Vertical drop crest and cistern, Q 4.5

Given data and assumptions

• $Q = 4.5$ m³/s, bed width $B = 3$ m, side slope 1:1, u/s FSL $= 106.50$ m, u/s bed $= 105.00$ m, d/s bed $= 103.50$ m.
• Full supply depth u/s $D_u = 1.500$ m; the d/s depth is taken equal to the u/s depth, $D_d = 1.500$ m (d/s FSL $= 103.5 + 1.5 = 105.000$ m); drop in water level $H_L = 1.500$ m.
• Specific gravity of masonry $G = 2.24$ (assumed, not given).

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD_u)D_u = (3 + 1\times1.500)(1.500) = 6.750\ \text{m}^2 \\ v &= Q/A = 0.667\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.023\ \text{m} \end{aligned}$$

2. Crest length, head and crest level

Discharge over a rectangular crest: $Q = C_d\sqrt{2g}\,L\,H^{3/2}$ with $C_d\sqrt{2g} = 0.41\times4.429 = 1.816$, where $H$ is the total head over the crest (including $h_a$). Take crest length equal to the bed width, $L = B$:

$$H = \left(\frac{Q}{1.816\,L}\right)^{2/3} = \left(\frac{4.5}{1.816\times3}\right)^{2/3} = 0.880\ \text{m}$$ $$\text{Crest level} = FSL_u + h_a - H = 106.500 + 0.023 - 0.880 = 105.64\ \text{m}$$

Height of crest above d/s bed $d_c = 105.64 - 103.50 = 2.14$ m.

3. Cistern (energy dissipation)

Discharge per metre length of crest $q = Q/L = 1.500$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 3.023$ 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 $y_2 = \frac{y_1}{2}(\sqrt{1+8F_1^2} - 1)$:

• Without any depression the conjugate depth (1.41 m) is less than the tail water depth $D_d = 1.500$ m, so the jump is already drowned and hydraulically no depression is needed. A nominal cistern depth is still provided to hold a water cushion under the jet: $X = 0.25$ m (cistern floor level $= 103.25$ m).
• With this $X$: $y_1 = 0.193$ m, $F_1 = 5.65$, $y_2 = 1.448$ m.

Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.448 - 0.193) = 6.28$ m.

4. Crest (drop) wall

Check the assumed top width of 0.75 m: for a rectangular crest $T = 0.55\sqrt{d_c} = 0.55\sqrt{2.14} = 0.80$ m $> 0.75$ m, so adopt $T = 0.80$ m. Height of wall from cistern floor to crest $H_w = 105.64 - 103.25 = 2.39$ m. Bottom width for a gravity masonry wall:

$$B_w = \frac{H_w + H}{\sqrt{G}} = \frac{2.39 + 0.88}{\sqrt{2.24}} = 2.19\ \text{m}$$

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

Summary

Answer: crest level 105.64 m, crest length 3.00 m, crest wall 0.80 m top and 2.19 m bottom width, cistern depth 0.25 m (level 103.25 m) and length 6.28 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 __|