Computer Fluid Dynamics E181107

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Computer Fluid Dynamics E181107 2181106 Transport equations, Navier Stokes equations Remark: foils with black background

1 Computer Fluid Dynamics E Transport equations, Navier Stokes equations Remark: foils with black background could be skipped, they are aimed to the more advanced courses Rudolf Žitný, Ústav procesní a pracovatelské techniky ČVUT FS 2010

2 Navier Stokes Equations ρu MOMENTUM transport Newton s law (mass times acceleration=force) Du ρ = pressure_ force + viscous_ stress + gravity + centrigugal_ forces mass acceleration Sum of forces on fluid particle

3 Pressure forces on fluid element surface Resulting pressure force acting on sides W and E in the -direction δ δ yδ p p δ δ yδ ( p ) 2 n = 1 y N T p δ δ yδ ( p + ) 2 n = 1 δy W B δ S δ E δ

4 Viscous forces on fluid element surface Resulting viscous force acting on all sides (W,E,N,S,T,B) in the -direction f = n τ f = nτ = nτ i i i i i y δ δ yδ ( τ ) 2 n = 1 W y δ δ yδ ( + + ) y y δ y δδ ( τ y + ) y 2 N n y = 1 δ δ δ y( τ + ) 2 T n = 1 δ δ yδ ( τ + ) 2 E n = 1 δy δ δ B δ δ δ y( τ ) 2 n = 1 δ y δ y δ δ ( τ y ) y 2 n y = 1 S

5 Balance of forces Du ρ δ δ y δ = pressure _ force + viscous _ stress + gravity Du p y ρ = S y Du y p y yy y ρ = S y y Du p ρ y ρ y = Du = p + i + S τ S y Cauchy s equation of momentum balances (in fact 3 equations) M [N/m 3 ]

6 Balance of ENERGY TOTAL ENERGY transport Total energy [J/kg] 1 ( ) E = i + u + v + w 2 DE ρ = heat + mechanical _ work Heat added to FE by diffusion only (convective transport is included in DE/) Kinetic energy [J/kg] Power of mechanical forces is velocity times force. Internal forces are pressure and viscous forces. Internal energy all form of energies (chemical, intermolecular, thermal) independent of coordinate system

7 Heat conduction q δ δ yδ ( q ) 2 n = 1 Heat transfer by conduction is described by Fourier s law y N δy T W B δ S δ E q δ δ yδ ( q + ) 2 n = 1 δ q = k T q q q y = k = k T T y T = k DE T T T ρ = iq = i( k T) = ( k ) + ( k ) + ( k ) y y

8 Mechanical work - pressure pu δ δ yδ ( pu ) 2 n = 1 y N T pu δ δ yδ ( pu + ) 2 n = 1 δy W B δ S δ E δ DE ρ = i( k T ) i( pu) = T T T = ( k pu) + ( k pv) + ( k pw) y y

9 Mechanical work - stresses u δ δ yδ ( τ u ) 2 n = 1 yv δ δ δ y( τ yv + ) 2 n = 1 DE ρ = i( k T ) i( pu) + i( τiu) = T = ( k pu + τ u + τ yv + τ w) + T + ( k pv + τ yu + τ yyv + τ yw) y y T + ( k pw + τ u + τ yv + τ w) y N δy W T B δ S δ u δ δ δ y( τ u + ) 2 n = 1 E δ u δ δ yδ ( τ u + ) 2 n = 1 The situation is more complicated because not only the work of normal but also shear stresses must be included.

10 Total energy transport This is scalar equation for total energy, comprising internal energy (temperature) and also kinetic energy. DE ρ = i( k T ) i( pu) + i( τiu) + SE DE ρ = i( k T pu + τiu) + SE [W/m 3 ]

11 Fourier Kirchhoff equation Kinetic energy can be eliminated from total energy equation 1 D( i + uiu) 2 ρ = i( k T ) i( pu) + i( τiu) + SE using Cauchy s equation multiplied by velocity vector (scalar product, this is the way how to obtain scalar equation from the vector equation) 1 D uiu ρ Du u = 2 u p u u S ρ = + τ i i i i + i Subtracting Eq.(2) from Eq.(1) we obtain transport equation for internal energy M (1) (2) Di ρ = i( k T ) p iu + τ : u + SE uism [W/m 3 ]

12 Fourier Kirchhoff equation Interpretation using First law of thermodynamics di = dq - p dv Di ρ = i( k T ) p iu + τ : u + SE uism Heat transferred by conduction into FE Epansion cools down working fluid Dissipation of mechanical energy to heat by viscous friction This term is ero for incompressible fluid

13 Dissipation term ui τ : u = τ ij = i j u u u τ + τy + τ + y τ u u u + τ + τ + y y y y y yy y u u u τ + τ + τ y y j Heat dissipated in unit volume [W/m 3 ] by viscous forces

14 Dissipation term 1 T τ : u = τ : ( u + ( u) ) = τ : e 2 This identity follows from the stress tensor symmetry 1 ( ( ) T e = u + u ) 2 Rate of deformation tensor Eample: Simple shear flow (flow in a gap between two plates, lubrication) y U=u (H) 1 1 u 1 ey = ey = ( uy + yu ) = = ɺ γ 2 2 y 2 τ : u = τ : e = τyey + τyey = τ ɺ yγ

15 Eample tutorial D=5cm L=5cm Rotating shaft at 3820 rpm H=0.1 mm y U=10 m/s Gap width H=0.1mm, U=10 m/s, oil M9ADS-II at 0 0 C µ=3.4 Pa.s, γ=10 5 1/s, τ= Pa, τγ= W/m 3 At contact surface S= m 2 the dissipated heat is 26.7 kw!!!!

