Lecture notes on fluid mechanics. Greifswald University 2007

Transcription

1 Lecture notes on fluid mechanics Greifswald University 2007 Per Helander and Thomas Klinger Max-Planck-Institut für Plasmaphysik 1

2 1 The equations of fluid dynamics Fluid mechanics is concerned with the mathematical description of the movement of liquids and gases. It is one of the classic topics of analytical mechanics, but is far from a closed subject. Because of its useful applications it is an area of active research. Intriguingly, although it is has been studied intensively for three centuries it still harbours some of Nature s most closely guarded secrets. To describe the motion of a fluid (think of air or water), we use the notation ρ(r, t) = density, p(r, t) = pressure, v(r, t) = velocity. Note that density and pressure are scalars, while the velocity is a vector quantity with three components, v = v xˆx + v y ŷ + v z ẑ. To describe how these quantities evolve we thus need five equations. These equations can be derived rigorously from kinetic gas theory, but we will not do so in this course. Here we use simpler, intuitive arguments to motivate the fluid equations. 1.1 Continuity equation The equation governing the evolution of ρ is called the continuity equation and is ρ + (ρv) = 0. (1) t To derive this equation, consider a fixed, stationary volume V in space. The mass of the fluid inside this volume is given by the volume integral m(t) = V ρdv. This mass can change as a result of fluid flowing out of (or into) V. The flux (i.e., the mass of fluid flowing in unit time) through a small surface ds = ˆndS, where ˆn is the unit normal, is equal to ρv ds, so the total flux out of the volume V is Φ = S ρv ds, 2

3 v(r,t) S n V Figure 1: A volume V with surface S in a fluid flow field v(r, t). where the integral is taken over the surface S bounding V. The unit vector n in ds = ˆndS is taken to point outward. The rate at which the mass changes in V is thus or dm dt = Φ, d ρdv = ρv ds = (ρv)dv, dt V S V where we have used Gauss s theorem to convert the surface integral to a volume integral. Hence V [ ρ t + (ρv) ] dv = 0 for an arbitrary (but fixed) volume V. Letting this volume be infinitesimally small, we obtain the continuity equation (1). This equation essentially says that mass is conserved: the density may change at a given point because the fluid goes elsewhere, but mass cannot be created or destroyed. 1.2 Equation of motion The motion of the fluid is governed by Euler s equation (1755), ( ) v ρ t + v v = p. (2) To derive this equation, we begin with Newton s law for a small volume V of fluid, m dv dt = F. 3

4 This time we let the volume V move with the fluid; in the derivation of the continuity equation V was fixed in space. The volume V is taken to be so small that the velocity is practically constant throughout V. Here m is the mass and F is the force m = V ρdv F = pˆnds S acting on the volume as a result of pressure from the ambient fluid. This pressure is equal to pˆnds on each surface element ds, where again ˆn is the unit normal vector (pointing outward). The total force is a surface integral taken over the surface S of V. According to Gauss s theorem so we have or, since V is small, F = pdv, V dv ρdv = pdv, dt V V ρ dv dt = p. (3) In this equation dv/dt denotes the acceleration of the small volume element V. This is not the same as the time derivative v/ t taken at a fixed point in space, because the volume V moves. The correct expression for its acceleration is found by noting that if r is the position of V at the time t, then its position at the time t + dt is r + vdt. The acceleration is thus dv dt = lim v(r + vdt, t + dt) v(r, t) = v dt 0 dt t + v v and is called the convective derivative of v. Using this expression in Eq. (3) gives Euler s equation (2). More generally, the rate of change, as measured in a local frame moving with the fluid, of any quantity T(r, t) (such as temperature, for example) is equal to dt dt = T t + v T. It is interesting to multiply the continuity equation (1) by v and add it to Euler s equation (2), giving (ρv) t + (ρvv + pi) = 0, (4) 4

