Diversion weir: Bligh and Khosla, 4000 m3/s

Data and assumptions

$Q = 4000$ m³/s, river width $300$ m, Bligh's $C = 12$, Lacey's $f = 1.2$, weir height above low water $3.0$ m, crest width $2.0$ m. Assumed: river bed (low water) level $= 100.0$ m, broad-crested weir with $C_d = 1.7$ ($Q = 1.7LH^{3/2}$), afflux $= 1.0$ m, specific gravity of concrete $G = 2.4$.

1. Waterway and flood levels

$$P = 4.75\sqrt{4000} = 300.4\ \text{m}\approx 300\ \text{m}\ (\text{equal to the river width})$$

Crest level $= 100.0 + 3.0 = 103.0$ m. Discharge per metre $q = 4000/300 = 13.33$ m³/s/m.

$$H = \left(\frac{4000}{1.7\times 300}\right)^{2/3} = 3.95\ \text{m},\quad \text{HFL}_{u/s} = 103.0 + 3.95 = 106.95\ \text{m},\quad \text{HFL}_{d/s} = 105.95\ \text{m}$$

2. Scour depth and cutoffs

$$R = 1.35\left(\frac{q^2}{f}\right)^{1/3} = 1.35\left(\frac{13.33^2}{1.2}\right)^{1/3} = 7.14\ \text{m}$$

• U/s pile bottom $= 106.95 - 1.5R = 96.23$ m, so $d_1 = 100 - 96.23 = 3.77$ m; adopt 3.8 m.
• D/s pile bottom $= 105.95 - 2R = 91.66$ m, so $d_2 = 8.34$ m; adopt 8.3 m.

3. Floor length by Bligh's principle

Static head $H = 103.0 - 100.0 = 3.0$ m. Creep length $L_c = CH = 12\times 3 = 36$ m.

$$b = L_c - 2(d_1 + d_2) = 36 - 2(3.8 + 8.3) = 11.8\ \text{m}$$

4. Khosla's check and design

Adopt a floor length $b = 12$ m between the piles: the 2 m wide crest body occupies the first 2 m, and the d/s floor below the vertical drop the remaining 10 m. Pile depths 3.8 m (u/s) and 8.3 m (d/s); floor thickness 1.0 m (u/s, under the crest) and 1.5 m (d/s) provisionally.

Khosla's constants

$\alpha = b/d$, $\lambda = \dfrac{1+\sqrt{1+\alpha^2}}{2}$, $\phi_E = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-2}{\lambda}$, $\phi_D = \dfrac{1}{\pi}\cos^{-1}\dfrac{\lambda-1}{\lambda}$ (in percent), total floor length $b = 12$ m.

Pressures at the key points

Pressure is expressed as a percentage of the head $H = 3.00$ m (no flow, pond at crest, d/s dry).

Mutual interference ($b' = b = 12$ m):

• at $C_1$ (due to the d/s pile): $C = 19\sqrt{\dfrac{8.3}{12}}\times\dfrac{3.8 + 8.3}{12} = +15.93\%$
• at $E_2$ (due to the u/s pile): $C = 19\sqrt{\dfrac{3.8}{12}}\times\dfrac{8.3 + 3.8}{12} = -10.78\%$

Floor thickness: at $C_1$, $+\dfrac{68.02 - 52.31}{3.8}\times 1.0 = +4.13\%$; at $E_2$, $-\dfrac{64.87 - 41.14}{8.3}\times 1.5 = -4.29\%$.

Corrections: int = mutual interference of piles, thk = floor thickness (no slope correction, since the floor is horizontal).

Floor thickness

Unbalanced head $h$ = hydraulic-gradient level $-$ floor top level; required thickness $t = \dfrac{h}{G-1}$ with $G = 2.40$. The HGL is taken as a straight line between the corrected values at $C_1$ ($x = 0$) and $E_2$ ($x = 12$ m).

The u/s part of the floor lies under the masonry crest and the pond water, whose weight balances the uplift, so a nominal 1.0 m is enough there. The critical point is just downstream of the vertical drop; provide 1.5 m there. It may be tapered towards the d/s end, but not below about 1.1 m.

Exit gradient

$$G_E = \frac{H}{d}\cdot\frac{1}{\pi\sqrt{\lambda}} = \frac{3.00}{8.30}\times\frac{1}{\pi\sqrt{1.379}} = 0.0980\ (\approx 1/10.2)$$

Permissible $= 0.1667$, so the floor is safe against piping.

Add u/s and d/s aprons (block apron with filter, length $\approx 1.5$-$2$ times the scour depth, say about 15 m).

Sketch

  HFL u/s ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
  Crest 103.0  _____ 2.0 m
 u/s apron    |     |   vertical drop     d/s apron
 ============ |_____|=========================
 100.0 bed    |<------ b = 12 m ------>|
   pile 3.8 m      floor 1.0 m | 1.5 m   pile 8.3 m

Answer: Waterway 300 m; crest 103.0 m; u/s cutoff 3.8 m; d/s cutoff 8.3 m; Bligh floor length $\approx 11.8$ m, adopted 12 m (Khosla exit gradient 0.098 $< 1/6$); floor thickness 1.0 m (nominal) under the crest and 1.5 m on the d/s floor below the drop.