3.7. SOMEEXAMPLESOFVORTICALFLOWS
41
3.7.3 Axisymmetricflow
Weturnnowtoalargeclassofvorticalflowswhichareprobablythesimplest
flowsallowingvortex stretching,namelytheaxisymmetricEulerflows. . These
aresolutionsofEuler’sequationsincylindricalpolarcoordinates(z,r,θ),under
theassumptionthatallvariablesareindependent ofthepolarangleθ. . Euler’s
equationsforthevelocityu=(u
z
,u
r
,u
θ
)incylindricalpolarcoordinatesare
∂u
z
∂t
+u·∇u
z
+
1
ρ
∂p
∂z
=0,
(3.50)
∂u
r
∂t
+u·∇u
r
u2
θ
r
+
1
ρ
∂p
∂r
=0,
(3.51)
∂u
θ
∂t
+u·∇u
θ
+
u
r
u
θ
r
+
1
ρr
∂p
∂θ
=0,
(3.52)
where
u·∇(·)=
u
z
∂z
+u
r
∂r
+
u
θ
r
∂θ
(·).
(3.53)
Wetakethedensity tobeconstant,so thesolenoidalconditionappliesinthe
form
∂u
z
∂z
+
1
r
∂ru
r
r
+
1
r
∂u
θ
∂θ
=0.
(3.54)
Thevorticityvectorisgivenby
z
r
θ
)=
1
r
∂ru
θ
∂r
1
r
∂u
r
∂θ
,
1
r
∂u
z
∂θ
∂u
θ
∂z
,
∂u
r
∂z
∂u
z
∂r
.
(3.55)
Thevorticityequationis
∂ω
∂t
+
u·∇ω
z
,u·∇ω
r
,u·∇ω
θ
+
u
θ
ω
r
r
ω·∇u
z
,ω·∇u
r
,ω·∇u
θ
+
u
r
ω
θ
r
=0.
(3.56)
Intheaxisymmetriccasewethushave
∂u
z
∂t
+
u
z
∂z
+u
r
∂r
u
z
+
1
ρ
∂p
∂z
=0,
(3.57)
∂u
r
∂t
+
u
z
∂z
+u
r
∂r
u
r
u
2
θ
r
+
1
ρ
∂p
∂r
=0,
(3.58)
∂u
θ
∂t
+
u
z
∂z
+u
r
∂r
u
θ
+
u
r
u
θ
r
=0,
(3.59)
∂u
z
∂z
+
1
r
∂ru
r
r
=0.
(3.60)
z
r
θ
)=
1
r
∂ru
θ
∂r
,−
∂u
θ
∂z
,
∂u
r
∂z
∂u
z
∂r
(3.61)
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42
CHAPTER3. VORTICITY
Iftheswirlvelocitycomponentu
θ
vanishes,thesystemsimplifiesfurther:
∂u
z
∂t
+
u
z
∂z
+u
r
∂r
u
z
+
1
ρ
∂p
∂z
=0,
(3.62)
∂u
r
∂t
+
u
z
∂z
+u
r
∂r
u
r
+
1
ρ
∂p
∂r
=0,
(3.63)
∂u
z
∂z
+
1
r
∂ru
r
r
=0.
(3.64)
z
r
θ
)=
0,0,
∂u
r
∂z
∂u
z
∂r
.
(3.65)
Notethattheonlynonzerocomponentofvorticityisω
θ
. Thevortexlinesare
thereforeallringswithacommonaxis,thez−axis. Thevorticityequationnow
hastheform
∂ω
θ
∂t
+u
z
∂ω
θ
∂z
+u
r
∂ω
θ
∂r
u
r
ω
θ
r
=0.
(3.66)
Thelastequationmayberewritten
D
Dt
ω
θ
r
=0,
D
Dt
=
∂t
+u
z
∂z
+u
r
∂r
.
(3.67)
Thus
ω
θ
r
isamaterialinvariantoftheflow.Wecaneasilyinterpretthemeaning
ofthisfact.Avortexringofradiusrhaslength2πr,andthevorticityassociated
withagivenringisaconstantω
θ
. But t thevorticity of alineisproportional
tothelinelength(recalltheincreaseofvorticitybylinestretching). Thusthe
ratio
ω
θ
2πr
mustbeconstantonagivenvortexring.Sincevortexringsmovewith
thefluid,
ω
θ
r
isamaterialinvariant.
