Question

conditions for most economical trapezoidal section​

Answer 1

The base of the most economical trapezoidal channel section is 6m and the side slope is 1H:2V, calculate the maximum discharge through the channel if the bed slope is 1 in 1000 and C = 50.

Answer 2

The most economical section of a trapezoidal channel will be a half-hexagon.

Let

b= Base width of the channel,

y= Depth of flow, and

θ= Angle made by the sides with horizontal.

Side slope =1 vertical to n horizontal.

Area of flow,

$A=\frac{(AB+CD)}{2} Xy= \frac{b(b+2ny)}{2} Xy=(b+ny)y$

$\frac{A}{y} = b+ny\\or, b= \frac{A}{y} -ny$

Wetted perimeter,

P= AD+AB+BC= AB+2BC (∵AD=BC)

$or, P= b+2\sqrt{BE^{2}+CE^{2} }= b+2\sqrt{n^{2} y^{2} +y^{2} }= b+2y\sqrt{n^{2} +1}$

Substituting the value of b, we get,

$P=\frac{A}{y} -ny+2y\sqrt{n^{2}+1 }$

The section of the channel will be most economical when its wetted perimeter (P) is minimum,

$i.e. \frac{dP}{dy} =0$

$or, \frac{d}{dy} [\frac{A}{y}-ny+2y\sqrt{n^{2}+1] } =0\\or, -\frac{A}{y^{2} } -n+2\sqrt{n^{2}+1]}=0\\or, \frac{A}{y^{2} } +n= 2\sqrt{n^{2}+1}$

Substituting the value of A, we get,

$\frac{(b+ny)y}{y^{2} } +n= 2\sqrt{n^{2}+1 } \\or, \frac{(b+ny)}{y} +n= 2\sqrt{n^{2}+1 } \\or, \frac{b+ny+ny}{y} =2\sqrt{n^{2}+1} \\or, \frac{b+2ny}{y}=2\sqrt{n^{2}+1}\\or, \frac{b+2ny}{2}=y\sqrt{n^{2}+1}$

[i.e. Half of the top width = One of the sloping sides]

Hydraulic radius, R=A/P

We know that,

A= (b+ny)y

$P=ib+2y\sqrt{n^{2}+1 } \\But,2y\sqrt{n^{2}+1 }=b+2ny\\or, P=b+(b+2ny)= 2(b+ny)$

Therefore, Hydraulic radius,

$R=\frac{(b+ny)y}{2(b+ny)} =\frac{y}{2}$

i.e., hydraulic radius equals half the flow depth.

Thus a circle with a center O and radius equal to the depth of flow will be tangential to the three sides of a most economical trapezoidal section; this condition stipulates that the most economical section of a trapezoidal channel will be a half-hexagon.

#SPJ3