高等流体力学 讲义Chap04

Chapter 04:Potential Flows1.Stream Functions2.Potential Functions3.The Complex Potential4.Blasius Integral Laws5.Kutta-Joukowski Theorem6.Conformal Mappings7.Joukowski Airfoil8.Method of Images9.Doublet Distributions10.Added Mass1. Stream FunctionsA stream function is a special scalar function that is useful when analyzing 2D, steady flows. As will be shown, the stream function has the following properties: 1. The stream function satisfies the continuity equation. 2. The stream function is a constant along a streamline.3. The flow rate between two streamlines is equal to the difference in the streamlines’ stream functions.First, let’s define the stream function.Define a scalar function, ψ, called a stream function , such that the continuity equation is automatically satisfied for 2D (planar and axi-symmetric), steady flows.For a steady, 2D, planar, incompressible flow in rectangular coordinates, define ψ=ψ(x , y ) such that:and x y u u y xψψ∂∂==−∂∂ (4.1)If the stream function is defined in this manner, then the continuity equation will automatically be satisfied:220y x u u x y x y y xψψ∂∂∂∂∇⋅=+=−=∂∂∂∂∂∂uFor a steady, 2D, planar, incompressible flow in polar coordinates, ψ=ψ(r , θ):1 and r u u r r θψψθ∂∂==−∂∂ (4.2)The continuity equation for a 2D, incompressible flow in polar coordinates is:()2211110r u ru r r r r r r r θψψθθθ∂∂∂∂+=−=∂∂∂∂∂∂Now let’s consider some of the additional properties of the stream function.The Stream Function is Constant Along a StreamlineLet’s determine the curve along which the stream function remains constant. We’ll only consider an incompressible flow in rectangular coordinates here for simplicity (the same result holds for polar coordinates and compressible flows).The total change in the stream function, d ψ, where ψ=ψ(x , y ), over some displacement (dx , dy ) is given by:(),y x x y d dx dy u dx u dy x y ψψψψψ∂∂=⇒=+=−+∂∂where the definition of the stream function has been used to write d ψ in terms of the velocities. To findthe slope of the curve along which ψ=constant, we let d ψ=0.constant 0y x y xd u dx u dyu dydx u ψψ===−+⇒=Notice that the slope of the curve along which the stream function is constant is exactly the same as the slope of a streamline. Thus, we conclude that the stream function is constant along a streamline !Example :A particular planar, incompressible flow can be described with the following stream function:Axy ψ=where A is a constant.a. Sketch the streamlines for the flow.b. Determine the velocity components for the flow.ψ=ψ1 2 ψ=ψ3where ψ1, ψ2, and ψ3 are constants.Note that these constants may varyfrom streamline to streamline.The Flow Rate Between Two Streamlines is Equal to the Difference in their Stream FunctionsNow let’s examine how the flow rate between two streamlines is related to the stream function. Consider the sketch below.The volumetric flow rate passing between the two streamlines, and thus crossing through a line drawn between the two streamlines can be found by calculating the volumetric flow rate through a small piece of the line and then integrating from one streamline to the other.()()()()ˆˆˆˆˆˆˆˆcos sin x x y y x y x x y y x y x y dy dx dQ d u u dA u u dA dAdA u dy u dx dy dx y xθθψψ⎛⎞=⋅=+⋅−=+⋅−⎜⎟⎝⎠∂∂=−=+∂∂u A e e e e e e eedQ d ψ∴=Integrating from streamline A to streamline B we have:AB B A Q ψψ=−The volumetric flow rate between two streamlines is equal to the difference in the streamline stream functions !Note that if ψB >ψA then the flow is from left to right. If ψB <ψA then the flow is from right to left.Example :The velocity field for a planar, incompressible flow is given by:22ˆˆ2()6x y x y xy =−−u e e a. Determine the stream function for this flow field if ψ(0,0)=0.b. Determine the volumetric flow rate across the line AB shown in the figure.