Introduction to Fluid Mechanics Introduction to Fluid Mechanics

Dr. BedientDr. Bedient

Civil and Environmental EngineeringCivil and Environmental Engineering

CEVE 101

Fluids:Fluids:Statics vs DynamicsStatics vs Dynamics

Atmospheric PressureAtmospheric Pressure

Pressure = Force per Unit Area

Atmospheric 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 Statics

Basic 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

UnitsUnits

SI - International SystemLength 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

UnitsUnits

English 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

Pressure

Force 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 = h

Thus 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 gage

Converting 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 PP

Rules 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 Surface

Basic 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 Surface

Basic 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