Derivation Of Conservation Equations

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Derivation of the basic equations of fluid flows. Noparticle in the fluid at this stage (next week).week) Conservation of mass of the fluid. Conservation of mass of a solute (applies to non-sinkingparticles at low concentration). Conservation of momentum. Application of these basic equations to a turbulent fluid.fluid

A few concepts before we get to the meat Tensor (Stress), Vectors (e.g. position, velocity) andscalars (e.g. T, S, CO2).We need to define a coordinate system, and an((infinitesimal)) element of volume.We assume a continuous fluid, andthat all the fields of interest aredifferentiable.differentiable

The Lagrangian framework is the framework in which thelaws of classical mechanics are often statedstated. Thevcoordinates of a point x (t ) describe the trajectory fromv vx0 x (t 0) . Density, ρ, can evolve along the trajectory.By the chain rule,rule along a parcel trajectory:vdρ dx ρ ρv ρ v u ρ dt dt x t const t xv const tvdρ ( x (t ), t ) v v u ρ (x , t )dt t D v u ÅConversion from Lagrangian to EulerianDt t t

Example:Let’s assume that we are in a river that feeds on glacialmelt. The water warms at a constant rate that is afunction of distance from the source. If we drift downriver (A la ‘Huckleberry Fin’), the temperature increaseswith time (DT/Dt 0). At one point along the river,however we may see no change in temperature with timehowever,( T/ t 0), as the water arriving there is always at thesame temperature. The heat flux is advective, (u T/ x 0).In short, the convective derivative is:D v u v w u Dt t x y z t

Mass conservation (Eulerian, differential approach):Accountingg for the changeg in mass inside a fixed,, constantsize volume:(x-Δx/2,y Δy/2,z Δz/2)(x Δx/2,y Δy/2,z Δz/2)Az ΔxΔy(x-Δx/2,y-Δy/2,z Δz/2)Ay ΔxΔzz AxΔyΔzyA Δ ΔAy ΔxΔz(x Δx/2,y Δy/2,z-Δz/2)Az ΔxΔyxMass ρVMass AxΔyzΔ(x-Δx/2,y-Δy/2,z-Δz/2)(x Δx/2,y-Δy/2,z-Δz/2) ( ρV ) Ax ρu Δx / 2 ρu Δx / 2 Ay ρv Δy / 2 ρv Δy / 2 Az ρw Δz / 2 ρw Δz / 2 t()(() ()) ρ111 ρu Δx / 2 ρu Δx / 2 ρv Δy / 2 ρv Δy / 2 ρw Δz / 2 ρw Δz / 2 tΔxΔyΔz() ρ ρv (ρu ) (ρv ) (ρw) (ρu ) 0 y z t t x()

Mass conservation (Eulerian, integral approach):Accounting for the change in mass inside a fixed,fixedconstant-volume volume (V0):dv vρdV ρu n dS dt V0 V0 v ρdV ρudV tV0V0 v ρ ρu 0 tWhere we used the divergencegtheorem:It states that the volume total of all sinks and sources, the volume integral ofthe divergence, is equal to the net flow across the volume's boundary (WIKI).

Reiteration (no sinks/sources):Mass conservation (Lagrangian, integral):DρdV 0 Dt VMass conservation (Eulerian,(E lintegral):l)dv vρdV ρu ndS dt V0 V0

Mass conservation:Note that:r ρ ( ρu ) 0 t tCan be written as:r1Dρ u 0ρDtDThe 2nd term is the fluid divergence (rate of outflow of volume per unitvolume). This can be nonzero only for compressible fluids. It is the rate ofloss of density due to compression/expansion.rFor bothFb th watert andd airi we can assume ththatt u 0 iin termstoff theirth idynamics (we need compressibility to pass sound ).

