Minimum impervious floor length, 10000 m3/s

Data and assumptions

$Q = 10000$ m³/s, $d_{50} = 0.33$ mm, flood level $= 207$ m, pond level $= 204$ m, d/s floor level $= 198$ m, safe exit gradient $= 1/6$. The design head for piping is the static head with no tail water, $H = 204 - 198 = 6$ m. The d/s cutoff is taken to a depth of $1.5R$ below the HFL.

1. Silt factor, waterway and scour depth

$$\begin{aligned} f &= 1.76\sqrt{d_{mm}} = 1.76\sqrt{0.33} = 1.011 \\ P &= 4.75\sqrt{Q} = 475.0\ \text{m},\qquad q = Q/P = 21.05\ \text{m}^3/\text{s/m} \\ R &= 1.35\left(\frac{q^2}{f}\right)^{1/3} = 10.26\ \text{m} \end{aligned}$$

2. Depth of d/s cutoff

Scour level $= 207 - 1.5R = 191.62$ m. Depth of the d/s pile below the floor:

$$d = 198 - 191.62 = 6.38\ \text{m}$$

3. Floor length from the exit gradient

$$G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = 1/6\ \Rightarrow\ \frac{1}{\pi\sqrt{\lambda}} = 0.1667\times\frac{6.38}{6} = 0.1773$$ $$\lambda = \left(\frac{1}{\pi\times 0.1773}\right)^2 = 3.223,\qquad \lambda = \frac{1+\sqrt{1+\alpha^2}}{2} \Rightarrow \alpha = \sqrt{(2\lambda-1)^2-1} = 5.353$$ $$b = \alpha d = 5.353\times 6.38 = 34.2\ \text{m}$$

Check: $G_E$ with $b = 34.2$ m is 0.1667 $= 1/6.0$.

Answer: Minimum total impervious floor length (between the end cutoffs) $\approx 34.2$ m, with the d/s cutoff 6.4 m deep. If a deeper d/s cutoff (2R) is used, the exit gradient is satisfied for any floor length, and the length is then fixed by the uplift thickness and the jump.

Conceptual section

  Flood level 207 ~~~~~~~~~~~~~~~~~~~~~~~
  Pond level  204 ~~~~~~~~~|gate|
 u/s block apron  ___crest__|    |__ glacis __ d/s jump floor
 ====================================================== 198
   u/s pile (1.0-1.5R)  <--- b = 34.2 m ---> d/s pile 6.4 m
                                   inverted filter + block apron