written 5.2 years ago by |

A section of a channel is said to be most economical when the cost of construction of the channel is minimum, but the cost of construction of channel depends on excavation and the lining, to keep the cost minimum the wetted perimeter should be minimum for a given discharge.

1] Most economical Rectangular channel:

Let b = width of channel,

d = depth of flow,

A = b x d - - - - - (1)

wetted perimeter, P = d + b + d

= b + 2d - - - - (2)

from equation (1), $b = \frac{A}{d}$

Substituting the value of 'b' in equation (2),

P = b + 2d

$= \frac{A}{d} + 2d$ - - - - (3)

for most economical section, P should be minimum for a given area.

OR $\frac{dP}{d(d)} = 0$

difference the equation (3) with respect to 'd' and equating the same to zero, we get,

$\frac{d}{d(d)} [ \frac{A}{d} + 2d] = 0 $

$\frac{-A}{d^2} + 2 = 0$

$A = 2d^2$

But A = b x d

$\therefore$ $b \times d = 2d^2$

b = 2d

Now hydraulic mean depth,

$m = \frac{A}{P}$

$= \frac{ b \times d}{b + 2d}$

$= \frac{2d \times d}{2d + 2d}$

$= \frac{2d^2}{4d}$

$[ m = \frac{d}{2}]$