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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 Conductfill out and returnNews articleMars 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,7thedition, 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 websitevideos, 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 setall must be turned infor full credit

No late homework accepteddue 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 credit foranswering + 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 1Tu March 4

Exam 2Tu 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

2ndfloor 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 forceat 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

331

f t

lbm

m

kg

vV

m

][)(22

atmpsiinlbfPaor

mN

AFp

RT

pRTp

const

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Properties of Fluids (2)

Specific weight

Specific gravity

33 ft

lbf

m

Ngg

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

ViscosityMovie

Viscosity of the oil is 104larger 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

tt

g

0lim

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Viscosity (3)Newtonian Fluids A velocity gradient is a rate

of shearing strain

As the stress increases (bypulling 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 fluidviscosity is larger for

high shearing strain In this course we will

deal with mostlyNewtonian 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 viscositybe careful with

unit conversions)

][ 1 sdy

dug

22

in

lbfor

m

N

22in

slbfor

m

sNm

s

ftor

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

Evis the Bulk Modulus with [N/m2]

Change in pressure required for a givenpercentage change in density or volume

ddp

ddpEv

/

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=2x109N/m2

CRTp

Cddp

pCd

CdEv

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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 RpAppRL ambdrop

Rp

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 tensionthe 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 RhRgmg

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 isVapor 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 Newtons 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

sxpzxp

maF

sy

yy

sin0

0

zs sin

sy

sy

pp

zxpzxp

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

yy

y

ay

p

zyxazyxy

p

zyxama

zxyyppzxy

yppF

22

xax

p

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Variation of pressure in a fluid (2) In z-direction, gravity also acts

on fluid element

gaz

p

zyxgazyxz

p

mamgyx

z

z

p

pyx

z

z

p

pF

z

z

z

22

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Variation of pressure in a fluid (3)

gaz

pa

y

pa

x

pzyx

,,

kzpj

ypi

xpp

kajaiaa zyx

gap

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Fluids at Rest (1)

If a=0

For an

incompressible fluid(density=constant)

g

gzp

y

p

x

p

gp

0

hp

zzgpp

gdzdp

BABA

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

dzRT

g

p

dp

RT

pgg

dz

dp

RTp

)0(

exp

ln

pCwhere

RT

gzCp

C

RT

gzp

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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 php g

0105.1000023.0 6, atmpsip Hgvap

Hgmmatm 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 h1is

negligible (gA

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Gage Pressure

Manometers do not determine absolutepressure (since patmis 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 apressure it is assumed to be a gage pressure

gatmA ppph g

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Manometers (2)

Inclined manometers increase the sensitivity of

the measurement

g sin2221 lpp

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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

ghApAF

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Inclined Surfaces (1)

For an inclined surface, the pressure varies

across the surface

Defining

pdAF

hdApdAFR g

sinyh

ydAdAyFR gg sinsin

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Area Centroid

Integral is areas Centroid

Thus, inclined surfaces can be treated as flat if

area centroid location is known The resulting force acts through another point,

yR

AyydA c

AhAyF ccR gg sin

Inclined Surfaces (2)

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Inclined Surfaces (2)

For moment of

resulting forceto act like actual

distributed force

dAypydAyF RR g sin2

g sincR AyF

Ay

I

Ay

dAyy

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/hAFR g

hhhhbh

bdyyAydAyy

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 higherpressure at greater depth

Force balance on only fluid in

ABCD

Buoyancy force only depends on

volume (Movie)

AB

AhF11

g CD

AhF22

g

g

ggg

B

objB

fluidB

F

AhhAhFAh

WFFF

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

gap

gazz

pp

rr

r

pp

1

rra 2

2rr

p

g

z

p

dzdrrdp g 2

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Rigid body rotation (2)

Along an isobar dp=0 dzdrrdp g 2

g

r

dr

dz 2

cg

rz

2

22

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Fl Vi li i (1)

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Flow Visualization (1)

Dye injectionin liquid canbe used tosee flow dye followsthe flow

Photographsshow streaksrepresenting

path of fluidparticles

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

Moviedifferences 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 flowspathlines, 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

amF

nsVV 0

dt

ds

s

V

dt

dn

n

V

sanadt

Vda sn

2Van V

s

Vas

F Al St li (1)

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F=ma Along Streamline (1)

Note: we are applying F=ma to a fluid particlethe 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

VVamF ss

sinsin,

gmgFg

s

F Al St li (2)

