Basic hydrodynamics. David Gurarie. 1 Newtonian fluids: Euler and Navier-Stokes equations

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1 Basic hydrodynamics David Gurarie 1 Newtonian fluids: Euler and Navier-Stokes equations The basic hydrodynamic equations in the Eulerian form consist of conservation of mass, momentum and energy. We denote by u = u (x; t) velocity field, and -densityfield, with D t = t + u - derivative (of any material quantity) along the flow direction. 1.1 Mass conservation The continuity (conservation) of mass requires the mass transfer rate along the flow D t be balanced by its change due to the volume compression/expansion by the flow u, t + (u) =D t +( u) =0 (1) f density is constant, (1) turns into incompressibility condition u =0 1. Momentum conservation The change of momentum D t u of a Newtonian fluid is balanced by a combination of body forces F b (distributed through the volume) and surface forces F s. The former include gravity, Coriolis (rotation), Coulomb, Lorentz (for electrically conducting fluids) etc. The latter have a general form F s = σ NdS where σ is the stress-tensor, N -normal to the surface element ds. The stresses is made of two components: pressure p and viscous (frictional) forces, depending on the rate of (velocity) deformation: u =( i u j ). The latter is made of the antisymmetric (rotation) component: u = ³ iu j ju i, and the symmetric rate of strain-tensor s ij = 1 ( iu j + j u i ). Only the latter (strain) would contribute to the viscous stresses. 1

2 Assuming isotropy (rotational symmetry) one can derives the following representation of the combined pressure + viscosity stresses 1 µ ν σ ij = pδ ij + νs ij + tr (s) δ ij () Lame coefficients ν, µ measure the dynamic viscosity, whiletr (s) = u gives divergence. Now the momentum conservation: t (u)+ [u (u)] = σ turns into the Navier-Stokes equation µ µ ν t (u i )+ [u(u i )] = i p + ν u i + i µ µ ν = i p + ν u i + u,or (4) i ( u),forconst ν, µ written here for the i-th component of momentum. Dividing by and using mass conservation (1) it takes the form D t u = p + ν u+ µ ν ( u) (5) where ν = ν, µ = µ are the kinematic viscosities. Two other simplifications arise for incompressible fluids (second viscous term drops) D t (u) = p + ν u and the ideal Euler fluid (no viscosity) D t u = u t +(u ) u + p =0 (6) The Euler momentum equation can be rewritten via (15) as µ u u t + + p u ( u) =0 (7) 1 Relation () results from the decomposition of a symmetric n n-matrix s into its scalar and traceless components s 0 + s 0 µ tr (s) + s () tr (s) s = n {z } s 0 n {z } s 0 f P n denotes the space of all such matrices, the isotropy of the linear strain-to-stress map Λ : s σ implies that Λ commutes with the natural action of Lie group SO (n) on space P n : s U su (conjugation). Any such map Λ of P n ( SO (n)-invariant) is well known to be of the form s 0 + s 0 λs 0 + µs 0 So Λ is determined by a pair of scalars λ,µ - Lame coefficients. n other words, Λ respects the scalar+traceless decomposition (), hence follows ().

3 n particular, steady-state potential flow ( u = 0) has Bernoulli form, W = u + p conserved along flow-lines u t = 0 ω = u = 0 u + p =Const Taking div of (7) we get µ u t ( u)+ + p u u u =0 Hence the ideal fluid pressure is a quadratic functional of velocity u p n o = 1 u + u u u = 1 ζ + u u ª u (8) Taking div of (7) we get the Euler equation for the vorticity field ζ t (u ζ) =ζ t +(u ) ζ (ζ ) u =0 (9) ntegral conservation laws Several integral conservation laws are easily verified for Euler fluids: Momentum: u dx = const is conserved due to u =0 Energy: E = ndeed, µ u t 1 u dx = const µ u = u u t = u + p µµ u + u (u ζ) = + p u -complete divergence. Total vorticity ζ dx = const ndeed, incompressibility of fields u; ζ implies (u ) f = (f u) and (ζ ) f = (fζ) for any f. Hence each component ζ k of ζ obeys by (9) t (ζ k )= (u k ζ ζ k u) - complete divergence

