Chapter 03 Handout

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Chapter 3: Pressure and Fluid Statics

Professor Seung-Jae MoonSchool of Mechanical Engineering

Hanyang University

Fall 2012

Chapter 3: Pressure and Fluid StaticsME33 : Fluid Flow 2

Pressure

Pressure is defined as a normal force exertedby a fluid per unit area.

Units of pressure are N/m2, which is called apascal (Pa).

Since the unit Pa is too small for pressuresencountered in practice, kilopascal (1 kPa =103 Pa) and megapascal (1 MPa = 106 Pa) are

commonly used.Other units include bar , atm, kgf/cm2, lbf/in2=psi.

1 bar = 105 Pa = 0.1 Mpa = 100 kPa

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

1 kgf/cm2 = 9.807 N/cm2 = 9.807 x 104

N/m2 = 9.807 bar = 9.7679 atm

1 atm = 14.696 psi

1 kgf/cm2 = 14.223 psi

1 torr =1 mmHg = 1/760 atm = 133.32 Pa



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Absolute, gage, and vacuum pressures

Actual pressure at a give point is called the absolute pressure and it is

measured relative to absolute vacuum (i.e., absolute zero pressure)

Most pressure-measuring devices are calibrated to read zero in the

atmosphere, and therefore they indicate gage pressure, Pgage=Pabs - Patm.

Pressure below atmospheric pressure are called vacuum pressure,

Pvac=Patm - Pabs.


Example 3-1 Absolute Pressure of a Vacuum Chamber

A vacuum gage connected to a chamber reads 5.8 psi at a

location where the atmospheric pressure is 14.5 psi.

Determine the absolute pressure in the chamber.

Solution

Pabs = Patm – Pvac = 14.5 – 5.8 = 8.7 psi



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Pressure at a Point

Pressure at any point in a fluid has the

same magnitude in all directions.

Pressure has a magnitude, but not a

specific direction, and thus it is a scalar

quantity.

PP

PP

P


Variation of Pressure with Depth

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

2 1

0

0

z zF ma

P x P x g x z

2 1 sP P P g z z

s= g: specific weight of the fluid, z: pressure head



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P=Patm+ gh or Pgage= gh

Variation of density with

elevation

P=constant



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.



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Scuba Diving and Hydrostatic Pressure


Pressure on diver at

100 ft?

Danger of emergency

ascent?

,2 3 2

,2 ,2

1998 9.81 100

3.28

1298.5 2.95

101.325

2.95 1 3.95

gage

abs gage atm

kg m mP gz ft

m s ft

atmkPa atm

kPa

P P P atm atm atm

Scuba Diving and Hydrostatic Pressure

1 1 2 2

1 2

2 1

3.954

1

PV PV

V P atm

V P atm

100 ft

1

2

Boyle’s law

If you hold your breath on ascent, your lungvolume would increase by a factor of 4, which

would result in embolism and/or death.



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Blaise Pascal (1623-1662, France)

수학자,물리학자,철학자

허약체질로집에갇혀서공부만했음.

12살에삼각형의내각의합은 180도임을

발견

18살에계산기발명

21살에파스칼의법칙발견

팡세와사람은생각하는갈대


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 2

1 2

1 2 1 1

F F F AP P

A A F A



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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


Example 3-2 Measuring Atmospheric Pressure with a Barometer

Determine the atmospheric pressure at a location

where the barometric reading is 740 mm Hg and

the gravitational acceleration is g =9.81 m/s2.

Assume the temperature of mercury to be 10oC, at

which its density is 13,570 kg/m3

.Solution

Patm = gh

= (13,570 kg/m3) x (9.81 m/s2) x(0.74 m)

= 98,500 Pa =98.5 kPa



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The Manometer

An elevation change of z ina fluid at rest corresponds toP/ g.

A device based on this iscalled a manometer.

A manometer consists of aU-tube containing one or more fluids such as mercury,water, alcohol, or oil.

Heavy fluids such asmercury are used if largepressure differences areanticipated.