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F=ma Along Streamline (2)

Force due to pressure:

2

s

s

pppp s

2

s

s

pppp s

2

s

s

pps

s

pynp

ynppynppF

s

ssps

2

)()(,

F Al St li (3)

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F=ma Along Streamline (3)

Net Force = ma

Note: we have not included shear forceswe 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 fluidparticle acceleration (RHS)

s

VV

s

pgF sin

s

VV

s

pg

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

VV

s

pg

sin

s

z

sin

0

szg

sVV

sp

dss

pdn

n

pds

s

pdp

s

V

s

VV

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

dzg

ds

dV

ds

dp

02

1 2 gdzdVdp

.2

1 2 ConstgdzdVdp

.2

1 2 ConstgzVdp

.2

1 2 ConstgzVp

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 ConstgzVp

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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 (higherpressure)

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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, gmgF gn

2V

amF nn

n

p

ysp

ysppysppF

n

nnpn

2

)()(,

2

n

n

ppn

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

dss

p

dp

2

cos V

n

pg

02

nzgV

np

02

gdzdnV

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

Constgzdn

V

p

0

2

gdzdnV

dp

.2

ConstgdzdnVdp

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

ConstgzdnV

p

.21 2 ConstgzVp

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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 pressurepressure is only converted from one

type to another (Static/Dynamic/Hydrostatic)

TpConstgzVp .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)

Hzg

Vp

2

2

g

.2

2

ConstgzV

pv

Example Stagnation Streamline

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ExampleStagnation Streamline

Free stream: high velocity (high dynamic pressure),low static pressure

Stagnation point: zero dynamic pressure, high staticpressure

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

Vpp

VpConstVp

stag

stagstag

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 concentrictubesone with a

forward facing tap

and the other with aside tap

43

4

2

3

2

2

1

ppV

pp

Vpp

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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 Bernoullis (see Sect. 3.8)

2

2

22

1

2

1

2

112

1

2

1gzVpds

t

VgzVp

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

gzVpgzVp

)(2 212 zzgV

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

90turn would require

infinite pressure gradient At a-a, pressure is

atmospheric across jet(straight streamlines)

p2>p1to cause streamline

curvature Contraction coefficient =

Aj/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 VAQ VAmassflow

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 lowpressure

If static pressuredrops below vapor

pressure of liquidboiling occurs

This cavitationleads 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

1VpVp

2211 AVAVQ

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

ppAQ

AA

AQ

AQ

AQpp

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Sluice Gates

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2

2

221

2

11

2

1

2

1gzVpgzVp 2211 AVAVQ

2

221

2

12

1)(

2

1VzzgV

2

12

21

2

2

1

222

2

2

21

2

1

2

2

21

1

)(2

1)(2

2

1)(

zz

zzgbzQ

z

zQzbzzg

bz

Q

bz

Qzzg

12 2gzbzQ 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 smallregionContinuum 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

ktzyxwjtzyxvitzyxuV ),,,(),,,(),,,(

kzyxwjzyxvizyxuV ),,(),,(),,(

jyxviyxuV ),(),(

ixuV )(

jtyxvityxuV ),,(),,(

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LagrangianParticle Velocity

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Lagrangian description offlow follows individualparticles

The position of a fluid particleis

The velocity of that fluidparticle is

This does not tell us what thevelocity will be at a pointaway from the fluid particle

ktzjtyitxr )()()(

dt

rdV

Acceleration - Eulerian

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In Eulerian representation of velocity

Since x, y, z do not depend on tEulerian =

fixed coordinates

dt

Vd

a

ktzyxwjtzyxvitzyxuV ),,,(),,,(),,,(

t

Vk

t

wj

t

vi

t

u

dt

Vd

AccelerationLagrangian (1)

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In Lagrangian representation of velocity, x, y,

and z depend on t as particle follows flow

dt

Vd

a

kttztytxwjttztytxvittztytxuV )),(),(),(()),(),(),(()),(),(),((

kdt

dz

z

w

dt

dy

y

w

dt

dx

x

w

t

w

jdtdz

zv

dtdy

yv

dtdx

xv

tv

idt

dz

z

u

dt

dy

y

u

dt

dx

x

u

t

u

dt

Vd

u

dt

dx

vdtdy

wdt

dz

AccelerationLagrangian (2)

