Vertical drop in 10 m wide canal, Q 20

Given data and assumptions

• $Q = 20$ m³/s, bed width $B = 10$ m, side slope 1:1, u/s FSL $= 105.00$ m, u/s bed $= 102.00$ m, d/s bed $= 100.00$ m.
• Full supply depth u/s = d/s $D = 3$ m (d/s FSL $= 103.00$ m); drop in water level $H_L = 2$ m.
• Bligh's coefficient $C = 8$; specific gravity of the masonry/concrete $G = 2.24$ (assumed).
• The printed d/s FSL (130 m) is clearly a misprint. The d/s depth is taken equal to the u/s depth (3 m), so d/s FSL $= 100 + 3 = 103$ m and $H_L = 105 - 103 = 2$ m.
• Bligh's safe hydraulic gradient of 1/8 means $C = 8$. The vertical drop is designed as a Sarda-type fall; the "design level" asked for is the crest level.

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD)D = (10 + 1\times3)(3) = 39.000\ \text{m}^2 \\ V_a &= Q/A = 0.513\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.013\ \text{m} \end{aligned}$$

2. Crest

Since $Q > 14$ m³/s, a trapezoidal crest is used, with crest length equal to the bed width, $L = 10$ m, and

$$Q = 1.99\,L\,H^{3/2}\left(\frac{H}{B_c}\right)^{1/6}, \qquad B_c = 0.55\sqrt{H + d}$$

where $H$ is the head over the crest (measured from the u/s FSL), $B_c$ is the top width of the crest and $d$ is the height of the crest above the d/s bed. Solving by trial:

$$H = 1.024\ \text{m}, \qquad \text{crest level} = 105.00 - 1.024 = 103.98\ \text{m}, \qquad d = 103.98 - 100.00 = 3.98\ \text{m}$$

Top width $B_c = 0.55\sqrt{1.024 + 3.98} = 1.23$ m. With a u/s batter of 1:3 and a d/s batter of 1:8, the base width $= 1.23 + 3.98\left(\tfrac13 + \tfrac18\right) = 3.05$ m.

3. Cistern

Energy head over the crest $E = H + h_a = 1.024 + 0.013 = 1.037$ m. Using the UPIRI formulas:

$$\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{1.037\times2} = 7.20\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(2.075)^{2/3} = 0.407\ \text{m} \end{aligned}$$

Cistern floor level $= 100.00 - 0.407 = 99.59$ m.

4. Impervious floor (Bligh's theory)

Maximum seepage head occurs with no flow, water up to the crest on the u/s side and the d/s side dry: $H_s = d = 3.98$ m. Required creep length $L = C\,H_s = 8\times3.98 = 31.81$ m.

• Cut-offs: u/s $d_1 = D/3 + 0.6 = 1.60$ m; d/s $d_2 = D/2 + 0.6 = 2.10$ m.
• Horizontal floor length $b = L - 2(d_1 + d_2) = 31.81 - 7.40 = 24.41$ m.
• Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 10.40$ m, which is more than $L_c$; adopt $L_d = 10.40$ m.
• U/s floor $= b - L_d - B_w = 24.41 - 10.40 - 3.05 =$ 10.96 m.
• Total floor length $= 10.96 + 3.05 + 10.40 = 24.41$ m.

Summary

Answer: crest (design) level 103.98 m (trapezoidal crest 10 m long, top width 1.23 m); cistern 7.20 m long and 0.41 m deep (floor at 99.59 m); u/s floor length 10.96 m (total floor 24.41 m, cut-offs 1.60 m u/s and 2.10 m d/s).

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 __|