Minimum hump height for critical depth, 3 m channel

Given data

$b=3$ m, $Q=15$ m³/s, $y_1=2$ m. $q=\dfrac{15}{3}=5$ m²/s, $V_1=2.5$ m/s.

$$E_1=2+\frac{2.5^2}{19.62}=2.3186\ \text{m}$$ $$y_c=\sqrt[3]{\frac{q^2}{g}}=\sqrt[3]{\frac{25}{9.81}}=1.3659\ \text{m},\qquad E_{min}=1.5y_c=2.0489\ \text{m}$$

(a) Minimum hump height for critical depth over the hump

$$\Delta z_{min}=E_1-E_{min}=2.3186-2.0489=0.270\ \text{m}$$

(b) Hump height for upstream depth 2.5 m

The upstream depth is raised, so the hump causes critical flow. Upstream energy:

$$E_1'=2.5+\frac{(5/2.5)^2}{19.62}=2.5+\frac{4}{19.62}=2.7039\ \text{m}$$ $$\Delta z=E_1'-E_{min}=2.7039-2.0489=0.655\ \text{m}$$

(c) Hump of 0.20 m

Since $0.20<\Delta z_{min}=0.270$ m, the flow is not choked: upstream depth remains 2.0 m. Depth over the hump (subcritical):

$$E_2=E_1-0.20=2.1186\ \text{m},\qquad y_2+\frac{25}{19.62\,y_2^2}=y_2+\frac{1.2742}{y_2^2}=2.1186$$ $$y_2=1.651\ \text{m}$$

Answer: (a) minimum hump height $=0.270$ m; (b) hump height $=0.655$ m; (c) upstream depth 2.00 m, depth over the 0.2 m hump $=1.65$ m.