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Lagrangian acceleration includes local acceleration

and change in velocity due to fluid particle motion

kzww

ywv

xwu

jz

vw

y

vv

x

vu

iz

uw

y

uv

x

uu

t

V

dt

Vd

VV

t

V

dt

Vd

kwjviuV

k

z

j

y

i

x

Material Derivative

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Material or Substantial derivative includeslocal and spatial variation in a quantity

VVt

V

Dt

VD

termconvective

termunsteady

VtDt

D

z

Tw

y

Tv

x

Tu

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

Tw

y

Tv

x

Tu

z

Tw

y

Tv

x

Tu

t

T

Dt

DT

0Dt

DT0

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)

systemsystem

sys bdbdmB cvvolumecontrol

cv bdbdmB

dt

bdd

dt

dB syssys

dt

bdd

dtdB cvcv

dt

dB

dt

dB syscv

Example: Mass, b=1 (B=m)

0dt

dmsys 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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III

cvsys

BBt

B

Dt

DB

22221111 bVAbVA

t

B

Dt

DBcvsys

AVbbmB

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

csout dAnVbbmB

cv

cv bdtt

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

cscvsys dAnVbbd

tDt

DB

bVt

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

sysdAnVbbd

tDt

DB

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3:57 (19thOlympics) 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=mb=B/m=1

cscv

sysdAnVbbd

tDt

DB

cscv

sysdAnVd

tDt

Dm

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

0

cscv

dAnVdt

dDt

D

Dt

Dmsys

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

0

cscv dAnVdt

0 cs

dAnV

inoutnetcs

mmmdAnV

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

AVm

A

dAnV

V cs

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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

0

cscv

dAnWdt

cvVVW

Deforming Control Volumes

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Deforming control volumes are both movingand unsteady

Local relative velocities used at all surfaces

0

cscv

dAnWdt

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Newtons 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

DtVDmamF syssyssys

Dt

VmDF

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

sysdAnVVdV

tF

cscv

sysdAnVbbd

tDt

DB

Dt

VmDF

sys

sys

Integral Linear Momentum

Equations

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cscv

sysdAnVVdV

tF

cscv

z

cscv

y

cscv

x

dAnVwwdtF

dAnVvvdt

F

dAnVuud

t

F

zyx nwnvnunV

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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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cscvsys

dAnWVdV

t

F

cvVVW

cs

cv

steady

cv

cvsysdAnWVWdVW

tF

)(0

cs

cv

cs

sysdAnWVdAnWWF

)(0

onconservatimass

cs

cv

cssys

dAnWVdAnWWF

cs

sysdAnWWF

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

dVD

Dt

maDF

)(

dVDt

rD

Dt

dVrD

Dt

dVDrFr

)0(0

VV

dVVDt

dVrDFr

Rotating Systems (2)

I t ti t l l

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Integrating over a control volume

Applying Reynolds Transport Theorem(b=rxV)

Moment of momentum equation

syssyssys

dVrDt

D

Dt

dVrDFr

sys

dVrDt

DFr

cscvsys

sysdAnVbbd

tbd

Dt

D

Dt

DB

cscv

dAnVVrdVrt

Fr

Moment of momentum equation

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For a steady system

Sign of rxV and T terms determined by right-

hand-rule

cs fluxmomentumangular

momentumangularunsteady

cvsystemontorquesofSum

dAnVVrdVr

t

T

cs dAnVVrT

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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

innetinnet

sysWQ

Dt

DE,,

syssys edE

gzV

upekeue 2

2

Energy Equation (2)

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Applying Reynolds Transport Theorem withB=E, b=e

cscv

innetinnet

sysdAnVeed

tWQ

DtDE

,,

gzV

upekeue 2

2

cscv

sysdAnVbbd

tDt

DB

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 inndirection

Rate change of distance is velocity

cs

shaftinnet sFWW

,

dAnpVWWcs

shaftinnet

)(,

inshaftinnetcscv

WQdAnVpeedt

,,

Energy Equation (3)