Tocomputeaxisymmetricflowwithoutswirlwecanintroduce thestream
functionψforthesolenoidalvelocityincylindricalpolarcoordinates:
u
z
=
1
r
∂ψ
∂r
, u
r
=−
1
r
∂ψ
∂z
.
(3.68)
ThisψisoftenreferredtoastheStokesstreamfunction. Then
ω
θ
=−
1
r
L(ψ), L≡
2
∂z2
+
2
∂r2
1
r
∂r
.
(3.69)
Inthesteadycase,thevorticityequationgives
1
r
∂ψ
∂r
∂z
1
r
∂ψ
∂z
∂r
1
r2
L(ψ)=0.
(3.70)
Thus afamilyof steadysolutions canbeobtainedbysolvinganyequationof
theform
L(ψ)=r
2
f(ψ),
(3.71)
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3.7. SOMEEXAMPLESOFVORTICALFLOWS
43
where f is s an n arbitrary y function, for r the stream function ψ. . The e situation
hereiscloselyanalogoustothesteady two-dimensionalcase, see theprevious
subsection.
Nowturningtoaxisymmetricflowwithswirl,theinstantaneousstreamline
andvortexlinescannowbehelicesandamuchlargerclassofEulerflowsresults.
Thesamestreamfunctionapplies.Theswirlvelocitysatisfies,from(3.59)
Dru
θ
Dt
=0,
D
Dt
=
∂t
+u
z
∂z
+u
r
∂r
.
(3.72)
Wecanunderstandthe meaningof (3.72)usingKelvin’stheorem. . First t note
thataringoffluidparticlesinitiallyonagivencircleCdefinedbyinitialvalues
ofz,r,willstayonthesamecircularringasitevolves.Theu
θ
componenttakes
theringintoitself,andthe(u
z
,u
r
,0)sub-fielddeterminesthetrajectoryC(t)of
thering,andthustheringevolvesasamaterialcurve. Sinceu
θ
isconstanton
thering,thecirculationonC(t)is2πru
θ
. ByKelvin’stheorem,thiscirculation
isamaterialinvariant,andweobtain(3.72).
Inthecaseofsteadyaxisymmeticflowwithswirlweseefrom(3.72)thatwe
maytake
ru
θ
=g(ψ),
(3.73)
where the e function n g is arbitrary. . Bernoulli’s s theorem for r steady flow w with
constantdensitygives
1
2
|u|
2
+
p
ρ
=H(ψ),
(3.74)
statingthat the Bernoulli function n H is s constant on n streamlines. . From m the
momentumequationintheform∇H−u×ω=0weget,fromthez-component
e.g.:
u
r
ω
θ
−u
θ
ω
r
=
∂H
∂z
.
(3.75)
Usingtheexpressionsforthecomponentsofvorticityandexpressingeverything
intermsofthestreamfunction,wegetfrom(3.73)and(3.75)
1
r2
∂ψ
∂z
+
1
r2
g
dg
∂ψ
∂z
=
dH
∂ψ
∂z
.
(3.76)
Eliminatingthecommonfactor
∂ψ
∂z
andrearranging,
L(ψ)=r
2
f(ψ)−g
dg
, f(ψ)=
dH
.
(3.77)
Thustwoarbitraryfunctions,f,gareinvolvedandanysolutionof(3.77)deter-
minesasteadysolutioninaxisymmetricflowwithswirl.
Problemset3
1. Considerafluidofconstantdensityintwodimensionswithgravity,and
supposethatthevorticityv
x
−u
y
iseverywhereconstantandequaltoω.Show
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44
CHAPTER3. VORTICITY
that the velocity y fieldhas s the form m (u,v) ) = (φ
x
y
y
−χ
x
) where e φ φ is
harmonicandχisanyfunctionofx,y(independentoft),satisfying∇
2
χ=−ω.
Showfurtherthat
∇(φ
t
+
1
2
q
2
+ωψ+p/ρ+gz)=0
whereψisthestreamfunctionforu,i.e.u=(ψ
y
,−ψ
x
),andq=u2+v2.
2. Showthat,for r anincompressiblefluid, , butone where the densitycan
varyindependentlyofpressure(e.g. saltyseawater),thevorticityequationis
Dt
=ω·∇u+ρ
−2
∇ρ×∇p.
Interpret thelast termontherightphysically. . (e.g. . whathappens s if linesof
constantparey=constantandlinesofconstantρarex−y=constant?). Try
tounderstandhowthetermactsasasourceofvorticity,i.e.causesvorticityto
becreatedintheflow.
