Download - Intro Fluids
Introduction to Fluid Mechanics Introduction to Fluid Mechanics
Dr. BedientDr. BedientCivil and Environmental EngineeringCivil and Environmental Engineering
CEVE 101
Fluids:Fluids:Statics vs DynamicsStatics vs Dynamics
Atmospheric PressureAtmospheric PressurePressure = Force per Unit AreaAtmospheric Pressure is the weight of the column of air above a unit area. For example, the atmospheric pressure felt by a man is the weight of the column of air above his body divided by the area the air is resting on
P = (Weight of column)/(Area of base)
Standard Atmospheric Pressure:
1 atmosphere (atm) 14.7 lbs/in2 (psi) 760 Torr (mm Hg) 1013.25 millibars = 101.3 kPascals
1kPa = 1Nt/m2
Fluid StaticsFluid StaticsBasic Principles: Fluid is at rest : no shear forces
Pressure is the only force acting
What are the forces acting on the block?
Air pressure on the surface - neglect
Weight of the water above the block
Pressure only a function of depth
UnitsUnitsSI - International System
Length MeterTime SecMass KgTemp 0K = 0C + 273.15Force Newton = Nt = 1 kg m / s2
Gravity 9.81 m/s2
Work = Fxd Joule = Nt-m
Power = F/t Watt = Joule/sec
UnitsUnitsEnglish Length in Ft
Time in Sec
lbm (slug) - 1 slug = 32.2 lbm
Force - lb
Gravity - 32.2 ft/sec2
Work = slug-ft/s2
Properties of FluidsProperties of FluidsDensity = (decreases with rise in T) mass per unit volume ( lbs/ft3 or kg/m3 )
for water density = 1.94 slugs/ft3 or 1000 kg/m3
Specific Weight = (Heaviness of fluid) weight per unit volume = g
for water spec wt = 62.4 lbs/ft3 or 9.81 kN/m3
Specific Gravity = SG Ratio of the density of a fluid to the density of water
SG = f / w SG of Hg = 13.55
Ideal Gas Law relates pressure to Temp for a gas
P = RT
T in 0K units
R = 287 Joule / Kg-0K
PressureForce per unit area:
lbs/in2 (psi), N/m2, mm Hg, mbar or atm
1 Nt/m2 = Pascal = Pa
Std Atm P = 14.7 psi = 101.33 kPa = 1013 mb
Viscosity fluid deforms when acted on by shear stress
= 1.12 x 10-3 N-s/m2
Surface tension - forces between 2 liquids or gas and liquid - droplets on a windshield.
Section 1: PressureSection 1: PressurePressure at any point in a static fluid not fcn of x,y,or z
Pressure in vertical only depends on of the fluid
P = h + Po
Gage pressure: relative to atmospheric pressure: P = hThus for h = 10 ft, P = 10(62.4) = 624 psf
This becomes 624/144 = 4.33 psi
P = 14.7 psi corresponds to 34 ft
10 ft
What is the pressure at point A? At point B?
G = 42.43 lbs/ft3
SG = 0.68 W = 62.4 lbs/ft3
At point A: PA = G x hG + PO
= 42.43 x 10 + PO
424.3 lbs/ft2 gage
At point B: PB = PA + W x hW
= 424.3 + 62.4 x 3
611.5 lbs/ft2 gageConverting PB to psi:
(611.5 lbs / ft2) / (144 in2/ft2)
= 4.25 psi
Pressure in a Tank Filled with Gasoline and Water
Measurement of PressureMeasurement of Pressure
Barometer (Hg) - Toricelli 1644
Piezometer Tube
U-Tube Manometer - between two points
Aneroid barometer - based on spring deformation
Pressure transducer - most advanced
QuickTime™ and aTIFF (LZW) decompressorare needed to see this picture.
