for some scalar field φ(x, t). The field φ(x, t) is called the potential, or velocity potential, for u. - PDF

3 IRROTATIONAL FLOWS, aka POTENTIAL FLOWS Irrotational flows are also known as potential flows because the velocity field can be taken to be the gradient of a 3.1 Velocity potential. That is, an irrotational

Please download to get full document.

View again

of 20
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.


Publish on:

Views: 17 | Pages: 20

Extension: PDF | Download: 0

3 IRROTATIONAL FLOWS, aka POTENTIAL FLOWS Irrotational flows are also known as potential flows because the velocity field can be taken to be the gradient of a 3.1 Velocity potential. That is, an irrotational flow has a velocity field u(x, t) that can be represented in the form u = φ, for some scalar field φ(x, t). The field φ(x, t) is called the potential, or velocity potential, for u. Note the sign convention, opposite to the usual sign convention for force F and force potential Φ. Note also: it will prove useful to include cases in which φ(x, t) is a multi-valued function of its arguments. A velocity field of the form u = φ is, indeed, irrotational it has vorticity ω = 0 because curl grad of any scalar field is zero: ω = u = ( φ) = 0. The converse that any irrotational velocity field u(x, t) can be written as φ for some φ(x, t) is also true, though less obvious. Given u(x, t) in a simply-connected domain D, we can construct φ(x, t) as a single-valued function. This will now be shown. We are given an irrotational velocity field u(x, t). Choose some fixed point x 0 D x 0 could be the origin of coordinates, or any other convenient choice then define φ(x, t) = x x 0 u(x, t). dl, where the integral is taken along any path within D that joins x 0 to x, with dl denoting the line element of the path at position x. This function φ(x, t) is an ordinary, single-valued function, being independent of the path of integration as long as D is simply-connected. The path-independence then follows from the fact that any two paths make up a closed curve around which the circulation C = u. dl must be zero, by Stokes theorem. (Since we are given that ω = u = 0 everywhere, we have, in particular, that u = 0 on a surface spanning the two paths.) 1 A simply-connected domain D is one in which any closed curve can be shrunk continuously to a point while staying in D. The simplest example of a domain that is not simply connected is a 2-D domain with an island, as in the picture on the left. A closed curve encircling such an island cannot be shrunk to a point in the way envisaged, and there is no surface spanning such a curve and lying within the domain. A domain that is not simply-connected, as in the picture, is called multiply-connected. In a multiply-connected domain, the path for the integral defining φ can go to one side or the other of an island; indeed it can wind round an island any number of times. Then φ can be multi-valued, and will be multi-valued whenever the circulation C for a closed curve encircling an island is nonzero: C = u. dl 0. An example in which such nonzero circulation C of crucial importance, making island multi-valued potentials φ useful, is the flow round a lifting aerofoil (bottom of p. 17, details in 3.8 below). Kelvin s circulation theorem suggests and experiments confirm that there is a significant set of circumstances in which irrotational, or potential, flows are good models of real flows. Very often this is because, to good approximation, the flow has started from rest and because, furthermore, viscosity is negligible on all the relevant closed curves within the fluid domain D. Examples: Airflow just ahead of an aircraft, and above and below the wings Flow of water toward a small drainage hole in the bottom of a large tank containing water previously at rest Flow towards the end of a straw (thin tube) through which air is being sucked Water wave motion, for waves propagating into water previously at rest All these examples depend of course on viscosity being small enough. Potential flow would be a bad, a grossly inaccurate, model for everyday (domestic-scale) flows of fluids like golden syrup or treacle. *There is one exception to the last statement: viscous flow in a narrow gap between two planes. 