Homework Two. Abstract: Liu. Solutions for Homework Problems Two: (50 points total). Collected by Junyu

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1 Homework Two Abstract: Liu. Solutions for Homework Problems Two: (50 points total). Collected by Junyu

2 Contents 1 BT Problem (8 points) (by Nick Hunter-Jones) 1 2 BT Problem 14.2 (12 points: ) (by Nick Hunter-Jones) 1 3 BT Problem 14.8 (ab: 8 points: 4+4) (By Nick Hunter-Jones) 3 4 BT Problem (6 points: 2+2+2) (by Nick Hunter-Jones and Sterl Phinney) 4 5 NONCOLLABORATION Problem (16 points: ) (by Nick Hunter-Jones) 6 1 BT Problem (8 points) (by Nick Hunter-Jones) Consider a velocity field v which specifies a flow with nonvanishing vorticity ω = v. We orientate the orthonormal basis in the tangent space at a point p in v, such that ˆx points in the direction of ω p. In the local frame at the point p, we can understand the motion of the basis vectors in the y-z plane as the gradient over small displacements. Consider the vector v z at a point y and at y + δy, and the vector v y at points z and at z + δz. The rotations of ŷ and ẑ are Ωŷ = v z(y + δy) v z (y) δy = v z y, and Ω ẑ = v y(z + δz) v y (z) = v y δz z, (1.1) where we chose a positive direction of rotation. As the local frame is orientated so that the ω points in the ˆx direction, we have that the vorticity is the sum of the angular velocities ω = ωˆx = ( v) x = v z y v y z = Ω ŷ + Ωẑ. (1.2) 2 BT Problem 14.2 (12 points: ) (by Nick Hunter-Jones) a) Consider the two-dimensional flow defined by the vector field v = (2xy, x 2 ). We see that the flow is 1 compressible : v = 2y, with zero vorticity : ω = v = 0. (2.1) 1 A quick note: while in three dimensions, the curl of a vector is a vector ( v) k = ɛ ijk jv k, in two dimensions the curl of a vector is a scalar ( v) = ɛ ij iv j = xv y yv x. 1

3 Integral curves of a vector field are curves whose tangents at any point p along the curve are given by the vector field at that point. For a curve parametrized by λ, d x(λ)/dλ = v(x(λ), and at a point along the curve, x/ λ = v x and y/ λ = v y. Solving, we find that the integral curves are hyperbolas x y = v x v y = 2y x b) Now consider the flow defined by v = (x 2, 2xy). The flow is x 2 2y 2 = c. (2.2) incompressible : v = 0, with vorticity : ω = v = 2y. (2.3) We can solve for the form of the integral curves of the vector field x y = v x v y = x 2y y = c x 2. (2.4) c) In 2d polar coordinates, the divergence and curl of a vector are v = ( ) v θ rvr + r θ, and v = 1 ( ) 1 v r rvθ r r r θ. (2.5) The 2d flow in polar coordinates v = (0, r) is incompressible : v = 0, with vorticity : ω = v = 2. (2.6) The integral curves of the vector field are lines of constant r, i.e. concentric circles, r/ φ = 0. d) The 2d flow in polar coordinates v = (0, 1/r) is incompressible : v = 0, with zero vorticity : ω = v = 0. (2.7) meaning zero vorticity v = 0 everywhere except the origin r 0. The integral curves are again, lines of constant r, concentric circles. Stokes theorem in 2d tells us 2π da( v) = dl v = rdφ 1 = 2π (2.8) r so we have v = 2πδ( r). S We can plot the vector fields v = (2xy, x 2 ), and v = (x 2, 2xy)

