Vertical drop design, Q 7.5 m3/s

Given data and assumptions

• $Q = 7.5$ m³/s, bed width $B = 1.5$ m, side slope 1:1, u/s FSL $= 156.10$ m, u/s bed $= 155.15$ m, d/s bed $= 154.45$ m.
• Full supply depth u/s $D_u = 0.950$ m, d/s $D_d = 0.950$ m (d/s FSL $= 155.400$ m); drop in water level $H_L = 0.700$ m.
• Bligh's coefficient $C = 6$; specific gravity of masonry/concrete $G = 2.25$.

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD_u)D_u = (1.5 + 1\times0.950)(0.950) = 2.327\ \text{m}^2 \\ v &= Q/A = 3.222\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.529\ \text{m} \end{aligned}$$

2. Crest length, head and crest level

Discharge over a rectangular crest: $Q = C_d\sqrt{2g}\,L\,H^{3/2}$ with $C_d\sqrt{2g} = 1.838$, where $H$ is the total head over the crest (including $h_a$). With $L = B = 1.5$ m the head needed would put the crest below the u/s bed. The crest is therefore fixed at the u/s bed level and its length increased (splayed wing walls):

$$\begin{aligned} H &= (FSL_u + h_a) - \text{crest} = 156.629 - 155.150 = 1.479\ \text{m} \\ L &= \frac{Q}{1.838\,H^{3/2}} = 2.27\ \text{m} \end{aligned}$$

Height of crest above d/s bed $d_c = 155.15 - 154.45 = 0.70$ m.

3. Cistern (energy dissipation)

Discharge per metre length of crest $q = Q/L = 3.307$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 2.179$ m. For a cistern of depth $X$ below the d/s bed, the jump must form with the conjugate depth $y_2$ equal to the tail water over the cistern, $y_2 = D_d + X$. Solving with $E_1 + X = y_1 + q^2/(2gy_1^2)$ and $y_2 = \frac{y_1}{2}(\sqrt{1+8F_1^2} - 1)$:

• Required $X = 1.076$ m. Provide $X = 1.10$ m (cistern floor level $= 153.35$ m).
• With this $X$: $y_1 = 0.443$ m, $F_1 = 3.58$, $y_2 = 2.032$ m.

Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(2.032 - 0.443) = 7.94$ m.

4. Crest (drop) wall

Top width (assumed) $T = 0.55$ m. Height of wall from cistern floor to crest $H_w = 155.15 - 153.35 = 1.80$ m. Bottom width for a gravity masonry wall:

$$B_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.80 + 1.48}{\sqrt{2.25}} = 2.19\ \text{m}$$

Adopt bottom width $B_w = 2.19$ m (not less than the top width), with the d/s face vertical or slightly battered.

5. Length of impervious floor (Bligh's theory)

Total creep length required $L = C\,H_L = 6\times0.700 = 4.20$ m. Cut-offs: u/s $d_1 = 0.6$ m, d/s $d_2 = 0.9$ m. Downstream floor length (Bligh) with $q$ in m²/s:

$$L_d = 2.21\,C\sqrt{H_L\sqrt{q/75}} = 2.21\times6\sqrt{0.700\sqrt{3.307/75}} = 5.08\ \text{m}$$

The d/s floor (measured from the toe of the wall) must also contain the cistern, so adopt $L_d = 7.94$ m. Creep already provided by cut-offs and wall $= 2(d_1 + d_2) + B_w = 5.19$ m, so the u/s floor needed is

$$b_u = L - [2(d_1+d_2) + B_w + L_d] = -8.93\ \text{m}$$

Bligh's creep is already satisfied, so a nominal u/s floor of 1.0 m is provided. Actual creep $= 14.13$ m $\ge 4.20$ m, safe. Total floor length $= b_u + B_w + L_d = 11.13$ m.

6. Thickness of the floor

Pressure head at the toe of the wall: creep to that point $= 4.39$ m, so residual head $h_1 = H_L\,\dfrac{L_{act} - 4.39}{L_{act}} = 0.483$ m. At the end of the cistern the creep is 12.33 m and $h_2 = 0.089$ m.

$$t_1 = \frac{4}{3}\frac{h_1}{G-1} = \frac{4}{3}\times\frac{0.483}{1.25} = 0.51\ \text{m}$$

Adopt floor thickness 0.51 m under the cistern near the wall, tapering to 0.30 m at the end of the floor (minimum 0.30 m).

Summary

Answer: crest level 155.15 m, crest length 2.27 m, crest wall 0.55 m top and 2.19 m bottom width, cistern depth 1.10 m and length 7.94 m, floor length 11.13 m, floor thickness 0.51 m.

Sketch:

 FSL u/s ~~~~~~~~~~~|
                    |  crest
 u/s bed ___________|__          FSL d/s ~~~~
        |  u/s floor |  \  drop        ~~~~~
     d1 |            |   \_____cistern________ d/s bed
        |            |<-Lc->|   d/s floor   | d2
                     |____ floor thickness __|