Vertical fall components, Q 1.0 m3/s

Given data and assumptions

• $Q = 1$ m³/s, bed width $B = 3$ m, rectangular canal, u/s FSL $= 100.75$ m, u/s bed $= 100.00$ m, d/s bed $= 99.00$ m.
• Full supply depth u/s = d/s $D = 0.75$ m (d/s FSL $= 99.75$ m); drop in water level $H_L = 1$ m.
• Bligh's coefficient $C = 6$; specific gravity of the masonry/concrete $G = 2.24$ (assumed).
• No side slope is given, so a rectangular canal ($m = 0$) is assumed. U/s FSL $= 100 + 0.75 = 100.75$ m, d/s FSL $= 99 + 0.75 = 99.75$ m, so $H_L = 1.0$ m. The vertical fall is designed as a Sarda-type fall.

1. Upstream flow conditions

$$\begin{aligned} A &= (B + mD)D = (3 + 0\times0.75)(0.75) = 2.250\ \text{m}^2 \\ V_a &= Q/A = 0.444\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.010\ \text{m} \end{aligned}$$

2. Crest

Since $Q < 14$ m³/s, a rectangular crest is used, with crest length equal to the bed width, $L = 3$ 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:

$$H = 0.344\ \text{m}, \qquad \text{crest level} = 100.75 - 0.344 = 100.41\ \text{m}, \qquad d = 100.41 - 99.00 = 1.41\ \text{m}$$

Top width $B_c = 0.55\sqrt{1.41} = 0.65$ m. Base width of the crest wall:

$$B_w = \frac{H + d}{\sqrt{G}} = \frac{0.344 + 1.41}{\sqrt{2.24}} = 1.17\ \text{m}$$

3. Cistern

Energy head over the crest $E = H + h_a = 0.344 + 0.010 = 0.354$ m. Using the UPIRI formulas:

$$\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.354\times1} = 2.98\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(0.354)^{2/3} = 0.125\ \text{m} \end{aligned}$$

Cistern floor level $= 99.00 - 0.125 = 98.87$ 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: $H_s = d = 1.41$ m. Required creep length $L = C\,H_s = 6\times1.41 = 8.43$ m.

• Cut-offs: u/s $d_1 = D/3 + 0.6 = 0.85$ m; d/s $d_2 = D/2 + 0.6 = 0.97$ m.
• Horizontal floor length $b = L - 2(d_1 + d_2) = 8.43 - 3.65 = 4.78$ m.
• Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 4.90$ m, which is more than $L_c$; adopt $L_d = 4.90$ m.
• U/s floor $= b - L_d - B_w = 4.78 - 4.90 - 1.17 = -1.29$ m, so a minimum u/s floor of 1.0 m is provided. Actual creep $= 10.72$ m $\ge 8.43$ m, safe.
• Total floor length $= 1.00 + 1.17 + 4.90 = 7.07$ m.

5. Floor thickness

Creep up to the d/s toe of the crest wall $= 2d_1 + b_u + B_w = 3.87$ m, so the residual uplift head there is $h_1 = H_s\,\dfrac{L - 3.87}{L} = 1.41\times\dfrac{10.72 - 3.87}{10.72} = 0.898$ m (with $L = 10.72$ m, the creep provided).

$$t_1 = \frac{4}{3}\,\frac{h_1}{G - 1} = \frac{4}{3}\times\frac{0.898}{1.24} = 0.97\ \text{m}$$

At the end of the cistern the creep is 6.85 m, $h_2 = 0.508$ m and $t_2 = 0.55$ m. Adopt a d/s floor 0.97 m thick near the crest wall, reducing to 0.55 m beyond the cistern (minimum 0.30 m); the u/s floor is given a nominal 0.30 m.

Summary

Answer: crest level 100.41 m (rectangular crest 3 m long, top width 0.65 m, base 1.17 m); cistern 2.98 m long and 0.13 m deep; impervious floor 7.07 m long (u/s 1.00 m, d/s 4.90 m) with cut-offs of 0.85 m (u/s) and 0.97 m (d/s); d/s floor thickness 0.97 m, reducing to 0.55 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 __|