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Notes on Fluid Dynamics

Notes on Fluid Dynamics
Notes on Fluid Dynamics Rodolfo Repetto Department of Civil, Chemical and Environmental Engineering University of Genoa, Italy rodolfo.repettounige.it phone number: +39 010 3532471 http://www.dicca.unige.it/rrepetto/ skype contact: rodolforepetto January 13, 2016 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 1 / 161Table of contents I 1 Acknowledgements 2 Stress in uids The continuum approach Forces on a continuum The stress tensor Tension in a uid at rest 3 Statics of uids The equation of statics Implications of the equation of statics Statics of incompressible uids in the gravitational eld Equilibrium conditions at interfaces Hydrostatic forces on at surfaces Hydrostatic forces of curved surfaces 4 Kinematics of uids Spatial and material coordinates The material derivative De nition of some kinematic quantities Reynolds transport theorem Principle of conservation of mass The streamfunction The velocity gradient tensor Physical interpretation of the rate of deformation tensor D Physical interpretation of the rate of rotation tensor Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 2 / 161Table of contents II 5 Dynamics of uids Momentum equation in integral form Momentum equation in di erential form Principle of conservation of the moment of momentum Equation for the mechanical energy 6 The equations of motion for Newtonian incompressible uids De nition of pressure in a moving uid Constitutive relationship for Newtonian uids The NavierStokes equations The dynamic pressure 7 Initial and boundary conditions Initial and boundary conditions for the NavierStokes equations Kinematic boundary condition Continuity of the tangential component of the velocity Dynamic boundary conditions Two relevant cases 8 Scaling and dimensional analysis Units of measurement and systems of units Dimension of a physical quantity Quantities with independent dimensions Buckingham's  theorem Dimensionless NavierStokes equations Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 3 / 161Table of contents III 9 Unidirectional ows Introduction to unidirectional ows Some examples of unidirectional ows Unsteady unidirectional ows Axisymmetric ow with circular streamlines 10 Low Reynolds number ows Introduction to low Reynolds number ows Slow ow past a sphere Lubrication Theory 11 High Reynolds number ows The Bernoulli theorem Vorticity equation and vorticity production Irrotational ows Bernoulli equation for irrotational ows 12 Appendix A: material derivative of the Jacobian Determinants Derivative of the Jacobian 13 Appendix B: the equations of motion in di erent coordinates systems Cylindrical coordinates Spherical polar coordinates 14 References Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 4 / 161Acknowledgements Acknowledgements These lecture notes were originally written for the course in \Fluid Dynamics", taught in L'Aquila within the MathMods, Erasmus Mundus MSc Course. A large body of the material presented here is based on notes written by Prof. Giovanni Seminara from the University of Genoa to whom I am deeply indebted. Further sources of material have been the following textbooks: Acheson (1990), Aris (1962), Barenblatt (2003), Batchelor (1967), Ockendon and Ockendon (1995), Pozrikidis (2010). I wish to thank Julia Meskauskas (University of L'Aquila) and Andrea Bon glio (University of Genoa) for carefully checking these notes. Very instructive lms about uid motion have been released by the National Committee for Fluid Mechanics and are available at the following link: http://web.mit.edu/hml/ncfmf.html Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 5 / 161Stress in uids The state of stress in uids Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 6 / 161Stress in uids The continuum approach The continuum approach I De nition of a simple uid The characteristic property of uids (both liquids and gases) consists in the ease with which they can be deformed. A proper de nition of a uid is not easy to state as, in many circumstances, it is not obvious to distinguish a uid from a solid. In this course we will deal with \simple uids", which Batchelor (1967) de nes as follows. \A simple uid is a material such that the relative positions of elements of the material change by an amount which is not small when suitable chosen forces, however small in magnitude, are applied to the material. . . . In particular a simple uid cannot withstand any tendency by applied forces to deform it in a way which leaves the volume unchanged." Note: the above de nition does not imply that there will not be resistance to deformation. Rather, it implies that this resistance goes to zero as the rate of deformation vanishes. Microscopic structure of uids The macroscopic properties of solids and uids are related repulsion to their molecular nature and to the forces acting between molecules. In the gure a qualitative diagram of the force between two molecules as a function of their distance d is d 0 shown. d d repulsion; d d attraction, where 0 0 attraction 10 d  10 m. 0 distance d Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 7 / 161 forceStress in uids The continuum approach The continuum approach II Let d be the average distance between molecules. We have gases d d ; 0 solids and liquids d d . 0 In solids the relative position of particles is xed, in uids (liquids and gases) it can be freely rearranged. Continuum assumption Molecules are separated by voids and the percentage of volume occupied by molecules is very small compared to the total volume. In most applications of uid mechanics the typical spatial scale L under consideration is much larger than the spacing between molecules d. We can then suppose that the behaviour of the uid is the same as if the uid was perfectly continuous in structure. This means that any physical property of the uid, say f , can be regarded as a continuous function of space x (and possibly time t) f = f (x; t):  In order for the continuum approach to be valid it has to be possible to nd a length scale L which is much smaller than the smallest spatial scale at which macroscopic changes take place and much larger than the microscopic (molecular) scale.  5 For instance in uid mechanics normally a length scale L = 10 m is much smaller than the  scale of macroscopic changes but still we have L  d. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 8 / 161Stress in uids Forces on a continuum Forces on a continuum I Two kind of forces can act on a continuum body: long distance forces; short distance forces. Long distance forces Such forces are slowly varying in space. This means that if we consider a small volume V the force is approximately constant over it. Therefore, we may write F = fV: As long distance forces are proportional to the volume of uid they act on, they are referred to as volume or body forces. In most cases of interest for this course F will be proportional to the mass of the element F =fV; 3 where  denotes density, i.e. mass per unit volume. The dimensions of  are  = ML (with M 3 mass and L length), and in the International System (SI) it is measured in kg m . The vector eld f is denominated body force eld. f has the dimension of an acceleration, or 2 2 force per unit mass f = LT (with T time), and in the SI it is measured in m/s . In general f depends on space and time: f = f(x; t). If we want to compute the force F on a nite volume V we need to integrate f over V ZZZ F = fdV: V Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 9 / 161Stress in uids Forces on a continuum Forces on a continuum II Short distance forces Such forces are extremely rapidly variable in space and they act on very short distances. This means that short distance forces are only felt on the surface of contact between adjacent portions of uid. Therefore, we may write  = tS: As short distance forces are proportional to the surface they act on, they are referred to as surface forces. The vector t is denominated tension. The tension t has the dimension of a force 2 1 2 2 per unit surface t = FL = ML T , and in the SI it is measured in Pa=N m . The vector t depends on space x, time t and on the unit vector n normal to the surface on which the stress acts: t = t(x; t; n). Convention: we assume that t is the force per unit surface that the uid on the side of the surface towards which n points exerts on the uid on the other side. Important note: t(n) =t(n). If we want to compute the force  on a nite surface S we need integrating t over S: ZZ  = tdS: S Note that, if S is a closed surface,  represents the force that the uid outside of S exerts on the uid inside. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 10 / 161Stress in uids The stress tensor The stress tensor I Cauchy's stress principle We now wish to characterise the state of stress at a point P of a continuum. To this end we consider a small tetrahedron of volume V centred in P. In the gure on the right e i denotes the unit vector in the direction of the axis x i (i = 1; 2; 3). The total surface force acting on the tetrahedron is t(n)S + t(e )S + t(e )S + t(e )S = 0: 1 1 2 2 3 3 In the above expression we have not displayed the dependence of t on x, as the value of x is approximately constant over the small tetrahedron. Moreover, t is xed. Note that if we wrote the momentum balance for the tetrahedron, volume forces would vanish more rapidly than surface forces as the volume tends to zero. Therefore, at leading order, only surface forces contribute to the balance. We note that S = e  nS: i i Therefore S t(n) t(e )e  n t(e )e  n t(e )e  n = 0; 1 1 2 2 3 3 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 11 / 161Stress in uids The stress tensor The stress tensor II or, in index notation,   S t (n) t (e )e n t (e )e n t (e )e n = 0: i i 1 1j j i 2 2j j i 3 3j j Note that, throughout the course we will adopt Einstein notation or Einstein summation convention. According to this convention, when an index variable appears twice in a single term of a mathematical expression, it implies that we are summing over all possible values of the index (typically 1, 2, 3). Thus, for instance f f f f i i i i f g = f g + f g + f g or f = f + f + f : j j 1 1 2 2 3 3 j 1 2 3 x x x x j 1 2 3 We can now write   t (n) = t (e )e + t (e )e + t (e )e n: i i 1 1j i 2 2j i 3 3j j Since neither the vector t nor n depend on the coordinate system, the term in square brackets in the above equation is also independent of it. Thus it represents a second order tensor, say (or in index notation  ). ij We can thus write t (n) = n; or, in vector notation, t(n) =n: (1) i ij j  is named the Cauchy stress tensor, or simply stress tensor.  represents the i component of ij ij the stress on the plane orthogonal to the unit vector e . j Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 12 / 161Stress in uids The stress tensor The stress tensor III Equation (1) implies that to characterise the stress in a point of a continuum we need a second order tensor, i.e. (given a coordinate system) 9 scalar quantities. We will show in the following (section 5) that  is symmetric ( = ), and therefore such scalar quantities reduce to 6. ij ij ji The terms appearing in the principal diagonal of the matrix  represent the so called normal ij stresses, those out of the principal diagonal are named tangential or shear stresses. It is always possible to choose Cartesian coordinates such that takes a diagonal form 0 1  0 0 I A 0  0 ; II 0 0  III and  ,  ,  are named principal stresses and they are the eigenvalues of the matrix I II III representing  . The corresponding directions are called principal directions. ij Obviously, the components of  depend on the coordinate system but the stress tensor does not ij as it is a quantity with a precise physical meaning. For any second order tensor it is possible to de ne 3 invariants, i.e. 3 quantities that do not depend on the choice of the coordinate system. A commonly used set of invariants is given by I = + + = tr = ; I =  +  +  ; I =   = det: 1 I II III jj 2 I II II III III I 3 I II III Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 13 / 161Stress in uids Tension in a uid at rest Tension in a uid at rest I The structure of in a uid at rest is a consequence of the de nition of simple uid put forward. We consider a small spherical domain in a uid at rest. Since the sphere is very small must be approximately constant at all points within the sphere. We locally choose the principal axes so that we can write as 0 1  0 0 I A  = 0  0 : II 0 0  III We can now write = + , where 1 2 0 1 0 1 1=3 0 0  1=3 0 0 jj I jj A A  = 0 1=3 0 ;  = 0  1=3 0 : 1 jj 2 II jj 0 0 1=3 0 0  1=3 jj III jj The tensor is spherical. It represents a normal compression on the sphere (see gure (a) 1 below). In fact on any portion S of normal n the force is given by S n = 1=3S n. 1 jj The second tensor is diagonal and the sum of the terms on the diagonal is zero. This means 2 that, excluding the trivial case in which all terms are zero, at least one term is positive and one is negative. Referring to the gure on the right this implies that this state of stress necessarily tend to change the shape of the small volume we are considering. This is not compatible with the de nition of simple uid given before, according to which such uid is not able to withstand a system of forces that tends to change its shape. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 14 / 161Stress in uids Tension in a uid at rest Tension in a uid at rest II Therefore, must be equal to zero in a uid at rest. Since uids are normally in a state of 2 compression we set  =p ; or, in vector form,  =pI; (2) ij ij where the scalar quantity p is called pressure, and I is the identity matrix. Note that, due to the minus sign in the above equation, p 0 implies compression. In general the pressure is a function of space and time p(x; t). p has the dimension of a force per unit area 2 1 2 (p = FL = ML T ) and in the SI is measured in Pa. Equation (2) implies that at a given point P of a uid at rest the force acting on a small surface passing from P is equal topn, i.e. it is always normal to the surface and its magnitude does not depend on the orientation of the surface. Note: in some textbooks (2) is assumed as an indirect de nition of a simple uid (Euler assumption). Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 15 / 161Statics of uids Statics of uids Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 16 / 161Statics of uids The equation of statics The equation of statics I Equation of statics in integral form Let V be a volume of uid within a body of uid at rest and let S be its bounding surface. We wish to write the equilibrium equation for this volume. From the equilibrium of forces we have ZZZ ZZ fdV + tdS = 0: (3) V S Equation (2) allows to rewrite the above expression as ZZZ ZZ fdV + pndS = 0; (4) V S which represents the integral form of the equation of statics. The above equation is often conveniently written in compact form as F +  = 0; (5) with F resultant of all body forces acting on V and  resultant of surface forces acting on S. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 17 / 161Statics of uids The equation of statics The equation of statics II Equation of statics in di erential form Using Gauss theorem equation (4) can be written as ZZZ frpdV = 0: V Since V is arbitrary the following di erential equation must hold p frp = 0; or, in index notation, f = 0; (6) i x i which is the equation of statics in di erential form. Equilibrium to rotation In principle, the above equation alone is not sucient to ensure equilibrium as we also have to impose an equilibrium balance to rotation. This can be written as ZZZ ZZ x fdV + px ndS = 0; (7) V S Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 18 / 161Statics of uids The equation of statics The equation of statics III or, in index notation, ZZZ ZZ  x f dV + p x n dS = 0: (8) ijk j k ijk j k V S Note:  is the alternating tensor. Its terms are all equal to zero unless when i, j and k are ijk di erent from each other, in which case  takes the values 1 or 1 depending if i, j and k are or ijk not in cyclic order. Thus, we have i j k  ijk 1 2 3 1 3 1 2 1 2 3 1 1 2 1 3 1 1 3 2 1 3 2 1 1 Applying Gauss theorem to equation (8) we have: ZZZ    x f (px ) dV = 0: ijk j k j x k V Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 19 / 161Statics of uids The equation of statics The equation of statics IV Carrying on the calculations: ZZZ    x f (px ) dV = ijk j k j x k V ZZZ   p x j  x f x p dV = ijk j k j x x k k V   (9) ZZZ p  x f x p dV = ijk j k j jk x k V   ZZZ ZZZ p  x f x dV p  dV = 0; ijk j k j ijk jk x k V V where  is the Kronecker delta ( = 0 if i =6 j and  = 1 if i = j). The above equation is ij ij ij automatically satis ed as the rst integral vanishes due to equation (6) and   = 0 by ijk ij de nition. