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Fluid Dynamics

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Fluid Dynamics is a branch of Fluid mechanics which stresses more on the fluid flow. Fluid resistance refers to forces that oppose the relative motion of an object through a fluid, a liquid or gas. Fluid mechanics is branch of physics that consisting fluid statics, kinematics and fluid dynamics. Fluid Dynamics gives the basics of how fluids behave in motion. It relates to velocity and acceleration when force is applied on it. 

A fluid may be a liquid, vapor, or gas. Motion of a fluid subjected to unbalanced forces or stresses. Almost all liquid flows and many gas flows are thus treated as incompressible. The motion continues as long as unbalanced forces are applied. Flow rate gives the rate at which the fluid flows. It is the ratio of the volume of fluid ($\Delta$ V) flowing in a given time ($\Delta$ t). In short, the fluid flow is all about substances such as liquid or gas that can flow having no fixed shape, and offers little resistance to an external pressure.
Fluid Flow Chart
It is given by
Fluid FlowFluid flow can be categorized into:
  • Streamline flow
  • Turbulent flow.
Streamline flow is an orderly flow of fluid. The flow rates of the fluid at any point will be the same for a given time whereas turbulent flow is a disorderly flow of fluid. The flow rate at any point varies with time.

Note: Fluid flow depends on the type of fluid.
For incompressible fluid $A_1$ $v_1$ = $A_2$ $v_2$ and for compressible fluid $p_1$ $A_1$ $v_1$ = $p_2$ $A_2$ $v_2$.
Where $A_1$ and $A_2$ are areas
           $p_1$ and $p_2$ are pressures
           $v_1$ and $v_2$ are volumes

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Fluid Dynamics Equations

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Fluid dynamics equations varies as per the fluid.
For incompressible fluid the equation is
Incompressible Fluid Equation
For a compressible fluid the equation is
Compressible Fluid Equation
where,

$\frac{P}{\rho g}$ = pressure energy per unit weight fluid or pressure head

$\frac{v^2}{2g}$ = kinetic energy per unit weight or kinetic head

z = potential energy per unit weight or potential head
P = pressure
$\rho$ = Density
K= constant
This is called Bernoulli's Equation.
We often see a leaf floating in water, ship sailing in water. How could this happen? This is what we call Buoyancy. It is the upward force experienced by any body. The point where the force of Buoyancy acts is called the Center of Buoyancy.

It is the product of density and volume given as
BuoyancyWhere $\rho$ is the density and v is the volume.
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The Archimedes Principle states that if a body is immersed in a liquid completely or partially it experiences an upward force equal to the weight of the fluid that the body displaces. It also gives the measure of buoyancy that is equal to weight of the displaced fluid.

Archimedes PrincipleThe buoyancy force depends on fluid displaced, if the weight of the body is more than that of the fluid displaced, the body will sink in water.
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This concept was given by Archimedes. It tells how much mass is concentrated in a unit volume. It is defined as
Formula for DensityIt is represented by symbol $\rho$. Its unit is kg/m3.
→ Read More Bernoulli equation is the statement that justifies the law of conservation of energy. It tells that energy per unit volume before flow is equal to energy per unit volume after the flow.
Bernoullis Principle
It is given as
Bernoullis PrincipleHere P1 and P2 are the given pressures before and after the flow,
v1 and v2 are velocities of the two fluids,
h1 and h2 are the height at which fluid flows,
$\rho$ is the density of the fluid which is constant.
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Fluid Dynamics Problems

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Lets go through some problems on Fluid dynamics:

Solved Examples

Question 1: Calculate the height of water column if it is moving with the same pressure of that of a 76 cm column of mercury.
Solution:
 
Since the pressure is same in both the cases
P = $\rho_1$ g h1 = $\rho_2$ g h2
Density of water $\rho_1$ = 103 kgm-3
Density of mercury $\rho_2$ = 13.6 $\times$ 103 kgm-3,
Height of mercury h2 = 76 cm = 0.76 m

The height h1 = $\frac{\rho_2 h_2}{\rho_1}$

                    = $\frac{13.6 \times 10^3 kg m^{-3} \times 0.76 m}{10^3 kg m^{-3}}$

                    = 10.34 m.

 

Question 2: An incompressible fluid has a diameter of 0.2 m in the beginning and 0.8 m at the end. If it has a volume of 180 m3 in the beginning what will be its volume in the end?
Solution:
 
Given: Diameter d1 = 0.2 m => r1 = 0.1 m
Area A1 = $\pi$ r12 = $\pi$ $\times$ (0.1)2 = 0.03142 m2
Diameter d2 = 0.8 m => r2 = 0.4 m
Area A2 = $\pi$ r22 = $\pi$ $\times$ (0.4)2 = 0.502 m2
For incompressible fluid A1v1 = A2v2

Volume v2 = $\frac{A_1 v_1}{A_2}$

               = $\frac{0.03142 m^2 \times 180 m^3}{0.502 m^2}$

               = 11.267 m3.