5 where I = δ jk is the unit tensor. This equation expresses conservation of momentum, just like the continuity equation describes mass conservation. In the continuity equation, the rate of change of density ρ was related to the divergence of the flux ρv. Here, the rate of change of the momentum density ρv is related to the divergence of the flux of momentum density, which is the tensor quantity ρvv + pi. The momentum can change locally because it flows elsewhere, but it cannot be created or destroyed. Euler s equation (2) does not take into account any internal friction of the fluid, i.e., viscosity. The force on each fluid element was taken to be given entirely by the pressure from neighbouring fluid elements. If viscosity is included, one obtains the Navier-Stokes equation, ( ) v ρ t + v v = p + µ 2 v. where µ is the viscosity coefficient. For many problems in fluid mechanics viscosity is unimportant, and we shall ignore it for the time being but return to this topic in a later chapter. Streamlines are defined as curves that are everywhere tangential to v. This means that a curve R(s) = [x(s), y(s), z(s)] is a streamline if dr/ds points in the same direction as v(r). This means that there is a scalar λ(s) such that dr ds = λ(s)v(r). These three differential equation can be summarised compactly by writing dx v x = dy v y = dz v z. 1.3 Incompressible fluids and ideal gases The continuity equation (1) and the three components of Euler s equation (2) are four equations for five unknowns (ρ,v, p), so we need a fifth equation to close the system. Unlike the other equations, this equation depends on the nature of the fluid. The simplest case is that of an incompressible fluid such as water, which is incompressible to a good approximation. In such a fluid the density is constant, ρ = constant, 5

6 This equation closes the system, and also simplifies the continuity equation, which because ρ/ t = 0 and ρ = 0 reduces to v = 0. For a compressible fluid, the required equation is in general obtained by considering the energy balance. The simplest case is obtained if any energy dissipation and heat exchange between different parts of the fluid may be neglected. This implies that the motion of any part of the fluid is adiabatic, and the entropy per unit mass, s, is constant for each fluid element, ds dt = s + v s = 0. t The relationship between s and the pressure p and density ρ is different for different fluids. At atmospheric pressures and normal temperatures most ordinary gases are so tenuous that they behave like ideal gases, where s p/ρ γ. For fully ionised plasmas, the adiabatic index is γ = 5/3, just like in a monoatomic gas, since the constituents (ions and electrons) each have three degrees of freedom. For a gas with diatomic molecules, such as air, γ = 7/5. Another limit of interest is obtained when γ = 1, so that p/n is constant. In an ideal gas, this corresponds to the case when the temperature is constant, and is realised when the heat conductivity is very large. Summary In summary, the governing equations of fluid mechanics are ρ = constant, for an incompressible fluid and v = 0, ( ) v ρ t + v v = p, ρ + (ρv) = 0, t (5) ( ) v ρ t + v v = p, (6) 6

7 ( ) p t + v = 0, (7) ργ for an ideal gas without heat conduction (or with very large heat conduction). To a good approximation, the former equations hold for water and the latter ones for air. 2 Hydrostatics If the fluid is subject to an external force, such as gravity, this needs to be added to the right-hand side of Euler s equation, which becomes ( ) v ρ t + v v = p + F, where F(r, t) is the force per unit volume acting on the fluid. For gravity F = ρgẑ, with g = 9.8 m/s 2 and ẑ pointing up. A fluid at rest in a gravitational field thus satsifies or, in an incompressible fluid, p + ρgẑ = 0, p = p 0 ρgz. This variation of pressure with height (or depth) is easily noticed when diving. More generally, time-independent solutions with flow must satisfy ρv v + p + ρgẑ = 0. or v2 + p + gẑ = v ( v). ρ The right-hand side of this equation is perpendicular to v, so multiplying the equation by a unit vector in the direction of v gives Bernoulli s equation ( d v 2 ds 2 + p ) ρ + gz = 0, where s is the length along v, and we have again assumed that the fluid in question is incompressible, so that ρ can be taken inside the derivative. Thus ρv p + ρgz = constant (8) 7

8 p smaller p larger Figure 2: The air above the roof moves faster and has therefore lower pressure than that underneath. along each streamline, so the pressure falls in regions with large flow. This is why roofs blow off during storms. Archimedes principle states that a solid body immersed in water exeriences an apparent loss of weight equal to that of the displaced volume of water. To prove this important statement, which was first published in the treatise On Floating Bodies by Archimedes ( BC), we again note that each small area element ds on the surface of the body experiences a force equal to df = pˆnds = pds, due to the pressure p of the ambient fluid. As usual, ˆn denotes the unit normal (pointing outward) and ds = ˆndS. Since in hydrostatic equilibrium p = ρgẑ, the total pressure force becomes F = pds = pdv = ẑ ρgdv = mgẑ, where we have used Gauß law to convert the surface integral to an integral over the volume of the body, and written m = ρdv 1 Recall that (A B) = A ( B) + B ( A) + A B + B A for any vector fields A and B. 8

9 for the mass of the displaced fluid. Archimedes principle implies that the water that a boat displaces has the same weight as the boat itself. This fact was used by Cao Chong ( AD), who was a son of the powerful Chinese warlord Cao Cao. He was renowned as a child prodigy, having the intelligence of an adult at the age of five. Another warlord had sent his father an elephant as a gift, and Cao Cao wanted to know its weight. Nobody could figure out how to weigh the elephant except young Cao Chong, who suggested to load the elephant on a boat on which the water level was marked. The elephant was then replaced by known weights until the boat was submerged to the same level. Unfortunately Cao Chong died of sickness at the age of thirteen. 9