2 For irrotational flow to be a good model we need F negligible in the vorticity equation Dω Dt = (ω. )u + F. where F includes, besides gravity, the force per unit mass from internal friction (viscosity). *In the simplest case of a uniformly viscous flow under gravity or other conservative force field, the viscous contribution to F can be shown to be equal to 2 u times a constant coefficient called the kinematic viscosity having dimensions of length 2 /time. In that case, F 2 ω (which happens to be zero for potential flows a fact that historically seems to have caused some confusion, for perhaps a century or so, until early in the twentieth century when pioneers in aerodynamics (especially Ludwig Prandtl) saw the importance of what happens to material curves very close to boundaries, where flow is hardly ever irrotational, and began to develop the boundary layer theory needed to understand this. For the time being, in these lectures, we can adopt the usual attitude to scientific model-making: simply adopt a set of assumptions, including, in this case, the assumption that ω = 0 everywhere, then ask how far the resulting model fits reality in various cases. (For further discussion of, and insight into, scientific model-making as such and such insight has great importance in today s and tomorrow s world see the animated image on my home page, and associated links).* Let us now look at some basic properties of irrotational flows. Mass conservation.u = 0 gives.( φ) = 0, i.e., 2 φ = 0, i.e., φ is a harmonic function, in the sense of satisfying Laplace s equation. The boundary condition for impermeable boundaries, also called the kinematic boundary condition, is that U.n = u.n = n.( φ) φ n, where U is the velocity of points on boundary as before.*this exemplifies what is called a Neumann boundary condition for Laplace s equation.* If the flow is irrotational, this is the only boundary condition we need. We can now apply techniques from Part IB Methods. For complex geometries, we can solve numerically see Part II(a) or (b) Numerical Analysis. 3.2 Some examples Because φ satisfies Laplace s equation 2 φ = 0, we can apply all the mathematical machinery of potential theory, developed in 1B Methods: A: Spherical geometry, axisymmetric flow 3 Recall the general axisymmetric solution of Laplace s equation obtained by separation-of-variable methods in the usual spherical polar coordinates, by trying φ = R(r) Θ(θ), etc., φ = } {A n r n + B n r (n+1) P n (cos θ), n=0 where A n, B n are arbitrary constants, P n is the Legendre polynomial of degree n, r is the spherical radius (r 2 = x 2 + y 2 + z 2 ), so that z = r cos θ, and θ is the co-latitude. (Recall that P 0 (µ) = 1, P 1 (µ) = µ, P 2 (µ) = 3 2 µ2 1 2, etc.) Let us look more closely at the kinds of flows represented by the first three nontrivial terms, those with coefficients B 0, A 1, and B 1. (i) all A s, B s zero except B 0 [Notation: e r = unit radial vector = x/ x = x/r]: φ = B 0 r = B 0 x, u = φ = B 0x x 3 = B 0e r r 2.( ) This represents source or sink flow, i.e., depending on the sign of the coefficient B 0, it represents radial outflow from a point source (B 0 0), or inflow to a point sink (B 0 0), at the origin. Mass appears or disappears, at the singularity at the origin, according as B 0 0. The velocity field is radially symmetric in both cases: source flow (B 0 0) sink flow (B 0 0) Let us check that incompressibility and mass conservation are satisfied: the outward mass flux across any surface containing the origin should be independent of the choice of surface. For simplicity, take a sphere of radius R. The outward mass flux is ρ φ = ρb 0 /R 2 per unit area, hence (ρb 0 /R 2 )(4πR 2 ) in total, = 4πρB 0, independent of R as expected. (ii) all A s, B s zero except A 1 : φ = A 1 r cos θ = A 1 z, u = φ = A 1 e z 4 where z is the