Jump upstream of a low dam

Given: $b = 5$ m, $n = 0.02$, $S_0 = 0.04$, uniform depth $y_n = 0.5$ m; a low dam raises the depth to 2 m at the dam.

Discharge

$$A = 2.5\ \text{m}^2,\ P = 6\ \text{m},\ R = 0.4167\ \text{m}$$ $$Q = \frac{1}{n}AR^{2/3}S_0^{1/2} = \frac{1}{0.02}(2.5)(0.4167)^{2/3}(0.2) = 13.95\ \text{m}^3/\text{s}$$ $$q = 2.789\ \text{m}^2/\text{s},\quad V = 5.58\ \text{m/s},\quad Fr = 2.52,\quad y_c = \left(\frac{2.789^2}{9.81}\right)^{1/3} = 0.926\ \text{m}$$

Uniform flow is supercritical ($y_n < y_c$; steep slope). Upstream of the dam the water is deep and subcritical, so a jump is possible.

Does a jump occur?

Sequent depth of 0.5 m:

$$y_2 = \frac{0.5}{2}\left(\sqrt{1 + 8(2.52)^2} - 1\right) = 1.549\ \text{m}$$

The depth just upstream of the dam (2 m) is greater than 1.549 m, so the dam backwater is deeper than the jump needs. A hydraulic jump does occur, forming upstream of the dam where the S1 backwater curve has depth 1.549 m.

Distance upstream of the dam

Direct step method along the S1 curve from $y = 2.0$ m (dam) to 1.549 m (5 steps, $S_0 = 0.04$):

$$L = 2.05 + 2.02 + 1.98 + 1.93 + 1.87 \approx 9.9\ \text{m}$$

The steep bed slope ($S_0 = 0.04$) dominates the friction slope, so the S1 curve is short.

Answer: A jump occurs (0.5 m to 1.55 m). Its downstream end is about 10 m upstream of the dam; the jump itself (length about $6(y_2 - y_1) \approx 6.3$ m) lies just upstream of this point.