Diameter of tube >> a fewmilimeters to avoid surfacetension.

1 2

2 atm

P P

P P gh


Example 3-3 Measuring Pressure with a Manometer

A manometer is used to measure the

pressure in a tank. The used fluid has a

specific gravity of 0.85, and the monometer

column height is 55 cm. If the local

atmospheric pressure is 96 kPa, determine

the absolute pressure within the tank.

Solution

= SG H2O = (0.85)(1000 kg/m3) = 850

kg/m3.

Pgage = gh = (850 kg/m3).(9.81

m/s2).(0.55 m) = 4,600 N/m2 = 4.6 kPa.

P = Patm + gh

= 96 kPa + 4.6 kPa = 100.6 kPa.



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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 – Pascal’s law.

Pressure can be determined by

adding and subtracting gh terms.


Measuring Pressure Drops

Manometers are well--suitedto measure pressure dropsacross valves, pipes, heatexchangers, etc.

Relation for pressure drop P1-P2 is obtained by starting atpoint 1 and adding or subtracting gh terms until wereach point 2.

If fluid in pipe is a gas, 2>> 1and P1 - P2= gh



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Example 3-4 Measuring Pressure with a Multifluid Manometer

The water in a tank is pressurized by air, and thepressure is measured by a multifluid manometer. The

tank is located on a mountain at an altitude of 1,400

m where the atmospheric pressure is 85.6 kPa.

Determine the air pressure in the tank if h1 = 0.1 m,

h2 = 0.2 m, and h3 = 0.35 m. Take the densities of

water, oil, and mercury to be 1000 kg/m3, 850 kg/m3,and 13600 kg/m3 respectively.

Solution


Other Pressure Measurement Devices

Bourdon tubes

Pressure transducers

Strain-gage pressure

transducers

Piezoelectric

transducers

Bourdon tubes Piezoelectric MEMS transducer



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Example 3-5 Effect of Piston Weight on Pressure in a Cylinder

The piston of a vertical piston-cylinder device

containing a gas has a mass of 60 kg and a

cross-sectional area of 0.04 m2. The local

atmospheric pressure is 0.97 bar, and the

gravitational acceleration is 9.81 m/s2. (a)

Determine the pressure inside the cylinder. (b)

If some heat is transferred to the gas and its

volume is doubled, do you expect the pressure

inside the cylinder to change?

Solution

(a)

(b) The volume change will

have no effect on the free-

body diagram drwan in part

(a), and the pressure inside

the cylinder will remain thesame.

If the gas behaves as an

ideal gas, the absolute

temperature doubles

when the volume is

doubled at constatn

pressure. (PV = nRT)


Fluid Statics

Fluid Statics deals with problems associatedwith fluids at rest.

In fluid statics, there is no relative motionbetween 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 pressure.

Variation of pressure is due only to the weight of thefluid. → The fluid statics is only relevant in presenceof gravity fields.

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



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Hoover Dam


Hoover Dam



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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).


Hydrostatic Forces on Plane Surfaces

On a plane surface, thehydrostatic forces form asystem of parallel forces.

For many applications,magnitude and location of application, which is called

center of pressure, mustbe determined.

Atmospheric pressure Patm

can be neglected when itacts on both sides of thesurface.



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Resultant Force

Absolute pressure at any point on the plate

P = P0 + gh = P0 + gysin

Resultant hydrostatic force FR acting on the surface

P0


Resultant Forces

First moment of area

y-coordinate of the centroid of the surface

Pressure at the centroid of the surface

Vertical distance of the centroid from the free surface of the liquid

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.



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Center of Pressure

Line of action of resultant force FR=PC Adoes not pass through the centroid of the

surface. In general, it lies underneath

where the pressure is higher.

Vertical location of Center of Pressure isdetermined by equation the moment of theresultant force to the moment of thedistributed pressure force.

yP : the distance of the center of pressure

from the x-axis.