ψB ˆˆsin x y θθ−e e sin θ cos θA: (1,0)xThe properties of stream functions described previously are enough to justify their use. There are additional reasons to use stream functions, however. Models of real flows can be produced by combining “building block” stream functions. The significance of this topic will be discussed in greater detail when examining the potential function (especially the complex potential); a topic covered later in the notes. For now, however, it is sufficient to present some “building block” stream functions and discuss how they can be combined to produce models of actual flows. First, let’s examine a few basic “building block” stream functions.Because the continuity equation is a linear PDE, the “building block” stream functions just presented can be combined together to produce new stream function flows.Proof :Let ψT =ψ1+ψ2 where ψ1 and ψ2 are stream functions that satisfy the continuity equation. The velocities determined from the new stream function are given by:()()12121212T x T y u y y y yu x x x xψψψψψψψψψψ∂+∂∂∂===+∂∂∂∂∂+∂∂∂=−=−=−−∂∂∂∂Substitute these velocities into the continuity equation:12121212y x u u x y x y y y x x x y x y y x y x ψψψψψψψψ∂∂⎛⎞∂∂∂∂∂∂⎛⎞∇⋅=+=+−+⎜⎟⎜⎟∂∂∂∂∂∂∂∂⎝⎠⎝⎠∂∂∂∂=+−−∂∂∂∂∂∂∂∂=u Thus, the new stream function, ψT , formed by the superpostion of the original stream functions also satisfies the continuity equation.Example :The doublet is formed by superposing a source and sink of equal strength separated by an infinitesimal distance.The stream function for the source/sink pair is given by (m >0):()2121222m m m ψθθθθπππ=−=−Re-arrange and take the tangent of both sides:()212121tan tan 2tan tan 1tan m θθπψθθθθ−⎛⎞=−=⎜⎟+⎝⎠(4.3) Note that a trig identity has been used in deriving the previous expression. From the figure we observe that:1sin tan cos r r a θθθ=− and 2sin tan cos r r aθθθ=+Substitute these expressions into equation (4.3) and simplify:22222222222222222222sin sin 2cos cos tan sin sin 1cos cos sin cos sin sin cos sin cos sin 1cos 2sin cos cos sin cos r r r a r a r r m r a r a r ar r ar r a r r aar r a r a r r a θθπψθθθθθθθθθθθθθθθθθθθθ⎛⎞⎛⎞−⎜⎟⎜⎟+−⎛⎞⎝⎠⎝⎠=⎜⎟⎛⎞⎛⎞⎝⎠+⎜⎟⎜⎟+−⎝⎠⎝⎠−−−−=+−−−=−+−2222sin ar r a θ−=− 1222sin tan 2m ar r a θψπ−−⎛⎞∴=⎜⎟−⎝⎠(4.4) Stream function for a source/sink pair of equal strength each located a distance a from the origin along the x -axis.xThe streamlines for the stream function given in equation (4.4) are shown in the following figure.0lim aψ→=m →∞) such that the ratio ma /πxψNote :1. Doublets have an orientation. The stream function for a doublet oriented in the y -direction is given by:doublet oriented along y-axiscos K r θψ=x xshown for K >0Example :The flow of a frictionless fluid (there can be slip at solid surfaces) around a non-rotating cylinder can be modeled as a uniform stream superimposed with a doublet:ψflow around a non-rotating cylinder = ψuniform stream + ψdoubletIf the cylinder radius is R , determine the velocity of the fluid on the surface of the cylinder as a function of angular position, θ.x doublet with strength k uniform stream with velocity, U ∞U ∞Note that inside the cylinder the streamlines look like:Notes :1. Stream