Chapter 3: Pressure and Fluid StaticsFluid Statics Fall 2009 University of Palestine College of...

Chapter 3: Pressure and Fluid Statics

Fall 2009

University of PalestineCollege of Engineering & Urban PlanningApplied Civil Engineering

Lecturer:Eng. Eman Al.Swaity

Chapter 3: Pressure and Fluid StaticsESOE 505221 Fluid Mechanics 2

FLUID STATICS

Hydrostatics is the study of pressures throughout a fluid at rest and the pressure forces on finite surfaces. As the fluid is at rest, there are no shear stresses in it. Hence the pressure at a point on a plane surface always acts normal to the surface, and all forces are independent of viscosity. The pressure variation is due only to the weight of the fluid. As a result, the controlling laws are relatively simple, and analysis is based on a straightforward application of the mechanical principles of force and moment. Solutions are exact and there is no need to have recourse to experiment.

EGGD3109 Fluid Mechanics


Pressure

Pressure is defined as a normal force exerted by a fluid per unit area(even imaginary surfaces as in a control volume).

Units of pressure are N/m2, which is called a pascal (Pa). Since the unit Pa is too small for pressures encountered in practice, kilopascal (1 kPa = 103 Pa) and megapascal (1 MPa = 106 Pa) are commonly used. [ML-1T-2]Other units include bar, atm, kgf/cm2, lbf/in2=psi.



Pressure

1 bar = 105 Pa = 0.1 MPa = 100 kPa

1 atm = 101,325 Pa = 101.325 kPa = 1.01325 bars

1 kgf/cm2 = 9.807 N/cm2 = 9.807 104 N/m2 = 9.807 104 Pa = 0.9807 bar = 0.9679 atm

1 atm = 14.696 psi.

1 kgf/cm2 = 14.223 psi.



Pressure at a Point

By considering the equilibrium of a small triangular wedge of fluid extracted from a static fluid body, one can show that for any wedge angle θ, the pressures on the three faces of the wedge are equal in magnitude:



Pressure at a Point

Pressure at any point in a fluid is the same in all directions.

Pressure has a magnitude, but not a specific direction, and thus it is a scalar quantity.

Proof on the Blackboard



Pressure at a Point



Pressure at a Point

This result is known as Pascal's law, which states that the pressure at a point in a fluid at rest, or in motion, is independent of direction as long as there are no shear stresses present.

Pressure at a point has the same magnitude in all directions, and is called isotropic .



Variation of Pressure with Depth

Therefore, the hydrostatic pressure increases linearly with depth at the rate of the specific weight of the fluid.




In the presence of a gravitational field, pressure increases with depth because more fluid rests on deeper layers.

To obtain a relation for the variation of pressure with depth, consider rectangular element

Force balance in z-direction gives

Dividing by x and rearranging gives

∆z is called the pressure head

2 1

0

0z zF ma

P x P x g x z

2 1 sP P P g z z







Pressure in a fluid at rest is independent of the shape of the container.

Pressure is the same at all points on a horizontal plane in a given fluid.



Absolute, gage, and vacuum pressures

Actual pressure at a given point is called the absolute pressure.

Most pressure-measuring devices are calibrated to read zero in the atmosphere, and therefore indicate gage pressure, Pgage = Pabs - Patm.

Pressure below atmospheric pressure are called vacuum pressure, Pvac=Patm - Pabs.



Absolute, gage, and vacuum pressures



Hydrostatic Pressure Difference Between Two Points

For a fluid with constant density,

If you can draw a continuous line through the same fluid from point 1 to point 2, then p1 = p2 if z1 = z2.

•By this rule p1 = p2 and p4 = p5

•p2 does not equal p3 even though they are at the same elevation, because one cannot draw a line connecting these points through the same fluid. In fact, p2 is less than p3 since mercury is denser than water.




Any free surface open to the atmosphere has atmospheric pressure, p0.

The shape of a container does not matter in hydrostatics




Pressure in layered fluid.



