PART II: 2D Potential Flow

Transcription

1 AERO301:Spring2011 II(a):EulerEqn.& ω = 0 Page1 PART II: 2D Potential Flow II(a): Euler s Equation& Irrotational Flow We have now completed our tour through the fundamental conservation laws that apply to fluidmechanicsingeneral,andarenowabouttoembarkonproblemsspecifictoincompressible aerodynamics. Westartbyfocusingon2Dproblems,andworkourwaytowardcalculating L and M for airfoils. Recall that incompressible aerodynamics is relevant when M 0. This allows us to consider ρtobeconstant.inaddition,wearealsoconcernedwiththecasewhen Re.Thiseliminates viscosity from the governing equations, which is a dramatic simplification, although the remaining equations are still extraordinarily complex. With these assumptions our governing equations are

2 AERO301:Spring2011 II(a):EulerEqn.& ω = 0 Page2 v = u x + v y + w z = 0 ρ D v Dt ( v = ρ t + u v x + v v y + w v ) z = p+ρ g p+ 1 2 ρ V 2 = const along a streamline Ourfirsttaskistodevelopsomemeansofdealingwiththenonlinearnatureofthemomentum and Bernoulli s equation.(as written they are near-impossible to solve by hand.)

3 AERO301:Spring2011 II(a):EulerEqn.& ω = 0 Page3 Vorticity Todothis,weinvoketheconceptofvorticity,ω,whichisthetendencyoffluidelements torotate. Takethe / xofthe(2dforsimplicity) v-momentumequationandsubtract / y of the 2D u-momentum equation and using the continuity(mass) equation to eliminate some terms y [ ( ) u ρ t + u u x + v u + p ] y x ρg x + [ ( ) v ρ x t + u v x + v v + p ] y y ρg y can be manipulated to give [ ( v ρ t x u ) + u ( v y x x u ) + v ( v y y x u )] = 0 y wherethevorticity ωis

4 AERO301:Spring2011 II(a):EulerEqn.& ω = 0 Page4 ω = v x u y From this, we can immediately see that Dω = 0 Dt Wecouldgothroughanequivalentproceduretothe2Ddevelopmentin3Dandwewould findthat ω = vandthat D ω/dt = 0inaninviscidfluidin3Daswellas2D. Itturnsoutthat ω = constisaveryusefulrestriction.why? Atsomeinitialtime,weimaginethat u=v=0everywhereandtherefore ω = 0 everywhereatthatinitialtimeandatallfuturetimes. Soinasense,ourmomentum equations reduce to a statement that the vorticity everywhere is zero.

5 AERO301:Spring2011 II(a):EulerEqn.& ω = 0 Page5 Wehavethesameconditionifwehaveauniformupstreamvelocity, u= U, v=0. Afluidflowwhich ω = 0everywhereiscalledirrotational. Twoexamples(checkwhether v=0andwhether v=0): SolidBodyRotation: v r = 0, v θ = C r u= C y, v= C x Free(Bathtub)Vortex: v r = 0, v θ = C/r

6 AERO301:Spring2011 II(a):EulerEqn.& ω = 0 Page6 So,byworkingwithourmomentumequations,weareabletodeveloptheideaofvorticity and, if the initial(or perhaps upstream) vorticity is zero our governing equations become v=0 u x + v y = 0 v=0 v x u y = 0 in 2D in 2D p+ 1 2 ρ V 2 = const along a streamline What is useful about this arrangement is that 1. The continuity and(angular) momentum equations are both linear and therefore solvable. 2. Continuity and momentum are decoupled from the pressure.

7 AERO301:Spring2011 II(a):EulerEqn.& ω = 0 Page7 So the solution procedure becomes 1. Solve for the velocity field using continuity and momentum. 2. Plug the velocities into Bernoulli s equation to get the pressure field. 3.Integratethepressurefieldonthebody(wing,tail,elevator,whatever)tofindthenet forceandmomentonthebody. Todothis,weneedtointroducejustafewmoreconcepts...

