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ME 3250Fluid Dynamics I
Spring 2014
Prof. Mike Renfro
AUST 110
TuTh 2:00-3:15 PM
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Handouts
Course Policy
Syllabus
Academic Conduct – fill out and return News article – Mars Orbiter
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Course Information (1)
Text: Munson, B.R., Rothmayer, A.P., Okiishi,
T.H., and Huebsch, W.W., Fundamentals of
Fluid Mechanics, 7th edition, Wiley (2013).
Office: UTEB 472
Phone: 486-2239
E-mail: [email protected] Office Hours: M 10-11, Tu 10-11, Th 3:30-
4:30 (or by appointment)
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Course Information (2)
HuskyCT Site
Updated homework assignments, announcements
Link to publisher website – videos, sample quizzes
Lecture notes
Old test solutions
Grades
Pre-req: ME 2233 (Thermo I) and its pre-reqs
(calculus, physics, etc.)
Pre-req quiz (5%) on Tu 1/28, 30 minutes in class
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Grade Distribution (1)
Homework (10%)
2-3 problems assigned each lecture due followingThursday
2-3 problems graded per set – all must be turned infor full credit
No late homework accepted – due at start of class
Format/units (see Mars Lander handout)
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Grade Distribution (2)
i>clicker 2 questions (15%)
3-4 clicker questions each class will be used to:
Review material from previous lecture
Test new material presented during lecture See if concepts need additional attention
Grading for clicker questions is ½ credit foranswering + ½ credit for answering correctly
You may usually talk through your answers withother students unless I specifically say otherwise
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Attendance
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Grade Distribution (3)
Exams (35%)
Exam 1 – Tu March 4
Exam 2 – Tu April 22
1 crib sheet can be brought to each exam
Final Exam (25%)
Tu May 6, 1-3pm (tentative)
Comprehensive
2 crib sheets can be brought to exam
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Grade Distribution (4)
FLUENT Projects (10%)
One or two projects using FLUENT software
2nd floor ME computer lab in E-II
Project includes brief 1 page report and data
analysis
Final grades are relative to class performance
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Academic Honesty
Anything you turn in for a grade must be
entirely your own work
Cheating includes obvious copying of exams
or homework but also listing the books or a
friends answer or working backwards from this
answer
As long as you turn in your own work, I
encourage working with others on homework
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Intro to Fluid Dynamics (1)
What is a fluid? (Chap. 1)
When a small force is applied to a solid, the solid
“strains” (displacement) until the “stresses” in the
solid balance the force – at equilibrium a solid is atrest and if the force is removed the solid recovers
stress
F
strain
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Intro to Fluid Dynamics (2)
When a force is applied to a fluid (liquid or gas),
the stresses lead to continuous straining (a rate of
strain) via fluid motion (velocity) – at equilibrium
a fluid flows Fluids flow even for infinitesimally small stresses
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Intro to Fluid Dynamics (3)
Fluid Statics = A study of forces caused by
stationary fluids
Fluid Dynamics = A study of fluid (gas or
liquid) motion due to applied forces
Fluid Mechanics = A study of forces caused byfluid motion
These are often used synonymously
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Applications of Fluid Mechanics (1)
Study the behavior of fluids at rest
Fluid statics analysis (Chap. 2)
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Applications of Fluid Mechanics (2)
Study the forces that fluids impart on systems
Study the global behavior of fluid in a system
Integral (control volume) analysis (Chap. 5)
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Applications of Fluid Mechanics (3)
Study the local behavior of fluid flow
Differential analysis (Chap. 6)
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Properties of Fluids (1)
Pressure
Density
Ideal gas
Most liquids
33
1
f t
lbm
m
kg
vV
m
][)(22
atm psiinlbf Paor
m N
A F p
RT
p RT p
const
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Properties of Fluids (2)
Specific weight
Specific gravity
33 ft
lbf
m
N g g
water
SG
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Viscosity (1)
You should be familiar with concepts of
density, temperature, pressure, and velocity
Viscosity is one of the most important
properties (with density) that make fluids
behave differently
Viscosity – Movie
Viscosity of the oil is 104 larger than water –
same density
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Viscosity (2)
What is viscosity?
All fluids strain when a stress is applied
The viscosity, m, is the stress, , required to
achieve a given rate of strain,(we will discuss fluid strain and stress
more in Chap. 6)
g
dy
du
t t
g
0lim
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Viscosity (3) – Newtonian Fluids A velocity gradient is a rate
of shearing strain
As the stress increases (by pulling the upper platefaster), the shearing strainincreases (the fluid flows
with a steeper velocitygradient)
Newtonian fluids have alinear relationship between and strain rate (du/dy)
This linear coefficient is theviscosity, m
dy
dum
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Viscosity (4) – Non-Newtonian
Fluids
Non-Newtonian fluidsdo not have a linearrelationship betweenstress and strain rate
Non-Newtonian Movie Corn starch forms a
shear thickening fluid – viscosity is larger for
high shearing strain In this course we will
deal with mostly Newtonian fluids
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Viscosity (5) - Units
Rate of shearing strain
Stress is a force per unit area
Viscosity
Also called dynamic viscosity
Kinematic viscosity
(other units exist for viscosity – be careful with
unit conversions)
][ 1 sdy
dug
22
in
lbf or
m
N
22in
slbf or
m
s N m
s
ft or
s
m 22
m
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Viscosity (6) – Temperature Effects
Viscosity stronglydepends on T
For liquids viscosity
decreases with T (effectsof molecular interactionsdecrease)
For gases viscosity
increases with T (effectsof molecular collisionsincrease)
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Viscosity (7) – Walls and No-slip
At walls, viscosity causes the fluid to stick to
the wall the wall and fluid have the same
velocity (otherwise there would be infinite
strain and stress)
This is called the “No-slip condition” (movie)
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30 minutes
Pre-req Quiz
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Additional Fluid Properties (1)
Bulk modulus
If p is increased, how much does decrease (as a percentage)?
since
Ev is the “Bulk Modulus” with [N/m2]
Change in pressure required for a given percentage change in density or volume
d dp
d dp E v
/
m
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Bulk Modulus
For ideal gases undergoing an isothermal
process
Thus, bulk modulus for air is of the order 105
N/m2
For liquid water Ev=2x109 N/m2
C RT p
Cd dp
pC d
Cd E v
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Additional Fluid Properties (2)
Liquid Surfaces Gases characterized by weak intermolecular forces
(flying through space unaffected by neighbors)
Liquids characterized by significant intermolecularforces
In liquid neighboring molecules pull
equally in all directions
At surface, net force is down(balanced by pressure)
Surface also stretches molecules
(surface tension)
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Surface Tension (1)
Droplet Formation
Liquids in free space form spherical drops due to
surface tension
Surface tension is property of fluid, [N/m] =
force per unit length of surface
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Surface Tension (2)
Pressure in a drop
Balance of forces on cross
section of drop
22 R p A p p R L ambdrop
R p
2
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Surface Tension (3)
Bending of surface around an object provides
tension that must be “broken”
Surface acts like membrane
Movie
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Surface Tension (4)
At solid surface, attractive force between the
solid and liquid molecules can be large enough
to overcome surface tension – the solid surface
“wets” (fig A)
Or, too weak and the surface does not wet (fig
c)
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Surface Wetting
Wetting pulls liquid up small capillary tubes
Angle of liquid surface causes net upward
force to balance weight of liquid
g cos22 Rh R g mg
g
Rh
cos2
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Additional Fluid Properties (3)
Vapor Pressure
“Fast” molecules from liquidwith sufficient velocity toescape surface evaporate
“Slow” molecules from gasare captured by surface forcesand condense
Equilibrium occurs whensufficient vapor exists(evaporation=condensation)
This equilibrium pressure is“Vapor pressure”, pv
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Vapor Pressure
Vapor pressure, pv [N/m2] or [atm] is a
property of a liquid and is strongly T
dependent
At boiling T, vapor pressure = 1 atm
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Fluid Statics
If fluid has no velocity (left), or is movingtogether as a system (right) then there is nostraining motion
Each fluid element moves rigidly with itsneighbor
No shear force on fluid00 m
dy
duand
dy
du
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Balance of Forces (1)
If there is no shear force, only forces on fluid
element are pressure and weight
Can balance forces on any control volume
(fluid element) using Newton’s Second Law
(F=ma)
Wedge chosen
as an example
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Balance of Forces (2)
Consider case where acceleration is zero
Regardless of angle, pressure acts equally in all directions,even with acceleration (see further analysis on text p. 39)
However, pressure can vary from point to point
s x p z x p
ma F
s y
y y
sin0
0
z s sin
s y
s y
p p
z x p z x p
0
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Variation of pressure in a fluid (1) Now consider a finite sized
cubic fluid element with
pressure variations
y
y
y y
y
a y
p
z y xa z y x y
p
z y xama
z x y y p p z x y
y p p F
22
xa x
p
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Variation of pressure in a fluid (2) In z-direction, gravity also acts
on fluid element
g a z
p
z y x g a z y x z
p
mamg y x
z
z
p
p y x
z
z
p
p F
z
z
z
22
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Variation of pressure in a fluid (3)
g a z
pa
y
pa
x
p z y x
,,
k z p j
y pi
x p p ˆˆˆ
k a jaiaa z y xˆˆˆ
g a p
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Fluids at Rest (1)
If a=0
For an
incompressible fluid(density=constant)
g
g z p
y
p
x
p
g p
0
h p
z z g p p
gdz dp
B A B A
g
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Fluids at Rest (2)
For a compressible fluid
(e.g., ideal gas)
If T, R, and g areconstant with height then
dz RT
g
p
dp
RT
pg g
dz
dp
RT p
)0(
exp
ln
pC where
RT
gz C p
C
RT
gz p
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Standard Atmosphere
Considerable variations in atmospheric
properties with height
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Pressure Measurements
Barometers measure
atmospheric pressure
through balance of
pressure and liquidmercury weight
vapatm ph p g
0105.1000023.0 6
, atm psi p Hg vap
Hg mmatm 7601
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Manometers (1)
Manometers use fluid
statics to measure
relative pressure
between two points Balance of forces
If fluid A is a gas, then
pressure rise at h1 is
negligible (gA<<g2)
22321 h p p ph p atm A A g g
22h p p atm A g
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Gage Pressure
Manometers do not determine absolute pressure (since patm is not measured)
Gage pressure is pressure relative to
atmospheric Gage pressure can be negative for absolute
pressures below atmospheric
