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CHAPTER 1

1

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CHAPTER 2

2

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CHAPTER 3

3

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Bernoullis Equation

An other application of momentum eq is the Bernoullis Eq given as:

Consider x component of Momentum eq

or inviscid flo! !ith no "od# force it reduces to

$

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Bernoullis Equation

%

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&ovem"er 21' 2(11 )

Bernoullis Equation

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*

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+

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,

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1(

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-he equation

or rotational flo!

or irrigational flo!

11

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EXEMPLE 3.1

e"ruar# 1$' 2((% AE 2+(: Chapter 1: .ntroductor#-houghts

12

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&ovem"er 21' 2(11 13

INCOMPREIBLE !LO" IN A #$CT

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&ovem"er 21' 2(11 1$

INCOMPREIBLE !LO" IN A #$CT

For constant density the above eq reduces to

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&ovem"er 21' 2(11 1%

APPLICATION O! INCOMPREIBLE !LO" IN A #$CT

%enturi

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&ovem"er 21' 2(11 1)


%enturi

/ Consider flow entering a CD Nozzle as shown below/ This type of nozzles is called a VENTU!

/ "tation where area is #ini#u# is called throat/ !f ratio of inlet to throat area is $nown and pressure difference

%&'%( is e)peri#entally #easured then any velocity can be

#easured

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0sing Bernoulli eq

1*

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&ovem"er 21' 2(11 1+


&Lo' (ee) 'in) Tunnel*

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&ovem"er 21' 2(11 1,

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&ovem"er 21' 2(11 2(

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0sing manometer the pressure difference is measured as:

21

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&ovem"er 21' 2(11 22

E*+,%-E ./&

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&ovem"er 21' 2(11 23

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&ovem"er 21' 2(11 2$

E)a#ple ./0

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2%

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&ovem"er 21' 2(11 2)


Pitot Tu+e

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&ovem"er 21' 2(11 2*


Pitot Tu+e

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Pitot Tu+e

2+

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&ovem"er 21' 2(11 2,


Pitot Tu+e

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&ovem"er 21' 2(11 3(


Pitot Tu+e

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31

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32

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C(

33

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C( e,a-(le

3$

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C( E,a-(le

3%

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3)

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3*

1hich is -aplace equation2 Thus !2 !C flow is described by -aplace equation

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3+

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3,

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.f .C flo! is . also then

$(

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La(lae Equation

$1

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$2

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E,a-(le o/ a))ition o/ solutions o/ LP#

$3

-%D 3 linear partial differential equation

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Boun)ar0 Con)itions

$$

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$%

At infinity

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$)

At wall ( on surface )

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$*

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$+

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$NI!ROM !LO"

$,

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%(

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%1

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%2

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%3

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%$

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%%

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%)

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%*

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%+

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%,

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)(

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)1

$ i/ !l &1 t El t !l *

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1231% AE 2+(: Chapter 3 4art A5 )2

$ni/or- !lo' &1stEle-entar0 !lo'*

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)3

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)$

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)%

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))

oure !lo' &2n) Ele-entar0 !lo'*

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1231% AE 2+(: Chapter 3 4art A5 )*

oure !lo' &2n)Ele-entar0 !lo'*

By definition, a source flowis one inwhich the streamlines are radial lines

emanating from the origin. Also,(3.59a)

where cis a constant. i.e., thevelocity varies inversely as thedistance from the origin

In a source flow the velocity is direc-ted away from the origin. In a sinkflow, the velocity is directed towardthe origin. In fact, sin flow is anegative source flow

A source is really a line along thez-

a!is. "he constant ccan #e o#tainedfrom considerations of mass flowacross the surface of a cylinder(r,,z) and of de$th l.

