Hydraulic jump stilling basin (25 m³/s/m)

Assumptions (not given in the question): $C_d=0.75$ in $q=\tfrac23C_d\sqrt{2g}H^{3/2}$; losses on the spillway face neglected; the tail water depth in the river is the normal depth (wide channel, Manning); basin length $L=5\,y_2$.

Step 1: Head over crest

$$25=2.2147\,H^{3/2}\Rightarrow H=5.03\ \text{m}$$

Step 2: Tail water (Manning, wide channel)

$$y_n=\left(\frac{qn}{\sqrt{S_0}}\right)^{3/5}=\left(\frac{25\times0.028}{\sqrt{0.001}}\right)^{0.6}=6.41\ \text{m}$$

Step 3: Conjugate depth at the toe (floor at bed level)

$E=50+5.03=55.03$ m, so $y_1=0.766$ m, $V_1=32.6$ m/s, $Fr_1=11.90$ and

$$y_2=\frac{y_1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)=12.52\ \text{m}$$

Since $y_n=6.41$ m is less than $y_2$, the jump would be swept away and the floor must be depressed.

Step 4: Depression of floor

Require $y_n+z=y_2(z)$ where $E=50+H+z$:

Depression of the floor: $z=6.50$ m (say 6.5 m).

Step 5: Length and type

$$L=5\,y_2=5\times12.91=64.6\ \text{m}\ \Rightarrow\ \text{adopt }65\ \text{m}$$

$Fr_1=12.96>4.5$ and $V_1=34.5$ m/s $>18$ m/s, hence a USBR Type II basin: chute blocks of height $y_1=0.724$ m and a dentated end sill $0.2y_2=2.58$ m high. Protect the d/s bed with riprap.

Answer: $y_1=0.724$ m, $y_2=12.91$ m, $Fr_1=12.96$; depress the floor by 6.5 m; basin length about 65 m.