4 and the vector fields v = (0, r), and v = (0, 1/r) BT Problem 14.8 (ab: 8 points: 4+4) (By Nick Hunter-Jones) a) Consider an airfoil in a steady flow. Ignoring vorticity and gravity, Bernoulli s law tells us that B t = B b on the top of the wing and bottom of the wing P t ρv2 t = P b ρv2 b P = 1 2 ρ(v2 b v2 t ) ρv v. (3.1) The lift due to the pressure difference is F = da P = da ρv v. (3.2) Assuming the invariance of the flow along the length of the airfoil, we evaluate the line integral ( ) F = Lρv dl v = Lρv dx v t dx v b = LvρΓ, (3.3) t b where the circulation Γ, is the integral around a closed curve S, Γ = S dl v. In order to conserve momentum, the force up on the wing must push down on the air that is generating the lift, which is why we want the air directed downwards at the end of the wing. b) Equal and opposite to the lift on the plane is a downward force on the air. Consider the circulation near the edge of the wing. The air from the high pressure region underneath the wing will spill over to the low pressure region on top of the wing. This circular motion generates a vortex that moves behind the aircraft. To see why this must be the case consider the surface above the boundary layer around the wing. The circulation around S is the flux of vorticity through S, Γ = S dσ ω, which essentially counts the vortex lines through S. The line integrals along either side of the wing will not differ to much for the steady flow, so F Γ would be small. For the lift to be large, we must have a large Γ and thus many vortex flux lines penetrating the surface. Wingtip vortices are essential to getting substantial lift. 3

5 c) Birds in flight also have wingtip vortices. To experience lift, there must also be a downwards force on the air. The air near the wingtip will flow circularly over the wing to the low pressure region on top, which generates vortices moving behind the bird, with vortex lines along the air flow. Looking at the bird from behind, the left vortex will spin clockwise and the right vortex counter clockwise. Therefore, to get an additional lift from the upward motion of the vortex, you want to be on the outside, behind and at a wide angle, of the bird in front of you. In the V formation, the birds seem to make use of this additional lift. d) We can sail into the wind by catching it in a sail. This works the same as with the airfoil; by catching the wind we create a low pressure over the outside of the sail and experience a lift normal to the sail, i.e. forward. As long as we are not angled too much, we can tack back and forth upwind. e) Moving through the water, by curving or undulating their bodies or fins, animals can direct the flow such that they experience lift and can propel themselves upstream. 4 BT Problem (6 points: 2+2+2) (by Nick Hunter-Jones and Sterl Phinney) Sterl used turmeric powder in a bowl of water to do the experiment. The three photos are three frame grabs from the video of the experiment. Spoon is moving right to left. The first two are from a single spoon pass, about a second apart. The spoon was moving slowly. The third frame is from a second spoon pass, when we were moving the spoon more quickly, so the vortices were spinning faster and being shed to larger distances. The light wasn t great, but that turned out useful since the turmeric particles in the vortices and wake streaked into particle path-lines because they moved during the exposure time. 4

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7 i) When we move the spatula through the water, vortices are generated due to the rapid deceleration of the fluid in a thin boundary layer along the surface, which introduces circulation and vorticity into the flow. At the edges of the spatula, where the boundary layer extents into the bulk flow and thickens, the shearing in the boundary layer gives rise to vorticity in the bulk flow. ii) The vortices created by the movement of the spatula through the water is in some sense self stabilizing. There are a number of stable vortex pair configurations, but this is fairly easy to observe in a sink. The object creates a vortex/anti-vortex pair, equal and opposite vortices, propagating through the water with a constant speed which is typically the magnitude and direction of the outside motion. In the frame of the vortex pair, the fluid between the vortex pair is at rest. 2 5 NONCOLLABORATION Problem (16 points: ) (by Nick Hunter-Jones) Consider an ideal fluid flow which has constant entropy, zero vorticity, and is incompressible. As v = 0, then we can define the velocity potential v = Ψ, such that the zero vorticity condition is always satisfied. a) Taking the divergence of Ψ, so that 2 Ψ = 0, we can equivalently say that any function Ψ which satisfies the Laplace equation, represents an irrotational incompressible flow. For a flow around a body, the velocity normal to the surface must vanish as the fluid is not flowing into or out of the body, and at the surface ˆn v = ˆn Ψ = 0. Thus, irrotational incompressible velocity potentials are harmonic functions (satisfying the Laplace equation) with Neumann boundary conditions Ψ/ n = 0 on the boundary. As the Laplace equation and the Neumann boundary condition are linear in Ψ, linear combinations of solutions are also solutions. b) Consider the flow due to the velocity potential Ψ s (r) = m/r, satisfying 2 Ψ s (r) = 1 ( r 2 r 2 r r As v = Ψ = mˆr/r 2, the flux of the fluid flow out of a sphere S is 3 S da v ˆn = π π ) m = 0 for r 0. (5.1) r 2π dθ dφ r 2 sin θ m = 4πm. (5.2) 0 r2 If m > 0, the flux is out of the surface and Ψ s describes a source of fluid flow. If m < 0, then a flux is into the surface and Ψ s describes a sink. 2 Vortices are important phenomena in many areas of physics, including nonperturbative effects in field theories, in superconductors, etc. But it might be of interest to note that the proliferation of vortex/antivortex pairs plays the vital role the Kosterlitz-Thouless transition, a topological phase transition, for which John Kosterlitz and David Thouless were awarded 2016 Nobel prize. 3 Equivalently, R da v ˆn = R dv v = R dv 2 Ψ = 4πm as 2 (1/r) = 4πδ(r). 6