Cartesian co-ordinate parallel to the symmetry axis, and e z = (0, 0, 1) a unit axial vector. So this simply represents uniform flow in axial or z direction, a u field like this: z (iii) all A s, B s zero except B 1 : φ = B 1 cos θ r 2 = B ( 1z ez, u = B r 3 1 r 3ze ) r ( ) 3 r 4 Note that apart from the different coefficient B 1 this is / z of case (i), eqs. ( ) above. (To check this, hold x and y constant in r 2 = x 2 +y 2 +z 2, then 2rdr = 2zdz r/ z = z/r (r 1 )/ z = z/r 3, etc.) So the solution ( ) can be regarded as the limiting case of a superposition note that we can use superposition, because Laplace s equation is linear a superposition of two copies of the solution ( ) above with their origins separated by a infinitesimal distance δz and with equal and opposite coefficients B 0 = ±B 1 /δz. Specifically, φ = B 1 z/r 3 can be regarded as the result of adding together the two potentials φ = B 1 /δz {x 2 + y 2 + (z δz)2 } 1/2 and φ = +B 1 /δz {x 2 + y 2 + (z 1 2 δz)2 } 1/2. and then taking the limit δz 0. The resulting solution φ = B 1 z/r 3 is often, therefore, called a dipole, or occasionally a doublet (and case (i), the single mass source or sink, is correspondingly called a monopole ) (rarely a singlet ). The velocity field of the dipole looks like this, when B 1 0: (iv) Uniform flow past a sphere r = a; this turns out to be a combination of (ii) and (iii) above: Irrotational and incompressible, so 2 φ = 0 in r a. Uniform flow at, so φ Ur cos θ as r, i.e., tends toward case (ii) as r. 5 No normal flow across surface of sphere, so φ/ r = 0 on r = a. Now cos θ (= P 1 (cos θ)) is the only Legendre polynomial cos θ. Therefore it is worth trying a solution that superposes (ii) and (iii) above, i.e. of form φ = (A 1 r + B 1 r 2 ) cos θ If this is to fit the boundary condition at, we must choose A 1 = U. If it is also to fit the boundary condition at r = a, then we must choose φ/ r = (A 1 2B 1 r 3 ) cos θ = 0 θ, hence A 1 = 2B 1 a 3, hence B 1 = a 3 U/2. Therefore ) solution is φ = U (r + a3 cos θ. 2r 2 The corresponding velocity field φ it is convenient to calculate its components in spherical polars, to check that we have satisfied the boundary condition at r = a is ( ) ) φ u = φ = r, 1 φ r θ, 0 = (U cos θ(1 a3 r ), a3 (1 + )U sin θ, 0 3 2r3 [As a consistency check, you can verify that these expressions for the spherical components of φ correspond to a linear combination of the velocity fields derived in (ii) and (iii) above when A 1 = U and B 1 = a 3 U/2.] *Experiment shows that this is not a good model for steady flow round a solid sphere, essentially because rotational fluid from near the surface is carried very quickly into the wake behind the sphere. But it is a good model when the solid sphere is replaced by a spherical bubble. It can also be a good model for a rapidly oscillating flow past a solid sphere, such as would be induced by a sound wave whose wavelength a. This can be modelled by the above solution, unchanged except to make U a function of time that oscillates sufficiently rapidly about zero (so that fluid particles travel distances a during one oscillation, which will be the case if the oscillation frequency U/a).* 6 B: Cylindrical (circular 2-D) geometry Much the same pattern as before, except that, as always in 2-D potential theory, the solutions involve logarithms as well as powers of the radial coordinate r; and potentials can be multi-valued. Also, there is no particular direction corresponding to the axis of symmetry in case A. Recall the general solution obtained by separation of variables in 2-D cylindrical polars r, θ: φ = A 0 log r + B 0 θ + { An r n cos(nθ + α n ) + B n r n cos(nθ + β n ) } n=1 [*or replace cosine terms by equivalent complex forms, terms e inθ times r n or r n ). Powers, or other analytic functions, of a complex variable z = x + iy or its complex conjugate z = x iy are solutions of Laplace s equation in two dimensions.