Fluid Mechanics-61341

An-Najah National UniversityCollege of Engineering

Chapter [2]

Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics Dr. Sameer Shadeed1

Dr. Sameer Shadeed

Chapter [2]

Fluid Statics

Fluid Statics Problems

Fluid statics refers to the study of fluids at rest ormoving in such a manner that no shearingstresses exist in the fluid

These are relatively simple problems since no

Dr. Sameer Shadeed2

These are relatively simple problems since novelocity gradients exist. Thus, viscosity does notplay a role

Applications include the hydraulic pressure,manometry, dams, and fluid containment (tanks)

Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Pressure (P) is the force per unit area, where the force isperpendicular to the area

P (N/m2 or Pa) =A (m2)

Pressure

1 kN/m2 = 1 kPa

1 kPa = 0.145 psi

Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics Dr. Sameer Shadeed3

Pressure in a fluid acts equally in all directions

Pressure in a static liquid increases linearly with depth

increase in depth (m)

pressure increase

p= γ h

Pressure at a Point

Pressure is a scalar quantity that is defined atevery point within a fluid

Dr. Sameer Shadeed4 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

Force balance in the x-direction:

Pressure at a Point

Force balance in the z-direction:

Vertical force on DA Vertical force on lower boundary

Total weight of wedge element

= specific weight

Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics5 Dr. Sameer Shadeed

From last slide:

Divide through by to get

Pressure at a Point

Now shrink the element to a point:

This can be done for any orientation, so

The result shows that pressure at any pointin a fluid at rest has a single value,independent of direction as long as thereare no shearing stresses (or velocitygradients) present in the fluid

Pressure at a Point

Dr. Sameer Shadeed7

gradients) present in the fluid

For fluids in motion with shearing stresses,this result is not exactly true, but is still avery good approximation for most flows

Pressure Transmission

In a closed system, pressure changes from onepoint are transmitted throughout the entire system(Pascal’sPascal’s LawLaw).

Hydraulic LiftFluid Mechanics-2nd Semester 2010- [2] Fluid Statics8 Dr. Sameer Shadeed

outout in

Pressure Variation with Elevation

Static fluid: All forces must balance as there are no accelerations

Look at force balance in direction of D ll

From the previous figure, note that

Shrink cylinder to zero length:

From the previous slide:

The pressure-elevation relation derived on the previous slide,

is perfectly general (applies also to variable gg)

But if gg is constant, the above equation is easy to integrate:

The quantity is known as the piezometric pressure

and is called the piezometric head

For an incompressible fluid (gg is constant), pressure andelevation at one point can thus be related to pressure andelevation at another point as:

Constant22

hzzppp )( 1221

Absolute and Gage Pressure

What is the pressure at the faucet?

What do you do if you want more pressure at the faucet?Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics15 Dr. Sameer Shadeed

Blood Pressure

26.8 k Pa

Example 1

Solution:

What is the water pressure at a depth of 35 ft?

Solution: With the information given, all we can calculate is the pressure difference between points 1 and 2

Example 2

What is the gage pressure at point 3?

Solution: Two step solution:1) Calculate2) Calculate

Example 3

s.g.=0.8

(relative to atmospheric pressure at point 1)

Pressure Measurement

MechanicalMechanical PressurePressure GagesGages

The BourdonBourdon pressure and AneroidAneroid barometerare typical mechanical devices for measuring gageand absolute pressures, respectively

Pressure Measurement

LiquidLiquid PressurePressure GagesGages

ManometerManometer:: gravimetric device based upon liquid leveldeflection in a tube

Mercury barometer: evacuated glass tube with openend submerged in mercury

The Manometer

Simple, accurate device for measuring small tomoderate pressure differences

Rules of ManometryManometry:

pressure change across a fluid column ofgg

Dr. Sameer Shadeed22

height h is ggh

pressure increases in the direction of gravity,decreases in the direction opposing gravity

two points at the same elevation in acontinuous static fluid have the samepressure

The Manometer

yx 32211

)(01 gagepp atm

The Manometer

Find the location of the surface in the manometer

Solution: The distance Dhis the height of the liquid in the manometer above the heavier liquid in the tank

Example 4

heavier liquid in the tank

Find the gage pressure at the pipe center

Example 5

Solution: Manometer equation from the pipe center to the open end of the manometer

)4.62)(1(5.0)4.62)(2(1)4.62)(1(5.2

Example 6

Example 6 (Solution)

Find the specific weight of the fluid which filled part CD of the tube

cmhcmh

186.102)5.0(4

Solution:

Example 7

B C4 Manometer Equation

DliqA phhp )05.0(

3/4995)9810(10186.0

)05.010186.0(

)05.0(

)(gageopp DA

Differential Manometer

Used for measuring pressure differences betweenpoints along a pipe

21 )( phlhlp wmw

hppp mw )(12

Find the change in piezometric pressure and in piezometrichead between points 1 and 2.