8 c) Consider the velocity potential Ψ us (r) = Ux m/r, defining a velocity field v = Ψ us = U ˆx + m x. (5.3) r3 We see that far away r, there is a uniform flow in the ˆx direction. The stagnation point, where the velocity vanishes, occurs at z = y = 0 and, from v x = U + mx/r 3, at x = r s = m/u, (5.4) where we define the stagnation point on the x-axis r s. The flow has axial symmetry along the x direction and the streamlines sourced at the stagnation point define a surface which separates the fluid from that flowing from x direction and the fluid sourced at the origin. Think about a cigar, where the tip sits at the stagnation point, with a flow outside and a source inside the cigar. d) If we were to replace the streamlines sourced at the stagnation point, which separates the flow inside and outside a cigar as described above, with a solid surface, i.e. like a solid cigar, the flow outside the cigar would be identical to that defined by Ψ us. e) The source potential Ψ s from part b, can be differentiated to find the dipole potential Ψ s / x = mx/r 3, which itself defines a flow. 4 We now consider the linear combination of a uniform flow and a dipole Ψ ud = Ux + µx/r 3 = U(r + µ/ur 2 ) cos θ, which defines the velocity field ( v = U 1 2µ ) ( Ur 3 cos θ ˆr U 1 + µ ) Ur 2 sin θ ˆθ. (5.5) As a sanity check, r, v U cos θ ˆr U sin θ ˆθ = U ˆx. The radial component of the velocity vanishes on the sphere at radius a = (2µ/U) 1/3, i.e. a surface where the velocity normal to that surface is zero. The ideal inviscid flow defined by the dipole in a uniform v-field is that of the flow around a sphere in a uniform field. f) For 2d flows, where incompressibility v = x v x + y v y = 0, we can define a stream function Φ(x, y) where x Φ = v y and y Φ = v x, which automatically satisfies the incompressibility condition. Equivalently, for a 3d flow with axial symmetry (no φ dependence) v = v(r, θ), we can define a stream function Φ(r, θ), where 1 Φ r 2 sin θ θ = v r, and 1 Φ r sin θ r = v θ, (5.6) such that v = 0 is identically satisfied. Thus for the Ψ ud flow, the stream function defined by 1 Φ ud r 2 sin θ θ = U (1 a3 r 3 ) cos θ, and 1 Φ ud r sin θ r ) = U (1 + a3 2r 2 sin θ, (5.7) 4 This is called the dipole potential because the combined flow from a source m/r 1 and a sink m/r 2, as the distance between them, oriented on the x axis, is taken to be small, gives the potential Ψ = mɛ x(1/r) +... = mɛx/r

9 we find that Φ ud (r, θ) = U 2 ) (r 2 a3 sin 2 θ. (5.8) r At r = a, we have a surface of streamlines, separating the dipole inside the sphere, and the uniform flow moving around the sphere. Contour plots nicely show the spherical region where the streamlines vanish in the flow, created by the dipole, and the uniform flow at longer distances