Physics > Fluid Mechanics > 2.0 Fluid pressure
Fluid Mechanics
1.0 Introduction
1.1 Ideal liquid
1.2 Density of a liquid $\left( \rho \right)$
1.3 Relative density of a liquid $(RD)$
1.4 Density of a mixture of two or more liquid
1.5 Density variation with temperature
1.6 Density variation with pressure
2.0 Fluid pressure
2.1 Atmospheric pressure
2.2 Pressure variation with depth
2.3 Measurement of pressure
2.4 Pressure difference in accelerating fluids
3.0 Pascal's law
4.0 Buoyant force
5.0 Flow of fluids
6.0 Viscosity
7.0 Stoke's law
8.0 Intermolecular forces
9.0 Angle of contact
2.2 Pressure variation with depth
1.2 Density of a liquid $\left( \rho \right)$
1.3 Relative density of a liquid $(RD)$
1.4 Density of a mixture of two or more liquid
1.5 Density variation with temperature
1.6 Density variation with pressure
2.2 Pressure variation with depth
2.3 Measurement of pressure
2.4 Pressure difference in accelerating fluids
Pressure at any level in the atmosphere is equal to the total weight of the air above unit area. So, the pressure decreases with increase in altitude.
Similarly, pressure at any level in the liquid is equal to the total weight of the liquid above a unit area. So, the hydrostatic pressure increases with increase in depth.
Consider a section of liquid of height $h$ and cross sectional area $A$ as shown in the figure.
Let pressure at point $A$ and point $B$ is $P_1$ and $P_2$ respectively.
Mass of the liquid section is, $$m = \rho Ah$$
From FBD, $$\begin{equation} \begin{aligned} {P_1}A + mg = {P_2}A \\ \left( {{P_2} - {P_1}} \right)A = \left( {\rho Ah} \right)g \\ {P_2} - {P_1} = \rho gh \\ {P_2} = {P_1} + \rho gh \\\end{aligned} \end{equation} $$
So, hydrostatic pressure for a point at depth $h$ below the surface of liquid of density $\rho $ is given by, $$P = {P_O} + \rho gh$$
Note:
The pressure $\left( {P = {P_O} + \rho gh} \right)$ is known as absolute pressure at that point
The pressure difference between hydrostatic pressure and atmospheric pressure is known as gauge pressure $$P - {P_O} = \rho gh$$ Relation between absolute pressure and gauge pressure can be written as, $$\begin{equation} \begin{aligned} \\ {\text{Gauge pressure = Absolute pressure - Atmospheric pressure}} \\\end{aligned} \end{equation} $$
Pressure acts equally in all directions
Pressure always acts normal to the boundaries of the fluid
Pressure will be same at all points in liquid lying at the same level
Consider a section of liquid of cross-sectional area $A$ as shown in the figure.
From FBD at equilibrium, $$\begin{equation} \begin{aligned} {P_1}A = {P_2}A \\ {P_1} = {P_2} \\\end{aligned} \end{equation} $$
So, pressure at all points in a liquid is same lying at the same level
As we know pressure at a point depends on the depth $h$ below the free surface of the liquid, density of the liquid and acceleration due to gravity $g$
So, pressure at any point is independent of the amount of liquid and shape of the container.
Therefore, the pressure will be same at the same level for a liquid filled in a vessel of different shapes to same height.
The relation $\left( {P = {P_O} + \rho gh} \right)$ is a straight line $$\begin{equation} \begin{aligned} P = {P_O} + \rho gh \\ y = c + mx \\\end{aligned} \end{equation} $$ On comparing the above equation we get, $$c = {P_O}\quad m = \rho g$$