16 Fourier Kirchhoff equation Internal energy can be epressed in terms of temperature as di=c p dt or di=c v dt. Especially simple form of this equation holds for liquids when c p =c v and divergence of velocity is ero (incompressibility constraint): DT ρc = i( k T) + τ : u + Si [W/m 3 ] An alternative form of energy equation substitute internal energy by enthalpy DH ρ p = i( k T ) p iu + + i( τiu) + Si t where total enthalpy is defined as p 1 H = i + + uiu ρ 2 Thermal energy Pressure energy Kinetic energy

17 Eample tutorial Calculate evolution of temperature in a gap assuming the same parameters as previously (H=0.1 mm, U=10 m/s, oil M9ADS-II). Assume constant value of heat production term W/m 3, uniform inlet temperature T 0 =0 o C and thermally insulated walls, or constant wall temperature, respectively. Parameters: density = 800 kg/m 3, c p =1.9 kj/(kg.k), k=0.14 W/(m.K). Approimate FK equation in 2D by finite differences. Use upwind differences in convection terms T 0 =0 y DT ρc = i( k T) + τ : u

18 Summary Mass conservation (continuity equation) ρ + ( ρu) = 0 t Momentum balance (3 equations) Du ρ = p + iτ + S M Energy balance DT ρc = i( k T) + τ : u + Si State equation F(p,T,ρ)=0, e.g. ρ=ρ0+βt Thermodynamic equation di=c p.dt

19 Constitutive equations Constitutive equations represent description of material properties Kinematics (rate of deformation) stress (dynamic response to deformation) 1 ( ( ) T e = u + u ) 2 rate of deformation is symmetric part of gradient of velocity) Gradient of velocity is tensor with components u i j u = i j σ = pδ + τ Viscous stresses affected by fluid flow. Stress is in fact momentum flu due to molecular diffusion Second viscosity [Pa.s] Dynamic viscosity [Pa.s] = iu + 2 ( II ) e τ λδ µ

20 Constitutive equations = iu + 2 ( II ) e τ λδ µ Rheological behaviour is quite generally epressed by viscosity function 3 3 µ ( II ), where II e : e eije ji = = i= 1 j= 1 and by the coefficient of second viscosity, that represents resistance of fluid to volumetric epansion or compression. According to Lamb s hypothesis the second (volumetric) viscosity can be epressed in terms of dynamic viscosity µ 2 λ = µ 3 This follows from the requirement that the mean normal stresses are ero (this mean value is absorbed in the pressure term) trace τ = τ + τ + τ = 3λ iu + 2 µ iu = 0 yy This is second invariant of rate of deformation tensor scalar value (magnitude of shear rate)

21 Constitutive equations iu τ = 2 µ ( II )( e δ ) 3 The simplest form of rheological model is NEWTONIAN fluid, characteried by viscosity independent of rate of deformation. Eample is water, oils and air. More complicated constitutive equations eist for fluids ehibiting yield stress (fluid flows only if stress eceeds a threshold, e.g. ketchup, tooth paste, many food products), generalied newtonian fluids (viscosity depends upon the actual state of n 1 deformation rate, eample are power law fluids µ = K( 2II ) ) thiotropic fluids (viscosity depends upon the whole deformation history, eamples thiotropic paints, plasters, yoghurt) viscoelastic fluids (ehibiting recovery of strains and relaation of stresses). Eamples are polymers.

22 Unknowns / Equations There are 13 unknowns: u,v,w, (3 velocities), p, T, ρ, i, τ, τ y, (6 components of symmetric stress tensor) And the same number of equations Continuity equation 3 Cauchy s equations Energy equation State equation Thermodynamic equation 6 Constitutive equations ρ + ( ρu) = 0 t Du ρ = p + iτ + S Di ρ = i( k T ) p iu + τ : u + SE uism p/ρ=rt di=c p.dt iu τ = 2 µ ( II)( e δ ) 3 M

23 Navier Stokes equations Using constitutive equation the divergence of viscous stresses can be epressed iu T 2 iτ = 2 i( µ ( II)( e δ )) = i( µ ( II )( u + ( u) )) i( µ ( II )( iu) δ ) 3 3 This is the same, but written in the inde notation (you cannot make mistakes when calculating derivatives) ij u j u i 2 u k = ( µ ) + ( µ ) ( µ ( ) δ ij ) = 3 i i i i j i k u j u i 2 u i = ( µ ) + ( µ ) ( µ ( )) = 3 i i i j j i 2 2 u j µ u i u i 2 µ u i u i = ( µ ) + + µ ( + µ ) = 3 i i i j j i j i j i 2 u j µ u i µ u i 2 µ u = ( µ ) i i i j j i j i i T µ ( II) 2 iτ = i( µ ( II) u) + µ ( II) i( u) + ( iu) ( iu) µ ( II) 3 3 These terms are small and will be replaced by a parameter s m These terms are ZERO for incompressible fluids

24 Navier Stokes equations General form of Navier Stokes equations valid for compressible/incompressible Non-Newtonian (with the eception of viscoelastic or thiotropic) fluids Du = p + i( ( II) u) + s + S m M ρ µ Special case Newtonian liquids with constant viscosity Du = p + 2 u + S ρ µ M Written in the cartesian coordinate system u u u u p u u u ρ ( + u + v + w ) = + µ ( + + ) + ρg t y y v v v v p v v v ρ ( + u + v + w ) = + µ ( + + ) + ρg t y y y w w w w p w w w ρ ( + u + v + w ) = + µ ( + + ) + ρg t y y y

Chapter 2: Fluid Dynamics Review

7 Chapter 2: Fluid Dynamics Review This chapter serves as a short review of basic fluid mechanics. We derive the relevant transport equations (or conservation equations), state Newton s viscosity law leading