Bump followed by hydraulic jump, find v1, v2, y4

Given: horizontal channel, $y_1 = 1$ m (upstream, subcritical), $y_2 = 0.3$ m (supercritical, after the bump). The jump takes the flow from section 3 (start of jump, still at $y_2 = 0.3$ m because friction is neglected) to section 4. Friction is neglected, and sections 1 and 2 are at the same bed level.

$v_1$ and $v_2$

Continuity: $v_1y_1 = v_2y_2$, so $v_2 = v_1/0.3$.

Energy (no loss between 1 and 2):

$$y_1 + \frac{v_1^2}{2g} = y_2 + \frac{v_2^2}{2g}$$ $$1 - 0.3 = \frac{v_1^2}{2g}\left[\left(\frac{1}{0.3}\right)^2 - 1\right] \;\Rightarrow\; v_1^2 = \frac{2(9.81)(0.7)}{10.111} = 1.358$$ $$v_1 = 1.165\ \text{m/s},\qquad v_2 = 3.885\ \text{m/s}$$

$y_4$ (depth after the jump)

The supercritical depth at the jump toe is $y_3 = y_2 = 0.3$ m (the same flow, no losses before the jump):

$$Fr_3 = \frac{3.885}{\sqrt{9.81 \times 0.3}} = 2.265$$ $$y_4 = \frac{y_3}{2}\left(\sqrt{1 + 8Fr_3^2} - 1\right) = 0.15\left(\sqrt{1 + 41.04} - 1\right) = 0.822\ \text{m}$$

Check by specific force (per unit width): $\dfrac{q^2}{g\,y_3} + \dfrac{y_3^2}{2} = \dfrac{1.358}{9.81 \times 0.3} + 0.045 = 0.506\ \text{m}^2$ and $\dfrac{1.358}{9.81 \times 0.822} + \dfrac{0.822^2}{2} = 0.168 + 0.338 = 0.506\ \text{m}^2$. The two are equal.

Answer: $v_1 = 1.17$ m/s; $v_2 = 3.89$ m/s; $y_4 = 0.82$ m.