5.1 Fluid momentum equation Hydrostatics Archimedes theorem The vorticity equation... 42

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1 Chapter 5 Euler s equation Contents 5.1 Fluid momentum equation Hydrostatics Archimedes theorem The vorticity equation Kelvin s circulation theorem hape of the free surface of a rotating fluid Fluid momentum equation o far, we have discussed some kinematic properties of the velocity fields for incompressible and irrotational fluid flows. We shall now study the dynamics of fluid flows and consider changes in motion due to forces acting on a fluid. We derive an evolution equation for the fluid momentum by considering forces acting on a small blob of fluid, of volume and surface, containing many fluid particles Forces acting on a fluid The forces acting on the fluid can be divided into two types. Body forces, such as gravity, act on all the particles throughout, F v = ρgd. urface forces are caused by interactions at the surface. For the rest of this course we shall only consider the effect of fluid pressure. 39

2 40 5. Hydrostatics Collisions between fluid molecules on either sides of the surface produce a flux of momentum across the boundary, in the direction of the normal n. The force exerted on the fluid into by the fluid on the other side of is, by convention, written as F s = pnd, where p(x) > 0 is the fluid pressure. n 5.1. Newton s law of motion Newton s second law of motion tells that the sum of the forces acting on the volume of fluid is equal to the rate of change of its momentum. ince Du/Dt is the acceleration of the fluid particles, or fluid elements, within, one has We now apply the divergence theorem, ρ Du Dt d = pnd + ρgd. ρ Du Dt d = ( p+ρg)d, and notice that both integrands must be identical, since is arbitrary. o, the evolution of fluid momentum is governed by Euler s equation ρ Du [ ] u Dt = ρ t +(u )u = p+ρg. (5.1) This equation neglects viscous effects (tangential surface forces due to velocity gradients) which would otherwise introduce an extra term, µ u, where µ is the viscosity of the fluid, as in the Navier-tokes equation ρ Du Dt = p+ρg+µ u. (5.) For the rest of the course we shall only consider perfect fluids which are idealised fluids, inviscid and incompressible with constant mass density. 5. Hydrostatics We first consider the case of a fluid at rest, such that u = 0. Euler s equation is then reduced to the equation of hydrostatic balance, where C is a constant. p = ρg p(x) = ρg x+c, (5.3) Example 5.1 The density of mass in the ocean can be considered as constant, ρ 0, and the gravity g = gê z. (The coordinate z is the upward distance from sea-level.)

3 Chapter 5 Euler s equation 41 From Euler s equation one has dp dz = ρ 0g p(z) = p 0 ρgz. Hence the pressure increases linearly with depth (z < 0). Taking typical values for the physical constant, g 10ms, ρ kgm 3 and a pressure of one atmosphere at sea-level, p 0 p atm = 10 5 Pa = 10 5 Nm gives p(z) 10 5 (1 0.1z); the pressure increases by one atmosphere every 10m. (Notice that the change in pressure force on a surface, between the ocean surface z = 0 and a given depth z = d, is equal to (p p 0 ) = ρ 0 gd, which is the weight of a column of water of height d and section.) 5.3 Archimedes theorem g ρ 0 g p The force on a body in a fluid is an upthrust equal to the weight of fluid displaced. Consider a solid body of volume and surface totally submerged in a fluid of density ρ 0. The total force on the body caused by the fluid surrounding it is F = pnd, where p is the fluid pressure. g n z O F = ρ 0 g (Notice that the pressure distribution on the surface is the same whether the fluid contains a solid or not.) o, using successively the divergence theorem and the equation of hydrostatic balance, P = ρg, we find F = pd = ρ 0 gd = ρ 0 g. The buoyancy force is equal the weight of the mass of fluid displaced, M = ρ 0, and points in the direction opposite to gravity. If the fluid is only partially submerged, then we need to split it into parts above and below the water surface, and apply Archimedes theorem to the lower section only. Consider a solid body of volume and density ρ s partially submerged in a fluid of density ρ 0 > ρ s. Let 1 be the volume of solid above the fluid surface and the volume underneath. ince, the solid is in equilibrium, its weight is balanced by the buoyancy force, so that ρ s g = ρ 0 g. ρ s ρ 0 g ρ 0 1 ρ s g Hence, the fractions of the volume of solid immersed in the fluid and not immersed are = ρ s ρ 0 and 1 = 1 = ρ 0 ρ s ρ 0, respectively. For an iceberg of density ρ s 0.915kgm 3 floating in salted water of density ρ s 1.05kgm 3, / 89.3% and 1 / 10.7%. Question: A glass of water with an ice cube in it is filled to rim. What happens as ice melts?

4 4 5.4 The vorticity equation 5.4 The vorticity equation In the expression of the acceleration of a fluid particle, Du Dt = u t +(u )u, the nonlinear term can be rewritten using the vector identity ( ) u u ( u) = (u )u (u )u = where u = u u. o, we can rewrite Euler s equation (5.1) as and take its curl u t + u t +(u )u = 1 ρ p+g, ( ) ( u p u ω = ρ ( u ) +g (ρ constant), ω (u ω) = 0, t ω +(u )ω (ω )u+( u)ω ( ω)u = 0. t ) u ω, For incompressible flows u = 0 and, in addition, ω = 0 as ω = u. Hence, Dω Dt = ω +(u )ω = (ω )u. (5.4) t This is the vorticity equation. It shows that the vorticity of a fluid particle changes because of gradients of u in the direction of ω Properties of the vorticity equation. i. If ω = 0 everywhere initially, then ω remains zero. Thus, flows that start off irrotational remain so. ii. In a two-dimensional planar flow, u = (u(x,y),v(x,y),0), the vector vorticity has only one non-zero component, ω = ( v/ x u/ y)ê z, so that (ω )u = ω d u(x,y) = 0. dz Hence, the vorticity equation, reduced to Dω Dt = ω +(u )ω = 0, (5.5) t shows that the vorticity of a fluid particle remains constant. If, in addition the flow is steady, ω/ t = 0 then the vorticity is constant along streamlines.

