Fluid Statics
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Transcript of 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