*] Again, consider just the first few terms: (i) all A s, B s zero except A 0 : φ = A 0 log r u = φ = A 0e r r = A 0x r 2, 7 Represents radial source flow if A 0 0 flow radially outward from a 2-D mass source (line source in 3-D). Total outflow across circle of radius R is 2πR (A 0 /R) = 2πA 0 (independent of R), 2-D version of case A(i) above. Radial sink flow (if A 0 0): (ii) all A s, B s zero except B 0 : φ = B 0 θ u = φ = B 0e θ r. Note that φ is multi-valued recall earlier remarks about islands in 2-D whereas u is singlevalued, as it must be in order to make physical sense. This represents circular flow, with flow speed B 0 /r. The circulation C = u. dl round any circle of radius R is nonzero. It is 2πR (B 0 /R) = 2πB 0 (independent of R), = κ, say. This flow has u = 0 except at R = 0, where there is a δ-function singularity in the vorticity, of strength κ. [This follows from Stokes theorem applied to an arbitrarily small circle, or other closed curve, surrounding the origin.] The flow is sometimes described as a line vortex or a point vortex on the understanding that we are imagining the physical domain to be a 2-D space. A superposition of such flows may serve as a simple model of more general vorticity distributions. (iii) all A s, B s zero except A 1 : φ = A 1 r cos(θ + α 1 ) = A 1 r (cos θ cos α 1 sin θ sin α 1 ) = A 1 (x cos α 1 y sin α 1 ), Represents uniform flow with velocity U = A 1 in direction θ = α 1. Like spherical case A(ii). 8 (iv) all A s, B s zero except B 1 : φ = B 1 cos(θ + β 1 ) r. This represents a 2-D dipole pointing in the direction θ = β 1. (Related to pair of mass-source solutions as before.) (v) Uniform flow, with circulation, past cylinder (r = a): Irrotational and incompressible, so 2 φ = 0 in r a. Uniform flow U, plus an irrotational flow with circulation κ at, so we require φ Ur cos θ + κθ 2π No normal flow across cylinder, so we also require as r φ r = 0 on r = a. 9 Solution: ) φ = U cos θ (r + a2 r + κθ 2π ; κ (= 2πB 0) is an arbitrary constant! Velocity field (in cylindrical polars): ( ) ( ) ) φ u = r, 1 φ = U cos θ (1 a2, U sin θ (1 + a2 r θ r 2 r 2 Stagnation points (a) (u = 0) on r = a and sin θ = (b) where cos θ = 0, and r is s.t. 1 + a2 r 2 in r a if κ/(4πua) 1 ). κ a 2πUa r = 0 ) + κ. 2πr κ, possible when 0 κ/(4πua) 1, and 4πaU (quadratic equation for a/r, one real root 3.3 Pressure in time-dependent potential flows with conservative forces Laplace s equation (linear) gives the solution for the flow, u; but if we want to determine the pressure field p, then we must also use the momentum equation (Euler s equation, nonlinear): [ ] u ρ t + (u. )u = p ρ Φ We confine attention to conservative (potential) forces, and again use (u. )u = ω u + ( 1 2 u 2 ), remembering that ω = 0 in potential flows, and that u t = ( ) φ t φ =. t Hence So we have now found, in summary, { φ t u 2 + p } ρ + Φ = 0 φ t u 2 + p ρ + Φ = H(t), say, independent of x. 10 Two forms of Bernoulli s theorem both applying to the flow of an inviscid, incompressible fluid under conservative (potential) forces Φ per unit mass, and both being corollaries of the momentum equation written as u t + 1 ( u 2) + ω u = 1 p + Φ. 2 ρ The two forms, deduced in 2.4 and just above, respectively say that: IF the flow is steady, i.e. u t then the quantity obeys H = 1 2 u 2 + p ρ + Φ u. ( 1 2 u 2 + p ρ + Φ) = 0 = 0, irrotational, i.e. u = 0, H = φ t u 2 + p ρ + Φ ( φ t u 2 + p ) ρ + Φ = 0 and hence 1 2 u 2 + p ρ + Φ = H = constant on φ t u 2 + p ρ + Φ = H(t), independent streamlines of spatial position x We may also summarize, and slightly generalize, the ways of representing the velocity field in various circumstances, as follows: Stream function and velocity potential ( 1.8 & 3.1) IF the flow is incompressible, i.e.. u = 0, irrotational, i.e. u = 0, then there exists *a vector potential A a scalar potential φ with u = A * with u = + φ *In 2-D (or axisymmetric) flow A has only one component, from which we define* a stream function ψ (or Ψ) (φ being called the velocity potential) 11 u ψ (or Ψ) so u + φ ψ (or Ψ) is constant on streamlines In a multiply connected domain in 2-D: ψ may be multi-valued if islands/holes φ may be