10 Problems 1. How does the pressure, temperature and density vary with height in a stationary (v = 0) isentropic ( s = 0) atmosphere? Answer: ρ(z) = ρ 0 ( 1 z h) 1/(γ 1), p(z) = p 0 ( 1 z h) γ/(γ 1), ( T(z) = T 0 1 z ). h h = γp 0 (γ 1)gρ 0. For air at normal pressure and temperature at z = 0, the height of the atmosphere becomes h = 28 km, so that the temperature falls by about one degree every 100 m. 2. Generalise Bernoulli s equation (8) to the case where the fluid is not incompressible but obeys Eq. (7). Answer: ρv γp + ρgz = constant. γ 1 3. A child is having a bath and plays with a boat, which is carying a stone whilst floating on the water. If child tips the stone into the water, does the water level rise, fall, or remain unchanged? 4. Estimate the time it takes to empty a bath tub, using Bernoulli s law. Assume that the body of water is a rectangular slab of 1 m 2 cross section and 50 cm depth, and that the water leaving the tub forms a cylindrical jet with 3 cm 2 diameter. The pressure in this jet is close to the air pressure. 10

11 3 Potential flow The simplest class of fluid motions are those satisfying v = 0, which we shall call irrotational. By a theorem in vector analysis, any such flow field can be written as the gradient of a scalar function v = φ(r, t). If the fluid is incompressible, v = 0, then this function satisfies Laplace s equation 2 φ = 0, and Euler s equation is satisfied if the pressure is given by To see this, we note that since 2 p ρ = φ t φ 2. (9) 2 ( ) v 2 = v ( v) + v v, 2 Euler s equation can for an incompressible fluid be written as ( ) ( ) v ρ t v ( v) = p + ρv2. 2 Irrotational and incompressible flows are thus found by first solving Laplace s equation and then constructing the pressure from Eq. (9). Of course, Laplace s equation occurs in many different areas of mathematical physics. For instance, this equation describes the electrostatic potential in vacuum, and then v corresponds to (minus) the electric field. Irrotational, incompressible flows are called potential flows. Simple examples are, for instance, the potential from a point charge or a line charge φ = C r, φ = C lnr, 2 Recall that (A B) = A ( B) + B ( A) + A B + B A for any vector fields A and B. 11

12 where r = x 2 + y 2 + z 2, R = x 2 + y 2, and C is a constant. These functions satisfy 2 φ = 0 except at the origin, where 2 φ is equal to a delta function. The streamlines of v are directed radially outward, away from the source of fluid at the origin, just like an electric field points away from the charges that produces it. Another example is the field from a line vortex, φ = Cϑ, where ϑ is the usual polar angle, defined so that x = R cos ϑ, This potential gives rise to a flow field y = R sinϑ. v = C ˆϑ r, where ˆϑ = r ϑ is the unit vector pointing in the direction of ϑ. This flow field is irrotational, v = 0, everywhere but at the origin, where the curl is a delta function. To construct a more interesting and nontrivial example, we note that since φ = C/r satisfies Laplace s equation, then its derivatives also do so. For example φ = z ( C r ) = Cz r 3 also satisfies this equation. In spherical coordinates (r, θ, ϕ), The corresponding velocity is φ = C cos θ r 2, v = φ = C (2ˆrcos r 3 θ + ˆθ ) sinθ. (10) In electrodynamics this solution to Laplace s equation is known as a dipole. Its streamlines are depicted in Fig. 3. In fluids dynamics, it describes the flow field around a moving solid sphere immersed in the fluid. To see this, we note that the boundary condition on the fluid flow on the surface of a body moving with a velocity u (v u) ˆn = 0, 12

13 Figure 3: Streamlines of the flow (10). where ˆn is the unit vector perpendicular to the surface of the body. If the body is a sphere centred at the origin and moving with the velocity u = uẑ, we thus require v r = uẑ r. (11) The right-hand side is equal to u z r = u cos θ, and if the flow is given by (10) then the left-hand side is v r = 2C cos θ r 3. If the radius of the sphere is a, then the boundary condition (11) is thus satisfied if C = ua3 2, and we conclude that the (irrotational) flow velocity around a moving sphere is given by v = ua3 ( 2r 3 2ˆr cos θ + ˆθ ) sin θ. (12) Of course, it is now trivial to find the flow field around a stationary sphere, simply by transferring to a frame moving with the velocity u. Since ( u = uẑ = u ˆrcos θ ˆθ ) sinθ, 13

