USBR stilling basin design (80 m³/s)

Assumptions: spillway discharge coefficient $C_d=0.7$ in $q=\tfrac23C_d\sqrt{2g}H^{3/2}$; energy losses on the spillway face are neglected; the basin has the same width as the spillway (8 m); the tail-water depth is the normal depth in the river channel (rectangular, 8 m wide).

Step 1: Unit discharge and head over crest

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

Step 2: Tail-water depth (Manning)

River section 8 m wide, $S=1/500$, $n=0.016$. Solving $Q=\frac1nAR^{2/3}S^{1/2}$ for $Q=80$ m³/s gives normal depth

$$y_n=2.63\ \text{m}$$

Step 3: Conditions at the toe (without depression)

Total head above the river bed $E=(96-65)+H=33.86$ m.

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

Since $y_n=2.63\ \text{m}<y_2$, the jump would be swept away; the floor must be depressed.

Step 4: Depress the floor by $z$

With the floor $z$ lower, $E=31+z+H$ and tail water depth $=y_n+z$; the condition $y_n+z=y_2$ is solved by trial:

So depress the floor by $z=4.66$ m: basin floor RL $=65-4.66=60.34$ m.

Step 5: Select the USBR basin

$Fr_1=14.45>4.5$ and $V_1=27.4$ m/s $>18$ m/s, so the USBR Type II basin (with chute blocks, dentated end sill and no baffle blocks) is chosen; a Type III basin is not suitable above about 18 m/s.

 crest 96 ____
          \ \ spillway face
           \ \_ chute blocks    dentated sill
 65 river ===\_|#|#|__________|^|^|^|=== river
 floor 60.34 <----- L = 32 m ------>

Answer: $y_1=0.365$ m, $y_2=7.29$ m, $Fr_1=14.45$; USBR Type II basin, floor depressed by 4.66 m to RL 60.34 m, length 32 m, chute blocks 0.4 m, dentated end sill 1.46 m high.