Usually, if no information is given on a pressure it is assumed to be a gage pressure
g atm A p p ph g
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Manometers (2)
Inclined manometers increase the sensitivity of
the measurement
g sin2221 l p p
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Bourdon Pressure Gages
Metal tube straightens as pressure increases movingneedle
Set to zero for atmospheric pressure Gage pressuremeasurement
Most common pressure sensor
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Forces on Surfaces (1)
For fluids in motion both pressure and shear
forces act on surfaces
For static fluids (including systems in motion
as a whole) there is no shear force
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Forces on Surfaces (2)
Net force is for isobaric
surfaces (normal to gravity, horizontal)
Atmospheric pressure cancels since acting on
both sides of surface (usually)
Gage pressure
ghA pA F
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Inclined Surfaces (1)
For an inclined surface, the pressure varies
across the surface
Defining
pdA F
hdA pdA F R g
sin yh
ydAdA y F R g g sinsin
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Area Centroid
Integral is area’s “Centroid”
Thus, inclined surfaces can be treated as flat if
area centroid location is known The resulting force acts through another point,
yR
A y ydA c
Ah Ay F cc R g g sin
Inclined Surfaces (2)
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Inclined Surfaces (2)
For moment of
resulting forceto act like actual
distributed force
dA y pydA y F R R g sin2
g sinc R Ay F
A y
I
A y
dA y y
c
x
c
R 2
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Tabulated Moments of Inertia
See p. 59-60for
transformations
to use tabulated
Ixy data
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Example: Plane surface (wall)
Centroid is located at h/2
Thus, force is
Force actsthrough 2/3 point
2/hA F R g
hhhhbh
bdy y A ydA y y
h
c
R32
32
))(2/(
3
20
22
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Example: Curved surface
Direct integration difficult for arbitrary surface
Free body diagram used to compute resultant
force
F1 is gh acting through center of surface AB
F2 acts through 2/3 point of surface AC
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Buoyancy (1)
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Buoyancy (1) . Submerged surfaces are subject to
an upward force due to higher pressure at greater depth
Force balance on only fluid in
ABCD
Buoyancy force only depends on
volume (Movie)
AB
Ah F 11
g CD
Ah F 22
g
g
g g g
B
obj B
fluid B
F
Ahh Ah F Ah
W F F F
1212
12
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Buoyancy (2)
Since buoyancy only depends on volume and
not density of the object, its force acts through
the centroid of the object (see p. 70 for formal
proof) However, the weight of the object acts through
the center of mass which is different from the
centroid if the object is not homogeneous
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Stability
If center of mass is below centroid, object will be stable
If center of mass is above centroid, object will be unstable
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Rigid body rotation (1)
Last time we derived for accelerating systems
For a rotating system
g a p
g a z z
p p
r r
r
p p
ˆ
ˆ1
ˆ
r r a ˆ2
2 r r
p
g
z
p
dz dr r dp g 2
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Rigid body rotation (2)
Along an isobar dp=0 dz dr r dp g 2
g
r
dr
dz 2
c g
r z
2
22
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Fl Vi li i (1)
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Flow Visualization (1)
Dye injectionin liquid can be used to“see” flow – dye followsthe flow
Photographsshow streaksrepresenting
path of fluid particles
Fl Vi li i (2)
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Flow Visualization (2)
Streaklines, Pathlines, and Streamlines
Streaklines = instantaneous location of fluid
particles that once passed through a specified point
inject dye continuously at fixed points and take snapshotat later time
Pathlines = path that particles follow
inject dye briefly at fixed points and take time-lapsed
photo for a period of time
Movie – differences in pathlines and streaklines
St li
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Streamlines
Streamlines = lines in the flow that are locally
tangent to the velocity of the fluid
St li (2)
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Streamline (2)
Streamlinesdetermined bymeasuringinstantaneousvelocity and
integrating tofind tangent lines
Harder tomeasure than
streaklines Most useful to
mathematicallydescribe flow
St li (3)
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Streamlines (3)
For steady flows – pathlines, streaklines, andstreamlines are identical - Movie
I i id Fl (1)
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Inviscid Flow (1)
In real fluids, if there is fluid motion with non-
uniform velocity then there will be strain and
shear forces
However, it is often true that these shear forcesare much smaller than forces due to pressure
gradients or gravity
In these cases the fluid is assumed to beinviscid (m=0)
I i id Fl (2)
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Inviscid Flow (2)
Inviscid flows are not strongly affected by drag atsurfaces and can flow around sharp corners
Viscid flows are slowed by drag at the surface much
more strongly
I i id Fl (3)
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Inviscid Flow (3)
Changes in overallvelocity or geometryof a problem canchange the
importance ofviscous forces
Some regions of a
flow may be inviscidwhile others showstrong viscouseffects
St li A l i
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Streamline Analysis
Fluid particlessubject to
Since streamline is
tangent to V
am F
n sV V ˆ0ˆ
dt
ds
s
V
dt
dn
n
V
sanadt
V d a sn
ˆˆ
2V an
V s
V a s
F Al St li (1)
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F=ma Along Streamline (1)
Note: we are applying F=ma to a fluid particle – the fluid particle follows a
pathline but we are using derivatives along the streamline to represent the
fluid acceleration the flow must be steady
Force due to gravity along streamline is
s
V V am F s s
sinsin,
g mg F g
s
F Al St li (2)
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F=ma Along Streamline (2)
Force due to pressure:
2
s
s
p p p p s
2
s
s
p p p p s
2
s
s
p p s
s
p yn p
yn p p yn p p F
s
s s p s
2
)()(,
F Al St li (3)
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F=ma Along Streamline (3)
Net Force = ma
Note: we have not included shear forces – we are assuming the flow is
inviscid (pressure forces are more important than viscous forces)
In static fluids the pressure gradient was balanced by gravity
In moving fluids, any imbalance in pressure and gravity (LHS) causes fluid particle acceleration (RHS)
s
V V
s
p g F sin
s
V V
s
p g
sin
F ma Along Streamline (4)
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F=ma Along Streamline (4)
Along streamline
Since along a streamline dn=0, for any derivative
partial and ordinary derivates in s are the same
(Note: analysis is limited to along a streamline)
Finally,
s
V V
s
p g
sin
s
z
sin
0
s z g
sV V
s p
ds s
pdn
n
pds
s
pdp
s
V
s
V V
2
2
F ma Along Streamline (5)
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F=ma Along Streamline (5)
Integrating along a streamline
If we assume fluid is incompressible (negligible density change)
02
2
ds
dz g
ds
dV
ds
dp
02
1 2 gdz dV dp
.2
1 2 Const gdz dV dp
.2
1 2 Const gz V dp
.2
1 2 Const gz V p
Bernoulli Equation
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Bernoulli Equation
Can only be applied to
Steady flow
Inviscid flow
Incompressible flow
Flow along a streamline
.21 2 Const gz V p
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Stagnation Flow
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Stagnation Flow
Stagnation point occurswhere flow is diverted around
two sides of an object
Dividing streamline includesstagnation point
Movie
Flow decelerates toward
stagnation point (higher pressure)
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Pressure Along Dividing Streamline
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Pressure Along Dividing Streamline
F=ma Normal to Streamline (1)
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F=ma Normal to Streamline (1)
Force due to gravity across streamline is
Force due to pressure
coscos, g mg F g n
2V am F nn
n
p
y s p
y s p p y s p p F
n
nn pn
2
)()(,
2
n
n
p pn
F=ma Normal to Streamline (2)
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F=ma Normal to Streamline (2)
Across streamline
Since normal to streamline ds=0, for any derivative
partial and ordinary derivates in n are the same
(Note: analysis is limited to normal to a streamline)
n
z
cos
dnn
p
dnn
p
ds s
p
dp
2
cos V
n
p g
02
n z g V
n p
02
gdz dnV
dp
F=ma Normal to Streamline (3)
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F=ma Normal to Streamline (3)
Integrating normal to streamline
If we assume fluid is incompressible
.
2
Const gz dn
V
p
0
2
gdz dnV
dp
.2
Const gdz dnV dp
Steady Inviscid Incompressible Flow
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Steady, Inviscid, Incompressible Flow
Along streamline:
Across streamline:
Pressure changes along streamline accelerates fluid
particles Pressure changes normal to streamline turns fluid
particles (changes streamline direction)
.
2
Const gz dnV
p
.21 2 Const gz V p
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Movie
Bernoulli Equation –
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Meaning of Terms
Along the streamline:
Static pressure dynamic pressure hydrostatic pressure total pressure
Each term has units of pressure (N/m2)
Static pressure = actual local pressure in the flow (thermodynamic pressure)
Dynamic pressure = pressure change due to velocity
Hydrostatic pressure = pressure change due to height
Total pressure = sum of all parts = constant
The Bernoulli Equation conserves pressure – pressure is only converted from one
type to another (Static/Dynamic/Hydrostatic)
T pConst gz V p .2
1 2
Bernoulli Equation
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Bernoulli Equation
Dividing by specific weight:
Pressure head velocity head elevation head total head
Each term has units of height (m)
Instead if we multiply by specific volume:
Elevation head is equivalent to potential energy per unit mass
Velocity head is equivalent to kinetic energy per unit mass
Pressure head is equivalent to flow work (pv) per unit mass
The Bernoulli Equation is a form of energy conservation (with no thermal orviscous losses or work or heat additions)
H z g
V p
2
2
g
.2
2
Const gz V
pv
Example – Stagnation Streamline
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Example – Stagnation Streamline
Free stream: high velocity (high dynamic pressure),low static pressure
Stagnation point: zero dynamic pressure, high static pressure
E.g., sticking your hand out the window of a movingcar
z
Stagnation Pressure
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Stagnation Pressure
Stagnation pressure = pressure that will beachieved if a fluid is brought to rest
Neglecting hydrostatic pressure, stagnation
pressure is simply static + dynamic pressure
2
22
2
1
2
1.
2
1
V p p
V pConst V p
stag
stag stag
Measuring Static/Stagnation Pressure
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Measuring Static/Stagnation Pressure
A side wall tap measures thestatic pressure by convertingstatic pressure tohydrodynamic pressure(pressure head) – dynamic
pressure at point 3 is zero (noslip)
Tap facing into flow convertsstatic and dynamic pressureto hydrostatic pressure -dynamic pressure at point 2 is
NOT zero
Pitot-static Tube
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Pitot static Tube
Two concentrictubes – one with a
forward facing tap
and the other with aside tap
43
4
2
3
2
2
1
p pV
p p
V p p
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Why does the pressure drop between 2 and 1?
Hydraulic grade line
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Hydraulic grade line
Energy line istotal head offluid(measured by
pitot tube)
Hydraulicgrade line is
pressure andelevation
head only(measured bystatic tube)
Why does HGL decrease in this system?
