Chapter 3

Mechanics of Fluids : Chap. 3: Fluid StaticsDepartment of Mechanical Engineering MEHB334

FLUID STATICSFLUID STATICS


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

Px = Pz = Pn

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

=> PASCALs Law

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


In a closed system, due to Pascals Law, the pressure change produced at one point in the system will be transmitted throughout the entire systemPrinciple of the Hydraulic Lift

2

2

1

1

AF

AF =


A 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.

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Balancing the force in s direction : -

singdsdP =

But sin = dz/dl, so : -

gdzdP =

For incompressible fluid, is constant : -

Constant gz P =+ = Piezometric Pressure (Pascal)

Constant z gP =+ = Piezometric Head (meter)

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If If BB > > AA which distribution is correct ?which distribution is correct ?

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( )Datum2211 z gP z gP z gP +=+=+

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gh P P ah +=

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

The equation has to be applied to each The equation has to be applied to each fluid separatelyfluid separately

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Troposphere : Troposphere : RTPg

gdzdP == Function of Temperature Function of Temperature

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Applying the equation from 1 to 2 : -

2m21m1 gz P gz P +=+P1 = Patm ; z2-z1 = h ; P2 = Pvap

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( ) ( )abggh P-P f2fm21 =Static pressure change from

A to B

Static pressure change due to

the system

Static pressure change due to difference in

elevation

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

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Mechanics of Fluids : Chap. 3: Fluid StaticDepartment of Mechanical Engineering MEHB334

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Importance of Hydrostatic Force CalculationsExamples : -

hh

dF=PdA y

x

y

dA

edge view

normal view

elemental area

centroid

Area, A


-,

.

AsinygAhgF__

h ==

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


3.8.1 Hydrostatic Force on Plane Surfaces3.8.1 Hydrostatic Force on Plane Surfaces

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

Ay

Iy _

xxcp =

Ay

Ix _

xycp =

- And xcp, is given by : -

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


3.8.1 Hydrostatic Force on Plane Surfaces3.8.1 Hydrostatic Force on Plane Surfaces- Centroid and Second Moment of Area, Ixx of regular shapes : -

This is given in This is given in Appendix . No Appendix . No Need to RememberNeed to Remember

w

h

hinge

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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 Surfaces3.8.1 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=hdA

dF

dF

dFy

x ds

Vertical projection

v

C v

Area, A v

x

y

l

H


3.8.2 Hydrostatic Force on Curve Surfaces3.8.2 Hydrostatic Force on Curve Surfaces



- Horizontal Component of the Force, Fx : -

proj_vertproj_vert_CentroidVV

_

x AAhgF ==_

P

- Line of action of Fx : -

vv

_v_xx

v_cp

Ah

Iy =

- Vertical Component of the Force, Fy : -

SurfaceAbove Fluid of Weight== surface_abovey gF

- 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 : - 2y

2x FFF +=

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

surface_above_imaginaryy gF =



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

DCB



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


3.8.2 Hydrostatic Force on Curve Surfaces3.8.2 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 = 80 cm.

A B

CDCOG

Mg

Vol4

Vol1

Vol2

Vol3

b

f


3.9 3.9 BouyancyBouyancy- 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 : -

Vol displaced of Weight =

== bouyancy1f FgMg

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

Archimedes Principle


3.9 3.9 BouyancyBouyancy

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 anddimension 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.



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?



HydrometryHydrometry .. Hydrometer .. Hydrometer is a device use to measure the is a device use to measure the specific gravity of a liquidspecific gravity of a liquid

Based on the buoyancy Based on the buoyancy principleprinciple

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


3.10 Stability of Immersed Body3.10 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 Body3.11 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 Chapter 3End of Chapter 3


Chapter 3

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