Pascal's Law

Pascal's Law

Statement: It states that the pressure or intensity of pressure at a point in a static fluid is equal in all directions. This is proved as:

Fig. Force on a Fluid Element

 The fluid element is of very small dimensions i.e., d_x, d_y and d_s. Consider an arbitrary fluid element of wedge shape in a fluid mass at rest as shown in Fig. Let the width of the element perpendicular to the plane of paper is unity and P_x, P_y and P_z are the pressures or intensity or pressure acting on the face AB AC and BC respectively. Let ∠ABC = Ɵ. 

Then the forces acting on the element are:

1.     Pressure forces normal to the surfaces, and

2.     Weight of element in the vertical direction.

The forces on the faces are:

Force on the face AB                        = P_x × Area of face AB

                                                          = P_x × d_y × 1

Similarly force on the face AC         = P_y × d_x × 1

Force on the face BC                        = P_z × d_s × 1

 Weight of element                            = (Mass of element) × g

                                                          = (Volume of ρ) × g = ((AB × AC × 1)/2) × ρ × g,

Where ρ = density of fluid.

Resolving the forces in x-direction, we have

                 P_x × d_y × 1 - P (d_s × 1) sin (90० - Ɵ) = 0

                              P_x × d_y × 1 - P_z d_s × 1 cosƟ = 0.

          But from Fig.                             d_s cosƟ = AB = d_y

           ∴   P_x × d_x × 1-P_z × d_s × 1 - P_z × d_s × 1 = 0

                                                                     P_x = P_z

   Similarly, resolving the forces in y-direction, we get

P_y × d_x × 1 – P_z × d_s × 1 cos(90० - Ɵ)- (d_x × d_y)/2 × 1 × ρ × g = 0

                                     P_y × d_x – P_z d_s sinƟ – (d_xd_y)/2 × ρ × g = 0

But d_s sinƟ = d_x and also the element is very small and hence weight id negligible.

            ∴   P_yd_x – P_z × d_x = 0

                                     P_y = P_z

The above equation shows that the pressure at any point in x, y, z directions is equal.

Since the choice of fluid element was completely arbitrary, which means the pressure at any point is the same in all directions.

Problem 1. A hydraulic press has a ram of 20 cm diameter and a plunger of 3 cm diameter. It is used for lifting a weight of 30 KN. Find the force required at the plunger.

Solution. Given: 

Dia. Of ram,                    D = 20 cm =0.2 m

∴ Area of ram,               A = π/4 D² = 0.0314 m²

Dia. Of plunger                  d = 3 cm = 0.03 m

 ∴ Area of plunger,   a = π/4 (0.03)² = 7.068 × 10^-4  m²

Weight lifted,          W = 30 KN = 30000 N.

Pressure intensity developed due to plunger = Force/Area = F/A

By Pascal’s Law, this pressure is transmitted equally in all directions

Hence pressure transmitted at the ram = F/A

                        ∴  Force acting on ram = pressure intensity × Area of ram

                                                              = F/a × A = (F × 0.0314)/ (7.068 × 10^-4 ) N

But force acting on ram = Weight lifted = 30000N

                            30000 = (F × 0.0314)/ (7.068 × 10^-4 )

                                    F = (30000 × 7.068 × 10^-4)/ (0.0314) = 675.2 N.

Problem 2. An oil sp. gr. 0.9 is contained in a vessel. At a point the height of oil is 40 m. Find the corresponding height of water at the point.

Solution. Given:

Sp. gr. Of oil,                            S₀ = 0.9

Height of oil,                            Z₀  = 40 m

Density of oil,                           ρ₀ = Sp. gr. Pf oil × Density of water = 0.9 ×1000 = 900 kg/m³

Intensity of pressure,                    P = ρ₀ × g × Z₀ = 900 × 9.81 × 40 N/m²

 ∴ Corresponding height of water   = P/Density of water × g

                                                         = (900 × 9.81 × 40)/ (1000 × 9.81) = 36 m of water.

 Absolute, Gauge, Vacuum, Atmospheric Pressure

 Atmospheric Pressure is defined as the pressure exerted by atmosphere on the surface of earth.

P_atm = 1.013 bar = 1.013 × 10⁵ N/m² = 10.33 m of water = 760 mm of Hg

Gauge Pressure is the pressure measured with the help pg pressure measuring instrument in which the atmospheric pressure is taken as datum.

Fig. Relationship Between Pressures

Vacuum Pressure is defined as the pressure below the atmospheric pressure. The vacuum pressure is also known as negative gauge pressure.

Absolute Pressure is measured above the absolute zero.

Absolute Pressure = Atmospheric Pressure + Gauge Pressure

Vacuum Gauge Pressure = Atmospheric Pressure – Absolute Pressure.