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“Modified” Bernoulli Equation
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Modified Bernoulli Equation
Bernoulli equation can only be used for steady,inviscid, incompressible flow
“Modified” forms of equation can be used
carefully for special cases of compressible,viscous, or unsteady flows
E.g., unsteady Bernoulli’s (see Sect. 3.8)
2
2
22
1
2
1
2
112
1
2
1 gz V pds
t
V gz V p
s
s
Application of Bernoulli Equation
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Application of Bernoulli Equation
Free Jets
Confined Flows
Venturi and orifice flowmeters
Sluice gates
Free Jets
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Free Jets
If streamlines are straight at jet exit (free jet,
R=∞) then no pressure gradient across jet,
p2=p1 V1=0
2
2
221
2
112
1
2
1
gz V p gz V p
)(2 212 z z g V
ghV 22
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Vena Contracta (1)
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Vena Contracta (1)
If a jet exit is too sharp,the streamlines cannotturn and flow is smallerthan hole diameter
90° turn would require
infinite pressure gradient At a-a, pressure is
atmospheric across jet(straight streamlines)
p2>p1 to cause streamline
curvature Contraction coefficient =
A j/Ah
Vena Contracta (2)
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Vena Contracta (2)
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Confined Flows
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Co ed ows
If a flow is completely enclosed, then no pointmust be at atmospheric pressure (or zero
velocity) – less information known
Must generally use VAQVAmassflow
Repeat for static taps on
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Repeat for static taps on
both ends of the manometer
Cavitation
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High velocity causesa locally low pressure
If static pressuredrops below vapor
pressure of liquid – boiling occurs
This “cavitation”leads to gas bubbles
flowing in liquiduntil pressureincreases
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Orifice and Venturi Meters
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Restriction in flow byorifice or Venturi tube
(specially shaped nozzle)
causes increase invelocity and decrease in
pressure
For horizontal flow:222
2
112
1
2
1V pV p
2211 AV AV Q
Flow Meters
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Flow rate is directly related to static pressure
drop across orifice and known geometry
In practice, these must be calibrated to account
for non-uniform flow, viscous effects, etc.
2
1
2
212
2
1
2
2
2
22
1
2
2
21
1
2
122
1)(
A
A
p p AQ
A A
AQ
AQ
AQ p p
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Sluice Gates
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2
2
221
2
11
2
1
2
1 gz V p gz V p 2211 AV AV Q
2
221
2
12
1)(
2
1V z z g V
2
12
21
2
2
1
222
2
2
21
2
1
2
2
21
1
)(2
1)(2
2
1)(
z z
z z g bz Q
z
z Q z b z z g
bz
Q
bz
Q z z g
12 2 gz bz Q If z1>>z2
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***ADD**
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Add physical interpretation of acrossstreamline equation and a couple of examples
E.g. Pressure in a vortex
Chapter 4 - Fluid Kinematics
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p
Fluid kinematics = study of fluid motionwithout concern for forces driving flow
What is the velocity, acceleration?
In reality, a point in space does not have avelocity (it is unlikely there is a moleculeexactly at that point)
When we talk velocity of a fluid particle wemean the average over all molecules in a smallregion – Continuum Hypothesis
Eulerian - Velocity Field
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Eulerian description of flow describes the velocity at
all points at all times Velocity field is a 3 component vector in 4
dimensions Movie
Some simpler cases often occur: Steady 3-D flow
Unsteady, 2-D flow
Movie Steady, 2-D flow
Movie
Steady, 1-D flow
k t z y xw jt z y xvit z y xuV ˆ),,,(ˆ),,,(ˆ),,,(
k z y xw j z y xvi z y xuV ˆ),,(ˆ),,(ˆ),,(
j y xvi y xuV ˆ),(ˆ),(
i xuV ˆ)(
jt y xvit y xuV ˆ),,(ˆ),,(
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Lagrangian – Particle Velocity
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Lagrangian description offlow follows individual particles
The position of a fluid particleis
The velocity of that fluid particle is
This does not tell us what thevelocity will be at a pointaway from the fluid particle
k t z jt yit xr ˆ)(ˆ)(ˆ)(
dt
r d V
Acceleration - Eulerian
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In Eulerian representation of velocity
Since x, y, z do not depend on t – Eulerian =
fixed coordinates
dt
V d
a
k t z y xw jt z y xvit z y xuV ˆ),,,(ˆ),,,(ˆ),,,(
t
V k
t
w j
t
vi
t
u
dt
V d
ˆˆˆ
Acceleration – Lagrangian (1)
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In Lagrangian representation of velocity, x, y,
and z depend on t as particle follows flow
dt
V d
a
k t t z t yt xw jt t z t yt xvit t z t yt xuV ˆ)),(),(),((ˆ)),(),(),((ˆ)),(),(),((
k dt
dz
z
w
dt
dy
y
w
dt
dx
x
w
t
w
jdt dz
z v
dt dy
yv
dt dx
xv
t v
idt
dz
z
u
dt
dy
y
u
dt
dx
x
u
t
u
dt
V d
ˆ
ˆ
ˆ
u
dt
dx
vdt dy
wdt
dz
Acceleration – Lagrangian (2)
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Lagrangian acceleration includes local acceleration
and change in velocity due to fluid particle motion
k z ww
ywv
xwu
j z
vw
y
vv
x
vu
i z
uw
y
uv
x
uu
t
V
dt
V d
ˆ
ˆ
ˆ
V V
t
V
dt
V d
k w jviuV ˆˆˆ
k
z
j
y
i
x
ˆˆˆ
“Material” Derivative
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Material or Substantial derivative includeslocal and spatial variation in a quantity
V V t
V
Dt
V D
termconvective
termunsteady
V t Dt
D
z
T w
y
T v
x
T u
t
T
Dt
DT
Steady/Unsteady Flow
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In steady flow, partial derivatives with respectto time are zero
However, spatial variations can still cause a
derivative for a fluid particle to be non-zero
z
T w
y
T v
x
T u
z
T w
y
T v
x
T u
t
T
Dt
DT
0 Dt
DT 0
t
T
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Control Volumes / Systems
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Just like in Thermo we sometimes consider systems (controlmass, Lagrangian) and sometimes control volumes (Eulerian)
We will start with relationships for systems…
m=constant
F=ma
… and derive from them relationships for control volumes
Steady Control Volume Unsteady Control VolumeControl Mass
Extensive/Intensive
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Consider an intensive property b (units/kg) B=mb (units)
system system
sys bd bdm B cvvolumecontrol
cv bd bdm B
dt
bd d
dt
dB sys sys
dt
bd d
dt dB cvcv
dt
dB
dt
dB syscv
Example: Mass, b=1 (B=m)
0dt
dm sys mdt
dmcv
Relating CV and CM (1)
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Consider a control mass that moves from (1) to (2) int
Define control volume as common region plus regionI
Relating CV and CM (2)
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II I
cv sys
B Bt
B
Dt
DB
22221111 bV AbV A
t
B
Dt
DBcv sys
AVbbm B
Rate change of property B for a system equals
change for a control volume plusinflow/outflow to control volume
Relating CV and CM (3)
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In general: velocity may not be normal to surface
property and velocity may vary across surface
cs
out dAnV bbm B ˆ
cv
cv bd t t
B
Reynolds Transport Theorem
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Formal relationship between changes in a control
mass and control volume This is similar to a material derivative except it
applies to a finite sized control volume and materialderivative applies at a point
RTT becomes material derivative as volume goes tozero
cscv
sys dAnV bbd t Dt
DBˆ
bV t
b
Dt
Db
Deforming Control Volumes
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Reynolds Transport Theorem can be applied to
any CV even if it moves
V is relative velocity
= V1-V0
cscv
sysdAnV bbd
t Dt
DBˆ
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3:57 (19th Olympics) versus 4:02 (20th
Olympics) speed skating (3000 m)
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19th - Salt Lake City (4200 ft): p = 650 mm Hg
20th - Turin, Italy (810 ft): p = 739 mm Hg
Density atm = 1.23 kg/m3
Find difference in air resistance
Conservation of Mass (1)
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Mass is conserved for a system, which bydefinition is a control surface that follows a
specific set of mass
Apply Reynolds Transport Theorem with B=m b=B/m=1
cscv
sysdAnV bbd
t Dt
DBˆ
cscv
sysdAnV d
t Dt
Dmˆ
Conservation of Mass (2)
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Since mass is conserved, msys=constant For a closed control surface (a system) –
Lagrangian description
For an open control volume (Eulerian
description), this becomes
0ˆ
cscv
dAnV d t
d Dt
D
Dt
Dm sys
0
Integral Continuity Equation
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For steady flows this becomes
The term
since mass flow is out if V is in same direction as n
0ˆ
cscvdAnV d t
0ˆ cs
dAnV
inout net
cs
mmmdAnV
ˆ
Uniform Flow
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Recall from Thermo 1, for uniform flow
We define average velocity at an inlet or outlet so that
We usually mean average velocitywhen we speak about the velocity at
inlet or exit
VAm
AV m
A
dAnV
V cs
ˆ
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Moving Control Volumes
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If the controlvolume is moving,the coordinatesystem is fixed tothe control volume
The velocity offlow across thecontrol surface isevaluated relativeto this coordinate
system
0ˆ
cscv
dAnW d t
cvV V W
Deforming Control Volumes
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Deforming control volumes are both movingand unsteady
Local relative velocities used at all surfaces
0ˆ
cscv
dAnW d t
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Newton’s Second Law
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but since mass is conserved for a system
Sum of forces on a system in any direction
equals change in momentum for the system in
that direction
Dt V Dmam F sys sys sys
Dt
V m D F
sys
sys
Linear Momentum Equations (1)
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Applying Reynolds Transport Theorem withB=mV, b=V
This is a vector equation (it is really 3equations in x, y, z directions)
cscv
sysdAnV V d V
t F ˆ
cscv
sysdAnV bbd
t Dt
DBˆ
Dt
V m D F
sys
sys
Integral Linear Momentum
Equations
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cscv
sysdAnV V d V
t F ˆ
cscv
z
cscv
y
cscv
x
dAnV wwd t F
dAnV vvd t
F
dAnV uud
t
F
ˆ
ˆ
ˆ
z y x nwnvnunV ˆˆˆˆ
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Moving Reference Frame
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Moving control volumescan be used
