1 Incompressible viscous fluid flow. The Navier- Stokes equations. The purpose of this section is to give a brief summary of the Navier-Stokes equations for a flow of an incompressible viscous fluid. 1.1 Kinematic definitions. The position vector of a fluid particle at time t is denoted as r = (x, y, z) = xi + yj + zk, where i, j, k are the usual Cartesian unit vectors. The velocity of a fluid particle is denoted as u = (u, v, w) = ui + vj + wk. Alternatively, we can write r =xiei, u = uiei, where i = 1, 2, 3, ei is the unit vector along the Cartesian axis xi and the summation convention over repeated indices is used so that, for example, 3 X xiei = xiei. (1.1) i=1 There are two common approaches to describing the flow. Lagrangian specification: all the relevant quantities (e.g. position vector, velocity, acceleration etc.) for a chosen fluid particle are functions of time and the position of the particle at some initial time, for instance at t = 0. For example, if r0 is the position vector of a fluid particle at time t = 0, then the position vector of the same particle at time t will be r = r (r0, t) . Example 1. Given the position vector of a fluid particle in Lagrangian specifi- cation, its velocity and acceleration are calculated as follows: ∂r (r , t) u = 0 , (1.2) ∂t ∂u (r , t) ∂2r (r , t) a = 0 = 0 . (1.3) ∂t ∂t2 Eulerian specification: each quantity related to the fluid flow is a function of time t and the position in space, r = (x, y, z). For instance, velocity and density in the flow are written as u = u (r, t) = u (x, y, z, t) , ρ = ρ (r, t) = ρ (x, y, z, t) , (1.4) respectively. Example 2. Given the velocity field in Eulerian specification, u = u (r, t) , (1.5) find the acceleration of the fluid particles in the flow. 1 We change to the Lagrange variables and use the results (1.2) and (1.3). Then u = u (r, t) = u (r (r0, t) , t) . (1.6) Also, for the acceleration, ∂u (r (r , t) , t) ∂u (x (r , t) , y (r , t) , z (r , t) , t) a = 0 = 0 0 0 (1.7) ∂t ∂t ∂u (r, t) ∂x (r , t) ∂u (r, t) ∂y (r , t) ∂u (r, t) ∂z (r , t) ∂u (r, t) = 0 + 0 + 0 + (1.8) ∂x ∂t ∂y ∂t ∂z ∂t ∂t ∂u (r, t) = + (u · ∇) u. (1.9) ∂t with ∂ ∂ ∂ ∇ = i + j + k . (1.10) ∂x ∂y ∂z This result is often written as Du a = (1.11) Dt where the differential operator D ∂ = + (u · ∇) (1.12) Dt ∂t is the so-called material time derivative. In this context the partial time derivative in (1.12), ∂/∂t, is called the local derivative with respect to time and the operator (u · ∇) is the convective derivative. Example 3. Given the density and velocity distributions in the flow, ρ = ρ (r, t) and u = u (r, t) , (1.13) find the rate of change of density in a fluid particle. The answer is Dρ ∂ρ = + (u · ∇) ρ. (1.14) Dt ∂t 1.2 Conservation of mass. Conservation laws in a moving substance are conveniently formulated for a material volume, i.e. a volume containing the same fluid particles at all times. Consider a small material volume, δV. The mass in δV does not change, D (ρδV ) = 0, (1.15) Dt Note that 1 D (ρδV ) Dρ 1 D (δV ) = + ρ , (1.16) δV Dt Dt δV Dt 2 and, in coordinates, 1 D (δV ) 1 D (δxδyδz) = (1.17) δV Dt δxδyδz Dt 1 D (δx) 1 D (δy) 1 D (δz) = + + . (1.18) δx Dt δy Dt δz Dt Then, since 1 D (δx) δu = , (1.19) δx Dt δx with similar expressions for the other two terms in (1.18), we have, in the limit δV → 0, δu ∂u → , (1.20) δx ∂x and therefore 1 D (δV ) ∂u ∂v ∂w lim = + + = divu. (1.21) δV →0 δV Dt ∂x ∂y ∂z The limit, δV → 0, applied to (1.15) with (1.16) yields Dρ + ρdivu =0. (1.22) Dt The equivalent forms of (1.22) are ∂ρ + (u · ∇) ρ + ρdivu =0; (1.23) ∂t ∂ρ + div (ρu) =0. (1.24) ∂t The equations (1.22), (1.23), (1.24) are all different expressions of the mass con- servation. Fluid is called incompressible if D (δV ) = 0, (1.25) Dt or divu =0. (1.26) The last equation is known as the continuity equation for an incompressible fluid. Then the equation of mass conservation becomes ∂ρ + (u · ∇) ρ = 0. (1.27) ∂t The fluid is called homogeneous if ρ ≡ const. (1.28) For a homogeneous incompressible flow the mass conservation equation is satisfied as long as the velocity field satisfies the continuity equation (1.26). 