In general

Solution: Manometer equation from point 2 to point 1

Example 8

In general

(piezometric pressure)

(piezometric head)

Example 9

Pressure Forces on Plane Surfaces Surfaces exposed to fluids experience a force due to thepressure distribution in the fluid The resultant force on vertical, rectangular surfaces canbe found using a graphical interpretation known as thepressurepressure prismprism

The differential force is

Integrating to get the total force on the gate yields:

dAldAhpdAdF sin

dAlF sin

Pressure Forces on Plane Surfaces

From basic mechanics, we recall for first moment of area,

Recognizing that

AldAl c

αsinAlγF c

AhFhαsinl ccc Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics Dr. Sameer Shadeed

In general, the resultant force on an area equals the pressure atthe centroid of the area (ghc) times the area (A)

To complete the analysis, we must compute the location of thecenter of pressure where the resultant force F can be assumedto act

Pressure Forces on Plane Surfaces

dAldAllldFFlp

sin)sin( 2

AlAlIAlIFl

AlIIdAl

cccccp

/bygmultiplyin)(sin

thatgRecognizin

Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics Dr. Sameer Shadeed

Example 10

Example 11

Find the normal force required to open the rectangular gateif it is hinged at the top. The gate is 5 m wide and θ =30o

Example 12

Solution: First find the total hydrostatic force acting on the plate

MNmmmmN

45.25510/9810 3

ccp 92.11

2574.11

12/55)5.2)30cos/8((

MNFMNmF

31.1))30cos/8(92.11((45.25

Given: Gate AB is 4 ft wide, hinged at A and Gage Greads -2.17 psi. Find the horizontal force at B to hold gate.

Example 13

OilSG=0.75

Water18ft

First convert negative pressure in tank to ft of water

h 54.62

14417.2

AhF coil

33706434.6275.

Total hydrostatic force acting on the gate from oil

Ibo 33706434.6275.

AhF cwater

1497664)3518(4.62

ccwaterp 3.10

12/6410

ftl oilp 4)( Total hydrostatic force acting on the gate from water

4ft3.3ft

FwFoil

Boilwater

6433703.314976

Resultant pressure forces on curved surfaces are more difficult todeal with because the incremental pressure forces, which are normal tothe surface, vary continually in direction There are two ways to approach the problem. One is to use directintegration and the second method is to utilize the basic mechanicsconcepts of a free body and the equilibrium equations

Pressure Forces on Curved Surfaces

Fig. 2.11

Example 14

Example 15

The familiar laws of bouncy (ArchamedesArchamedes’’ principleprinciple)and flotation are usually stated (1) a body immersed in afluid is buoyed up by a force equal to the weight of fluiddisplaced; and (2) a floating body displaces its own weightof the liquid in which it floats

Buoyancy and the Stability of Floating Bodies

The resultant buoyant force on a submerged orpartially submerged object in a static fluid is given byArchimedes’Archimedes’ principleprinciple as:

)( objectsubmergedofvolumeFB

)( isplaceddiquidlofvolumeFB

The buoyant force is equal to the weight of thefluid displaced by the object and is in a directionopposite the gravitational force

The line of action of the buoyantforce passes through the centroidof the displaced volume, oftencalled the center of buoyancy(COBCOB)

The stability of submerged objects(balloon and sub-marine) isdetermined by the center ofgravity (GG):

Stable: GG is below COBCOB

Unstable: GG is above COBCOBDr. Sameer Shadeed71 Fluid Mechanics-2nd Semester 2010- [2] Fluid Statics

For floating objects, stability is complicated by the fact that theCOBCOB changes with rotation

The stability of floating objects (Ship) is determined by themetacentric (MM) (the point of intersection of the vertical linethrough B’B’ with the centerline of the ship):

Stable: GG is below MM

Unstable: GG is above MM

Example 16

Solution:

Example 17

Example 18

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