Is a hydraulic jump stilling basin required?

A hydraulic jump basin is needed if the natural tail water depth is less than the conjugate (sequent) depth $y_2$ needed to hold the jump at the toe.

Given: $Q=100$ m³/s, spillway width $B=10$ m, height of crest above d/s bed $P=30$ m, $S_0=1/500$, $n=0.018$, $C_d=0.75$. The river downstream is assumed rectangular with the same width.

Step 1: Unit discharge and head

$$q=\frac{100}{10}=10\ \text{m}^3/\text{s/m},\qquad 10=\tfrac23(0.75)(4.429)H^{3/2}\Rightarrow H=2.73\ \text{m}$$

Step 2: Depth and velocity at the toe

$E=P+H=30+2.73=32.73$ m (losses neglected).

$$y_1+\frac{q^2}{2gy_1^2}=32.73\Rightarrow y_1=0.397\ \text{m},\quad V_1=25.2\ \text{m/s},\quad Fr_1=12.76$$

Step 3: Required tail water (sequent depth)

$$y_2=\frac{y_1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)=6.97\ \text{m}$$

Step 4: Available tail water (Manning)

For the 10 m wide channel carrying 100 m³/s with $S_0=0.002$ and $n=0.018$:

$$Q=\frac1nAR^{2/3}S_0^{1/2}\ \Rightarrow\ y_n=2.75\ \text{m}$$

(Check: $A=27.5$ m², $R=1.77$ m, $Q=100.0$ m³/s.)

Step 5: Decision

$$y_n=2.75\ \text{m}<y_2=6.97\ \text{m}\quad(y_n/y_2=0.39)$$

The available tail water is much smaller than the required sequent depth, so the jump would be swept far downstream and the d/s bed would be scoured.

Answer: a stilling basin is required. The floor must be depressed (or a sill/baffle arrangement provided) to give a tail water depth of about $y_2=6.97$ m; since $Fr_1=12.76>4.5$ and $V_1>18$ m/s, a USBR Type II basin of length about $4.3y_2=30.0$ m is suitable.