Continuum Mechanics Lecture 7 Theory of 2D potential flows

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1 Continuum Mechanics ecture 7 Theory of 2D potential flows Prof.

2 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

3 2D potential flows From the Helmholtz decomposition, we have 2D flows are defined by z ( ) =0 and v z =0. 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. v = φ + A v x = x φ v x = y ψ v y = y φ v y = x ψ where the velocity potential satisfies the aplace equation. where the stream function satisfies the aplace 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.

4 Isopotential curves and stream lines The velocity field is defined equivalently by two scalar fields v x = x φ = y ψ v y = y φ = x ψ They are conjugate functions that satisfy the Cauchy-Riemann relations. They are also harmonic functions (aplace equation), with however different B. C. φ =0 in S with φ n =0 on the boundary ψ =0 in S with ψ t =0on the boundary or ψ = constant along Isopotential curves are defined by Stream lines are defined by dφ =dx x φ +dy y φ =0 dψ =dx x ψ +dy y ψ =0 or or dy dx = v x v y dy dx = v y v x Isopotential curves and stream lines are orthogonal to each other. x ψ x φ + y ψ y φ =( v y )v x +(+v x )v y =0

5 Complex potential and complex derivative We define the complex potential where z = x + iy and i 2 = 1. F (z) =φ(x, y)+iψ(x, y) 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. uckily, since the velocity potential and the stream function are conjugate, the complex velocity potential is differentiable. We define the complex velocity w(z) = df dz where the complex derivative is defined as df dz = F x = 1 i F y We obtain w(z) = x φ + i x ψ = y ψ i y φ = v x iv y In cylindrical coordinates, we have v r = r φ = 1 r θψ v θ = 1 r θφ = r ψ and the complex velocity writes w(z) =v x iv y =(v r iv θ )exp iθ

6 Uniform flow Complex potential F (z) =U exp iα z Complex velocity w(z) =U exp iα Velocity field v x = U cos(α) v y = U sin(α) Potential lines Streamlines Velocity potential Stream function φ = U cos(α)x + U sin(α)y ψ = U cos(α)y U sin(α)x

7 Stagnation flow F (z) =Cz 2 Streamlines are hyperbolae. φ = C(x 2 y 2 ) ψ =2Cxy v x =2Cx v y = 2Cy In polar coordinates φ = Cr 2 cos(2θ) ψ = Cr 2 sin(2θ) This potential can also be used to describe a flow past a corner.

8 Flow past an edge Complex potential F (z) =c z = cr 1/2 exp iθ/2 Velocity potential φ = c r cos θ 2 Stream function ψ = c r sin θ 2 Velocity field v r = c 1 cos θ 2 r 2 v θ = c 1 sin θ 2 r 2 The velocity field becomes infinite at the tip of the edge.

9 Flow around a source or a sink Complex potential F (z) = m 2π log (z z 0) Complex velocity w(z) = m 1 2π z z 0 φ = m 2π log r ψ = mθ 2π Velocity field v x = m x x 0 2π r 2 v y = m 2π v r = y y 0 r 2 In polar coordinates m 2πr v = r v r + v r The velocity divergence is zero everywhere r =0 for r>0. We apply the divergence theorem to a circle centered on the singularity: v ndl = v r 2πr = m 1 Rmax m 2 dr S 2 v2 dxdy = R min 4π r = m2 4π log Rmax The kinetic energy in the flow is R min In a real flow, the singularity is usually embedded inside the boundary condition.

10 Flow around a point vortex Complex potential F (z) = i Ω 2π log (z z 0) Complex velocity w(z) = i Ω 1 2π z z 0 φ = Ωθ 2π ψ = Ω 2π log r Velocity field v x = Ω y y 0 2π r 2 v y = Ω x x 0 2π r 2 In polar coordinates v θ = Ω 2πr e z =0 The velocity curl is zero everywhere v = r v θ + v θ r We apply the curl theorem on a circle centered on the singularity: v tdl = v θ 2πr = Ω 1 Rmax m 2 The kinetic energy in the flow is S 2 v2 dxdy = R min 4π There is a direct analogy with the energy of dislocations in a solid. for r>0. dr r = m2 4π log Rmax R min

11 Superposition principle and boundary conditions ike for the Navier equation for thermoelastic equilibrium problems, the aplace 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!

