---
title: "Sarda fall for Bagmati branch canal, Q 25"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:sarda-fall-bagmati-q25"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/sarda-fall-bagmati-q25
site: BCT Engineering
---

# Sarda fall for Bagmati branch canal, Q 25

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2080 Chaitra, 4×2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/sarda-fall-bagmati-q25
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Eastern main canal of Bagmati Irrigation Project requires 1.2 m Sarda fall for a branch canal carrying a discharge of 25 m³/s with a flow depth of 1.8 m and channel bed width of 20 m. If a safe exit gradient for soil is 1/5, design following components: i) Dimension of crest (ii) Cistern (iii) Upstream and downstream cutoff (iv) Length of impervious floor.

## Short answer

Using Khosla's theory for exit gradient 1/5: trapezoidal crest at 101.05 m (top width 0.95 m, base 1.98 m); cistern 4.81 m long, 0.24 m deep; u/s cut-off 1.20 m, d/s 1.50 m; impervious floor length 15.5 m (u/s 6.3 m, d/s 7.2 m).

## Answer

#### 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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For the Sarda fall, what is the crest length and how is H found?
   A. L = B = 20 m; H from Q = 1.835 L H^(3/2)
   B. L = B + 0.6 D = 21.08 m; H = (Q/(1.835 L))^(2/3) = 0.748 m
   C. L = B + 6D = 30.8 m; H = Q/L
   D. L = B/0.6; H = Q/(L x D)
   Answer: B. Sarda crest length is B + 0.6D = 21.08 m and the rectangular crest formula gives H = 0.748 m. (Key point: Sarda crest length and head)

2. Why is Khosla's theory used here instead of Bligh's?
   A. Because a safe exit gradient 1/5 is given, and Khosla relates exit gradient to floor length and cut-off depth
   B. Because Bligh's theory cannot be used for any fall
   C. Because the discharge is above 20 m^3/s
   D. Because Khosla needs no cut-off depth
   Answer: A. The exit gradient G_E = (H/d)(1/(pi sqrt(lambda))) is Khosla's criterion, so the safe gradient fixes lambda and the floor length. (Key point: Khosla exit gradient method)

3. How is the total floor length found after lambda = 10.13?
   A. b = d / alpha
   B. b = alpha x d, with alpha = b/d = 19.24, so b = 28.9 m
   C. b = H_s x lambda
   D. b = G_E x d
   Answer: B. alpha = b/d is obtained from lambda, so b = alpha x d = 19.24 x 1.5 = about 28.9 m, adopted as 29 m. (Key point: Floor length from lambda)

4. How was the scour level used to decide the d/s cut-off?
   A. Scour depth R = 1.35 (q^2/f)^(1/3) = 1.51 m and scour level = FSL_d - 1.5R, which came only 0.22 m below the cistern floor, so practical minimum cut-offs were adopted
   B. Scour depth equals the drop of 1.2 m
   C. Scour depth was taken as zero
   D. Scour level was taken above the FSL
   Answer: A. Lacey's regime depth gave a scour level only slightly below the cistern floor, so minimum practical cut-offs of 1.5 m d/s and 1.0 m u/s were used. (Key point: Cut-off depth by Lacey scour)

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