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.