1
MEK 307 Fluid Mechanics
A. Ozan Celik www.hku.hk
2
Basics of fluid flow
Basics of fluid flow, kinematics
Mechanics
Statics
Dynamics
Kinematics
Kinetics
Kinematics: deals with motion apart from considerations
of mass, force or energy
3
Dimensions and units (Chapter 1.6)
SI BG
F=ma
Basic dimensions: length [L], mass [M], force [F], time [T]
[F] = [M][LT-2]
lb = Slug.ft/s2
Slug = lb.s2/ft
N = kg.m/s2
base units
derived units
(derivative dimensions)
SI: International System of Units
BG: British Gravitational System
We use sets of units to properly define, express certain physical properties/
phenomenon
4
Dimensions and units, example:
SI BG
Energy =Force x Distance
= [M][LT-2][L]
=(kg.m/s2)(m)
=Nm
=Joule
= [F][L]
=lb.ft
(1.3558 J )
5
Unit homogeneity:
2+2=?4
Manning’s Equation:
Q=(1/n)A.R2/3.S1/2
L3/T=?[L2][L]2/3
??
Where:
Q = Flow Rate [L3/T]
A = Flow Area, [L2]
n = Manning’s Roughness
Coefficient
R = Hydraulic Radius, [L]
S = Channel Slope, [L/L]
Has units:
[L1/3/T]
(1 meter = 3.2808399 feet)
For those who are interested: http://en.wikipedia.org/wiki/Robert_Manning_(engineer)
6
Properties of Fluids (Chapter 2.2)
Density, ρ = mass per unit volume, [M/L3]
Specific volume, v = 1/ ρ, volume occupied by unit mass [L3/M]
Specific weight, γ = weight per unit volume, [F/L3] (depends on latitude)
Specific gravity, SG (s) = ρ/ ρwater
http://www.agen.ufl.edu/~chyn
density of fluids vary with
temperature
7
Properties of Fluids (Chapter 2)
Compressibility of Liquids
v
vE
pp
v
vvdp
dv
v
dv
dpvE 12
1
12
Bulk Modulus (volume modulus of elasticity) Units?
We consider liquids to be incompressible but
why?
pressure volume,specific pv
8
Properties of Fluids
Fluids Solids
Deforms continuously or
flows when subject to
shear force
Can resist a shear force at
rest (with some
displacement but tend to
retain shape)
Shear force: Force component tangent to a surface
Ft
F Fn Shear Stress: τ = Ft /A
Pressure: p= Fn /A
A
9
The strain in a solid is independent of the time over
which the force is applied and (if the elastic limit
is not reached) the deformation disappears when the
force is removed.
A fluid continues to flow for as
long as the force is applied and will not recover its
original form when the force is removed.
Properties of Fluids
10
Properties of Fluids
Shear force, Ft: Force component tangent to a surface
Normal Force, Fn: Force component normal to a surface
Shear Stress: τ= Ft/A Pressure: p=Fn/A
Ft
F Fn
A
Deformation Compressibility
(dynamic)
Viscosity, μ
Some liquids are
more viscous than
others
3120 psi needed to
compress a unit
volume of water
1%
11
Properties of Fluids (Chapter 2.6)
Viscosity is the means by which the fluid resists friction
(shear stress)
Constant of
proportionality, viscosity
(dynamic, absolute)
For fluids, the rate of strain
(angular deformation) is
proportional to the applied
stress.
Newtonian Fluids: μ is constant
12
Viscosity (Chapter 2)
For gases μ as T
For liquids μ as T
μ => units?
13
Properties of Fluids
Laminar shear flow velocity profile Uniform flow velocity profile
No Shear
In practice we are concerned with flow past solid (sometimes rough)
boundaries; pipe walls, river bed etc. and there will always be shear stress
14
Shear Stress and Viscosity (Chapter 2)
Viscosity, μ is a measure of resistance to relative motion of
adjacent fluid layers (angular deformation)
τ
τ
τ = μdu/dy
FLOW
y
u du
u
u+du dy
Velocity profile
15
Shear Stress and Viscosity (Chapter 2)
Viscosity, μ Fluids resist to shear lightly
τ
Ideal fluids Viscosity, μ=0 Only normal stresses
present
Newtonian fluids Viscosity, μ=const. (for const. temperature)
τ=0
du/dy=0
u=0, fluid is at rest Only normal stresses
present
u=u(y)=const.