Pascal’s Law

Two points at the same elevation in a continuous fluid at rest are at the same pressure, called Pascal’s law,

Pressure applied to a confined fluid increases the pressure throughout by the same amount. In picture, pistons are at same height:

Ratio A2/A1 is called ideal mechanical advantage

1 2 2 21 2

1 2 1 1

F F F AP P

A A F A



Pascal’s Law



Pascal’s Law



Pressure Measurement and Manometers

Piezometer tube The simplest manometer is a tube, open at the top, which is attached to a vessel or a pipe containing liquid at a pressure (higher than atmospheric) to be measured. This simple device is known as a piezometer tube.

This method can only be used for liquids (i.e. not for gases) and only when the liquid height is convenient to measure. It must not be too small or too large and pressure changes must be detectable.




U-tube manometer This device consists of a glass tube bent into the shape of a "U", and is used to measure some unknown pressure. For example, consider a U-tube manometer that is used to measure pressure pA in some kind of tank or machine.

Finally, note that in many cases (such as with air pressure being measured by a mercury manometer), the density of manometer fluid 2 is much greater than that of fluid 1. In such cases, the last term on the right is sometimes neglected.




Differential manometerA differential manometer can be used to measure the difference in pressure between two containers or two points in the same system. Again, on equating the pressures at points labeled (2) and (3), we may get an expression for the pressure difference between A and B:




Inverted U-tube Differential manometers


ghhgPhgP

PP

datumAt

liquidBBAA

)()(

21

x- x

21



Inverted U-tube Differential manometers-Example


22

21

2

kN/m 8.18N/m 18835

15.081.98001.081.910003.081.9100019620

)()(

N/m 1962081.910002

21

x-x

B

B

liquidBBAA

A

P

P

ghhgPhgP

P

PP



Inclined-tube manometerAs shown above, the differential reading is proportional to the pressure difference. If the pressure difference is very small, the reading may be too small to be measured with good accuracy. To increase the sensitivity of the differential reading, one leg of the manometer can be inclined at an angle θ, and the differential reading is measured along the inclined tube.




An elevation change of z in a fluid at rest corresponds to P/g.A device based on this is called a manometer.A manometer consists of a U-tube containing one or more fluids such as mercury, water, alcohol, or oil.Heavy fluids such as mercury are used if large pressure differences are anticipated.1 2

2 atm

P P

P P gh

The Manometer




Mutlifluid Manometer

For multi-fluid systems

Pressure change across a fluid column of height h is P = gh.

Pressure increases downward, and decreases upward.

Two points at the same elevation in a continuous fluid are at the same pressure.

Pressure can be determined by adding and subtracting gh terms.




Example:U-tube manometer containing mercury was used to find the negative pressure in the pipe, containing water. The right limb was open to the atmosphere. Find the vacuum pressure in the pipe, if the difference of mercury level in the two limbs was 100 mm and height of water in the left limb from the centre of the pipe was found to be 40 mm below.




2

22

33

2211

/ 27.8610073.13.

..

/ 73.13/ 13734

01.0*81.9*106.1304.0*81.9*101

0

21

mKNabsP

atmPgaugePabsP

mKNmNP

P

ghghP

PP

pipe

pipepipepipe

pipe

pipe

pipe





23

23

2.

.

23

/38.602*81.9*10001080

/73.2073.1001080

/73.10081.9*13600*755.0

/806.0*81.9*106.13

)(0

21

mkNghPP

mkNP

mkNP

PPP

mkNP

hgP

PP

waterairL

abs

atm

atmairabs

air

mercurymercuryair

The atmospheric pressure is 755 mm of mercury (sp. Gravity = 13.6), calculatei) Absolute pressure of air in the tank,ii) Pressure gauge reading at L.

General Example


Measuring Pressure Drops

Manometers are well--suited to measure pressure drops across valves, pipes, heat exchangers, etc. Relation for pressure drop P1-P2 is obtained by starting at point 1 and adding or subtracting gh terms until we reach point 2. If fluid in pipe is a gas, 2>>1 and P1-P2 gh

(Mistyped on page 73)



The Barometer

Atmospheric pressure is measured by a device called a barometer; thus, atmospheric pressure is often referred to as the barometric pressure.

