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Introduction to Fluid Mechanics

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Title: Introduction to Fluid Mechanics


1
Introduction to Fluid Mechanics
CEVE 101
  • Dr. Bedient
  • Civil and Environmental Engineering

2
FluidsStatics vs Dynamics
3
Atmospheric 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

4
Fluid 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

5
Units
SI - International System Length Meter Time Se
c Mass Kg Temp 0K 0C 273.15 Force Newto
n Nt 1 kg m / s2 Gravity 9.81 m/s2 Work
Fxd Joule Nt-m Power F/t Watt
Joule/sec
6
Units
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
7
Properties of Fluids
  • Density r (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 g (Heaviness of fluid)
  • weight per unit volume g rg
  • 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 rf / rw SG of Hg 13.55

8
Ideal Gas Law relates pressure to Temp for a gas
P rRT 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 m 1.12 x 10-3 N-s/m2 Surface tension -
forces between 2 liquids or gas and liquid -
droplets on a windshield.
9
Section 1 Pressure
Pressure at any point in a static fluid not fcn
of x,y,or z Pressure in vertical only depends on
g of the fluid
P gh Po
Gage pressure relative to atmospheric pressure
P gh 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
10
Pressure in a Tank Filled with Gasoline and Water
What is the pressure at point A? At point B?
gG 42.43 lbs/ft3 SG 0.68
gW 62.4 lbs/ft3
At point A PA gG x hG PO
42.43 x 10 PO
424.3 lbs/ft2 gage
At point B PB PA gW 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
11
Measurement of Pressure
Barometer (Hg) - Toricelli 1644 Piezometer
Tube U-Tube Manometer - between two
points Aneroid barometer - based on spring
deformation Pressure transducer - most advanced
12
Manometers - measure DP
  • 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

13
Manometers
Simple Example
P g x h PO
Find the pressure at point A in this open u-tube
monometer with an atmospheric pressure Po PD g
W x hE-D Po Pc PD PB PC - g Hg x hC-B PA
PB
14
Section 2 Hydrostatics And the Hoover Dam
  • For a fluid at rest, pressure increases
    linearly with depth. As a consequence, large
    forces can develop on plane and curved surfaces.
    The water behind the Hoover dam, on the Colorado
    river, is 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
    thickness varies from 45 feet at the top to 660
    feet at the base.

15
Hydrostatic Force on a Plane Surface
Basic Concepts and Naming
Pressure g h g 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
?h
16
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
For a Rectangular Gate Ixc 1/12 bh3 Ixyc
0 For a circle Ixc p r4 / 4 Ixyc 0
?h
17
Hydrostatic Force on a Plane Surface
The Center of Pressure YR lies below the centroid
- since pressure increases with depth
FR g A YC sinq or FR g A Hc YR (Ixc / YcA)
Yc XR (Ixyc / YcA) Xc but for a rectangle
or circle XR Xc For 90 degree walls FR g A
Hc
18
Hydrostatics Example Problem 1
What is the Magnitude and Location of the
Resultant force of water on the door?
gW 62.4 lbs/ft3 Water Depth 6 feet Door
Height 4 feet Door Width 3 feet
19
Hydrostatics Example Problem 1
Important variables HC and Yc 4 Xc 1.5 A
4 x 3 12 Ixc (1/12)bh3 (1/12)x3x43
16 ft4
Magnitude of Resultant Force FR gW A HC FR
62.4 x 12 x 4 2995.2 lbs
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
20
Section 3 Buoyancy
Archimedes 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 g
x Vdisplaced
FB gW x Vdisp
Vdisp
FB
W FB
FB
21
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?
FB
gW 62.4 lbs/ft3
22
Buoyancy Example Problem 1
Displaced Volume of Water Vdisp-W 4/3 x p x
R3 Vdisp-W 33.51 ft3
Buoyancy Force FB gW x Vdisp-w FB 62.4 x
33.51 FB 2091.024 lbs up
Sum of the Forces SFy 0 500 - 2091.024 T T
1591.024 lbs down
23
Will It Float?
Ship Specifications Weight 300 million
pounds Dimensions 100 wide by 150 tall by
800 long
Given Information gW 62.4 lbs/ft3
24
Assume Full Submersion FB Vol x gW 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
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