5 Chapter 5 Euler s equation 43 iii. ortex stretching. The stretching of a vortex leads to the increase of its vorticity. Consider, for example, an incompressible steady flow in a converging cone, function of the radial distance only in spherical polar coordinates, u ϕ u r r u(r) = u r (r)ê r +u ϕ (r)ê ϕ. O The component u r represents a radial inflow and u ϕ the swirling of the fluid. ince (u r (r)ê r ) = 0, only the swirling motion contributes to a non-zero vorticity ω(r) = u = (u ϕ (r)ê ϕ ) = ω r ê r +ω θ ê θ. From mass conservation in an incompressible spherically symmetric inflow, we find u = 1 d ( r ) r u r = 0 ur = k dr r, where k > 0 is constant. ince the flow is steady, ω/ t = 0, the evolution equation for the radial component of the vorticity, ω r, becomes Dω r Dt dω r = u r dr = ω du r r dr d dr ln ω r u r = 0 ω r = α, constant. u r Thus, w r = αu r = αk r, which demonstrates that the vorticity ω r increases as u r increases; the initial vortex is stretched by the inflow. This is the reason for the bath-plug vortex. A small amount of background vorticity is amplified by the flow converging into a small hole. (This mechanism can be interpreted as the conservation of angular momentum of fluid particles.) 5.5 Kelvin s circulation theorem The circulation around a closed material curve remains constant in an inviscid fluid of uniform density, subject to conservative forces. Hence, dt = d u dl = 0, (5.6) dt C(t) if C(t) is a closed curve formed of fluid particles following the flow. Proof. Let C(t) be a closed material curve, hence formed of fluid particles, of parametric representation x(s,t) with s [0,1], say. Using this parametric representation, the rate of

6 hape of the free surface of a rotating fluid change of the circulation around C(t) can be written as dt = d dt = = 0 0 u(x(s,t),t) x(s,t) ds = 0 s 0 {[ ( ) ] u x t + t u x s +u x s t {[ ] u t +(u )u x } u +u ds, s s [ d u(x,t) x dt s } ds, ] ds, since x/ t = u is the velocity of the fluid particle at x(s,t). o, using Euler s equation in the form u ( t +(u )u = pρ ) +g x, we find dt = 0 [ ( pρ ) +g x x s + s Thus, since the curve C(t) is closed, dt = as required. C(t) ( u )] ( u ds = p )ds. 0 s ρ +g x ( u d p ) ρ +g x = 0, Recall that the circulation around a closed curve C is equal to the flux of vorticity through an arbitrary surface that spans C. o, from Kelvin s circulation theorem, dt = d = dt C(t)u dl d ω n d = 0; dt (t) this demonstrates that the flux of vorticity through a surface that spans a material curve is constant. Thus, many properties of the vorticity equation could equivalently be derived from the circulation theorem (e.g. vortex stretching, persistence of irrotationality.) 5.6 hape of the free surface of a rotating fluid The surface of a rotating liquid placed in a container is not flat but dips near the axis of rotation. This phenomenon, which can be observed when stirring coffee in a mug, results from a radial pressure-gradient balancing the centrifugal force acting within the fluid. Consider, for example, a cylindrical container partially filled with fluid and mounted on a z horizontal turntable. When the flow reaches Ω a steady state, u/ t = 0, the fluid rotates uniformly, with a constant angular velocity Ω about the vertical z-axis. (The fluid rotates with the container as a solid body.) h 0 h(r) In order to calculate the height of the free surface of fluid, z = h(r), we shall solve Euler s O r equation, ( ) u ρ t u ω = ρu ω = (p+ 1 ) ρ u +ρg, (5.7)

7 Chapter 5 Euler s equation 45 in cylindrical polar coordinates. The velocity and vorticity fields of the fluid in uniform rotation are u = u θ ê θ and ω = ω z ê z, where u θ = rω and ω z = Ω. Hence, the nonlinear term in Euler s equation u ω = u θ ω z ê r = rω ê r. The radial component of the vector equation (5.7) (i.e. its scalar product with ê r ) is therefore ρrω = p r ρ du θ dr = p r ρrω, p r = ρrω. (5.8) This shows that the pressure must vary with the radius inside the fluid, in order to balance the centrifugal force. Moreover, the pressure must also satisfy the vertical hydrostatic balance (i.e. balance between pressure and gravity). From the vertical component of the vector equation (5.7) (i.e. its scalar product with ê z ) we find 0 = p ρg p(r,z) = ρgz +A(r), z where A(r) is a function to be determined using appropriate boundary conditions. At the free surface the fluid pressure should match the atmospheric pressure. Hence, p = p atm at z = h(r), so that p atm = ρgh(r)+a(r) A(r) = p atm +ρgh(r). o, substituting the expression of the pressure, in equation (5.8) leads to p(r,z) = p atm +ρg(h(r) z), p r = ρgdh dr = ρrω h(r) = h 0 + r Ω g, where h 0 is the height of the free surface at r = 0, on the rotation axis. This result shows that the free surface of a uniformly rotating fluid in a cylindrical container is a paraboloid. uch rotating cylindrical containers filled with highly-reflecting liquids (e.g. mercury or ionic liquid coated with silver) have been used to build mirrors with a large (up to 6m) smooth reflecting paraboloid surface, which could be used as primary mirrors for telescopes.

8 hape of the free surface of a rotating fluid