If dsis an element of arc which su#-

tends an angle don the circle ofradius r, the corres$onding elementalsurface area on the cylinder of de$th lis%

"herefore, the mass flow across dSis

"hus the total mass flow across thesurface of the cylinder is

0,/ == VrcVr

)( rdlldsdS ==

lrdVSdVmd r==

2 2

0 0( / ) 2 .rm V lrd c r lrd c l

= = = &

oure !lo' &2n) Ele-entar0 !lo'*

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oure !lo' &2n)Ele-entar0 !lo'*

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1231% AE 2+(: Chapter 3 4art A5 ),

oure !lo' &ontinue)*

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1231% AE 2+(: Chapter 3 4art A5 *(


&olume flow rate '

ow the volume flow rate $er unit

de$th is defined as the sourcestrength . "hen, sing

or (3.*+)

om$aring (3.59a) with (3.*+) gives

e shall now determine , , and for this flow

lcmv 2/ ==

2 .r

vrV

l =

&

rVr

2

=

2/=c

/rV c r=


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1231% AE 2+(: Chapter 3 4art A5 *1


&olume flow rate '

ow the volume flow rate $er unit

de$th is defined as the sourcestrength . "hen, sing

or (3.*+)

om$aring (3.59a) with (3.*+) gives

e shall now determine , , and for this flow

(a) Velocity potential : e have,

$on integrating, these give%

(3.*)

"hus, e/ui$otentials are the curves

i.e. r = const

i.e., they are circles with center at theorigin (see fig 3.+0)

lcmv 2/ ==

2 .r

vrV

l =

&

rVr

2

=

2/=c

01

,2 ====

V

rrV

rr

rln2

=

constr=

= ln2

/rV c r=


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(a) Stream Function : e have,

$on integrating, these give%

(3.+)

"he e/uations of the streamlines are

i.e., = const

i.e., they are radial lines originating at

the origin (see fig 3.+0)

ote% streamlines and e/ui$otentialsare mutually $er$endicular, as they

should #e

(c) Circulation :

1ince the flow is irrotational, the

circulation is 2ero. i.e., ' 0

0,2

1

==

==

VrrVr r

2

=

const==

2

#ou+let !lo' &3r) Ele-entar0 !lo'*

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#ou+let !lo' &3r)Ele-entar0 !lo'*

onsider a source-sin $air (of

strength ) distance la$art as

shown in 4ig 3.+a) 1tream function atPdue to the

source is

and due to the sin is

"hus the total stream function at

Pis

where

ow let

this is shown in fig 3.+#

hen you tae this limit, the source-

sin $air is called a doublet. ithout

this limit, it is not a dou#let, it is 6ust asource-sin $air

"he $roduct lis called the strengthof the doubletdenoted #y a$$a

112

=

222

=

=

=+= 2)(2 2121

12 =

constlwhilel =0

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1231% AE 2+(: Chapter 3 4art A5 *$

l cos4

b 5 r ' l cos4

a 5 l sin4

"in d4 5 d4 5 a6b

#ou+let !lo' &ont)*

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#ou+let !lo' &ont)*

"he stream function for the dou#let is

(3.75)

4rom fig 3.+#,

"herefore,

i.e., (3.7)

In a similar fashion, the velocitypotential for a dou#let is given #y

(3.77)

Note% If the sin is $laced to the left ofthe source, then the signs in

e/uations (3.7) and (3.77) will #e

reversed

= == dconstll 2lim0

sin , cosa l b r l = =

cos

sinsin

lr

l

b

add

==

==

cos

sin

2lim

0 lr

l

constl

r

sin

2=

r

cos

2=

2

cos

2rV

r r

= =

2

1 sin

2V

r r

= =

#ou+let !lo' &ont)*

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1231% AE 2+(: Chapter 3 4art A5 *)

#ou+let !lo' &ont)*

8/n of the streamlines is ' const

i.e.,

i.e.,

this is an e/n of a circle with diameter

don the vertical a!is and with center

at d+ directly a#ove the origin. 1ee

4ig 3.+5

By convention, we designate the

direction of the dou#let #y an arrow

drawn from the sin to the source

8/n of the e/ui$otentials is ' const

i.e.,

i.e.,

this is an e/n of a circle with diameter

don thex-a!is and with center at d+

to the right of the origin. :irection of dou#let is donated #y arrow

drawn from sin to source . :irection and

sign will change #y changing $lacement

c

r

=

sin

2

sinsin

2d

cr ==

cconstr

==

cos

2

coscos

2d

cr ==

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1231% AE 2+(: Chapter 3 4art A5 **

