A Dam is constructed cross a trapezoidal channel having bottom width $b=25m_{?}$ and $2$:$1_{?}$ side slope,

longitudinal slope of $12\frac{cm}{k}{m}_{,}n={0\mathrm{.}025}_{?}$ and carries a discharge of $250\frac{m^3}{s_{?}}.$

$$ Choose one of these values of $y_n(5,90-4\mathrm{.}94-6\mathrm{.}7),$ theoretically.

$$ Choose one of these values of $y_c(1\mathrm{.}3-4\mathrm{.}07-2\mathrm{.}05),$ theoretically.

$$ It's required to define the type of slope.

Where:

Slope is critical if $y_n=y_c,$ Slope is mild if $y_n>y_c,$ Slope is steep if $Q=\frac{1}{n}\frac{A^{\frac{5}{3}}}{P^{\frac{1}{2}}}{S}^{\frac{1}{2}},A=y_n(b+2y_n),P=b+4\mathrm{.}4y{n}_n*\frac{Q^2}{g}=\frac{A^3}{T}20kmb=12(m{)}_{?}3$:$210c\frac{m}{k}m,n=0\mathrm{.}0340\frac{m^3}{s}3\mathrm{.}8m1\mathrm{.}01y_n{y}_n(2\mathrm{.}5-3\mathrm{.}67-1\mathrm{.}1)y_c(1\mathrm{.}0-5\mathrm{.}2-3\mathrm{.}4)Kn=4000\frac{m^3}{s}Zc=13m^{\frac{5}{2}}{y}_n=y_c{y}_n>y_{c}Q=\frac{1}{n}{A}^{\frac{5}{3}}{S}^{\frac{1}{2}}{S}^{\frac{1}{2}},A=y_n(b+1\mathrm{.}5y_n),P=b+3\mathrm{.}61y_n,\frac{Q^2}{g}=\frac{A^3}{(T)}$

$\frac{x}{y}=\frac{1}{S_o}\frac{1-(\frac{Z_c}{z}{)}^2}{1-(\frac{K_n}{K}{)}^2},K=\frac{A}{n}{R}^{\frac{2}{3}},T=12+3y,Z=\sqrt[\phantom{2}]{\frac{A^3}{T}},y=y1-y2$

$x=(\frac{x}{y})av,xy{y}_n$

$Q=\frac{1}{n}{A}^{\frac{5}{3}}{S}^{\frac{1}{2}}{S}^{\frac{1}{2}},A=y_n(b+1\mathrm{.}5y_n),P=b+3\mathrm{.}61y_n,\frac{Q^2}{g}=\frac{A^3}{(T)}$

$\frac{x}{y}=\frac{1}{S_o}\frac{1-(\frac{Z_c}{z}{)}^2}{1-(\frac{K_n}{K}{)}^2},K=\frac{A}{n}{R}^{\frac{2}{3}},T=12+3y,Z=\sqrt[\phantom{2}]{\frac{A^3}{T}},y=y1-y2$

$x=(\frac{x}{y})av,xy{y}_n$

$Q=\frac{1}{n}\frac{A^{\frac{5}{3}}}{P^{\frac{1}{2}}}{S}^{\frac{1}{2}},A=y_n(b+2y_n),P=b+4\mathrm{.}4y{n}_n*\frac{Q^2}{g}=\frac{A^3}{T}$

A weir $is$ constructed cross a trapezoidal channel having length $20km,$ bottom width $b=12(m{)}_{a}nd$

$3$:$2$ side slope, longitudinal slope $of10c\frac{m}{k}m,n=0\mathrm{.}03$ and carries a discharge $of40\frac{m^3}{s},$ depth

$of$ water upstream the weir $is3\mathrm{.}8m,$ assume that $BWC$ begin with $1\mathrm{.}01y_{n}as$ water depth. Its required

$to$ apply following,

$$ Choose one $of$ these values $of{y}_n(2\mathrm{.}5-3\mathrm{.}67-1\mathrm{.}1),$ theoretically.

$$ Choose one $of$ these values $of{y}_c(1\mathrm{.}0-5\mathrm{.}2-3\mathrm{.}4),$ theoretically.

$$ It's required $to$ define the type $of$ slope.

$IfKn=4000\frac{m^3}{s}$ and $Zc=13m^{\frac{5}{2}},$ calculate Maximum length according water profile

Where:

Slope $is$ critical $if{y}_n=y_c,$ Slope $is$ mild $if{y}_n>y_c,$ Slope $is$ steep $if{y}_n$

$Q=\frac{1}{n}{A}^{\frac{5}{3}}{S}^{\frac{1}{2}}{S}^{\frac{1}{2}},A=y_n(b+1\mathrm{.}5y_n),P=b+3\mathrm{.}61y_n,\frac{Q^2}{g}=\frac{A^3}{(T)}$

$\frac{x}{y}=\frac{1}{S_o}\frac{1-(\frac{Z_c}{z}{)}^2}{1-(\frac{K_n}{K}{)}^2},K=\frac{A}{n}{R}^{\frac{2}{3}},T=12+3y,Z=\sqrt[\phantom{2}]{\frac{A^3}{T}},y=y1-y2$

$x=(\frac{x}{y})av,xy$