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 20 / 161Statics of uids Implications of the equation of statics Implications of the equation of statics Let us now consider the equation of statics (6). In order to integrate this equation we need an equation of state for the uid, stating how the density  depends on the other physical properties of the uid, and in particular p. However, some general conclusions can be drawn by simple inspection of the equation. As a rst consideration we note that not all f(x) and p(x) allow for a uid to be at rest. In particular the relationship f(x) =rp implies that f(x) admits a potential W , so that f(x) =rW: In the particular case in which  = const, f has to be conservative. If f is conservative we have that f =r. In this case we have r =rp: Applying the curl to the above expression we nd      r (r) =rrp ) rrrr =rrp:   The above relationship implies that level surfaces of  and  must coincide. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 21 / 161Statics of uids Statics of incompressible uids in the gravitational eld Statics of incompressible uids in the gravitational eld We assume  = const. In this case we say that the uid behaves as if it was incompressible. f is the gravitational body force eld. We consider a system of Cartesian coordinates (x ; x ; x ), with x vertical upward directed axis. 1 2 3 3 The gravitational eld can therefore be written as f = (0; 0;g). With the above assumptions equation (6) can be easily solved to get p =gx + const; 3 and, after rearrangement, we obtain Stevin law p x + = const; (10) 3 3 3 where is the speci c weight of the uid ( = FL , measured in N m in the SI). The quantity h = x + p= is called piezometric or hydraulic head. Stevin law implies that, in an 3 incompressible uid at rest, h is constant. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 22 / 161Statics of uids Equilibrium conditions at interfaces Equilibrium conditions at interfaces I Surface tension The fact that small liquid drops form in air and gas bubbles form in liquids can be explained by assuming that a surface tension acts at the interface between the two uids. If we draw a curve across the interface we assume that a force per unit length of magnitude  exists, acting on the surface containing the interface and in the direction orthogonal to the curve. The dimension of  1 2 is  = FL = MT ans in the SI is measured in N 1 m . Drop of water on a leaf. The existence of such a force can be explained considering what happens at molecular level, close to the interface: due to the existence of the interface, there is no balance of molecular forces acting on particles very close to the interface.  can be positive (traction force on the surface) or negative (compression force on the surface), depending on the two uids in contact. In particular we have:  0 immiscible uids;  0 miscible uids. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 23 / 161Statics of uids Equilibrium conditions at interfaces Equilibrium conditions at interfaces II Pressure jump across a curved surface We consider an equilibrium interface between two uids. This implies that  = const on the surface. We consider a curved surface. Let O be a point on the surface and let us adopt a system of coordinates centred in O and such that the (x y) plane is tangent to the surface. The equation of the surface is F (x; y; z) = z(x; y) = 0: (11) Note that  and its rst derivatives are zero at (x; y) = (0; 0). Close to O the approximate expression of the normal vector n is   rF   n =  ; ; 1 ; jrFj x y correct to the rst order in the small quantities =x, =y. The resultant of the tensile force on a small portion of the surface S containing O is given by I  n dx; C with n normal to the surface and dx a line element of the closed curve C bounding the surface S.   Recalling the equation of the surface (11) we can write dx = (dx; dy; dx + dy). x y Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 24 / 161Statics of uids Equilibrium conditions at interfaces Equilibrium conditions at interfaces III If the surface is at, n is uniform and the above integral is zero. If the surface is curved the resultant is directed, at leading order, along z and has magnitude I    dy + dx: x y C Green's theorem states that   ZZ I g f dxdy = fdx + gdy: x y C S In the present case the above equation can be speci ed so that   f = ; g = : y x Therefore we get: I ZZ     2 2 2 2        dy dx = + dS + S: 2 2 2 2 x y x y x y C O S We nally nd     2 2   1 1  + = + ; 2 2 x y R R 1 2 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 25 / 161Statics of uids Equilibrium conditions at interfaces Equilibrium conditions at interfaces IV where R and R are radii of curvature of the surface along two orthogonal directions. Note that 1 2 1 1 it can be shown that + is independent on the orientation chosen. R R 1 2 The above equation implies that, in order for a curved interface between two uids to be in equilibrium, a pressure jump p must exist across the surface so that   1 1 p = + : (12) R R 1 2 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 26 / 161Statics of uids Hydrostatic forces on at surfaces Hydrostatic forces on at surfaces I We adopt in this section the following assumptions:  = const; f = (0; 0;g) gravitational eld. We wish to compute the force on a at solid surface. Magnitude of the force The magnitude of this force is given by ZZ ZZ jj = pndS = pdS: S S Note that, by de nition,  is the force that the surface exerts on the uid. Thus the force of the uid on the surface is equal to. We consider a plane inclined by an angle with respect to a horizontal plane and introduce a coordinate , with origin on the horizontal plane where p = 0, laying on the surface and oriented along the line of maximum slope on the surface. We therefore can write, using equation (10), p = sin: Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 27 / 161Statics of uids Hydrostatic forces on at surfaces Hydrostatic forces on at surfaces II Substituting in the de nition ofjj we obtain ZZ ZZ jj =  sindS = sin dS = sinS; (13) S S whereS is the static moment of the surface S with respect to the axis y, de ned as ZZ S = dS: (14) S S can be written asS = S, with  being the  coordinate of the centre of mass of S. Thus G G we can write jj = sin S = z S = p S; (15) G G G where z is a vertical coordinate directed downwards and with origin on the horizontal plane p = 0 (see the gure of the previous page), and p is the pressure in the centre of mass of S. G Equation (15) states that the magnitude of the force exerted by an incompressible uid at rest in the gravitational eld on a at surface is given by the product of the pressure p at the G centre of mass of the surface and the area of the surface S. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 28 / 161Statics of uids Hydrostatic forces on at surfaces Hydrostatic forces on at surfaces III Application point of the force We now wish to determine where the force  is applied. To this end we impose the equilibrium to rotation with respect to the y axis, given by the intersection of the planes  = 0 and z = 0, ZZ ZZ ZZ 2 2  jj = pdS = sin dS = sin  dS = sinI; (16) C S S S where we have introduced the moment of inertiaI of the surface with respect to the axis y ZZ 2 I =  dS: (17) S Finally, we recall that 2 I =I + S; (18) 0 G whereI is the moment of inertia of the surface with respect to an axis parallel to y and passing 0 through the centre of mass of the surface. Thus, substituting (18) into (16) and recalling equations (13) and (14) we obtain I I 0 0  = + = + ; (19) C G G  S S G which is often more convenient to use than (16). Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 29 / 161Statics of uids Hydrostatic forces of curved surfaces Hydrostatic forces of curved surfaces In the case of forces on curved surfaces it is not possible to take the normal vector n out of the following integral ZZ  = pndS; S as n changes from point to point on S. In this case it is necessary to specify explicitly n and solve the integral. An alternative, often more convenient, method consists of selecting a closed control volume, bounded by the curved surface and by a suitable number of at surfaces. In this case the calculation of the forces on the at surfaces is straightforward and the force on the curved surface can be determined employing the integral form of the statics equation (4), provided it is possible to compute the volume of the control volume. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 30 / 161Kinematics of uids Kinematics of uids Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 31 / 161Kinematics of uids Spatial and material coordinates Spatial and material coordinates I The kinematics of uids studies uid motion per se, with no concern to the forces which generate the motion. All kinematic notions that will be introduced in the present chapter are valid for any uid described as a continuum. A very good reference for kinematics of uid is Aris (1962); the present section is largely based on this textbook. Understanding how to study uid motion from the kinematic point of view is a prerequisite to study the dynamics of uids, which will be considered in the following chapter. The basic mathematical idea is that, within the continuum approach, uid motion can be described by a point transformation. Material coordinates Let us consider a uid particle which at time t is located in the position = ( ; ; ). The 0 1 2 3 same particle at time t is at position x = (x ; x ; x ). Without loss of generality we can set 1 2 3 t = 0. The motion of the particle in the time interval 0; t is described by the function 0 x ='(; t); or, in index notation, x =' ( ; ; ; t); (20) i i 1 2 3 which, at any time t, tells us the position in space of the particle that was in at t = 0. If in the equation (20) we x we obtain the particle trajectory.  are named material or Lagrangian coordinates as a particular value of identi es the material particle that at t = 0 was in. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 32 / 161Kinematics of uids Spatial and material coordinates Spatial and material coordinates II Particle velocity We can de ne the particle velocity as '(; t) u = : t Particle acceleration We can de ne the particle acceleration as 2 u(; t) '(; t) a = = : 2 t t Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 33 / 161Kinematics of uids Spatial and material coordinates Spatial and material coordinates III Spatial coordinates We assume that the motion is continuous, that at a given time a single particle cannot occupy two di erent positions and, conversely, that a single point in space cannot be occupied simultaneously by two particles. This implies that equation (20) can be inverted to obtain  = (x; t); or, in index notation,  =  (x ; x ; x ; t): (21) i i 1 2 3 Equation (21) gives the initial position (at t = 0) of a material particle that at time t is in x. Mathematically, the condition of invertibility of (20) can be expressed as J 0 (see Aris, 1962), where the Jacobian J is de ned as   (x ; x ; x ) 1 2 3 J = det : (22) ( ; ; ) 1 2 3 Knowledge of equation (20) or (21) is enough to completely describe the ow. The ow, however, can also be studied by describing how any uid property, sayF (e.g. density, pressure, velocity, . . . ) changes in time at any position in space. F = f (x; t): S This approach is referred to as spatial approach or Eulerian approach. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 34 / 161Kinematics of uids Spatial and material coordinates Spatial and material coordinates IV Alternatively, we can describe the evolution of a uid propertyF associated with a given uid particle. In this case we write F = f (; t): M Note that a given value of identi es the particle that in t = 0 was in. This approach is referred to as material approach or Lagrangian approach. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 35 / 161Kinematics of uids The material derivative The material derivative I The time derivative of a generic physical property of the uidF has a di erent meaning in Eulerian and Lagrangian coordinates. Eulerian coordinates: F f (x; t) S = local derivative: t t This represents the variation in time ofF at a given point in space. Such a point can, in general, be occupied by di erent particles at di erent times. Lagrangian coordinates: DF f (; t) M = material derivative: Dt t This represents the time evolution ofF associated with a given material particle. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 36 / 161Kinematics of uids The material derivative The material derivative II Since the physical meaning of the two derivatives is di erent it is customary in uid mechanics to denote them with di erent symbols.     time derivative at constant x; t t x   D   time derivative at constant: Dt t  Note: If the uid propertyF is the position of a material particle (F =' ) we have i D' D' i u = ; or in vector notation u = ; (23) i Dt Dt which is the velocity of the uid particle. In general it is more convenient in uid mechanics to adopt a spatial (Eulerian) description of the ow. However, for the de nition of some physical quantities the material derivative is required. For instance the acceleration a is de ned as Du a = ; Dt while u=t =6 a, as it represents the rate of change of velocity at a xed point in space, i.e. it is not referred to a material particle. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 37 / 161Kinematics of uids The material derivative The material derivative III It is then often necessary to de ne the material derivative in terms of spatial coordinates. Using (20) we can write f (x; t) = f '(; t); t = f (; t); S S M from which it follows DF f (; t) f f ' f f M S S i S S = = + = + u : i Dt t t x t t x i i Thus we nd DF F F DF F = + u ; or, in vector form, = + urF: (24) i Dt t x Dt t i IfF is a vector quantity we obtain DF F F DF F i i i = + u ; or in vector form, = + (ur)F: (25) j Dt t x Dt t j Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 38 / 161Kinematics of uids De nition of some kinematic quantities De nition of some kinematic quantities I Trajectories or particle paths Equation (20) can be seen as a parametric equation of a curve in space, with t as parameter. The curve goes through the point when t = 0. It represents the particle path or pathline or particle trajectory. Particle trajectories can be obtained from a spatial description of the ow by integration of the spatial velocity eld dx = u(x; t); x(0) =: (26) dt Steady ow The velocity eld in spatial coordinates is described by the vector eld u(x; t). If u does not depend on time the ow is said to be steady. Note that steadiness of ow does not imply that each material particle has a constant velocity in time as u(; t) might still depend on time. Streamlines Given a spatial description of a velocity eld u(x; t), streamlines are curves which are at all points in space parallel to the velocity vector. Mathematically, they are therefore de ned as dx u = 0; (27) Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 39 / 161Kinematics of uids De nition of some kinematic quantities De nition of some kinematic quantities II with dx an in nitesimal segment along the streamline. The above expression can also be written as dx dx dx 1 2 3 = = : (28) u u u 1 2 3 The unit vector dx=jdxj can be written as dx=ds, where the curve parameter s is the arc length measured from an initial point x = x(s = 0). Equation (27) then implies that 0 dx u = : (29) ds juj Particle paths and streamlines are not in general coincident. However they are in the following cases. dx Steady ow. In this case the equation for a pathline is = u(x). The element of the arc dt length along the pathline is ds =jujdt, which, substituted in the above expression, yields dx u = ; ds juj which shows that, for a steady ow, the di erential equation for pathlines and streamlines are the same. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 40 / 161Kinematics of uids De nition of some kinematic quantities De nition of some kinematic quantities III Unsteady ow the direction of which does not change with time. In this case we can write u(x; t) = f (x; t)u (x) with u the velocity eld at the initial time. In this case the argument 0 0 used for steady ows still holds. Uniform ow A ow is said to be uniform if u does not depend on x. u = u(t): This is a very strong requirement. Sometimes the ow is called uniform if u does not change along the streamlines. Plane ow A ow is said to be plane or twodimensional if it is everywhere orthogonal to one direction and independent of translations along such direction. In a plane ow it is therefore possible to choose a system of Cartesian coordinates (x ; x ; x ) so 1 2 3 that u has the form u = (u ; u ; 0); 1 2 and u and u do not depend on x . 