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Design a regime channel for discharge of 50cumec and silt factor 1.1. Use Laceys theory.
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Solution :

Given :

$Q=50$ cumec

Lacey's silt factor $F=1.1$

Design steps by Lacey's theory :

1) Calculate velocity of flow,

$\begin{aligned} \qquad{V} &=\left(\frac{Q F^{2}}{140}\right)^{(1/6)}\\ &=\left(\frac{50\times 1.1^{2}}{140}\right)^{(1/6)}\\ &=0.8695 m/s \end{aligned}$

Where,

$F=$ silt factor

2) Calculation of area of cross-section of canal,

$\begin{aligned}\qquad A &= \frac{Q}{V} =\frac{50}{0.8695}\\ &=57.50 m^2 \end{aligned}$

3) Calculation of perimeter,

$\qquad P=4.75 \sqrt{Q}$

$ \quad \quad \quad= 4.75 \sqrt{50}$

$ \quad \quad \quad= 33.5876 m$

4) Calculate hydraulic radius,

$\begin {aligned} \qquad R &=\frac{5}{2} \times \frac{{V}^{2}}{{~F}}=\frac{5}{2} \times \frac{{0.8695}^{2}}{1.1}\\ &=1.7183 m \end{aligned}$

5) Calculate bed slope,

$\begin{aligned} \qquad {S} &=\frac{{F}^{(5/3)}}{3340 \times {Q}^{(1/6)}} \\ &=\frac {{1.1}^{(5/3)}}{3340 \times {50}^{(1/6)}} \\ &=\frac{1}{5420} \end{aligned}$

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