PC can be taken to be zero since there is only Hg vapor above point C, and it is very low relative to Patm. Change in atmospheric pressure due to elevation has many effects: Cooking, nose bleeds, engine performance, aircraft performance.

C atm

atm

P gh P

P gh



The Barometer

Standard atmosphere is defined as the pressure produced by a column of mercury 760 mm (29.92 inHg

or of water about 10.3 m ) in height at 0°C (Hg = 13,595 kg/m3) under standard gravitational acceleration (g = 9.807 m/s2).

1 atm = 760 torr and 1 torr = 133.3 Pa



Fluid Statics

Fluid Statics deals with problems associated with fluids at rest. In fluid statics, there is no relative motion between adjacent fluid layers. Therefore, there is no shear stress in the fluid trying to deform it. The only stress in fluid statics is normal stress

Normal stress is due to pressureVariation of pressure is due only to the weight of the fluid → fluid statics is only relevant in presence of gravity fields.

Applications: Floating or submerged bodies, water dams and gates, liquid storage tanks, etc.



Hoover Dam



Hoover Dam



Hoover Dam

Example of elevation head z converted to velocity head V2/2g. We'll discuss this in more detail in Chapter 5 (Bernoulli equation).



Pressure Distributions-Flat Surfaces









Pressure Distributions-Curved Surfaces



Hydrostatic Forces on Plane Surfaces

On a plane surface, the hydrostatic forces form a system of parallel forcesFor many applications, magnitude and location of application, which is called center of pressure, must be determined.Atmospheric pressure Patm can be neglected when it acts on both sides of the surface.



Resultant Force

The magnitude of FR acting on a plane surface of a completely submerged plate in a homogenous fluid is equal to the product of the pressure PC at the centroid of the surface and the area A of the surface



Resultant Force



Center of Pressure

Line of action of resultant force FR=PCA does not pass through the centroid of the surface. In general, it lies underneath where the pressure is higher.Vertical location of Center of Pressure is determined by equation the moment of the resultant force to the moment of the distributed pressure force.

Ixx,C is tabulated for simple geometries. Derivation of FR and examples on blackboard

,xx Cp C

c

Iy y

y A



The centroidal moments of inertia for some common geometries



Submerged Rectangular Plate

What is the yp for case (a)?



Submerged Rectangular Plate

What is the yp for case (a)?



Example: Hydrostatic Force Acting on the

Door of a Submerged CarA heavy car plunges into a lake during an accident and lands at the bottom of the lake on its wheels. The door is 1.2 m high and 1 m wide,and the top edge of the door is 8 m below the free surface of the water.Determine the hydrostatic force on the door and the location of the pressure center, and discuss if the driver can open the door.

Pave = PC = ghC = g(s + b/2)= 84.4 kN/m2

FR = PaveA = (84.4 kNm2) (1 m 1.2 m) = 101.3 kN

yP = 8.61 m



Example: Hydrostatic Force Acting on the

Door of a Submerged Car

Pave = PC = ghC = g(s + b/2)= 84.4 kN/m2

FR = PaveA = (84.4 kNm2) (1 m 1.2 m) = 101.3 kN

yP = 8.61 m

Discussion A strong person can lift 100 kg, whose weight is 981 N or about 1 kN. Also, the person can apply the force at a point farthest from the hinges (1 m farther) for maximum effect and generate a moment of 1 kN · m. The resultant hydrostatic force acts under the midpoint of the door, and thus a distance of 0.5 m from the hinges. This generates a moment of 50.6 kN · m, which is about 50 times the moment the driver can possibly generate. Therefore, it is impossible for the driver to open the door of the car. The driver’s best bet is to let some water in (by rolling the window down a little, for example) and to keep his or her head close to the ceiling. The driver should be able to open the door shortly before the car is filled with water since at that point the pressures on both sides of the door are nearly the same and opening the door in water is almost as easy as opening it in air.