Mass balance for conserved scalar:Addingng momolecularcu ar diffusion:ffus onrr C V t dV S (Cu K C ) ndSWhere V is the volume of the control volume and S its surface, andusing Fick’s law. By the help of the divergence theorem:rC C V t (Cu K C ) dV 0Since the volume is arbitrary, this can be true if and only if:r C (Cu ) (K C ) t

Momentum balance (Navier-Stokes):Newton’s 2nd law of motion states that the time rate of changeg ofmomentum of a particle is equal to the force acting on it. This law isLagrangian, the “time rate of change” is with respect to a reference systemfollowing the particle.vdvvρudV ρ gdV T dS dt V (t )V (t ) V ( t )Where g is the body force per unit mass (e.g. gravity) and T is the surfaceforce per unit surface area bounding V.Iff theh volumelis smallll enough,h theh integrandsd can beb takenk out off theh integral:l d vdd vv (ρuδV )ρudV ρudVdt V (t )dt V (t ) dtvvd vdu v d (ρδδV )du(ρuδV ) ρδV u ρδVdtdtdtdt

Momentum balance (Navier-Stokes):Th bodyTheb d forcefisi similarlyi il l treated:tt dvv ρ gdV ρgδVV (t )Defining a stress tensor (expanded on the next slide):vvT Τ nA d applyingAndl i ththe divergencediththeorem:v T dS V ( t ) ΤdV ΤδVV (t )vDuv ρ ρg ΤDt

Surface forcing:For an inviscid fluid,, the surface forceexerted byy the surroundinggvvfluid is normal to the surface, i.e. T p n , and p is called thepressure force.In general,stress force, S,v is also present. For viscousv viscousv vvT Τ nfluids: T p n S . Bydefinition, and we now havevvT pI , where S n and I is the identity tensor.For Newtonian fluids,v Σ μ u2And the resultant NavierNavier-StokesStokes equations for incompressible fluids are:vDuv2vρ ρg p μ uDt

Rotational symmetry: τ xx τ xy τ xz τ xx τ xy τ xz T τ yx τ yy τ yz τ xy τ yy τ yz τ τ ττττzyzz yzzz zx xz1 p (τ xx τ yy τ zz )3

Total stress tensor, Newtonian fluid: p 2 μ u x u v T μ y x u w μ z x u v u w μ μ ij yxzx u j ui v w v p 2μμ pδ ij μ y xx z y ij v w w μ p 2 μi,, j {1,j { , 2,, 3} {x,} { , y, z}} z z y Stokes, 1845:1 Σij linear function of velocity gradients.1.gradients2. Σij should vanish if there is no deformation of fluid elements.3. Relationship between stress and shear should be isotropic.222 T T Tuuv p ˆi ( T ) xx yx zx 2 μ μ 2 μ2 x y z x x y y xv22 u w p ( u ) p2 μ 2 μ μ u μ 2u z z x x x x

Navier-Stokes equations:vDu p μ 2 v vv u g, u 0Dtρ ρ u1 p μ 2u 2u 2u u u u u v w 2 2 2 ρ x ρ x y z t x y z v1 p μ 2 v 2 v 2 v v v v u v w 2 2 2 ρ y ρ x y z t x y z w w w w1 p μ 2 w 2 w 2 w u v w 2 2 2 gρ z ρ x z t x y z y u v w 0 x y zCoriolis is added when moving the framework to an acceleratingframework. Have to add boundary & initial conditions.

Navier-Stokes equations (Boussinessq approximation):Separate balance of fluid at rest from moving fluid.p p0 ( z ) p ' ( x , y , z , t )ρ ρ 0 ρ ' ( x, y , z , t )First order balance (hydrostatic):1 p00 gρ 0 z2 d order2ndd bbalance:l:vDuv2vρ0 p ' μ u gρ 0 , u 0Dt

Example: steady flow under gravity down an inclined plane.zgαFrom: AchesonAcheson, 1990 u w 0 x z21 p u0 ν 2 g sin αρ 0 x zBCs: z 0 : w u 01 p0 g cos α uρ 0 zz h: μ 0, p pa z

Example: steady flow under gravity down an inclined plane.zgαSolution:p pa ρ 0 g (h z ) cos αgu z (2h z )sin α2νQ: what ν should we use?

