Verify jump downstream of sluice gate

Given: $b = 5$ m, $Q = 20\ \text{m}^3/\text{s}$ ($q = 4\ \text{m}^2/\text{s}$), gate opening $a = 0.67$ m, jet depth $y_2 = 0.4$ m (so $C_c = 0.4/0.67 = 0.597$), downstream uniform depth $y_t = 2.5$ m, $\alpha = \beta = 1$.

Upstream depth (free flow). Energy equation across the gate:

$$V_2 = \frac{4}{0.4} = 10\ \text{m/s},\quad E = 0.4 + \frac{10^2}{2g} = 5.497\ \text{m}$$ $$y_1 + \frac{(4/y_1)^2}{2g} = 5.497 \;\Rightarrow\; y_1 = 5.47\ \text{m}$$

a) Does a jump occur?

The jet leaving the gate is supercritical:

$$Fr_2 = \frac{10}{\sqrt{9.81 \times 0.4}} = 5.05$$

The downstream uniform flow ($y_t = 2.5$ m, $V = 1.6$ m/s) is subcritical ($Fr = 0.32$). Flow must change from supercritical to subcritical, which is possible only through a hydraulic jump, so a jump occurs.

Its position depends on the sequent depth of the jet:

$$y_2' = \frac{0.4}{2}\left(\sqrt{1 + 8(5.05)^2} - 1\right) = 2.66\ \text{m}$$

Since the tailwater $2.5\ \text{m} < 2.66$ m, the tailwater is slightly too low, so the jump does not form at the vena contracta. It is swept a short distance downstream of the gate. The supercritical depth increases gradually (M3 type) until its sequent depth equals 2.5 m, and the jump then forms. The jet at the gate remains a free jet at 0.4 m, and the discharge equation for free flow applies.

b) Downstream depth increased to 3 m

Now $y_t = 3\ \text{m} > y_2' = 2.66$ m. The tailwater is too high, so the jump moves upstream, reaches the gate, and becomes a submerged (drowned) jump that covers the jet. The percent submergence is

$$\frac{3 - 2.66}{2.66} = 12.7\%$$

The gate is therefore submerged: the flow at the gate is no longer free, and the upstream depth rises. An approximate analysis keeps the jet velocity of 10 m/s and finds the effective pressure depth $y_x$ at the jet section from the momentum equation to the downstream section ($y_t = 3$ m, $V_t = 1.333$ m/s):

$$\frac{\gamma b y_x^2}{2} = \frac{\gamma b y_t^2}{2} + \rho Q(V_t - V_2) = 220\,725 - 173\,333 = 47\,392\ \text{N}$$ $$y_x = 1.39\ \text{m}$$

Energy from upstream to the jet section (with the pressure depth $y_x$):

$$y_1' + \frac{(4/y_1')^2}{2g} = 1.39 + 5.097 = 6.49\ \text{m} \;\Rightarrow\; y_1' = 6.47\ \text{m}$$

Answer: (a) A jump occurs since $Fr_2 = 5.05$ and the downstream flow is subcritical; because 2.5 m < 2.66 m it forms just downstream of the gate. (b) With 3 m tailwater the jump is submerged against the gate, and the upstream depth rises from about 5.47 m to about 6.5 m (approximate analysis).