---
title: "Sarda fall: Q 8, FSL 218.30"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:sarda-fall-q8-bed10"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/sarda-fall-q8-bed10
site: BCT Engineering
---

# Sarda fall: Q 8, FSL 218.30

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2081 Chaitra, 8 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-q8-bed10
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Differentiate between head regulator and cross-regulator. Design a vertical drop (Sarda type fall) for the following conditions. Full supply discharge: u/s = d/s = 8 m³/sec; full supply level: u/s 218.30 m, d/s 216.80 m; Bed width: u/s = d/s = 10 m; full supply depth: u/s = d/s = 1.8 m; drop: 1.2 m; side slope of canal: 1.5:1. Design the floor on Bligh's theory taking coefficient of creep = 8.0.

## Short answer

Head regulator controls off-take flow; cross regulator raises parent canal water level. For the Sarda fall: rectangular crest at 217.70 m (top width 0.90 m, base 2.20 m); cistern 4.77 m long, 0.235 m deep; impervious floor length 16.19 m (u/s 6.48 m, d/s 7.50 m) with u/s cut-off 1.20 m, d/s 1.50 m.

## Answer

#### Head regulator and cross regulator (difference)
| Point | Head regulator | Cross regulator |
|---|---|---|
| Location | At the head (off-take) of a canal/distributary | Across the parent canal, d/s of an off-take |
| Direction of flow control | Controls flow entering the off-take channel | Controls flow continuing in the parent canal |
| Purpose | Regulate supply, exclude silt, measure discharge | Raise water level to feed off-take; regulate d/s flow |
| Flow | Perpendicular or at an angle to the parent | Along the parent canal |
| Design discharge | Off-take discharge | Parent discharge minus off-take discharge |

#### Design of the Sarda-type vertical drop
#### Given data and assumptions
- $Q = 8$ m³/s, bed width $B = 10$ m, side slope 1.5:1, u/s FSL $= 218.30$ m, u/s bed $= 216.50$ m, d/s bed $= 215.00$ m.
- Full supply depth u/s = d/s $D = 1.8$ m (d/s FSL $= 216.80$ m); drop in water level $H_L = 1.5$ m.
- Bligh's coefficient $C = 8$; specific gravity of the masonry/concrete $G = 2.24$ (assumed).
- Bed levels follow from the FSLs and depth: u/s bed $= 218.30 - 1.8 = 216.50$ m and d/s bed $= 216.80 - 1.8 = 215.00$ m, giving a drop of 1.5 m. The stated drop of 1.2 m does not agree with the given levels, so the levels are used.

#### 1. Upstream flow conditions
$$
\begin{aligned} A &= (B + mD)D = (10 + 1.5\times1.8)(1.8) = 22.860\ \text{m}^2 \\ V_a &= Q/A = 0.350\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.006\ \text{m} \end{aligned}
$$

#### 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:
$$
H = 0.602\ \text{m}, \qquad \text{crest level} = 218.30 - 0.602 = 217.70\ \text{m}, \qquad d = 217.70 - 215.00 = 2.70\ \text{m}
$$
Top width $B_c = 0.55\sqrt{2.70} = 0.90$ m. Base width of the crest wall:
$$
B_w = \frac{H + d}{\sqrt{G}} = \frac{0.602 + 2.70}{\sqrt{2.24}} = 2.20\ \text{m}
$$

#### 3. Cistern
Energy head over the crest $E = H + h_a = 0.602 + 0.006 = 0.608$ m. Using the UPIRI formulas:
$$
\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.608\times1.5} = 4.77\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(0.912)^{2/3} = 0.235\ \text{m} \end{aligned}
$$
Cistern floor level $= 215.00 - 0.235 = 214.76$ 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: $H_s = d = 2.70$ m. Required creep length $L = C\,H_s = 8\times2.70 = 21.59$ m.
- Cut-offs: u/s $d_1 = D/3 + 0.6 = 1.20$ m; d/s $d_2 = D/2 + 0.6 = 1.50$ m.
- Horizontal floor length $b = L - 2(d_1 + d_2) = 21.59 - 5.40 = 16.19$ m.
- Minimum d/s floor (from the d/s face of the crest wall) $= 2(D + 1.2) + H_L = 7.50$ m, which is more than $L_c$; adopt $L_d = 7.50$ m.
- U/s floor $= b - L_d - B_w = 16.19 - 7.50 - 2.20 = $ **6.48 m**.
- Total floor length $= 6.48 + 2.20 + 7.50 = 16.19$ m.