VV 22

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For steady, uniform flow

inshaftinnetcscv WQdAnVgz

Vp

udgz

V

ut ,,

22

22

hpvup

u

inshaftinnet

cscv

WQdAnVgzV

hdgzV

ut

,,

22

22

inshaftinnetinout WQgzV

hAVgzV

hAV ,,

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

inshaftinnetinout

inout

inout

WQgzgzVV

hhm,,

22

22

inshaftinnetinout WQgzV

hmgzV

hm ,,22

qgzgz

VV

pvpvuu inoutinout

inoutinout 22

22

lossBernoulliquugzgzVV

pvpv inoutinoutinout

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)

inshaftin

in

in

in

out

out

out

out wlossgzVp

gzVp

,22

slininin

outoutout hhz

g

V

g

pz

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

Movierotation/translation/angulardeformation

Translation

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Translationfluid elements translate at localfluid velocity zwyvxuV

tux

tvy

twz

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:

zdtyxx

ud

zyx

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 aand as rotation of x and yaxis

aa

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 aand

If OA=-OBthen the fluid element will only rotate andnot deform

If OA=+OBthen the fluid element will only deformand not rotate

x

v

ttOA

a

0lim

yu

ttOB

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

vOBOAz

2

1

2

zv

yw

x21

xw

zu

y21

Rotation and Vorticity

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Rotation rate is a vector:

Vorticity is defined as twice the rotation rate - Movie

y

u

x

vz

2

1

z

v

y

wx

2

1

x

w

z

uy

2

1

zyx zyx

wvu

zyx

zyx

V

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

vOBOAz

22

g

z

v

y

wxg

x

w

z

uyg

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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

0

cscv

sysdAnVd

tDt

Dm

Mass Conservation (2)

0

sysdAnVd

Dm

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As volume0, all flows become uniform and

volume integral becomes homogeneous

cscv

tDt

0

22

22

22

xyz

z

wwxy

z

z

ww

zxy

y

vvzx

y

y

vv

zyx

x

uuzy

x

x

uuzyx

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

zyxt

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

rr

vr

rt

zr

zvvrvV zr

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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 ycan be found it automaticallysatisfies mass conservation

0

V

t

y

v

x

uV

0

uy

yv

x

y0

xyyxy

v

x

u yy

Stream Function (2)

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What is the stream function?

For constant y

Line of constant stream function is a

streamline

udyvdxdyy

dxx

d

yyy

0yduv

dxdy

Stream Function (3)

Since flow cant cross a streamline, flow between two

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differentially close streamlines: Difference in stream

function is flow rate

between streamlines

yyy

ddxx

dyy

vdxudydq

12 yyy dvdxudydqQ

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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

gpVVt

V

What is the Potential Function?

vdyudxdydxd

ff

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)

yx 0fd

v

u

dx

dy

0yduv

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

MovieExample: 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:

xUu

facos

yUv

fasin

)(cos

)(sin

yfUx

xfUy

af

af

aaf cossin xyU

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

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Half-body can be approximated as uniformflow + source (Movie)

Stagnation point occurs at r=b (angle

f cosUrUx

rm

ln2

f

f

f

f

sin1

2cos

ln2

cos

Ur

v

rmU

rv

rm

Ur

r

U

mb

b

mUvr

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:

yyy2

sin m

Ursourceuniform

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

rv

f

1

Circulation

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Define the circulation for a closed curve in anyflow as:

For vortex: sdVccw

rdvsdV

Kdrr

K 2

f2

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 rrmrmrm

f

2/1222 cos2 ararrrsource

2/1221 cos2 ararrrsink

f

cos2

cos41ln

4 22

arar

arm

r

am

arar

arm

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

rK

rma

f coscos

fcosKr

Flow over a Circular Cylinder (1)

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Lets 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

KU

r

KUr

UaK 2

f cos12

2

Urr

a

Flow over a Circular Cylinder (2)

f cos12

2

Ura

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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

Ur

a

rv

Pressure Distribution over Cylinder

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Since potential flow analysis assumes inviscidincompressible flow, Bernoullis equation can

be used

Pressure on cylinder surface (r=a):22

2

1

2

1 vpUp surface

)sin41(2

1sin4

2

1

2

1 22222 UpUUppsurface

sin2sin1 2

2

UUr

av

Pressure Distribution over Cylinder

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Comparison ofinviscid and

experimental

pressure

distributions:

)sin41(2

1 22 Uppsurface

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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

sysdAnVVdV

tF

Momentum Equation (2)

In x- directionzyxg

yzxF x

yxzxxx

x

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Momentum Equation (3)