12 Flow around a doublet We superpose a source and a sink F (z) = m 2π (log (z z 0) log (z + z 0 )) very close to each other Taylor expanding, we find F (z) m exp iα = µ expiα π z z Parameter µ is called the doublet strength. For α =0, we find φ = µx r 2 ψ = µy r 2 cos θ = µ r = µ sin θ r The velocity field is given by It is a dipole field v = µ r r 3 z 0 = exp iα v r = r φ = µ cos θ r 2 Potential and streamlines are circles. and v θ = 1 r θφ = µ sin θ r 2 The general form around an arbitrary center z0 is: F (z) = µ expiα z z 0

13 Flow past a cylinder We superpose a uniform flow and a doublet. F (z) =U z + µ z We find φ = U x + µ x r 2 ψ = U y µ y r 2 µ The streamline ψ =0 is the circle r = We reverse engineer the process. For a cylinder a radius a, if we define U µ = U a 2 then the potential flow around the cylinder is F (z) =U z + a2 z The velocity field is given by v r = U 1 a2 cos θ v θ = U 1+ a2 sin θ r 2 S S 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. r 2

14 Force acting on the cylinder Using the second Bernoulli theorem (curl-free, incompressible, no gravity), we know that the quantity H = P is uniform. ρ v2 We thus have p ρv2 = p ρu 2 The force acting on the cylinder is given by F = pndl On the cylinder, we have v r =0 and v θ = 2U sin θ p = p ρu 2 The pressure field on the cylinder is thus 1 4 sin 2 θ Using n = (cos θ, sin θ) F x = 1 2 ρu 2 a F y = 1 2 ρu 2 a 2π 0 2π 0 we find: 1 4 sin 2 θ cos θdθ 1 4 sin 2 θ sin θdθ sin θ 4 2π 3 sin3 θ 3 cos θ 4 3 cos3 θ 0 2π 0 =0 =0 Exercise: compute the torque on the cylinder (use the cylinder axis). It is also zero!

15 We superpose a uniform flow, a doublet and a vortex. F (z) =U z + a2 iω z z 2π log a Streamlines are given by ψ = U 1 a2 r 2 y Ω r 2π log a Flow past a cylinder with vorticity The cylinder r=a is still a proper boundary condition. v r = U 1 a2 r 2 cos θ v θ = U 1+ a2 r 2 sin θ + Ω 2πr On the cylinder, we have to stagnation point given by or one stagnation point away from the cylinder if At the boundary, we have v θ = 2U (sin θ sin θ s ) Using the Bernoulli theorem and integrating the pressure field on the boundary, we can compute the force on the cylinder (exercise) sin θ s = Ω < 4πU a F x =0 Ω 4πU a F y = ρu Ω

16 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!

17 F = pndl We use curvilinear coordinates along the body t = (cos θ, sin θ) n =(sinθ, cos θ) The force components are F x = p sin θdl F y = The complex force: Blasius formulae p cos θdl In Cartesian coordinates, we have dx = cos θdl dy =sinθdl The complex force is defined as F x if y = F x n t p (dy + idx) = i pdz F y Bernoulli theorem: p = p ρu ρv2 with v 2 = w(z)w(z) Boundary condition: v n =0 v x dy v y dx =0 and w dz = wdz We finally get the force for an arbitrary shaped body: F x if y = i 2 ρ w 2 (z)dz

18 We consider an arbitrary closed contour in the complex plane. We define the complex circulation as C = w(z)dz Using the same definition as before along the contour, we have C = (v x dx + v y dy)+i (v x dy v y dx) where the Cartesian coordinates are related to the curvilinear ones by We finally get C = The complex circulation dx = dl cos θ v tdl + i Γ is the physical circulation and Q is the physical mass flux. dy = dl sin θ v ndl = Γ + iq On the contour defining the body shape, the mass flux is zero and we have C = Γ = v tdl

19 A conformal mapping is a differentiable complex function M that maps the complex plane z into another complex plane Z. We have Z = M(z) and z = m(z) with m = M 1 If a flow is defined by a potential function Conformal mapping We need to build more complex profile than just a cylinder. We use for that a mathematical trick called conformal mapping. f(z) F (Z) =f(m(z)) in the z plane, then the function 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. The new complex velocity writes W (Z) = df dz = df dm dz dz = w(z)m (Z) The complex circulation is conserved by conformal mapping C = W (Z)dZ = w(z)m (Z)dZ = w(z)dz l