Second moment of area (area moment of

inertia)

Parallel axis theorem

Ixx,C: second moment of area about

the x-axis passing through the

centroid

Ixx,C is tabulated for simple

geometries.


Center of Pressure

, xx C

p C

c

I y y

y A

Solving for yP:

P0=0:

Vertical distance of the center of the pressure

from the free surface:



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Example 3-6 Hydrostatic Force Acting on the Door of a Submerged Car

A 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.

Solution

Properties: = 1000 kg/m3, g = 9.81 m/s2.

Dimensions : b = 1.2 m, s = 8 m, w = 1 m.

Average gage pressure on the door centroid

PatmPatm+ gy Door


Resultant hydrostatic force on the door

FR = Pavg A = Pavgbw

= 84.4 x 1 x 1.2 = 101.3 kN

Pressure center

FRyP = gIxx,O = MO

Ixx,O = Ixx,C + yC2 A = wb3/12 + (s + b/2)2wb

= 1x1.23/12 + (8 + 0.6)2x1x1.2

= 88.896 m4

yP = gIxx,O /FR = 1000 x 9.81 x 88.896 / (101.3 x 103)

= 8.61 m

Example 3-6 Hydrostatic Force Acting on the Door of a Submerged Car

O

s

b

w

yPFR

MO



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Hydrostatic Forces on Submerged 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. (Newton’s third law)


Hydrostatic Forces on Curved Surfaces

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 theenclosed 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 = (FH2+FV

2)1/2

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



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Example 3-7 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.

N = 0



Solution: = 1000 kg/m3

(a) Horizontal force on vertical surface

Fx – FH = 0

FH = Fx = Pavg A = ghC A = g(s + r /2) A

= 1000 x 9.81 x (4.2 + 0.8/2) x 0.8 x 0.1

= 36,100 N = 36.1 kN

Vertical force on horizontal surface

Fy = Pavg A = ghC A = ghbottom A

= 1000 x 9.81 x 5 x 0.8 x 0.1 = 39200 N = 39.2 kN

Weight of the fluid block per m length

Fy = Pavg A = gV = g(R2 – R2 / 4).(1 m)

= 1000 x 9.81 x 5 x 0.82 x (1 – / 4) x 1

= 1300 N = 1.3 kN



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Fy – W – FV = 0

FV = Fy – W = 39.2 – 1.3 = 37.9 kN

Resultant force

(b) Moment about A

FRRsin – WcylR = 0

Wcyl = Fr sin = 52.3sin46.4o = 37.9 kN


FR

A

Wcyl


Buoyancy and Stability

Buoyant force is a force that tends

to lift body.

FB – Ftop + Fbottom = 0

FB = Fbottom – Ftop

= f g(s+h) A – f gsA

= f ghA = f gV.

Buoyancy is due to the fluid

displaced by a body. FB = f gV.

Archimedes pr incipal : 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.

FB



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Buoyancy and Stability

Buoyancy force FB is

equal only to the

displaced volume

f gVdisplaced.

Three scenarios

possible

body< fluid: Floating

body body= fluid: Neutrally

buoyant

body> fluid: Sinking body


Example 3-8 Measuring Specific Gravity by a Hydrometer

If you have a seawater aquarium, you have probably

used a small cylindrical glass tube with some lead-

weight at its bottom to measure the salinity of the

water by simply watching how deep the tube sinks.

Such a device that floats in a vertical position and is

used to measure the specific gravity of a liquid is

called a hydrometer. The top part of the hydrometer

extends above the liquid surface, and the divisions

on it allow one to read the specific gravity directly.

The hydrometer is calibrated such that in pure water

it reads exactly 1.0 at the air-water interface. (a)

Obtain a relation for the specific gravity of a liquid as

a function of distance z from the mark

corresponding to pure water and (b) determine the

mass of lead that must be poured into a 1-cm-

diameter, 20-cm-long hydrometer if it is to float

halfway (the 10-cm mark) in pure water.

