Fluid Statics

Fluid StaticsFluid Statics

4th SEMESTER

BS ME (2009-2013)

4th SEMESTER

BS ME (2009-2013)

Fluid StaticsFluid Statics

It is the study of fluids at rest There will be no shearing stresses The only forces that develop on the surface

of particles will be due to pressure Pressure distribution is due to weight of

fluid

It is the study of fluids at rest There will be no shearing stresses The only forces that develop on the surface

of particles will be due to pressure Pressure distribution is due to weight of

fluid

Fluid Statics

What is Pressure Anyway???What is Pressure Anyway???

Definition of PressureDefinition of Pressure

Pressure is defined as the amount of force exerted on a unit area of a substance:

2

force NP Pa

area m

Direction of Fluid Pressure on Boundaries

Direction of Fluid Pressure on Boundaries

Furnace duct Pipe or tube Heat exchanger

Dam

Pressure is a Normal Force(acts perpendicular to surfaces)It is also called a Surface Force

Absolute and Gauge PressureAbsolute and Gauge Pressure

Absolute pressure: The pressure of a fluid is expressed relative to that of perfect vacuum (P = 0)

Gauge pressure: Pressure expressed as the difference between the pressure of the fluid and that of the surrounding atmosphere.

Usual pressure measuring devices record gauge pressure.

To calculate absolute pressure

Absolute pressure: The pressure of a fluid is expressed relative to that of perfect vacuum (P = 0)

Gauge pressure: Pressure expressed as the difference between the pressure of the fluid and that of the surrounding atmosphere.

Usual pressure measuring devices record gauge pressure.

To calculate absolute pressure

abs atm gaugeP P P

Units for PressureUnits for Pressure

Units Definition

1 pascal (Pa) 1 kg m-1 s-2

1 bar 1 x 105 Pa

1 atmosphere (atm) 101,325 Pa

1 torr 1 / 760 atm = 1mm Hg

760 mm Hg 1 atm

14.7 lbf / in2 (psi) 1 atm

1 psi (lbf/in2) = 6894.8 Pa (N/m2) = 6.895x10-3 N/mm2 = 6.895x10-2 bar

Pressure At A PointPressure At A Point

Consider a Wedge of Fluido No Shearing StressesNo Shearing Stresses

Summation of Forces Must Summation of Forces Must Equal ZeroEqual Zero

AndAnd

From GeometryFrom Geometry

HenceHence

o There is no pressure change There is no pressure change in the horizontal directionin the horizontal direction

o There is a vertical change in There is a vertical change in pressure proportional to the pressure proportional to the density, gravity, and depth density, gravity, and depth changechange

Pressure Variation in a StaticPressure Variation in a Static

Let’s determine the pressure distribution in a fluid at rest in which the only body force acting is

due to gravity

Let’s determine the pressure distribution in a fluid at rest in which the only body force acting is

due to gravity

Equilibrium Condition

The sum of the forces acting on the fluid must

be equal to zero

Let Pz and Pz+z denote the pressures at the base and top of the cube, where the elevations are z and z+z respectively.

What Are The z-direction Forces?

x

y

z

( )mg S z g

z zPS

zPS

Pressure Distribution for a Fluid at RestPressure Distribution for a Fluid at Rest

A force balance in the z direction gives:

For an infinitesimal element (Δz0)

dPg

dz

0z z z zF PS PS S zg

z z zP Pg

z

Pressure Variation in a Static Fluid, contd.Pressure Variation in a Static Fluid, contd.

Incompressible Fluid γ=constant Incompressible Fluid γ=constant

Pressure in a continuously distributed uniform static fluid varies only Pressure in a continuously distributed uniform static fluid varies only with vertical distance and is independent of the shape of the container.with vertical distance and is independent of the shape of the container.

Pressure Variation in a Static Fluid, contd.Pressure Variation in a Static Fluid, contd.