8 AERO301:Spring2011 II(b): φ, ψand Γ Page8 II(b): Scalar Velocity-Field Specifiers: ψ, φ and Γ Howdoweactuallysolvetheequationsonthepreviouspage? We define two new scalar quantities: Streamfunction: ψ Velocity Potential: φ Velocity Potential, φ Thereisavectoridentityinmathematicsthatsays: thecurlofthegradientofascalar is zero ( φ)=0 Ifthefluidisirrotationalthen v=0

9 AERO301:Spring2011 II(b): φ, ψand Γ Page9 Puttingtheseideastogether,ifaflowisirrotationalthenthevelocityfieldcanbeexpressedasthegradientofascalar: u = φ/ x v= φ or v = φ/ y We call φ the velocity potential w = φ/ z The attractiveness about this approach is that instead of needing to keep track of three scalarvelocitycomponents, u, v,and w,wewrapthemallupinasinglescalar, φ. Some examples to illustrate this concept: UniformFlow: u= U, v=0

10 AERO301:Spring2011 II(b): φ, ψand Γ Page10 Stagnation-PointFlow: u=kx, v= ky Note that lines of constant φ are perpendicular to streamlines! Whentheflowfieldisirrotational,therebyallowing φtobedefined,thenweneedtosolveonly oneequation(itwouldhavebeenthreefor(u,v,w)).thisiswhyintheoreticalaerodynamics, there is such an importance placed on whether or not the flow is irrotational or rotational. Becauseirrotationalflowscanbedescribedbyavelocitypotential φ,thenwecallsuchflows potential flows.

11 AERO301:Spring2011 II(b): φ, ψand Γ Page11 Butwriting vintermsof φisonlyusefulifwecanfindanequationthatwecansolvefor φ. Wehavealreadyusedthemomentumequationindefining φ thefactthat φexistsmeans that we conserve angular momentum. However,wehavenotyetusedcontinuity, v=0,soletusputourdefinition v= φinto that equation. We obtain 2 φ = 2 φ x φ y φ z 2 = 0 This equation is the famous Laplace s Equation, and it is the governing equation for incompressible, irrotational flows. Fortunately,itisonlyalinear,2nd-orderPDEsoitssolutionsarewellknown.(Infact,its probably the most-studied equation in all of mathematics.)

12 AERO301:Spring2011 II(b): φ, ψand Γ Page12 To find a unique solution to this differential equation we need boundary conditions. We have physicalboundaryconditionsonthevelocities...soletustranslatethesetoconditionson φ: Condition Velocities Freestream u= U, v=0 Velocity Potential No Penetration No Slip The overarching philosophy is that once we have a PDE and boundary conditions we simply superpose(add together) some simple flows(e.g., uniform flow, vortex flow etc.) that individually satisfy the equation and together satisfy the boundary conditions to get much more complicated(and relevant) solutions.

13 AERO301:Spring2011 II(b): φ, ψand Γ Page13 Streamfunction, ψ Another approach, besides using φ, is to define a quantity ψ, called the streamfunction, thatsatisfies v=0automaticallyin2d: u= ψ y and v= ψ x Note that the velocities are formed with derivatives of the other direction and that there is a minus sign associated with the v component. Check continuity Notethatthisapproachcanonlyworkfor2Dflows.

