ME 3250 - Lectures

8/21/2019 ME 3250 - Lectures

http://slidepdf.com/reader/full/me-3250-lectures 1/409

ME 3250Fluid Dynamics I

Spring 2014

Prof. Mike Renfro

AUST 110

TuTh 2:00-3:15 PM

8/21/2019 ME 3250 - Lectures


Handouts

Course Policy

Syllabus

Academic Conduct – fill out and return News article – Mars Orbiter

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures



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

8/21/2019 ME 3250 - Lectures


Attendance

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures



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

8/21/2019 ME 3250 - Lectures



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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures



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

8/21/2019 ME 3250 - Lectures



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

8/21/2019 ME 3250 - Lectures


Applications of Fluid Mechanics (1)

Study the behavior of fluids at rest

Fluid statics analysis (Chap. 2)

8/21/2019 ME 3250 - Lectures



Study the forces that fluids impart on systems

Study the global behavior of fluid in a system

Integral (control volume) analysis (Chap. 5)

8/21/2019 ME 3250 - Lectures



Study the local behavior of fluid flow

Differential analysis (Chap. 6)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Properties of Fluids (2)

Specific weight

Specific gravity

33 ft

lbf

m

N g g

water

SG

8/21/2019 ME 3250 - Lectures


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

http://localhost/var/www/apps/conversion/tmp/scratch_7/V1_1.mov







8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


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





8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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)



8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


30 minutes

Pre-req Quiz

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures



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)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Surface Tension (3)

Bending of surface around an object provides

tension that must be “broken”

Surface acts like membrane

Movie



8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures



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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Standard Atmosphere

Considerable variations in atmospheric

properties with height

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Manometers (2)

Inclined manometers increase the sensitivity of

the measurement

g sin2221 l p p

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Bourdon Pressure Gages

Metal tube straightens as pressure increases movingneedle

Set to zero for atmospheric pressure Gage pressuremeasurement

Most common pressure sensor

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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


8/21/2019 ME 3250 - Lectures



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

8/21/2019 ME 3250 - Lectures


Tabulated Moments of Inertia

See p. 59-60for

transformations

to use tabulated

Ixy data

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Buoyancy (1)

8/21/2019 ME 3250 - Lectures


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



8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Stability

If center of mass is below centroid, object will be stable

If center of mass is above centroid, object will be unstable

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Fl Vi li i (1)

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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



8/21/2019 ME 3250 - Lectures


Streamlines

Streamlines = lines in the flow that are locally

tangent to the velocity of the fluid

St li (2)

8/21/2019 ME 3250 - Lectures


Streamline (2)

Streamlinesdetermined bymeasuringinstantaneousvelocity and

integrating tofind tangent lines

Harder tomeasure than

streaklines Most useful to

mathematicallydescribe flow

St li (3)

8/21/2019 ME 3250 - Lectures


Streamlines (3)

For steady flows – pathlines, streaklines, andstreamlines are identical - Movie

I i id Fl (1)



8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures



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)

8/21/2019 ME 3250 - Lectures



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)

8/21/2019 ME 3250 - Lectures



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)

8/21/2019 ME 3250 - Lectures



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

8/21/2019 ME 3250 - Lectures


Bernoulli Equation

Can only be applied to

Steady flow

Inviscid flow

Incompressible flow

Flow along a streamline

.21 2 Const gz V p

8/21/2019 ME 3250 - Lectures


Stagnation Flow

8/21/2019 ME 3250 - Lectures


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)



8/21/2019 ME 3250 - Lectures


Pressure Along Dividing Streamline

8/21/2019 ME 3250 - Lectures


Pressure Along Dividing Streamline

F=ma Normal to Streamline (1)

8/21/2019 ME 3250 - Lectures



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


8/21/2019 ME 3250 - Lectures



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


8/21/2019 ME 3250 - Lectures



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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Movie

Bernoulli Equation –



8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Why does the pressure drop between 2 and 1?

Hydraulic grade line

8/21/2019 ME 3250 - Lectures


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?

