As per the Pascal's law, pressure or intensity of pressure at a point in a static fluid (fluid being in rest) is equal in all directions.
Showing a fluid element
Consider an arbitrary fluid element of wedge shape having very small dimensions i.e. dx, dy and ds as shown in figure.
Let us assume the width of the element perpendicular to the plane of paper to be unity and let Px, Py and Pz be the pressure intensities acting on the face AB, AC and BC respectively.
Let ∠ ABC = θ. Then the forces acting on the element are:
The forces on the faces are:
Force on the face AB = PX × Area of face AB
FAB = px × dy × 1 ………... (1)
Similarly, force on the face AC (FAC)= py × dx × 1 ……... (2)
Force on the face BC (FBC)= pz × ds × 1 ……. (3)
Element's weight = (Mass of element) × g
where ρ = density of fluid.
Resolving the forces in x-direction, we have
px × dy × 1 – pz (ds × 1) sin (90° – θ) = 0
px × dy × 1 – pz ds ×cos θ = 0 ……... (4)
But from fig.
ds cosθ = AB = dy ……. (5)
Thus, from equation (1) and (2):
∴ px × dy × 1 – pz × dy × 1 = 0
px = pz ………… (3)
Similarly, resolving the forces in y-direction, we get
But ds sin θ = dx and the element is very small and hence dxdy will be negligible i.e. weight of fluid element can be neglected.
∴ pydx – pz × dx = 0
py = pz ……… (4)
From equations (3) and (4)
px = py = pz
The equation above illustrates that the pressure at any point in x, y, and z directions is equal. As the choice of the fluid element was completely random and arbitrary, it means that the pressure at any point is the same in all directions.
Hydraulic lift, hydraulic break, etc.
Pressure at any point in a fluid at rest is found out by the Hydro-static Law. As per this law, the rate of increase of 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
Forces on the fluid element
ΔA = Cross-sectional area of element
ΔZ = Height of fluid element
p = Pressure on face AB
Z = Distance of fluid element measured from the free surface.
The forces acting on the fluid element are:
w = Weight density of fluid.
Equation (1) 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.
Now, by integrating the above equation (1) for liquids:
p = ρgZ ……… (2)
where p is the pressure above atmospheric pressure and Z is the height of the point from free surface.
Here Z is called pressure head.
Pressure at a depth “h”
Showing a point A location within the fluid
∫dp = ∫ρgdh
P = ρgh + C …… (1)
At h = 0, P = Patm
Thus, C = Patm
P = ρgh + Patm ……… (2)
Therefore, PGauge = ρgh (N/m2 or Pascal)
ρ1gh1 = ρ2gh2 valid for all fluids.
ρ1h1 = ρ2h2 valid for all fluids.
Hydrostatic forces on an inclined plane submerged surface in a liquid
Imagine a plane surface of any shape being immersed in a liquid in such a way that the plane surface makes an angle θ with the free surface of the liquid as shown in the figure below.
The force is given by:
Hence, we can conclude that the force is independent of the angle of inclination (θ). Thus, the same expression could be used for the force calculation of Horizontal and vertical submerged bodies.
Centre of Pressure (h*):
It is the point where whole of the hydrostatic force is assumed to be acting.
Plane vertical surface (θ = 90°)
Therefore, centre of pressure for a vertically submerged surface is given by
Vertical submerged surface
Plane horizontal surface (θ = 0°)
Horizontal submerged surface
centre of pressure for the vertical submerged surface is given by:
Thus, the centre of pressure of a body submerged parallel to free surface will be equal to the distance of centroid of the body from the free surface.
Hydrostatic force on a curved surface
AC = curved surface
FY = vertical component of FR
FX = Horizontal component of FR
FR = Resultant force on a curved surface.
The horizontal component of force on a curved surface:
The horizontal component of force acting on a curved surface equals the hydrostatic force on the vertical projected area of the curved surface.
A = Projected Area
= depth of centroid of an area.
This force acts at the center of the pressure of the corresponding area.
The vertical component of force on a curved surface