Assumptions: w = 1000 kg/m3



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(a) Solution

Whydro = FB,w = wgVsub = wgAz0

In a fluid lighter than water ( f < w)

Whydro = FB,f = f gVsub = f gA(z0 + z)

In a fluid heavier than water ( f > w)

Whydro = FB,f = f gVsub = f gA(z0 - z)

(b) FB

= Wlead

w

gVsub

= mg

m = wVsub = w( R2hsub) = 1000(x 0.0052 x 0.1)

= 0.00785 kg = 7.85 g

Example 3-8 Measuring Specific Gravity by a Hydrometer


Example 3-9 Weight Loss of an Object in Seawater

A crane is used to lower weights into the sea ( f = 1025 kg/m3) for an underwater construction

project. Determine the tension in the rope of the

crane due to a rectangular 0.4 m x 0.4 m x 3 m

concrete block ( concrete = 2300 kg/m3) when it is

(a) suspended in the air and (b) completely

immersed in water.

(a) V = 0.4 x 0.4 x 3 = 0.48 m3

FT, air - W = 0

FT, air = W = concretegV

= 2300 x 9.81 x 0.48

(b) FT, water + FB - W = 0

FB = f gV = 1025 x 9.81 x 0.48

= 4800 N = 4.8 kN

FT, water = W - FB = 10.8 kN – 4.8 kN

= 6.0 kN (T)



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Example: Galilean Thermometer

Galileo's thermometer is made of a sealedglass cylinder containing a clear liquid.

Suspended in the liquid are a number of weights, which are sealed glass containerswith colored liquid for an attractive effect.

As the liquid changes temperature it changesdensity and the suspended weights rise andfall to stay at the position where their density isequal to that of the surrounding liquid.

If the weights differ by a very small amount andordered such that the least dense is at the topand most dense at the bottom they can form atemperature scale.


Example: Floating Drydock

Auxiliary Floating Dry Dock Resolute

(AFDM-10) partially submerged

Submarine undergoing repair work on

board the AFDM-10

Using buoyancy, a submarine with a displacement of 6,000 tons can be lifted!



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Example: Submarine Buoyancy and Ballast

Submarines use both static and dynamic depth

control. Static control uses ballast tanks

between the pressure hull and the outer hull.

Dynamic control uses the bow and stern planes

to generate trim forces.



Normal surface trim SSN 711 nose down after accident

which damaged fore ballast tanks



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Damage to SSN 711

(USS San Francisco)after running a ground on

8 January 2005.



Ballast Control Panel: Important station for controlling depth of submarine



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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.



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Examples of Archimedes Principle


The Golden Crown of Hiero II, King of

Syracuse

Archimedes, 287-212 B.C.

Hiero, 306-215 B.C.

Hiero learned of a rumor where

the goldsmith replaced some of

the gold in his crown with silver.Hiero asked Archimedes to

determine whether the crown was

pure gold.

Archimedes had to develop a

nondestructive testing method.



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The Golden Crown of Hiero II, King of

SyracuseThe weight of the crown andnugget are the same in air:Wc = cVc = Wn = nVn.

If the crown is pure gold, c= nwhich means that the volumesmust be the same, Vc=Vn.

In water, the buoyancy force isB = H2OV.

If the scale becomes

unbalanced, this implies thatthe Vc ≠ Vn, which in turnmeans that the c ≠ n.

Goldsmith was shown to be afraud!


Hydrostatic Bodyfat Testing

What is the best way to measurebody fat?

Hydrostatic Bodyfat Testing using Archimedes Principle!

Process

Measure body weight W = bodyV

Get in tank, expel all air, andmeasure apparent weight Wa

Buoyancy force B = W - Wa = H2OV. This permitscomputation of body volume.

Body density can be computed body = W/V.

Body fat can be computedfrom formulas.




Hydrostatic Bodyfat Testing in Air?

Same methodology as

Hydrostatic testing in water.

What are the ramifications of

using air?

Density of air is 1/1000th of

water.

Temperature dependence of air.

Measurement of small volumes.

Used by NCAA Wrestling (there

is a BodPod on PSU campus).

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