No slip condition: Fluid in the immediate contact
with a solid boundary has the same velocity as the
solid boundary
16
Surface Tension (Chapter 2)
Exists whenever there is a density discontinuity
(Interfaces between a gas and a liquid; two different liquids…etc)
Cohesion: Attraction between like molecules, enables a fluid to
resist tensile stresses
Adhesion: Attraction between unlike molecules, enables a fluid to
stick to another body
http://www.bcscience.com
17
Vapor Pressure (Chapter 2)
Pressure of a vapor in equilibrium with
its non-vapor phases
Vapor pressure depends on
temperature
To avoid boiling, pressure on top of
liquid should be higher than vapor
pressure
http://www.unit5.org/
http://www.chem.purdue.edu
Towards the
perfect boiled egg
on top of Mount
Everest
On top of Mount Everest the pressure is about
260 mbar (26.39 kPa, 0.25 atm)
69°C
You need water temperature higher than 70°C to
cook an egg (for a hard yolk)
Conclusion: you can not boil an egg
on top of Mount Everest. http://blog.khymos.org/
18
Vapor Pressure, Cavitation (Chapter 2)
Cavitation damage on an impeller of a BW5000 pump
www.lightmypump.com
19
Fluid Statics (Chapter 3)
Fluid at rest (no shear force)
Normal Force, Fn: Force component normal to a surface
(Fn :surface force component)
Pressure: P = Fn / A
Fn
A: area
20
Fluid Statics
Absolute and gage pressures
Absolute pressure = pressure measured relative to absolute zero (perfect
vacuum)
Gage pressure = pressure measured relative to local atmospheric
pressure
Pressure usually doesn’t affect liquid properties and atmospheric pressure
usually appears on both sides of an equation. So we commonly use gage
pressure in problems dealing with liquids.
21
Fluid Statics (Chapter 3)
Pressure at a point!
A
Variation of pressure in a static Fluid:
Pressure increases linearly with depth
is same in all directions
A dz
dp
22
Fluid Statics (Chapter 3)
Variation of Pressure in a static Fluid
Pressure increases linearly with depth (for an
incompressible fluid)
p = -γz + patm
z h
patm
Pascal’s law: a surface of equal pressure for a
liquid at rest is a horizontal plane. It is a surface
everywhere normal to the direction of gravity.
23
Fluid Statics (Chapter 3)
Pressure expressed in height of fluid
[L] = [F]/[L]2 x [F]/[L]3 = [L]
h is the height of fluid at the bottom of which the pressure will be p
Also called pressure head
1 Atm = 76 cm Hg
= 10.33 m H2O
24
Fluid Statics
Pressure head, an important property
h2 h1
1) Identify a reference point (datum)
2) Identify elevation of points 1 and 2
1
2
z1
z2
3) Sum of elevation of a point and pressure head at that point
is constant
h1 + z1 = h2 + z2 = const.
that is
p1/ γ + z1 = p2/ γ + z2 = const.
25
Fluid Statics
Remarks
p = γh or h = p/γ when
• Specific weight, γ = const.
• fluid is at rest
• fluid is incompressible
• the pressure of air or vapor acting on the free
surface is neglected (for now)
• distances (h) measured vertically downward
from the free surface
z h
p
26
Fluid Statics
Pressure head, examples
pA
pB
pC
pD
pA = pB = pC = pB = γh
γ γ γ γ h
Which of these pressures is the highest?
