Fluid Mechanics and Hydraulics

P_{H}= Total pressure on the projected are of the curved surface on the vertical plane.
P_{H} is equal to the total pressure exerted by the liquid on an imaginary vertically immersed plane surface which is the vertical projection of the curved surface, it will act at the center of pressure of the lane surface.
P_{v}= The weight o the liquid contained in the portion extending vertically above the curved surface upto the free surface of the liquid.
P_{V} will act through the centre of gravity of the volume of liquid contained in the portion extending above the curved surface upto the free surface of the liquid (represented by the profile ABCDEFA in the present case).

Momentum correction factor (\beta)=\frac{1}{A V^{2}} \int \mathrm{v}^{2} d A

{Sink Flow :}
The sink flow is the flow in which fluid moves radially inwards towards a point where it disappears at a constant rate. This flow is just opposite to the source flow.
Figure shows a sink flow in which the fluid moves radially inwards towards point \mathrm{O}, where it disappears tat a constant rate.
The pattern of stream lines and equipotential lines of a sink flow is the same as that of source flow.
All the equations derived for a source flow shall hold to good for sink flow also except that in sink flow equations, q is to be replaced by (-q)

Boundary layer thickness (Normal Boundary layer thickness disturbance thickness.
Distance from the boundary surface in which the velocity reaches 99 \% of the stream velocity is the boundary layer thickness.

Turbulent boundary layer on smooth plate.
\frac{\delta}{x}=\frac{0.376}{\left(R_{e x}\right)^{1 / 5}}, \quad R_{e x}=\frac{v_{0} x}{v}
[Valid for R_{e x} \gt 5 \times 10^{5} and R_{e x} \lt 10^{7} ]
=\frac{0.22}{\left(R_{e x}\right)^{1 / 6}}, 10^{7} \lt R_{e x} \lt 10^{9}

Specific Energy (E)=y+\frac{v^{2}}{2 g}
\Rightarrow \quad E=y+\alpha \cdot \frac{Q^{2}}{2 g A^{2}}\left[\right. as \left.\frac{Q}{A}=v\right]

For critical flow condition,
\begin{aligned} \frac{d E}{d y} & =0 \\ \Rightarrow \quad \frac{d E}{d y} & =1+\frac{\alpha Q^{2}}{2 g} \frac{d}{d y}\left(A^{-2}\right) \\ \Rightarrow \quad \frac{d E}{d y} & =1+\frac{\alpha Q^{2}}{2 g}\left(-\frac{2}{A^{3}}\right) \frac{d A}{d y} \end{aligned}
'dA' can be written as 'Tdy' \Rightarrow \frac{d A}{d y}=T
\Rightarrow \quad \frac{\mathrm{dE}}{\mathrm{dy}}=1-\frac{\alpha \mathrm{Q}^{2} \mathrm{~T}}{\mathrm{gA} \mathrm{A}^{3}}
\Rightarrow 1-\frac{\alpha Q^{2} \mathrm{~T}}{g A^{3}}=0
\Rightarrow \quad \frac{\alpha Q^{2}}{g}=\frac{\mathrm{A}^{3}}{\mathrm{~T}}

At critical flow, A=A_{C};
\mathrm{T}=\mathrm{T}_{\mathrm{C}}
\Rightarrow \quad \frac{\alpha Q^{2}}{g}=\frac{A_{C}^{3}}{T_{C}}

Equating pressure at \mathrm{A}-\mathrm{A}^{\prime}
\begin{aligned} & P_{a t m}+\rho_{B} \gamma_{w} \times \frac{25}{100}=P_{x}-\rho_{A} \gamma_{w} \times \frac{75}{100} \\ & \therefore \quad P_{x}=P_{a t m}+\rho_{B} \gamma_{w} \times 0.25+\rho_{A} \gamma_{w} 0.75 \end{aligned}
\begin{gathered} =101.3+13.6 \times 9.81 \times 0.25+0.8 \times 9.81 \times 0.75 \\ =140.54 \mathrm{kN} / \mathrm{m}^{2} \end{gathered}

Given,
\begin{aligned} & B=1.0 \mathrm{~m} \\ & y_{1}=0.2 \mathrm{~m} \\ & y_{2}=1.0 \mathrm{~m} \\ & g=10 \mathrm{~m} / \mathrm{s} \end{aligned}

The value of specific force
\begin{aligned} & =\frac{\mathrm{P}+\mathrm{M}}{\gamma_{\mathrm{w}}}=\text { constant } \\ & =\frac{\mathrm{P}_{1}+\mathrm{M}_{1}}{\gamma_{\mathrm{w}}}=\frac{\mathrm{P}_{2}+\mathrm{M}_{2}}{\gamma_{\mathrm{w}}} \\ & =A \overline{\mathrm{y}}_{1}+\frac{\mathrm{Q}^{2}}{\mathrm{Ag}} \end{aligned}

For horizontal, frictionless rectangular channel
\therefore \quad \frac{2 q^{2}}{g}=y_{1} y_{2}\left(y_{1}+y_{2}\right)
\begin{aligned} Q & =\sqrt{\frac{1 \times 0.2 \times(1.2) \times 10}{2}} \quad(B=1 \mathrm{~m}) \\ & =1.095 \mathrm{~m}^{3} / \mathrm{s} / \mathrm{m} \\ Q & =1.095 \mathrm{~m}^{3} / \mathrm{s} \end{aligned}
Now, specific force =0.2 \times 1 \times \frac{0.2}{2}+\frac{1.095^{2}}{1 \times 0.2 \times 10} =(0.62) \mathrm{m}^{3}