Vertical drop crest level and width, Q 30

Given data and assumptions

• $Q = 30$ m³/s, bed width $B = 15$ m, side slope 2:1, u/s FSL $= 315.00$ m, u/s bed $= 312.00$ m, d/s bed $= 310.00$ m.
• Full supply depth u/s $D_u = 3.000$ m; d/s depth taken equal to the u/s depth, $D_d = 3.000$ m (d/s FSL $= 313.000$ m); drop in water level $H_L = 2.000$ m.
• Crest length equals the canal bed width, $L = B$. Rectangular crest with $C_d = 0.415$; specific gravity of masonry $G = 2.24$ (assumed).

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD_u)D_u = (15 + 2\times3.000)(3.000) = 63.000\ \text{m}^2 \\ v &= Q/A = 0.476\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.012\ \text{m} \end{aligned}$$

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 $h_a$). Take crest length equal to the bed width, $L = B$:

$$H = \left(\frac{Q}{1.838\,L}\right)^{2/3} = \left(\frac{30}{1.838\times15}\right)^{2/3} = 1.058\ \text{m}$$ $$\text{Crest level} = FSL_u + h_a - H = 315.000 + 0.012 - 1.058 = 313.95\ \text{m}$$

Height of crest above d/s bed $d = 313.95 - 310.00 = 3.95$ m.

3. Cistern (energy dissipation)

Discharge per metre length of crest $q = Q/L = 2.000$ m²/s. Energy available above the d/s bed $E_1 = FSL_u + h_a - \text{d/s bed} = 5.012$ 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 (1.89 m) is less than the tail water depth $D_d = 3.000$ 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 $= 309.75$ m).
• With this $X$: $y_1 = 0.201$ m, $F_1 = 7.10$, $y_2 = 1.918$ m.

Length of cistern (hydraulic jump length) $L_c = 5(y_2 - y_1) = 5(1.918 - 0.201) = 8.59$ m.

4. Crest (drop) wall

For a rectangular crest, top width $T = 0.55\sqrt{d} = 0.55\sqrt{3.95} = 1.09$ m; adopt 1.10 m. Height of wall from cistern floor to crest $H_w = 313.95 - 309.75 = 4.20$ m. Bottom width for a gravity masonry wall:

$$B_w = \frac{H_w + H}{\sqrt{G}} = \frac{4.20 + 1.06}{\sqrt{2.24}} = 3.51\ \text{m}$$

Adopt bottom width $B_w = 3.51$ m (not less than the top width), with the d/s face vertical or slightly battered.

Summary

Answer: crest level 313.95 m, crest length 15.00 m, crest wall 1.10 m top and 3.51 m bottom width, cistern depth 0.25 m and length 8.59 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 __|