27
Fluid Statics
Pressure head, examples
water
gasoline
A B C
Which of these points
(A, B, C) will
experience the highest
pressure?
pA = pB = pC = γgasolineh
h
28
Fluid Statics
Pressure head, examples
Pressure at the same depth is the same, moving sideways
does not change the pressure
Pressure distribution
29
Fluid Statics
Pressure head, examples
γ2
γ1 A
B
pB = ?
pB =
h1
h2
γ1h1
γ2h2
+
30
Fluid Statics
Pressure head, examples
γ1
γ2
p1
p2
p2 - p1 = ?
h1
h2
h3
h4
p1 + γ1(h1-h2) = p2+ γ1(h4-h3) + γ2(h3-h2)
p2 - p1 = γ1(h1-h2) - γ2 (h3-h2) - γ1(h4-h3)
31
Fluid Statics
Absolute and gage pressures
Absolute pressure = pressure measured relative to absolute zero (perfect
vacuum)
Gage pressure = pressure measured relative to local atmospheric
pressure
Throughout the course, unless otherwise specified, pressure will mean
gage pressure.
32
Fluid Statics
Absolute and gage pressures, example
γ2
γ1 A
B
pB (gage) =
h1
h2
γ1h1
γ2h2
+
γ2
γ1 A
B
h1
h2
pB (abs.) =
γ1h1
γ2h2
+
+patm
patm
pB (gage)
pB (abs.) =
pB (gage) + patm
pB (abs.)
33
Fluid Statics
Pascal's Principle
Any external pressure applied to a fluid
is transmitted undiminished throughout
the liquid and onto the walls of the
containing vessel
http://www.ac.wwu.edu/~vawter
This is the basic idea of hydraulic machines!
34
Fluid Statics
Summary of last lecture
Pressure increases linearly with depth
dp/dz=-γ =>
liquid with free surface pb=γh
h
Pressure at a point is the same in
all directions
A
Pressure is constant on a horizontal plane
pb
35
Fluid Statics
Pressure head, Example
An open tank contains water 1.4 m deep covered by a 2-m-thick layer of
oil (s = 0.855). What is the pressure head at the bottom of the tank, in
terms of a water column?
ooi hp
wwoowob hhppp
ooi hp
36
Fluid Statics
Measurement of pressure, barometer
p0 = γy + pvapor
pa = p0
pa = patm
po = pa = patm= γy + pvapor
patm= γy
/37
Fluid Statics
Measurement of pressure, other methods
Bourdon gage
Pressure transducers
Converts energy
from the pressure
system to
displacement of a
mechanical system
http://www.pchemlabs.com
http://www.pcb.com http://www.honeywell.com http://www.fiso.com
38
Fluid Statics
Measurement of pressure, manometer
Piezometer column
pA = γh + RM(sM/sF)γ +patm
Simple manometer
Liquids only
Liquids and gases
We can omit portions of the
same fluid with the same end
elevations (pressure doesn’t
change on a horizontal
plane).
39
Fluid Statics
Measurement of pressure, manometer
Vacuum
pA = γh - RM(sM/sF)γ +patm
40
Fluid Statics
Measurement of pressure, manometer
Vacuum
pA /γF = -((sM/sF)Rm+h)
41
Fluid Statics
Measurement of pressure, differential manometer
pA /γ - hA - RM(sM/sF) +hB = pB/ γ
pA /γ - pB/ γ = hA - hB + RM(sM/sF)
Liquid A and B have the same
density!
pA /γ - pB/ γ = zB – zA+RM - RM(sM/sF)
m
F
M Rs
sz
p
1
m
F
M Rs
sz
p
1
42
Fluid Statics
Manometer problem
The air pressure in a tank is measured
by an oil manometer. For a given oil-
level difference between the two
columns, determine the absolute
pressure in the tank.
(density of oil is given to be = 850
kg/m3)
43
Fluid Statics
Differential manometer problem
Fresh and seawater flowing in
parallel horizontal pipelines are
connected to each other by a
double U-tube manometer.
Determine the pressure
difference between the two
pipelines.
hw=0.6 m, hHg=0.1 m, hsea=0.4 m
The densities of seawater and mercury
are given to be sea = 1035 kg/m3 and Hg
= 13,600 kg/m3. The density of water w
=1000 kg/m3.