14 AERO301:Spring2011 II(b): φ, ψand Γ Page14 Examples: UniformFlow: u= U, v=0 Stagnation-PointFlow: u=kx, v= ky Tofindthegoverningequationfor ψ,substitutefor uand vdefinitionsinthe(2d)momentumequation v=0: 2 ψ = 2 ψ x ψ y 2 = 0

15 AERO301:Spring2011 II(b): φ, ψand Γ Page15 Unlike φthatcanbedefinedin3d, ψonlyworksin2d.however, ψhassomeothernice properties: ψcanbeusedinrotational(thereforewecanconsiderviscous)flowswhile φisstrictly for irrotational flows. Lines of constant ψ are streamlines The ψbetweentwostreamlinesisequaltothevolumefluxperunitdepthbetween the streamlines(huh?) Again, boundary conditions for ψ: Condition Velocities Velocity Potential Freestream u= U, v=0 No Penetration No Slip

16 AERO301:Spring2011 II(b): φ, ψand Γ Page16 Polar Coordinates Wehaveintroducedourselvesto φand ψincartesiancoordinates,butitisofteneasierto use polar coordinates. In polar coordinates: v r v θ v z = φ r 1 r φ θ φ z And 2D continuity is r (r v r) + v θ θ = 0 so ( vr v θ ) = ( 1r ψ θ ψ r ) These transformations will be useful when we consider vortices etc. later on.

17 AERO301:Spring2011 II(b): φ, ψand Γ Page17 Circulation, Γ Everythingwehaveusedtodescribevelocityfields( v, ω, φ, ψ)arelocalquantities they describe conditions at a point. Yet local definitions fall short and sometimes we need a global quantification. Forinstance,inthecaseofthe(bathtub)vortexweknowthatthevorticityiszeroeverywhere exceptattheoriginwhereitisinfinite.therefore,weonlyknowthestrengthofthevortex intermsoftheconstant, C,thatisusedinthevortex sdefinition: v θ = C/r. Becausewewillsoonbedealingwithmorecomplicatedflows,weneedamoregeneral definition of overall vortex strength.(vortex strength will soon be related to lift.) Let us define circulation: Γ = v ˆnds C

18 AERO301:Spring2011 II(b): φ, ψand Γ Page18 where visthevelocityand dsisadifferentiallengthontheboundaryofagivencontour region C. The integral must be taken counterclockwise around a closed path. The concept of circulation can be misleading. In aerodynamics, it simply means that the line integral about C is finite it does not necessarily mean that the fluid elements are moving in circles in the flowfield. Examples: UniformFlow u= U, v=0

19 AERO301:Spring2011 II(b): φ, ψand Γ Page19 VortexFlow(whereCincludestheorigin) v r = 0, v θ = C/r Therefore,weusuallyredefineavortexflowtobe v θ = Γ/2πrandcall Γthevortex strength. Vortex Flow(where C doesn t include the origin) Circulation also has an intimate connection with vorticity. From Stokes theorem we have C v ˆnds= ( v) ˆndS S

20 AERO301:Spring2011 II(b): φ, ψand Γ Page20 Hence, the(negative) circulation about a curve C is equal to the vorticity integrated over anyopensurfaceboundedby C. Thus,iftheflowisirrotationaleverywherewithinthecontourintegral( v=0)then itfollowsthat Γ = 0everywhere. For a vortex in particular, because the vorticity is zero everywhere except the origin, then any region that does not include it has zero circulation. Any region that does include the origin includes an infinitesimally small point with infinite vorticityand 0 =Γ.

21 AERO 301: Spring 2011 II(c): Elementary Flows Page 21 II(c): Elementary 2-D Flows and Superposition Nowthatweknowabout φ, ψ,and Γ,howdoweusethem? Ourapproachwillbetofindseveralsimpleflows,eachofwhichsatisfy 2 φ = 0and 2 ψ = 0, and superpose them so that the satisfy boundary conditions. We can do this because the governing equation(laplace s Eqn.) is linear. The simple flows are building blocks for any possible inviscid, incompressible flow, and if we areclever,thenwewillneverneedanythingbutthesimplepieces(andweareclever). Again, the boundary conditions we are concerned with are Uniform Freestream and No Penetration of Solid Surfaces What are the simple building blocks?...

