A rectangular channel 4m wide has depth of water1.5m. The slope of the bed of channel is 1 in 1000. C=55. It is desired to increase the discharge keeping i & C constant.Find the new dimensions and increase in discharge.

Given,

• b = 4m

• d = 1.5m

• i = ${1 \over1000}$

• C = 55

$i.e. Area = b * d = 4 * 1.5 = 6m^2$

$ Wetted$ $ Perimeter, P = d + b + D = 1.5 + 4 + 1.5 = 7m$

$ m = {A \over P} = {4 \over 7} = 0.857$

$ Q = AC\sqrt {mi} = 6 * 55 \sqrt{0.857 * {1 \over 1000}} = 9.66 m^3/s$

$ For$ $max$ $discharge$ $for$ $a$ $given$ $area,$ $slope$ $of$ $bed$ $and$ $roughness. $

Let,

$b^1 = new$ $width$ $of$ $channel$

$d^1 = new$ $depth$ $of$ $flow$

$ Area, A = b^1 * d^1,$ $ where$ $A = 6m^2$

$B = b^1 * d^1$

$Max$ $discharge,$ $b^1 = 2d^1$

$6 = 2d^1 * d^1$

$i.e. d^{{1}^{2}} = {6 \over 2} = 3$

$i.e. d^1 = \sqrt 3 = 1.732$

$b^1 = 2*1.732 = 3.464$

$New$ $dimension,$ $b^1 = 3.464m$ $and$ $d^1 = 1.732m$

$ Wetted$ $ Perimeter, p^1 = d^1 + b61 + d^1 = 1.732 + 3.464 + 1.732 = 6.928$

$ Hydraulic$ $ mean$ $ depth, m^1 = {A \over P^1} = {6 \over 6.928} = 0.866m$

$\left[\begin{array}{lll}m^1 = {d^{-1} \over 2} = {1.73 \over 2} = 0.866m\end{array}\right]$

$ Max$ $ discharge,$, $Q^1 = AC\sqrt {m^1i} = 6*55*\sqrt {0.866*{1 \over 1000}} = 9.71m^3/s$

$ Increase$ $in$ $ discharge,$, $Q^1 - Q = 9.71 - 9.66 = 0.05m^3/s$ $Ans.$