44
Fluid Statics
Piston problem
F1=200 N, F2=?
45
Fluid Statics
Measurement of pressure, inclined manometer
Example
http://www.gnw.co.uk/
The actual vertical
rise is still the same as it is in a vertical
tube, but here it is much easier to read
accurately when the
scale has been expanded!
46
Fluid Statics
Example
47
Fluid Statics
Forces on plane areas, applications: Panama Canal locks
http://sems1.cs.und.edu/~sems/
48
Fluid Statics
Forces on plane areas, applications: Foundation failure
http://www.helitechonline.com/
49
Fluid Statics
Forces on plane areas, applications: Gates
http://sems1.cs.und.edu/~sems/
Feeder Gates for Canal
Gate Valves for
Spillway Control
http://users.owt.com/chubbard/gcdam/html/photos
50
Fluid Statics
Forces on plane areas, applications: Dams
Hoover Dam
A beaver dam
http://commons.wikimedia.org/wiki/User:Leoboudv http://snailstales.blogspot.com
51
Fluid Statics
Forces on plane areas, horizontal surface
If fluid is at rest:
No tangential forces exist
All forces are normal to the surfaces
If the pressure is uniformly distributed over the area A (e.g.
submerged horizontal area for liquids and gases) , then:
pAdAp pdA F
A
p = γh h F
F: static fluid force, acts at the
centroid of area A
If fluid is gas: pressure
variation with depth over a
surface can be neglected!
52
Fluid Statics
Forces on plane areas, vertical plane
The deeper the plane is submerged, the
closer the resultant (F) moves to the
centroid of the surface
When the fluid at rest is liquid, then the
pressure distribution is not uniform
F: static fluid force, always acts
below the centroid of area A
53
Fluid Statics
Forces on plane areas, magnitude of F
If the width (into the page), x is constant, then F = pA = 0.5(pMJ+pNK)(MNx)
If x varies, dA=xdy, p = γh, h=ysinθ, then the force dF on the
horizontal strip:
dAyhdApdAdF sin
54
Fluid Statics
Forces on plane areas, magnitude of F
dAyhdApdAdF sin
Integrating
AyydAsin Fd F c sin
Ah F c
A
Total force on any plane area (submerged in liquid) can
be found by multiplying the specific weight by the
product of the area and the depth of its centroid.
55
Fluid Statics
Forces on plane areas, center of pressure
The point where the line of action of the resultant force goes through
Pressure prism
is the centroid of the pressure prism (volume).
56
Fluid Statics
Forces on plane areas, center of pressure
hp = 2/3 MN
Pressure prism
concept is useful
to determine the
center of pressure
for simple
geometries.
57
Fluid Statics
Forces on plane areas, center of pressure
If the shape of the area is not regular we have to
take moments and integrate: Axis of
moment
dAyydF sin2
dAydAyyypdAFyA
p 2sin)sin(
58
Fluid Statics
Forces on plane areas, center of pressure
dAyydF sin2
dAydAyyypdAFyA
p 2sin)sin(
AyF c sin
Ay
I
Ay
Iy
c
O
c
Op
sin
sin
AyII
dAyI
ccO
A
O
2
2
59
Fluid Statics
Forces on plane areas, location of center of pressure
Ay
Iy
Ay
IAy
Ay
Iy
c
cc
c
cc
c
Op
2
Ic : moment of inertia of an area about its centroidal axis
Parallel axis theorem
60
Fluid Statics
Forces on plane areas, center of pressure, two methods
1) The point where the line of action of the resultant force goes through is
the centroid of the pressure prism (volume).
F
Ay
Iyy
c
ccp
2) Consider the moments
61
Fluid Statics
Forces on plane areas, Steps
Ahc FAy
Iyy
c
ccp
1) Determine the area (A) of the submerged surface
2) Find the centroid of the area, yc, and then hc
3) Calculate the magnitude of the resultant hydrostatic force, F
4) Calculate the line of action of the hydrostatic force, yp
62
Fluid Statics
Forces on plane areas, applications: Gates