22 AERO 301: Spring 2011 II(c): Elementary Flows Page Uniform flow Ourfirstelementaryflowrepresentsauniformflow,i.e.aflowwithnoobject. u= U v=0 φ = U x ψ = U y Incompressible? Irrotational? Do φand ψsatisfylaplace sequation? Satisfy uniform freestream boundary condition? 2. Source flow Oursecondelementaryflow,iscalledasourceflow. Considerinjectingmassintoa2D systemattheorigin,andthisinjectedfluidflowsradiallyoutwardfromtheorigin. Ifthe volumefluxperunitdepthoftheinjectedmassis Λ,thevelocitiesare:

23 AERO 301: Spring 2011 II(c): Elementary Flows Page 23 v r = Λ 2π r v θ = 0 Incompressible?(Hard way = Cartesian. Easy way = polar.) v= 1 r r (r v r)+ 1 r v θ θ Irrotationality?(Hard way = Cartesian. Easy way = polar.) ω = 1 r r (r v θ) 1 r v r θ Wouldthisscrewuptheuniformfreestreamboundaryconditionwereitaddedtoa uniform flow? Whatisthecirculationaroundanyclosedcontour(thatdoesnotpassthroughthe origin)? Whatare φand ψ? φ = Λ 2π lnr and ψ = Λ 2π θ

24 AERO 301: Spring 2011 II(c): Elementary Flows Page Vortex Flow The vortex flow we already considered earlier is another elementary flow. We established that it is incompressible and irrotational(except at the origin where it is infinite). Because of its connection to circulation, we define its strength using Γ and write φ = Γ 2π θ and ψ = Γ 2π lnr Does this affect the freestream boundary condition?

25 AERO301:Spring2011 II(d): Γ = 0Flows Page25 II(d): Flows About Bodies without Circulation Superposition example#1: Uniform + source flow Having established some elementary flows, we superpose them to make more interesting and relevant flowfields. Imagine we first superpose a source at the origin with a uniform freestream flow. What does this produce? Find: velocity fields, stagnation point(s), and the stagnation streamline(i.e., the body shape).

26 AERO301:Spring2011 II(d): Γ = 0Flows Page26 ( ) Λ Stagnationpoint: (r,θ)=,π 2πU Thevalueofstreamfunctionforthestagnationstreamline ψ stag = Λ/2. Thisisimportant becauseitgivesusanequationforthebodysurface

27 AERO301:Spring2011 II(d): Γ = 0Flows Page27 Thebodyshapethatresultsfromthisisshown on the right.(all positions have been nondimensionalized by the distance from the source at the origin to the stagnation point.) All the fluid outside the body/stagnation streamline originates upstream, and the fluid inside the body streamline originates at the source. Stagnation Streamline Stagnation Streamline Stagnation S treamli ne This means that the body streamline really does act asabodysurfaceandwehavemodelledsomething thatlooksverymuchliketheleadingedgeofawing However,thebodystreamlinesnevercloseastheytraveltoward x + andwehavenot doneagoodjobyetofmodellingthetrailingedge. Tofixthis,weaddanotherbasicflowtotheproblem:asink(i.e.,asourcewithanegative source strength) with a Λ that s equal and opposite to the strength of the existing source. Thecombinationwilladdnonetmasstothesystem.

28 AERO301:Spring2011 II(d): Γ = 0Flows Page28 Superposition example#2: Uniform + source + sink y y (x,y) x θ 1 θ 2 x l l l l Noticethatneitherthesourcenorthesinkareattheorigin. The combined streamfunction field is ψ = U y+ Λ 2π θ 1 Λ 2π θ 2 where ( ) θ 1 = tan 1 y x+l and ( ) θ 2 = tan 1 y x l Becauseofthewaythesystemissetupit seasiertousecartesiancoordinatesthistime.