Reynolds decomposition of the N-S equationsAssume a turbulent flow. At any given point in space weseparate the mean flow (mean can be in time, space, orensemble) and deviation from the mean such that:p p p ' , u u u ' , u v v' , u w w'p p, p ' 0, etc'Substituting into the continuity equation (linear): u v w u ' v' w' 0; 0 x y z x y z

Substituting into x-momentum Navier-Stokes equation: u u u u1 p μ 2u 2u 2u u v w 2 2 2 y zρ x ρ x y z t x (u ' u ') (v' u ') (w' u ') x yz The evolution of the mean is forced by correlations of fluctuating properties.The correlation terms time density are the “Reynolds stresses”.Substituting into a scalar conservation equationequation: C C C C K 2C 2C 2C u v w 2 2 2 t x y zρ x y z (u ' C ') (v' C ') (w' C ') x y z

Note that the Reynolds stress tensor is symmetric (as isthe viscous stress tensor): u ' u ' u ' v' u ' w' τ ρ 0 u ' v' v' v' v' w' u'w'v'w'w'w' The closure problem: to develop equations for the evolution of the Reynoldsstresses theselves, higher order correlation are needed (e.g. w’u’u’) and soon. For this reason theories have been devised to describe τij in terms ofth mean flow.theflFor more, see: m/W k /Introduct on to turbulence/Reynolds averaged equatto turbulence/Reynolds averaged equationsons

One solution to the closure problem is to link the Reynolds’ stress to mean-flowQuantities. For example: U ρ 0 w' u ' K eddydd z T ρ 0 w' T ' K eddy zThis type of formulation is appealing because it:a. Provide for down-gradient flux.b Is reminiscent ofb.f molecularll diffusiond ffandd viscosity.c. Provide closure to the equations of the the mean fields.This type of formulation is problematic because:a. Keddy is a property of the flow and not the fluid.b. Keddy is likely to vary with orientation, unlike molecular processes.How is Keddy related to the turbulence?

Assume a gradient in a mean property (momentum, heat, solute, etc’.Remember: no mean gradientÆno flux). Assume a fluctuating velocity field:l’ is the distance a parcel travels before it loses its identity. The rate of upwardvertical turbulent transfer of Ψ is down the mean gradient: ψ ψ ψ w' ψ l ' K eddy w' l ' z z z

How is Keddy related to the turbulence? U ρ 0 w' u ' ρ 0 K eddy z T ρ 0 w' T ' ρ 0γ T zTennekesTk andd LumleyL l (1972) approachh thithis problemblffrom didimensionalil analysisl ibased on assuming a single length scale-l and a single velocity scale ω w’2 1/2. ρ 0 w' u ' cρ 0ω ; c O(1)2The eddies involved in momentum transfer have vorticities,vorticities ω/l; thisvorticity is maintained by the mean shear (l is the length scales of theeddies, e.g. the decorrelation scale). Uω /l c; c O(1) z

It follows that:K e ddydd U lω l z2In analogy with momentum flux, for heat we have: T ρ 0 c p w' T ' ρ 0 c pγ T zIt is most commonly assumed, and verified that γT Keddy.

Eddy-diffusion: perspective from a dye patch (figures from lecture notes ofBill Young, UCSD)Dye patch dominant scale of eddies. Dashed circle denotes initial position of patch.

Dye patch dominant scale of eddiesDye patch dominant scale of eddies

Cheat sheet:1 G1.Gradientdioff a scalarl ((a vector):) φ ˆ φ ˆ φˆ φ i j k x y z2. Divergence of a vector (a scalar): ϕ x ϕ y ϕ z ϕ x y zv3 Divergence of a Tensor,3.Tensor T12 T(face)(direction): Txx Tyx Tzx ˆ T i y z x Txz Tyz Tzz ˆ k z y x T T Tˆj xy yy zy x y z

Cheat sheet (cotinued):4 Laplacian4.L l i off a vector (a( vector):)222 ϕ ϕ ϕx xxˆ ϕ ( ϕ ) i 2 2 2 y z x 2ϕ y 2ϕ y 2ϕ y 2ϕ z 2ϕ z 2ϕ z ˆ ˆj k 2 2 2222 x z yz y x2vv

Derivation of the basic equations of fluidflows. No particle in the fluid at this stage (next week). Conservation of mass of the fluid. Conservation of mass of a solute (applies to non-sinking particles at low concentration). Conservation of momentum. Application of these basic equations to a turbulent fluid.

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