%orte, !lo' &t Ele-entar0 !lo'*

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1231% AE 2+(: Chapter 3 4art A5 *+

%orte, !lo' & Ele-entar0 !lo'*

4rom the definition of Vorte Flow,

we have

i.e., streamlines are concentric circles

centered at the origin

(a) Circulation! % By definition,

"herefore,

1u#stituting for V , gives

Also,

(3.;0*)

and

(3.;05)

rCVVr == ,0

= C sdV erdsdeVV , ==

drVerdeVsdV == )()(

==2

0

2 CdC

2

=C

rV

=

2

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1231% AE 2+(: Chapter 3 4art A5 *,

%orte, !lo' &ont)*

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%orte, !lo' &ont)*

(a) Velocity "otential x,y) % By

definition, we have

Integrating these e/uations gives,

(3.;;+)

8/uation of the e/ui$otentials is%

' constant, i.e.,

i.e., = const "hus, e/ui$otentials are radial lines

originating from the origin

(a) Stream Function x,y) % By

definition, we have

Integrating these e/uations gives,

(3.;;)

8/uation of the streamlines is%

' constant, i.e.,

i.e., r = const "hus, streamlines are circles with

center at the origin

rV

rV

r r

2

1,0 ==

==

2

=

const=

2

rV

rV

r r

2,01 ==

==

rln2

=

constr=

ln2

Ta+le o/ Ele-entar0 !lo's

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1231% AE 2+(: Chapter 3 4art A5 +1

Ta+le o/ Ele-entar0 !lo's

1ince elementary flows form the #asis of more com$licated flows, we

summari2e their results in "a#le 3.; #elow

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AE 24Essentials o/ Aero)0na-is

Ca(ter 3 &Part B*

!lo' Co-+inations

$ni/or- !lo' 'it a oure an) in5

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+$


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1231% AE 2+(: Chapter 3 4art B5 +%


"he velocity field is o#tained from

(3.*)

(3.)

ote that the velocity com$onents

of the resultant flow are the

sum of the corres$onding velocity

com$onents of the individual flows

#uestion% Is there (or are there) any

stagnation $oint(s) in the resultant

flow >

$nswer% 4ind out #y setting the

velocity com$onents to 2ero.

i.e., (3.7) (3.9)

4rom (3.9), ' . "hen from (3.7)

"hus, a stagnation $oint e!ists atB%

rV

rV

r

2cos

1

+=

=

sin=

= Vr

V

),( VVr02cos =

+ rV

0sin = V

cos( ) 0,2 2V rr V

+ = =

( , ) ( 2 , )r V =


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"hus the stagnation $oint is at a

distance

directly u$stream of the source

%bserve% if the source strength increases, the stagnation $oint moves

farther away from the source. If thefree stream velocity of the

uniform flow increases, the stagnation

$oint moves closer to the source. "his

is clearly intuitively o#vious

ow let=s find the e/n of the stream-line which contains the stagnation

$oint. 1u#stitute the coordinates of

the stagnation $oint in ' const

"his streamline is shown as the curve

ABCin fig 3.++

=V

DB2

V

constV

V =

+

=

2sin

2

const==2


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&mportant conclusion%

e now that no flow occurs across a

streamline. "hus, any streamline can#e re$laced with a solid surface of the

same sha$e.