29 AERO301:Spring2011 II(d): Γ = 0Flows Page29 Again, find the stagnation point(s), and the stagnation(body) streamline.

30 AERO301:Spring2011 II(d): Γ = 0Flows Page30 1. ψ =U y+ Λ 2π 2.Stagnationpoints: (x,y)= [ ( ) ( )] y y tan 1 tan 1 x+l x l ( ±l ) 1+ Λ π U l,0 3.Streamfunctionofthestagnationstreamline: ψ stag = 0 4.It strickytoplotstreamlinesbecauseofthe tan 1 functions.aquick-and-dirtymathematica plot is below. This assumes that the lengths have been nondimensionalized by landthat Λ=2π U l. psi = y + lambda(arctan[x+1, y] - ArcTan[x-1, y]) For[points={}; xx = -3; incx = 0.05, xx 3 + incx/2, xx += incx, For[yy = -1.5; incy = 0.1, yy incy/2,yy += incy, points = Append[points, {xx, y/.findroot[(psi/.{lambda - 1, x - xx}) == 0, {y, yy}]}]]] ListPlot[points, PlotRange- {{-3, 3}, {-1.5, 1.5}}, AspectRatio- 0.5]

31 AERO301:Spring2011 II(d): Γ = 0Flows Page31 NowconsideraspeciallimitingcaseofaRankineovalin which l 0.Ifwesimplylet l=0,thesourceandthesink cancel out and there s nothing but freestream flow. But, if wesimultaneouslylet Λ as l 0thenweretainsome effect. (r,θ) θ 2 θ 1 Toimaginethissituation,consider θ 1 and θ 2 as l 0.Bothbecomeapproximatelyequalto θbut θ 2 isalittlebiggerthan θand θ 1 isalittlebitsmaller. Asbefore,wehave ψ =U y+ Λ 2π (θ 1 θ 2 ) θ 1 2l θ h θ 2 Consideringthediagram, h r(θ 2 θ 1 )as l 0 and h=2lsinθ 1 2lsinθas l 0.Thus ψ = U y κ sinθ 2πr }{{} doublet where κ = 2lΛ=constant as l 0 and Λ

32 AERO301:Spring2011 II(d): Γ = 0Flows Page32 κ(kappa) is the doublet strength Thespecialcombinationofasourceandasinkiscalledadoublet,andistheequivalentof a dipole in electric field theory. The velocities that result from a doublet are: Noticethatthesedependonθanddecreasewithr 2.Doubletflowdoesnothavecircular symmetry(as do vortices and sources) and its effect decreases more rapidly as the distance from the origin increases. Thevelocitypotentialofadoubletis φ doublet = κ cosθ 2πr. (Get this by integrating the velocities.) Streamlines of doublet flow trace out figure-of-eights centered at the origin.

33 AERO301:Spring2011 II(d): Γ = 0Flows Page33 Superposition example#3: Uniform + doublet Let us find velocities, stagnation point(s), the stagnation streamfunction value and the body shape for uniform freestream flow superposed with a doublet of strength κ. κ Stagnation points: r = and θ = 0 or π 2π U Stagnation streamline value and location: κ θ = 0 or π or r= 2π U Adoubletsuperposedonauniformflowgivestheflowoveracylinder. Takingthecylinder sradiustobe Rwecaneliminate κandwritethestreamfunctionas ] ψ = U r [1 R2 r 2 sinθ

34 AERO301:Spring2011 II(d): Γ = 0Flows Page34 The streamlines about the cylinder are given to the right.(the coordinates are nondimensionalized by R.) Thisisaclosedbodythatisactually somethingofinterest.thegoalistoknow theliftanddragsoweneedthepressure distribution on the body. Use Bernoulli s equation: p(r,θ)+ 1 2 ρ {[v r (r,θ)] 2 +[v θ (r,θ)] 2} = p ρ U2 Leading Edge 1 C p Trailing Edge -π -π/2 π/2 π θ -1 C p = 1 4sin 2 θ Onthecylinder ssurface. -2-3