1 2 3 Axisymmetric ow A ow is said to be axisymmetric if, chosen a proper system of cylindrical coordinates (z; r;') the velocity u = (u ; u ; u ) is independent of the azimuthal coordinate ', and u = 0. z r ' ' Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 41 / 161Kinematics of uids Reynolds transport theorem Reynolds transport theorem I LetF(x; t) be a property, either a scalar or a vector, of the uid and V (t) a material volume entirely occupied by the uid. A material volume is a volume which is always constituted by the same particles. We can de ne the integral ZZZ F (t) = F(x; t)dV: (30) V (t) We wish to evaluate the material derivative of F . Since V (t) depends on time the derivative D=Dt can not be taken into the integral. However, if we work with material coordinates, the volume remains unchanged in time and equal to the value V it had at the initial time. We can 0 thus write ZZZ ZZZ D D F(x; t)dV = F(; t)JdV ; 0 Dt Dt V (t) V 0 where dV = JdV , with J Jacobian of the transformation, de ned by (22). We can now write 0 ZZZ DF DJ J +F dV : 0 Dt Dt V 0 It can be shown (see section 12) that DJ = (r u)J: (31) Dt Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 42 / 161Kinematics of uids Reynolds transport theorem Reynolds transport theorem II The above integral can then be written as   ZZZ DF +F(r u) JdV ; 0 Dt V 0 and going back to the spatial coordinates x, we nd   ZZZ DF +F(r u) dV: Dt V (t) Finally, recalling (24), we have ZZZ ZZZ   D F F(x; t)dV = +r (Fu) dV: (32) Dt t V (t) V (t) This result is known as Reynolds transport theorem. The above expression can be also be written as ZZZ ZZZ ZZ D F F(x; t)dV = dV + Fu ndA; (33) Dt t V (t) V (t) S(t) Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 43 / 161Kinematics of uids Reynolds transport theorem Reynolds transport theorem III with S(t) being the bounding surface of the volume V (t) and n the outer normal to this surface. Equation (33) shows that the material derivative of a variableF integrated over a material volume V (t) can be written as the integral of F=t over the volume V (t) plus the ux ofF through the surface S(t) of this volume. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 44 / 161Kinematics of uids Principle of conservation of mass Principle of conservation of mass Let us consider a material volume V with bounding surface S. The principle of conservation of mass imposes that: the material derivative of the mass of uid in V is equal to zero. The mass of the uid in V is given by ZZZ dV: V Therefore we have ZZZ D dV = 0: Dt V Recalling (32) we have: ZZZ  +r (u)dV = 0: (34) t V since the volume V is arbitrary the following di erential equation holds   +r (u) = 0; or, in index notation, + (u ) = 0: (35) j t t x j This equation is known in uid mechanics as continuity equation. In the particular case in which the uid is incompressible, i.e. the density  is constant, the above equation reduces to u j r u = 0; or, in index notation, = 0: (36) x j This implies that the velocity eld of an incompressible uid is divergence free. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 45 / 161Kinematics of uids The streamfunction The streamfunction I A di erential form df = p(x; y)dx + q(x; y)dy; R is said to be an exact di erential if df is path independent. This happens when f f df = dx + dy: x y Therefore, in this case f f p = ; q = ; x y and this implies p q = : (37) y x Plane ow of an incompressible uid Let us consider a plane ow on the (x ; x ) plane so that the velocity has only two components u 1 2 1 and u . Let us also assume that the uid is incompressible. The continuity equation (36) reduces 2 to u u 1 2 + = 0: x x 1 2 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 46 / 161Kinematics of uids The streamfunction The streamfunction II The above expression implies that the following di erential d =u dx + u dx ; 2 1 1 2 is exact, as the condition (37) is satis ed. Then we have u = ; u = ; (38) 1 2 x x 2 1 and the scalar function (x ; x ; t) is de ned as 1 2 Z = (u dx + u dx ): (39) 0 2 1 1 2 In the above expression is a constant and the line integral is taken on an arbitrary path joining 0 the reference point O to a point P with coordinates (x ; x ). We know that, as d is an exact 1 2 di erential, the value of does not depend on the path of integration but only on the initial 0 and nals points. The function has a very important physical meaning. The ux of uid volume across the line joining the points O and P (taken positive if the ux is in the anticlockwise direction about P) is given by the integral Z (u dx + u dx ): 2 1 1 2 This means that the ux through any curve joining two points is equal to the di erence of the value of at these points. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 47 / 161Kinematics of uids The streamfunction The streamfunction III Therefore the value of is constant along streamlines as, by de nition, the ux across any streamline is zero. For this reason the function is named streamfunction. The advantage of having introduced the streamfunction is that we can describe the ow using a scalar function rather than the vector function u. Axisymmetric ow of an incompressible uid Let us now consider an axisymmetric ow of an incompressible uid. Let us assume a system of cylindrical coordinates (z; r;'). The corresponding velocity components are (u ; u ; u ). Due to the axisymmetry of the ow we know that u = 0 and that z r ' ' u and u do not depend on '. In this case the continuity equation (36) reads r z u 1ru z r r u = + = 0: z r r We can again de ne a streamfunction as Z 1 1 u = ; u = ; = r(u dr u dz): (40) r z 0 z r r z r r The streamfunction for an incompressible axisymmetric ow can also be expressed in terms of other orthogonal systems of coordinates, e.g. spherical polar coordinates, see equation (121). Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 48 / 161Kinematics of uids The velocity gradient tensor The velocity gradient tensor I Let us consider two nearby points P and Q with material coordinates and + d. At time t their position is x(; t) and x( + d; t). We can relate the position of the two particles with the following relationship x i 2 x ( + d; t) = x (; t) + d + O(d ); i i j  j 2 2 where O(d ) represents terms of order d or smaller that will be neglected. The small displacement vector d at the time t has become dx = x( + d; t) x(; t) and it takes the expression x i dx = d: (41) i j  j x i De nition: the quantity is a tensor which is named displacement gradient tensor. This  j tensor is fundamental in the theory of elasticity. In uid mechanics it is more signi cant to reason in terms of velocities (u = Dx=Dt). The relative velocity of two particles with material coordinates and + d can be written as   u D x i i du = d = d: (42) i j j  Dt  j j Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 49 / 161Kinematics of uids The velocity gradient tensor The velocity gradient tensor II Inverting (41) we can rewrite the above expression as u  u i k i du = dx = dx: (43) i j j  x x k j j The above equation expresses the relative velocity in terms of the current relative position. u i De nition: the quantity (orru in vector form) is a tensor that is named velocity gradient x j tensor. In generalru is non symmetric. Any tensor can be decomposed into a symmetric and an antisymmetric part. In particular we can write     u 1 u u 1 u u i i j i j = + + ; or in vector form ru = D + : (44) x 2 x x 2 x x j j i j i Above we have de ned   u 1 u i j D = + ; rate of deformation tensor; (45) ij 2 x x j i   u 1 u i j = ; rate of rotation tensor: (46) ij 2 x x j i Both the above tensors play a vey important role in uid mechanics. Their physical meaning is explained in the two following sections. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 50 / 161Kinematics of uids Physical interpretation of the rate of deformation tensor D Physical interpretation of the rate of deformation tensor I We now wish to interpret the physical meaning of the rate of deformation tensor D. Let us consider how a small material element of uid deforms during motion. Let P and Q be two close material particles with coordinates and + d, whose positions at time t are x(; t) and x( + d; t). Let the length of the small segment connecting P and Q at time t be ds. Recalling (41) we can write x x i i 2 ds = dx dx = d d : i i j k   j k 2 Let us now take the material derivative of ds .   D u x x u u x i i i i i i 2 ds = + d d = 2 d d : j k j k Dt       j k j k j k Note that we have used the fact that d and d do not change in time as they are material j k segments. Moreover, we could swap j and k as they both are dummy indexes. Recalling (41), (42) and (43) we know that u u x i i i d = dx; d = dx: j j k i  x  j j k Therefore, we can write 1 D D u 2 i ds = ds ds = dx dx = D dx dx: i j ij i j 2 Dt Dt x j Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 51 / 161Kinematics of uids Physical interpretation of the rate of deformation tensor D Physical interpretation of the rate of deformation tensor II In the above expression we have used the fact the antisymmetric terms in u=x vanish upon i j summation and, therefore, only the symmetric part of the velocity gradient tensor (i.e. D ) ij survives. The above expression can also be rewritten as 1 D dx dx i j ds = D : (47) ij ds Dt ds ds th The term dx=ds is the i component of a unit vector in the direction of the segment PQ. i Therefore equation (47) states that the rate of change of the length of the segment (as a fraction of its length) is related to its direction through the deformation tensor D. We can also observe that if D = 0 the segment PQ remains of constant length. Therefore we can state that if D = 0 the motion is locally and instantaneously rigid. The tensor D is therefore related to deformation of material elements. Meaning of the terms on the main diagonal of D dx Let PQ be parallel to the coordinate axis x . In this case = e , with e unit vector in the 1 1 1 ds direction of x . Then equation (47) simpli es to 1 1 D dx = D : 1 11 dx Dt 1 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 52 / 161Kinematics of uids Physical interpretation of the rate of deformation tensor D Physical interpretation of the rate of deformation tensor III Thus the element D represents the rate of longitudinal strain of an element parallel to x . 11 1 Obviously, the same interpretation applies to the other two terms on the main diagonal of D, i.e. D and D . 22 33 Meaning of the terms out of the main diagonal of D We now consider two segments PQ and PR, where R is a 0 0 material particle with material coordinates + d . Let ds be the length of the segment PR and  the angle between the segments PQ and PR. We then have 0 0 ds ds cos = dx dx : i i Taking the material derivative of the above expression, using again (43), we have D u u i i 0 0 0 0 0 (ds ds cos) = du dx + dx du = dx dx + dx dx : i i j i i i i j Dt x x j j As i and j are dummy indexes they can be interchanged, and we can then write     0 0 1 D 1 D D u u dx dx dx dx j j j 0 i i i cos ds + ds sin = + = 2D : ij 0 0 0 ds Dt ds Dt Dt x x ds ds ds ds i j Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 53 / 161Kinematics of uids Physical interpretation of the rate of deformation tensor D Physical interpretation of the rate of deformation tensor IV 0 Now suppose, as an example, that dx is parallel to the axis x and dx to the axis x . This 1 2 0 0 implies that dx =ds = , dx=ds = and  = ==2. Then we have i1 i j2 12 i D 12 = 2D : 12 Dt This implies that the term D (with i6= j) can be interpreted as one half of the rate of decrease ij of the angle between two segments parallel to the x and x axes, respectively. i j Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 54 / 161Kinematics of uids Physical interpretation of the rate of rotation tensor Physical interpretation of the rate of rotation tensor I We now consider the tensor de ned by equation (46). We rst note that an antisymmetric tensor can be related to a vector by the following relationship 1 =  ; (48) ij ijk k 2 where the coecient1=2 has been introduced for convenience. and have the following forms 0 1 0 1 0 3 2 1 1 A A = 0 ; = : 3 1 2 2 0 2 1 3 Comparing the above equation with the de nition of given in (46) we obtain       u u u u u u 3 2 1 3 2 1 = ; = ; = : 1 2 3 x x x x x x 2 3 3 1 1 2 Thus is the curl of the velocity u k =r u; or in index form = : (49) i ijk x j In uid mechanics the vector is known as vorticity. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 55 / 161Kinematics of uids Physical interpretation of the rate of rotation tensor Physical interpretation of the rate of rotation tensor II To show the physical meaning of vorticity let us recall Stokes theorem ZZ ZZ I (r u) ndS =  ndS = u dl; l S S which holds for any open surface S bounded by a closed curve l. We now choose a plane surface S with normal n, bounded by a small circle l of radius r centred at x. Let r be a unit vector connecting the point x to any point on the circle l. Let moreover l be a unit vector tangential to the circle. We thus have l = n r. The average of the projection of the angular velocity of points on l in the normal direction n is I I I ZZ 1 1 1 1 1 n (r u)dl = u (n r)dl = u ldl =  ndS  n: 2 2 2 2r 2r 2r 2S 2 l l l S As this result is valid for any n, this shows that the vorticity =r u can be interpreted as twice the angular velocity of the uid. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 56 / 161Dynamics of uids Dynamics of uids Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 57 / 161Dynamics of uids Momentum equation in integral form Momentum equation in integral form Let us consider a material volume V with bounding surface S. Newton's rst principle states that: the material derivative of the momentum of the uid in V is equal to the resultant of all external forces acting on the volume. The momentum of the uid in V is given by ZZZ udV: V Therefore we have (in index notation): ZZZ ZZZ ZZ D u dV = f dV + t dS: (50) i i i Dt V V S Recalling (33) we have: ZZZ ZZ ZZZ ZZ (u )dV + u u n dS = f dV + t dS: (51) i i j j i i t V S V S This is the integral form of the momentum equation and is often written in compact form as I + W = F + ; (52) with I named local inertia and W being the ux of momentum across S. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 58 / 161Dynamics of uids Momentum equation in di erential form Momentum equation in di erential form I Let us now consider the expression ZZZ ZZZ ZZZ D  F F FdV = (F)+ (Fu )dV = F + +F (u )+u dV; j j j Dt t x t t x x j j j V V V withF any function of space and time. Recalling (35) this simpli es to ZZZ ZZZ ZZZ D F F D FdV =  +u dV =  FdV: (53) j Dt t x Dt j V V V In the particular case in which the generic functionF is the velocity u we have ZZZ ZZZ D D udV =  udV: (54) Dt Dt V V Using equations (1) and (54), equation (50) can be written as ZZZ ZZZ ZZ Du i  dV = f dV +  n dS: i ij j Dt V V S Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 59 / 161Dynamics of uids Momentum equation in di erential form Momentum equation in di erential form II Using Gauss theorem we get ZZZ Du i  f  dV = 0: i ij Dt x j V Since V is arbitrary the following di erential equation must hold    u u u i i ij  + u f = 0; or, in vector form,  +(ur)ufr = 0: (55) j i t x x t j j This is known as Cauchy equation. This equation holds for any continuum body. In order to specify the nature of the continuum a further relationship is needed, describing how the stress tensor depends on the kinematic state of the continuum. This relationship is called constitutive law and will be discussed in section 6. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 60 / 161Dynamics of uids Principle of conservation of the moment of momentum Principle of conservation of the moment of momentum I Given a material volume V , the material derivative of the moment of momentum of the uid in V is equal to the resultant of all external moments acting on V . The above principle is expressed mathematically as follows. ZZZ ZZZ ZZ D x udV = x fdV + x tdS; (56) Dt V V S or, in index notation, 0 1 ZZZ ZZZ ZZ D A  x u dV x f dV x t dS = 0: (57) ijk j k j k j k Dt V V S We use again equation (54) and note that Dx=Dt = u . Moreover, the de nition of the operator j j  implies that ijk ZZZ  u u dV = 0: ijk j k V Thus we have, also using Gauss theorem and equation (1), 2 3 ZZZ   Du  x j k kl 4 5  x f x  dV = 0; ijk j k j kl Dt x x l l V Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 61 / 161Dynamics of uids Principle of conservation of the moment of momentum Principle of conservation of the moment of momentum II and after rearrangement 2 3 ZZZ   Du  k kl 4 5  x  f   dV = 0; ijk j k jl kl Dt x l V The term in brackets in the above equation is zero for equation (55). Therefore we obtain ZZZ    dV = 0: ijk jl kl V Since in the above expression V is arbitrary the following di erential equation must hold:    = 0; ijk jl kl or   = 0: ijk kj The above equation implies:  = ; (58) kj jk which imposes that the stress tensor must be symmetrical. