Fluid Mechanics and Hydraulics

Minimum velocity at bottom $V=\sqrt{5g\ell}$ $=\sqrt{5\times 10\times 0.5}$$=5:m/s$

Water
flowed by the pipe in lh = $\pi r^2h$
= $\dfrac{22}{7} \times$ $\dfrac{7\times 7}{100\times100}$ $\times 5000 m^3 =77m^3$
Volume
of expected water in the tank = $\frac{50 \times 44 \times 7}{100} = 154
m^3$ 
Required
time= $154/77 = 2 hrs$.

Gravity does cause things to fall with increasing speed, but as they speed up viscous force by air increases increases. Since according to stokes law $F _r \propto v$.  Eventually the acceleration stops and the raindrop reaches a constant terminal velocity.

In deep water temperature decreases, viscosity of water increases with decrease in temperature. So, viscosity of water at constant temperature is more in deep water as temperature is less in deep waters.

Fluid friction increases with the depth of a flowing liquid.

DPD of cell A=5A=5 bars
DPD of cell B=5−5=0B=5−5=0 bars
DPD of cell C=10−4=6C=10−4=6 bars
DPD of cell D=8−3=5D=8−3=5 bars
As cell B has the lowest value of DPD, therefore water will move from cell B to cells A, C and D.

A suction at the end of the tube draws air from further up the tube, and that in turn draws air from further up the tube and so on. So the result is that a pulse of high pressure air travelling down the tube is reflected as a pulse of low pressure air traveling up the tube. So the pressure wave has been reflected at the open end, with a change in phase of 180. In the open-open pipe, there is such a reflection at both ends. So the above argument is false.

Mechanical advantage of a hydraulic press is the ratio of the area of the press cylinder to the area of the pump cylinder. Since Area = pi * r^2, and the radius of the press cylinder (R) is equal to the diameter of the pump cylinder (d = 2r), then R = 2r. The ratio of areas is (pi * R^2) / (pi * r^2) = (2r)^2 / r^2 = 4.

acceleration due to gravity $g = 10m/sec^2$

The viscous drag force can be written as below:
$F =  \eta \ A \ (\cfrac{dV}{dX})$ where, F is viscous drag force (or tangential force)
$\eta$ is coefficient of viscosity
$A$ is surface area of layers in contact
$\dfrac{dV}{dX} $ is velocity gradient
$\eta$ is property of liquid.
So, we can see that drag is directly proportional to the velocity gradient, surface area and nature of liquid. Drag force acts in opposite direction of liquid flow. 

Given,

Area, $A=\dfrac{\pi {{d}^{2}}}{4}=\dfrac{\pi }{4}{{\left( 0.02 \right)}^{2}}$

Volume rate of flow, $\dot{V}=\dfrac{volume}{time}=\dfrac{20\,L}{60}=\dfrac{{{10}^{-3}}}{3}\,{{m}^{3}}{{s}^{-1}}$

Volume rate of flow = Cross-Section Area x Velocity of Flow

$ \dot{V}=Av $

$v=\dfrac{{\dot{V}}}{A}=\dfrac{{{10}^{-3}}}{3}\times \dfrac{4}{\pi {{\left( 0.02 \right)}^{2}}}=1.06\,m{{s}^{-1}}=106\,cm{{s}^{-1}}$

Hence, velocity of water leaves hose is$106\,cm{{s}^{-1}}$.

Poiseuille's law states Q = (pi * P * r^4) / (8 * eta * L). Solving for eta: eta = (pi * P * r^4) / (8 * Q * L). Plugging in P=20, r=1.5*10^-3, Q=10^-7, L=0.1, we get eta = (3.14 * 20 * (1.5*10^-3)^4) / (8 * 10^-7 * 0.1) = 4 * 10^-3 Pa-s.

The Newton's viscous force acting between two liquid surfaces with relative velocity $dv$ and distance $dy$ between the layers is given as $-\eta A\dfrac{dv}{dy}$

Shearing stress  $=\eta \dfrac{dv}{dy}=1.5 \times 10^{-3} N /m^{2}$

The profile of the velocity of fluid as shown in the figure The velocity of the layer of fluid ,which in contact with metal plates (fixed), is zero. As we move towards the centre from either plate the velocity of the layer of fluid increases and it becomes maximum at the location of moving plate. This maximum value is same as that of the velocity of plate