Lecture 06 Fluid Statics and Pascals Law

Chapter 2 Fluid staticsChapter 2 Fluid staticsChapter 2 Fluid staticsChapter 2 Fluid statics

PressurePressurePressurePressurePressure is defined as the amount of force exerted on a unit area of a substance. P i N l F ( tPressure is a Normal Force (acts perpendicular to surfaces) It is also called a Surface ForceIt is also called a Surface ForceIn algebraic form this definition may be stated as:stated as:

PamN

areaforceP === 2

PressurePressurePressurePressure

Pressure acts uniformly in all directions on a small volume of fluid. P i l fi ld ( t)Pressure is a scalar field p = p(x; y; z; t)The value of p varies in space. Pressure at any point in a stationary fluid is independent of direction.In a fluid confined by solid boundariesIn a fluid confined by solid boundaries, pressure acts perpendicular to the boundaryy

Pascal's lawPascal's lawPascal s lawPascal s law

For a static fluid, as shown by the following analysis, pressure is independent direction.Consider the equilibrium of a small qfluid element in the form of a triangular prism ABCDEFg p


pressure in the x direction Pxpressure in the y direction Py , and p y y ,pressure normal to any plane inclined at any angle θ to the horizontal Psat any angle θ to the horizontal Ps


Px is acting at right angle to ABEFPy at right angle to CDEFy g gPs at right angle to ABCD.


There can be no shearing forces for a fluid at restThere will be no accelerating forces,


The sum of the forces in any directionThe sum of the forces in any direction must be zero.The forces acting areThe forces acting are

pressures on the surrounding gravity force.

Pascal's lawPascal's lawForce due to PForce due to Pxx = P= Pxx X Area ABFEX Area ABFE

PP δδ δδ= P= Px x δδyyδδzzHorizontal component of force due to Horizontal component of force due to

PPss = = -- (P(Pss x Area ABCD) sinx Area ABCD) sinθθ== -- PP δδssδδzz (δ(δyy/δ/δs)s)= = -- PPssδδssδδzz (δ(δyy/δ/δs) s) = = --PPssδδyyδδzz


PPyy has no component in the x directionhas no component in the x directionAt equilibriumAt equilibriumAt equilibriumAt equilibrium

PPxxδδy y δδzz --PPssδδyyδδz = 0z = 0i e Pi e P = P= P (1)(1)i.e. Pi.e. Pxx = P= Ps s (1)(1)


Similarly in the y direction force due to PSimilarly in the y direction force due to PSimilarly in the y direction, force due to PSimilarly in the y direction, force due to Pyy= P= Pyyδδxxδδzz

Force due to PsForce due to Ps== (P(P X Area ABCD)cos(X Area ABCD)cos(θθ))Force due to PsForce due to Ps= = -- (P(Ps s X Area ABCD)cos(X Area ABCD)cos(θθ) ) = = -- PPssδδssδδz (z (δδx/x/δδs) s) = = -- PPssδδxxδδzz


Force due to weight of elementForce due to weight of elementForce due to weight of element Force due to weight of element = = -- mg mg = = -- ρρVg Vg = = -- ρρ ((δδxxδδyyδδz/2) gz/2) gρρ (( yy ) g) g


At equilibriumAt equilibriumPPyyδδxxδδzz -- PPssδδxxδδz z -- ρρ ((δδxxδδyyδδz/2) g = 0z/2) g = 0yy ss ρρ (( yy ) g) g

δδx, x, δδy, and y, and δδz are very small quantitiesz are very small quantitiesδδ δδ δδ negligiblenegligibleδδxxδδyyδδz negligiblez negligible


PPyyδδxxδδzz -- PPssδδxxδδz z -- ρρ ((δδxxδδyyδδz/2) g = 0z/2) g = 0yy ss ρρ (( yy ) g) g


PPyyδδxxδδzz -- PPssδδxxδδz = 0z = 0yy ss


PPyy -- PPss = 0= 0yy ss


Therefore the equation reduces toTherefore the equation reduces toPy = Ps Py = Ps (2)(2)yy ( )( )


Combine equations (1) and (2)Combine equations (1) and (2)



Px = Py = PsPx = Py = Ps



Px = Py = PsPx = Py = Ps

i t i t i i ll di tii.e. pressure at a point is same in all directions. This is Pascal's law. This applies to fluid at rest.

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Lecture 06 Fluid Statics and Pascals Law