lecture 3

Lecture 3

FLUID STATICS

Mechanics of Fluids 1: Lecture 3: Fluid StaticsDepartment of Mechanical Engineering MEHB223

Department of Mechanical Engineering MEHB223 Mechanics of Fluids 1: Lecture 3: Fluid Statics

Chapter Summaryn Fluid Pressure at a Point; Absolute, Vacuum

and Gauge Pressure n Pressure Variation with Elevationn Pressure Measuring Devices – Barometer,

Open-End Manometer, Differential Manometern Hydrostatic Paradoxn Hydrostatic Forces on Submerged Surfaces –

Plane surface, Curved Surfacesn Buoyancyn Stability of Immersed and Floating Bodies


n Fluids, in general, exerts both normal and shearing forces on surfaces in contact

n Shearing forces are produced only with relative motion

n Without relative motion ~ only normal force => called Pressure Forces

n No Relative Motion Means : -– Stationary– Moving with the same constant velocity– Moving with the same varying velocity

3.1 Introduction


3.2 Fluid Pressure At A Point

By balancing the forces in the vertical and horizontal directions : ~

Px = Pz = Pn

=> Pressure at a point is the same in all directions

=> PASCAL’s Law

=> The conclusion is also applicable if there is relative motion


3.3 Pressure TransmissionIn a closed system, due to Pascal’s Law, the pressure change produced at one point in the system will be transmitted throughout the entire systemPrinciple of the Hydraulic Lift


Example 3.1A hydraulic jack has dimensions as shown. If one exerts a force or 100N on the handle of the jack, what load F2, can the jack support? Neglect lifter weight.


n Pressure normally measured as pressure difference

n The pressure above absolute vacuum : Absolute Pressure

n Relative Pressure above atmospheric Pressure : Gage Pressure

n Relative Pressure below atmospheric Pressure : Vacuum Pressure

3.4 Absolute, Vacuum and Gage Pressure


3.5 Pressure Variation With Elevation

Balancing the force in s direction : -

But sinα = dz/dl, so : -

For incompressible fluid, ρ is constant : -

= Piezometric Pressure (Pascal)

= Piezometric Head (meter)


n If γA > γB …which distribution is correct ?

3.5 Pressure Variation With Elevation ...

n If γB > γA …which distribution is correct ?.. Is this possible?

√ XX

Department of Mechanical Engineering MEHB223

X

Mechanics of Fluids 1: Lecture 3: Fluid Statics

n Therefore, for incompressible fluid, in any static fluid the piezometric head is constant throughout the fluid

n Thus at any point within the fluid : -


datum

1

2

3 zz

z

3

21


n Example : Pressure at a point, h below the surface of a stationary fluid : -


n Example : Case Involving 2 fluids. Determine the gage pressure at the bottom of the tank

n The equation has to be applied to each fluid separately


n Compressible Fluid (e.g. atmospheric air)


n Troposphere :

…… Function of Temperature

n Stratosphere :


n Barometer - use to measure atmospheric pressure

3.6 Pressure Measuring Devices

Applying the equation from 1 to 2 : -

P1 = Patm ; z2-z1 = h ; P2 = Pvap << Patm : -


3.6 Pressure Measuring Devices ...

n Piezometer - Consists of single vertical tube open at the top inserted into a pipe under pressure which rises in the tube depending on the pressure. If the top of the tube is open to atmosphere then it reads ‘gage’ pressure at that location in the pipe


n Open end U-tube Manometer : -3.6 Pressure Measuring Devices ...

If ρf << ρm :


n Differential manometer : -


Static pressure change from

A to B

Static pressure change due to

the system

Static pressure change due to difference in

elevation


n Example : Obtain the expression for pressure of the air in the tank .



n Example : The piston shown weigh 100N. In its original position, the piston is restrained from moving to the bottom of the cylinder by mean of a metal stop. Assuming there is neither friction no leakage between the piston and cylinder, what volume of oil (S=0.85) would have to be added to the 1 cm tube to cause the piston to rise 1 cm from its initial position?


