SUB:- FLUID MECHANICSSUB. CODE:- 2141906

NAME OF TOPIC :- PASCAL’S LAW ( FLUID STATICS )

Learning Objectives

1. Defining Pascal's Law

2. Understand what is Pascal's Law

3. Deriving the formula and applying it

4. Understand the applications of Pascal's Law

5. Learning how to applied Pascal's Law everyday life

INTRODUCTION

Blaise Pascal

French mathematician , physicist, inventor, writer and Christian philosopher

He was work in the natural and applied sciences where he made important contributions to study of fluids, and clarified the concepts of pressure and vacuum by generalizing the work of Evangelista Torricelli.

Pascal also wrote in defencse of the scientific method.

Pascal's law or the principle of transmission of fluid-pressure (also Pascal's Principle ) is a principle in fluid mechanics that states that pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such that the pressure variations (initial differences) remain the same. 

The law was established by French mathematician Blaise Pascal.

DEFINITIONPascal's principle is defined as :-A change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all

points in the fluid

This principle is stated mathematically as:

)( HgP

Consider an arbitrary fluid element of wedge shape ABC in afluid mass at rest as show in fig (a)

fig.(a) Pascal's law

STATE & PROOF

“ the intensity of pressure at any point in a liquid at rest, is the same in all directions.”

Let,

pressure acting on a face AB pressure acting on a face AC pressure acting on a face BC

force on a face AB

force on a face AC

force on a face BC

xpypzp

xPyP

zP

The forces acting on the element are(i) Force normal to the surface due to fluid pressure.

(ii) Force due to weight of fluid mass in vertical direction.

area of AB area of AC

area of BC

xPyP

zP

xp

yp

zp

1 dypP xx

)1( dxpP yy

)1( dspP zz

Weight of element = ( mass of element ) * ( gravity constant )

the element of the liquid is at rest, therefore sum of horizontal and vertical components of the forces equal to zero.

gvolume

gABAC 121

ABACareaABC

21,

gdxdyW 21

dyABdxAC ,

Resolving the forces horizontally ( in x-direction )

But, from triangle ABC,

....(a)

sin11sin

0sin

dspdypPP

PP

zx

Zx

zx

dyABds sin

zx

zx

ppdypdyp

Similarly, resolving the forces vertically ( in-y direction )

But from triangle ABC,

dxdygdxpdxp

gdydxdspdxp

WPP

zy

zy

zy

21cos

021cos11

0cos

dxACds cos

dxdygdxpdxp zy 21

The element is very small, negating weight of fluid

…(b)From equations (a) and (b)

hence, at any point in a fluid rest the pressure is exerted equally in all direction.

zyx ppp

zy

zy

pp

dxpdxp

0

Transmission of Pressure in a Liquid1. Liquids are practically incompressible.

2. The compression force causes pressure to act on the surface of the water. Pressure = Force (compression)

Surface area of liquid3. Pascal’s principle states that in a confined fluid, an externally applied pressure is transmitted uniformly in all direction.

4. In a hydraulic system, Pascal’s principle is applied as a force multiplier. The force multiplier of a hydraulic system can be represented by the equation:

Output force = Output piston area Input force Input piston area

Output piston area Input force Input piston area

APPLICATION OF PASCAL’S LAW

1. In measurement system (i.e. manometer, pressure gauge etc.)

2. In construction of machines such as hydraulic press, hydraulic jack, hydraulic lift, hydraulic crane, hydraulic riveter etc.

3. Force amplification in the braking system of most motor vehicles.

4. Used in artesian wells, water towers, and dams.

For example, if the lift cylinder were 25 cm in diameter and the small cylinder

were 1.25 cm in diameter, then the ratio of the areas is 400, so the hydraulic press

arrangement gives a multiplication of 400 times the force. To lift a 6000 newton

car, you would have to exert only 6000 N/400 = 15 N on the fluid in the small

cylinder to lift the car. However, to lift the car 10 cm, you would have to move the

oil 400 x 10cm = 40 meters. This is practical by pumping oil into this small

cylinder with a small compressor.

Problem Solving

Example 1:In hydraulic brake, a force of 80 N is applied to a piston with area of 4 cm2. a) What is the pressure transmitted throughout the liquid?b) If the piston at the wheel cylinder has an area of 8 cm2, what is the force exerted on it?

Solution (a) P = F/A = 80 N/4 cm2

= 20 N.cm2

(b) F = P x A = 20 N.cm2 x 8 cm2

= 160 N

Example 2:The figure shows a 10 N weight balancing a X N weight placed on a bigger

syringe. What is the value of X ?

F1/ A1 = F2/ A210 N / 1.5 cm2 = X N / 4.5 cm2

Therefore, X = 10 / 1.5 x 4.5 = 30 N

Example 3:The mass of X is 2 kg. It is placed at a piston A. The cross section areas of A and B

are 5 cm2 and 80 cm2 respectively.

(a) Calculate the force which acts on piston A

(b) Find the pressure which is exerted on piston B.

(c) Find the mass of Y which can be lifted by piston B.

(d) If piston A moves down by 20 cm, then piston B will go up by

Solution:- (a) Calculate the force which acts on piston A F = mg = 2 x 10 = 20 N (b) Find the pressure which is exerted on piston B. P = F/A = 20 N / 5 x 10-4 m2 = 40 000 N m-2 (c) Find the mass of Y which can be lifted by piston B. F1/A1 = F2/A2 2 x 10/ 5 x 10-4 = m x 10 / 80 x 10-4 m = 32 kg (d) If piston A moves down by 20 cm, then piston B will go up by 5 x 20 cm = 80 x l cm3 l = 1.25 cm

PREPAID BY:- chetan ahuja