4 Circulation (Kelvin): for any loop γ t that moves along the flow C (t) = u ds = const γ t We use the transport along the flow µ d u dt C = D t u ds = γ t γ t + p u ζ ds The grad-term gives 0 along any closed loop, while the H (u ζ) = RR Σ t (u ζ) NdS=0, due to the zero integrand (16). Helicity (Mofatt): (u ζ) dx = const One writes ³ u u t + + p u ζ =0 ζ t (u ζ) =0 multiplies the first equation by ζ the second one by u and sums together µ u (ζ u) t = + p ζ + ζ (u ζ) u { (u ζ)} The -nd term in the r.h.s. is 0, while the -rd is transformed into u ( R) = (R u) R ( u) = (R u) where R = u ζ and R ( u) =R ζ =0vanishes. Hence µ u (ζ u) t = + p ζ R u - complete divergence System of conservation laws The basic Euler gas-fluid dynamics written as mass-momentum conservation for (, u,p) is incomplete unless an equation of state is imposed, that would link pressure p to, u, and/or other parameters of state, s.a. entropy, temperature, etc. The simplest case is isentropic (barotropic) gas, where p = p () -agiven function of, e.g. C γ. One example is provided by the shallow water equations for (u,h), viewed as D compressible fluid of density equal to height h and pressure p = 1 gh, D t u = g h (10) t h + (uh) =0 4

5 More generally the internal state is described by specific entropy (per unit mass) s, that obeys the conservation law t (s)+ (us) =0 (11) -active tracer, and p = p (,s) - a function of pressure and density. The combined system for (, u,s) has the conservation law form u v t u v + u + p x uv + uv y v + p = 0 (1) s us vs for D, and similar expression for D. n the isentropic case the last equation drops, so the system order is reduced by 1. Euler system (1) has another conserved quantity, the total energy E, made of the kinetic one K = 1 u, and the potential (internal) one E (,s). ndeed, µ u t K = u + p + p ( u) (1) t E = [ue]+( u)(e E se s ) Hence, the total energy t (K + E)+ [u (K + E + p)] = ( u)(p + E E se s ) (14) -conserved provided the r.h.s. is zero. For incompressible fluids u =0, the kinetic energy is conserved, along with any function E (,s), as,s are passive scalars. n the compressible case we get the differential relation between p and E p =( + s s ) E E n particular,e = α s β yields p =(α + β 1) α s β. For shallow water, E = 1 gh yields p = 1 gh. 4 Appendix: vector algebra Commonly used vector identities for scalar/vector fields: V V V V (u v) =v ( u) u ( v) (a u) = a ( u),fora -const (fu) =f u + f u (u v) =(v ) u (u ) v + u ( v) v ( u) (a v) =+a ( v) (a ) v, fora -const ( u) = ( u) u u ( v) =u ( v) t (u ) v u ( v)+v ³ ( u) = (u v) (u ) v (v ) u (u ) u = u ( u) u (15) 5

6 Here v =( i v j ) denotes the complete derivative (Jacobian matrix) of v.f. v. Relation () follows from the formal determinant expansion by swapping rows 1 u 1 u u v 1 v v = v 1 v v 1 u 1 u u u 1 u u 1 v 1 v v To check () we call the components of fields u =(a, b, c);v =(f,g,h). Then u v =(bh cg, cf ah, ag bf) and the first component of the curl becomes (a y g b y f) (c z f a z h) (f y b g y a)+(h z a f z c) Here the second pair of brackets is obtained by interchanging fields u, v in the anti-symmetric expression (u v). Adding and subtracting af x the first pair is brought into the form fa x + ga y + ha z (a x + b y + c z ) f = 1-st comp. (v ) u ( u) v The second pair gives a similar expression with u, v interchanged, Q.E.D. Relations (V-V) are consequences of (V). To verify the latter we write u =(a; b; c) and v =(f; g; h). thenu ( v) is equal to i j k a b c h y g z f z h x g x f y =(b (g x f y ) c (f z h x );...;...) Adding and subtracting af x we arrange the 1-st component to be af x + bg x + ch x (af x + bf y + cf z )=u x v (u ) f By the same pattern the -nd and -rd components are whence follows (V). u y v (u ) g u z v (u ) h Corollary 1 f u denotes incompressible velocity and ζ = u -vorticity, then one has (u ζ) =ζ + u u (u ζ) =(ζ ) u (u ) ζ (16) 6

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