Inertial reference frame(not accelerating)
Non-inertial referenceframe (accelerating)
Accelerating referenceframes require caution
(relative velocities cannotsimply be used)
Steady Inertial Control Volume
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cscv sys
dAnW V d V
t
F ˆ
cvV V W
cs
cv
steady
cv
cv sysdAnW V W d V W
t F ˆ
)(0
cs
cv
cs
sysdAnW V dAnW W F ˆˆ
)(0
ˆˆ
onconservatimass
cs
cv
cs sys
dAnW V dAnW W F
cs
sysdAnW W F ˆ
Relative velocities used for
inertial control volumes
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Rotating Systems (1)
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For rotating systems, the moment of themomentum equation (angular momentum) is
often convenient
At a point, F=ma becomes Taking cross product of location of fluid
particle and F=ma
Dt
d V D
Dt
ma D F
)(
d V Dt
r D
Dt
d V r D
Dt
d V Dr F r
)0(0
V V
d V V Dt
d V r D F r
Rotating Systems (2)
I t ti t l l
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Integrating over a control volume
Applying Reynolds Transport Theorem(b=r xV)
Moment of momentum equation
sys sys sys
d V r Dt
D
Dt
d V r D F r
sys
d V r Dt
D F r
cscv sys
sysdAnV bbd
t bd
Dt
D
Dt
DBˆ
cscv
dAnV V r d V r t
F r ˆ
Moment of momentum equation
ˆ
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For a steady system
Sign of r xV and T terms determined by right-
hand-rule
cs fluxmomentumangular
momentumangular unsteady
cv systemontorquesof Sum
dAnV V r d V r
t
T
ˆ
cs dAnV V r T ˆ
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Energy Equation (1)
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From Thermo 1, energy of a system follows
where u is the internal energy from thermodynamics
innet innet
sysW Q
Dt
DE ,,
sys sys ed E
gz V
u pekeue 2
2
Energy Equation (2)
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Applying Reynolds Transport Theorem withB=E, b=e
cscv
innet innet
sysdAnV eed
t W Q
Dt DE ˆ
,,
gz V
u pekeue 2
2
cscv
sysdAnV bbd
t Dt
DBˆ
Work Work = force through a distance (s)
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Work force through a distance (s)
Can occur internal to system (as shaft work) or work at controlsurface (flow work)
Force at flow surface is pressure times area acting in – ndirection
Rate change of distance is velocity
cs
shaft innet s F W W
,
dAn pV W W cs
shaft innet
ˆ)(,
in shaft innet
cscv
W QdAnV peed t
,,ˆ
Energy Equation (3)
VV 22
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For steady, uniform flow
in shaft innet cscv W QdAnV gz
V p
ud gz
V
ut ,,
22
ˆ
22
h pvu p
u
in shaft innet
cscv
W QdAnV gz V
hd gz V
ut
,,
22
ˆ
22
in shaft innet inout W Q gz V
h AV gz V
h AV ,,
22
22
Steady, Uniform FlowVV 22
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For a 1 inlet, 1 outlet control volume
If flow is inviscid, shaft work will be zero
in shaft innet inout
inout
inout
W Q gz gz V V
hhm,,
22
22
in shaft innet inout W Q gz V
hm gz V
hm ,,
22
q gz gz
V V
pv pvuu inout inout
inout inout 22
22
loss Bernoulliquu gz gz V V
pv pv inout inout inout
inout 22
22
Extended Bernoulli Equation
VpVp22
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In units of length (head)
hL = head loss
hs = shaft work head (pump head)
in shaft in
in
in
in
out
out
out
out wloss gz V p
gz V p
,22
sl ininin
out out out hh z
g
V
g
p z
g
V
g
p
22
22
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Why Differential Analysis?
I l l i ll
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Integral analysis allows us tocompute overall (global) flow
behavior without concern for thedetailed flow inside a device
Integral analysis requires careful
integration at system boundaries(velocity profiles at exits must
be given or assumed)
Differential analysis is required
when we need to know thedetailed flow behavior at pointsinside a system (velocity profilesare computed directly) Recirculation zone will not show up in
integral analysis
Fluid Element Motion
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Differential equations for fluid flow can bederived by considering the motions and forces
of small fluid elements
Movie – rotation/translation/angulardeformation
Translation
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Translation – fluid elements translate at localfluid velocity z w yv xuV ˆˆˆ
t u x
t v y
t w z
Linear Deformation
zyx
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Volume of differential fluid element: Change in volume of fluid element in x direction
In general for 3-D:
zdt y x
x
ud
z y x
x
u
dt
d
1
V
z
w
y
v
x
u
dt
d
1
Rotation/Angular Deformation (1)
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Define angles a and as rotation of x and yaxis
a a
t
x
vtan
t
y
utan
Rotation/Angular Deformation (2)
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Rate of rotation of x and y axis
Note different sign convention for a and
If OA=-OB then the fluid element will only rotate andnot deform
If OA=+OB then the fluid element will only deformand not rotate
x
v
t t OA
a
0lim
yu
t t OB
0lim
Rotation/Angular Deformation (3)
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Rate of rotation of fluid elementdefined as average of OA and -OB
Likewise
y
u
x
vOBOA z
2
1
2
z v
yw
x21
xw
z u
y21
Rotation and Vorticity
ˆˆˆ
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Rotation rate is a vector:
Vorticity is defined as twice the rotation rate - Movie
y
u
x
v z
2
1
z
v
y
w x
2
1
x
w
z
u y
2
1
z y x z y x ˆˆˆ
wvu
z y x
z y x
V
ˆˆˆ
2
1
2
1
V
2
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Angular Deformation
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Rate of angular deformation (rate of shearing strain) offluid element defined as twice the average of OA and
+OB
Likewise
y
u
x
vOBOA z
22
g
z
v
y
w xg
x
w
z
u yg
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Mass Conservation (1)
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Previously we derived mass conservation for acontrol volume
For a differential element:
We apply control
volume equation to
element and let
dV0
0ˆ
cscv
sysdAnV d
t Dt
Dm
Mass Conservation (2)
0ˆ
sysdAnV d
Dm
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As volume0, all flows become uniform and
volume integral becomes homogeneous
cscv
t Dt
0
22
22
22
x y z
z
ww x y
z
z
ww
z x y
y
vv z x
y
y
vv
z y x
x
uu z y
x
x
uu z y x
t
cs
cs
cscv
Mass Conservation (3)
0
wvu
t
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For steady flow:
For incompressible fluids:
Incompressible flows have zero deformation(zero dilatation) – Velocity field is solenoidal
z y xt
0
V
t
0 V
0 V
Cylindrical Coordinates
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Mass conservation can be applied in anycoordinate system:
For cylindrical coordinates
Can look up form for gradient in cylindrical
coordinates:
0
V
t
0
11
z
vv
r r
vr
r t
z r
z vvr vV z r ˆˆ
ˆ
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Stream Function (1)
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For plane (2-D) steady incompressible flow,mass conservation becomes:
Define the stream function, y(x,y) such that
Thus, if y can be found it automaticallysatisfies mass conservation
0
V
t
y
v
x
uV
0
u y
y v
x
y 0
x y y x y
v
x
u y y
Stream Function (2)
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What is the stream function?
For constant y
Line of constant stream function is a
streamline
udyvdxdy y
dx x
d
y y y
0y d uv
dxdy
Stream Function (3)
Since flow can’t cross a streamline, flow between two
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differentially close streamlines: Difference in stream
function is flow rate
between streamlines
y y y
d dx x
dy y
vdxudydq
12 y y y d vdxudydqQ
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Potential Flow
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Potential Flow = Inviscid, Incompressible,Irrotational
Under these conditions the velocity potential,f, exists and satisfies:
Laplace equation is linear, so superposition ofsolutions can be used (this is generally not true
for fluids since full momentum equation isnonlinear)
02
f
g pV V t
V
What is the Potential Function?
vdyudxdydxd
f f
f
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Thus, if then
Compared to stream function:
Lines of constant f (equipotential lines) are
orthogonal to lines of constant y (streamlines)
y x 0f d
v
u
dx
dy
0y d uv
dxdy
Streamlines/Potential Lines
Streamlines and
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Streamlines andequipotential linesform grid describingflow
High velocities inregions wherestreamlines arecompressed
Low velocities wherestreamlines expand
How is Potential Flow Used? First, we find the potential function for several simple flows
(next slides)
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(next slides)
If we can then describe a new flow as a sum of simple flows,the total potential function = sum of simple potential functions
Movie – Example: flow over a surface can be approximated bya point source and uniform flow
Simple Flow 1: Uniform Flow
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Straight flow in one direction:
If a=0:
xU u
f a cos
yU v
f a sin
)(cos
)(sin
y f Ux
x f Uy
a f
a f
a a f cossin x yU
Uxf
Simple Flow 2: Source/Sink
Flow originating from a point with equal velocity in
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all directions and a volume flow rate, m
Source (m>0), Sink (m<0)
m=source strength
mrvr 2
f
r
v 10
r m
ln2
f
r
m
r vr
f
2
Example: Flow over a Half-body (1)
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Half-body can be approximated as uniformflow + source (Movie)
Stagnation point occurs at r=b (angle
f cosUr Ux
r m
ln2
f
f
f
f
sin1
2cos
ln2
cos
U r
v
r mU
r v
r m
Ur
r
U
mb
b
mU vr
2
2cos0
Flow over a Half-body (2)
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Can also find streamfunction for halfbody (see text):
Value of stream function at stagnation point:
Streamlines from stagnation point:
y y y 2
sin m
Ur sourceuniform
22
sin
2
mm
U
mU
y
y
sin2
2sin
2sin
2
U
mr
m
Ur
mUr
m
Simple Flow 3: Vortex
Flow circulating about a point
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Free vortex:
Free vortex is irrotational
Other vortices can be rotational No potential function
f K
r
K
r v
f
1
Circulation
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Define the circulation for a closed curve in anyflow as:
For vortex: sd V ccw
rd v sd V
K d r r
K 2
f 2
Simple Flow 4: Doublet (1)
A doublet is a combination of a source and sink
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As a0
1
212 ln
2ln
2ln
2 r r mr mr m
f
2/122
2 cos2 ar ar r r source
2/122
1 cos2 ar ar r r sink
f
cos2
cos41ln
4 22
ar ar
ar m
r
am
ar ar
ar m
aa
f
cos
cos2
cos4
4limlim
2200
Simple Flow 4: Doublet (2)
A doublet is a combination of a source and sink
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Streamlines:
Why is this useful?
