The Kutta-Joukowski Theorem

Continuum Mechanics Lecture 7 Theory of 2D potential flows

Prof. Romain Teyssier http://www.itp.uzh.ch/~teyssier

Continuum Mechanics 20/05/2013 Romain Teyssier Outline

- velocity potential and stream function - complex potential - elementary solutions - flow past a cylinder - lift force: Blasius formulae - Joukowsky transform: flow past a wing - Kutta condition - Kutta-Joukowski theorem

Continuum Mechanics 20/05/2013 Romain Teyssier 2D potential flows

From the Helmholtz decomposition, we have v = −→ φ + −→ −→A ∇ ∇× 2D flows are defined by ∂ ( )=0 and v =0 . z · z We have therefore −→A = ψez We consider in this chapter incompressible and irrotational flows. −→ v =0 ∆φ =0 ∇ · + B. C. −→ v =0 ∆ψ =0 v n =0 ∇× · We have two alternative but equivalent approaches.

vx = ∂xφ vy = ∂yφ where the velocity potential satisfies the Laplace equation.

vx = ∂yψ vy = ∂xψ where the stream function satisfies the Laplace equation. − In the potential case, the irrotational condition is satisfied automatically. In the stream function approach, this is the divergence free condition. Since both conditions are satisfied, both velocity fields are equal.

Continuum Mechanics 20/05/2013 Romain Teyssier Isopotential curves and stream lines

The velocity field is defined equivalently by two scalar fields

vx = ∂xφ = ∂yψ v = ∂ φ = ∂ ψ y y − x They are conjugate functions that satisfy the Cauchy-Riemann relations. They are also harmonic functions (Laplace equation), with however different B. C. ∆ φ =0 in S with −→ φ n =0 on the boundary L ∇ · ∆ ψ =0 in S with −→ ψ t =0 on the boundary L or ψ = constant along L ∇ ·

dy vx Isopotential curves are defined by d φ =d x ∂ x φ +d y ∂ y φ =0 or = dx − vy dy vy Stream lines are defined by d ψ =d x ∂ x ψ +d y ∂ y ψ =0 or = dx vx Isopotential curves and stream lines are orthogonal to each other.

∂ ψ∂ φ + ∂ ψ∂ φ =( v )v +(+v )v =0 x x y y − y x x y

Continuum Mechanics 20/05/2013 Romain Teyssier Complex potential and complex derivative

We define the complex potential F (z)=φ(x, y)+iψ(x, y) where z = x + iy and i 2 = 1 . − From complex derivation theory, we know that any complex function F is differentiable if and only if the two functions Φ and ψ satisfy the Cauchy-Riemann relations. Such complex functions are called analytic. Luckily, since the velocity potential and the stream function are conjugate, the complex velocity potential is differentiable.

dF We define the complex velocity w(z)= dz dF ∂F 1 ∂F where the complex derivative is defined as = = dz ∂x i ∂y

We obtain w(z)=∂ φ + i∂ ψ = ∂ ψ i∂ φ = v iv x x y − y x − y 1 1 In cylindrical coordinates, we have v = ∂ φ = ∂ ψ v = ∂ φ = ∂ ψ r r r θ θ r θ − r and the complex velocity writes w(z)=v iv =(v iv )exp iθ x − y r − θ −

Continuum Mechanics 20/05/2013 Romain Teyssier Uniform flow

Complex potential Velocity field iα F (z)=U exp− z

vx = U cos(α) Complex velocity vy = U sin(α) iα w(z)=U exp−

Potential lines Streamlines

Velocity potential φ = U cos(α)x + U sin(α)y

Stream function ψ = U cos(α)y U sin(α)x −

Continuum Mechanics 20/05/2013 Romain Teyssier Stagnation flow

F (z)=Cz2 Streamlines are hyperbolae. φ = C(x2 y2) vx =2Cx − ψ =2Cxy vy = 2Cy − In polar coordinates 2 φ = Cr cos(2θ) This potential can also be used ψ = Cr2 sin(2θ) to describe a flow past a corner.

Continuum Mechanics 20/05/2013 Romain Teyssier Flow past an edge

Complex potential F (z)=c√z = cr1/2 expiθ/2

Velocity potential θ φ = c√r cos 2 Stream function θ ψ = c√r sin 2 Velocity field c 1 θ v = cos r 2 √r 2 c 1 θ v = sin θ 2 √r 2 The velocity field becomes infinite at the tip of the edge.