6 cm

4 cm

4 cm

4 cm

1cm


n Pressure exerted by a fluid is only dependent on the vertical head and its density

n It is independent of the weight of the fluid present

3.7 Hydrostatic Paradox

The pressures are the same although the weight of the liquids are obviously different

paradox


3.8 Hydrostatic Force on Submerged Surfaces

Importance of Hydrostatic Force CalculationsExamples : -


hh

dF=PdA y

x

y

dA

edge view

normal view

elemental area

centroid

Area, A

3.8.1 Hydrostatic Force on Plane Surfaces

Hydrostatic Force = Pressure at the Centroid x Area

Atmospheric pressure is ignored since both sides are

open to atmosphere

Consider the magnitude of the hydrostatic force on one side

of the plane


centroid

Area, A

h

Fh

h

ycpCP


- The resultant hydrostatic force acts through the CENTRE OF PRESSURE (COP)- The slanted distance of COP from the centroid, ycp, is determined by : -

- And xcp, is given by : -

- If the shape is symmetrical about y axis xcp is zero


3.8.1 Hydrostatic Force on Plane Surfaces- Centroid and Second Moment of Area, Ixx of regular shapes : -

This is given in Appendix …. No Need to Remember


w

h

hinge

Fρgh


Example 1 : Find the magnitude of the hydrostatic force and its line of action from the hinge. Calculate the force F applied at the middle of the gate required to hold the gate in its vertical position


3.8.1 Hydrostatic Force on Plane SurfacesExample 2 : Find the magnitude of the force, P required to just start opening the 2m wide gate. Neglect the weight of the gate.


h

w

h

hinge

G


Example 3 : Find the magnitude of the force, G required to just start opening the gate. Neglect the weight of the gate.



Example 4 : An elliptical gate covers the end of a pipe 4m in diameter. If the gate is hinged at the top, what normal force, F is required to open the gate when water is 8m deep above the top of the pipe and the pipe is open to atmosphere on the other side? Neglect the weight of the gate.


h

elemental area, dAdF=γhdA

dF

dF

dFy

x ds

Vertical projection

v

C v

Area, A v

x

y

l

H

3.8.2 Hydrostatic Force on Curve Surfaces



- Horizontal Component of the Force, Fx : -

- Line of action of Fx : -

- Vertical Component of the Force, Fy : -

- Line of action of Fy is through the centroid of the Fluid Above the surface


Vertical projection

C v

Fy

FxyCPv

Centroid of fluid above

surface

CP of vertical projection

F


- The Resultant Hydrostatic Force, F is : -

Note the Horizontal and Vertical Component of the Force Acts From Different Points


FyCentroid of

imaginary fluid above surface

CP of vertical projectionFy

F


- Water underneath the surface ?

- The force will be exactly of the same magnitude but now acts in the opposite direction.

- Need to use imaginary surface in order to calculate vertical component



Example 1 : Surface AB is a circular arc with radius of 2m and a depth of a m into the paper. The distance EB is 4m. The fluid above surface AB is water and atmospheric pressure prevails on the free surface of water and on the bottom side of surface AB. Find the magnitude and line of action of the hydrostatic force acting on surface AB.


D

C

B

A

D’C’B’


Example 2 : Determine the hydrostatic force acting on this gate ?


3.8.2 Hydrostatic Force on Curve SurfacesExample 3 : What force must be exerted through the bolts to hold the dome in place? The metal dome and pipe weigh 6 kN. The dome has no bottom. Here ι = 80 cm.


A B

CDCOG

Mg

Vol4

Vol1

Vol2

Vol3

ρb

ρf

3.9 Bouyancy- Definition : Vertical Force on a body immersed in a stationary fluid.

It arises because the pressure varies with depth.- Consider a body partly immersed in a fluid : -

- Act through the centroid of the displace volume => Centre of Bouyancy

Archimedes Principle


3.9 Bouyancy

Example 1 :A metal part (2) is hanged by a thin cord from a floating wood block (1). The wood block has a specific gravity of 0.3 and dimension 50 x 50 x 10 mm. The metal part has volume 6600 mm3. Find the mass m2 of the metal part and the tension in the cord.


3.9 Bouyancy

Example 2 : The partially submerged wood pole is attached to the wall by a hinge as shown. The pole is in equilibrium under the action of weight and buoyant forces. Determine the density of the wood?


3.9 Bouyancy

n Hydrometry ….. Hydrometer is a device use to measure the specific gravity of a liquid

n Based on the buoyancy principle

n The depth of the hydrometer is dependent on the specific gravity of the liquid


3.10 Stability of Immersed Body- Stability depends on the relative position of Centre of Gravity (COG) and

Centre of Bouyancy (COB)

- If COB > COG = > Stable

- If COB < COG = > Unstable


3.11 Stability of Floating Body- The previous rule is not applicable to floating body because the

COB of displaced volume will change as the object is displaced : -

- Thus more involve d analysis is needed .


End of Lecture 3


lecture 3

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