Sarda fall: Q 8, FSL 218.30

Head regulator and cross regulator (difference)

Design of the Sarda-type vertical drop

Given data and assumptions

• $Q = 8$ m³/s, bed width $B = 10$ m, side slope 1.5:1, u/s FSL $= 218.30$ m, u/s bed $= 216.50$ m, d/s bed $= 215.00$ m.
• Full supply depth u/s = d/s $D = 1.8$ m (d/s FSL $= 216.80$ m); drop in water level $H_L = 1.5$ m.
• Bligh's coefficient $C = 8$; specific gravity of the masonry/concrete $G = 2.24$ (assumed).
• Bed levels follow from the FSLs and depth: u/s bed $= 218.30 - 1.8 = 216.50$ m and d/s bed $= 216.80 - 1.8 = 215.00$ m, giving a drop of 1.5 m. The stated drop of 1.2 m does not agree with the given levels, so the levels are used.

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD)D = (10 + 1.5\times1.8)(1.8) = 22.860\ \text{m}^2 \\ V_a &= Q/A = 0.350\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.006\ \text{m} \end{aligned}$$

2. Crest

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

$$Q = 1.835\,L\,H^{3/2}\left(\frac{H}{B_c}\right)^{1/6}, \qquad B_c = 0.55\sqrt{d}$$

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:

$$H = 0.602\ \text{m}, \qquad \text{crest level} = 218.30 - 0.602 = 217.70\ \text{m}, \qquad d = 217.70 - 215.00 = 2.70\ \text{m}$$

Top width $B_c = 0.55\sqrt{2.70} = 0.90$ m. Base width of the crest wall:

$$B_w = \frac{H + d}{\sqrt{G}} = \frac{0.602 + 2.70}{\sqrt{2.24}} = 2.20\ \text{m}$$

3. Cistern

Energy head over the crest $E = H + h_a = 0.602 + 0.006 = 0.608$ m. Using the UPIRI formulas:

$$\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.608\times1.5} = 4.77\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(0.912)^{2/3} = 0.235\ \text{m} \end{aligned}$$

Cistern floor level $= 215.00 - 0.235 = 214.76$ 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: $H_s = d = 2.70$ m. Required creep length $L = C\,H_s = 8\times2.70 = 21.59$ m.

• Cut-offs: u/s $d_1 = D/3 + 0.6 = 1.20$ m; d/s $d_2 = D/2 + 0.6 = 1.50$ m.
• Horizontal floor length $b = L - 2(d_1 + d_2) = 21.59 - 5.40 = 16.19$ m.
• Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 7.50$ m, which is more than $L_c$; adopt $L_d = 7.50$ m.
• U/s floor $= b - L_d - B_w = 16.19 - 7.50 - 2.20 =$ 6.48 m.
• Total floor length $= 6.48 + 2.20 + 7.50 = 16.19$ m.

5. Floor thickness

Creep up to the d/s toe of the crest wall $= 2d_1 + b_u + B_w = 11.09$ m, so the residual uplift head there is $h_1 = H_s\,\dfrac{L - 11.09}{L} = 2.70\times\dfrac{21.59 - 11.09}{21.59} = 1.312$ m (with $L = 21.59$ m, the creep provided).

$$t_1 = \frac{4}{3}\,\frac{h_1}{G - 1} = \frac{4}{3}\times\frac{1.312}{1.24} = 1.41\ \text{m}$$

At the end of the cistern the creep is 15.86 m, $h_2 = 0.716$ m and $t_2 = 0.77$ m. Adopt a d/s floor 1.41 m thick near the crest wall, reducing to 0.77 m beyond the cistern (minimum 0.30 m); the u/s floor is given a nominal 0.30 m.

Summary

Answer: crest level 217.70 m (rectangular crest 10 m long, top width 0.90 m, base 2.20 m); cistern 4.77 m long and 0.24 m deep; impervious floor 16.19 m long (u/s 6.48 m, d/s 7.50 m) with cut-offs of 1.20 m (u/s) and 1.50 m (d/s); d/s floor thickness 1.41 m, reducing to 0.77 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 __|