14 the irrotational flow around a stationary sphere is thus given by ( ) ( ) v a 3 u = ˆr r 3 1 cos θ + ˆθ a 3 2r 3 1 sinθ Two-dimensional potential flow Two-dimensional potential flows can be described elegantly in terms of complex variables. For a given flow v = φ = φ φ ˆx + x yŷ, it is useful to define the function ψ(x, y) through φ x = ψ y, (13) i.e. φ y = ψ x, (14) v = ψ ẑ. Since v is perpendicular to ψ, the curves of constant of ψ are streamlines. Now write z = x + iy, w = φ + iψ. Then w(z) is an analytic function, for its real and imaginary part satisfy the Cauchy- Riemann equations (13)-(14). Analytic functions have the property that the deriviative dw/dz does not depend on dz. For instance, taking dz = dx and dz = idy, respectively, yields and dw dz = dw dx = φ x + i ψ x, dw dz = dw idy = idw dy = ψ y i φ y. In either case, the Cauchy-Riemann equations (13)-(14) imply that dw dz = v x iv y. (15) Any analytical function thus describes an irrotational incompressible flow field in two dimensions. This is convenient for a number of purposes, and enables us for instance to 14

15 construct flows by using techniques of conformal mapping. To illustrate this by simple example, consider the function w(z) = z 2, i.e. w(x + iy) = (x 2 y 2 ) + 2ixy. This funtion maps the first quadrant onto the upper half plane, and thus describes fluid flow around a 90-degree corner, see Fig. 4. The mapping z w maps the steamlines in this figure onto straight horizontal lines in the (upper half of the) w-plane. The streamlines are given by curves of constant ψ = 2xy, and are thus hyperpolae. Since w (z) = 2z = 2(x + iy), the flow velocity is according to Eq. (15) equal to v = 2(xˆx yŷ), and vanishes at the origin. y x Figure 4: Streamlines near a corer. More generally, the function w(z) = z α 15

16 describes flow around a corner of angle π/α. If α < 1, then w (z) = αz α 1 becomes infinite at the origin, see Fig. 5, where α = 2/3. Figure 5: Streamlines near a corer. A less trivial example is furnished by the function ( ) w(z) = u z + a2, (16) z which maps the region outside the circle x 2 + y 2 = a 2 onto the entire complex plane. The streamlines are obtained by plotting curves of constant ( ) ψ(x, y) = Im w(z) = u y a2 y x 2 + y 2, as in Figure 6. For large z, the flow velocity is approximately v uˆx. We thus have an expression for the flow around a circular cylinder of radius a immersed in a steady flow with the velocity u. 4 Force on an immersed body Consider a stationary body immersed in a liquid at rest. If a force F is applied to the body, one may ask how large its acceleration a will be. If the liquid has zero density, then of course a = F/m, where m is the body mass. If the liquid has finite density, we expect the acceleration to be smaller since some of the force will be expended at 16

17 Figure 6: Streamlines of the flow (16) around a cylinder. accelerating the liquid. In general, the result will depend on the shape of the body. We shall consider the simplest case of a sphere immersed in an incompressible fluid. The flow pattern for this case was calculated in the previous section, and is given by Eq. (12). Note that the flow velocity of the liquid vanishes at infinity. The simplest way to find the relation between force and acceleration is to first calculate the kinetic energy of this flow, Substituting from Eq. (10) gives T fluid = T fluid = ρu2 a 6 8 r>a v 2 = a ρv 2 2 dv = ( ua 3 2πdr r 4 2r 3 and the total kinetic energy becomes Here a π 2πr 2 ρv 2 dr sin θdθ. 0 2 ) 2 ( ) 4 cos 2 θ + sin 2 θ π T = T body + T fluid = 0 ( ) cos 2 θ sin θdθ = 2πa3 ρu 2 3 2, ( m + 2πρa3 3 M = m + 2πρa3 3 ) u 2 2 = Mu2 2. is the sum of the body mass m and half (!) the mass of the displaced fluid. On the other hand, the work done by a force F acting on the body and pointing in the direction 17

18 of u is Fu. This work should equal the rate of change of kinetic energy and we conclude that dt dt = Mudu dt, F = M du dt. A spherical body thus has an apparent mass equal to its actual mass plus half the mass of the fluid it displaces. Problem 1. Calculate the apparent mass of a cylinder moving in the direction perpendicular to its axis in an infinite volume of incompressible fluid. 18