Continuum Mechanics 20/05/2013 Romain Teyssier Flow around a source or a sink

Complex potential Velocity field m m x x0 F (z)= log (z z0) v = − 2π − x 2π r2 Complex velocity m y y0 m 1 vy = − w(z)= 2π r2 2π z z0 − In polar coordinates m mθ m φ = log r ψ = v = 2π 2π r 2πr

v The velocity divergence is zero everywhere −→ v = ∂ v + r =0 for r>0. ∇ · r r r We apply the divergence theorem to a circle centered on the singularity:

v ndl = vr2πr = m L · 1 Rmax m2 dr m2 R The kinetic energy in the flow is v2dxdy = = log max 2 4π r 4π R S Rmin min In a real flow, the singularity is usually embedded inside the boundary condition.

Continuum Mechanics 20/05/2013 Romain Teyssier Flow around a point vortex

Complex potential Velocity field Ω Ω y y F (z)= i log (z z0) v = − 0 − 2π − x −2π r2 Complex velocity Ω x x0 Ω 1 vy = −2 w(z)= i 2π r − 2π z z0 − In polar coordinates Ωθ Ω Ω φ = ψ = log r vθ = 2π −2π 2πr

v The velocity curl is zero everywhere −→ v = ∂ v + θ e =0 for r>0. ∇× r θ r z We apply the curl theorem on a circle centered on the singularity:

v tdl = vθ2πr = Ω L · 1 Rmax m2 dr m2 R The kinetic energy in the flow is v2dxdy = = log max 2 4π r 4π R S Rmin min There is a direct analogy with the energy of dislocations in a solid.

Continuum Mechanics 20/05/2013 Romain Teyssier Superposition principle and boundary conditions

Like for the Navier equation for thermoelastic equilibrium problems, the Laplace equation for the potential and/or the stream function is a linear boundary value problem. When proper boundary conditions are imposed (no vorticity), the solution always exists and is unique.Two different solutions can be added linearly and the sum represent also a solution with the corresponding boundary conditions. The previous elementary solutions form a library that you can combine to build up more complex curl-free and divergence-free flows.

Streamlines are perpendicular to potential curves. The velocity component normal to a streamline is always zero. Therefore, each streamline can be used to define a posteriori the boundary condition.

You can therefore add up randomly complex potential to get any kind of analytical complex function. Then, you compute the streamlines. Then, you define the embedded body by picking any streamline. You finally get yourself a valid potential flow !

Continuum Mechanics 20/05/2013 Romain Teyssier Flow around a doublet

We superpose a source and a sink m F (z)= (log (z z ) log (z + z )) 2π − 0 − 0 iα very close to each other z0 = exp Taylor expanding, we find m expiα µ expiα F (z) = − π z − z Parameter µ is called the doublet strength. For α =0 , we find µx µ cos θ φ = = − r2 − r µy µ sin θ Potential and streamlines are circles. ψ = = r2 r µ cos θ 1 µ sin θ The velocity field is given by v = ∂ φ = and v = ∂ φ = r r r2 θ r θ r2 r It is a dipole field v = µ r3 expiα The general form around an arbitrary center z0 is: F (z)= µ − z z − 0 Continuum Mechanics 20/05/2013 Romain Teyssier Flow past a cylinder

We superpose a uniform flow and a doublet. µ F (z)=U z + ∞ x z y We find φ = U x + µ ψ = U y µ ∞ r2 ∞ − r2 µ The streamline ψ =0 is the circle r = U S S’ ∞ We reverse engineer the process. For a cylinder a radius a, if we define µ = U a2 ∞ then the potential flow around the cylinder is a2 F (z)=U z + ∞ z a2 a2 The velocity field is given by vr = U 1 cos θ vθ = U 1+ sin θ ∞ − r2 − ∞ r2 The flow has 2 stagnation points S and S’ given by r=a and θ=0 and π. The doublet is inside the embedded body, so there is no singularity in the flow.

Continuum Mechanics 20/05/2013 Romain Teyssier Force acting on the cylinder

Using the second Bernoulli theorem (curl-free, incompressible, no gravity), P 1 we know that the quantity H = + v 2 is uniform. 1 ρ 1 2 We thus have p + ρv2 = p + ρU 2 2 ∞ 2 ∞

The force acting on the cylinder is given by −→F = pndl − L On the cylinder, we have v r =0 and vθ = 2U sin θ − ∞ 1 The pressure field on the cylinder is thus p = p + ρU 2 1 4 sin2 θ ∞ 2 ∞ − Using n = (cos θ , sin θ ) we find: 2π 2π 1 2 2 4 3 Fx = ρU a 1 4 sin θ cos θdθ sin θ sin θ =0 −2 ∞ − ∝ − 3 0 0 2π 2π 1 2 2 4 3 Fy = ρU a 1 4 sin θ sin θdθ 3 cos θ cos θ =0 −2 ∞ − ∝ − 3 0 0 Exercise: compute the torque on the cylinder (use the cylinder axis). It is also zero !

