Flow over a 0.2 m step rise and drop

Given data

Rectangular channel $b=5$ m, upstream $y_1=5$ m, $V_1=4$ m/s, no losses.

$$q=V_1y_1=4\times5=20\ \text{m}^2/\text{s},\quad Q=100\ \text{m}^3/\text{s}$$ $$E_1=y_1+\frac{V_1^2}{2g}=5+\frac{16}{19.62}=5.8155\ \text{m}$$ $$Fr_1=\frac{4}{\sqrt{9.81\times5}}=0.571\ (\text{subcritical}),\quad y_c=\sqrt[3]{\frac{20^2}{9.81}}=3.442\ \text{m},\quad E_{min}=5.163\ \text{m}$$

(i) Step rise of 0.2 m

Energy equation (datum at the upstream bed): $E_1=\Delta z+E_2$

$$E_2=5.8155-0.2=5.6155\ \text{m}>E_{min}$$

so flow is possible, and since the approach flow is subcritical, the flow stays subcritical:

$$y_2+\frac{20^2}{19.62\,y_2^2}=y_2+\frac{20.387}{y_2^2}=5.6155\;\Rightarrow\;y_2=4.688\ \text{m}$$

Water surface elevation above the upstream bed $=0.2+4.688=4.888$ m. Change in water level $=4.888-5.000=-0.112$ m, i.e. the water surface falls by 0.112 m.

(ii) Step drop of 0.2 m

$$E_2=E_1+0.2=6.0155\ \text{m}$$

Subcritical root: $y_2+\dfrac{20.387}{y_2^2}=6.0155\Rightarrow y_2=5.286\ \text{m}$

Water surface elevation above the upstream bed $=5.286-0.2=5.086$ m; change in water level $=5.086-5=+0.086$ m, i.e. the water surface rises by 0.086 m.

 (i) rise:   y1=5 ____        (ii) drop:  y1=5 ____
                  \___ y2=4.69               ____/ y2=5.29
         step 0.2 |___|                    |___| drop 0.2

Answer: (i) $y_2=4.69$ m, water level falls by 0.11 m; (ii) $y_2=5.29$ m, water level rises by 0.09 m.

(Subcritical flow: the surface drops over a rise and goes up over a drop, the opposite of the bed movement.)