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

2. Fluids: Statics 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) lbs/in2 (psi) Torr (mm Hg) millibars = 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 Sec Mass Kg
Temp 0K = 0C
Force Newton = 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 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 = 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. Pressure at any point in a static fluid not fcn of x,y,or z
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. What is the pressure at point A? At point B?
Pressure in a Tank Filled with Gasoline and Water
What is the pressure at point A? At point B?
gG = lbs/ft3
SG = gW = 62.4 lbs/ft3
At point A: PA = gG x hG + PO
= x 10 + PO
424.3 lbs/ft2 gage
At point B: PB = PA + gW x hW
= 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 P = g x h + PO Simple Example:
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 γh

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 γh

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 Magnitude of Resultant Force:
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 = lbs
Location of Force:
YR = (Ixc / YcA) + Yc
YR = (16 / 4x12) + 4 = ft down
XR = Xc (symmetry) = 1.5 ft from the corner of the door

20. Archimedes Principle: Will it Float?
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 = lbs/ft3

22. Buoyancy Example Problem # 1 Displaced Volume of Water:
Vdisp-W = 4/3 x p x R3
Vdisp-W = ft3
Buoyancy Force:
FB = gW x Vdisp-w
FB = 62.4 x 33.51
FB = lbs up
Sum of the Forces:
SFy = 0 = T
T = 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/ft 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