Next slide…
r K
r ma
f coscos
f cos K r
Flow over a Circular Cylinder (1)
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Let’s combine uniform flow and a doublet tosee what flow that represents:
Choose: where a is the radius of acylinder
r
K f
cosUxf
f coscoscos2 r
r
K U
r
K Ur
U a K 2
f cos12
2
Ur r
a
Flow over a Circular Cylinder (2)
f cos12
2
Ur a
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Thus
This potential functionhappens to yield flowover a cylinder
r
f
cos12
2
U
r
a
r
vr
f sin1
12
2
U r
a
r v
Pressure Distribution over Cylinder
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Since potential flow analysis assumes inviscidincompressible flow, Bernoulli’s equation can
be used
Pressure on cylinder surface (r=a):22
2
1
2
1 v pU p surface
)sin41(2
1sin4
2
1
2
1 22222 U pU U p p surface
sin2sin12
2
U U r
av
Pressure Distribution over Cylinder
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Comparison ofinviscid and
experimental
pressure
distributions:
)sin41(2
1 22 U p p surface
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Momentum Equation (1)
Integral momentum equation for a control
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g qvolume:
Apply this to differential element:
Forces:
Normal stress,
(N/m2)
Shear stress, (N/m2)
Gravity
cscv
sysdAnV V d V
t F ˆ
Momentum Equation (2)
In x- direction z y x g
y z x F x
yx zx xx
x
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Momentum Equation (3)
In x- direction
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VAu
t
udAnV V d V
t cscv
ˆ
2222
2222
2222
z
z
uu y x
z
z
ww
z
z
uu y x
z
z
ww
y
y
uu z x y
y
vv y
y
uu z x y
y
vv
x
x
uu z y
x
x
uu
x
x
uu z y
x
x
uu z y x
t
u
u
A
V
Momentum Equation (4)
Dividing by volume and equating to force:
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Subtracting mass conservation times u:
22
1
22
1
22
1
22
1
221
221
z
z
u
u z
w
z
w z
z
u
u z
w
z
w
y
y
uu
y
v
y
v y
y
uu
y
v
y
v
x xuu
xu
xu x
xuu
xu
xu
t u g
y z x x
yx zx xx
u
z
w
z
uwu
y
v
y
uvu
x
u
x
uu
t u
t
u g
y z x x
yx zx xx
0
u
z
wu
y
vu
x
uu
t
Momentum Equation (5)
z
uw
y
uv
x
uu
t
u g
y z x x
yx zx xx
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The normal stress is due to both viscous stress() and pressure
z
uw
y
uv
x
uu
t
u g
y z x x
p x
yx zx xx
z
vw
y
vv
x
vu
t
v g
x z y y
p y
xy zy yy
xx xx p
z ww
ywv
xwu
t w g
x y z z p
z xz yz zz
g pV V t
V
Momentum Equation g pV V
t
V
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z
uw
y
uv
x
uu
t
u g
y z x x
p x
yx zx xx
z
vw
y
vv
x
vu
t
v g
x z y y
p y
xy zy yy
z ww
ywv
xwu
t w g
x y z z p z
xz yz zz
Inviscid Flow
g pV V t
V
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In inviscid flow there is no shear stress (=0)
Euler equation:
Steady inviscid flow:
(see text for proof that this reduces to Bernoulliequation)
Neglecting gravity:
g pV V t
V
1
g pV V
1
pV V
1
Irrotational Flows (1)
Define the velocity potential
ff f f
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Vorticity is:
Only irrotational flows have a velocity potential
v y
f u
x
f w
z
f f V
0
ˆˆˆ
ˆˆˆ
k
x y y x
j
z x x z
i
y z z y
k y
u
x
v j
x
w
z
ui
z
v
y
w
f f f f f f
Irrotational Flows (2) For incompressible, irrotational flow, mass
i i
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conservation is:
This is a much easier equation to solve thanmomentum and is useful if flow can beapproximated as irrotational and incompressible
We will see, only inviscid flows can beirrotational
2
2
2
2
2
220
z y xV
f f f f f
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Full Momentum Equation
g pV V t
V
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Need model for viscous forces () dy
dum
Newtonian Fluids (1)
id
dum
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is just part of the
full viscous
stress The complete
model:
dy
yx xy x
v
y
u m
zy yz y
w
z
v m
zx xz z
u
x
w m
x
u xx
m 2
y
v yy
m 2
z
w zz
m 2
Newtonian Fluids (2) The viscous stress tensor for Newtonian fluids:
uwvuu2
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If viscosity is constant:
z
w
y
w
z
v
z
u
x
w
y
w
z
v
y
v
x
v
y
u
z x x y x
2
2m
V V
V z
w
V y
v
V
x
u
z
w
y
w
y z
v
x z
u
x
w
z y
w
z
v
y
v
x
v
x y
u
z
u
z x
w
y x
v
y
u
x
u
m m m m 2
2
2
2
2
2
2
222
2
2
2
2
2
2
2
2
22
2
222
2
2
2
2
2
2
2
Navier-Stokes Equations
g pV V t
V
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Substituting viscous stress tensor yields NSE
g V V pV V t
V
m m
2
x g z
w
y
v
x
u
x z
u
y
u
x
u
x
p
z
uw
y
uv
x
uu
t
u m m m m
2
2
2
2
2
2
y g z
w
y
v
x
u
y z
v
y
v
x
v
y
p
z
vw
y
vv
x
vu
t
v m m m m
2
2
2
2
2
2
z g z
w
y
v
x
u
z z
w
y
w
x
w
z
p
z
ww
y
wv
x
wu
t
w m m m m
2
2
2
2
2
2
NSE Example: Flow between plates
Infinite plates in x and z, steady flow, gy=-g,
1 D fl ( 0) 0)(
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1-D flow (v=w=0) 0)( z xu
t u
2
2
0 y
u
x
p
m g
y
p
0 00
Flow between plates (2)
2
0 up
g y
p
0 )( x f gy p
uxgyp )(1
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Flow is parabolic
Flow rate per unit width
20
yu
x p
m
yu x g y
x p
)(1
m
)(21 2 xh y
x pu
m
2
21)(0)( h
x p xhhu
m
0)(0)0(
x g
y
u
22
2
1h y
x
pu
m
x
phh
h
x
pdyh y
x
pudyq
hh
m m m 332
1
2
12
33
3
0
22
0
x
phu
m 2
2
m ax
Keys to Success…
St t ith f ll NSE
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Start with full NSE Make reasonable assumptions and eliminate
terms
Solve simplified differential equations Apply boundary conditions
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Navier-Stokes Equations
R i
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Review
g V V pV V t
V
m m
2
x g z
w
y
v
x
u
x z
u
y
u
x
u
x
p
z
uw
y
uv
x
uu
t
u m m m m
2
2
2
2
2
2
y g z
w
y
v
x
u
y z
v
y
v
x
v
y
p
z
vw
y
vv
x
vu
t
v m m m m
2
2
2
2
2
2
z g z
w
y
v
x
u
z z
w
y
w
x
w
z
p
z
ww
y
wv
x
wu
t
w m m m m
2
2
2
2
2
2
Couette Flow (1)
Flow driven by moving plate
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1-D, steady, infinite in z and x
v=w=0
Same equations at pressure driven flow boundary conditions are different
2
2
0 y
u
x
p
m g y
p
0
U buu )(,0)0(
)( x f gy p
)()(2
1 2 xh y x g y x
pu
m
Couette Flow (2)
Ubuu )(0)0(
)()(2
1 2 xh y x g y x
pu
m
b
Uyby y x
pu 2
2
1
m
b
y
b
y p
U
b
b
y
U
u
12
2
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U buu )(,0)0(
0)( xh
)(2 x g x
pb
b
U
m
bb xU bU 12m
Pipe Flow (1) 1-D steady flow (vr =0, v=0), axisymmetric
(see p. 321 for full cylindrical NSE)21
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No gradients in z (besides pressure)
2
2
21
z
v
r
vr
r r z
p
z
vv z z z z
m m
r
vr
r r z
p z 1m
Pipe Flow (2)
Boundary conditions:
r
vr
r r z
p z 1m
0)0(
r
v z 0)( Rv z
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r
vr z f r
z
p z )(2
1 2
m 0)( z f
)(4
1 2 z g r z
pv z
m
2
4
1)( R
z
p z g
m 22
4
1 Rr
z
pv z
m
Pipe Flow (3)
22
4
1 Rr
z
pv z
m
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Movie – Parabolic velocity profiles in pipe
flow
z
p Rv
m 4
2
m ax
z
p R R R
z
prdr Rr
z
pQ
R
m
m
m 82422
4
1 444
0
22
28
max2 v
z
p R
A
QV
m
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Unsteady Flow (1)
Infinite plate, infinite fluid (in z and x), fluid
i i i ll 1 D fl )()(
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initially at rest, 1-D flow
At t=0, plate moves at velocity U
2
2
y
u
t
u
m
g y
p
0
00
0)()(
z x
0 wv
C gy p
Unsteady Flow (2)
Similarity solution: 2
2
y
u
t
u
m
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Assume solution is of form:
2
2
2
2
2
y
u
y
u
y
u
yt
u h
h
h
h
h
h
h
h
t f f
t t
y f
4
1)0(
4
t
y f u
h h
2:)(
f f h 2
21 )( C erf C f h
A “simple” problem…
Consider flow out of an infinitely long slot
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Problem is 2-D (w=0, d/dz=0), steady, gravity
negligible, incompressible
x
y
V
Governing Equations (1)
xgz
w
y
v
x
u
xz
u
y
u
x
u
x
p
z
u
wy
u
vx
u
ut
u
mmmm
2
2
2
2
2
2
0
z
w
y
v
x
u
t
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Simplify to…
g z y x x z y x x z w yv xut m m m m
y g z
w
y
v
x
u
y z
v
y
v
x
v
y
p
z
vw
y
vv
x
vu
t
v m m m m
2
2
2
2
2
2
z g z
w
y
v
x
u
z z
w
y
w
x
w
z
p
z
w
w y
w
v x
w
ut
w
m m m m
2
2
2
2
2
2
y
v
x
u
x y
u
x
u
x
p
y
uv
x
uu m m m
2
2
2
2
y
v
x
u
y y
v
x
v
y
p
y
vv
x
vu m m m
2
2
2
2
00
0
y
v
x
u
Governing Equations (2)
2
2
2
2
1 uu puvuu
0
y
v
x
u
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Three equations, three unknowns (u, v, p) u=u(x,y,,m, v=v(x,y,,m), p=p(x,y,,m)
Problem has a solution, but cannot simply
integrate equations CFD is an approach to estimate an answer to
governing equations
22 y x xp
y x
2
2
2
21
y
v
x
v
y
p
y
vv
x
vu
Computational Mesh
Computational fluid dynamics (CFD) solves
these differential equations on a grid
( ( ) ( )
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u=u(xi,yi,,m, v=v(xi,yi,,m), p=p(xi,yi,,m)
xi and yi are discrete – spacing x and y
Nx N number of nodes
Examples of Meshes
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Numerical Methods Finite Difference
Differential form of governing equations are discretized
and solved
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and solved
Finite Volume
On each cell, conservation laws are applied at a discrete
point of the cell [node].
