Question

A trapezoidal channel is to be designed to carry a discharge of 200 $m^3/sec$. The channel slope is 0.0004 and manning coefficient is 0.012. the side slope is 1: 1, if the rate of linings is 100/$m^2$, then the cost for most economical section per km length of channel is (in lakhs).

• A. 20.88

Answer 1

The correct option is A 20.88
Given, $Q=200m^3/sec,S=0.0004,n=0.012$

side slope = $1:1\Rightarrow m=1$

For most economical section, R = y/2

and B+2my = $2y\sqrt{1+m^2}$

$B+2y=2y\sqrt{1+1}$

B = 0.828y

Area of flow, A = (B+my)y

$=1.828y^2$

Using manning equation,

$Q=\frac{1}{n}A{R}^{2/3}{S}^{1/2}$

$200=\frac{1}{0.012}\times \left(1.828y^2\right){\left(\frac{y}{2}\right)}^{2/3}\sqrt{0.0004}$

y = 5.71 m

$\therefore$ B = $0.828\times 5.71=4.73m$

Perimeter of the economical section
= $B+2y\sqrt{1+m^2}$

$=4.73+2\times 5.71\sqrt{2}$

= 20.88 m

$Costperkm=20.88\times 100\times 1000$

= 2088031.9 = 2088032

= 20.88 lakhs