Sarda fall for Bagmati branch canal, Q 25

Given data and assumptions

• $Q = 25$ m³/s, bed width $B = 20$ m, side slope 1.5:1, u/s FSL $= 101.80$ m, u/s bed $= 100.00$ m, d/s bed $= 98.80$ m.
• Full supply depth u/s = d/s $D = 1.8$ m (d/s FSL $= 100.60$ m); drop in water level $H_L = 1.2$ m.
• Datum: u/s bed 100.00 m (assumed); d/s bed $= 100.00 - 1.2 = 98.80$ m and the depth is 1.8 m on both sides. Side slope 1.5:1 (not given, assumed).

Upstream flow conditions

$$\begin{aligned} A &= (B + mD)D = (20 + 1.5\times1.8)(1.8) = 40.860\ \text{m}^2 \\ V_a &= Q/A = 0.612\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.019\ \text{m} \end{aligned}$$

(i) Crest

Since $Q > 14$ m³/s, a trapezoidal crest is used, with crest length equal to the bed width, $L = 20$ 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:

$$H = 0.753\ \text{m}, \qquad \text{crest level} = 101.80 - 0.753 = 101.05\ \text{m}, \qquad d = 101.05 - 98.80 = 2.25\ \text{m}$$

Top width $B_c = 0.55\sqrt{0.753 + 2.25} = 0.95$ m. With a u/s batter of 1:3 and a d/s batter of 1:8, the base width $= 0.95 + 2.25\left(\tfrac13 + \tfrac18\right) = 1.98$ m.

(ii) Cistern

Energy head over the crest $E = H + h_a = 0.753 + 0.019 = 0.772$ m. Using the UPIRI formulas:

$$\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.772\times1.2} = 4.81\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(0.926)^{2/3} = 0.238\ \text{m} \end{aligned}$$

Cistern floor level $= 98.80 - 0.238 = 98.56$ m.

(iii) Upstream and downstream cut-offs

• U/s cut-off $d_1 = D/3 + 0.6 = 1.8/3 + 0.6 = 1.20$ m.
• D/s cut-off $d_2 = D/2 + 0.6 = 1.8/2 + 0.6 = 1.50$ m below the floor.
• Scour check: $q = 25/20 = 1.25$ m²/s, Lacey's $R = 1.35(q^2/f)^{1/3} = 1.57$ m ($f = 1$), so the d/s scour level $= 100.60 - 1.5R = 98.25$ m. The d/s cut-off reaches $98.56 - 1.50 = 97.06$ m, well below the scour level, so it is safe.

(iv) Length of impervious floor (Khosla, safe exit gradient $G_E = 1/5$)

Maximum seepage head (no flow, water up to the crest u/s, d/s dry): $H_s = 101.05 - 98.80 = 2.25$ m.

$$\frac{1}{\pi\sqrt{\lambda}} = G_E\,\frac{d_2}{H_s} = \frac{1}{5}\times\frac{1.5}{2.25} = 0.1335 \Rightarrow \sqrt{\lambda} = 2.384,\ \lambda = 5.68$$ $$\lambda = \frac{1 + \sqrt{1 + \alpha^2}}{2} \Rightarrow \alpha = \sqrt{(2\lambda - 1)^2 - 1} = 10.32, \qquad b = \alpha\,d_2 = 10.32\times1.5 = 15.5\ \text{m}$$

Provide a total impervious floor of 15.5 m: d/s floor (from the toe of the crest) $= 2(D + 1.2) + H_L = 2(3.0) + 1.2 = 7.2$ m, which contains the 4.81 m cistern; crest base 1.98 m; u/s floor $= 15.5 - 7.2 - 1.98 = 6.3$ m. Check: $G_E = \dfrac{H_s}{d_2\,\pi\sqrt{\lambda}} = 0.200 = 1/5$, safe.

Answer: (i) trapezoidal crest 20 m long at level 101.05 m (head 0.753 m), top width 0.95 m, base width 1.98 m; (ii) cistern 4.81 m long and 0.24 m deep (floor at 98.56 m); (iii) cut-offs 1.2 m (u/s) and 1.5 m (d/s); (iv) impervious floor 15.5 m (u/s 6.3 m, crest base 1.98 m, d/s 7.2 m).