Integral Control Volume Form of
Governing Equations
Taylor Series Expansion
!3!2!1
3
,
3
32
,
2
2
,
,,1
x
x
u x
x
u x
x
uuu
ji ji ji
ji ji
3322 xu xu xuuu
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Differential is converted to discrete algebraicexpression:
Error of order (x)2
!3
2!1
2
3
,
3
3
,
,1,1
x
x
u x
x
uuu
ji ji
ji ji
!3!2!1
,
3
,
2
,
,,1 x x x
uu
ji ji ji
ji ji
62
2
,
3
3
,1,1
,
x
x
u
x
uu
x
u
ji
ji ji
ji
x
uu
x
u ji ji
ji
2
,1,1
,
Second Derivatives
!3!2!1
3
,
3
32
,
2
2
,
,,1
x
x
u x
x
u x
x
uuu
ji ji ji
ji ji
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With an error of order (x)2
!42
!222
4
,4
42
,2
2
,,1,1
x
x
u x
x
uuuu
ji ji
ji ji ji
!3!2!1
3
,
3
32
,
2
2
,
,,1
x
x
u x
x
u x
x
uuu
ji ji ji
ji ji
12
2 2
,
4
4
2
,1,,1
,
2
2 x
x
u
x
uuu
x
u
ji
ji ji ji
ji
2
,1,,1
,
2
2 2
x
uuu
x
u ji ji ji
ji
Discretized Equations
Apply the derivative estimates to the
governing equations: 22
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governing equations:
Similar additional equation for v momentum
2
2
2
21
y
u
x
u
x
p
y
uv
x
uu
0
y
v
x
u
x
uu
x
u ji ji
ji
,,1
, 2
,1,,1
,
2
2 2
x
uuu
x
u ji ji ji
ji
21,,1,
2,1,,1,1,11,1,
,,1,1
, 222
122 y
uuu x
uuu x p p
yuuv
xuuu ji ji ji ji ji ji ji ji ji ji
ji ji ji
ji
01,1,,1,1
y
vv
x
uu ji ji ji ji
System of Algebraic Equations
Discretization turns 3 partial differential
equations into thousands of algebraic
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equations into thousands of algebraic
equations (3 for each mesh point i,j)
E.g.:
(2,2)
(4,2)
01,1,,1,1
y
vv
x
uu ji ji ji ji
01,23,22,12,3
y
vv
x
uu
01,43,42,32,5
y
vv
x
uu
NxN number of algebraic
equations
Boundary Conditions
Since the mesh is finite, derivatives must be
computed differently at the edge of the mesh
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computed differently at the edge of the mesh
(u-1,-1 doesn’t exist)
Boundary conditions must be specified at all
mesh edges
Boundary conditions
are always approximations
of realistic conditions
u,v=0
d/dx=0
d/dy=0
u=U, v=0
CFD Output – Infinite slot (1)
Solution is only available at grid points –
interpolation used for values between grid points
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CFD Output – Infinite slot (2)
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CFD Output – Vortex Generation
Many CFD simulations are unsteady to capture
transient features of flow
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transient features of flow
Cylinders in cross flow shed vortices that can
be visualized by streamlines
Movie – CFD – computed streamlines over
bluff-body
Errors in CFD
Numerical error – the iterative solution did not
find the correct answer to the algebraici
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find the correct answer to the algebraicequations
Discretization error – the derivative estimates
were not accurate enough (x too large) Governing equation error – a term was
removed that shouldn’t have been (gravity?)
Boundary condition error – the boundaryconditions do not reflect reality
CFD Validation/Accuracy (1) Numerical error can be
assessed by examiningresiduals
Residual y
vv
x
uu ji ji ji ji
1,1,,1,1
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Residuals would ideally be zero, but actuallyconverge to a small
constant value Once the residuals are
sufficiently small thealgebraic equationshave been solved butthey still may beinnaccurate
CFD Validation/Accuracy (1) Discretization error can be assessed by
recomputing a solution on a finer/different grid
The difference in the solutions is an estimate of
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The difference in the solutions is an estimate of
effect of grid (grid independence)
Errors in governing equations and boundary
conditions can be assessed by turning on/off
physical terms or by adding perturbations to
boundary conditions (sensitivity studies)
Finally, simulations for some cases should becompared to experiments (validation)
Post-Processing
Solution from CFD must be post-processed to
extract information of interest
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extract information of interest
E.g.: Flowlab assignment you will integrate
velocity profiles to compute mass and
momentum flux
i
iii r r uurdr udAndAV m 22
CFD Overview
Simplify equations as far as possible
Discretize equations using a finite grid (derivativesbecome differences)
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q g g ( become differences)
A few partial differential equations become
thousands+ of algebraic equations
Solve using numerical methods (ME elective course)
Validate solution to insure accuracy – Even with
commercial CFD codes this step MUST be done by
user CFD always gives a pretty answer – you must work
to make sure that answer is useful
Why Dimensional Analysis?
Imagine we are interested in the solution to
flow in a round pipe for 1000 differentcombinations of velocity fluid type (density
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ow a ou d p pe o 000 d e e tcombinations of velocity, fluid type (density,viscosity), and pipe diameter
Do we have to perform 1000 experiments?
1000 calculations?
Dimensional analysis determines how these
1000 cases are related to minimize the numberof independent calculations/experiments thatmust be performed
Problem Variables
For the pipe flow example, assuming the pipe
is smooth and the velocity profile does not
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y pchange with position in the pipe, the variablesare (experience required to generate this list)
Pipe diameter, D Centerline velocity, V
Viscosity, m
Density,
Pressure drop, p
Pipe length, L
),,,,( L DV f p m
Buckingham Pi Theorem (1)
“If an equation involving k variables is
dimensionally homogeneous it can be reduced
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dimensionally homogeneous, it can be reduced
to a relationship among k-r independent
dimensionless groups, where r is the minimum
number of dimensions required to describe thevariables”
Buckingham Pi Theorem (2)
k=6
p (N/m2), (kg/m3), m (N-s/m2), V (m/s), D
),,,,( L DV f p m
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p (N/m ), (kg/m ), m (N s/m ), V (m/s), D(m), L (m)
Both kg and N are not independent dimensions
since 1 N = 1 kg-m/s2; thus (N-s2/m4) r=3 (N, m, s)
p relationship can be expressed with k-r=3dimensionless Pi groups
Dimensionless Pi Groups (1)
To determine Pi groups
Select one variable for each of the independent
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Select one variable for each of the independent
dimensions (N, s, m) – these are called repeating
variables
E.g.: D (for m), V (for s), (for N) For the other non-repeating k-r variables construct
dimensionless parameters using only the non-
repeating variable and the repeating variables as
needed
Dimensionless Pi Groups (2)
p (N/m2) z y x
m D s
mV
m
s N
m
N p
4
2
2
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This is dimensionless for x=-1, y=-2, z=0
m (N-s/m2)
x=-1, y=-1, z=-1
L (m)
By inspection
12
V
p
z y x
m D s
mV
m
s N
m
s N
4
2
2 m
2VD
m
3 D
L
Result of Pi Group Construction
),,,,( L DV f p m
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Becomes
The Pi groups are not unique, we could form different
dimensionless ratios but in any case we would get a
relationship among a non-dimensional pressure drop
versus two non-dimensional groups
12
V
p
2
VD
m 3
D
L
D
LVD f
V
p,
2 m
Common Dimensionless Groups
Some groups are used so often that we have
named them: (many more than shown here)
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( y ) Reynolds number
Inertial forces/viscous forces
Froude number Inertia/gravity
Euler number
Pressure/momentum
Mach number
c=speed of sound
m
VDRe
gLV Fr
2V
p Eu
c
V Ma
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Reynolds Number
Re is the most important parameter in fluid
mechanics – it defines the relative importance
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pof viscosity to inertial forces
Low Re, viscosity dominates
High Re, inertia dominates Movie – Effect of Reynolds number on
viscous behavior
We will see later that Re also determines whenlaminar flows become turbulent
m
VDRe
A Specific Example – Pipe Flow
We derived previously for laminar pipe flow
Th
22
4
1
Rr z
p
v z
m
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Thus,
At pipe center
Integrating
D
LVD f
V
p,
2
m
z
p
Rr
v z
22
4m
z
p
R
V
2
4m
L
R
V p
2
4m
D
L
VD L
DV V
p
m
m
16
2/
422
D
L Eu
Re
16
Compiling Data
You perform many experiments of, for
example, pipe flow pressure drop versus
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p , p p p p
V,D,L,,m
Data should be reported in terms of non-
dimensional Pi groups – simplest, most useful
way to report relationships among variables
Following examples: 1 Pi group, 2 Pi groups, 3
Pi groups
Problems with 1 Pi Group
Assume the drag, F, on a spherical particle
falling through a viscous fluid is a function ofdi t d l it V d i it
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diameter, d, velocity, V, and viscosity, m
F[N]=f(d[m],V[m/s],m[N-s/m2])
k=4, r=3 (N,m,s) 1 Pi group
Experiment determines
F
.)( Consnothing f Vd
F
m
m
3Vd
F Vd F m 3
Terminal Velocity
The sphere will reach steady state (constant V)
when the drag force and buoyancy force equalthe weight terminal velocity
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the weight – terminal velocity
Stoke’s Law: mg
g g Vd
mg F F
sphere fluid
bouydrag
m 3Fdrag+Fb
g r r V f s
3
3
423 m
g r V f sterm
2
9
2
m
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Problems with 2 Pi Groups
The pressure drop in a fixed-length pipe
k=5, r=3 ),,,( DV f p m
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Data taken for water
and glycerin in a 0.5 in pipe
m
VD f
V
p2
0
500
1000
1500
2000
2500
0 5 10 15 20 25 30 35
Velocity (ft/s)
P r e s s u r e D
r o p ( l b f / f t ^ 2 )
Water
Glycerin
0
0.5
1
1.5
2
2.5
3
0 20000 40000 60000 80000 100000 120000
VD/
p / V ^ 2
Water
Glycerin
Problems with 2 Pi Groups (2)
Collapse of multiple data sets confirms
assumptions in Pi group development VDp
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For a 5-foot pipe:
m
VD f
V
p2
y = 18.235x-0.2456
R 2 = 0.9996
0
0.5
1
1.5
2
2.5
3
0 20000 40000 60000 80000 100000 120000
VD/
p / V ^ 2
Water
Glycerin
Power (Water)
246.0
2 24.18
m
VD
V
p
5
6
Water
Glycerin
Problems with 3+ Pi Groups
Now we also vary pipe length ),,,,( L DV f p m
LVD
f p
,2
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0
1
2
3
4
0 20000 40000 60000 80000 100000 120000
VD/
p / V ^ 2
Glycerin
L/D=240
L/D=60
L/D=120
D
f V
,2 m
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Problems with 1 Pi Group
Assume the drag, F, on a spherical particle
falling through a viscous fluid is a function ofdiameter d velocity V and viscosity m
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diameter, d, velocity, V, and viscosity, m
F[N]=f(d[m],V[m/s],m[N-s/m2])
k=4, r=3 (N,m,s) 1 Pi group
Experiment determines
F
.)( Consnothing f Vd
F
m
m
3Vd
F Vd F m 3
Terminal Velocity
The sphere will reach steady state (constant V)
when the drag force and buoyancy force equalthe weight terminal velocity
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the weight – terminal velocity
Stoke’s Law: mg
g g Vd
mg F F
sphere fluid
bouydrag
m 3 Fdrag+Fb
g r r V f s
3
3
423 m
g r V f sterm
2
9
2
m
Problems with 2 Pi Groups
The pressure drop in a fixed-length pipe
k=5, r=3 ),,,( DV f p m
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Data taken for water
and glycerin in a 0.5 in pipe
m
VD f
V
p2
0
500
1000
1500
2000
2500
0 5 10 15 20 25 30 35
Velocity (ft/s)
P r e s s u r e D
r o p ( l b f / f t ^ 2 )
Water
Glycerin
0
0.5
1
1.5
2
2.5
3
0 20000 40000 60000 80000 100000 120000
VD/
p / V
^ 2
Water
Glycerin
Problems with 2 Pi Groups (2)
Collapse of multiple data sets confirms
assumptions in Pi group development
VDp 3
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For a 5-foot pipe:
m
VD f
V
p2
y = 18.235x-0.2456
R 2 = 0.9996
0
0.5
1
1.5
2
2.5
3
0 20000 40000 60000 80000 100000 120000
VD/
p / V ^ 2
Water
Glycerin
Power (Water)
246.0
2 24.18
m
VD
V
p
5
6
Water
Glycerin
Problems with 3+ Pi Groups
Now we also vary pipe length ),,,,( L DV f p m
D
LVD f
V
p,
2
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0
1
2
3
4
0 20000 40000 60000 80000 100000 120000
VD/
p / V ^ 2
y
L/D=240
L/D=60
L/D=120
D
fV
,2 m
Why Modeling?
