^{7/2}$, so the laboratory jump dissipates only about 1.35 W.

Final Results.