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Crest width, cistern length of vertical drop, Q 1.55

Short answer

Crest top width 0.50 m, bottom width 1.39 m. Cistern depth 0.25 m (floor level 104.07 m) and length 4.67 m.

Design a crest width, cistern length and its level of a vertical drop structure for the data given below. Full supply discharge u/s and d/s = 1.55 cumecs; Drop height = 0.75 m; FSL u/s and d/s = 105.997 and 105.247; Full supply depth u/s and d/s = 0.929 m; Bed levels u/s and d/s = 105.068 and 104.318; Bed width u/s and d/s = 1.1 m. Top width of crest = 0.5 m for initial assumption, Cd = 0.415 for rectangular crest. The drop structure is of masonry with specific gravity 2.0. Side slope of the canal is 1:1. The Bligh's coefficient is 7.0 for sandy loam soil at foundation.

Given data and assumptions

  • Q=1.55Q = 1.55 m³/s, bed width B=1.1B = 1.1 m, side slope 1:1, u/s FSL =106.00= 106.00 m, u/s bed =105.07= 105.07 m, d/s bed =104.32= 104.32 m.
  • Full supply depth u/s Du=0.929D_u = 0.929 m, d/s Dd=0.929D_d = 0.929 m (d/s FSL =105.247= 105.247 m); drop in water level HL=0.750H_L = 0.750 m.
  • Bligh's coefficient C=7C = 7; specific gravity of masonry/concrete G=2G = 2.

1. Upstream flow conditions

A=(B+mDu)Du=(1.1+1×0.929)(0.929)=1.885 m2v=Q/A=0.822 m/s,ha=v22g=0.034 m\begin{aligned} A &= (B + mD_u)D_u = (1.1 + 1\times0.929)(0.929) = 1.885\ \text{m}^2 \\ v &= Q/A = 0.822\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.034\ \text{m} \end{aligned}

2. Crest length, head and crest level

Discharge over a rectangular crest: Q=Cd2g L H3/2Q = C_d\sqrt{2g}\,L\,H^{3/2} with Cd2g=1.838C_d\sqrt{2g} = 1.838, where HH is the total head over the crest (including hah_a). Take crest length equal to the bed width, L=BL = B:

H=(Q1.838 L)2/3=(1.551.838×1.1)2/3=0.838 mH = \left(\frac{Q}{1.838\,L}\right)^{2/3} = \left(\frac{1.55}{1.838\times1.1}\right)^{2/3} = 0.838\ \text{m} Crest level=FSLu+ha−H=105.997+0.034−0.838=105.19 m\text{Crest level} = FSL_u + h_a - H = 105.997 + 0.034 - 0.838 = 105.19\ \text{m}

Height of crest above d/s bed dc=105.19−104.32=0.88d_c = 105.19 - 104.32 = 0.88 m.

3. Cistern (energy dissipation)

Discharge per metre length of crest q=Q/L=1.409q = Q/L = 1.409 m²/s. Energy available above the d/s bed E1=FSLu+ha−d/s bed=1.713E_1 = FSL_u + h_a - \text{d/s bed} = 1.713 m. For a cistern of depth XX below the d/s bed, the jump must form with the conjugate depth y2y_2 equal to the tail water over the cistern, y2=Dd+Xy_2 = D_d + X. Solving with E1+X=y1+q2/(2gy12)E_1 + X = y_1 + q^2/(2gy_1^2) and y2=y12(1+8F12−1)y_2 = \frac{y_1}{2}(\sqrt{1+8F_1^2} - 1):

  • Required X=0.247X = 0.247 m. Provide X=0.25X = 0.25 m (cistern floor level =104.07= 104.07 m).
  • With this XX: y1=0.242y_1 = 0.242 m, F1=3.77F_1 = 3.77, y2=1.176y_2 = 1.176 m.

Length of cistern (hydraulic jump length) Lc=5(y2−y1)=5(1.176−0.242)=4.67L_c = 5(y_2 - y_1) = 5(1.176 - 0.242) = 4.67 m.

4. Crest (drop) wall

Top width (assumed) T=0.50T = 0.50 m. Height of wall from cistern floor to crest Hw=105.19−104.07=1.13H_w = 105.19 - 104.07 = 1.13 m. Bottom width for a gravity masonry wall:

Bw=Hw+HG=1.13+0.842=1.39 mB_w = \frac{H_w + H}{\sqrt{G}} = \frac{1.13 + 0.84}{\sqrt{2}} = 1.39\ \text{m}

Adopt bottom width Bw=1.39B_w = 1.39 m (not less than the top width), with the d/s face vertical or slightly battered.

Summary

ComponentValue
Crest level105.19 m
Crest length / head HH1.10 m / 0.84 m
Crest wall top / bottom width0.50 m / 1.39 m
Cistern depth / length0.25 m / 4.67 m

Answer: crest level 105.19 m, crest length 1.10 m, crest wall 0.50 m top and 1.39 m bottom width, cistern depth 0.25 m and length 4.67 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 __|

✓ Answer checked by Claude Opus 5.5 (high effort)

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