In $articular, since the streamline

ABCcontains the stagnation $oint, it

is a dividing streamline. i.e., itse$arates the fluid coming from the

free stream and the fluid emanating

from the source

All the fluid outsideABCis from the

free stream and all the fluid insideABCis from the source

"herefore, as far as the free stream is

concerned, the entire region inside

ABCcould be replaced with a

solid body of the same shape andthe free stream flow would not feel

the difference


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1231% AE 2+(: Chapter 3 4art B5 ++

$ o o t a ou e a )

"he streamline e!tends

downstream to infinity, forming a

semi'infinite body "herefore, if we want to construct the

flow over a solid semi-infinite #ody

descri#ed #y the curveABC, then all

we need to do is tae a uniform

stream with velocity and add to ita source of strength at $ointD.

he resulting flow will then

represent the flow over the

prescribed solid semi'infinite body

of shapeABC ow consider a source () and a sin

(? ) $laced res$ectively at a distan-ce bto the left and right of the origin.

!!!

2=

V

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+,


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1231% AE 2+(: Chapter 3 4art B5 ,(

ow su$erim$ose a uniform stream

with velocity as shown

"hen the stream function from thecom#ined flows is

e can now determine the velocity

flow field using the e/uations

#ut you will have to e!$ress

in terms of first. "his can #e done

with the geometry of fig 3.+3 "hen, #y setting velocity ' 2ero, we

can determine the stagnation $oints.

"here are two of them,AandB

"he location of the stagnation $oints

is found to #e

(3.7;)

"he e/n of the streamlines is

V

)(2

sin 21

+= rV

rV

rVr

=

=

,

1

),( 21

+==VbbOBOA

2


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1231% AE 2+(: Chapter 3 4art B5 ,1

"he e/n of the streamlines is

"he e/n of the s$ecific streamline

which $asses through the stagnation

$oints is o#tained #y setting a$$ro-

$riate stagnation values in the a#ove

e/n. ote that at $ointA%

and at $ointB%

ith these values, the constant in the

streamline e/n is 2ero. "hus the e/n

of the stagnation streamline is

"his is the e/n of an oval and is alsothe dividing streamline

"he oval is called a ankine %val

constrV =

+= )(2

sin 21

)( 21 ===

)0( 21 ===

0)(2

sin 21 =

+

rV

$ni/or- !lo' 'it a #ou+let&Nonli/tin6 !lo' O7er a Cirular C0lin)er*

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1231% AE 2+(: Chapter 3 4art B5 ,2

&Nonli/tin6 !lo' O7er a Cirular C0lin)er*

"he flow over a circular cylinder can

#e $roduced with a combination of

uniform flow and a doublet. "his isa classic $ro#lem in aerodynamics

onsider the dou#let $lus uniform

flow com#ination shown in 4ig 3.+*.

the direction of the dou#let is

u$stream, facing into the uniform flow. "he stream function for the com#ined

flow is

or

(3.9;)

If we let then,

(3.9+)

"he velocity field is o#tained #y

differentiating this e/n as follows

rrV

sin

2sin =

=

22

1sinrV

rV

VR 22

= 2

2

1)sin(r

RrV

=

= 22

1)cos(11

r

RrV

rrVr

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,3


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1231% AE 2+(: Chapter 3 4art B5 ,$


"hus,

(3.93)

Also,

i.e.,

or (3.9)

1tagnation $oints are located #y

setting (3.93) and (3.9) to 2ero%

(3.95)

(3.9*)

"he solution yields two stagnation

$oints ?A(R,0) andB(R,) in 4ig

3.+*

"he e/n of the streamline throughBcan #e o#tained #y su#stituting the

coordinates ofBinto the e/n '

const. "he result is% const ' 0

cos12

2

= V

r

RVr

+ )sin(1

2)sin(

2

2

3

2

Vr

R

r

RrV

rV = =V

sin1 2

2

+= V

r

RV

0cos12

2

=

V

rR

0sin12

2

=

+ V

r

R


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1231% AE 2+(: Chapter 3 4art B5 ,%


1imilarly, the e/n of the streamline

throughAcan #e o#tained #y

su#stituting the coordinates ofAintothe e/n ' const. "he result is

again% const ' 0

"hus, the streamline ' 0 $asses

through #oth stagnation $oints and it

is the dividing streamline. "he e/n ofthis streamline is o#tained from (3.9+)

as

(3.9)