In x- direction

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VAu

t

udAnVVdV

t cscv

2222

2222

2222

z

z

uuyx

z

z

ww

z

z

uuyx

z

z

ww

y

y

uuzxy

y

vvy

y

uuzxy

y

vv

x

x

uuzy

x

x

uu

x

x

uuzy

x

x

uuzyx

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

uz

w

z

wz

z

u

uz

w

z

w

y

y

uu

y

v

y

vy

y

uu

y

v

y

v

xxuu

xu

xux

xuu

xu

xu

tug

yzx x

yxzxxx

u

z

w

z

uwu

y

v

y

uvu

x

u

x

uu

tu

t

ug

yzx x

yxzxxx

0

u

z

wu

y

vu

x

uu

t

Momentum Equation (5)

z

uw

y

uv

x

uu

t

ug

yzx x

yxzxxx

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The normal stress is due to both viscous stress() and pressure

z

uw

y

uv

x

uu

t

ug

yzxx

px

yxzxxx

z

vw

y

vv

x

vu

t

vg

xzyy

py

xyzyyy

xxxx p

zww

ywv

xwu

twg

xyzzp

zxzyzzz

gpVVt

V

Momentum Equation gpVV

t

V

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z

uw

y

uv

x

uu

t

ug

yzxx

px

yxzxxx

z

vw

y

vv

x

vu

t

vg

xzyy

py

xyzyyy

zww

ywv

xwu

twg

xyzzp z

xzyzzz

Inviscid Flow

gpVVt

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:

gpVVt

V

1

gpVV

1

pVV

1

Irrotational Flows (1)

Define the velocity potential

ff f f

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Vorticity is:

Only irrotational flows have a velocity potential

vy

fu

x

f w

z

f fV

0

k

xyyx

j

zxxz

i

yzzy

ky

u

x

vj

x

w

z

ui

z

v

y

w

ffffff

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

zyxV

fffff

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Full Momentum Equation

gpVVt

V

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Need model for viscous forces () dydu

m

Newtonian Fluids (1)

id

dum

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isjust part of the

full viscous

stress The complete

model:

dy

yxxyx

v

y

um

zyyz y

w

z

vm

zxxzz

u

x

wm

x

uxx

m 2

y

vyy

m 2

z

wzz

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

zxxyx

2

2m

VV

Vz

w

Vy

v

V

x

u

z

w

y

w

yz

v

xz

u

x

w

zy

w

z

v

y

v

x

v

xy

u

z

u

zx

w

yx

v

y

u

x

u

mmmm 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

gpVVt

V

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Substituting viscous stress tensor yields NSE

gVVpVVt

V

mm

2

xgz

w

y

v

x

u

xz

u

y

u

x

u

x

p

z

uw

y

uv

x

uu

t

ummmm

2

2

2

2

2

2

ygz

w

y

v

x

u

yz

v

y

v

x

v

y

p

z

vw

y

vv

x

vu

t

vmmmm

2

2

2

2

2

2

zgz

w

y

v

x

u

zz

w

y

w

x

w

z

p

z

ww

y

wv

x

wu

t

wmmmm

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)( zxutu

2

2

0y

u

x

p

m g

y

p

0 00

Flow between plates (2)

2

0 up

gy

p

0 )(xfgyp

uxgyp )(1

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Flow is parabolic

Flow rate per unit width

20

yu

xp

m

yuxgy

xp

)(1

m

)(21 2 xhy

xpu

m

2

21)(0)( h

xpxhhu

m

0)(0)0(

xg

y

u

222

1hy

x

pu

m

x

phh

h

x

pdyhy

x

pudyq

hh

mmm 332

1

2

12

33

3

0

22

0

x

phu

m2

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

gVVpVVt

V

mm

2

xgz

w

y

v

x

u

xz

u

y

u

x

u

x

p

z

uw

y

uv

x

uu

t

ummmm

2

2

2

2

2

2

ygz

w

y

v

x

u

yz

v

y

v

x

v

y

p

z

vw

y

vv

x

vu

t

vmmmm

2

2

2

2

2

2

zgz

w

y

v

x

u

zz

w

y

w

x

w

z

p

z

ww

y

wv

x

wu

t

wmmmm

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

mgy

p

0

Ubuu )(,0)0(

)(xfgyp

)()(2

1 2 xhyxgyx

pu

m

Couette Flow (2)

Ubuu )(0)0(

)()(2

1 2 xhyxgyx

pu

m

b

Uybyyx

pu 2

2

1

m

b

y

b

yp

U

b

b

y

U

u

12

2

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Ubuu )(,0)0(

0)( xh

)(2 xgx

pb

b

U

m

bbxUbU 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

rrz

p

z

vv zzzz

mm

r

vr

rrz

p z1m

Pipe Flow (2)