3.Forsteadytwo-dimensionalflowofafluidofconstantdensity,wehave
ρu·∇u+∇p=0,∇·u=0.
Showthat,ifu=(ψ
y
,−ψ
x
),theseequationsimply
∇ψ×∇(∇
2
ψ)=0.
Thus, show that a solutionis s obtainedby givinga a function H(ψ) ) and d then
solving∇
2
ψ = = H
(ψ). Showalsothat t the pressure isgivenby
p
ρ
= H(ψ)−
1
2
(∇ψ)
2
+constant.
4. ProveErtel’s s theorem for r a a fluid of constant t density: : Iff f is s ascalar
materialinvariant,i.e. Df/Dt t = = 0,thenω·∇f f isalsoamaterial l invariant,
whereω=∇×uisthevorticityfield.
5. A A steady Beltrami flow is avelocity fieldu(x) for whichthe vorticity
isalwaysparalleltothevelocity,i.e. ∇×u=f(x)uforsomescalarfunction
f. Showthat t if asteadyBeltramifieldis alsothesteady velocity fieldof an
inviscidfluidofconstant density,thenecessarilyf f isconstant t onstreamlines.
Whatisthecorrespondingpressure? Showthatu=(Bsiny+Ccosz,Csinz+
Acosx,Asinx+Bcosy)issuchaBeltramifieldwithf =−1. (Thislastflow
anexampleofavelocityfieldyieldingchaoticparticlepaths. Thisistypicalof
3DBeltramiflowswithconstantf,accordingtoatheoremofV.Arnold.)
6. Another r formulaexhibitinga a vector fieldu = (u,v,w) ) whose curl l is
ω=(ξ,η,ζ),where∇·ω=0,isgivenby
u=z
1
0
tη(tx,ty,tz)dt−y
1
0
tζ(tx,ty,tz)dt,
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3.7. SOMEEXAMPLESOFVORTICALFLOWS
45
v=x
1
0
tζ(tx,ty,tz)dt−z
1
0
tξ(tx,ty,tz)dt,
w=y
1
0
tξ(tx,ty,tz)dt−x
1
0
tη(tx,ty,tz)dt.
Verifythisresult.(Notethatuwillnotingeneralbedivergence-free,e.g.check
ξ=ζ=0,η=x. Aderivationofthisformula,usingdifferentialforms,maybe
foundinFlanders’bookonthesubject.)
7. Inthis s problem m the e object is to find d a2D D propagating g vortex x dipole
structureanalogoustothatstudiedinsubsection3.6.2.Inthepresentcase,the
structure willmoveclockwise e onthe circle ofradius Rwithangularvelocity
Ω. Consider r a a rotatingcoordinate system m anda a circular r structure of radius
a, stationary and withcenter r at (0,R). . Relativeto o therotating system the
velocitytendstoΩ(−y,x)=Ω(−y,x)+ΩR(−1,0),y =y−R.Itturnsoutthat
(assumingconstantdensity),themomentum equationrelativetotherotating
framecanbereducedtothatinthenon-rotatingframeinthattheCoriolisforce
canbe absorbedintothegradientofamodifiedpressure, seealater chapter.
Thus weagaintake∇2ψ+k2ψ ψ = = 0,r < a. . Here e r =
(y)2+x2. Anew
termproportionaltoJ
0
(kr)mustnowbeincluded. Werequirethatu
θ
aAand
ω mustbecontinuousonr
=a. Showthat,relativetotherotatingframe,
ψ=
2RΩ
kJ
0
(ak)
sinθJ
1
(kr
)+
2Ω
k2J
0
(ka)
J
0
(kr
),
ifr
<a,
2
r
2
−ΩR(r−a2/r)+Ωa2lnr+C, ifr ≥a.
(3.78)
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46
CHAPTER3. VORTICITY
Chapter4
Potentialflow
Potentialorirrotationalflowtheoryisacornerstoneoffluiddynamics,fortwo
reasons. Historically,itsimportancegrewfromthedevelopmentsmadepossible
bythetheoryofharmonicfunctions,andthemanyfluidsproblemsthusmade
accessiblewithinthetheory. But t asecond, moreimportantpointisthatpo-
tential flowis s actually realizedinnature, , or r atleast approximated,inmany
situationsofpracticalimportance.Waterwavesprovideanexample.Herefluid
initiallyatrest is set inmotionby thepassage of a a wave. . Kelvin’s s theorem
insuresthattheresultingflowwillbeirrotationalwhenevertheviscousstresses
arenegligible.Weshallseeinalaterchapterthatviscousstressescannotingen-
eralbeneglectednearrigidboundaries. Butoftenpotentialflowtheoryapplies
awayfromboundaries,asineffectsondistantpointsoftherapidmovementsof
abodythroughafluid.