#### 5. Floor thickness
Creep up to the d/s toe of the crest wall $= 2d_1 + b_u + B_w = 11.09$ m, so the residual uplift head there is $h_1 = H_s\,\dfrac{L - 11.09}{L} = 2.70\times\dfrac{21.59 - 11.09}{21.59} = 1.312$ m (with $L = 21.59$ m, the creep provided).
$$
t_1 = \frac{4}{3}\,\frac{h_1}{G - 1} = \frac{4}{3}\times\frac{1.312}{1.24} = 1.41\ \text{m}
$$
At the end of the cistern the creep is 15.86 m, $h_2 = 0.716$ m and $t_2 = 0.77$ m. Adopt a d/s floor **1.41 m** thick near the crest wall, reducing to 0.77 m beyond the cistern (minimum 0.30 m); the u/s floor is given a nominal 0.30 m.

#### Summary
| Component | Value |
|---|---|
| Crest | rectangular, length 10 m, level 217.70 m, head $H$ = 0.602 m |
| Crest width (top / base) | 0.90 m / 2.20 m |
| Cistern (length / depth) | 4.77 m / 0.24 m, floor level 214.76 m |
| Floor (u/s / wall base / d/s) | 6.48 / 2.20 / 7.50 m, total 16.19 m |
| Cut-offs (u/s / d/s) | 1.20 m / 1.50 m |
| D/s floor thickness | 1.41 m, reducing to 0.77 m |

**Answer:** crest level 217.70 m (rectangular crest 10 m long, top width 0.90 m, base 2.20 m); cistern 4.77 m long and 0.24 m deep; impervious floor 16.19 m long (u/s 6.48 m, d/s 7.50 m) with cut-offs of 1.20 m (u/s) and 1.50 m (d/s); d/s floor thickness 1.41 m, reducing to 0.77 m.

Sketch:
```text
 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 __|
```

## Check yourself

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

1. How does a cross-regulator differ from a head regulator?
   A. A cross-regulator is built at the off-take of a canal to exclude silt
   B. A cross-regulator is built across the parent canal d/s of an off-take to raise the water level and control flow continuing down the parent canal
   C. A cross-regulator only measures discharge and cannot control levels
   D. They are the same structure with different names
   Answer: B. The head regulator controls flow entering the off-take; the cross-regulator sits across the parent canal and raises the level to feed the off-take. (Key point: Head vs cross regulator)

2. How is the crest length of the Sarda-type fall obtained?
   A. L = B + 0.6 D_u (10 + 0.6 x 1.8 = 11.08 m)
   B. L = B - 0.6 D_u
   C. L = B only
   D. L = 2B
   Answer: A. Sarda crest length is bed width plus 0.6 times the full supply depth, giving 11.08 m. (Key point: Sarda crest length)

3. The stated drop is 1.2 m but the FSLs differ by 1.5 m. What did the solution do?
   A. Used 1.2 m and ignored the levels
   B. Averaged the two to get 1.35 m
   C. Used the levels, so the head loss H_L = 1.5 m and bed drop 1.5 m
   D. Refused to solve the problem
   Answer: C. The solution derived bed levels from FSL minus depth and used the resulting 1.5 m drop. (Key point: Handling inconsistent data)

4. What total creep length is required by Bligh's theory for C = 8 and head 1.5 m?
   A. 12.0 m
   B. 5.3 m
   C. 1.5 m
   D. 9.6 m
   Answer: A. L = C x H_L = 8 x 1.5 = 12 m. (Key point: Bligh creep length)

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