Critical flow conditions and 3 m to 2 m contraction

Given data

$b_1=3$ m, $Q=3$ m³/s, $y_1=1$ m. Contracted width $b_2=2$ m. $q_1=1$ m²/s, $q_2=1.5$ m²/s.

$$E_1=1+\frac{1^2}{19.62}=1.0510\ \text{m},\qquad y_{c2}=\sqrt[3]{\frac{1.5^2}{9.81}}=0.6121\ \text{m},\quad E_{min2}=0.9182\ \text{m}$$

(a) Bed raised by 10 cm: depth in the contracted section

$$E_2=E_1-0.10=0.9510\ \text{m}>E_{min2}$$

so the flow is not choked (subcritical):

$$y_2+\frac{1.5^2}{19.62y_2^2}=y_2+\frac{0.11468}{y_2^2}=0.9510\;\Rightarrow\;y_2=0.744\ \text{m}$$

(b) Maximum rise without affecting the upstream depth

$$\Delta z_{max}=E_1-E_{min2}=1.0510-0.9182=0.1328\ \text{m}\approx13.3\ \text{cm}$$

(c) Rise of 30 cm ($>\Delta z_{max}$)

The flow is choked. Critical flow occurs in the contraction and the upstream depth rises to satisfy

$$E_1'=E_{min2}+0.30=0.9182+0.30=1.2182\ \text{m}$$ $$y_1'+\frac{1}{19.62y_1'^2}=1.2182\;\Rightarrow\;y_1'=1.182\ \text{m}$$

Water surface elevations (above the original upstream bed):

So the upstream water surface rises by 0.182 m (to 1.182 m), and the water surface at the contracted section is 0.912 m (critical depth 0.612 m), i.e. $1.182-0.912=0.270$ m below the new upstream surface.

Answer: (a) $y_2=0.74$ m; (b) $\Delta z_{max}=0.133$ m; (c) the upstream depth rises from 1.00 m to 1.18 m, and critical flow ($y=0.61$ m) occurs in the contraction, where the surface is at 0.912 m (about 0.07 m higher than in the 10 cm case).