Continuum Mechanics 20/05/2013 Romain Teyssier Flow past a cylinder with vorticity

We superpose a uniform flow, a doublet and a vortex. a2 iΩ z F (z)=U z + log ∞ z − 2π a Streamlines are given by a2 Ω r ψ = U 1 y log ∞ − r2 − 2π a The cylinder r=a is still a proper boundary condition. a2 vr = U 1 cos θ ∞ − r2 a2 Ω vθ = U 1+ sin θ + − ∞ r2 2πr Ω On the cylinder, we have to stagnation point given by sin θ = s 4πU a or one stagnation point away from the cylinder if Ω < 4πU a ∞ − ∞ At the boundary, we have vθ = 2U (sin θ sin θs) − ∞ − Using the Bernoulli theorem and integrating the pressure field on the boundary,

we can compute the force on the cylinder (exercise) Fx =0 Fy = ρU Ω − ∞

Continuum Mechanics 20/05/2013 Romain Teyssier The Magnus effect

Topspin tennis ball trajectory curves down.

Rotating pipes induce a force perpendicular to the wind direction

Warning: viscosity effects can’t be ignored !

Continuum Mechanics 20/05/2013 Romain Teyssier The complex force: Blasius formulae

F −→F = pndl x n − L t We use curvilinear coordinates along the body t = (cos θ, sin θ) n =(sinθ, cos θ) − F The force components are y

F = p sin θdl Fy = p cos θdl x − L L In Cartesian coordinates, we have dx = cos θdl dy =sinθdl

The complex force is defined as F iF = p (dy + idx)= i pdz∗ x − y − − L L

1 2 1 2 2 Bernoulli theorem: p = p + ρ U ρ v with v = w(z)w(z)∗ ∞ 2 ∞ − 2

Boundary condition: v n =0 v x dy v y dx =0 and w∗dz∗ = wdz · −

i We finally get the force for an arbitrary shaped body: F iF = ρ w2(z)dz x − y 2 L

Continuum Mechanics 20/05/2013 Romain Teyssier The complex circulation

We consider an arbitrary closed contour in the complex plane. We define the complex circulation as C = w(z)dz L

Using the same definition as before along the contour, we have C = (v dx + v dy)+i (v dy v dx) x y x − y L L where the Cartesian coordinates are related to the curvilinear ones by dx = dl cos θ dy = dl sin θ

We finally get C = v tdl + i v ndl = Γ + iQ · · L L

Γ is the physical circulation and Q is the physical mass flux.

On the contour defining the body shape, the mass flux is zero and we have

C = Γ = v tdl · L

Continuum Mechanics 20/05/2013 Romain Teyssier Conformal mapping

We need to build more complex profile than just a cylinder. We use for that a mathematical trick called conformal mapping. A conformal mapping is a differentiable complex function M that maps the complex plane z into another complex plane Z. 1 We have Z = M ( z ) and z = m ( Z ) with m = M − If a flow is defined by a potential function f ( z ) in the z plane, then the function F (Z)=f(m(Z)) is also analytic (it satisfies the Cauchy-Riemann relations). It is therefore a valid vector potential. The new streamlines and potential curves are the transform of the old one. dF df dm The new complex velocity writes W (Z)= = = w(z)m(Z) dZ dz dZ

The complex circulation is conserved by conformal mapping

C = W (Z)dZ = w(z)m(Z)dZ = w(z)dz L L l

Continuum Mechanics 20/05/2013 Romain Teyssier The Joukowski transform c2 Definition: z = Z + Z z Z c 2c 2c The circle of radius c becomes − the line segment [-2c, 2c]

Z = c expiθ z =2c cos θ Z z

The circle of radius a>c becomes an ellipse.

Z = a expiθ c2 c2 z =(a + ) cos θ + i(a )sinθ a − a

z z 2 The inverse transform is Z = + c2 2 2 − 1 z The derivative M ( z )= + has 2 singular points at z = 2c 2 2 ± 2 z c2 2 − Continuum Mechanics 20/05/2013 Romain Teyssier Acyclic flow past an ellipse

We use the Joukowski transform from a flow past a circular cylinder. The flow is acyclic: no circulation and no vortex component.

We assume that the flow at infinity is at an angle with the x-axis. The complex potential and velocity of the original flow are iα 2 iα 2 exp iα a i2α F (Z)=U Z exp− +a W (Z)=U exp− 1 exp ∞ Z ∞ − Z2 Using the Joukowski mapping Z = M ( z ) with a>c, we get the potential around an ellipsoidal cylinder.

2 2 2 c a iα iα a iα Using = z Z , we get f(z)=U z exp +M(z) exp− exp ∞ c2 − c2 Z − 2 iα iα c iα The original stagnation points Z = a exp become zs = a exp + exp− s ± ± a

Continuum Mechanics 20/05/2013 Romain Teyssier Acyclic flow past a plate

Leading edge

Trailing edge Zs

We use the previous results, taking c a → z z 2 f(z)=U z expiα 2i sin α + a2 ∞ − 2 2 − The stagnation points are on the x-axis z = 2a cos α s ± The complex velocity is given by w(z)=W (Z)M (z) The velocity at the leading and trailing edges is: Z = a W ( a)= 2iU sin α w( 2a) ± ± − ∞ ± →∞ (see flow past an edge). This is unphysical !

