Sarda fall: Q 12, U/S bed 103 m

Given data and assumptions

• $Q = 12$ m³/s, bed width $B = 10$ m, side slope 1.5:1, u/s FSL $= 104.50$ m, u/s bed $= 103.00$ m, d/s bed $= 101.50$ m.
• Full supply depth u/s = d/s $D = 1.5$ m (d/s FSL $= 103.00$ m); drop in water level $H_L = 1.5$ m.
• Bligh's coefficient $C = 6$; specific gravity of the masonry/concrete $G = 2.24$ (assumed).
• D/s depth = u/s depth = 1.5 m, so d/s FSL $= 103.0$ m and the drop is 1.5 m.

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD)D = (10 + 1.5\times1.5)(1.5) = 18.375\ \text{m}^2 \\ V_a &= Q/A = 0.653\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.022\ \text{m} \end{aligned}$$

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:

$$H = 0.760\ \text{m}, \qquad \text{crest level} = 104.50 - 0.760 = 103.74\ \text{m}, \qquad d = 103.74 - 101.50 = 2.24\ \text{m}$$

Top width $B_c = 0.55\sqrt{2.24} = 0.82$ m. Base width of the crest wall:

$$B_w = \frac{H + d}{\sqrt{G}} = \frac{0.760 + 2.24}{\sqrt{2.24}} = 2.00\ \text{m}$$

3. Cistern

Energy head over the crest $E = H + h_a = 0.760 + 0.022 = 0.782$ m. Using the UPIRI formulas:

$$\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.782\times1.5} = 5.41\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(1.173)^{2/3} = 0.278\ \text{m} \end{aligned}$$

Cistern floor level $= 101.50 - 0.278 = 101.22$ 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: $H_s = d = 2.24$ m. Required creep length $L = C\,H_s = 6\times2.24 = 13.44$ m.

• Cut-offs: u/s $d_1 = D/3 + 0.6 = 1.10$ m; d/s $d_2 = D/2 + 0.6 = 1.35$ m.
• Horizontal floor length $b = L - 2(d_1 + d_2) = 13.44 - 4.90 = 8.54$ m.
• Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 6.90$ m, which is more than $L_c$; adopt $L_d = 6.90$ m.
• U/s floor $= b - L_d - B_w = 8.54 - 6.90 - 2.00 = -0.37$ m, so a minimum u/s floor of 1.0 m is provided. Actual creep $= 14.80$ m $\ge 13.44$ m, safe.
• Total floor length $= 1.00 + 2.00 + 6.90 = 9.90$ m.

5. Floor thickness

Creep up to the d/s toe of the crest wall $= 2d_1 + b_u + B_w = 5.20$ m, so the residual uplift head there is $h_1 = H_s\,\dfrac{L - 5.20}{L} = 2.24\times\dfrac{14.80 - 5.20}{14.80} = 1.452$ m (with $L = 14.80$ m, the creep provided).

$$t_1 = \frac{4}{3}\,\frac{h_1}{G - 1} = \frac{4}{3}\times\frac{1.452}{1.24} = 1.56\ \text{m}$$

At the end of the cistern the creep is 10.62 m, $h_2 = 0.633$ m and $t_2 = 0.68$ m. Adopt a d/s floor 1.56 m thick near the crest wall, reducing to 0.68 m beyond the cistern (minimum 0.30 m); the u/s floor is given a nominal 0.30 m.

Summary

Answer: crest level 103.74 m (rectangular crest 10 m long, top width 0.82 m, base 2.00 m); cistern 5.41 m long and 0.28 m deep; impervious floor 9.90 m long (u/s 1.00 m, d/s 6.90 m) with cut-offs of 1.10 m (u/s) and 1.35 m (d/s); d/s floor thickness 1.56 m, reducing to 0.68 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 __|