Anexampleofpotentialflowinabarotropicfluidisprovidedbythetheory
of sound. . There e the potentialisnotharmonic,buttheirrotationalproperty
is acquiredby the smallness ofthenonlinear term u·∇uinthemomentum
equation. Thelatterthusreducesto
∂u
∂t
+
1
ρ
∇p≈0.
(4.1)
Sincesoundproducesverysmallchangesofdensity,herewemaytakeρtobewill
approximatedbytheconstantambientdensity. Thusu=∇φwith
∂φ
∂t
=−p/ρ.
4.1 Harmonicflows
Inapotentialflowwehave
u=∇φ.
(4.2)
WealsohavetheBernoullirelation(forbodyforcef=−ρ∇Φ)
φ
t
+
1
2
(∇φ)
2
+
dp
ρ
+Φ=0.
(4.3)
47
48
CHAPTER4. POTENTIALFLOW
Figure4.1: AdomainV,boundedbysurfacesS
i,o
where
∂φ
∂n
isprescribed.
Finally,wehaveconservationofmass
ρ
t
+∇·(ρ∇φ)=0.
(4.4)
Themostextensiveuseofpotentialflowtheoryistothecaseofconstantdensity,
where∇·u=∇2φ=0. Theseharmonicflowscanthusmakeuseofthehighly
developedmathematicaltheoryofharmonicfunctions.intheproblemswestudy
hereweshallusuallyconsiderexplicitexampleswhereexistenceisnotanissue.
Ontheotherhandthequestionofuniquenessofharmonicflowsisanimportant
issuewediscussnow. Atypicalproblemisshowninfigure4.1.
Aharmonicfunctionφhasprescribednormalderivativesoninnerandouter
boundaries S
i
,S
o
of anannular region n V. . The e difference e u
d
= ∇φ
d
of two
solutionsofthisproblemwillhavezeronormalderivativesontheseboundaries.
Thatthedifference must infactbezero throughout V V canbeestablished d by
notingthat
∇·(φ
d
∇φ
d
)=(∇φ
d
)
2
d
2
φ
d
=(∇φ
d
)
2
.
(4.5)
Theleft-handsideof(4.5)integratestozerooverV tozerobyGauss’theo-
remandthehomogeneousboundaryconditionsof
∂φ
d
∂n
. Thus
V
(∇φ
d
)2dV =0,
implyingu
d
=0.
Implicitinthisproofistheassumptionthatφ
d
isasingle-valuefunction.A
functionφissingle-valuedinV ifandonlyif
C
dφ=0onanyclosedcontour
CcontainedinV.Inthreedimensionsthisisinsuredbythefactthatanysuch
contourmaybeshrunktoapointinV.Intwodimensions,thesameconclusion
appliestosimply-connecteddomains.Innon-simplyconnectdomainsuniqueness
ofharmonicflowsin2DSisnotassured. Noteforaharmonicflow
C
dφ=
C
u·dx=Γ
C
,
(4.6)
so that t a a potential which is s not single valuedis s associated with a non-zero
circulationonsomecontour. Since e there is novorticitywithinthedomainof
harmonicity,we must lookoutsideofthisdomaintofindthevorticitygiving
risetothecirculation.
4.1. HARMONICFLOWS
49
Example 4.1: : The e point vortex of problem 1.2is anexample of a flow
harmonic in a non-simply connected domain which h excludes s the e origin. . If
u=
1
(−y/r
2
,x/r
2
) thenthepotential is
θ
+constant andthe circulation
onansimplyclosedcontourorientedcounter-clockwise is1. . This s defines the
pointvortexofunitcirculation. Herethevorticityisconcentratedattheorigin,
outsidethedomainofharmonicity.
Example 4.2 2 Steady two-dimensional l flow harmonic flow with velocity
(U,0) at t infinity, , past a circular r cylinder r of f radius a a centered at t the origin,
isnotunique. Theflowofexample2.4plusanarbitrarymultipleofthepoint
vortexflowofexample4.1willagainyieldaflowwiththesamevelocityatin-
finity,andstilltangenttotheboundaryr=a:
φ=Ux(1+a
2
/r
2
)+
Γ
θ.
