Question 3 of 4: Hydraulic Jump and Bump in a Wide Channel

Problem 3: Hydraulic Jump and Bump in a Wide Channel (25 points)

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 wide (per-unit-width) horizontal channel carries supercritical flow (1) that undergoes a hydraulic jump to subcritical flow (2), then rises over a smooth 10-cm bump, passing through the critical depth exactly at the crest (3), before the bed drops back to its original level downstream (4). No losses occur except at the jump.

Find. (a) $V_4$; (b) $y_4$; (c) $V_1$; (d) $y_1$.

Approach. Critical flow at the crest fixes the unit discharge $q$ and specific energy $E_3=1.5y_3$. Since sections 2 and 4 sit at the same (original) bed elevation, energy conservation over the loss-free bump gives $E_2=E_4=E_3+h$; this cubic in $y$ has one subcritical root ($y_2$) and one supercritical root ($y_4$). The hydraulic jump relation then links $y_1$ (supercritical, same $q$) to $y_2$.

1. Unit discharge and specific energy at the crest. Critical flow means $V_3=\sqrt{gy_3}$, so $$q = V_3 y_3 = \sqrt{(9.81)(0.30)}\,(0.30) = \boxed{q\approx0.5147\ \text{m}^2/\text{s}},\qquad E_3=1.5y_3=\boxed{E_3=0.450\ \text{m}}.$$
2. Specific energy at 2 and 4 (bed level 0, crest at $+h$). With no losses between 2, 3 and 4, the total head is constant, so $E_2=E_4=E_3+h=0.450+0.10=\boxed{0.550\ \text{m}}$.
3. (a)–(b) Solve $y+q^2/(2gy^2)=0.550$ for its two positive roots. This cubic ($y^3-0.550y^2+0.01350=0$) has a subcritical root $y_2\approx0.4949$ m and a supercritical root $$\boxed{y_4\approx0.1950\ \text{m}} \;\Rightarrow\; V_4=\frac{q}{y_4}=\frac{0.5147}{0.1950}=\boxed{V_4\approx2.639\ \text{m/s}}.$$ (the flow chokes at the critical crest and continues accelerating on the downhill side, exactly as flow accelerates past a sonic throat.)
4. Subcritical depth just upstream of the bump. The other root, $y_2\approx0.4949$ m, gives $V_2=q/y_2=1.040\ \text{m/s}$ ($Fr_2=0.47<1$, confirming subcritical, consistent with sitting downstream of the jump).
5. (c)–(d) Hydraulic jump from 1 to 2. The jump relation $y_2/y_1=\tfrac12\!\left(-1+\sqrt{1+8Fr_1^2}\right)$ with $Fr_1^2=q^2/(gy_1^3)$ and the known $y_2=0.4949$ m solves (bisection) to $$\boxed{y_1\approx0.1653\ \text{m}} \;\Rightarrow\; V_1=\frac{q}{y_1}=\frac{0.5147}{0.1653}=\boxed{V_1\approx3.114\ \text{m/s}}\ \ (Fr_1\approx2.45).$$ The upstream flow is indeed supercritical, consistent with a hydraulic jump occurring there.