Continuum Mechanics 20/05/2013 Romain Teyssier Flow past a plate with circulation

from P. Huerre’s lectures

Continuum Mechanics 20/05/2013 Romain Teyssier Flow past a plate with circulation

On the original circular cylinder, we have: iα iα 2 exp iΩ Z F (Z)=U Z exp− +a log ∞ Z − 2π a 2 iα a i2α iΩ W (Z)=U exp− 1 exp ∞ − Z2 − 2ΠZ Ω The stagnation points are now defined by sin (θ α)= s − 4πU a ∞ We still have an infinite number of solution, depending on the value of the point vorticity. For a particular value of the circulation, the stagnation point will coincide with the trailing edge, therefore removing the singularity.

Ωc = sin α4πU a − ∞ For a given body shape, we always choose the critical circulation as defining the unique physical solution.

Continuum Mechanics 20/05/2013 Romain Teyssier The Kutta condition

Initially, we have zero circulation

Starting vortex produces vorticity

Kelvin’s theorem

«A body with a sharp trailing edge which is moving through a fluid will create about itself a circulation of sufficient strength to hold the rear stagnation point at the trailing edge.»

Continuum Mechanics 20/05/2013 Romain Teyssier The Joukowski profiles

We consider now the more general case of a circular cylinder for which the center has been offset from the origin. iα iα 2 exp iΩ Z b F (Z)=U (Z b)exp− +a log − ∞ − Z b − 2π a 2 − iα a i2α iΩ W (Z)=U exp− 1 exp ∞ − (Z b)2 − 2π(Z b) − − Recipe: using the Kutta condition, we impose the singular trailing edge to be a stagnation point. By adjusting b, we remove the singularity at the leading edge.

Continuum Mechanics 20/05/2013 Romain Teyssier Critical circulation for Joukowski profiles

The trailing edge Z =+ c is imposed to be a stagnation point. 2 iα a i2α iΩ U exp− 1 exp =0 ∞ − (c b)2 − 2π(c b) − − iβ Since b is the center of the cylinder, we can define the angle c b = a exp− −

Ωc = 4πU a sin (α + β) − ∞

Continuum Mechanics 20/05/2013 Romain Teyssier Flow past an arbitrarily shaped cylinder

We now consider the inverse problem: we know the shape of the cylinder and we would like to find the conformal mapping to a circular cylinder. Any analytic complex function can be expanded in its Laurent series around the origin. We restrict ourselves to mapping for which points at infinity are invariants. ∞ an 1 M(z)=z + where n+1 zn an = M(z)z dz n=0 2πi l The general flow around the circular cylinder is given by the potential iα iα 2 exp iΩ Z F (Z)=U Z exp− +a log ∞ Z − 2π a Injecting the mapping for Z and Taylor expanding around infinity, we get:

iα iΩ z ∞ bn f(z)=U z exp− log + and ∞ −2π a zn n=0 iα iΩ ∞ nbn w(z)=U exp− ∞ −2πz − zn+1 n=1 The general flow is uniform to leading order, then a vortex flow to next order, then a doublet flow to higher order, and so on... The circulation on the new body is C = w(z)dz = W (Z)dZ = Ω l L

Continuum Mechanics 20/05/2013 Romain Teyssier The Kutta-Joukowski theorem

We now compute the force acting on the arbitrarily shaped body. i We have the Clausius formula F iF = ρ w2(z)dz x − y 2 l

2 2 2iα iα iΩ ∞ cn The kinetic energy is expanded as w (z)=U exp− U exp− + ∞ − ∞ πz zn n=2 dz dz We have (residue theorem) =2 i π and = 0 for n 2 z zn ≥ L L

iα The force is for any profile Fx iFy = iρU exp− Ω − ∞ We recover the force acting on the circular cylinder. General results:

- no drag Fx =0

- without circulation F y =0 (d’Alembert’s paradox). 2 The force on a general Joukowski profile is Fy =4πρU a sin (α + β) ∞

Continuum Mechanics 20/05/2013 Romain Teyssier Lift coefficient

The lift coefficient is a dimensionless number that measures the performance of a wing profile (L is the length of the wing section). F C = y y 1 2 2 ρU L ∞ For a Joukowski profile with small attack angle and small bending angle,

a C =8π sin (α + β) 2π(α + β) y L

The theory disagrees more and more with the experiment: we have neglected viscous effects. It breaks down completely above 10 degrees. This is because the zero streamline is detaching from the wing.

Continuum Mechanics 20/05/2013 Romain Teyssier