Horizontally immersed surface

gh *A




Vertically immersed surface

dxbgxp.b.dx

p.b.dxp

gx ρp

..Rbody on the pressure Total The

dA strip on the pressure Total

strip thealongpoint any at Pressure

AxgxAgdxbxgR )(.




Center of Pressure

ogIdxbxgdxbgx

p.b.dx.(x)M

p.b.dx.(x)

p.b.dx

gx ρp

.

body allfor

M

o,-o surface freeabout pressure theofmoment The

strip on the pressure Total

strip thealongpoint any at Pressure

22




xAx

Ih

AxgP

xAx

I

Ax

I

Axg

gI

P

gIh

gIhP

G

Gooo

o

Center of Pressure




Example 1




Example 2




Example 3







Inclined Immersed Surface

AxgP

Using the same procedures as in Vertical surface

sinl




xAx

Ih

Axg

G

2sin

P














Hydrostatic Forces on Curved Surfaces

FR on a curved surface is more involved since it requires integration of the pressure forces that change direction along the surface.Easiest approach: determine horizontal and vertical components FH and FV separately.




Horizontal force component on curved surface: FH=Fx. Line of action on vertical plane gives y coordinate of center of pressure on curved surface.Vertical force component on curved surface: FV=Fy+W, where W is the weight of the liquid in the enclosed block W=gV. x coordinate of the center of pressure is a combination of line of action on horizontal plane (centroid of area) and line of action through volume (centroid of volume).Magnitude of force FR=(FH

2+FV2)1/2

Angle of force is = tan-1(FV/FH)




1) Liquid above surface




1) Liquid above surface Horizontal component of force on surface: By considering the equilibrium of the liquid mass contained in ABC, we get FH = F = resultant force of liquid acting on vertically projected area (BC) and acting through the centre of pressure of F. Vertical component of force on surface By considering the equilibrium of the liquid mass contained in ADEC, we get FV = W = weight of liquid vertically above the surface (ADEC) and through the centre of gravity of the liquid mass.Resultant force FR pointing downward, and making an angle α with horizontal




2) Liquid below surface

The space above the surface ADCB may be empty or contain other fluid.

Imagine that the space (ADCB) vertically above the curved surface is occupied with the same fluid as that below it (disregard what actually is filling that space). Then the surface AB could be removed without disrupting the equilibrium of the fluid. That means, the force acting on the underside of the surface would be balanced by that acting on the upper side under this imaginary condition. Therefore we may use the same arguments as in the preceding case:




2) Liquid below surface

Horizontal component of force on surface: FH = F = resultant force of liquid acting on vertically projected area (AB) and acting through the centre of pressure of F.

Vertical component of force on surface FV = W = weight of imaginary liquid (i.e., same liquid as on the other side of the surface) vertically above the surface (ADCB) and through the centre of gravity of the liquid mass. Resultant force FR pointing upward, and making an angle α with horizontal



Example: A Gravity-Controlled Cylindrical Gate

A long solid cylinder of radius 0.8 m hinged at point A is used as an automatic gate. When the water level reaches 5 m, the gate opens by turning about the hinge at point A. Determine (a) the hydrostatic force acting on the cylinder and its line of action when the gate opens and (b) the weight of the cylinder per m length of the cylinder.

= 36.1 kN

= 39.2 kN



Example: A Gravity-Controlled Cylindrical Gate

= 1.3 kN

The weight of the cylinder is



Example: Curved Surfaces


P1

P2

PH

PV




kN 5.7309.444.29PP

44.09 1*381.92.39 2

1)(

2

1P

29.4 1*381.9)(P

unit widthFor

kN 2.39481.910Mat pressure

kN 81.9181.910Lat pressure

21

2

1

3

3

H

LM

L

P

kNAreapp

kNAreap










o

VVV

V

V

HHH

H

H

PPP

NP

NP

PPP

kNAreagh

kNAreagh

6.6443.36

8.76tan

kN 8.76PPP

kN 61.672661.41

kN 26260005.14

5.181.910

kN 61.41416105.12

5.181.9850

kN 43.3655.1698.52

16.55 5.1*1.55.181.910 2

1)(

2

1P

52.98 5.1*3381.910 2

1)(

2

1P

1-

2V

2H

21

23

2

2

1

21

322

311


Buoyancy and Stability

Buoyancy is due to the fluid displaced by a body. FB=fgV.

