Question

A trapezoidal Lacey channel carries $35m^3/sec$ discharge with a sit factor of $0.9$. The side slopes are $1H:2V$. The depth of flow will be ___ m.

• A. $3.45m$
• B. $2.36m$
• C. $1.83m$
• D. $6.5m$

Answer 1

The correct option is C $1.83m$

Given, $Q=35m^3/sec,f=0.9,\text{side slopes}=1H:2V$

Using Lacey’s theory,

(i) Velocity

$={\left(\frac{Q{f}^2}{140}\right)}^{1/6}={\left(\frac{35\times {0.9}^2}{140}\right)}^{1/6}=0.766m/s$

(ii) Area

$A=\frac{Q}{V}=\frac{35}{0.766}=45.692m^2$

(iii) Wetted perimeter

$P=4.75\sqrt{Q}=4.75\sqrt{35}=28.10m$

For trapezoidal channel,

$A=(B+my)y=(B+\frac{y}{2})y$

where $m=\frac{1}{2}$

$\Rightarrow 45.692=(B+0.5y)y....\left(i\right)$

Also,
$P=B+\sqrt{1+m^2}\times y\times 2$

$\Rightarrow P=B+\sqrt{5}\times y$

$\Rightarrow P=B+2.236y$

$\Rightarrow 28.10=B+2.236y$

$\Rightarrow B=28.10-2.236y$

Putting value of B in (i) $45.692=(28.10-2.236y+0.5y)y$

$\Rightarrow 1.736y^2-28.10y+45.692=0$

On solving
$y=1.832,14.35m$

Neglecting y = 14.35 m as it is not feasible

Flow depth,
$\therefore y=1.834m$