Boundary conditions:

r

vr

rrz

p z1m

0)0(

r

vz 0)( Rvz

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r

vrzfr

z

p z)(2

1 2

m0)( zf

)(4

1 2 zgrz

pvz

m

2

4

1)( R

z

pzg

m 22

4

1Rr

z

pvz

m

Pipe Flow (3)

224

1Rr

z

pvz

m

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MovieParabolic velocity profiles in pipe

flow

z

pRv

m4

2

m ax

z

pRRR

z

prdrRr

z

pQ

R

m

m

m 82422

4

1 444

0

22

28

max2 v

z

pR

A

QV

m

http://localhost/var/www/apps/conversion/tmp/scratch_7/ME250-Fluids/Lectures/V6_6.movhttp://localhost/var/www/apps/conversion/tmp/scratch_7/ME250-Fluids/Lectures/V6_6.mov8/21/2019 ME 3250 - Lectures

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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

gy

p

0

00

0)()(

zx

0 wv

Cgyp

Unsteady Flow (2)

Similarity solution: 22

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

tff

tt

yf

4

1)0(

4

t

yfu

hh

2:)(

ff h2

21 )( CerfCf 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

gzyxxzyxxzwyvxut mmmm

ygz

w

y

v

x

u

yz

v

y

v

x

v

y

p

z

vw

y

vv

x

vu

t

vmmmm

2

2

2

2

2

2

zgz

w

y

v

x

u

zz

w

y

w

x

w

z

p

z

w

wy

w

vx

w

ut

w

mmmm

2

2

2

2

2

2

y

v

x

u

xy

u

x

u

x

p

y

uv

x

uu mmm

2

2

2

2

y

v

x

u

yy

v

x

v

y

p

y

vv

x

vu mmm

2

2

2

2

00

0

y

v

x

u

Governing Equations (2)

2

2

2

2

1 uupuvuu

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 yxxp

yx

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)

xiand yiare discretespacing x and y

NxN 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

ux

x

ux

x

uuu

jijiji

jiji

3322 xuxuxuuu

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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

ux

x

uuu

jiji

jiji

!3!2!1

,

3

,

2

,

,,1xxx

uu

jijiji

jiji

62

2

,

3

3

,1,1

,

x

x

u

x

uu

x

u

ji

jiji

ji

x

uu

x

u jiji

ji

2

,1,1

,

Second Derivatives

!3!2!1

3

,

3

32

,

2

2

,

,,1

x

x

ux

x

ux

x

uuu

jijiji

jiji

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With an error of order (x)2

!42

!222

4

,4

42

,2

2

,,1,1

x

x

ux

x

uuuu

jiji

jijiji

!3!2!1

3

,

3

32

,

2

2

,

,,1

x

x

ux

x

ux

x

uuu

jijiji

jiji

12

2 2

,

4

4

2

,1,,1

,

2

2 x

x

u

x

uuu

x

u

ji

jijiji

ji

2,1,,1

,

2

2 2

x

uuu

x

u jijiji

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 jiji

ji

,,1

, 2

,1,,1

,

2

2 2

x

uuu

x

u jijiji

ji

21,,1,

2,1,,1,1,11,1,

,,1,1

, 222

122 y

uuux

uuuxpp

yuuv

xuuu jijijijijijijijijijijijijiji

01,1,,1,1

y

vv

x

uujijijiji

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

uujijijiji

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,-1doesnt 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 OutputInfinite slot (1)

Solution is only available at grid points

interpolation used for values between grid points

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CFD OutputInfinite slot (2)

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CFD OutputVortex 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

MovieCFDcomputed streamlines over

bluff-body

Errors in CFD

Numerical errorthe iterative solution did not

find the correct answer to the algebraici

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find the correct answer to the algebraicequations

Discretization errorthe derivative estimates

were not accurate enough (x too large) Governing equation errora term was

removed that shouldnt have been (gravity?)

Boundary condition errorthe boundaryconditions do not reflect reality

CFD Validation/Accuracy (1) Numerical error can be

assessed by examiningresiduals

Residualy

vv

x

uu jijijiji

1,1,,1,1

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Residuals would ideallybe 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 rruurdrudAndAVm 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 accuracyEven with

commercial CFD codes this step MUST be done by

user CFD always gives a pretty answeryou 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

),,,,( LDVfp 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

),,,,( LDVfp 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) zyx

mDs

mV

m

sN

m

Np

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

zyx

mDs

mV

m

sN

m

sN