(4.7)
4.1.1 Twodimensions: : complexvariables
Intwodimensionsharmonicflowscanbestudiedwiththepowerfulapparatusof
complexvariabletheory. Wedefinethecomplexpotentialasananalyticfunction
ofthecomplexvariablez=x+iy:
w(z)=φ(x,y)+iψ(x,y).
(4.8)
Wewillsuppresstinourformulasinthecasewhentheflowisunsteady. Ifwe
identifyφwiththepotentialofaharmonicflow,andψwiththestreamfunction
oftheflow,thenbyourdefinitionsofthesequantities
(u,v)=(φ
x
y
)=(ψ
y
,−ψ
x
),
(4.9)
yieldingtheCauchy-Riemannequationsφ
x
y
y
=−ψ
x
.Thederivativeof
wgivesthevelocitycomponentsintheform
dw
dz
=w
(z)=u(x,y)−iv(x,y).
(4.10)
Noticethat theCauchy-Riemannequationsimplythat∇φ·∇ψ=0atevery
point where thepartials aredefined, implyingthat the streamlines arethere
orthogonaltothelinesofconstantpotentialφ.
Example 4.3: : The e uniform m flow w at an angle α to the horizontal, with
velocityQ(cosα,sinα)isgivenbythecomplexpotentialw=Qze−iα.
Example 4.4: : Incomplexnotationtheharmonicflowofexample4.2may
bewritten
w=U(z+a
2
/z)−
logz
(4.11)
wheree.g. wetaketheprinciplebranchofthelogarithmfunction.
As aresult oftheidentificationofthecomplexpotentialwithananalytic
functionofacomplexvariable,theconformalmapbecomesavaluabletoolin
50
CHAPTER4. POTENTIALFLOW
Figure4.2:Flowontoawedgeofhalf-angleα.
theconstructionofpotentialflows. Forthisapplicationwemaystartwiththe
physicalofz-plane,wherethecomplexpotentialw(z)isdesired. Aconformal
mapz → → Z Z transforms s boundaries and d boundary y conditions s andleads s to o a
problemwhichcanbesolvedtoobtainacomplexpotentialW(Z). Underthe
mapvaluesofψarepreserved,sothatstreamlinesmapontostreamlines.
Example4.5:Theflowontoawedge-shapedbody(seefigure4.2).Consider
intheZplanethecomplexpotentialofauniformflow,−UZ,U>0.Theregion
aboveuppersurfaceofthewedgetotheleft,andtheandthepositivex-axisto
theright,ismappedontotheupperhalf-planeY >)bythefunctionZ=z
π
π−α
.
Thusw(z)=−Uz
π
π−α
.
Example4.6:Themapz(Z)=Z+
b
2
Z
mapsthecircleofradiusa>binthe
Z-planeontotheellipseofsemi-majoraxis
a
2
+b
2
a
andsemi-minor(y)-axis
a
2
−b
2
a
inthez-plane. Andtheexterior r ismappedontotheexterior. . Uniformflow
withvelocity(U,0)atinfinity,pastthecircularcylinder |Z|=a,has complex
potentialW(Z)=U(Z+a
2
/Z). InvertingthemapandrequiringthatZ Z ≈z
forlarge|z|givesZ=
1
2
(z+
z2−4b2). Thenw(z)=W(Z(z))isthecomplex
potentialforuniformflowpasttheellipse.Noticehowthemapsatisfies
dz
dZ
→1
asz→∞Thisinsuresthatthatinfinitymapsbytheidentityandsotheuniform
flowimposedonthecircularcylinderisalsoimposedontheellipse.
4.1.2 Thecircletheorem
Wenowstatearesultwhichgivesthemathematicalrealizationofthephysical
actof“placingarigidbodyinanidealfluidflow”,atleastinthetwo-dimensional
case.
Theorem3 Letaharmonicflowhave complex x potentialf(z), analyticinthe
domain|z|≤a. Ifacircularcylinderofradiusaisplaceattheorigin,thenthe
newcomplexpotentialisw(z)=f(z)+
f
a2
¯z
.
Toshowthisweneedtoestablishthattheanalyticalpropertiesofthenew
flowmatchthoseoftheold,inparticularthat theanalyticproperties andthe
singularitiesintheflowareunchanged.Thenweneedtoverifythatthesurface
ofthecircleisastreamline.Takingthelatterissuefirst,notethatonthecircle
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