Contraction with lowered bed, 10 m³/s

Given data

$b_1=3.5$ m, $Q=10$ m³/s, $y_1=1.75$ m; contracted $b_2=2.25$ m, bed lowered by $0.97$ m; no energy loss.

$$q_1=\frac{10}{3.5}=2.8571\ \text{m}^2/\text{s},\quad V_1=\frac{2.8571}{1.75}=1.633\ \text{m/s},\quad Fr_1=\frac{1.633}{\sqrt{9.81\times1.75}}=0.394$$ $$E_1=1.75+\frac{1.633^2}{19.62}=1.8859\ \text{m}$$

The upstream flow is subcritical.

Contracted section

$$q_2=\frac{10}{2.25}=4.4444\ \text{m}^2/\text{s},\quad y_{c2}=\sqrt[3]{\frac{q_2^2}{g}}=1.2628\ \text{m}$$

The bed is lowered, so energy above the new bed increases:

$$E_2=E_1+0.97=2.8559\ \text{m}$$

Subcritical root of $y_2+\dfrac{q_2^2}{2gy_2^2}=y_2+\dfrac{1.0067}{y_2^2}=2.8559$:

$$y_2=2.7198\ \text{m}\quad(>y_{c2},\ \text{subcritical})$$

Water level elevations (datum: upstream bed)

• Upstream: $1.7500$ m
• Contracted section: $-0.97+2.7198=1.7498$ m

Difference in water level $=1.7500-1.7498=0.0002$ m, which is practically zero.

Answer: depth in the contracted section $=2.720$ m; the water surface elevation at the contracted section is the same as upstream (difference ≈ 0.000 m, i.e. less than 1 mm). The increase in velocity head is exactly balanced by the lowering of the bed.