multi-valued if islands/holes with net outflow: u. n ds 0 with net circulation: u. dl Applications of irrotational, time-dependent Bernoulli Inviscid, irrotational models can give useful insight into certain kinds of unsteady fluid behaviour and their timescales, relevant to laboratory and industrial fluid systems. The simplest examples include accelerating flows in tubes: Fast jet generator, version 1 In the situation sketched below we neglect gravity. The tube contains water, supplied from the container on the left. The flow starts from rest at time t = 0. The pressure p at x = 0, the left-hand end of the tube, is controlled by a feedback mechanism, not shown in the sketch; thus p at x = 0 is prescribed as a given function of time: p x=0 = p atm + p 0 (t) ; p 0 (t) 0 for t 0 and p 0 (t) = 0 for t 0. Assume that the flow is unidirectional ( x) in the tube and that the water emerges from the far end x = L as a free jet in which the pressure p p atm, the ambient atmospheric pressure. (Not perfectly accurate, but experiment shows that it is a reasonable first-guess model of what actually happens; p p atm in the emerging jet, because sideways accelerations, hence sideways p components, are small.) Because the flow starts from rest it is irrotational, and therefore has a velocity potential. Because the flow x within the tube, the velocity potential and its time derivative must take the simple forms φ = u(t) x + χ(t) and φ = u(t) x + χ(t), 12 where χ(t) is an arbitrary function of time t alone, and where u is the flow velocity along the tube. Note that φ cannot depend on y or z; otherwise the flow not x. So u cannot depend on y or z either; nor can u depend on x, otherwise 2 φ 0 (.u 0, mass not conserved). Apply time-dependent Bernoulli: H constant along tube, H x 0 = H x=0 = constant; so u x + χ u2 + p ρ = u 0 + χ u2 + p 0(t) ρ. Notice the implication that p = p(x, t) and moreover that p = (ρ u, 0, 0). (Pressure varies linearly with x; pressure gradient is uniform along the tube, as it has to be because fluid acceleration is uniform along the tube.) Taking p = 0 at x = L, we have (with several terms cancelling or vanishing) so u L E.g. if p 0 = positive constant for t 0, then + 0 ρ u = p 0(t) ρl, u = 1 ρl = u 0 + p 0(t) ρ u = p 0 ρl t. t 0 p 0 (t ) dt. So the flow keeps accelerating as long as this model remains applicable, i.e., as long as friction (viscosity) remains unimportant and as long as the excess pressure p 0 is maintained (by the pressurized container and the assumed control mechanism). Notice that χ(t) has disappeared from the problem. Indeed we could have taken χ(t) = 0 w.l.o.g. at the outset, on the grounds that only φ is of physical interest and that χ(t) = 0. This would have changed the value of H, but not the thing that matters the constancy of H Fast jet generator, version 2, Same problem as before, except that pressure now controlled via force on piston to be p atm +p 0 (t) here; pressure here (x = 0) is now unknown: so 13 Assume that the flow is irrotational everywhere. Then u(x, t) = φ(x, t) everywhere, H is same constant everywhere. Within the tube, φ = u(t)x as before, now taking χ(t) = 0 w.l.o.g. In the closeup view on the right, the flow outside a small hemisphere H of radius a (shown dotted) is approximately the same as simple mass-sink flow ( 3.2 A, p. 25), with φ = πa2 u(t) + χ 1 (t). (Yes, 2π not 4π : why?) Here 2πr r 2 = x 2 + y 2 + z 2, and χ 1 (t) is not arbitrary. (Why? The whole flow has a single velocity potential φ(x, t), and we are not at liberty to introduce discontinuities!) Indeed, solutions of 2 φ = 0 are very smooth. Detailed solution, beyond our scope here, (e.g. using the whole infinite series on p. 25), shows that the smooth function φ(x, t) is fairly well approximated if we pick χ 1 (t) such that φ = 0 on the hemisphere H. Then χ 1 (t) = πa2 u(t) = a u(t). 2πa 2 In summary, u(t) x within the tube, and φ(x, t) = πa 2 u(t) a u(t) ( ) to the left of H. 2
Related Search
Similar documents
View more...
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks