How is the top width of the crest wall obtained?

T = 0.55 sqrt(H + d_c), about 0.90 m

Why is the cistern depth only a nominal 0.25 m?

The tail water depth already exceeds the conjugate depth, so the jump is drowned and only a nominal cushion is kept

The conjugate depth is greater than tail water, so a deep cistern is needed

The cistern depth must equal the crest head H

2071 Bhadra · 8 marks

Sarda fall: Q 9, drop 1.2 m

Short answer

Rectangular crest at 104.97 m (top width 0.77 m, base 1.80 m); cistern 4.73 m long and 0.23 m deep, floor level 102.77 m.

Design the crest and cistern of a drop structure (Sarda type) for a discharge of 9 cumecs and a drop height of 1.2 m: FSL u/s and d/s = 105.7 m and 104.5 m; Bed Level u/s and d/s = 104.2 m and 103.0 m; Bed Width u/s and d/s = 8 m; Side Slope of Channel = 1:1.

Given data and assumptions

$Q = 9$ m³/s, bed width $B = 8$ m, side slope 1:1, u/s FSL $= 105.70$ m, u/s bed $= 104.20$ m, d/s bed $= 103.00$ m.
Full supply depth u/s = d/s $D = 1.5$ m (d/s FSL $= 104.50$ m); drop in water level $H_L = 1.2$ m.
Specific gravity of the masonry/concrete $G = 2.24$ (assumed).
Only the crest and the cistern are asked for, so the floor is not designed.

1. Upstream flow conditions

\begin{aligned} A &= (B + mD)D = (8 + 1\times1.5)(1.5) = 14.250\ \text{m}^2 \\ V_a &= Q/A = 0.632\ \text{m/s}, \qquad h_a = \frac{V_a^2}{2g} = 0.020\ \text{m} \end{aligned}

2. Crest

Since $Q < 14$ m³/s, a rectangular crest is used, with crest length equal to the bed width, $L = 8$ 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 = 0.727\ \text{m}, \qquad \text{crest level} = 105.70 - 0.727 = 104.97\ \text{m}, \qquad d = 104.97 - 103.00 = 1.97\ \text{m}

Top width $B_c = 0.55\sqrt{1.97} = 0.77$ m. Base width of the crest wall:

B_w = \frac{H + d}{\sqrt{G}} = \frac{0.727 + 1.97}{\sqrt{2.24}} = 1.80\ \text{m}

3. Cistern

Energy head over the crest $E = H + h_a = 0.727 + 0.020 = 0.747$ m. Using the UPIRI formulas:

\begin{aligned} L_c &= 5\sqrt{E\,H_L} = 5\sqrt{0.747\times1.2} = 4.73\ \text{m} \\ X &= \tfrac14\,(E\,H_L)^{2/3} = \tfrac14\,(0.896)^{2/3} = 0.232\ \text{m} \end{aligned}

Cistern floor level $= 103.00 - 0.232 = 102.77$ m.

Summary

Component	Value
Crest	rectangular, length 8 m, level 104.97 m, head $H$ = 0.727 m
Crest width (top / base)	0.77 m / 1.80 m
Cistern (length / depth)	4.73 m / 0.23 m, floor level 102.77 m

Answer: crest level 104.97 m (rectangular crest 8 m long, top width 0.77 m, base 1.80 m); cistern 4.73 m long and 0.23 m deep.

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 __|

✓ Answer checked by Claude Opus 5.5 (high effort)

Sarda fall: Q 12, bed level 103 m · Irrigation and Drainage Engineering, chapter 7
Crest width, cistern length of vertical drop, Q 1.55 · Irrigation and Drainage Engineering, chapter 7 · asked 2×
Sarda fall: Q 12, U/S bed 103 m · Irrigation and Drainage Engineering, chapter 7
Vertical drop crest and cistern, Q 4.5 · Irrigation and Drainage Engineering, chapter 7
Vertical drop crest level and width, Q 30 · Irrigation and Drainage Engineering, chapter 7
Sarda fall: Q 12, FSL 501.40 · Irrigation and Drainage Engineering, chapter 7
Slope-area discharge, 10 km reach, 104.77 m · Engineering Hydrology, chapter 4

All Irrigation and Drainage Engineering chapter 7 questions →