The Joukowski transform Definition: z = Z + c2 Z The circle of radius c becomes the line segment [-2c, 2c] c Z 2c 2c z Z = c exp iθ z =2c cos θ Z z The circle of radius a>c becomes

20 The Joukowski transform Definition: z = Z + c2 Z The circle of radius c becomes the line segment [-2c, 2c] c Z 2c 2c z Z = c exp iθ z =2c cos θ Z z The circle of radius a>c becomes an ellipse. Z = a exp iθ z =(a + c2 a ) cos θ + i(a c2 a )sinθ The inverse transform is Z = z 2 + z 2 2 c 2 The derivative M (z) = z 2 z 2 2 c 2 has 2 singular points at z = ±2c

21 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 F (Z) =U Z exp iα +a 2 expiα W (Z) =U Z exp iα 1 a2 Z 2 expi2α Using the Joukowski mapping ellipsoidal cylinder. Z = M(z) with a>c, we get the potential around an Using c 2, we get Z = z Z The original stagnation points f(z) =U Z s = ±a exp iα z a2 c 2 expiα +M(z) exp iα a2 c 2 expiα become z s = ± a exp iα + c2a exp iα

22 Acyclic flow past a plate eading edge Z s Trailing edge z s We use the previous results, taking c a f(z) =U z exp iα z z 2 2i sin α 2 + a 2 2 The stagnation points are on the x-axis The complex velocity is given by z s = ±2a cos α 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!

23 Flow past a plate with circulation from P. Huerre s lectures

24 Flow past a plate with circulation On the original circular cylinder, we have: F (Z) =U Z exp iα +a 2 expiα Z W (Z) =U exp iα 1 a2 Z 2 expi2α The stagnation points are now defined by iω 2π log iω 2ΠZ sin (θ s α) = Z a Ω 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. Ω c = sin α4πu a For a given body shape, we always choose the critical circulation as defining the unique physical solution.

25 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.»

26 The Joukowski profiles We consider now the more general case of a circular cylinder for which the center has been offset from the origin. F (Z) =U (Z b)exp iα +a 2 expiα iω Z b Z b 2π log a W (Z) =U exp 1 iα a 2 iω (Z b) 2 expi2α 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.

27 Critical circulation for Joukowski profiles The trailing edge Z =+c U exp iα 1 a2 (c b) 2 expi2α is imposed to be a stagnation point. Ω c = 4πU a sin (α + β) iω 2π(c b) =0 Since b is the center of the cylinder, we can define the angle c b = a exp iβ

28 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 aurent series around the origin. We restrict ourselves to mapping for which points at infinity are invariants. a n M(z) =z + z n a n = 1 where M(z)z n+1 dz 2πi n=0 The general flow around the circular cylinder is given by the potential F (Z) =U Z exp iα +a 2 expiα iω Z Z 2π log a Injecting the mapping for Z and Taylor expanding around infinity, we get: f(z) =U z exp iα iω z 2π log b n + a z n and n=0 w(z) =U exp iα iω 2πz nb n z n+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

29 We now compute the force acting on the arbitrarily shaped body. We have the Clausius formula F x if y = i 2 ρ w 2 (z)dz The kinetic energy is expanded as We have (residue theorem) The force is for any profile and We recover the force acting on the circular cylinder. General results: - no drag F x =0 The Kutta-Joukowski theorem - without circulation F y =0 (d Alembert s paradox). The force on a general Joukowski profile is w 2 (z) =U 2 exp 2iα iα iω U exp πz + dz z =2iπ F x if y = iρu exp iα Ω l dz z n = 0 for n 2 F y =4πρU 2 a sin (α + β) n=2 c n z n

30 ift coefficient The lift coefficient is a dimensionless number that measures the performance of a wing profile ( is the length of the wing section). C y = F y 1 2 ρu 2 For a Joukowski profile with small attack angle and small bending angle, C y =8π a sin (α + β) 2π(α + β) 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.

Exercise 9: Model of a Submarine

Exercise 9: Model of a Submarine Fluid Mechanics, SG4, HT3 October 4, 3 Eample : Submarine Eercise 9: Model of a Submarine The flow around a submarine moving at a velocity V can be described by the flow caused by a source and a sink with