Hydraulic jump stilling basin at spillway toe

Given: $Q=80$ m³/s, width $B=8$ m, crest RL 96.00 m, river bed RL 65.00 m, tail water RL 71.00 m (depth $y_{TW}=6$ m), $C_d=0.7$, bed slope 1:500, $n=0.016$ (not needed because the tail water is given), $L/y_2=5.1$.

Step 1: Unit discharge and head on crest

$$q=\frac{80}{8}=10\ \text{m}^3/\text{s/m},\qquad q=\tfrac23C_d\sqrt{2g}\,H^{3/2}\Rightarrow10=\tfrac23(0.7)(4.429)H^{3/2}\Rightarrow H=2.86\ \text{m}$$

Step 2: Conditions at the toe (floor at river-bed level)

Fall of crest $=96-65=31$ m, so the specific energy at the toe (losses neglected) $E=31+H=33.86$ m.

$$y_1+\frac{q^2}{2g\,y_1^2}=E\ \Rightarrow\ y_1=0.390\ \text{m},\quad V_1=25.6\ \text{m/s},\quad Fr_1=\frac{V_1}{\sqrt{gy_1}}=13.10$$ $$y_2=\frac{y_1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)=7.04\ \text{m}$$

The tail water depth (6.0 m) is less than $y_2=7.04$ m, so the jump would be swept away. The floor must be depressed.

Step 3: Depress the floor by $z$

With the floor lowered by $z$: $E=31+z+H$ and tail-water depth over the floor $=6+z$. Condition $6+z=y_2(z)$; by trial:

Depression $z\approx1.10$ m, so floor RL $=65-1.10=63.90$ m (say RL 63.90 m).

Step 4: Length of basin

$$L=5.1\,y_2=5.1\times7.10=36.2\ \text{m}\ \Rightarrow\ \text{adopt }37\ \text{m}$$

Step 5: Appurtenances

• Chute blocks at the toe, height and width $\approx y_1=0.384$ m (say 0.4 m), alternating with equal gaps.
• End sill of height $\approx0.2y_2=1.42$ m; $Fr_1=13.42>4.5$ and $V_1=26.0$ m/s $>18$ m/s, so a USBR Type II layout (chute blocks and dentated end sill) is used.
• Training walls up to $1.2y_2$ above floor, and riprap protection downstream of the end sill.

 crest 96.0 ____
           \ \
            \ \__[chute blocks]__________[sill]__
 bed 65.0 ===\ _ _ _ _ _ _ _ _ _ _ _ _ _ _/===== TWL 71.0
 floor 63.90     |<------ L = 37 m ------>|

Answer: $H=2.86$ m, $y_1=0.384$ m, $y_2=7.10$ m, $Fr_1=13.42$; floor depressed by 1.10 m to RL 63.90 m, basin length 37 m.