Theories of Seepage, Spillways and Miscellaneous

\begin{aligned} y &=0.23m \\ V&=8.1m/s \\ q&=Vy=0.23 \times 8.1 \\ &= 1.863 m^3/s-m\\ S_0&=\frac{1}{800} \\ D&=0.015 \end{aligned}
y_n= Normal depth of flow
R = y for wide rectangular channel
By Manning's equation
\begin{aligned} q&=\frac{y_n}{n}R^{2/3}S_0^{1/2} \\ 1.863&=\frac{y_n^{5/3}}{0.015} \times \left ( \frac{1}{1800} \right )^{1/2} \\ y_n&= 1.108m \end{aligned}
y_1 is conjugate depth of y_0,
\begin{aligned} \frac{y_1}{y_n}&=\frac{-1+\sqrt{1+8Fr_n^2}}{2} \\ \because &\;\;\left ( Fr_n^2=\frac{q^2}{gy_n^3} \right ) \\ y_1&=\frac{-1+\sqrt{1+8 \times \frac{1.863^2}{9.81 \times 1.108^3}}}{2} \times 1.108\\ &=0.418m\simeq 0.42m \end{aligned}

Discharge velocity,
\begin{aligned} V&=ki\\ V&=k\frac{h_1-h_2}{L}\\ &=80 \left ( \frac{55.2-34.1}{2500} \right )\\ &=0.6752 m/day\\ &\text{Porosity}, n =0.25\\ &\therefore \;\;\text{Seepage velocity,}\\ V_s&=\frac{V}{n}=\frac{0.6752}{0.25}\\ &=2.7008 m/day\\ \text{Time taken}&=\frac{L}{V_s}\\ &=\frac{2500}{2.7008}=925.65 days \end{aligned}

Irrigation canal below the bed level of a natural stream
\rightarrow Super passage

Non-modular outlet :These are the outlets whose discharge depends on the difference in water levels in the distributing channel and the water < course. The discharge of such outlets, therefore, varies with the variation of the water levels in the distributing channel and the water course.
Semi-modular outlet : These are the outlets whose discharge varies with the variation of the water level in the distribution channel but it is independent of the water level in the water course, so long as the minimum working head required for their working is available.
Modular outlet : These are the outlets whose discharge is independent of the water levels in the distributing channel and the water course, within reasonable working limits. In other words modular outlets maintain a constant discharge irrespective of variation of the water levels in the distributing channel and the water course.

\begin{aligned} \phi_{\mathrm{R}} &=\frac{2.75}{4} \times 100=68.75 \% \\ \phi_{\mathrm{P}} &=100-\phi_{\mathrm{R}}=31.25 \% \\ \text { Now, } \phi_{p} &=\frac{\text { Pressure head at point } \mathrm{P}}{\text { Total head }} \times 100 \\ 31.25 &=\frac{h}{4} \times 100 \\ \therefore \quad h &=1.25 \mathrm{m} \end{aligned}