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**Pascal’s Law **

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 P_{x}, P_{y} 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:

- Pressure forces normal to the surfaces, and
- Weight of element in the vertical direction.

The forces on the faces are:

Force on the face AB = P_{X} × Area of face AB

F_{AB} = p_{x} × dy × 1 ………... (1)

Similarly, force on the face AC (F_{AC})= p_{y} × dx × 1 ……... (2)

Force on the face BC (F_{BC})= p_{z} × 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.

∴ p_{y}d_{x} – p_{z} × d_{x} = 0

p_{y} = p_{z} ……… (4)

From equations (3) and (4)

p_{x} = p_{y} = p_{z}

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.**Examples:**

Hydraulic lift, hydraulic break, etc.

Hydraulic lift

- In a hydraulic lift, a smaller force is required to lift a larger weight but still, the conservation of energy is not violated because the smaller force moves by a larger distance whereas larger wt. moves by smaller distance And hence work done in both the cases are same and hence conservation of energy is followed.

**Hydrostatic Law**

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:

- Pressure force on AB = p × ΔA and acting perpendicular to face AB in the downward direction.
- Pressure force on

acting normal to face CD , vertically upward direction. - Weight of fluid element = Density × g × Volume = ρ × g × (ΔA × ΔZ).
- Pressure force on surfaces BC and AD are equal and opposite. For equilibrium of fluid element, we have

or ∂p/∂Z = ρ × g [cancelling ΔAΔZ on both sides]

∴ ∂p/∂Z = ρ × g ......(1)

where,

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.**Note.2:**

- Hydrostatic law can be applied to both compressible and incompressible fluids.

**Pressure at a depth “h”****Showing a point A location within the fluid**

Now use

∫dp = ∫ρgdh

P = ρgh + C …… (1)

At h = 0, P = Patm

Thus, C = Patm

P = ρgh + Patm ……… (2)

Therefore, PGauge = ρgh (N/m^{2} or Pascal)**Note 3:**

- As we move vertically down is a fluid the pressure increase as +ρgh. As we move vertically up in a fluid the pressure decreases as –ρgh.
- There is no charge g pressure in the horizontally same level.
- For the conversion of one fluid column to another fluid column, we can use the following :

ρ_{1}gh_{1} = ρ_{2}gh_{2} valid for all fluids.

ρ_{1}h_{1} = ρ_{2}h_{2} valid for all fluids.

**The Hydrostatic Paradox**

- The pressure at any point depends only upon the depth below the free surface and the unit weight of the liquid.

The size and shape of the container do not affect the pressure exerted by the fluid. Hence, the pressure at the bottom of all containers will be the same if they are filled with the same liquid up to the same height.The hydrostatic paradox

**Hydrostatic Forces**

- When a fluid, in contact with a surface is exerts a normal force on the surface, it is known as the hydrostatic force.

**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 Forces on Curved Surfaces**

Hydrostatic force on a curved surface

AC = curved surface

F_{Y} = vertical component of F_{R}

F_{X} = Horizontal component of F_{R}

F_{R} = 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.

Horizontal force:

Where,

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**

- The vertical component of the hydrostatic force on a curved surface is given by the weight of the fluid contained by the curved surface up to the free surface of the liquid.
- It will act at the center of gravity of the volume of liquid contained in a portion extended above the curved surface up to the free surface of the liquid.

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