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Sometimes this is really how we design
things…
Tacoma Narrows Bridge
Bridge collapse in WA (1940) due to wind-
induced oscillations
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Movie
Modeling
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New Tacoma Narrows bridge to have twosupsension structures side-by-side
Scale models used to test expensive designs
New bridge tested in 9mx9m NRC wind tunnel How can we trust the results from the model?
Model Similarity (1)
We are interested in knowing a particular
variable for a large design (drag, pressure
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drop, oscillation frequency, etc.)
IF we know what other variables affect the
variable of interest then we can generalize therelationship in terms of Pi groups
,...),( 321 f
Model Similarity (2)
If we build a model that has the same value of
all dependent PI groups, then the dependent Pi
d ill b
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group measured will be accurate
If:
Then:
real ,2mod,2 real ,3mod,3
real ,1mod,1
,...),( 321 f
Example: I want to know the pressure drop that will
occur in a 3-foot diameter, 1 mile long oil pipe
while pumping 500 lbm/s of oil (thisdetermines the size of pump required)
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determines the size of pump required)
I have a ¼” pipe to experiment with
Choose pipe length so that L/D is the same
Choose fluid and velocity so that Re is thesame
D L f
V p Re,
2
Example: (cont.)
D
L f
V
pRe,
2
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Model Similarity (3)
In practice it may not be possible to have all
dependent Pi groups match reality – must use
i i ff
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experience to capture most important effects
Validation of model must be used to insure
that all variables were included whendeveloping Pi groups
Movie – Physical model of plume
disbursement in city
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Pipe Flow
We have already derived the
equation for pressure driven
laminar flow in a round pipewith no gradients in the flow
22
4
1 Rr
z
pv z
m
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with no gradients in the flow
direction
However, this geometry isimportant enough to
consider in greater detail
Two additional effects
Turbulence
Entrance length
2
maxvV
What is Turbulence?
Turbulent flows contain fluctuations in
velocity
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These fluctuations are coherent leading to
structures in the flow
Movie – Turbulence in a stirred bowl, small
eddies dissipate first
Turbulence at a Point
For laminar flow u’=0 (no fluctuations)
)(')( t uut u
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Turbulent flow – fluctuations around 25%
Transition to Turbulent Flow
Movie – Turbulence in pipe flow
Original experiment of Osborne Reynolds
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Importance of Reynolds Number
Experiments of Reynolds showed
Laminar flow Re<2100
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Turbulent flow Re>4000
Transitional flow 2100<Re<4000
Laminar part of the time
Turbulent part of the time
Weakly turbulent
Turbulence occurs when inertial forces aremuch greater than viscous forces
m
VDRe
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Entrance Region
Entrance length is region in a pipe where flow
adjusts to “fully-developed” profile
ll d l d fil i l i f d
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Fully developed profile is solution we found
previously (no gradients in x)
Entrance Region
Entrance length is region in a pipe where flow
adjusts to “fully-developed” profile
F ll d l d fil i l i f d
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Fully developed profile is solution we found
previously (no gradients in x)
Entrance Length
Correlations for entrance length:
Laminar flow:Re060
l e
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Turbulent flow:
Laminar flows: l=0.5-125 D (Re=10-2100)
Turbulent flows: l=18-50 D (Re=4000-2x106)
Re06.0 D
6/1Re4.4 Dl e
Pressure Drop in Pipe Flow
Pressure drop is linear in fully developed region
Entrance region has extra loss due to higher velocity
gradients at wall
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gradients at wall
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Friction Factor We found a non-dimensional relationship for
laminar pipe flow (based on centerline V):
D
L Eu
Re
16
D
L
VDV p
m 16
2
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Based on average velocity:
The friction factor, f is defined such that:for laminar flow:
DRe
DVD
D
LV f p 2
2
1
D
LV
D
L
DV V p 22
Re
32
2164
m
Re
64 f
Head Loss in Terms of f
Recall from modified Bernoulli’s equation
lh z V p
z V p
2
2
221
2
11
22
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Friction factor is directly related to loss in pipe
flow
l g g
2122 g g
D
L
g
V f
phl
2
2
g
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Effect of Wall Roughness
Wall roughness
also increasesfriction factor
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friction factor
e, is height of
average perturbation of
wall shape
Friction Factor
The friction factor f is defined such that
D D
LVD g
V
p e
m
,,
2
1 2
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The friction factor, f, is defined such that
Thus,
For laminar flow f=64/Re
D
LV f p 2
2
1
D g f e Re,
Moody Chart
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Head Loss
Moody chart used to estimate f for a turbulent
flow
Head loss for pipes in a system added together
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Head loss for pipes in a system added together
to determine total pressure drop
D
L g
V f ph L2
2
g
Lh z g
V p z
g
V p 1
2
111
2
11
22 g g
Minor Losses (1)
Additional elements in a pipe flow system can
cause pressure loss
Loss coefficient K
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Loss coefficient, K L
2
21 V
p K
L
g
V K h L L
2
2
minor ,
Minor Losses (2)
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Moody Chart
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Minor Losses
Loss coefficients
are tabulated formay conditions
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may conditions
Specified by
manufacturers offlow products
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Turbulent Flow
The average defined by:
)(')( t uut u
T t
udt T
u 1
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t T
0)(' t u
0'2 u
''vu
Turbulent Transport Correlation in velocity fluctuations can lead to
enhanced transport of momentum
Enhanced transport appears to be extra
viscosity
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viscosity
0'' vu
Turbulent Viscosity
In turbulent flows shear stress (from nonlinear
udu/dx terms):
y
uvu
y
ut xy
m m m ''
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Where the turbulent or eddy viscosity is
If we can define h then we can solve for theaverage velocity profile
y yy
yu
vut
/
'' h m
Turbulence Modeling
There are no exact solutions in turbulent flows
– a model for h must be assumed
For different problems a variety of models for
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For different problems a variety of models for
h have been proposed/tested
None are perfect – for some cases they work
and others they fail
Turbulence modeling is an active research area
A Turbulent Model Example
Prandtl proposed the so-called mixing length
model: ul
2 h
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Where the mixing length, l, is defined as the
distance to a wall (in bounded flows)
In other problems l is defined differently
y
Pipe Flow Fluctuations in
velocity carry
extra momentumtoward wall
Hi h
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Higher meanvelocity gradients
at wall Friction higher
for turbulentflows
n
R
r
U
u /1
1
Effect of Reynolds Number As Re increases, exponent of velocity profile
increases larger friction at wall
Movie – Laminar versus turbulent pipevelocity profile
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velocity profile
n
R
r
U
u /1
1
External Flow over Objects
When anexternal flow
encounters anobject, the no-slip condition
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pcauses steepvelocity
gradients at thesurface
The regionover which theobject is felt bythe flow iscalled the
boundary layer
Far from the object the flow remains
uniform and viscous effects can beneglected since there are no appreciable
velocity gradients
External Flow over Objects
When anexternal flow
encounters anobject, the no-slip condition
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pcauses steepvelocity
gradients at thesurface
The regionover which theobject is felt bythe flow iscalled the
boundary layer
Far from the object the flow remains
uniform and viscous effects can beneglected since there are no appreciable
velocity gradients
Boundary Layer
The velocity profile in the
boundary layer goes from zeroat the wall to the free stream
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value at infinity
The thickness of the boundarylayer is defined as, =
y(u=0.99), where the velocity
is 99% of the free stream value
Solution of Boundary Layer
Navier Stokes equations for 2-D B-L:
x g z
w
y
v
x
u
x z
u
y
u
x
u
x
p
z
uw
y
uv
x
uu
t
u m m m m
2
2
2
2
2
2
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Neglecting (1) unsteady effects, (2) gradients
and flow in z-direction, (3) gravity
Plus mass conservation:
y g z
w
y
v
x
u
y z
v
y
v
x
v
y
p
z
vw
y
vv
x
vu
t
v m m m m
2
2
2
2
2
2
2
2
2
2
y
u
x
u
x
p
y
uv
x
uu
m m
2
2
2
2
y
v
x
v
y
p
y
vv
x
vu
m m
0
y
v
x
u
Blasius Assumptions
Blasius (1908) – student of Prandtl
Noted that boundary layers are small (/L<<1)
Therefore v<<u, d/dx<<d/dy
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, y
Thus dp/dy must be small pressure is constant
across a boundary layer
The pressure of the free stream is imposed on thesurface
small small small
small
small
y
v
x
v
y
p
y
vv
x
vu
2
2
2
2
22
m m
Flow Over a Flat Plate
Since the free
stream for flow overa flat plate is
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uniform velocity,
dp/dx=0
2
2
2
2
2
y
u
x
u
x
p
y
uv
x
uu
small zero
m m
2
2
yu
yuv
xuu
m
Non-dimensional BL Equations (1)
Can non-dimensionalize equation choosing:
x*=x/L, u*=u/U, y*=y/, v*=v/V (to be
2
2
y
u
y
uv
x
uu
m 0
y
v
x
u
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x x/L, u u/U, y y/, v v/V (to be
determined)
From mass conservation:
The non-dimensional terms du*/dx* and
dv*/dy* will be of the same order of
magnitude if V=U/L
Thus, v*=vL/U
0*
*
*
*
y
vV
x
u
L
U
0*
*
*
*
y
v
x
u
Non-dimensional BL Equations (2)
Becomes:
2
2
y
u
y
uv
x
uu
m
2*
*2
2*
**2
*
**
2
y
uU yuv
LU
xuu
LU
m