8/21/2019 ME 3250 - Lectures


“Modified” Bernoulli Equation

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Application of Bernoulli Equation

Free Jets

Confined Flows

Venturi and orifice flowmeters

Sluice gates

Free Jets

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Vena Contracta (1)

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


Vena Contracta (2)

8/21/2019 ME 3250 - Lectures


Confined Flows

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Repeat for static taps on

both ends of the manometer

Cavitation

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Orifice and Venturi Meters

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Sluice Gates

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


***ADD**

8/21/2019 ME 3250 - Lectures


Add physical interpretation of acrossstreamline equation and a couple of examples

E.g. Pressure in a vortex

Chapter 4 - Fluid Kinematics

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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),,(







8/21/2019 ME 3250 - Lectures


Lagrangian – Particle Velocity

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Control Volumes / Systems

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


Consider a control mass that moves from (1) to (2) int

Define control volume as common region plus regionI


8/21/2019 ME 3250 - Lectures


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


8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


3:57 (19th Olympics) versus 4:02 (20th

Olympics) speed skating (3000 m)

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Moving Control Volumes

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Deforming control volumes are both movingand unsteady

Local relative velocities used at all surfaces

0ˆ

cscv

dAnW d t

8/21/2019 ME 3250 - Lectures


Newton’s Second Law

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Moving Reference Frame

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Rotating Systems (1)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Energy Equation (1)

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Why Differential Analysis?

I l l i ll

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Differential equations for fluid flow can bederived by considering the motions and forces

of small fluid elements

Movie – rotation/translation/angulardeformation

Translation



8/21/2019 ME 3250 - Lectures


Translation – fluid elements translate at localfluid velocity z w yv xuV ˆˆˆ

t u x

t v y

t w z

Linear Deformation

zyx

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


Define angles a and as rotation of x and yaxis

a a

t

x

vtan

t

y

utan


8/21/2019 ME 3250 - Lectures


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


8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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



8/21/2019 ME 3250 - Lectures


Angular Deformation

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Mass Conservation (1)

8/21/2019 ME 3250 - Lectures


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


0ˆ

sysdAnV d

Dm

8/21/2019 ME 3250 - Lectures


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


0

wvu

t

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Stream Function (1)

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Potential Flow

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


(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

http://localhost/var/www/apps/conversion/tmp/scratch_7/ME250-Fluids/Lectures/V6_3.mov


8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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)



8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Free vortex:

Free vortex is irrotational

Other vortices can be rotational No potential function

f K

r

K

r v

f

1

Circulation

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Streamlines:

Why is this useful?

Next slide…

r K

r ma

f coscos

f cos K r

Flow over a Circular Cylinder (1)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Comparison ofinviscid and

experimental

pressure

distributions:

)sin41(2

1 22 U p p surface

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Momentum Equation (1)

Integral momentum equation for a control

8/21/2019 ME 3250 - Lectures


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


In x- direction z y x g

y z x F x

yx zx xx

x

8/21/2019 ME 3250 - Lectures



In x- direction

8/21/2019 ME 3250 - Lectures


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


Dividing by volume and equating to force:

8/21/2019 ME 3250 - Lectures


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


z

uw

y

uv

x

uu

t

u g

y z x x

yx zx xx

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Full Momentum Equation

g pV V t

V

8/21/2019 ME 3250 - Lectures


Need model for viscous forces () dy

dum

Newtonian Fluids (1)

id

dum

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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)(

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Start with full NSE Make reasonable assumptions and eliminate

terms

Solve simplified differential equations Apply boundary conditions

8/21/2019 ME 3250 - Lectures


Navier-Stokes Equations

R i

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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


8/21/2019 ME 3250 - Lectures


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



8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Unsteady Flow (1)

Infinite plate, infinite fluid (in z and x), fluid

i i i ll 1 D fl )()(

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

( ( ) ( )

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Numerical Methods Finite Difference

Differential form of governing equations are discretized

and solved

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


CFD Output – Infinite slot (2)

8/21/2019 ME 3250 - Lectures


CFD Output – Vortex Generation

Many CFD simulations are unsteady to capture

transient features of flow

8/21/2019 ME 3250 - Lectures


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



8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


( 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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Reynolds Number

Re is the most important parameter in fluid

mechanics – it defines the relative importance

8/21/2019 ME 3250 - Lectures


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



8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Problems with 2 Pi Groups

The pressure drop in a fixed-length pipe

k=5, r=3 ),,,( DV f p m

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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?

