Minimum length of concrete apron of stilling basin

Given: crest length $B=15$ m, $Q=115$ m³/s, height of spillway $P=4$ m, $S_0=0.001$, $n=0.025$, basin width = 15 m. Losses on the spillway face are assumed negligible.

Step 1: Unit discharge and critical depth

$$q=\frac{115}{15}=7.667\ \text{m}^3/\text{s/m},\qquad y_c=\left(\frac{q^2}{g}\right)^{1/3}=1.816\ \text{m}$$

Specific energy at the crest above the apron: $E_0=P+1.5y_c=4+1.5(1.816)=6.72$ m.

Step 2: Supercritical depth $y_1$ at the toe

$$y_1+\frac{q^2}{2gy_1^2}=6.72\Rightarrow y_1=0.706\ \text{m},\quad V_1=10.87\ \text{m/s},\quad Fr_1=4.13$$ $$y_2=\frac{y_1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)=3.78\ \text{m}$$

Step 3: Tail water depth $y_0$ (Manning, rectangular 15 m wide)

$$Q=\frac1nAR^{2/3}S_0^{1/2}\ \Rightarrow\ y_0=3.43\ \text{m},\quad V=2.24\ \text{m/s},\quad Fr_0=0.386$$

Since $y_0=3.43\ \text{m}<y_2=3.78\ \text{m}$, the jump does not form at the toe. It moves downstream by a distance $\Delta x$ along the apron, to the section where the supercritical depth $y_1'$ has the conjugate depth $y_0$:

$$y_1'=\frac{y_0}{2}\left(\sqrt{1+8Fr_0^2}-1\right)=0.823\ \text{m}\quad(Fr_1'=3.28)$$

Step 4: Distance $\Delta x$ (gradually varied flow)

Friction makes the depth rise from $y_1=0.706$ m to $y_1'=0.823$ m. By the direct step method, $\Delta x=\dfrac{\Delta E}{S_0-\overline{S_f}}$, with $S_f=\left(\dfrac{nV}{R^{2/3}}\right)^2$, summed over many small steps:

$$\Delta x=14.4\ \text{m}$$

Step 5: Length of jump

For $Fr_1'=3.28$, the curve of $L_j/y_2$ against $Fr$ gives $L_j/y_2\approx5.4$ (here $y_2=y_0$ is the tail water depth):

$$L_j=5.4\times3.43=18.5\ \text{m}$$

Step 6: Minimum length of apron

$$L=\Delta x+L_j=14.4+18.5=32.9\ \text{m}$$

Answer: the minimum length of the concrete apron is about 33 m (distance to the start of the jump 14.4 m + jump length 18.5 m).