PRESSURE VARIATION IN A FLUID AT REST

The pressure at any point in a fluid at rest is obtained by Hydrostatic law which states that the rate of the increase in pressure in a vertically downward direction must be equal to the specific weight of the fluid at that point.

Consider a small fluid element as shown in figure:

Figure  Forces on a fluid element 

Let    ΔA = Cross-sectional area of element

          ΔZ = Height of fluid element

          p = Pressure on face AB

          Z = Distance of fluid element from free surface

The force acting on the fluid element are :

1. Pressure force on AB = p × ΔA and acting perpendicular to face AB in the downward direction.
2. Pressure force on CD ={ p + (jp/jZ )ΔZ } × ΔA, acting perpendicular to face CD, vertically upward direction.
3. Weight of fluid element = Density × g × volume = ρ × g × (ΔA × ΔZ).
4. Pressure forces on surfaces BC and AD are equal and opposite. For equilibrium of fluid element, we have

         pΔA – { p+(jp/jZ)ΔZ}×ΔA + ρ×g×(ΔA×ΔZ) = 0

or     pΔA – pΔA – (jp/jz) ΔZΔA + ρ×g×ΔA×Z = 0 

or     - (jp/jZ)ΔZΔA + ρ×g×ΔAΔZ =0

or     (jp/jZ)ΔZΔA = ρ×g×ΔAΔZ = 0

or     (jp/jZ) = ρ×g

                   jp/jZ = ρ×g = w      (ρ×g = w)

where, w = Weight density of fluid.

It states that rate of increase of pressure in a vertical direction is equal to weight density of the fluid at that point. This is Hydrostatic Law.

By integrating the above equation for liquids, we get

          ʃ dp = ʃ ρgZ

          p = ρgZ   

where p is the pressure above atmospheric pressure and Z is the height of the point from free surfaces.

From equation, we have          Z = (p/ρ×g)

Here Z is called pressure head.

Problem. An open tank contains water up to a depth of 2m and above it an oil of sp.gr. 0.9 for a depth of 1m. find the pressure intensity (a) at the interface of the two liquids, and (b) the bottom of the tank.