Sarda fall: Q 12, FSL 501.40

Given data and assumptions

• $Q = 12$ m³/s, bed width $B = 10$ m, side slope 1:1, u/s FSL $= 501.40$ m, u/s bed $= 500.00$ m, d/s bed $= 498.80$ m.
• Full supply depth u/s = d/s $D = 1.4$ m (d/s FSL $= 500.20$ m); drop in water level $H_L = 1.2$ m.
• Bligh's coefficient $C = 7$; specific gravity of the masonry/concrete $G = 2.24$ (assumed).
• D/s bed $= 500.20 - 1.40 = 498.80$ m, which agrees with the 1.20 m drop of the bed.

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD)D = (10 + 1\times1.4)(1.4) = 15.960\ \text{m}^2 \\ V_a &= Q/A = 0.752\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.029\ \text{m} \end{aligned}$$

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:

$$H = 0.753\ \text{m}, \qquad \text{crest level} = 501.40 - 0.753 = 500.65\ \text{m}, \qquad d = 500.65 - 498.80 = 1.85\ \text{m}$$

Top width $B_c = 0.55\sqrt{1.85} = 0.75$ m. Base width of the crest wall:

$$B_w = \frac{H + d}{\sqrt{G}} = \frac{0.753 + 1.85}{\sqrt{2.24}} = 1.74\ \text{m}$$

3. Cistern

Energy head over the crest $E = H + h_a = 0.753 + 0.029 = 0.782$ m. Using the UPIRI formulas:

$$\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.782\times1.2} = 4.84\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(0.938)^{2/3} = 0.240\ \text{m} \end{aligned}$$

Cistern floor level $= 498.80 - 0.240 = 498.56$ 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: $H_s = d = 1.85$ m. Required creep length $L = C\,H_s = 7\times1.85 = 12.93$ m.

• Cut-offs: u/s $d_1 = D/3 + 0.6 = 1.07$ m; d/s $d_2 = D/2 + 0.6 = 1.30$ m.
• Horizontal floor length $b = L - 2(d_1 + d_2) = 12.93 - 4.73 = 8.20$ m.
• Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 6.40$ m, which is more than $L_c$; adopt $L_d = 6.40$ m.
• U/s floor $= b - L_d - B_w = 8.20 - 6.40 - 1.74 = 0.06$ m, so a minimum u/s floor of 1.0 m is provided. Actual creep $= 13.87$ m $\ge 12.93$ m, safe.
• Total floor length $= 1.00 + 1.74 + 6.40 = 9.14$ m.

5. Floor thickness

Creep up to the d/s toe of the crest wall $= 2d_1 + b_u + B_w = 4.87$ m, so the residual uplift head there is $h_1 = H_s\,\dfrac{L - 4.87}{L} = 1.85\times\dfrac{13.87 - 4.87}{13.87} = 1.199$ m (with $L = 13.87$ m, the creep provided).

$$t_1 = \frac{4}{3}\,\frac{h_1}{G - 1} = \frac{4}{3}\times\frac{1.199}{1.24} = 1.29\ \text{m}$$

At the end of the cistern the creep is 9.71 m, $h_2 = 0.554$ m and $t_2 = 0.60$ m. Adopt a d/s floor 1.29 m thick near the crest wall, reducing to 0.60 m beyond the cistern (minimum 0.30 m); the u/s floor is given a nominal 0.30 m.

Summary

Answer: crest level 500.65 m (rectangular crest 10 m long, top width 0.75 m, base 1.74 m); cistern 4.84 m long and 0.24 m deep; impervious floor 9.14 m long (u/s 1.00 m, d/s 6.40 m) with cut-offs of 1.07 m (u/s) and 1.30 m (d/s); d/s floor thickness 1.29 m, reducing to 0.60 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 __|