Question

What are the best dimensions for a trapezoidal canal having side slopes 1(vert.) on 3 (horiz.) and n of 0.020 if it is to carry 40m/s uniformly on a slope of 0.009? Water depth= m Width of channel bed B = m

Answer 1

TOP WIDTH (T) FLOW DEPTH (y) 1 m BOTTOM WIDTH (B)

A channel is said to be the best channel if it is a economical channel.

Conditions for the trapezoidal channel to be economical are as follows

у Hydraulic Depth(R) 2

&

Top Width(T) = 2yVm2 +1

$\mathbf{Where \, T = B+2my}$

Above two conditions to be satisfied for the trapezoidal channel to be economical .

where y = depth of flow in trapezoidal channel

Where A = area of cross-section

n= manning's co-efficient

S_o = Slope of bed

Q= Discharge in channel

where B = Bed Width

m = side slope ( ie 1 vertical:m horizontal)

So From Second condition for economical trapezoidal channel

Where, B + 2my = 2yVm2 +1

$\mathbf{Where, \, B+2*3*y = 2y\sqrt{3^2+1}}$

$\mathbf{Where, \, B=0.32455*y}$

$\mathbf{Now \,Area \,of \,a \,trapezoidal \,section \,(A)=(B+my)y}$

$\mathbf{ \,(A)=(0.32455*y+3*y)y=3.32455*y^2}$

Now , By manning's Eqn

$\mathbf{Q=\frac{A*R^{\frac{2}{3}}\sqrt{S_o}}{n}}$

$\mathbf{Q=\frac{(3.32455*y^2)*(y/2)^{\frac{2}{3}}\sqrt{S_o}}{n}.......(\because \,R=\frac{y}{2})}$

$\mathbf{Q=\frac{3.32455*(y)^{\frac{8}{3}}\sqrt{S_o}}{n*2^{\frac{2}{3}}}.......(\because \,R=\frac{y}{2})}$

$\mathbf{40=\frac{3.32455*(y)^{\frac{8}{3}}\sqrt{0.009}}{0.020*2^{\frac{2}{3}}}.......(\because \,R=\frac{y}{2})}$

$\mathbf{After \,solving\, We \,get\,,y=1.6859\,m}$

$\mathbf{B=0.32455*y=0.32455*1.6859=0.5472\,m}$

Hence water depth (y) = 1.6859 m Bed width (B) = 0.5472 m

For any query further,Please feel free to ask me in the comment section below..i will be much happy to help you ...and please dont forget to like .