functions can also be defined for steady, compressible, 2D flows. For example, in rectangular coordinates:00 and x y u u y xρρψψρρ∂∂==−∂∂ where ρ0 is a reference densityThe continuity equation for these conditions is:()()()000y x u u x y x y y x ρρρρψψρρρρρ∂∂⎛⎞⎛⎞∂∂∂∂∇⋅=+=+−=⎜⎟⎜⎟∂∂∂∂∂∂⎝⎠⎝⎠u2. Stream functions also exist for axi-symmetric, incompressible flows (referred to as Stokes’ stream functions): ψ=ψ(r , z )11 and r z u u r z r rψψ∂∂=−=−∂∂ The continuity equation for these conditions is:()10r zru u r r z ∂∂+=∂∂3. Stream functions cannot be defined for arbitrary 3D flows.2. Potential FunctionsThe velocity field for an irrotational flow can be written as the gradient of a potential function, φ:φ=∇u (4.5) since, from a vector identity:φ∇×∇=0and because an irrotational flow is defined as one with zero vorticity, i.e.:∇×=ω0Now let’s ensure that the continuity equation is satisfied for an incompressible fluid (compressible potential flows will be considered in a separate set of notes dedicated specifically to compressible flows):0 0 φ∇⋅=⇒∇⋅∇=u20 φ∴∇= (4.6) This is Laplace’s Equation!, a well studied, linear, elliptic partial differential equation that appears in many other disciplines such as electromagnetics and conduction heat transfer.The momentum equations for a potential flow simplify to Bernoulli’s equation since the flow is everywhere irrotational (refer to an earlier set of notes concerning Bernoulli’s equation):()()12p G F t t φφφρ∂++∇⋅∇+=∂ (4.7) where G is a conservative body potential (e.g. for gravity, G = gz where g and ˆz epoint in opposite directions) and F (t ) is a function of time. For an steady potential flow, Eqn. (4.7) simplifies to:()12constant p G φφρ+∇⋅∇+= (4.8)Note that the momentum equation (Eqn. (4.7) or (4.8)) need not be solved to determine the fluidkinematics. Solving Eqn. (4.6) subject to appropriate boundary conditions is sufficient to determine the flow velocity field. This occurs because we placed two restrictions on the flow field: the continuityequation and the irrotationality assumption. The momentum equation can be solved to determine the fluid pressure field once the velocity field is known.The appropriate boundary conditions for Laplace’s equation are either Dirichlet (the functions value is specified), Neumann (the functions gradient is specified), or mixed. At solid surfaces the appropriate boundary condition for the flow is that the flow velocity normal to the surface is equal to the surface velocity, i.e.:ˆˆ⋅=⋅u n U n (4.9)where u is the fluid velocity, U is the boundary velocity, and ˆnis the normal vector to the boundary. This is a Dirichlet or kinematic boundary condition. Note that the no-slip condition is not satisfied for potential flows. This occurs because potential flows have no viscous force contributions since the viscous terms in the Navier-Stokes equations (i.e. momentum equations) drop out due to the irrotationality assumption. As a result, the Navier-Stokes equations, which are normally 2nd order PDEs, simplify to the 1st order Euler’s equations (and can be simplified further to Bernoulli’s equation). Hence, only a single boundary condition must be specified.Neumann boundary conditions are specified at free surfaces, i.e. surfaces where the pressure is defined. These are sometimes called dynamic boundary conditions. Bernoulli’s Eqn (Eqns. (4.7) or (4.8)) are used to relate free pressure boundary conditions to the velocity field.Notes :1. Incompressible potential