Consider a small fluid element as shown in

Let      a = cross sectional area of element
           dz = height on fluid element
           P = pressure on face AB
           Z = distance of fluid element from free surface

The net upward pressure force

 = (p + dp) a-p.a
= p.a + dp. a-p.a
= dp.a

Weight of the prism

W = volume x γ = dz x a x ( ).g

For equilibrium of the prism

+ ↑ ∑  = 0

dp.a – dz x a. ρg = 0

dp.a = dz.a.ρg

dp = ρ.g.dz

dp/dz = ρ.g

This is hydrostatic law which state, “The rate of increase of pressure in the vertically downward direction, at point in a static fluid, must be equal tot eh specific weight of the fluid.”

Case I: Incompressible fluids, ρ= constant
 
∫ dp = ∫ ρgdz

∫ dp = ρg ∫ dz

P2 - P1 = ρ . g (Z2 - Z1)

P = ρgh

This shows that the pressure increases linearly vertically downwards in an incompressible fluid. It is shown by triangular distribution.

Case II: Compressible Fluids. ρ ≠ constant.

Assuming that the pressure and the density are reached by the perfect gas equation.

P = ρ R T

ρ = P/ RT

Use hydrostatic equation          dp/dz = ρg
                                                 dp/dz = P/RT.g
                                                 dp = dz/T. g/ R

This equation is called as the aerostatic law.

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