Question 10.7: Design a Sarda-type fall using the following data: full supp......
Design a Sarda-type fall using the following data: full supply discharge = 28 {m}^{3}{s}^{-1} ; upstream FSL = 100.00 m AOD; downstream FSL = 98.50 m AOD; upstream bed level = 99.00 m AOD; downstream bed level = 97.50 m AOD; upstream bed width = 6.00 m; downstream bed width = 6.00 m; drop, H_{\mathrm{d}r} = 1.50 m; safe exit gradient of the subsoil = 1 in 5.
CREST DESIGN
Adopt a rectangular crest (Fig. 10.18). B = 0.55d^{1/2} ; for a trapezoidal crest, B=0.55\ (H+d)^{1/2}\mathop{\mathrm{and}}\ H+d=D_{1}+99.00-97.50=2.5\,\mathrm{m} . Hence B (trapezoidal crest)0.87 m. Choose a rectangular crest, width B = 0.80 m. Adopting a crest length of L =6 m, equation (10.51) gives the head over the crest,
\text { length of cistern, } L_{ c }=2.5 L_{ d } \text {, } (10.51)
H = 0.43 m.
Check for B: the crest level = 100.00 – 0.43 = 99.57 m AOD. Therefore d = 99.57 – 97.50 = 2.07 m and B = 0.55 d = 0.79 m. Therefore a crest width of 0.8 m is satisfactory.
The base width, B_{1} = 1.25 m (equation (10.50)
\text { depth of cistern, } d_{ c }=1 / 2\left(E_{ c } H_{ dr }\right)^{1 / 2} \text {; } (10.50)
with S_{\mathrm{s}}= 2.0). The velocity of approach (assuming a 1:1 trapezoidal channel) = 2.8/(6+1)1 = 0.4 {m}\,{s}^{-1} . Therefore the upstream total energy level (TEL) = 100.000 + 0.008 = 100.008 m AOD, and E = 100.008- 99.57 = 0.438 m.
The depth of the cistern, d_{\mathrm{c}} = 0.19 m (equation (10.55)),
\text { volume of basin, } V=Q H_{ dr } / 150\left( m ^3\right) \text {, } (10.55)
and the length of the cistern, L_{ c } \simeq 4 \,m (equation (10.54)).
\cot \alpha=y_{ c } / L_{ d 1} \text {. } (10.54)
The RL of the cistern bed = 97.50 – 0.19 = 97.31 m AOD.
IMPERVIOUS FLOOR DESIGN (WORKED EXAMPLE 9.1 PROVIDES DETAILED CALCULATIONS)
The maximum seepage head, H_{s}=d (with no water downstream, and the upstream water level at crest level). Adopting nominal upstream and downstream cut-off depths of 1m and 1.5 m respectively, the base length of the impervious floor for the exit gradient of 1 in 5 is approximately 13 m. The length of the impervious floor downstream of the crest is approximately 6 m (equation 10.56)).
\text { width of basin, } W_{ B }=V /\left[L_{ B }\left(D_2+d_{ c }\right)\right] \text {, } (10.56)
The upstream floor thickness (nominal thickness of 0.3 m) at the toe of the crest is approximately 1.5 m, and at 5 m from the toe it is approximately 0.14 m; adopt a minimum of 0.3 m.
Appropriate energy-dissipating devices (for large discharges) and upstream and downstream bed protection works must be provided. The detailed layout of the design is shown in Fig. 10.17.