#uestion% hat is the sha$e of this

streamline >

$nswer% @oo at e/n (3.9). It is satisfied

for r' for all values of thus solution of

this e/n is% r = R' const

"his e/n re$resents a circle in $olar

coordinates with radiusRand center

at the origin

"herefore, e/n (3.9), i.e., the divid-ing streamline, re$resents a circle of

radiusRas shown in 4ig 3.+* a#ove

01)sin(2

2

=

=

r

RrV


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Cther solutions of the streamline e/n

(3.9) are ' and ' 0.

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1231% AE 2+(: Chapter 3 4art B5 ,*


1ince the flow is $erfectly symmetric

a#out hori2ontal and vertical a!is,

thus the $ressure distri#ution over the u$$er nad lower surfaces of the

cylinder is com$letely #alanced. "hus,

there is no net normal force (lift )

com$onent in the vertical direction

1imilarly $ressure distri#ution is #alanced

on front and rear halves thus no net a!ial(drag) 4orce is generated.

In other words, there is no net lift and

no net drag. "hat is why we call this

case the Dnonlifting flow over a

circlar c!linderE. "his is nown as

Dd*$lembert*s paradoE

Because in real life we are aware that

there always is drag associated with

an o#6ect immersed in a moving fluid

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1231% AE 2+(: Chapter 3 4art B5 ,+


$nswer% Because the flow is viscous

only within the #oundary layer which

is very thin. Cutside the #oundarylayer the flow is inviscid. 1o all of this

analysis will a$$ly to the region of

flow outside the #oundary layer. "his

is also im$ortant in aerodynamics

@et us calculate the velocity distri#u-tion and the $ressure distri#ution on

the surface of the cylinder

"he velocity distri#ution on the

surface of the cylinder is o#tained

from the velocity e/ns (3.93) and(3.9) with r = R, resulting in

(3.99)

(3.;00)

"herefore,

(3.;00a)

the minus sign is included to mae

the e/n consistent with the sign

convention for (i.e., $ositive

counterclocwise, ne!t slide)

0=rV

sin2 = VV

2 sinV V V = =

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&ovem"er 2(11 AE 2+(: Chapter 3 ,,

2 sinV V V = =

Velocity in direction of

decreasing angle thus it is

negative on upper surface

Velocity in direction of

increasing angle thus it is

positive on lower surface

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"he velocity is ma!imum at i.e., at the to$ and the #ottom of the cylinder

&ovem"er 2(11 AE 2+(: Chapter 3 1((

2 =


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1231% AE 2+(: Chapter 3 4art B5 1(1

& 6 0 *

"he $ressure coefficient C" is

(3.3*) A$$lying Bernoulli=s e/uation to an

ar#itrary $oint and a $oint in the free

stream, we have

from which we have

(3.3)

1u#stituting (3.3) into (3.3*) gives

(3.37)

But, from (3.;00a) we have

1u#stituting this into (3.37) gives

(3.;0;)

#

""C"

2

212

21

+=+ V"V"

2 212

( )" " V V =

2

1

=

VVC"

sin2=VV

2sin41="C

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&ovem"er 2(11 AE 2+(: Chapter 3 1(2

E,a-(le 3.8 9 (a6e 22

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1231% AE 2+(: Chapter 3 4art B5 1(3

+ample ,-. / page 001:

onsider the nonlifting flow over a

circular cylinder. alculate the loca-tions on the surface of the cylinder

where the surface $ressure e/uals

the freestream $ressure

Solution:

At , we have from (3.;0;)%

i.e.,

"hese results are shown in 4ig 3.30

At stagnation, C"' ;, and

"he minimum $ressure occurs at the

to$ and #ottom i.e., at

At the to$ and #ottom, C"' -3, and

the corres$onding $ressure is

= ""212 sin,0sin41 === "C

0000 330,210,150,30=

+= #""

090= #" 3

Li/tin6 !lo' O7er a C0lin)er

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1231% AE 2+(: Chapter 3 4art B5 1($

In the uniform-flow-$lus-dou#let com-

#ination we $roduced the flow around

a circular cylinder with 2ero lift (anddrag).