Compressible Fluid γ=Variable Assuming a Perfect Gas Compressible Fluid γ=Variable Assuming a Perfect Gas

For Isothermal ProcessFor Isothermal Process

For Non-Isothermal ProcessFor Non-Isothermal Process

WhereWhereoTT00 is Sea Level is Sea Level

TemperatureTemperatureoB is Lapse RateB is Lapse Rate

Measurement of PressureMeasurement of Pressure

Absolute PressureAbsolute Pressure

Measured Relative to a Measured Relative to a Perfect VacuumPerfect Vacuum

Gage PressureGage Pressure

Measured Relative to Measured Relative to the Local Atmospheric the Local Atmospheric PressurePressure

Vacuum or SuctionVacuum or Suction

Negative Gage PressureNegative Gage Pressure

Mercury BarometerMercury Barometer

Mercury Barometer It is Used for the It is Used for the

Measurement of Measurement of Atmospheric PressureAtmospheric Pressure

Mercury Barometer It is Used for the It is Used for the

Measurement of Measurement of Atmospheric PressureAtmospheric Pressure

HereHere

pp11=0, =0, zz11 = h = h, , pp22=p=paa, , zz22 = h = h

ppaa = = γγMMhh

ManometryManometry

Piezometer TubePiezometer Tube

OrOr

It is a Simple Device But has Following DisadvantagesIt is a Simple Device But has Following DisadvantagesoNo suitable for Vacuum Gage PressureNo suitable for Vacuum Gage PressureoPressure Should be Relatively SmallPressure Should be Relatively SmalloFluid Should be Liquid NOT GasFluid Should be Liquid NOT Gas

o Pressure in a continuously Pressure in a continuously distributed uniform static fluid distributed uniform static fluid varies only with vertical varies only with vertical distance and is independent of distance and is independent of the shape of the container. the shape of the container.

o The pressure is the same at all The pressure is the same at all points on a given horizontal points on a given horizontal plane in the fluid. plane in the fluid.

o The pressure increases with The pressure increases with depth in the fluid.depth in the fluid.

Manometry, contd.Manometry, contd.

U Tube ManometerU Tube Manometer

Manometry, contd.Manometry, contd.

Differential U Tube ManometerDifferential U Tube Manometer

The U-tube manometer may The U-tube manometer may also be used to measure the also be used to measure the difference in pressure betweendifference in pressure betweentwo containers or two points two containers or two points in a given systemin a given system

Manometry, contd.Manometry, contd.

Inclined Tube ManometerInclined Tube Manometer

The Inclined Tube Manometer is used to measure The Inclined Tube Manometer is used to measure small pressure changes between two containers or small pressure changes between two containers or two points in a given systemtwo points in a given system

Bourdon Tube GageBourdon Tube Gage

Example: Thermodynamic Properties

Example: Thermodynamic Properties

When we say a car’s tire is When we say a car’s tire is filled “to 32 lb,” we mean filled “to 32 lb,” we mean that its internal pressure is 32 that its internal pressure is 32 lbf/inlbf/in22 above the ambient above the ambient atmosphere. If the tire is at atmosphere. If the tire is at sea level, has a volume of 3.0 sea level, has a volume of 3.0 ftft33, and is at 75°F, estimate , and is at 75°F, estimate the total weight of air, in lbf, the total weight of air, in lbf, inside the tire.inside the tire.

Given Data:p = 32 psi (gage) = 32 + 14.7 = 46.7 psiaV = 3.0 ft3

t = 75 oF

Find:Find:Total weight of air in lbf = ?Total weight of air in lbf = ?

Solution: Determination of WeightSolution: Determination of Weight

γ = wt / V =gp

RT

=gpV

wtRT

Given Data:p = 32 psi (gage) = 32 + 14.7 = 46.7 psiaV = 3.0 ft3

t = 75 oF

. ( . )=

( )wt

322 467 144 31716 75 460

= .wt lbf071

Example: Effect of ViscosityExample: Effect of Viscosity

The belt in Fig. moves at a steady velocity V and skims the top of a tank of oil of viscosity, as shown. Assuming a linear velocity profile in the oil, develop a simple formula for the required belt-drive power P as a function of (h, L, V, b,μ ). What belt-drive power P, in watts, is required if the belt moves at 2.5 m/s over SAE 30W oil at 20°C, with L 2 m, b 60 cm, and h 3 cm?

Solution: Effect of ViscositySolution: Effect of Viscosity

Given Data:V = 2.5 m/secμ of SAE 30W oil at 20oC = 0.29 N-s/m2

L = 2 mb = 60 cmh = 3 cm

AVF

h

A bLbLV

Fh

P FV

Power P = ?

bLVP

h

2

. . ..