3 1.3 Rate of change of momentum in a material volume. The task is to simplify the expression, d ρudV. (1.29) dt ˆ V (t) Since mass in conserved in a small material volume , ρδV = const, we have, d Du ρudV = ρ dV. (1.30) dt ˆ ˆ Dt V (t) V (t) 1.4 Motion near a fluid particle. At a fixed moment in time, t, consider two particles at a small distance δr from each other. Particle 1. Position vector r = (x1, x2, x3) , velocity u (r) = (u1, u2, u3) . Particle 2. Position vector r + δr, velocity u (r + δr) , with δr = (δx1, δx2, δx3) Here we use Cartesian coordinates with the unit basis vectors ei, i = 1, 2, 3. Our aim is to express the velocity of the second particle in terms of the velocity of the first particle and various spatial derivatives of the velocity field calculated at the point r. Consider, for example, the x1-component, u1 (r + δr) = u1 (x1 + δx1, x2 + δx2, x3 + δx3) (1.31) ∂u1 (x1, x2, x3) ∂u1 (x1, x2, x3) ∂u1 (x1, x2, x3) 2 = u1 (x1, x2, x3)+ δx1+ δx2+ δx3+O |δr| . ∂x1 ∂x2 ∂x3 (1.32) This result can be written shorter using the summation convention, ∂u1 (r) u1 (r + δr) = u1 (r) + δxi + ... (1.33) ∂xi Similar relations hold for u2 (r + δr) , u3 (r + δr) and also for the complete vector velocity, u (r + δr) = uj (r + δr) ej (1.34) " # ∂uj (r) ∂uj (r) = uj (r) + δxi + ... ej = uj (r) ej + δxiej + ... (1.35) ∂xi ∂xi ∂uj (r) ∂ui (r) = u (r) + δxiej + ... = u (r) + δxjei + ... (1.36) ∂xi ∂xj Note the change of subscripts in the last expression. Using the identity, ∂u 1 ∂u ∂u ! 1 ∂u ∂u ! i = i + j + i − j , (1.37) ∂xj 2 ∂xj ∂xi 2 ∂xj ∂xi 4 and the notation, ! ! 1 ∂ui ∂uj 1 ∂ui ∂uj eij = + , ξij = − (1.38) 2 ∂xj ∂xi 2 ∂xj ∂xi we can re-write (1.36) as u (r + δr) = u (r) + eijδxjei + ξijδxjei + ... (1.39) translation deformation rotation We conclude that the local motion near a fluid particle can be described as a su- perposition of three motions: a pure translation with velocity u (r), deformation characterized by the symmetric second-order tensor, eij, and rotation described by the antisymmetric second-order tensor, ξij. The tensor eij is known as the rate-of- strain tensor. 1.5 Forces in fluids. We distinguish between long-range or body forces (such as gravity) and short-range forces due to molecular interactions inside the fluid (pressure, friction). Another force that often appears in applications is surface tension (a short-range force acting on the boundary between two fluids). According to Newton’s Second Law, the rate of change of momentum of the fluid in a material volume equals the total force acting on this volume of fluid. The task for us is to learn how to describe the forces in the fluid. Long-range forces do not pose much problem. In the case of gravity, for example, the force on the fluid with density ρ contained in the volume V (t) is ρgdV, (1.40) ˆ V (t) where g is the acceleration due to gravity. Short-range forces. At a fixed moment in time, consider a volume of fluid with surface area A. Choose a small element of the surface area, δA, with the outward unit normal nb. Let δF be the force exerted on δA by the fluid outside of our volume. For a small δA, δF ≈ ΣδA, (1.41) where Σ is the stress vector. Note that Σ = Σ (r, t, nb) , i.e. the stress vector in general depends on the orientation of the area element. We are looking to find a quantity which would be independent of the position of the surface area. In Newton’s Second Law, Du ρ dV = ρgdV + ΣdA, (1.42) ˆ Dt ˆ ˆ V V A 5 let us assume that the typical linear size of the fluid volume, δr, say, is small. Then Du ρ dV = O δr3 , ρgdV = O δr3 , dF = O δr2 . (1.43) ˆ Dt ˆ ˆ V V A The balance of terms in (1.43) requires then ΣdA=0. (1.44) ˆ A From the last relation, it can be shown that the stress vector components are pro- portional to the unit normal, more precisely if we write Σ = Σiei then the individual components Σi can be written in terms of a second-order stress tensor, σij, as Σi = σjinj or Σi = σijnj. (1.45) We have used here the fact (which follows from conservation of angular momen- tum) that the stress tensor is symmetric, σij = σji.