8/21/2019 ME 3250 - Lectures


Sometimes this is really how we design

things…

Tacoma Narrows Bridge

Bridge collapse in WA (1940) due to wind-

induced oscillations

8/21/2019 ME 3250 - Lectures


Movie

Modeling

http://localhost/var/www/apps/conversion/tmp/scratch_7/tacnarr.mpeg

http://localhost/var/www/apps/conversion/tmp/scratch_7/tacnarr.mpeg

http://www.wsdot.wa.gov/projects/sr16narrowsbridge/

8/21/2019 ME 3250 - Lectures


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

http://www.wsdot.wa.gov/projects/sr16narrowsbridge/

8/21/2019 ME 3250 - Lectures


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


If we build a model that has the same value of

all dependent PI groups, then the dependent Pi

d ill b

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures



In practice it may not be possible to have all

dependent Pi groups match reality – must use

i i ff

8/21/2019 ME 3250 - Lectures


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



8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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



8/21/2019 ME 3250 - Lectures


Turbulent flow – fluctuations around 25%

Transition to Turbulent Flow

Movie – Turbulence in pipe flow

Original experiment of Osborne Reynolds



8/21/2019 ME 3250 - Lectures


Importance of Reynolds Number

Experiments of Reynolds showed

Laminar flow Re<2100

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Fully developed profile is solution we found

previously (no gradients in x)

Entrance Length

Correlations for entrance length:

Laminar flow:Re060

l e

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


gradients at wall

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Effect of Wall Roughness

Wall roughness

also increasesfriction factor

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Head Loss

Moody chart used to estimate f for a turbulent

flow

Head loss for pipes in a system added together

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Loss coefficient, K L

2

21 V

p K

L

g

V K h L L

2

2

minor ,

Minor Losses (2)

8/21/2019 ME 3250 - Lectures


Moody Chart

8/21/2019 ME 3250 - Lectures


Minor Losses

Loss coefficients

are tabulated formay conditions

8/21/2019 ME 3250 - Lectures


may conditions

Specified by

manufacturers offlow products

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Turbulent Flow

The average defined by:

)(')( t uut u

T t

udt T

u 1

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


viscosity

0'' vu

Turbulent Viscosity

In turbulent flows shear stress (from nonlinear

udu/dx terms):

y

uvu

y

ut xy

m m m ''

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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



8/21/2019 ME 3250 - Lectures


velocity profile

n

R

r

U

u /1

1

External Flow over Objects

When anexternal flow

encounters anobject, the no-slip condition

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


, 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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


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 *

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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:

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


Inviscid Flow Over a Sphere (1) Flow stagnates on

front of

sphere/highest pressure

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


Viscid Flow Over Cylinder (2)

Flow from A-C high-medium pressure

(favorable pressure gradient)

Flow from C-F medium to high pressure

8/21/2019 ME 3250 - Lectures


(adverse pressure gradient)

Boundary layer separates

Flow Separation

Separated flows are no

longer thin boundarylayers

Flow recirculates behind

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


CD depends on Re, geometry, surface

roughness, etc – because flow separation

changes with these parameters

AU

Drag C D

2

21

Example Drag Coefficients

8/21/2019 ME 3250 - Lectures


Drag Coefficient

The asymmetric pressure distribution caused

by separation leads to drag A drag coefficient is defined as

8/21/2019 ME 3250 - Lectures


CD depends on Re, geometry, surface

roughness, etc – because flow separation

changes with these parameters

AU

Drag C D

2

21

Example Drag Coefficients

8/21/2019 ME 3250 - Lectures


More Examples

8/21/2019 ME 3250 - Lectures


CD versus Re

8/21/2019 ME 3250 - Lectures


Impact of Aerodynamic Designs

Wind tunnel testing has enabled more

aerodynamic (lower CD auto designs)

8/21/2019 ME 3250 - Lectures


Lift All objects have

drag, but liftrequires anasymmetricgeometry or flow

Flow over an

8/21/2019 ME 3250 - Lectures


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)

8/21/2019 ME 3250 - Lectures


Airfoils (2) Pressure

below airfoil

is higher that pressure aboveairfoil

8/21/2019 ME 3250 - Lectures


airfoil

Edge of winggenerateswing tipvortex

Movie



8/21/2019 ME 3250 - Lectures


8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


BL

Surface friction – this form of drag due to boundary layer velocity gradients (dominant

for streamlined objects without separation) –

higher for turbulent BL

8/21/2019 ME 3250 - Lectures


Fluid Mechanics

Application of mass conservation and

momentum balance to solve for flow (velocityand pressure)

Since many fluid flows are complicated and

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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 ?

8/21/2019 ME 3250 - Lectures


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

8/21/2019 ME 3250 - Lectures


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%)

8/21/2019 ME 3250 - Lectures


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%)

ME 3250 - Lectures

Documents