Hydrostatic Pressure 

Hydrostatic pressure is the pressure resulting from fluids at rest; it increases with depth. Its magnitude depends on the fluid's density and the force of gravity acting on it. Hydrostatic pressure is important in several fields: water systems, engineering, and biology.

1.0What is Hydrostatics?

Hydrostatic is the term given to fluids at rest. It refers to the understanding of forces and pressures within stationary fluids, especially liquids. Hydrostatics pertains to the effects of gravity on fluids that do not move.

2.0Principle of Hydrostatics 

The principle of hydrostatic states that a fluid at rest experiences an increase in pressure with depth because of the weight of the fluid above. It acts equally in all directions and depends on fluid density and gravitational pull. The intensity of hydrostatic pressure is independent of the shape of the container since it depends only on depth, fluid density, and gravity.

3.0Hydrostatic Pressure 

Hydrostatic pressure in 12th chemistry refers to the pressure exerted by a liquid at rest due to the weight of the liquid above it. Hydrostatic pressure is dependent on the depth of the liquid and the density of the liquid, as well as the gravitational force. 

While in Physics Hydrostatics pressure is part of fluid mechanics dealing with fluids at rest. Science primarily examines the forces and pressures a fluid may exert upon an object and surfaces that are in equilibrium.

Hydrostatic Law of Pressure

According to the hydrostatic law of pressure, the pressure exerted at a given point by the fluid is proportional to the fluid's density and the acceleration due to gravity; it depends on the depth, or height, of the fluid above the given point where it exerts pressure. Mathematically, it can be expressed as: 

$P=\rho gh$

Here:

• P = Hydrostatic Pressure (in pascals, Pa)
• ρ = Density of the fluid (in kg/m³)
• g = Acceleration due to gravity (9.8 m/s²)
• h = Depth of the fluid (in meters)
• The S.I. unit of Hydrostatic Pressure is Pascal. 

4.0Derivation of Hydrostatic Pressure Formula

To Prove: $P=\rho hg$

Solution: Imagine a column of water with the total volume V and A as the base surface area. 

We know that: 

$Weight(W)=Mass(m)\times Accelerationduetogravity(g)$

$W=mg$

And 

$Density=\frac{mass}{Volume}$

$\rho =\frac{m}{V}orm=\rho \times V$

Therefore, 

$W=\rho Vg$

We know that Volume is basically the product of surface area and height. Hence, 

$V=Ah$

$W=\rho Ahg$

$Pressure=\frac{Force}{Area}$

$P=\frac{\rho Ahg}{A}$

$P=\rho hg$

5.0Hydrostatic Pressure and Fluid Pressure

Both pressures are similar but hydrostatic pressure is particularly where the fluid is stationary under gravity.

6.0Solved Examples

Problem 1: A pressure gauge is placed at the bottom of a tank filled with a liquid of unknown density. The height of the liquid column is 12 meters, and the pressure at the bottom is measured to be 1.2×10⁵ Pa. Calculate the density of the liquid. (Atmospheric pressure is 1.01×10⁵Pa)

Solution: The pressure at the bottom of the tank is the sum of atmospheric pressure and hydrostatic pressure.

$P_{total}=P_{atm}+P_{hydrostatic}$ 

$P_{hydrostatic}=1.2\times 10^5-1.01\times 10^5$

$P_{hydrostatic}=0.19\times 10^{5}Pa$

Now, for solving the density of fluid: 

$P_{hydrostatic}=\rho gh$

$\rho =\frac{P_{hydrostatic}}{gh}$

$\rho =\frac{0.19\times 10^5}{9.8\times 12}$

$=\frac{19000}{117.6}\approx 161.56kg/m^3$

Problem 2: A vertical pipe is open at both ends and contains water. It is 20 meters high. Calculate the velocity of the water flowing from the top of the pipe if the water is at rest at the bottom. Assume the water surface at the top of the pipe is open to the atmosphere.

Solution: For solving this question we can use Bernoulli’s equation here. Bernoulli's principle says the total pressure remains constant, meaning: 

$P_{top}+\frac{1}{2}{v}_{top}^2+\rho g{h}_{top}=P_{bottom}+\frac{1}{2}\rho v_{bottom}^2+\rho g{h}_{bottom}$

It is given that the velocity at the bottom is zero (v_bottom =0), and P_top = P_bottom, the equation will now become: 

$\rho g{h}_{top}=\frac{1}{2}\rho v_{top}^2$

$v_{top}=\sqrt{2g{h}_{top}}$

$=\sqrt{2\times 9.8\times 20}=\sqrt{392.4}\approx 19.8m/s$

Problem 3: A steel tank is submerged 30 meters underwater. The tank has a spherical shape with a radius of 2 meters. Calculate the total force acting on the spherical surface due to the hydrostatic pressure.

Solution: The formula for force acting on a spherical surface submerged in a fluid is given by the formula: 

$F=P_{average}A$

Where

$P_{average}=\rho gh$ and A = Area of sphere = $4\pi r^2$

$Area=4\times \frac{22}{7}\times 2=50.27m^2$

$P_{average}=\rho gh$

$P_{average}=1000\times 9.8\times 30=294000Pa$

$F=P_{average}A$

$F=294000\times 50.27$

$=1,47,79,380N=14.78MN$