Manometers - measure Manometers - measure PPRules of thumb:
When evaluating, start from the known pressure end and work towards the unknown end At equal elevations, pressure is constant in the SAME fluid When moving down a monometer, pressure increases When moving up a monometer, pressure decreases Only include atmospheric pressure on open ends
ManometersManometers
Find the pressure at point A in this open u-tube monometer with an atmospheric pressure Po
PD = W x hE-D + Po
Pc = PD
PB = PC - Hg x hC-B
PA = PB
Simple Example:
P = x h + PO
For a fluid at rest, pressure increases linearly with depth. As a For a fluid at rest, pressure increases linearly with depth. As a consequence, large forces can develop on plane and curved surfaces. consequence, large forces can develop on plane and curved surfaces. The water behind the Hoover dam, on the Colorado river, is The water behind the Hoover dam, on the Colorado river, is approximately 715 feet deep and at this depth the pressure is 310 psi. approximately 715 feet deep and at this depth the pressure is 310 psi. To withstand the large pressure forces on the face of the dam, its To withstand the large pressure forces on the face of the dam, its thickness varies from 45 feet at the top to 660 feet at the base. thickness varies from 45 feet at the top to 660 feet at the base.
Section 2: HydrostaticsSection 2: HydrostaticsAnd the Hoover Dam
Hydrostatic Force on a Plane SurfaceBasic Concepts and Naming
Pressure = h = spec gravity of water
h = depth of water
C = Center of Mass of Gate
CP = Center of Pressure on Gate
Fr = Resultant Force acts at CP
γh
Hydrostatic Force on a Plane SurfaceBasic Concepts and Naming
C = Centroid or Center of Mass
CP = Center of Pressure
Fr = Resultant Force
I = Moment of Inertia
γh
For a Rectangular Gate:Ixc = 1/12 bh3
Ixyc = 0
For a circle:
Ixc = r4 / 4
Ixyc = 0
Hydrostatic Force on a Plane Surface
The Center of Pressure YR lies below the centroid - since pressure increases with depth
FR = A YC sin
or FR = A Hc
YR = (Ixc / YcA) + Yc
XR = (Ixyc / YcA) + Xc
but for a rectangle or circle: XR = Xc
For 90 degree walls:
FR = A Hc
Hydrostatics Example Problem # 1
What is the Magnitude and Location of the
Resultant force of water on the door?
W = 62.4 lbs/ft3
Water Depth = 6 feet
Door Height = 4 feet
Door Width = 3 feet
Hydrostatics Example Problem #1
Magnitude of Resultant Force:
FR = W A HC
FR = 62.4 x 12 x 4 = 2995.2 lbs
Important variables:HC and Yc = 4’
Xc = 1.5’
A = 4’ x 3’ = 12’
Ixc = (1/12)bh3
= (1/12)x3x43 = 16 ft4
Location of Force:YR = (Ixc / YcA) + Yc
YR = (16 / 4x12) + 4 = 4.333 ft down
XR = Xc (symmetry) = 1.5 ft from the corner of the door
Section 3: BuoyancyArchimedes Principle: Will it Float?
The upward vertical force felt by a submerged, or partially submerged, body is known as the buoyancy force. It is equal to the weight of the fluid displaced by the submerged portion of the body. The buoyancy
force acts through the centroid of the displaced volume, known as the center of buoyancy. A body will sink until the buoyancy force is equal to
the weight of the body.
FB = x Vdisplaced
= Vdisp
FB
FB
W = FB
FB = W x Vdisp
Buoyancy Example Problem # 1
A 500 lb buoy, with a 2 ft radius is tethered to the bed of a lake. What is the tensile force T in the cable?
W = 62.4 lbs/ft3
FB
Buoyancy Example Problem # 1
Displaced Volume of Water:
Vdisp-W = 4/3 x x R3
Vdisp-W = 33.51 ft3
Buoyancy Force:
FB = W x Vdisp-w
FB = 62.4 x 33.51
FB = 2091.024 lbs up
Sum of the Forces:
Fy = 0 = 500 - 2091.024 + T
T = 1591.024 lbs down
Will It Float?Ship Specifications:
Weight = 300 million pounds
Dimensions = 100’ wide by 150’ tall by 800’ long
Given Information: W = 62.4 lbs/ft3
Assume Full Submersion:
FB = Vol x W FB = (100’ x 150’ x 800’) x 62.4 lbs/ft3 FB = 748,800,000 lbs
Weight of Boat = 300,000,000 lbs The Force of Buoyancy is greater than the Weight of the Boat
meaning the Boat will float!
How much of the boat will be submerged?
Assume weight = Displaced Volume
WB = FB
300,000,000 = (100’ x H’ x 800’) x 62.4 lbs/ft3
H = Submersion depth = 60.1 feet