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 62 / 161Dynamics of uids Equation for the mechanical energy Equation for the mechanical energy I Let us now consider Cauchy equation (55) and multiply it by u . Since i is now a repeated index i we obtain the following scalar equation 2  Du  Du 1 u i ij i i u u f u = 0; )  u f u + = 0: i i i i i i i ij ij Dt x 2 Dt x x j j j Reorganising the above expression and using the fact that the tensor  is symmetric, we have ij   2 Du  u 1 1 u i j i  =u f + u +  ; i i i ij ij 2 Dt x 2 x x j j i or, recalling the de nition of the rate of deformation tensor D , given by (45), ij 2  1 Du i  =u f + u D  : i i i ij ij ij 2 Dt x j Integrating the above equation over an arbitrary volume V and using (53) we get ZZZ ZZZ ZZ ZZZ D 1 2 u dV = u f dV + u n dS D  dV: i i i ij j ij ij i Dt 2 V V S V Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 63 / 161Dynamics of uids Equation for the mechanical energy Equation for the mechanical energy II Note that we can de ne the kinetic energyE associated with the uid in V as k ZZZ 1 2 E = u dV: k i 2 v Thus we obtain ZZZ ZZ ZZZ D E = u f dV + u t dS D  dV; (59) k i i i i ij ij Dt V S V or in vector form ZZZ ZZ ZZZ D E = u fdV + u tdS D :dV: (60) k Dt V S V The above equation states that the rate of change of the kinetic energy of the uid in the material volume V is equal to the power associated to the resultant of all external forces minus the internal power used to deform the uid within V . The last term in equation (60) is therefore associated with internal energy dissipation. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 64 / 161The equations of motion for Newtonian incompressible uids The equations of motion for Newtonian incompressible uids Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 65 / 161The equations of motion for Newtonian incompressible uids De nition of pressure in a moving uid De nition of pressure in a moving uid I In section 2 we showed that, in a uid at rest, the stress tensor takes the simple form  =p ; ij ij where the scalar quantity p is the static pressure. In the case of a moving uid the situation is more complicated. In particular: the tangential stresses are not necessarily equal to zero; the normal stresses might depend on the orientation of the surface they act on. This implies that the simple notion of pressure as a normal stress acting equally in all directions is lost. We wish now to nd a proper de nition for the pressure in the case of a moving uid. 1 1 A natural choice is to consider  = tr, which we know to be an invariant under rotation of ii 3 3 1 the axes. A simple physical interpretation of  is available. Let us consider a small cube with ii 3 side dl centred in x. As the cube is small we can assume that is constant within it. Taking a system of Cartesian coordinates (x ; x ; x ) with axes parallel to the sides of the cube the average 1 2 3 value of the normal component of the stress over the surface of the cube is 1 1 2 (2 + 2 + 2 ) dl =  : 11 22 33 ii 2 6dl 3 1 As the is an invariant of, the numerical value of  is independent of the orientation of the ii ii 3 cube. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 66 / 161The equations of motion for Newtonian incompressible uids De nition of pressure in a moving uid De nition of pressure in a moving uid II 1 The quantity  reduces to the static uid pressure when the uid is at rest, and its mechanical ii 3 signi cance makes it an appropriate generalisation of the elementary notion of pressure. Therefore, we adopt the following de nition of pressure 1 1 p =  ; or; p = tr: (61) ii 3 3 Important note Incompressible uids For an incompressible uid the pressure p is an independent, purely dynamical variable. In the rest of this course we will deal exclusively with incompressible uids. Compressible uids In the case of compressible uids we know from classical thermodynamics that we can de ne the pressure of the uid as a parameter of state, making use of an equation of state. Thermodynamical relations refer to equilibrium conditions, so we can denote the thermodynamic pressure as p . e The connection between p and p is not trivial as p refers to dynamic conditions, in which e elements of uid in relative motion might not be in thermodynamic equilibrium. A thorough discussion of this subject can be found in Batchelor (1967). Here it suces to say that, for most applications, is it reasonably correct to assume p = p . e Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 67 / 161The equations of motion for Newtonian incompressible uids De nition of pressure in a moving uid De nition of pressure in a moving uid III For the discussion to follow it is convenient to split to the stress tensor  into an isotropic part ij p , and a deviatoric part d which is entirely due to uid motion. We thus write ij ij  =p + d : (62) ij ij ij The tensor d accounts for tangential stresses and also normal stresses whose sum is zero. ij Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 68 / 161The equations of motion for Newtonian incompressible uids Constitutive relationship for Newtonian uids Constitutive relationship for Newtonian uids I We derive the constitutive relationship under the following assumptions. 1 The tensor d is a continuous function ofru. 2 Ifru = 0 then d = 0, so that =pI, i.e. the stress reduces to the stress in static conditions. 3 The uid is homogeneous, i.e. does not depend explicitly on x. 4 The uid is isotropic, i.e. there is no preferred direction. 5 The relationship between d andru is linear. Both the tensors d andru have nine scalar components. The linear assumption means that each component of d is proportional to the nine components ofru. Hence, in the most general case there are 81 scalar coecients that relate the two tensors, in the form u k d = A ; (63) ij ijkl x l where A is a fourthorder tensor which depends on the local state of the uid but not directly ijkl on the velocity distribution. Note that since d is symmetrical so it must be A in the indices i ij ijkl and j. It is convenient at this stage to recall the decomposition of the velocity gradient tensor (44) into a symmetric and an antisymmetric part u i = D + : ij ij x j Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 69 / 161The equations of motion for Newtonian incompressible uids Constitutive relationship for Newtonian uids Constitutive relationship for Newtonian uids II The assumption of isotropy of the uid implies that the tensor A has to be isotropic. A tensor ijkl is said to be isotropic when its components are unchanged by rotation of the frame of reference. It is known from books on Cartesian tensors (e.g. Aris, 1962) that all isotropic tensors of even order can be written as the sum of products of tensors, with being the Kronecker tensor. In the case of a fourthorder tensor we can write 0 00 A =  +  +   ; ijkl ik jl il jk ij kl 0 00 where ,  and  are scalar coecients. Since A is symmetrical in i and j it must be ijkl 0  =: 0 If  = the tensor A is also symmetrical in the indices k and l. This implies that ijkl A = 0; ijkl kl as is antisymmetric. The fact that d can not depend on is reasonable as, on the ground kl ij kl of intuition, we do not expect that a motion locally consisting of a rigid body rotation induces stress in the uid. Note that this also implies that the assumption 2 has to be rewritten as D = 0) d = 0. We now have that equation (63) reduces to 00 00 d =  D +  D +   D =D +D +  D : ij ik jl kl il jk kl ij kl kl ij ji ij kk Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 70 / 161The equations of motion for Newtonian incompressible uids Constitutive relationship for Newtonian uids Constitutive relationship for Newtonian uids III u k Recalling that D = =r u, the above expression takes the form kk x k 00 d = 2D + r u  : (64) ij ij ij Finally, we recall that, by de nition, d makes no contribution to the mean normal stress, ij therefore 00 d = (2 + 3 )r u = 0; ii and, since this expression holds for any u, we nd 00 2 + 3 = 0: (65) From (62), (64) and (65) we nally obtain the constitutive equation for a Newtonian uid in the form     1 1  =p + 2 D r u  ; or, in vector form,  =pI + 2 D (r u)I : ij ij ij ij 3 3 (66) Notice that for an incompressible uid we haver u = 0 by the continuity equation (36), therefore the constitutive law simpli es in this case to  =p + 2D ; or, in vector form,  =pI + 2D: (67) ij ij ij Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 71 / 161The equations of motion for Newtonian incompressible uids Constitutive relationship for Newtonian uids Constitutive relationship for Newtonian uids IV De nitions 1 1  is named dynamic viscosity. It has dimensions  = ML T , and in the IS it is 2 measured in N s m . It is often convenient to de ne a kinematic viscosity as   = : (68)  2 1 2 1 The kinematic viscosity has dimensions  = L T , and in the IS is measured in m s . Inviscid uids A uid is said to be inviscid or ideal if  = 0. For an inviscid uid the stress tensor reads  =p ; or, in vector form,  =pI; (69) ij ij i.e. it takes the same form as for a uid at rest. Note that inviscid uids do not exist in nature. However, in some cases, real uids can behave similarly to ideal uids. This happens in ows in which viscosity plays a negligible e ect. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 72 / 161The equations of motion for Newtonian incompressible uids The NavierStokes equations The NavierStokes equations We now wish to derive the equations of motions for an incompressible Newtonian uid. We consider the Cauchy equation (55) and substitute into it the constitutive relationship (67). We obtain    u u i i  + u f p + 2D = 0: (70) j i ij ij t x x j j Let us consider the last term of the above expression. We can write it as    2 2  u u 1 u u i j i j 2D = 2 + = + : ij 2 x x 2 x x x xx j j j i i j j u u j j For the continuity equation, we have = 0) = 0. We can then write equation (70) as x x x j i j 2 u u 1 p u u 1 i i i 2 +u f +  = 0; or, in vector form, +(ur)uf + rpr u = 0: j i 2 t x x x t  j i j (71) Recalling the de nition of material derivative (25) the above equation can also be written as 2 Du 1 p u Du 1 i i 2 f +  = 0; or, in vector form, f + rpr u = 0: (72) i 2 Dt x x Dt  i j These are called the NavierStokes equations and are of fundamental importance in uid mechanics. They govern the motion of a Newtonian incompressible uid and have to be solved together with the continuity equation (36). Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 73 / 161The equations of motion for Newtonian incompressible uids The dynamic pressure The dynamic pressure We now assume that the body force acting on the uid is gravity, therefore we set in the NavierStokes equation (71) f = g. When  is constant the pressure p in a point x of the uid can be written as p = p +g x + P; (73) 0 where p is a constant and p +g x is the pressure that would exist in the uid if it was at rest. 0 0 Finally, P is the part of the pressure which is associated to uid motion and can be named dynamic pressure. This is in fact the departure of pressure from the hydrostatic distribution. Therefore, in the NavierStokes equations, the term grp can be replaced withrP. Thus we have: r u = 0; u 1 2 + (ur)u + rPr u = 0: (74) t  If the NavierStokes equations are written in terms of the dynamic pressure gravity does not explicitly appear in the equations. In the following whenever gravity will not be included in the NavierStokes this will be done with the understanding that the pressure is the dynamic pressure (even if p will sometimes be used instead of P). Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 74 / 161Initial and boundary conditions Initial and boundary conditions Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 75 / 161Initial and boundary conditions Initial and boundary conditions for the NavierStokes equations Initial and boundary conditions for the NavierStokes equations We know from the previous section that the motion of an incompressible Newtonian uid is governed by the NavierStokes equations (71) and the continuity equation (36), namely u 1 2 + (ur)u f + rpr u = 0; t  r u = 0: Initial conditions To nd an unsteady solution of the above equations, we need to prescribe initial conditions, i.e. the initial (at time t = 0) spatial distribution within the domain of pressure and velocity p(x; 0); u(x; 0): (75) Boundary conditions Equations (71) and (36) have also to be solved subjected to suitable boundary conditions. We will discuss in the following the boundary conditions that have to be imposed at the interface between two continuum media. We will then specify these conditions to the following, particularly relevant, cases: solid impermeable walls; free surfaces, e.g. interfaces between a liquid and a gas. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 76 / 161Initial and boundary conditions Kinematic boundary condition Kinematic boundary condition I The kinematic boundary condition imposes that at a boundary of the domain the normal velocity of the surface v = v n (with v velocity of the boundary and n unit vector normal to the surface) n is equal to the normal velocity of uid particles on the surface u = u n. Thus we have n u = v at the boundary: (76) n n Let us determine v . Let F (x; t) = 0 be the equation of the surface and n the normal to this n surface, de ned as rF n = : (77) jrFj Let us consider a small displacement of the surface in the time interval dt. The di erential dF taken along the direction normal to F = 0 in the time interval dt has to be equal to zero for F = 0 to still represent the equation of the surface. Thus F F dF = dn + dt = 0: (78) n t In the above expression dn represents the displacement of the interface along the normal direction in the time interval dt. The normal component of the velocity of the surface is dn v = : (79) n dt Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 77 / 161Initial and boundary conditions Kinematic boundary condition Kinematic boundary condition II Comparing (79) and (78) we obtain F=t v = : n F=n F Equation (77) implies n njrFj =rF n)jrFj = . Therefore the above equation can be n written as F=t v = : (80) n jrFj Substituting (80) into (76) we nd F=t rF = u n = u ; jrFj jrFj from which, recalling (24) F DF + urF = = 0: (81) t Dt The above equation states that the F = 0 is a material surface, i.e. it is always constituted by the same uid particles. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 78 / 161Initial and boundary conditions Continuity of the tangential component of the velocity Continuity of the tangential component of the velocity Given a boundary surface between two continuum media experience shows that the tangential component of the velocity is continuous across the interface. Let us denote with subscripts a and b the two continuum media. We thus have u = u at the boundary; (82) a t b t where subscript t indicates the tangential components of u. This condition can be justi ed by the observation that a discontinuity of the tangential velocity would give rise to the generation of intense (in nite) stress on the surface, which would tend to smooth out the discontinuity itself. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 79 / 161Initial and boundary conditions Dynamic boundary conditions Dynamic boundary conditions Let us now consider an interface between two uids. Since the boundary is immaterial, i.e. it has no mass, the elements that constitute the interface have to be in equilibrium to each other. This implies that: the tangential component of the stress has to be continuous across the interface; a jump in the normal component of the stress is admissible, which has to be balanced by the surface tension, according to equation (12). Thus, recalling (67) we can write   1 1 (p I + 2 D )n (p I + 2 D )n = + n; at the interface (83) a a a b b b R R 1 2 where subscripts a and b denote the uid at the two sides of the interface, and the normal unit vector n points from the uid a to b. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 80 / 161Initial and boundary conditions Two relevant cases Two relevant cases Let us now consider two cases of particular relevance in uid mechanics. Fluid in contact with a solid impermeable wall When a uid is in contact with a solid the boundary conditions described above take a very simple form. The kinematic boundary condition (81) and the conditions imposing the continuity of the tangential component of the velocity (82), imply that the velocity of the uid at the wall u has to be equal to the velocity of the wall u . Thus we have w u = u at the wall: (84) w This is named noslip boundary condition. In the particular case in which the solid is not moving we obtain u = 0 at the wall: (85) There is no need to impose the dynamic boundary conditions (83), unless the problem for the solid deformation is also solved, i.e. it is assumed that the solid is deformable. Interface between a liquid and a gas (free surface) Typically there is no need to solve the problem for the gas motion. This has the following consequences. Conditions (82) are no longer needed. In equation (83) the stress on the gas side (b) reduces to the contribution of the pressure p n. gas Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 81 / 161Scaling and dimensional analysis Scaling and dimensional analysis Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 82 / 161Scaling and dimensional analysis Units of measurement and systems of units Units of measurement and systems of units I A very comprehensive book on scaling and dimensional analysis, which pays particular attention to problems in uid mechanics, is Barenblatt (2003). This section is based on this book. Measurement of a physical quantity is a comparison of a certain quantity with an appropriate standard, or unit of measurement. We can divide the units for measuring physical quantities into two categories: fundamental units; derived units. This has the following meaning. Let us consider a certain class of phenomena (e.g. mechanics). Let us list a number of quantities of interest and let us adopt reference values for these quantities as fundamental units. For instance we can choose mass, length and time standards as fundamental units. Once fundamental units have been decided upon it is possible to obtain derived units using the de nition of the physical quantities. For instance, we know that density is mass per unit volume. We can therefore measure the density of a certain body by comparing it with the density of a body that contains a unit of mass in a volume equal to the cube of a unit of length. Important note. Given a certain class of phenomena there is a minimum number of fundamental units necessary to measure all quantities within that class. However, a system of units needs not be minimal, i.e. we may choose as fundamental units more units than we strictly need. De nition. A set of fundamental units that is sucient for measuring all physical properties of the class of phenomena under consideration is called a system of units. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 83 / 161Scaling and dimensional analysis Units of measurement and systems of units Units of measurement and systems of units II A system of units consisting of one fundamental unit (e.g. the metre) is sucient to describe geometric objects. Two fundamental units (e.g. the metre and the second) are sucient to describe kinematic phenomena. Three fundamental units (e.g. the metre, the second and the kilogram) are sucient to describe dynamic phenomena. . . . In the International System of Units SI the fundamental units for studying dynamic phenomena are: the kilogram kg for mass (equal to the mass of the International Prototype Kilogram, preserved at the Bureau of Weights and measures in Paris); the metre m for length (the length of the path travelled by light in vacuum during a time interval of 1=299; 792; 458 s); the second s for time (the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyper ne levels of the ground state of the caesium 133 atom). Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 84 / 161Scaling and dimensional analysis Units of measurement and systems of units Units of measurement and systems of units III De nition. Two systems of units are said to belong to the same class of systems of units if they di er only in the magnitude of the fundamental units, but not in their physical nature. For instance if we choose to describe a mechanical problem adopting as fundamental units one 3 3 kilometre (= 10 m), one metric ton (= 10 kg) and one hour (= 3600 s) we have a system of units in the same class as the SI (metrekilogramsecond). If we regard the metrekilogramsecond as the original system in its class, then the corresponding units on an arbitrary system in the same class are obtained as follows unit of length = m=L; unit of mass = kg=M; unit of time = s=T; (86) where L, M and T are positive numbers that indicate the factor by which the fundamental units change by passing from the original system to another system in the same class. The class of systems of units based on length, mass and time is called LMT class. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 85 / 161Scaling and dimensional analysis Dimension of a physical quantity Dimension of a physical quantity I De nition of dimension The function that determines the factor by which the numerical value of a physical quantity changes upon passage from the original system of units to an arbitrary system within a given class is called dimension function or dimension of that quantity. The dimension of a quantityF is denoted by F. For example, if the units of length and time are changed by factors L and T , respectively, then 1 the unit of velocity changes by a factor LT . According to the above de nition we can say that 1 LT is the dimension of velocity. De nition. Quantities whose numerical value is independent of the choice of the fundamental units within a given class of systems of units are called dimensionless. Important principle. In any equation with physical meaning all terms must have the same dimensions. If this was not the case an equality in one system of units would not be an equality in another system of units within the same class. Thus, from Newton law we nd that the dimension of force (F ) in the LMT class is 2 F = ma = LMT ; with m mass and a acceleration. If, on the other hand, we adopt the LFT class (lengthforcetime), then the dimension of mass is 1 2 m = L FT . Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 86 / 161Scaling and dimensional analysis Dimension of a physical quantity Dimension of a physical quantity II The dimension function is a powerlaw monomial We will prove this using the LMT class of systems of units. We know that the dimension of a physical quantity a within this class depends on L, M and T only. a =(L; M; T ): Suppose we have chosen an original system (e.g. metrekilogramsecond). Moreover, we choose two further systems in the same class, say 1 and 2, so that, upon passage from the original system to these new systems, the fundamental units decrease by factors L , M , T and L , M , 1 1 1 2 2 T , respectively. 3 Let a be the numerical value of the quantity in the original system. This value will become, by de nition of dimension, a = a(L ; M ; T ) in the rst new system and a = a(L ; M ; T ) in 1 1 1 1 2 2 2 2 the second one. Thus we have a (L ; M ; T ) 2 2 2 2 = : (87) a (L ; M ; T ) 1 1 1 1 All systems of units within a given class are equivalent, i.e. there are no preferred systems. This implies that we may assume system 1 as the original system of the class. System 2 can then be obtained by decreasing the fundamental units by L =L , M =M and T =T . This implies that 2 1 2 1 2 1 the numerical value a of the considered quantity can now be written as 2   L M T 2 2 2 a = a  ; ; : 2 1 L M T 1 1 1 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 87 / 161Scaling and dimensional analysis Dimension of a physical quantity Dimension of a physical quantity III Therefore we have   a L M T 2 2 2 2 = ; ; : (88) a L M T 1 1 1 1 Comparing equations (87) and (88) we obtain the following functional equation for    (L ; M ; T ) L M T 2 2 2 2 2 2 = ; ; : (89) (L ; M ; T ) L M T 1 1 1 1 1 1 To solve this equation we proceed as follows. We rst di erentiate both sides of (89) with respect to L and then set L = L = L, M = M = M and T = T = T , nding 2 2 1 2 1 2 1 (L; M; T ) 1 L = (1; 1; 1) = ; (L; M; T ) LL L where = (1; 1; 1) is a constant. The solution of the above equation is L (L; M; T ) = L C (M; T ): 1 Substituting this expression into (89), we nd the following functional equation for C 1   C (M ; T ) M T 1 2 2 2 2 = C ; : 1 C (M ; T ) M T 1 1 1 1 1 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 88 / 161Scaling and dimensional analysis Dimension of a physical quantity Dimension of a physical quantity IV We now di erentiate this equation with respect to M and then set M = M = M and 2 2 1 T = T = T . 2 1 C (M; T ) 1 1 M = C (1; 1) = ; 1 C (M; T ) M M M 1 where, again, = C (1; 1) is a constant. Solving for C we obtain 1 1 M C = M C (T ): 1 2 Proceeding in a similar way we nally nd C (T ) = C T : 2 3 Thus the solution is  = C L M T : 3 The constant C has to be equal to 1 as L = M = T = 1 means that the fundamental units 3 remain unchanged, so that the value of the quantity a also has to remain unchanged and, therefore, it must be (1; 1; 1) = 1. We then nally have  = L M T : (90) and we can easily verify that this is actually a solution of our original functional equation (89). Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 89 / 161Scaling and dimensional analysis Quantities with independent dimensions Quantities with independent dimensions I De nition The quantities a ; a ;:::; a are said to have independent dimensions if the monomial 1 2 k a a ::: a has a dimension function equal to 1 (i.e. it is dimensionless) only for 1 2 k = = = = 0. Example 3 1 Let us consider, for example, the quantities density ( = ML ), velocity (u = LT ) and 2 force (f = MLT ). Let us now construct the monomial = u f . We require this monomial to be dimensionless, thus 3 2 =  u f = M L L T M L T = + 3 + + 2 = M L T = 1: The above equation implies + = 0; 3 + + = 0; 2 = 0: This above system has no solution unless = = = 0. This means that the quantities , u and f have independent dimensions. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 90 / 161Scaling and dimensional analysis Quantities with independent dimensions Quantities with independent dimensions II Theorem Within a certain class of systems of units, it is always possible to pass from an original system of units to another system, such that any quantity, say a , in the set of quantities a ;:::; a with 1 1 k independent dimensions, changes its numerical value while all the others remain unchanged. Proof Let us consider a system of units PQ::: . Let us consider a set of quantities with independent dimensions, whose values in a chosen original system of units are a ;:::; a . Upon change of the 1 k 0 0 system of units to an arbitrary one, their numerical value becomes a ;:::; a , such that 1 k 0 0 1 1 k k a = a P Q :::; ::: a = a P Q :::; 1 k 1 k where the powers ; ;::: (i = 1;:::; k) are determined by the dimensions of each quantity. i i We want to nd a system of units such that 0 0 0 1 1 a = a P Q :::; a = a ; ::: a = a : 1 2 k 1 2 k We thus have a system of equations 1 1 2 2 k k P Q  = A ; P Q  = 1; :::; P Q =1 : 1 Taking the logarithm of the above equations we obtain ln P + ln Q + = ln A ; ln P + ln Q + = 0; :::; ln P + ln Q + = 0: 1 1 1 2 2 k k Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 91 / 161Scaling and dimensional analysis Quantities with independent dimensions Quantities with independent dimensions III This system has a solution unless the lefthand side of the rst equation is a linear combination of the remaining ones, so that ln P + ln Q + = c ( ln P + ln Q +::: ) + + c ( ln P + ln Q +::: ); 1 1 2 2 2 2 k k k k with c ;:::; c constants. However, this implies, going back to the exponent form, that 2 k     c c 2 k 1 1 2 2 k k P Q  = P Q ::: P Q ; or c c 2 k a = a ::: a : 1 2 k This contradicts the fact that a ;:::; a have independent dimensions and the theorem is 1 k therefore proved. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 92 / 161Scaling and dimensional analysis Buckingham's  theorem Buckingham's  theorem I Any physical study (experimental or theoretical) consists in nding one or several relationships between physical quantities in the form a = f (a ;:::; a ; b ;:::; b ): (91) 1 k 1 m In the above expression a denotes the quantity of interest. On the righthand side of the above equation we have separated the physical quantities into two groups. The k quantities a ;:::; a have independent dimensions; 1 k the m quantities b ;:::; b can be expressed in terms of the dimensions of a ;:::; a . 1 m 1 k Thus we can write 1 1 m m b = a ::: a ; ::: b = a ::: a : m 1 1 k 1 k Note: it must be that the dimension of a is dependent on the dimensions of a ;:::; a , so that 1 k a = a ::: a : 1 k Indeed, if a had dimensions independent from the dimensions of the variables a ;:::; a , for the 1 k theorem proved above it would be possible to pass from the original system of units to another system, such that the numerical value of a would change and the numerical values of a ;:::; a 1 k and b ;:::; b would remain unchanged. This would indicate the need to include further 1 m quantities on the righthand side of equation (91). Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 93 / 161Scaling and dimensional analysis Buckingham's  theorem Buckingham's  theorem II We now introduce a  = ; a ::: a 1 k b b 1 m  = ; :::  = : m 1 1 1 m m a ::: a a ::: a 1 k 1 k We can thus write equation (91) as 1 1 m m f (a ;:::; a ;  a ::: a ;:::;  a ::: a ) f (a ;:::; a ; b ;:::; b ) 1 1 m 1 k 1 m k 1 k 1 k  = = ; a ::: a a ::: a 1 k 1 k or  =F(a ;:::; a ;  ;:::;  ): 1 k 1 m  and  (i = 1;:::; m) are dimensionless, therefore they don't change their numerical value i upon changing of the system of units. Now, suppose that we change the system of units so that a changes its value and a ;:::; a remain unchanged. In the above equation a would be the 1 2 k 1 only variable to change and this indicates that the functionF can not depend of a . The same 1 argument holds for all the a ;:::; a variables. Therefore the above equation can be written as 1 k  =F( ;:::;  ): (92) 1 m Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 94 / 161Scaling and dimensional analysis Buckingham's  theorem Buckingham's  theorem III We have therefore proved that equation (91) is equivalent to equation (92), which involves only dimensionless variables. Note, moreover, that (92) involves a smaller number of variables than (91). In particular, the number of variables involved has decreased by k, i.e. by the number of variables involved in (91) that have independent dimensions. This fact is of fundamental importance and it is one of the main reasons for which working with dimensionless quantities is typically desirable. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 95 / 161Scaling and dimensional analysis Dimensionless NavierStokes equations Dimensionless NavierStokes equations I When dealing with theoretical modelling of physical phenomena it is convenient to work with dimensionless equations. The main reasons for that are: according to the  theorem the number of parameters involved in the problem decreases if one passes from a dimensional to a dimensionless formulation; in dimensionless form (if proper scalings are adopted) it is much easier to evaluate the relative importance of di erent terms appearing in one equation. Let us consider the NavierStokes equation and assume that the body force is gravity. Equations (71) and the continuity equation (36), can then be written as u 1 2 + (ur)ug + rpr u = 0; (93) z t z  z z 4 2 1 3 where the vector g, representing the gravitational eld, has magnitude g and is directed vertically downwards. We recall the physical meaning of all terms: 1 : convective terms; 2 : gravity; 3 : pressure gradient; 4 : viscous term. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 96 / 161Scaling and dimensional analysis Dimensionless NavierStokes equations Dimensionless NavierStokes equations II We now wish to scale the above equation. Suppose that L is a characteristic length scale of the domain under consideration and U a characteristic velocity. We can then introduce the following dimensionless coordinates and variables x u t    x = ; u = ; t = : L U L=U Above and in what follows superscript stars indicate dimensionless quantities. We still have to scale the pressure. We might consider two di erent situations: 1 In equation (93) 3 balances with 4 . In this case we can write p  p = : U=L 2 If, on the other hand, in (93) the pressure gradient 3 balances with the convective terms 1 , we can scale the pressure as follows p  p = : 2 U Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 97 / 161Scaling and dimensional analysis Dimensionless NavierStokes equations Dimensionless NavierStokes equations III Low Reynolds number ows Let us consider the rst case. Making equation (93) dimensionless using the above scales we obtain    u Re      2  Re + (u r )u + z +r p r u = 0; (94)  2 t Fr where z is the upward directed vertical unit vector. In the above equation we have de ned two dimensionless parameters. UL Re = : Reynolds number. It represents the ratio between the magnitude of inertial  (convective) terms and viscous terms. It plays a fundamental role in uid mechanics. U Fr =p : Froude number. It represents the square root of the ratio between the gL magnitude of inertial (convective) terms and gravitational terms. It plays a fundamental role when gravity is important, e.g. in free surface ows. If we now consider the limit Re 0 the dimensionless NavierStokes equation (94) reduces to the so called Stokes equation, i.e.   2  r p r u = 0: (95) This equation is much simpler than the NavierStokes equation as it is linear. In section 10 we will derive some analytical solutions of equation (95). Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 98 / 161Scaling and dimensional analysis Dimensionless NavierStokes equations Dimensionless NavierStokes equations IV Large Reynolds number ows Let us now consider the case in which the pressure gradient balances the convective terms. The dimensionless NavierStokes equation takes the form  u 1 1      2  + (u r )u + z +r p r u = 0: (96)  2 t Fr Re In the limit Re1 the viscous term in equation (96) tends to zero. Thus we are led to think that, at large values of Re, the uid behaves as an ideal or inviscid uid. This argument, however, has to be used with care as dropping o the viscous term from (96) means to neglect the term containing the highest order derivatives in the equation. Therefore, if the viscous term in (96) is neglected it is not possible to impose all boundary conditions. To resolve this contradiction we need to assume that at the boundaries thin boundary layers form, within which viscous terms in the NavierStokes equations have the same magnitude as convective terms. If boundary layers keep very thin everywhere in the ow domain the uid out of the boundary layers, in the core of the domain, actually behaves as if it was inviscid. This point will be discussed in more detail in section 11. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 99 / 161Unidirectional ows Unidirectional ows Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 100 / 161Unidirectional ows Introduction to unidirectional ows Introduction to unidirectional ows We consider the ow of an incompressible Newtonian uid in the gravitational eld. We thus have equations (93) and the continuity equation (36), namely u 1 2 + (ur)u g + rpr u = 0; t  r u = 0: We consider a unidirectional ow, i.e. a ow in which the velocity has everywhere the same direction (say the direction of the axis x) and it is independent of x. Thus, assuming Cartesian coordinates (x; y; z), we have u = u(y; z; t); 0; 0: (97) It is easy to check that with the velocity eld (97) all the non linear terms in the NavierStokes equations vanish. Thus the governing equations in this case are linear and therefore much more amenable for analytical treatment. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 101 / 161Unidirectional ows Some examples of unidirectional ows Some examples of unidirectional ows I Let us consider the unidirectional ow shown in the gure. The direction of ow is inclined by an angle with respect to a horizontal plane. Referring to the gure we consider the system of Cartesian coordinates (x; y; z), with x direction of ow. The corresponding velocity components are u(y; z); 0; 0. In this case the NavierStokes equations take the form   2 2 u 1p u u + g sin +  + = 0; (98) 2 2 t x y z 1p + g cos = 0; (99) y p = 0: (100) z Equations (99) and (100) simply impose that the pressure distribution is hydrostatic on the crosssection of the ow (planes with x = const). This also implies that, as in hydrostatics (10), the piezometric (or hydraulic) head h is constant on such planes and is thus a function of x only. We can thus write p(x; y) h(x) = x sin + y cos + ; Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 102 / 161Unidirectional ows Some examples of unidirectional ows Some examples of unidirectional ows II from which h 1p = sin + : x x We can then rewrite equation (98) as   2 2 u h u u + g  + = 0: (101) 2 2 t x y z Since u does not depend on x it follows that alsoh=x is independent of x. Hence, we can write h =j(t); x where j is function of time only. Upon substitution of j, equation (101) takes the form   2 2 u u u gj + = 0: (102) 2 2 t y z In the particular case of steady ow this simpli es to 2 2 u u g + = j: (103) 2 2 y z  Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 103 / 161Unidirectional ows Some examples of unidirectional ows Some examples of unidirectional ows III CouettePoiseuille ow Let us now consider a particular case of the ow described above, i.e. the ow within a gap formed by two at parallel walls, each one of which is moving in the x direction with a given velocity, say u (lower wall) 1 and u (upper wall). Moreover, we assume that j has a 2 constant prescribed value. We wish to study the motion of a uid within this gap. In this case u = u(y); 0; 0. Equation (103) can be written as 2 d u g = j; 2 dy  The solution of the above equation is gj 2 u = y + c y + c : 1 2 2 The constants c and c can be determined imposing the noslip boundary conditions at the 1 2 walls, i.e. u(0) = u ; u(a) = u : 1 2 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 104 / 161Unidirectional ows Some examples of unidirectional ows Some examples of unidirectional ows IV We nally nd gj u u 2 1 u = (a y)y + y + u : (104) 1 2 a From the above solution we can easily compute the volume ux per unit length q as Z a gj u a u a 2 1 3 q = udy = a + + : (105) 12 2 2 0 We now consider a few particular cases. Poiseuille ow: j =6 0, u = u = 0. 1 2 In this case the ow is driven by a hydraulic head gradient alone. The velocity distribution (104) reduces to gj u = (a y)y; (106) 2 i.e. the velocity pro le is parabolic. The maximum velocity is located at the centre of the channel (y = a=2) and is equal to gj 2 u = a ; max 8 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 105 / 161Unidirectional ows Some examples of unidirectional ows Some examples of unidirectional ows V and the volume ux per unit length is gj 2 3 q = a = u a: max 12 3 2 The average velocity u is equal to u . max 3 Let us now compute the shear stress on the wall. The stress tensor has the form 0 1 du 0  0 B C dy B C B C du  = pI: B C  0 0 A dy 0 0 0 Thus we easily nd that the tangential stress  exerted by each wall is given by j  = a: 2 Quite surprisingly the shear stress is not dependent on the viscosity of the uid. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 106 / 161Unidirectional ows Some examples of unidirectional ows Some examples of unidirectional ows VI Couette ow: j = 0, u = 0, u 6= 0. 1 2 In this case the ow is driven by the movement of the upper wall and the hydraulic head gradient is zero. The velocity distribution (104) reduces to u 2 u = y: (107) a Moreover, we nd u a u 2 2 q = ;  = : 2 a Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 107 / 161Unidirectional ows Some examples of unidirectional ows Some examples of unidirectional ows VII Unidirectional freesurface ow We now consider a steady freesurface ow over an inclined plane, as shown in the gure. The velocity vector can be written as u = u(y); 0; 0 and equation (103) reduces to 2 d u g = j; 2 dy  subjected to the noslip condition at y = 0 and the dynamic boundary conditions at y = h, i.e. du u(0) = 0; = 0: dy y=h The corresponding velocity distribution and ux per unit length are Z h gj gj 3 u = (2h y)y; q = udy = h : 2 3 0 If, as in fact is normally the case, the ux q and the slope j are xed (rather than h) we obtain the following expression for h s 3q 3 h = : gj Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 108 / 161Unidirectional ows Some examples of unidirectional ows Some examples of unidirectional ows VIII Axisymmetric Poiseuille ow Let us now consider a steady, completely developed ow in a straight pipe with circular crosssection of radius R. Let the pipe axis be in the z direction and let the ow be axisymmetric. In cylindrical coordinates (z; r;') the velocity vector takes the form u = u(r); 0; 0, with u velocity component in the z direction. With these coordinates equation (103) takes the form     2 d 1 d gj 1 d du gj + u = ; ) r = : 2 dr r dr  r dr dr  The above equation has to be solved subjected to the noslip boundary condition at r = R and a regularity condition in r = 0. We then have du gj gj 2 2 r = r + c ; ) u = r + c log r + c : 1 1 2 dr 2 4 Regularity at r = 0 imposes c = 0. Moreover, enforcing the noslip boundary condition yields 1 gj 2 c = R . The solution is 2 4  gj 2 2 u = R r : (108) 4 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 109 / 161Unidirectional ows Some examples of unidirectional ows Some examples of unidirectional ows IX This is known as Poiseuille ow. The velocity pro le is a paraboloid. The volume ux Q is given by Z Z R 2 gj 4 Q = urd'dr = R : (109) 8 0 0 Written in cylindrical coordinates (z; r;') the stress tensor for this ow eld takes the form 0 1 du 0  0 B C dr B C B C du  = pI: B C  0 0 A dr 0 0 0 We then easily compute the tangential stress  on the wall, which reads j  = R: 2 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 110 / 161Unidirectional ows Unsteady unidirectional ows Unsteady unidirectional ows I Flow over a periodically oscillating plate Let us now consider one example of unsteady unidirectional ow. In this case we need solving equation (102). We consider the ow in the region y 0 induced by periodic motion along the x axis of a rigid at wall located at y = 0. The velocity of the wall u can be written as w   u 0 it u = u cos(t) = e + c:c: ; w 0 2 where c:c: denotes the complex conjugate. Since there is no imposed pressure gradient equation (102) reduces to 2 u u  = 0: (110) 2 t y We seek a separable variable solution in the form it u = f (y)e + c:c: (111) where f (y) is a complex function. Substituting (111) into (110) we obtain 2 i d f f = 0: 2  dy Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 111 / 161Unidirectional ows Unsteady unidirectional ows Unsteady unidirectional ows II p 1 Remembering that i =p (1 + i), the solution of the problem is 2 2 0 1 0 13 6 B C B C7 y(1 + i) y(1 + i) it 6 B C B C7 u(y; t) = c exp r + c exp r e + c:c: 1 2 4 A A5   2 2 As the solution should not be divergent for y1 we require c = 0. Moreover, the noslip 2 boundary condition at the wall imposes u 0 c = : 1 2 Thus the solution of (110) is 0 1 0 1 0 1 B C B C B C u y(1 + i) y y 0 it B C B C B C u(y; t) = exp r e + c:c: = u exp r cos tr : (112) 0 A A A 2    2 2 2 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 112 / 161Unidirectional ows Unsteady unidirectional ows Unsteady unidirectional ows III 6 2 The solution is sketched in the gure for u = 1 m/s,  = 10 m /s (water) and for two 0 1 1 di erent values of , (a) = 1 s , (b) = 0:1 s . (a) 1 (b) 1 0.5 0.5 0 0 −0.5 −0.5 −1 −1 0 0.002 0.004 0.006 0.008 0.01 0 0.002 0.004 0.006 0.008 0.01 y m y m It is important to notice that the velocity does not spread to in nity in the y direction for long times. The solution (112) suggests that a characteristic length scale l of the layer of uid interested by motion is given by r  l : Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 113 / 161 u m/s u m/sUnidirectional ows Axisymmetric ow with circular streamlines Axisymmetric ow with circular streamlines I We present here another case in which the NavierStokes equations take a linear form. Let us consider a ow such that all streamlines are circles centred on a common axis of symmetry. Moreover, adopting cylindrical coordinates (z; r;'), we assume that the velocity, which is purely azimuthal, only depends of the radial coordinate r and, possibly, on time t. Thus we have u = 0; 0; w(r; t). We nally assume axisymmetry, so that  0. This ow is strictly related ' with unidirectional ows. The continuity equation and the and NavierStokes equations in cylindrical coordinates are reported in the appendix 13 (equations (166), (167), (168) and (169)). It is immediate to verify that, when the velocity eld takes the form u = 0; 0; w(r; t), and the ow is axisymmetric the above equations reduce to p = 0 (113) z 2 w 1p + = 0 (114) r  r     w 1 w w  r = 0; (115) 2 t r r r r and the continuity equation is automatically satis ed. Equation (113) implies that the pressure does not depend on z, and equation (114) that the radial variation of p supplies the force Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 114 / 161Unidirectional ows Axisymmetric ow with circular streamlines Axisymmetric ow with circular streamlines II necessary to keep the uid element moving along a circular path. Finally, equation (115) is linear and it is the analogous of equation (102), for a unidirectional ow. Steady ow between two concentric rotating cylinders Let us consider two concentric cylinders with radius R and 1 R , respectively (R R ). Each of the cylinders rotates 2 2 1 with a given constant angular velocity ( and ). The gap 1 2 between the cylinders is lled with uid. The ow is steady and equation (115) reduces to   d dw w r = 0; dr dr r with the boundary conditions w = R (r = R ); w = R (r = R ): 1 1 1 2 2 2 The above equation can be rewritten as   1 d dw 2 r rw = 0: r dr dr Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 115 / 161Unidirectional ows Axisymmetric ow with circular streamlines Axisymmetric ow with circular streamlines III We then have dw dw c w c 1 1 2 r rw = c ; ) = + ; ) w = + c r: 1 2 2 dr dr r r 2r Enforcing the boundary conditions we nally nd   2 2 1 R R 1 2 1 2 1 2 w = + r : (116) 2 2 2 2 r R R R R 1 2 1 2 We now consider a few particular cases. = 0, the inner cylinder is at rest. 1 In this case the solution (116) simpli es to 2 2 2 R (r R ) 2 2 1 w = : (117) 2 2 r(R R ) 1 2 The shear stress  on the outer cylinder (r = R ) is 2   2 dw w 2 R 2 1  = j = = ; r' r=R 2 2 2 dr r R R r=R 2 1 2 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 116 / 161Unidirectional ows Axisymmetric ow with circular streamlines Axisymmetric ow with circular streamlines IV and the couple per unit length of cylinder m necessary to keep the outer cylinder in rotation is 2 m = 2R : 2 This device (rotational cylinder rheometer) is often used to measure the viscosity of a uid, as, by rearrangement of the above formula, it is possible to obtain the value of the dynamic viscosity  by measuring the couple m required to keep the outer cylinder in motion. R = 0, = , ow inside a single rotating cylinder. 1 1 2 From (116) we immediately get w = r ; 2 which is a rigid body rotation. R 1, 0, ow around a single rotating cylinder. 2 2 From (116) we obtain 2 R 1 1 w = : (118) r This is the so called \free vortex" velocity distribution. Notice that the vorticity associated with this velocity eld is everywhere zero. In this case the couple per unit length m transmitted to the uid by the rotating cylinder is 2 m = 4R  ; 1 1 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 117 / 161Unidirectional ows Axisymmetric ow with circular streamlines Axisymmetric ow with circular streamlines V and it is not zero. This implies a continuum growth of the angular momentum of the uid. This is not in contrast with the assumption of steady ow since the total angular momentum associated with the velocity distribution (118) is in nite. R R 2 1 = 0,  1. 1 R 1 In this case we can write R R 2 1 " = ; R = R (1 +"); 2 1 R 1 with " 1. Let us now de ne a new coordinate  as r R r R 1 1  = = ; ) r = R (1 +"); 1 R R "R 2 1 1 with 0 1. Substituting the above expression into (117) and expanding in terms of " we nd   3 1 2 w = R  + R   " +O(" ) 2 1 2 1 2 2 It appears that in the limit of small gap (compared with the radius) the velocity tends to a linear distribution, i.e. to the Couette ow (107). Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 118 / 161Low Reynolds number ows Low Reynolds number ows Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 119 / 161Low Reynolds number ows Introduction to low Reynolds number ows Introduction to low Reynolds number ows In section 8 we have shown (page 98) that for low values of the Reynolds number, the equations of motion reduce, at leading order, to the following linear equations 2 r u =rp; (119a) r u = 0: (119b) The above equations, being linear, are much more amenable to analytical treatment than the original NavierStokes equations. In the following of this section we consider the classical solution obtained by Stokes in 1851 for the slow ow past a sphere. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 120 / 161Low Reynolds number ows Slow ow past a sphere Slow ow past a sphere I For the ow around a sphere of radius a a sensible de nition for the Reynolds number is aU Re = ;  where U is the magnitude of the velocity far from the sphere. We consider a ow such that Re 1. Moreover, let the pressure far from the sphere be equal to p . 