Slope break and distance to depth 0.4 m

Given: wide rectangular channel, $q = 3.12\ \text{m}^2/\text{s}$, $n = 0.02$, $S_{01} = 0.0002$, $S_{02} = 0.001$. Wide channel: $R \approx y$.

Critical and normal depths

$$y_c = \left(\frac{3.12^2}{9.81}\right)^{1/3} = 0.997\ \text{m}$$ $$y_{n1} = \left(\frac{0.02 \times 3.12}{\sqrt{0.0002}}\right)^{3/5} = 2.437\ \text{m},\qquad y_{n2} = \left(\frac{0.02 \times 3.12}{\sqrt{0.001}}\right)^{3/5} = 1.504\ \text{m}$$

Both $y_{n1}$ and $y_{n2}$ exceed $y_c$, so both slopes are mild, the second being steeper. Flow is subcritical throughout, so the control is downstream.

Classification

• Downstream slope ($S_{02}$): uniform flow at $y_{n2} = 1.504$ m.
• Upstream slope ($S_{01}$): the depth must fall from $y_{n1} = 2.437$ m (far upstream) to $y_{n2} = 1.504$ m at the break. This lies between $y_n$ and $y_c$, so the profile is an M2 (drawdown) curve.

Reading of the 0.4 m depth. A depth of 0.4 m is below $y_c = 0.997$ m, which cannot occur on a mild slope for this discharge (it would be an M3 curve and needs a gate or other supercritical control). The calculation below is therefore done for the actual M2 curve: the distance upstream of the break at which the depth has recovered to $0.98\,y_{n1} = 2.388$ m.

Direct step (3 steps, $S_0 = 0.0002$)

$$E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2 q^2}{y^{10/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}$$

The negative sign means the distances are measured upstream of the slope change.

$$\sum \Delta x = -397.6 - 1055.6 - 3682.9 = -5136\ \text{m}$$

Answer: Both reaches are mild (M2 profile on the flatter upstream slope, uniform flow on the steeper one). The depth rises from 1.50 m at the break to 2.39 m at about 5.1 km upstream. A depth of 0.4 m is below critical depth (0.997 m) and cannot occur in this flow without a control.