flows are often referred to as ideal fluid flows since is the fluid is incompressible and viscous forces are negligible.2. Potential functions can be defined for 3D flows (as long as they’re irrotational). Recall that stream functions could only be specified for 2D flows.3. The governing equation for potential flows (Laplace’s equation) is a linear PDE so that the principle of superposition can be used to combine potential flow solutions. The approach is similar to that discussed previously for stream functions.4. Potential functions and stream functions are intimately related. This will become clear in the following section of notes concerning the complex potential function.5. Streamlines (ψ = constant) and equi-potential lines (φ = constant) are perpendicular everywhere in the flow. Consider the curves along which ψ = constant (a streamline) and φ = constant:00d d d d ψψφφ==∇⋅==∇⋅x xwhere d x is a small distance along the curves. Re-write these relations in terms of the velocities:00yy x xxx y yu dy d u dx u dy dx u u dyd u dx u dy dx u ψφ∇⋅==−+⇒=∇⋅==+⇒=−x xFrom analytical geometry, two curves will be perpendicular to each other if the slopes of the linesmultiplied together equal -1. Hence, we see that the streamlines and equi-potential lines will always be perpendicular to each other. The resulting mesh of streamlines and equi-potential lines is known as the flow net .3. Complex Variable Methods for Investigating Planar, Ideal, Irrotational FlowsA good mathematics reference for this topic is: Churchill, R.V. and Brown, J.W., Complex Variables and Applications , McGraw-Hill.Let’s define the complex potential, f (z ): ()f z i φψ=+where z = x +iy = r exp(i θ) where 0 ≤ θ < 2π and exp(i θ)=cos(θ)+i sin(θ)φ is the velocity potential ψ is the stream functionWhy do this? Because it allows us to present information in a compact manner and because we can use tools from complex variable mathematics to analyze fluid flows.Notes:1. A few complex variables preliminaries:a. z = x +iy = r exp(i θ) where 0 ≤ θ < 2π and exp(i θ)=cos(θ)+i sin(θ) b . |z | = (x 2+y 2)1/2 = r c. arg(z ) = tan -1(y /x ) = θ d. log(z ) = ln(r ) + i θ e. A function f of the complex variable z is analytic on an open set if it has a derivative at each point in that set. (counter-example: f (z ) = |z |2 is not analytic anywhere since its derivative exists only at z = 0.)2. If a function, f (z )=a (x ,y )+ib (x ,y ), is analytic in a domain D then its component functions, a and b , are harmonic conjugates in D .Proof:A function, h , is harmonic if it has continuous partial derivatives of the first and second order and satisfies Laplace’s equation:20h ∇=If a function, f (z )=a (x ,y )+ib (x ,y ), is analytic in D , then the first order partial derivatives of its component functions, a and b , must satisfy the Cauchy-Riemann equations throughout D .and a b a b x y y x∂∂∂∂==−∂∂∂∂ Differentiating:222222222222anda b a bx y x y x xa b a by x y x y y ∂∂∂∂==−∂∂∂∂∂∂∂∂∂∂==−∂∂∂∂∂∂ But from advanced calculus:2222 and a a b bx y y x x y y x∂∂∂∂==∂∂∂∂∂∂∂∂ Substituting and simplifying:2222222222 and 0 and 0a ab bx y y x a b ∂∂∂∂=−=−∂∂∂∂⇒∇=∇=Thus, a and b are harmonic.θry xIf two functions, a and b , are harmonic in a domain D and their first-order partial derivatives satisfy the Cauchy-Riemann equations throughout D , b is said to be a harmonic conjugate of a .3. Any analytic function, f (z )=φ(x ,y )+i ψ(x ,y ), is a valid planar, incompressible, irrotational flow field.Proof: R ecall that for an irrotational flow, the velocity may be written as the gradient of a potential function, φ:from a vector identity: 0 for any φφφ=∇×=∇×∇=⇒=∇ωu 0u For the flow to satisfy the continuity equation for an incompressible fluid:20 0 is a harmonic function!