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Fluid Dynamics - Math 6750 Vorticity and Potential Flows - Week 8

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Fluid Dynamics - Math 6750 Vorticity and Potential Flows - Week 8 Fluid Dynamics - Math 6750 Vorticity and Potential Flows - Week 8 1 Vorticity and potential flow The nonlinear term of the convective derivative can be expressed using a vector identity as 1 u · ru = r juj2 − u × (r × u): 2 We recall that the vorticity field is defined by ! = r × u and that if the vorticity field is identically zero, then the fluid is irrotational. From Calculus, we know that a field is irrotational if and only if it is a potential flow, i.e. ! = 0 , u = rΦ: Definition 1 (Potential flow.). A potential flow is a flow that can be represented as u = rΦ, where Φ is a scalar function called the velocity potential. Given an irrotational vector field u defined on some fluid region D, we may define a potential φ: D −! R as the line integral Z P φ(P ) = u · dx; O where O 2 D is some arbitrary reference point. For simply-connected fluid region D, the potential φ is independent of the path between O and P . Consequently, φ is a well-defined function and we have u = rφ. If u is irrotational and incompressible, then the corresponding velocity potential φ must be harmonic since r · u = r · rφ = ∆φ = 0: Definition 2 (Circulation.). The circulation of u on C (C simple, smooth, oriented, closed contour) is I ΓC = udγ: C Here the integral is understood as a line integral along C. Remark 1. Stokes' theorem applied to the velocity field yields Z ΓC = n · !dS; S where S is the surface enclosed by C. The right-hand side is sometimes referred to as the flux of vorticity through S. Example 1. If u is a potential flow, then ΓC = 0 on any well behaved closed curve. Remark 2. Since r × rΦ = 0 for any gradient field, the vorticity of a potential flow vanishes and irrotational flow are most always refer to as potential flow. 1 Example 2. Consider a rigid body rotation with angular velocity Ω. The flow field is given by u = (0; 0; Ω) × (x; y; z) = (−Ωy; Ωx; 0) and ! = r × u = 2(0; 0; Ω) = 2Ω 6= 0: Example 3. Consider the two-dimensional irrotational vortex with Ω = (0; 0; α=r2). Then αy αx u = − ; ; 0 r2 r2 and one can verify that ! = 0. Example 4. Consider the stagnation point flow u = (αx; −αy; 0): One can verify that ! = 0 and there exists a velocity potential φ such that u = rφ, with α φ(x; y) = (x2 + y2); 2 up to some additive constant. k Example 5. Consider the line vortex flow u = r eθ, with eθ the unit vector in polar coordinates defined by eθ = (− sin θ; cos θ; 0): Using the gradient operator in polar coordinates, the velocity potential φ, if it exists, satisfies 1 k @ φ = 0; @ φ = ;@ φ = 0: r r θ r z Up to some additive constant, we find that φ = kθ. This doesn't contradict the fact we 3 mentioned above since u is defined on R n f0g which is not simply-connected. For closed curves that don't go around the origin, the line integral is zero. But for the unit circle centered at the origin, the line integral is I u · dx = 2πk: C 2 Inviscid vortical flow A barotropic fluid is an example of a compressible fluid. Definition 3. A barotropic fluid is defined by specifying the pressure as a given function of the density, i.e. p = p(ρ). Theorem 1 (Kelvin). Let C(t) be a simple closed material curve in an inviscid fluid with body force −ρrΦ (conservative). If either ρ =constant or the fluid is barotropic, then the circulation is invariant under the flow: D Γ = 0 (1) Dt C(t) 2 Proof. We parameterise C(t) as C(t) = fx(s; t): 0 ≤ s ≤ 1g: Then ! d d Z d Z 1 @x(s; t) Γ(t) = u(x; t) dx = u(x(s; t); t) · ds dt dt C(t) dt 0 @s Z 1 Du @x(s; t) Z 1 @u(x(s; t); t) = · ds + u(x(s; t); t) · ds 0 Dt @x 0 @s Z 1 Du @x(s; t) Z 1 1 @ = · ds + fu(x(s; t); t) · u(x(s; t); t)g ds: 0 Dt @x 0 2 @s The second integral vanishes since C(t) is a closed curve. Rewriting the material derivative of u using Euler equation, we see that d Z 1 p @x(s; t) Γ(t) = −∇ + χ · ds dt 0 ρ @s Z p = −∇ + χ · dx = 0; C(t) ρ since we are integrating a conservative vector field over a closed curve. Remark 3. Kelvin's theorem allows for a direct Lagrangian description of the dynamics of vorticity, even though vorticity is defined as an Eulerian variable. Remark 4. If a body is moving through an inviscid, incompressible fluid, or if a uniform flow of such a fluid passes around a body, then the vorticity far from the body will be zero. Then, from Kelvin's theorem, the vorticity in the fluid will be zero everywhere. Remark 5. The theorem does not require the fluid region to be simply connected. The Euler equation enters the proof in evaluating the line integral around C, so the theorem continues to hold even if the viscous effect happens to be important elsewhere in the flow. Definition 4. A vortex line is a curve which is everywhere tangent to the vorticity !