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.
It is given by
Fluid 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
For incompressible fluid the equation is
$\frac{P}{\rho g}$ = 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
It is the product of density and volume given as
Where $\rho$ is the density and v is the volume.
→ Read More
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.
The 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.
→ Read More
This concept was given by Archimedes. It tells how much mass is concentrated in a unit volume. It is defined as
It is represented by symbol $\rho$. Its unit is kg/m^{3}.
→ 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. Here P_{1} and P_{2} are the given pressures before and after the flow,
v_{1} and v_{2} are velocities of the two fluids,h_{1 }and h_{2 }are the height at which fluid flows,
$\rho$ is the density of the fluid which is constant.
→ Read More 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 h_{1 }= $\rho_2$ g h_{2}Density of water_{ }$\rho_1$ = 10^{3 }kgm^{3}_{}Densit_{}y of mercury $\rho_2$ = 13.6 $\times$ 10^{3} kgm^{3},
Height of mercury h_{2} = 76 cm = 0.76 m
The height h_{1} = $\frac{\rho_2 h_2}{\rho_1}$
= $\frac{13.6 \times 10^3 kg m^{3} \times 0.76 m}{10^3 kg m^{3}}$
= 10.34 m.
_{}
Solution:
Since the pressure is same in both the cases
P = $\rho_1$ g h_{1 }= $\rho_2$ g h_{2}Density of water_{ }$\rho_1$ = 10^{3 }kgm^{3}_{}Densit_{}y of mercury $\rho_2$ = 13.6 $\times$ 10^{3} kgm^{3},
Height of mercury h_{2} = 76 cm = 0.76 m
The height h_{1} = $\frac{\rho_2 h_2}{\rho_1}$
= $\frac{13.6 \times 10^3 kg m^{3} \times 0.76 m}{10^3 kg m^{3}}$
= 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 d_{1} = 0.2 m => r_{1} = 0.1 m
Area A_{1} = $\pi$ r_{1}^{2} _{}= $\pi$ $\times$ (0.1)^{2} = 0.03142 m^{2}
Diameter d_{2} = 0.8 m => r_{2} = 0.4 m
Area A_{2} = $\pi$ r_{2}^{2} = $\pi$ $\times$ (0.4)^{2} = 0.502 m^{2}
For incompressible fluid A_{1}v_{1} = A_{2}v_{2}
Volume v_{2} = $\frac{A_1 v_1}{A_2}$
= $\frac{0.03142 m^2 \times 180 m^3}{0.502 m^2}$
= 11.267 m^{3}.
Solution:
Given: Diameter d_{1} = 0.2 m => r_{1} = 0.1 m
Area A_{1} = $\pi$ r_{1}^{2} _{}= $\pi$ $\times$ (0.1)^{2} = 0.03142 m^{2}
Diameter d_{2} = 0.8 m => r_{2} = 0.4 m
Area A_{2} = $\pi$ r_{2}^{2} = $\pi$ $\times$ (0.4)^{2} = 0.502 m^{2}
For incompressible fluid A_{1}v_{1} = A_{2}v_{2}
Volume v_{2} = $\frac{A_1 v_1}{A_2}$
= $\frac{0.03142 m^2 \times 180 m^3}{0.502 m^2}$
= 11.267 m^{3}.