φφφ∇⋅=⇒∇⋅∇=∇=∴u Also recall that the stream function, ψ, is defined for 2D flows such that continuity for an incompressible fluid is automatically satisfied:22 andso that 0continuity is automatically satisfied!x y u u y xx y x yψψψψ∂∂==−∂∂∂∂∇⋅=−=∂∂∂∂∴u If the flow is also irrotational, the stream function must also satisfy:22222-0-0is a harmonic function!y xu u x yx yψψψψ∂∂=∇×=⇒=∂∂∂∂⇒−=−∇=∂∂∴ωu 0In addition,Cauchy-Riemann equationsand are harmonic conjugates!x y u x y u y xφψφψφψ∂∂⎫==⎪∂∂⎪⎬∂∂⎪==−⎪∂∂⎭∴ From Note 2, the components of any analytic function are harmonic conjugates. Thus, since thegoverning equations for the fluid are Laplace’s equation and since φ and ψ are harmonic conjugates, then any analytic f (z )=φ+i ψ will be a valid flow field.Thus, by choosing various forms of f (z ) that are analytic, we can produce various valid(incompressible and irrotational) flow fields. Whether or not the flow fields are interesting from an engineering perspective is another matter.3. Some “building block” flows and their complex potentials:z =x +iy and z 0=x 0+iy 0 0 ≤ θ < 2π()()1/22210000 and tan y y r x x y y x x θ−⎛⎞−⎡⎤=−+−=⎜⎟⎣⎦−⎝⎠4. Fluid velocities are found via differentiation of the complex potential:()x y dff z u iu dz′==−()()() where ,,also, 1(an identical result occurs if we consider )x yx y f df z f z x y i x y x dz x f i u iu x x xz x df f df z u iu dz y dz y φψφψ∂∂==+∂∂∂∂∂⇒=+=−∂∂∂∂=∂∂∂∴=−=∂∂For example:()2222()log() (source/sink at origin)211exp()cos sin 22exp()22cos and sin 22or in polar coordinates using some geometry and trig.:an x y r x y mf z z df m m m i mi dz z r i rr m mu u r ru u u u θπθθθππθππθθππ=−====−∴==+=+-1-1tan d tan tan 1tan y y r r x x r u u u u u u u u u u θθθθθθ⎛⎞+⎜⎟⎛⎞⎛⎞⎝⎠⎛⎞=+⇒=⎜⎟⎜⎟⎜⎟⎝⎠⎛⎞⎝⎠⎝⎠−⎜⎟⎝⎠ Substituting in for our values of u x and u y and simplifying:()2222tan and tan 1tan 0 21tan 0 and 2r r r r r u u m u u u u u r u mu u rθθθθθθθθπθπ⎛⎞+⎜⎟⎛⎞⎝⎠⎛⎞+==⇒+=⎜⎟⎜⎟⎝⎠⎛⎞⎝⎠−⎜⎟⎝⎠⇒==In general the relation between the velocity components expressed in rectangular and polar coordinates is given by:cos sin sin cos x r y r u u u u u u θθθθθθ=−=+()()()()()()cos sin sin cos cos sin cos sin cos sin x y r r rdfu iu u u i u u dzu i iu i u iu i θθθθθθθθθθθθθ=−=−−+=−−−=−−5. We can use the principle of superposition to combine complex potentials and form new complex potentials since if two functions are analytic in a domain D , then their sum is also analytic.u x u y u r u θ tan-1(u y /u x ) tan -1(u θ/u r )θuθ6. A few example flows created by superposition:Flow over a Rankine half-body :Combine the complex potentials for a uniform stream and a source (m >0):()log 2mf z Uz z π=+Flow over a Rankine oval :Combine the complex potentials for a uniform stream, a source, and a sink:()log()log()22m mf z Uz z a z a ππ=+−−+where m >0 and a ∈ℜ > 0.Flow around a non-rotating cylinder of radius R:Combine the complex potentials for a uniform stream and a doublet:()cf z Uz z=+where the constant c is found by not allowing any flow through the cylinder walls, i.e .