(x; t). The vortex line which pass through some simple closed curve in space are said to form the boundary of a vortex tube. A consequence of the Kelvin's theorem is the Helmholtz's theorem. Theorem 2 (Helmholtz). 1. The strength of a vortex tube does not vary with time. A vortex tube is a collection of vortex line (a line whose tangent is parallel to the voracity) through a closed contour moving through the fluid. 2. Vortex lines move with the fluid. A vortex cannot end in the fluid. It must form a ring, end at a boundary or go to infinity. 3. Fluid elements initially free of vorticity remain fluid of vorticity. 3 Proof. Let us define a vortex surface as a surface such that ! is everywhere tangent to the surface. Suppose that at t = 0, the vortext line is the intersection of two vortex surfaces S1 and S2. On each of these surfaces, !(x; 0) · ni = 0 where ni is normal to Si, i=1,2: We claim that both S1;S2 remain as vortex surfaces as time advances, i.e. as the points on Si move with the fluid, the surface they comprise at any time t is a vortex surface. Choose ∗ any arbitrary closed material curve C1(t) that encloses a surface S1 (t) ⊂ S1(t). It follows from Stokes theorem that Z Z u · dx = ! · n dS = 0: ∗ C1(0) S1 (0) ∗ This quantity remains zero by virtue of Kelvin's circulation theorem. Since C1(t) and so S1 (t) was arbitrary, this shows that ! · n1 = 0 on S1(t) and an identical argument shows that ! · n2 = 0 on S2(t). We proceed to prove the second statement. Consider any two arbitrary cross sections A1;A2 of a vortex tube and consider the vortex tube bounded by these two cross sections. Since u is incompressible, r · ! = r · (r · u) = 0 and it follows from the divergence theorem that Z Z Z 0 = r · ! dx = ! · n dS = ! · n dS: V @V A1[A2 Let C(t) be a closed material curve which lie on the wall of the vortex tube, enclosing a closed region A(t). Then Z Z !(x; t) · n(x; t) dS = u(x; t) · dx: A(t) C(t) It follows from Kelvin's circulation theorem that Γ remains constant as time proceeds. Example 6 (Vorticity generation near a wall). Consider a one dimensional shear flow near a wall. The viscous shear stress is given by τx = µ (@xv − @yu) = −µωz: Recall that for a Newtonian fluid and incompressible flow, the vorticity equation has the form (ν = µ/ρ) D ! = ! · ru + νr2!: (2) Dt Remark 6. If the flow is steady, then in dimensionless form the vorticity equation is 1 u · r! = ! · ru + r2!: Re The left-hand side represents the advection of vorticity by u, and the second term on the right- hand side represents the transport of vorticity by diffusion. The first time on the right-hand side is the production term associated with the stretching of vortex lines. 4 We now state some other versions of the vorticity equation for inviscid flows. Theorem 3. For an inviscid fluid of constant density, the vorticity equation is D ! = ! · ru: (3) Dt For a barotropic inviscid fluid, the vorticity equation is D ! Dρ ! = ! · ru + : (4) Dt ρ Dt Proof. Assuming that the body force is a potential force, f = −∇Φ, the conservation of linear momentum equation can be written as Du 1 + rp + rΦ = 0: Dt ρ Taking the curl of the previous equation and using the fact that the curl of a gradient field is zero yield Du r × = 0: Dt Expressing the nonlinear term with the vorticity and using the conservation of mass equation Dρ Dt + ρr · u = 0, we find 1 0 = r × @ u + r × (u · ru) = @ ! + r × rjuj2 − u × ! t t 2 = @t! − ! · ru + u · r! − ur · ! + !r · u D! ! Dρ = − ! · ru − : Dt ρ Dt Here we used the vector identity r × (a × b) = b · ra − a · rb + ar · b − br · a and the facts that the divergence of vorticity is zero as well as the curl of a gradient field.
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