Boundary Layer Theory, Drag and Lift

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}

In case of discrete particle settling and Stoke's law valid, at terminal velocity, since there is no change in velocity, the net force on the body is zero. Hence,


F_D=\left ( \frac{\pi}{6}D^3\rho _sg-\frac{\pi}{6}D^3\rho _sg \right )=\frac{\pi}{6}gD^3(\rho _s-\rho _f)
For density of medium (\rho _f) and mass density of sphere (\rho _s) constant,
Drag force (F_D)\propto D^3
For particle P, diameter (D_P)=0.5 mm
For particle Q, diameter (D_Q)=1 mm
\Rightarrow \frac{(F_D)_Q}{(F_D)_P}=\frac{(D_Q)^3}{(D_P)^3}=\left ( \frac{1}{0.5} \right )^3=8
\Rightarrow (F_D)_Q=8 \times (F_D)P

Given:
\begin{aligned} \frac{u}{u_\infty }&=\sin \left ( \frac{\pi}{2} \frac{y}{\delta }\right )\\ u_\infty &=0.3 m/s\\ \delta &=1m\\ \frac{du}{dy}&=\frac{d}{dy}u_\infty \cdot \sin\left ( \frac{\pi}{2} \frac{y}{\delta } \right )\\ &=\frac{u_\infty \cdot \pi}{2\delta } \cos \left ( \frac{\pi}{2} \frac{y}{\delta } \right )\\ \text{At y=0 and} & \; \delta=1 \\ \left.\begin{matrix} \frac{du}{dy} \end{matrix}\right|_{y=0}&=\frac{0.3 \pi}{2(1)} \cos \left ( \frac{\pi}{2}\frac{0}{1} \right )\\ &=0.47 s^{-1} \end{aligned}

Given flow is a couette flow with pressure gradient
\mu =0.01N-s/m^2
h=5mm
if \tau _{y=0}=0, \; \text{find}\; \frac{dp}{dx}
The velocity distribution, for the couette flow with pressure gradient is given by
u=\frac{wy}{h}-\frac{1}{2\mu }\frac{dp}{dx}[hy-y^2]
where w is the velocity of the top plate and h is the gap between the plates
\begin{aligned} \frac{du}{dy}&=\frac{w}{h}-\frac{1}{2\mu }\frac{dp}{dx}[h-2y] \\ \left.\begin{matrix} \frac{du}{dy} \end{matrix}\right|_{y=0} &=\frac{w}{h}-\frac{1}{2\mu }\frac{dp}{dx}h \\ \tau _{y=0}&=\mu \left [ \frac{w}{h}-\frac{1}{2\mu }\frac{dp}{dx}h \right ] \\ \tau _{y=0}&=\frac{w\mu }{h}-\frac{1}{2}\frac{dp}{dx}h \\ \text{when}\; \tau _{y=0}&=0, \; \text{we will have}\\ \frac{1}{2}\frac{dp}{dx}h&=\frac{w\mu }{h}\\ \frac{dp}{dx}&=\frac{2w\mu }{h^2} \\ &=\frac{2 \times 1 \times 10^{-2}\times 0.01}{(5 \times 10^{-3})^2} \\ &=8 N/m^2 \; \text{per} \; m \end{aligned}

Power required to overcome the drag = Drag force \times velocity
\begin{aligned} &=520 \times \frac{120 \times 1000}{60 \times 60}= 17333.33 \mathrm{watt} \\ &\quad (1 \mathrm{mhp}=735.5 \mathrm{watt})\\ &=\frac{17333.33}{735.5}\text{ metric horse power }\\ &=23.567 \text{ metric horse power} \end{aligned}