Bump followed by hydraulic jump, find bump height

Given: horizontal channel, $y_1 = 1$ m (upstream, subcritical), $y_3 = 0.3$ m (supercritical, after the bump), then a hydraulic jump to $y_4$. Friction is neglected. Section 1 and section 3 are at the same bed level.

$v_1$ and $v_3$

Continuity: $v_1y_1 = v_3y_3$, so $v_3 = v_1(1/0.3) = 3.333\,v_1$.

Energy (smooth flow over the bump, no loss between 1 and 3):

$$y_1 + \frac{v_1^2}{2g} = y_3 + \frac{v_3^2}{2g}$$ $$1 - 0.3 = \frac{v_1^2}{2g}\left[\left(\frac{1}{0.3}\right)^2 - 1\right] \;\Rightarrow\; v_1^2 = \frac{2(9.81)(0.7)}{10.111} = 1.358$$ $$v_1 = 1.165\ \text{m/s},\qquad v_3 = 3.885\ \text{m/s}$$

$y_4$ (after the jump)

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

Bump height $h$

The flow changes from subcritical to supercritical smoothly over the bump, so it passes through critical depth at the crest (section 2). With $q = v_1y_1 = 1.165\ \text{m}^2/\text{s}$:

$$y_c = \left(\frac{1.165^2}{9.81}\right)^{1/3} = 0.517\ \text{m},\qquad E_c = 1.5\,y_c = 0.776\ \text{m}$$ $$E_1 = 1 + \frac{1.165^2}{2g} = 1.069\ \text{m}$$ $$h = E_1 - E_c = 1.069 - 0.776 = 0.293\ \text{m}$$

Answer: $v_1 = 1.17$ m/s; $v_3 = 3.89$ m/s; $y_4 = 0.82$ m; $h \approx 0.29$ m (critical flow assumed at the crest).