35 AERO301:Spring2011 II(d): Γ = 0Flows Page35 The pressure distribution is interesting but we really care about net forces. The net lift and drag(perunitspan)aregivenby 2π L = D = 0 2π 0 p sinθ Rdθ p cosθ Rdθ Returningtothe C p distribution, p= 1 2 ρ U2 C p+ p. Wedonotcareabout p becauseitwillalwayscancelasweintegrateabouttheclosed body. 1 2 ρ U2 isconstantandcanbemovedoutsideoftheintegral. So,wehave c l = c d = L 1 2 ρ U2 2R = 1 2 D 1 2 ρ U2 2R = 1 2 2π 0 2π 0 C p sinθ dθ C p cosθ dθ Evaluatingtheseusingorthogonality... c l = 0 and c d = 0 D Alembert s Paradox.

36 AERO301:Spring2011 II(e): Γ 0Flows Page36 II(e): Flows About Bodies with Circulation When we have inviscid flow over a doublet we have top/bottom and fore/aft symmetry and, therefore,noliftandnodrag. Togeteitherweneedtobreakthesymmetry. Wecando thisbyaddingavortexattheorigin. Thisissimply ψ = U r ) (1 R2 r 2 sinθ + Γ 2π lnr Thevelocitiesare......andthestagnationpointsare...

37 AERO301:Spring2011 II(e): Γ 0Flows Page37 or r= R ( r=r, θ = sin 1 Γ ) 4π U R Γ 4π U R + ( ) Γ 2 1 4π U R if, θ = π 2 Γ 4π U R 1 if Γ 4π U R > 1 Addingthevortexattheorigindoesnotchangethebodyshape!

38 AERO301:Spring2011 II(e): Γ 0Flows Page38 Γ = 0 Γ = 2π U R Γ = 4π U R Γ = 4.4π U R

39 AERO301:Spring2011 II(e): Γ 0Flows Page39 When Γ > 4π U Randthestagnationstreamlinemovesoffthebodythereisfluidoutside the r=rcylinderthatistrappedintheneighborhoodofthecylinderinsidethestagnation streamline that simply circles the cylinder. Restrictingourselvestocaseswhen Γ 4π U R,whatisthepressuredistributionaboutthe body? C p (R,θ)=1 4sin }{{ 2 θ } A 2Γ sinθ π U R }{{} B Γ 2 4π 2 U }{{ 2 R2 } C Aisthepreviousresult B is a non-uniform pressure contribution due to the vortex Cisauniformpressurecontributionduetothevortexthatdoesnotaffecttheliftordrag

40 AERO301:Spring2011 II(e): Γ 0Flows Page40 So,wehave and c l = c d = L 1 2 ρ U2 2R = 1 2 D 1 2 ρ U2 2R = 1 2 2π 0 2π 0 C p sinθ dθ C p cosθ dθ c l = Γ U R c d = 0 L = ρ U Γ Thisisanimportantresult. Itstatesthattheliftperunitspanona2Dbodyisdirectly proportional to the circulation around the body.

41 AERO301:Spring2011 II(e): Γ 0Flows Page41 The Kutta Joukowski Theorem So,despitethefactthatcirculationisaconceptwehaveinvented,wecanuseitasameans of calculating lift per unit span in any situation, for any object without a detailed investigation of the pressure field on the object. This is called the Kutta Joukowski Theorem. Saidanotherway:Ina2Dinviscidflow,theliftperunitspanonanobjectisgivenby L = ρ U Γ Just keep in mind that circulation theory is an alternative way of looking at the generation ofliftonanaerodynamicbody. Inreality, thetruephysicalsourcesofliftaredueto the pressure and shear stress distributions exerted on the surface. The Kutta-Joukowski Theorem is merely an alternative way of expressing the consequences of these stresses. Thepoweroftheapproachisthatitismucheasiertocalculatethecirculationaroundan object rather than the individual surface stresses.