.P watt

2029 06 2 25

725003

Example: Hydrostatic PressureExample: Hydrostatic Pressure

In Fig. the tank contains water and immiscible oil at 20°C. What is h in cm if the density of the oil is 898 kg/m3?

Given Data:γ of water at 20oC = 9789 N/m3

Findh in the Fig = ?

Solution: Hydrostatic PressureSolution: Hydrostatic Pressure

( ) ( )water oil

p h h

6 12100 100

9789 h 12100 100

. 981 898 0

. . .h

898 981 176202 1057125

100

.

.h cm

704895 1008

898 981

Given Data:γ of water at 20oC = 9789 N/m3

Example: The AtmosphereExample: The Atmosphere

A gage on an airplane measures a local pressure of 54 kPa. The lapse rate is 0.006 K/m. Effective sea level temperature is 10 oC. Effective sea level pressure is 100 kPa. Estimate the plane’s altitude.

Given Data:p = 54 kPaB = 0.006 K/mpo = 100 kPato = 10 oC = 283 K

/g RB

o

o

Bzp p

t

1

FindAltitude of plane= ?

Solution: Altitude of PlaneSolution: Altitude of Plane

Given Data:p = 54 kPaB = 0.006 K/mpo = 100 kPato = 10 oC = 283 K

/RB g

o

o

t pz

B p

1

. / .

.z

287 0006 981

283 541

0006 100

.

. .z 0176

4716667 1 054

z m4847

Example: ManometryExample: Manometry

In Fig. both ends of the manometer are open to theatmosphere. Estimate the specific gravity of fluid X.

Given Data:Density of SAE 30 oil at 20oC = 891 kg/m3Density of Water at 20oC = 998 kg/m3

Solution: ManometrySolution: Manometry

p = γh

(0.1–0.09)(891) +(0.07–0.05) (998)+(0.04–0.06) ρx = 0

ρx =1443.5 kg/m3

sx =1443.5 / 998 = 1.45

(0.1) (891*9.81) +(0.07)(998*9.81)

+(0.04) (ρx*9.81) –(0.06) (ρx*9.81) –

(0.05) (998*9.81) – (0.1) (891*9.81) = 0

Hydrostatic Force on a Plane SurfaceHydrostatic Force on a Plane Surface

For fluids at rest, the force must be perpendicular to the surface since there are no shearing stresses present.

where

first moment of the area with respect to the x axis

Hydrostatic Force on a Plane SurfaceHydrostatic Force on a Plane Surface

or more simply as

Magnitude of the force is independent of •Angle And depends only on •Specific weight of the fluid, •Total area•Depth of the centroid of the area below the surface

At what point does this force act??

since

second moment of the area (moment of inertia), Ix

Hydrostatic Force on a Plane SurfaceHydrostatic Force on a Plane SurfaceIt can be written

Parallel axis theorem can be used to express Ix as

Above equation shows that resultant force does not pass through the centroid but is always below it, since Ixc/yc A > 0.

The x coordinate, for the resultant force can be determined

where Ixy is the product of inertia with respect to the x and y axes

Some Common ShapesSome Common Shapes

Pressure PrismPressure Prism

Pressure PrismPressure Prism

Hydrostatic Force on a Curved Surface

Hydrostatic Force on a Curved Surface

For curved surfaces, the term dA makes integration tedious and no general relations can be found

A simple approach can be utilized

ExampleExampleThe 6-ft-diameter drainage conduit of Fig is half full of water at rest. Determine the magnitude and line of action of the resultant force that the water exerts on a 1-ft length of the curved section BC of the conduit wall.

F1 = γhc A = (62.4 lb/ft3)(3/2 ft )(3 ft2) = 281 lb = FH

W = γ vol = (62.4 lb/ft3)(9π/4 ft2 )(1 ft) = 441 lb= FV

Buoyancy and Archimedes’ PrincipleBuoyancy and Archimedes’ Principle

Summing forces in z direction

A is the horizontal area of the upper (or lower) surface

desired expression for the buoyant force

Taking momentsor

Buoyancy and Archimedes’ PrincipleBuoyancy and Archimedes’ Principle

buoyant force passes through the centroid of the displaced volume

StabilityStability