()()2220cos sin ()cos sin 1cos cos cos 0cos r r r u r R c c f z Ur i Ur r r c u U r r r c u r R U c UR Rφψθθθθφψθθθθθ==⎡⎤⎡⎤=++−⎢⎥⎢⎥⎣⎦⎣⎦∂∂===−∂∂⇒===−⇒=so that the complex potential becomes:2()R f z U z z ⎛⎞=+⎜⎟⎝⎠ (4.11)Notes :1. Real (viscous) flow over a sphere (a golf ball in the figure below) is shown below. The streamlines for flow over a cylinder look much the same.The streamlines over the front half of a cylinder are similar to those predicted by the potential flow analysis. In fact, the velocity and pressures field on the front half of the cylinder are also accurate (the pressures will be discussed in a moment.) The flow field downstream of the cylinder is not accurately predicted. The discrepancy between the potential flow analysis and real life occurs due to the formation of a viscous boundary layer on the cylinder surface. The boundary layer separates near the top/bottom points of the cylinder and forms a wake. Assuming irrotational flow in theboundary layer and wake are poor assumptions. However, outside the boundary layer and wake, thepotential flow assumption is reasonable. We’ll discuss boundary layers in a later section of notes.2. The pressure distribution on the cylinder surface can be predicted using Eqn. (4.11) and Bernoulli’s equation:2()R f z U z z ⎛⎞=+⎜⎟⎝⎠From Eqn. (4.10), the flow velocity field is:()()()()()()()()()()222222222222exp 11exp 21exp 2exp exp exp 1cos 1sin exp r df R R u iu i U U dz z r i R R U i U i i i r r R R U i i r r θθθθθθθθθθ⎛⎞⎛⎞−−==−=−⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎡⎤⎡⎤=−−=−−−⎢⎥⎢⎥⎣⎦⎣⎦⎡⎤⎛⎞⎛⎞=−++−⎢⎥⎜⎟⎜⎟⎝⎠⎝⎠⎣⎦On the cylinder surface (r = R ):()2sin r r R r R u u U θθ====−The pressure distribution on the cylinder surface is found via Bernoulli’s equation, and expressed in terms of a dimensionless pressure coefficient, c p :()221214sin s p p pc U θρ∞−==− (4.12)Notes :a. The total drag (F D ) and lift (F L ) on the cylinder may be found by integrating the pressure distribution over the entire cylinder surface:2020cos sin D L F p Rd F p Rd θπθθπθθθθθ=====−=−∫∫Either by actually evaluating Eqn. (4.13) or noting that the velocity field is symmetric over thefront and back and upper and lower surfaces, the drag and lift forces on the cylinder are both zero. Of course in real flows we know that the drag is not zero. The fact that the potential flow model predicts zero drag while real flows have non-zero drag is known as d’Alembert’sParadox . We, of course, now know that the discrepancies are explained by the formation of a boundary layer and boundary layer separation. d’Alembert’s paradox will be discussed again when reviewing Blasius’ integral law.pb. Eqn. (4.12) is compared to experimental data in the plot below.Again, the potential flow analysis predicts the pressure distribution reasonably well over the upstream part of the cylinder but does a poor job over the back half due to boundary layer separation.Flow around a rotating cylinder of radius R:Combine the complex potentials for a uniform stream, a doublet, and a free vortex:()log 2c i f z Uz z z πΓ=+− where the constant c is found by not allowing any flow through the cylinder walls, just as in the previous example. Note that the addition of a vortex will not change the value of c since a vortex only produces tangential flow and not radial flow. As a result, the complex potential becomes:2()log 2R i f z U z z z π⎛⎞Γ=+−⎜⎟⎝⎠Notes :1. The corresponding velocity field is:()()()()2222exp 121exp 2exp 2r df R i u iu i U dz z z R i U i i rr θθπθθπ⎛⎞Γ−−==−−⎜⎟⎝⎠⎛⎞Γ=−−−−⎜⎟⎝⎠ Using the previous results for a non-rotating cylinder:()()22221cos 11sin 2r R u U r R u U r r θθθπ⎛⎞=−⎜⎟⎝⎠⎛⎞Γ=−++⎜⎟⎝⎠ (4.14) On the cylinder surface (r = R ):()012sin 2r r R r R u u U Rθθπ===Γ=−+ (4.15) Using Bernoulli’s equation, the pressure coefficient over the surface is:221114sin 4sin 22p c UR UR θθππΓΓ⎛⎞=−+−⎜⎟⎝⎠(4.16) The corresponding drag, F D , and lift, F L , are:0D L F F U ρ==−Γ(4.17)Notes :a. The drag again is zero and is not unexpected due to the fore/aft symmetry of the velocity field.b. The lift is non-zero and is related to the flow circulation. This type of lift is referred to as Magnuslift . Both drag and lift for potential flows will be discussed in detail when reviewing Blasius’integral law and the Kutta-Joukowski theorem.c. The photo below shows the flow past a rotating golf ball. The flow is from left to right and the golfball rotates in a clockwise manner (Γ < 0). From Eqn. (4.17), the lift on the golf ball will be in thepositive vertical direction.In real (i.e. viscous) flows the lift on a rotating object comes primarily from deflection of thedownstream wake (the fluid momentum is directed downward resulting in an upward force on theball) rather than from the unbalanced pressure distribution on the object. The Magnus effect is oftenmistakenly referred to as the primary source of the lift force.Notes…:7. Flow in and around corners of varying angles can be modeled using the following complex potential:() where and are constants n f z Az A n =a. This produces flows between boundaries intersecting at an angle π/n (only flows with n ≥1/2 are of interest):b. The potential and stream functions are given by:()()()()()()n() exp exp cos sin cos and sin n n n n n n f z Az A r i Ar in Ar n iAr n Ar n Ar n θθθθφθψθ====+⎡⎤⎣⎦∴==c. The fluid speed at the origin is:()1100000lim lim ()lim exp (1)lim 01lim 11n n r r r r r f z nAr i n nAr n A n n θ−−→→→→→′==−=>⎧⎪⇒==⎨⎪∞<⎩u uyyn =3 n =2 n=1 n =2/3d. In a real fluid (one with viscosity), the flow along the surface streamline would:for nfor n<1: separate after reaching the corner unless the corner angle is small4. Blasius Integral LawsConsider the 2D, incompressible, inviscid, steady, irrotational flow around an arbitrary closed body:Using COLM, determine the lift, L , and drag, D , acting on the body:()()outside outside outside outside x x y C C d y x y C C d D pdy u u dy u dx L pdx u u dy u dx ρρρρρ⋅⋅−−=−−+=−∫∫∫∫u A u AFrom Bernoulli’s equation (neglecting gravity):()()22221122 x y x y p u u c p c u u ρρ++=⇒=−+where c is a constant. Substituting and re-arranging:()()()()outsideoutside 2221222212x y x x y C x y x y y C D cdy u u dy u dy u u dx L cdx u u dx u u dy u dx ρρρρ⎡⎤=−++−−⎣⎦⎡⎤=−+−−⎣⎦∫∫Noting that:outside outside 0C C cdy cdx ==∫∫and simplifying:()()outsideoutside 22122212x y x y C x y x y C D u u dy u u dx L u u dx u u dy ρρρρ⎡⎤=−−+⎣⎦⎡⎤=−−−⎣⎦∫∫As shown below, the previous drag and lift relations may be written in terms of the complex potential.u x()()()()()()()()()outside outside outside outside 22222222222211222222222x y C C x x y y C x y x y x y x y C x y x y x y D df i dz i u iu dx idy dz iu iu u u dx idy i u dx u dx u u dy i u dy u dy u u dx u u dy u u dx i u u dx u ρρρρρρρρ=⎛⎞=−+⎜⎟⎝⎠=−−+⎡⎤=−++−−⎣⎦=−−+−−−−∫∫∫∫ ()outside x y C L u dy =⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦∫Substituting the expressions for lift, L , and drag, D , found previously:outside 22C df i dz D iL dz ρ⎛⎞=−⎜⎟⎝⎠∫ BLASIUS’ INTEGRAL LAWHow is this result used? Typically, it is applied using a theorem from complex variables referred to as the Residue Theorem (Churchill and Brown, p. 169) which states:()1()2Res ()k nz z k C w z dz i w z