Archimedes principal : The buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume.





Archimedes’ Principle states that the buoyant force has a magnitude equal to the weight of the fluid displaced by the body and is directed vertically upward.

•Buoyant force is a force that results from a floating or submerged body in a fluid.•The force results from different pressures on the top and bottom of the object•The pressure forces acting from below are greater than those on top

Now, treat an arbitrary submerged object as a planar surface:

Arbitrary Shape

V

Forces on the Fluid




Balancing the Forces of the F.B.D. in the vertical Direction:

VAhhW 12

W is the weight of the shaded areaF1 and F2 are the forces on the plane surfacesFB is the buoyant force the body exerts on the fluid

Then, substituting:

Simplifying,

The force of the fluid on the body is opposite, or vertically upward and is known as the Buoyant Force.The force is equal to the weight of the fluid it displaces.

g




We find that the buoyant force acts through the centroid of the displaced volume.

The location is known as the center of buoyancyThe location is known as the center of buoyancy.




We can apply the same principles to floating objects:

If the fluid acting on the upper surfaces has very small specific weight (air), the centroid is simply that of the displaced volume, and the buoyant force is as before.



Buoyancy force FB is equal only to the displaced volume fgVdisplaced.

Three scenarios possible

1. body<fluid: Floating body

. body=fluid: Neutrally buoyant

3. body>fluid: Sinking body

EGGC3109 Fluid Mechanics


Example 1:


h0.75

1.25

G

B

A wooden block of width 1.25 m, depth 0.75And length 3.0 m is floating in water. Specific weightOf wood is 6.4kN/m3 find:

Position of center of buoyancy

m

h

h

KNW

gVW

FW

dis

B

244.02

0.489buoyancy ofCenter

489.0

)3*25.1*(*81.9*10001018

184.6*0.3*25.1*75.03

.


Example 2:



Stability

What about a case where the ball is on an inclined floor?

It is not really appropriate to discuss stability for this case since the ball is not in a state of equilibrium. In other words, it cannot be at rest and would roll down the hill even without any disturbance.



Stability of Immersed Bodies


Stable Equilibrium: if when displaced returns to equilibrium position.

Unstable Equilibrium: if when displaced it returns to a new equilibrium position.

Stable Equilibrium: Unstable Equilibrium:

C > CG, “Higher” C < CG, “Lower”


Stability of Immersed Bodies

Rotational stability of immersed bodies depends upon relative location of center of gravity G and center of buoyancy B.

G below B: stable

G above B: unstable

G coincides with B: neutrally stable.



Stability of Floating Bodies

If body is bottom heavy (G lower than B), it is always stable.

Floating bodies can be stable when G is higher than B due to shift in location of center buoyancy and creation of restoring moment.

Measure of stability is the metacentric height GM. If GM>1, ship is stable.

The metacenter may be considered to be a fixed point for most hull shapes for small rolling angles up to about 20°.



Stability of Floating Bodies



Metacentre and Metacentric Height




In Water



h

bdhV

bdI

12

section crossr rectangulaFor 3


Example

mBGBMGM

mV

IBM

mOBOGBG

OG

mB

h

h

gVW

FW

dis

B

019.005.0069.0

069.015.03.0

12/15.0

05.015.02.0

2.0

15.02

0.3O

3.0

)1*50.0*(*81.9*100010472.1

3

3

.


The Golden Crown of Hiero II, King of Syracuse

The weight of the crown and nugget are the same in air: Wc = cVc = Wn = nVn.If the crown is pure gold, c=n which means that the volumes must be the same, Vc=Vn.In water, the buoyancy force is B=H2OV.If the scale becomes unbalanced, this implies that the Vc ≠ Vn, which in turn means that the c ≠ n

Goldsmith was shown to be a fraud!


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