Hydraulic jump stilling basin (28 m³/s/m)

Given: $q=28$ m³/s/m, crest 55 m above the d/s river bed, bed slope 1:1000, $n=0.028$ (gravel bed), $C_d=0.75$, coefficient of tail-water depth 1.2 (the tail water over the floor must be at least $1.2y_2$), coefficient of length 4.5 ($L=4.5y_2$), relative density of sediment 2.65 (used for the riprap). Energy losses on the spillway face are neglected.

Step 1: Head over the crest

$$28=\tfrac23(0.75)(4.429)H^{3/2}=2.2147\,H^{3/2}\Rightarrow H=5.43\ \text{m}$$

Step 2: Tail water in the river (wide channel, Manning)

$$y_n=\left(\frac{q\,n}{\sqrt{S_0}}\right)^{3/5}=\left(\frac{28\times0.028}{\sqrt{0.001}}\right)^{0.6}=6.86\ \text{m}$$

Step 3: Jump at the toe (floor at bed level)

$E=55+5.43=60.43$ m:

$$y_1=0.819\ \text{m},\quad V_1=34.2\ \text{m/s},\quad Fr_1=12.07,\quad y_2=13.57\ \text{m}$$

Required tail water $1.2y_2=16.28$ m $\gg y_n=6.86$ m, so the floor must be depressed by $z$.

Step 4: Depression of the floor

The condition is $y_n+z\ge1.2\,y_2(z)$, with $E=55+H+z$:

So depress the floor by $z\approx10.13$ m below the river bed (say 10.5 m).

Step 5: Length and details

$$L=4.5\,y_2=4.5\times14.16=63.7\ \text{m}\ \Rightarrow\ \text{adopt }64\ \text{m}$$

• $Fr_1=13.58>4.5$ and $V_1=37.0$ m/s: a Type II (chute blocks, dentated sill) arrangement is used; chute-block height $\approx y_1=0.757$ m; end sill $0.2y_2=2.83$ m.
• Downstream protection: velocity over the sill $V=q/y_n=4.08$ m/s. By Isbash's formula with $C=0.86$ and $S_s=2.65$, the stone size is $d=\dfrac{V^2}{2gC^2(S_s-1)}=0.69$ m, so use riprap of about 0.7-0.8 m size on a filter layer.

Answer: $y_1=0.757$ m, $y_2=14.16$ m, $Fr_1=13.58$; floor depressed by 10.13 m; basin length 64 m; riprap about 0.7 m stones.