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Boundary layer thickness depends on LRe-1/2
2*
*22
2*
*2
2*
**
*
**
y
u L
UL y
u
U
L
y
uv
x
uu
m
m
2*
*22
1
*
**
*
** Re
y
u L
y
uv
x
uu
Boundary Layer Thickness
Numerical solutions show C=5 for laminar flow
For Re>105 the BL will become turbulent
2/12/1 ~Re LUL
CLCL
m
2/1Re5 L
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Displacement Thickness (1)
Because the velocity slows at the plate, there is less
mass flux through the boundary layer than in theouter flow
Th fl t b di l d b di t *
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The flow appears to be displaced by a distance *
0
* udyU U
Displacement Thickness (2)
If the shape of the boundary layer profile is
known, the displacement thickness can befound
* 1 dyU
u
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From numerical solutions
Displacement thickness is equivalent to
boundary layer thickness but is not arbitrarily
defined (99%U)
0
U
34.0Re721.1 2/1* L
Drag on Plate
Shear stress at the plate can also be calculated
from velocity profile (du/dy at wall)
From numerical solutions:
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Friction coefficient defined as:
LU U LU w
m
m
22/3
332.0332.0
2/1
2
Re664.0
21
U
c w
f
Drag/Friction Coefficient
For pipes
Where pA is the drag in the pipe
D
LU f p 2
2
1
Re
64
2
1 2
f
D
L
U
p
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Where pA is the drag in the pipe
For a flat plate drag is the integral of shear
stress across the plate:
2/1
0 22
Re328.1
2
1
2
1
L
w D
bLU
bdl
bLU
Drag C
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Inviscid Flow Over a Sphere (1) Flow stagnates on
front of
sphere/highest pressure
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Flow is highest
velocity/lowest pressure at top of
cylinder
Flow achieves highest
pressure again at rearstagnation point
Inviscid Flow Over a Sphere (2) What is the net
force on cylinder?
No viscous stress
Pressure
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Pressure
distribution is
symmetricright/left and
top/bottom
Inviscid flow hasno net force
(drag/lift)
Viscid Flow Over Cylinder (1)
Fluid near wall subject to viscous drag
Higher pressure from outer flow imposed
across boundary layer
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Viscid Flow Over Cylinder (2)
Flow from A-C high-medium pressure
(favorable pressure gradient)
Flow from C-F medium to high pressure
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(adverse pressure gradient)
Boundary layer separates
Flow Separation
Separated flows are no
longer thin boundarylayers
Flow recirculates behind
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Flow recirculates behindseparation point, pressure is less thanwould be achieved byinviscid flow
Net drag caused by thisseparated flow
Effect of Re
At low Re, viscosity isimportant everywhere(wide boundary layer)
Flow happens to resembleinviscid flow
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inviscid flow
At moderate Re, the flowis laminar but effects ofviscosity are apparent inseparation
At high Re, the flow isturbulent
Drag Coefficient
The asymmetric pressure distribution caused
by separation leads to drag A drag coefficient is defined as
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CD depends on Re, geometry, surface
roughness, etc – because flow separation
changes with these parameters
AU
Drag C D
2
21
Example Drag Coefficients
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Drag Coefficient
The asymmetric pressure distribution caused
by separation leads to drag A drag coefficient is defined as
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CD depends on Re, geometry, surface
roughness, etc – because flow separation
changes with these parameters
AU
Drag C D
2
21
Example Drag Coefficients
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More Examples
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CD versus Re
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Impact of Aerodynamic Designs
Wind tunnel testing has enabled more
aerodynamic (lower CD auto designs)
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Lift All objects have
drag, but liftrequires anasymmetricgeometry or flow
Flow over an
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airfoil can be
asymmetric eitherthrough the wingshape or by tiltingthe wing relativeto the incoming
flow (angle ofattack)
AU
Lif t C L
2
2
1
Airfoils (1)
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Airfoils (2) Pressure
below airfoil
is higher that pressure aboveairfoil
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airfoil
Edge of winggenerateswing tipvortex
Movie
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Why are Golf Balls Dimpled? (1)
Reynolds number: (~30 m/s, D~0.05 m,
v=1.46e-5 m2/s) Re=105 – almost exactly at transition to
t b l t b d l
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turbulent boundary layer
Dimples on golf balls can help trip boundarylayer and cause turbulence sooner
Why are Golf Balls Dimpled? (2)
Turbulent BL
separates later – causes lower drag
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Two Major Sources of Drag
Separation – this form of drag is due to large
pressure gradients across object (dominantdrag for blunt objects) – lower for turbulent
BL
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BL
Surface friction – this form of drag due to boundary layer velocity gradients (dominant
for streamlined objects without separation) –
higher for turbulent BL
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Fluid Mechanics
Application of mass conservation and
momentum balance to solve for flow (velocityand pressure)
Since many fluid flows are complicated and
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y pcan’t be directly solved, simplified approachesare usually used
ME 250 was a study of various tools/simpleanalysis approaches to understand fluid flow
and its interactions with surfaces
Navier Stokes Equations The most complete description of fluid flow is the
full NSE – mass conservation, momentum balance
at all points in the flow for Newtonian fluids g V V pV V
t
V
m m
2 0
V
t
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x g z
w
y
v
x
u
x z
u
y
u
x
u
x
p
z
uw
y
uv
x
uu
t
u m m m m
2
2
2
2
2
2
y g z
w
y
v
x
u
y z
v
y
v
x
v
y
p
z
v
w y
v
v x
v
ut
v
m m m m
2
2
2
2
2
2
z g z
w
y
v
x
u
z z
w
y
w
x
w
z
p
z
ww
y
wv
x
wu
t
w m m m m
2
2
2
2
2
2
0
z
w
y
v
x
u
t
Appendix A - CFD
For a general complicated flow, simple tools
cannot recover local velocity fields CFD can solve NSE for complex flows
CFD h ld l b d h NEED
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CFD should only be used when you NEED
local velocity information AND the flowcannot be solved by a simpler approach
Chapter 6 – Solve NSE Directly
Start with full NSE
Make reasonable assumptions and eliminateterms
S l i lifi d diff i l i
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Solve simplified differential equations
Apply boundary conditions
Use this approach when you have a simple (1-
D) flow and NEED to know local velocity
profiles
Chapter 3, 6 – Neglect Viscosity
For flows far from walls and at high Re, the
effects of viscosity are negligible Potential and stream function solutions can be
added together (superposition) 02 f
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added together (superposition)
Or velocity field is known (uniform)
Bernoulli’s equation gives pressure from
velocities
Use this approach when you NEED localinformation AND the flow is inviscid
0 f
Chapter 2 – Fluid Statics
If velocity is zero everywhere, NSE reduce to
Pressure field is directly related to gravity (oracceleration of system)
g p
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acceleration of system)
Buoyancy is upward force on submergedobject due to density of fluid
Manometer analysis is common application offluid statics
Use this approach when V=0
Chapter 5 – Integral Analysis (1)
Integral analysis applies to a control volume
for a system Integral analysis provides NO information
about the flow inside the system
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y
Reynolds transport theorem used to deriverelationships for system mass and momentum balance
cscv
sys dAnV bbd t Dt
DBˆ
Chapter 5 – Integral Analysis (2)
Mass conservation
Momentum equation
0ˆ
cscv
dAnV d
t
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Be careful with integration and signs
Use this approach when you need ONLY
information for control volume and you havesufficient information at boundaries
cscv
sysdAnV V d V
t
F ˆ
Dimensional Analysis
Dimensional analysis is useful for finding how
different fluid problems are related to oneanother
Useful for integral and differential analysis
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g y
Useful outside of fluid mechanics Use this approach in general to understand
actual parameters affecting problem
Dimensional analysis does not provide specificinformation only an understanding ofgoverning parameters
Losses, Friction, Drag (1)
Loss factors describe overall system behavior
– like integral analysis this is not useful if you NEED local information
Use loss friction drag factors to design pipe
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Use loss, friction, drag factors to design pipe
systems, understand overall pressure losses Pipe friction – use Moody chart to determine f
f
D LV
p
2
21
Losses Use loss tables to compute K for components
Use drag coefficient tables to compute CD for
K
V
p
2
21
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objects
Use loss coefficients when you need ONLY
overall pressure drop for a system or dragforce
AU
F C Drag
D2
2
1
Other Things to Know..
What IS viscosity? What effects does it have
on flow? How does pressure affect flow?
H d fl id d f d t l it ?
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How do fluids deform due to velocity?
How do pitot static tubes work?
What is the vena contract effect?
Turbulent Flow At high Re, all flows become turbulent
Eddies (vortices) carry momentum in alldirections
Faster mixing of momentum, flatter velocityfil
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profiles
Turbulent viscosity used as a model toapproximate effects of eddy mixing
You do not have sufficient background in
turbulent flows yet to calculate local velocityfields (but can still use loss factor charts)
Exam Coverage (1)
Fluid Kinematics (10%)
Acceleration, dilitation, angular deformation,vorticity
Fluid Statics (10%)
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Fluid Statics (10%)
Manometers, pressure distribution on submergedsurfaces
Inviscid Flows (10%)
Bernoulli equation, stream functions
Exam Coverage (2)
Integral Analysis (25%)
Mass and momentum balance
Differential Anlaysis (25%)