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1(%


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1231% AE 2+(: Chapter 3 4art B5 1()

"he resulting stream function for the

flow of 4ig 3.3+ is F i.e.,

GHHHHH (3.;;7)

"he resulting $attern given #y (3.;;7)is setched at the right of 4ig 3.3+

If r = Rin (3.;;7), then ' 0 for all

values of . "hus, the dividing

streamline is the circle of radiusR

ote that the streamlines are nolonger symmetrical a#out the

hori2ontal a!is. "hus, there will #e a

lift force #ut still no drag force

"he velocity field can #e o#tained

from the stream function of (3.;;7).

"he result is%

(3.;;9)

(3.;+0)

21 +=

R

r

r

RrV ln

21)sin(

2

2

+

=

rV

r

RV

2sin1

2

2

+=

cos12

2

= V

r

RVr


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1231% AE 2+(: Chapter 3 4art B5 1(*

"he stagnation $oints are o#tained #y

setting the velocity com$onents to

2ero, thus

(3.;+;)

GHHHHHH (3.;++) 8/n (3.;+;) gives, r = R. "hen e/n

(3.;++) gives

(3.;+3)

1ince is $ositive, e/n (3.;+3) tellsus that must lie in the th/uadrant

"hus, there are two stagnation $oints

on the #ottom half of the cylinder as

shown #y $oints ; and + in 4ig 3.33a.

"hese results are valid only when

#ecause otherwise, e/n (3.;+3) is

meaningless

02

sin12

2

=

+=

rV

r

RV

0cos12

2

=

= V

r

RVr

=

RV

4sin 1

RV

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1231% AE 2+(: Chapter 3 4art B5 1(+

If

there is only one stagnation $oint on

the surface of the cylinder, namely,$oint (R, -+) la#eled as $oint 3 in

4ig 3.33#

4or the case of

we return to e/n (3.;+;). It was satis-

fied #y r = RF however, it is alsosatisfied #y

1u#stituting into (3.;++)

and solving for r, gives

(3.;+)

"hus, there are two solutions and

therefore two stagnation $oints

Cne is inside and the other is outside

the cylinder , and #oth on the vertical

a!is, as shown #y $oints and 5 in

4ig 3.33c. @et=s e!$lain this a littlemore

RV= 4

RV> 4

2 =2 =

2

2

44R

VVr

=


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1231% AE 2+(: Chapter 3 4art B5 1(,

"he resulting stream function for the "hus, for steady flow we would write

Li/tin6 !lo' O7er a C0lin)er &Potos*

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1231% AE 2+(: Chapter 3 4art B5 11(

"he following e/uations re$resent a

flow field

"hus, for steady flow we would write

E,a-(les 3.14: 3.11 9 (a6e 22

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1231% AE 2+(: Chapter 3 4art B5 111


flow field


;utta

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1231% AE 2+(: Chapter 3 4art B5 112


flow field


#ra6 7ersus Re0nol)s Nu-+er

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1231% AE 2+(: Chapter 3 4art B5 113


flow field


#ra6 7ersus Re0nol)s Nu-+er

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1231% AE 2+(: Chapter 3 4art B5 11$


flow field


#ra6 7ersus Re0nol)s Nu-+er &Poto*

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1231% AE 2+(: Chapter 3 4art B5 11%


flow field



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1231% AE 2+(: Chapter 3 4art B5 11)


flow field



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1231% AE 2+(: Chapter 3 4art B5 11*


flow field


Pressure #istri+ution O7er a C0lin)er

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1231% AE 2+(: Chapter 3 4art B5 11+


flow field


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11,

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12(

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121

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4. CHAPTER 3- 1-12-11

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