0 We make our problem dimensionless using the following scales x u (p; p ) 0     x = ; u = ; (p ; p ) = ; 0 a U U=a where the symbol denotes dimensionless variables. In the following we adopt a dimensionless approach but skip the to simplify the notation. Let i be the unit vector in the direction of the ow very far from the sphere (see the gure below). The ow is axisymmetrical about the direction i. We consider a system of polar spherical coordinates (r;;'), centred in the centre of the sphere, with the zenithal and ' the azimuthal coordinates, respectively. The corresponding velocity components are u = (u ; u ; u ). The r ' direction i coincides with the axis = 0;. Our dimensionless problem can be written as 2 r u =rp; (120a) r u = 0; (120b) u = 0 (r = 1); (120c) u i (r1) (120d) p p (r1) (120e) 0 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 121 / 161Low Reynolds number ows Slow ow past a sphere Slow ow past a sphere II As a consequence of the axisymmetry of the ow we have = 0; u = 0; ' ' and the continuity equation takes the form  1 1 2 r u + (sinu ) = 0: r 2 r r r sin This allows us to introduce the so called Stokes streamfunction , de ned as 1 1 u = ; u = : (121) r 2 r sin r sin r Given a vector b = (b ; b ; b ) we have r '       r b 1 b ' b r r r b = (b sin) + (rb ) + (rb ) ; ' ' r sin ' r sin ' r r r with r, and' unit vectors along the three coordinate directions. It is then easy to show that     u =r 0; 0; = curl 0; 0; : (122) r sin r sin Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 122 / 161Low Reynolds number ows Slow ow past a sphere Slow ow past a sphere III For future convenience we use the notation (curl) rather than (r) for the curl operator. Recalling the vector identity 2 r u =r(r u) curl curl u; (123) equation (120a) can be written as curl curl u =rp: Further taking the curl of the above expression we can eliminate the pressure, to get 3 curl u = 0: Using (122) the above expression can written in terms of the streamfunction as   4 curl 0; 0; = 0: (124) r sin It is not dicult to show that     2 D 2 curl 0; 0; = 0; 0; ; r sin r sin 2 where the operator D is de ned as 2 2 1 cot 2 D = + : 2 2 2 2 r r r Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 123 / 161Low Reynolds number ows Slow ow past a sphere Slow ow past a sphere IV Therefore, equation (124) reduces to 4 D = 0: (125) The boundary conditions, written in terms of , take the following form. Condition on the sphere surface (r = 1) Using the de nition of the streamfunction (121) the condition (120c) can be written as = = 0 (r = 1): (126) r Condition at in nity (r1) To write the condition at in nity (120d) as a function of we note that 2 2 1 r sin 2 u = cos = ) = r cos sin ) = + g (r); r 1 2 r sin 2 2 2 1 r sin 2 u = sin = ) = r sin ) = + g (): 2 r sin r r 2 Comparing the above expressions we nd 2 2 r sin = + C: (127) 2 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 124 / 161Low Reynolds number ows Slow ow past a sphere Slow ow past a sphere V We now seek a separable variable solution, thus writing = f (r)g(). The boundary condition 2 (127) suggests to choose g() = sin , so that 2 (r;) = f (r) sin : After some algebra it can be shown that   2 d 2 2 2 D = f sin : 2 2 dr r and hence we nally have to solve the following equation   2 2 d 2 f = 0: (128) 2 2 dr r The general solution of this homogeneous equation is 4 2 1 f = ar + br + cr + dr : 1 The boundary condition at in nity shows that it must be a = 0 and b = . The condition at the 2 3 1 sphere surface imposes c = and d = . Thus the Stokes solution for the ow past a sphere is 4 4   1 3 1 2 1 2 = r r + r sin : (129) 2 4 4 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 125 / 161Low Reynolds number ows Slow ow past a sphere Slow ow past a sphere VI The two velocity components are immediately found from (121) and read     3 1 3 1 u = 1 + cos; u = 1 + + sin: (130) r 3 3 2r 2r 4r 4r Finally, the pressure can be calculated from equation (120a) and is found to be 3 cos p = p : (131) 0 2 2r We can now compute the drag force F that the ow exerts on the sphere. This quantity is of particular practical interest. Obviously, F is in the i direction, so we just have to compute the following scalar quantity ZZ F = ( cos sin) dS; rr r r=1 which is the force magnitude. Note that the above expression is dimensionless; to nd the dimensional force we have to multiply it by aU. We have   u 3 cos r  j = p + 2 =p + ; rr r=1 0 r 2 r=1     u 1u 3 r  j = r + = sin: r=1 r r r r 2 r=1 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 126 / 161Low Reynolds number ows Slow ow past a sphere Slow ow past a sphere VII The contribution to the drag from the normal stress is given by Z Z   2  3 cos p + cos sindd' = 2; 0 2 0 0 and the contribution from the tangential stress is Z Z 3 2  3 sin dd' = 4: 2 0 0 The dimensionless drag force on the sphere is then F = 6; (132) and, going back to dimensional quantities, Stokes drag force = 6aU: (133) Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 127 / 161Low Reynolds number ows Lubrication Theory Lubrication theory I This technique provides a good approximation to the real solution when the domain of the uid is long and thin. For simplicity let us assume that the ow is two dimensional (all derivatives with respect to the third coordinate, say z, may be neglected) and that the height of the domain is h(x) and a typical streamwise length is L. The uid velocity at the vessel walls is zero (noslip condition) but the uid velocity at the surface of the cell equals the cell velocity (U). Therefore changes in the xvelocity u are on the order of U, that isjuj U, andju=yjju=yj U=h , where h is a characteristic value of h(x). 0 0 The change in uid velocity as we move through a distance L in the xdirection is likely to be at most U, and thereforeju=xj U=L. The continuity equation, u v + = 0; x y implies thatjv=yj U=L; hencejvj h U=L. 0 Scaling We nondimensionalise        x = Lx ; y = h y ; h(x) = h h (x ); u = Uu ; v = h Uv =L; p = p p ; 0 0 0 0      where p is an appropriate scale for the pressure (to be chosen). Note that x , y , u , v and p 0 are all order 1. (Note also that the ow has a low Reynolds number, so we expect to scale the pressure gradient with the viscous terms.) Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 128 / 161Low Reynolds number ows Lubrication Theory Lubrication theory II Neglecting gravity and assuming a steady solution, the nondimensional governing equations are     2  2  2  u u h p p u u 0 2   0 2  Re u + v = + + ; (134)    2 2 x y UL x x y     2  2  2  v v h p p v v 0 3   0 3  Re u + v = + + ; (135)    2 2 x y ULy x y   u v + =0; (136)   x y where  = h =L 1 and Re = UL=. 0  We may immediately cancel the viscous terms that have a repeated x derivative since they are  much smaller than the viscous terms with a repeated y derivative. Balancing the pressure derivative and viscous terms in the xcomponent equation (134) leads to the scaling 2 p =UL=h . 0 0 Multiplying equation (135) by  and simplifying, equations (134) and (135) can be written as      2  p u u u 2    Re u + v = + ; (137)    2 x y x y  p 0 = ; (138)  y 2 3 where we have neglected terms of order  and terms of order  Re relative to the leadingorder terms. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 129 / 161Low Reynolds number ows Lubrication Theory Lubrication theory III Solution procedure 2 The quantity  Re is called the reduced Reynolds number. We assume it is not too large, which places an upper bound on the possible ux.  We may immediately solve (138) to nd that the pressure is a function of x only, that is, the pressure is constant over the height of the gap.   The governing equations are thus (137) and (136), where p is a function of x only and  these must be solved subject to noslip boundary conditions for u at the walls. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 130 / 161Low Reynolds number ows Lubrication Theory Lubrication theory IV Series expansion for small reduced Reynolds number 2 In the case that the reduced Reynolds number is small,  Re 1 we can use a series expansion method to nd the velocity, by setting  2   2  2  u =u + Re u +  Re u +:::; 0 1 2  2   2  2  v =v + Re v +  Re v +:::; 0 1 2  2   2  2  p =p + Re p +  Re p +:::: 0 1 2    noting that all the p 's are independent of y, and then solving for u (from equation (137)), v 0 0 i   (from equation (136)), u (from equation (137)), v (from equation (136)), etc in that order. An 1 1 equation for the pressure can be obtained by integrating the continuity equation over the gap height.    In many cases it is suciently accurate to nd just the rst terms u and v (or even just u ). 0 0 0 Generalisation Note that we could generalise this approach to include: dependence upon the third spatial dimension; timedependence of the solution; gravity; . . . . Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 131 / 161Low Reynolds number ows Lubrication Theory Lubrication theory V Example of solution We consider the domain shown in the gure. For simplicity, we assume twodimensional ow. We wish to solve the ow in the gap 0 y h(x), with 0 x L. The ow is subject to the following boundary conditions: noslip at y = 0 and y = h(x); given R h 0 ux per unit length F = udy at x = 0; 0 given pressure p = 0 at x = L. We assume that h = h(0) is a typical value of 0 the thickness of the domain in the ydirection and assume that  = h =L 1. We can, therefore, apply the lubrication theory. 0 We scale the variables as follows x y u v     x = ; y = ; u = ; v = ; L h U U 0 with U = F=h . 0 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 132 / 161Low Reynolds number ows Lubrication Theory Lubrication theory VI 2 Assuming that  Re 1, we need to solve the following dimensionless equations (see equations (137), (138) and (136)) 2   u p = 0; (139) 2  y x  p = 0; (140)  y   u v + = 0; (141)   x y subject to the boundary conditions    u = v = 0 (y = 0); (142)      u = v = 0 y = h (x ); (143) Z 1    u dy = 1 (x = 0); (144) 0   p = 0 (x = 1): (145) Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 133 / 161Low Reynolds number ows Lubrication Theory Lubrication theory VII   Equation (140) imposes that p cannot depend on y . As a consequence equation (139) can be  integrated with respect to y and, also using the boundary conditions (142) and (143), we obtain   1 dp    2   u (x ; y ) = y h y : (146)  2 dx    In the above expression the term dp =dx is still an unknown function of x . Using the boundary condition (144) and (146) we nd that  dp =12: (147)  dx  x =0  We now integrate the continuity equation (141) with respect to y Z  Z  h   h  u v u         + dy = v (h ) v (0) + dy = 0;      x y x o o where we have used the noslip boundary conditions (142) and (143). 1 Using Leibniz rule and, again, the noslip boundary conditions (142) and (143) we obtain the following second order equation for the pressure    d dp 3 h = 0:   dx dx Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 134 / 161Low Reynolds number ows Lubrication Theory Lubrication theory VIII From the above equation and using (147) we obtain  dp 12 = ;  3 dx h which we can plug into equation (146) to obtain the following expression for the velocity in the  x direction  6    2   u (x ; y ) = y h y : 3 h  The y component of the velocity can be obtained from the continuity equation (141) and reads   3 2  y y dh    v (x ; y ) =6 + : 4 3  h h dx Finally, the pressure distribution can be obtained by integrating (128) and using the boundary condition (145). We note that we managed to obtain an analytical expression for the velocity without having to   specify the shape of the domain h (x ). 1 b(z) b(z) Z Z f (x; z) b(z) a(z) dx = f (x; z)dx f (b; z) + f (a; z) : z z z z a(z) a(z) Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 135 / 161High Reynolds number ows High Reynolds number ows Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 136 / 161High Reynolds number ows The Bernoulli theorem The Bernoulli theorem I As a rst tool to study high Reynolds number ows we introduce the Bernoulli theorem. As it will appear in the following, provided that some assumptions hold, this theorem is a very powerful tool to solve practical problems by very simple means. Let us recall the following vector identity 1 u (r u) = r(u u) (ur)u: (148) 2 Plugging it into the NavierStokes equation, we can rewrite (71) as   u p 1 2 2 u f +r + juj r u = 0; (149) t  2 where is the vorticity de ned by (49). Let us now assume that f is conservative. We can then write f =r , with scalar potential function. Let, moreover assume that the ow is steady, so that u=t = 0. Under the above assumptions we can write (149) as 0 2 rH = u +r u; where we have de ned 1 p 0 2 H = juj + + : (150) 2  Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 137 / 161High Reynolds number ows The Bernoulli theorem The Bernoulli theorem II If the uid is inviscid (or, more realistically, if viscosity plays a negligible role in the ow under consideration), we have 0 rH = u: Projecting the above equation in the direction of ow we obtain 0 urH = 0: 0 The above equation implies that H is constant along the streamlines, which is a remarkably simple result. Particular case: gravitational body force eld In the particular case in which the body force is gravity we have = gz; with z a vertical and upwards directed coordinate. In this case we then have 1 p 0 2 H = juj + + gz: 2  0 It is customary, in uid mechanics and hydraulics to work with the quantity H = H =g, so that 2 p juj H = z + + : (151) 2g H is named total head and it represents the total mechanical energy per unit weight of the uid. Note that if the uid is at rest H reduces to the pressure head, de ned in section 3. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 138 / 161High Reynolds number ows The Bernoulli theorem The Bernoulli theorem III Bernoulli theorem can be stated as follows. If the following conditions are satis ed: the uid is incompressible, the body force is gravity (or more in general it is conservative), the ow is steady, the e ects of viscosity are negligible, then the total head H is constant along streamlines. Note that the value of H can di er from one streamline to another. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 139 / 161High Reynolds number ows Vorticity equation and vorticity production Vorticity equation and vorticity production I Vorticity equation We wish to determine an equation for the vorticity =r u, see equation (49). The importance of the vorticity equation for studying large Reynolds number ows will become clear in the following. We take the curl of the NavierStokes equation (71), in which we assume that the body force eld is conservative, so that we can write f =r . We then obtain     u u u 1 2 r +r u +r + rpr u = 0: (152) t 2  In the above equation we have used the vector identity (148). As the curl of a gradient is zero the second, fourth and fth terms in equation (152) vanish. Therefore we can write, using the index notation,   2 u u k k   ( u )  = 0; ijk ijk klm l m ijk 2 x t x x x j j j l or 2 i i (  u ) = 0: ijk klm l m 2 t x x j l We note that by de nition of the alternating tensor we have  = . Moreover, the following ijk kij formula holds   =    : (153) kij klm il jm im jl Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 140 / 161High Reynolds number ows Vorticity equation and vorticity production Vorticity equation and vorticity production II Therefore, we can write 2 i i (u u ) = 0: i j j i 2 t x x j j From the continuity equation we have that u=x = 0. Moreover, the divergence of a curl is j j zero, and therefore, =x = 0. The above equation then simpli es to j j 2 u i i i i + u  = 0; (154) j j 2 t x x x j j j or, in vector form, 2 + (ur) (r)ur = 0: (155) t This equation is called vorticity equation and it is of fundamental importance in uid mechanics. The rst and second terms in equation (155) represent advective transport of vorticity. The last term represents viscous (di usive) transport of vorticity. The third term does not have a counterpart in the NavierStokes equations. It accounts for changes of vorticity due to deformation of material elements of the uid. Note, that pressure and body force do not appear in the vorticity equation. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 141 / 161High Reynolds number ows Vorticity equation and vorticity production Vorticity equation and vorticity production III Changes of vorticity in a volume V Let us now study how the amount of vorticity changes in a uid volume V . A suitable measure of the amount of vorticity is the enstrophy, de ned as ()=2. Multiplying equation (154) by i we obtain 2 2 2 u i i i i + u  = 0: j i j i 2 t 2 x 2 x x j j j After some algebraic manipulation this can be written as "   2 2 2 2 2 u i i i i i + u  = 0: j i j 2 t 2 x 2 x x 2 x j j j j Taking the integral of the above expression over a material volume V and applying the Reynolds transport theorem we obtain ZZZ ZZZ ZZZ ZZZ   2 2 2 2 D u i i i i dV = dV +  dV  dV: (156) i j 2 Dt 2 x x 2 x j j j V V V V The term on the left hand side represents the time variation of the enstrophy associated with the volume V . Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 142 / 161High Reynolds number ows Vorticity equation and vorticity production Vorticity equation and vorticity production IV The rst term on the right hand side induces changes of vorticity in V , as a response to the velocity distribution. Note, however, that this is not a source term: if at some time the vorticity within V is zero this term can not produce new vorticity. Making use of Gauss theorem the second term on the right hand side can be transformed into a ux term across the surface S bounding V as follows ZZZ ZZ 2 2  i 2  dV = n dS: j i 2 x 2 2 x j j V S Therefore it does not produce nor dissipate vorticity. Finally the last term represents viscous dissipation of vorticity and always contributes to decrease the amount of vorticity within the volume V . Equation (156) shows that the vorticity can not be generated within a body of uid. It then follows that vorticity can only be generated at the boundary of the domain. A typical source of vorticity is, for instance, the noslip condition (84), which holds in correspondence of solid walls. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 143 / 161High Reynolds number ows Vorticity equation and vorticity production Vorticity equation and vorticity production V Generation of vorticity due to an impulsively started solid body To understand the generation and transport of vorticity let us consider an example: a uid occupying an in nite region and initially (at time t = 0) at rest is set in motion by a solid body immersed in the uid that, at t = 0, impulsively starts moving with velocity U. Suppose that we study this ow in a frame moving with the solid body. We can think that the development of motion in the uid takes place in three di erent phases. 1 At the initial time (t = 0) the uid starts moving and the ow is irrotational, i.e. the vorticity is zero everywhere. In fact the vorticity is initially con ned in an in nitesimally thin layer at the wall and, within that layer the vorticity is theoretically in nite. 2 For t 0 the vorticity starts be transported away from the wall. Transport occurs both for viscous di usion and advection. If di usion was the only transport mechanism the thickness p of the boundary layer in which the ow is not irrotational would be of order t at time t. At the very initial stage advection is expected to have a relatively small e ect as, initially, the normal component of the relative velocity of the uid with respect to the wall is expected to be small. Thus for small times the thickness of the boundary layer will be of the order of p t. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 144 / 161High Reynolds number ows Vorticity equation and vorticity production Vorticity equation and vorticity production VI 3 For larger times two di erent scenarios might occur. The body is thin and oriented in the direction of motion In this case the normal component of the relative velocity close to the wall will remain small even for large times. In this case a steady ow might be reached in which longitudinal advection and di usion are balanced. If L is the longitudinal spatial dimension of the body the characteristic time for a uid particle to travel in the region close to the body is of order L=U. In this case the vorticity keeps con ned within a boundary layer at the wall with p thickness  of order L=U. It follows that  1 1 / p =p : L UL= Re In this case it is said that boundary layer separation does not occur. The above considerations suggest that, at large values of the Reynolds number, if no boundary layer separation occurs, the motion is irrotational in most of the domain. We will see in the next subsection that the absence of vorticity allows for great simpli cation of the governing equations. The body is thick or not oriented in the direction of motion In this case advection in the direction normal to the body is strong and the region with vorticity grows rapidly. In this case it is said that boundary layer separation occurs. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 145 / 161High Reynolds number ows Irrotational ows Irrotational ows I Potential function of the velocity We have seen in the last section that at large values of the Reynolds number it might happen that in most of the ow domain motion remains irrotational. We now wish to study if the assumptions of incompressible uid, and irrotational ow, i.e. r u = 0; r u = 0: (157) allow for any simpli cations of the problem. Note that the conditions (157) are of purely kinematic nature even if they are consequence of the dynamic behaviour of the uid. Let us consider a closed reducible curve C and let us take the line integral of the velocity along this curve. We have, by Stokes theorem, I ZZ ZZ u dx = (r u) ndS =  ndS = 0: C S S We now consider any two points, say O and P, on C. They split C into two curves, C and C , 1 2 with both C and C joining O to P. We then have 1 2 Z Z u dx = u dx: C C 1 2 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 146 / 161High Reynolds number ows Irrotational ows Irrotational ows II This implies that the integral between O and P does not depend on the path of integration but only on the starting and ending points. We can then de ne a potential function (x) of the velocity eld, such that Z P (x) =(x ) + u dx: (158) 0 0 Equation (158) implies that we can write u =r: (159) If we recall the continuity equation for an incompressible uid, i.e.r u = 0, and plug (159) into it we nd 2 r  = 0; (160) which implies that the potential function  has to be harmonic. In other words the velocity potential function satis es the Laplace equation. If we solve the problem for the function  we can then easily nd the velocity u using equation (159). It is clear that the mathematical problem for an irrotational ow is much easier than that for a rotational ow, for the following main reasons: equation (160) is linear, whereas the NavierStokes equations are nonlinear; in the case of an irrotational ow it is sucient to solve the problem for a scalar function rather than a vector function; Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 147 / 161High Reynolds number ows Irrotational ows Irrotational ows III the problem for the pressure is decoupled from the problem for the velocity eld. How to compute the pressure in an irrotational ow will be discussed in the following. Owing to the properties of equation (160) we can state that the velocity distribution has the following properties. As equation (160) is elliptic the solution for  and all its derivatives is smooth except, at most, on the boundary. The function  is singlevalued if the considered domain is simply connected. In the following we will only consider the case of simply connected regions. Conditions for  to be uniquely determined Let us now consider the boundary conditions we need to impose for the solution for  to be unique. We rst note that the following vector identity holds r (u) =r u + ur = u u; from which we can write, also using the divergence theorem, ZZZ ZZZ ZZ u udV = r (u)dV = u ndS: (161) V V S Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 148 / 161High Reynolds number ows Irrotational ows Irrotational ows IV Let u =r and u =r be two solutions of equation (160). The di erence (u u ) is also 1 1 2 2 1 2 a solution, owing to the linearity of the equation. Recalling equation (161) we can write ZZZ ZZ 2 ju uj dV = (  )(u u ) ndS: 1 2 1 2 1 2 V S The above expression shows that u and u coincide if 1 2 (u u ) n = 0 on the boundary S, i.e. if the normal components of the velocity are the 1 2 same (Neumann conditions); or if  = on S (Dirichlet conditions); 1 2 of if (u u ) n = 0 on a portion of S and  = on the remaining part of S. 1 2 1 2 It is important to notice that for an irrotational ow it is not possible to impose the noslip condition at solid walls as only the normal component of the velocity is required. However, close to rigid walls a boundary layer exists, in which the ow is rotational. To determine the ow in the boundary layer the NavierStokes equations have to be solved. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 149 / 161High Reynolds number ows Bernoulli equation for irrotational ows Bernoulli equation for irrotational ows I We have shown that the potential function  can be obtained by using the irrotationality of the ow and the continuity equation. We now consider the momentum equation (NavierStokes equation) and study how it simpli es in the case of irrotational ow. We anticipate that the use of the momentum equation will allow us to determine the pressure. Recalling the vector identities (123) and (148) the NavierStokes equation (71) can then be written as   2 u juj 1 +r u + rp f = 0 t 2  Assuming that the the body force eld is conservative we can write f =r . If the ow is irrotational we have = 0 and u =r. Thus we can write   2  juj p r + + + = 0: t 2  This equation can be solved to get 2  juj p + + + = F (t): (162) t 2  Notice that, without loss of generality, we can introduce a function  such that Z  = Fdt; Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 150 / 161High Reynolds number ows Bernoulli equation for irrotational ows Bernoulli equation for irrotational ows II which allows to eliminate the unknown function F (t) from (162) to nally obtain 2  juj p H = + + + = c; (163) t 2  with c arbitrary constant. This is the Bernoulli theorem for irrotational ows. Notice that the last 0 three terms ofH in the above equation represent H as de ned by equation (150). It is important to stress that Bernoulli theorem holds in a stronger form in the case of irrotational ows. In particular 0 H is constant also in unsteady ow conditions (this is not true for H in rotational ows); 0 H is constant in the whole domain and not only along streamlines as it is the case for H in rotational ows; the theorem holds exactly for real viscous uids, provided the vorticity is everywhere zero. Important note: it is important to stress that even if in irrotational ows the viscous terms in the NavierStokes equation vanish, the viscous stresses are not necessarily zero. In fact it is the divergence of the stress tensor that vanishes, not the stress itself. In other words viscous stresses might exist in irrotational ows. However they do not contribute to the momentum equation. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 151 / 161Appendix A: material derivative of the Jacobian Appendix A: material derivative of the Jacobian Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 152 / 161Appendix A: material derivative of the Jacobian Determinants Determinants De nition: A permutation i; j;:::; p of the rst n integers 1; 2;::: n is called even or odd according as the natural order can be restored by an even or odd number of interchanges. De nition: The determinant of a n n matrix A with elements a is ij X det A = a a ::: a ; (164) np 1i 2j where the summation is taken over all permutations i; j;:::; p of the integer numbers 1; 2;:::; n, and the sign is positive for even permutations and negative for odd ones. Therefore, for instance, the determinant of a 3 3 matrix A is det A =a a a + a a a + a a a 11 22 33 12 23 31 13 21 32 a a a a a a a a a : 12 21 33 11 23 32 13 22 31 Note that in each term of the sum there is only one element from each row and each column. Derivative of a determinant If the elements of a n n matrix A are function of a variable s, so that a (s), the derivative with ij respect to s of det A is the sum of n determinants obtained by replacing one row of A by the derivatives of its elements. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 153 / 161Appendix A: material derivative of the Jacobian Derivative of the Jacobian Derivative of the Jacobian I We consider the Jacobian 0 1 x x x 1 1 1 B C    1 2 3 B C B C x x x 2 2 2 B C J = det : B C    B 1 2 3 C A x x x 3 3 3    1 2 3 We wish to compute its material derivative DJ=Dt. Let us consider an element of the above matrix. We have   D x Dx u i j i = = : Dt   Dt  j j j In the above expression we could interchange the order of di erentiation because D=Dt is di erentiation with constant by de nition of material derivative. If we regard u as a function of i (x ; x ; x ) we can write 1 2 3 u u x u x u x u x i i 1 i 2 i 3 i k = + + = :  x  x  x  x  j 1 j 2 j 3 j k j Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 154 / 161Appendix A: material derivative of the Jacobian Derivative of the Jacobian Derivative of the Jacobian II We know that the derivative of a determinant of a 3 3 matrix is the sum of three determinants, each of a matrix in which one row is di erentiated. Thus to compute DJ=Dt we have to sum up three terms the rst of which is 0 1 0 1 u u u u x u x u x 1 1 1 1 k 1 k 1 k B C B C    x  x  x  1 2 3 1 2 3 k k k B C B C B C B C x x x x x x B 2 2 2 C B 2 2 2 C detB C = detB C: B C B C       1 2 3 1 2 3 B C B C A A x x x x x x 3 3 3 3 3 3       1 2 3 1 2 3 With k = 1 we have 0 1 u x u x u x 1 1 1 1 1 1 B C x  x  x  1 1 1 2 1 3 B C B C x x x u B 2 2 2 C 1 det = J: B C B C    x 1 2 3 1 B C A x x x 3 3 3    1 2 3 With k = 2; 3 we have u =x times the determinant of a matrix with two identical rows, which 1 k is therefore equal to zero. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 155 / 161Appendix A: material derivative of the Jacobian Derivative of the Jacobian Derivative of the Jacobian III Computing the other two term of the DJ=Dt we thus nally nd   DJ u u u 1 2 3 = + + J = (r u)J: (165) Dt x x x 1 2 3 Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 156 / 161Appendix B: the equations of motion in di erent coordinates systems Appendix B: the equations of motion in di erent coordinates systems Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 157 / 161Appendix B: the equations of motion in di erent coordinates systems Cylindrical coordinates Cylindrical coordinates Let us consider cylindrical coordinates (z; r;'), with corresponding velocity components (u ; u ; u ). z r ' Continuity equation u 1 1u z ' + (ru ) + = 0 (166) r z r r r ' NavierStokes equations     2 2 u u u u u 1p u 1 u 1 u z z z ' z z z z + u + u + +  + r + = 0: (167) z r 2 2 2 t z r r ' z z r r r r ' 2 u u u u u u 1p r r r ' r ' + u + u + + + z r t z r r ' r  r     2 2 u 1 u 1 u u 2 u r r r r '  + r + = 0: (168) 2 2 2 2 2 z r r r r ' r r ' u u u u u u u 1 p ' ' ' ' ' r ' + u + u + + + + z r t z r r ' r r '     2 2 u 1 u 1 u 2 u u ' ' ' r '  + r + + = 0: (169) 2 2 2 2 2 z r r r r ' r ' r Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 158 / 161Appendix B: the equations of motion in di erent coordinates systems Spherical polar coordinates Spherical polar coordinates I Let us consider spherical polar coordinates (r;;') (radial, zenithal and azimuthal), with corresponding velocity components (u ; u ; u ). r ' Continuity equation  1 1 1 u ' 2 r u + (sinu ) + = 0: (170) r 2 r r r sin r sin ' NavierStokes equations 2 2 u u u u u u u u 1p r r r ' r ' + u + + + + r t r r r sin ' r r  r      2 1 u 1 u 1 u r r r 2  r + sin + + 2 2 2 2 2 r r r r sin r sin '  2u 2 (u sin) 2 u ) r ' = 0: (171) 2 2 2 r r sin r sin ' 2 u u u u u u u u u cot 1 p ' r ' + u + + + + + r t r r r sin ' r r r      2 1 u 1 u 1 u 2  r + sin + + 2 2 2 2 2 r r r r sin r sin ' Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 159 / 161Appendix B: the equations of motion in di erent coordinates systems Spherical polar coordinates Spherical polar coordinates II  2 u u 2 cos u ) r ' + = 0: (172) 2 2 2 2 2 r r sin r sin ' u u u u u u u u u u cot 1 p ' ' ' ' ' r ' ' + u + + + + + + r t r r r sin ' r r r sin'      2 1 u 1 u 1 u ' ' ' 2  r + sin + + 2 2 2 2 2 r r r r sin r sin '  2 u 2 cos u u r ' + + = 0: (173) 2 2 2 2 2 r sin ' r sin ' r sin Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 160 / 161References References D. J. Acheson. Elementary Fluid Dynamics. Oxford University Press, 1990. R. Aris. Vectors, tensors, and the basic equations of uid mechanics. Dover Publications INC., New York, 1962. G. I. Barenblatt. Scaling. Cambridge University Press, 2003. G. K. Batchelor. An Introduction to Fluid Dynamics. Cambridge University Press, 1967. H. Ockendon and J. R. Ockendon. Viscous Flow. Cambridge University Press, 1995. ISBN 0521458811. C. Pozrikidis. Fluid Dynamics: Theory, Computation, and Numerical Simulation. Springer, softcover reprint of hardcover 2nd ed. 2009 edition, Nov. 2010. ISBN 1441947191. Rodolfo Repetto (University of Genoa) Fluid dynamics January 13, 2016 161 / 161
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