π===∑∫A residue is the coefficient in front of the 1/(z -z 0) term (the b 1 term in the series below) in the Laurent series expansion of an analytic complex function about the point z 0 (Churchill and Brown, p. 144):()()0010()n n n n n n b w z a z z z z ∞∞===−+−∑∑ where the coefficients a n and b n are given by()()11001()1() and 22n n n n C C f z dz f z dz a b i i z z z z ππ+−+==−−∫∫ The details of the expressions above won’t concern us here and are only presented for completeness. Blasius’ Integral Law is used in deriving the Kutta-Joukowski theorem given in the following section which relates the lift (and drag) around any arbitrary, closed object to the circulation, Γ, caused by the object.5. Kutta-Joukowski TheoremNow consider the flow around an arbitrary closed body (centered at the origin) in a uniform stream of horizontal velocity, U . Far from the body (z →∞) the complex potential will be of the following form (a Laurent series expansion):1()log 2n n n b m i f z Uz z zπ∞=−Γ⎛⎞=++⎜⎟⎝⎠∑ (4.18) Note that the coefficients, a n , for the terms involving z n (n ≥ 2) in the Laurent series (refer to the previous set of notes on the Blasius integral law) are all zero since we are considering external flows (recall that the velocity field is given by df /dz so that terms involving z n where n ≥ 2 will approach ∞ as z →∞).Furthermore, since we are concerned only with closed bodies, the net source term, m , should also be zero. We, however, will continue to include the source term until the end of this analysis.Given the complex potential above, let’s apply Blasius’ Integral Law to determine the lift and drag about an arbitrary object:11222231211222n n n df m i U nb z dz z df m i U m i U O dz z z z πππ∞−+=−Γ⎛⎞=++−⎜⎟⎝⎠−Γ−Γ⎛⎞⎛⎞⎛⎞⎛⎞=+++⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠⎝⎠∑ Using the Residue Theorem to evaluate Blasius’ Integral Law:outside 2202i Res 2i 22222z C df df m i D iL i dz i i U dz dz mU i U ρρρπππρρ=−Γ⎛⎞⎛⎞⎛⎞−==⋅=⋅⋅⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠=−+Γ∫Thus, we see that for a closed object (m =0): 0 and D L U ρ==−Γ KUTTA-JOUKOWSKI THEOREM (4.19) For an object that is not closed (e.g. a Rankine half-body), we have:and D mU L U ρρ=−=−ΓNotes:1. The result given above indicates that there is no drag around an arbitrary closed object in an a steady, incompressible, irrotational, inviscid flow. In real life of course there is always some drag on an object. The conflict between the derived value of zero drag and the real-life value of non-zero drag is referred to as D’Alembert’s Paradox . There is no paradox, in fact, if one realizes that it is viscous effects (skin friction drag and form, aka pressure, drag resulting from the formation of a wake (which in turn is a result of boundary layer separation)) which produces drag on an object.2. Bodies of semi-infinite extent (e.g. a Rankine half-body) do have drag on them due to the fact that the net source term, m , is not zero. The drag is a result of a non-zero flux of horizontal momentum out through the control surface.3. The Kutta-Joukowski theorem states that the lift on an object is directly proportional to the net circulation, Γ, caused by the object. This is an important observation that is especially useful inaerodynamics when calculating the lift on an airfoil. As will be shown later, the circulation around an airfoil is dependent on the free stream velocity so that the lift turns out to be proportional to thecirculation squared.Γ U。