Newtonian 6 Fluids and Navier-Stokes Equations
6.1 Newton's Viscosity Law Derivation
Need a relationship between Tij and ij to evaluate the governing equations
6.1.1 Development of Tij
For a Newtonian fluid we can decompose Tij into Tij = Aij + d ij { {
fluid @ rest
fluid in motion
where d ij
deviatoric stress tensor
The deviatoric stress tensor is non-isotropic and is only concerned with the stresses due to fluid motion Definition: isotropic => transforms onto itself or put another way, the components are always the same regardless of rotation, e.g. ij , ijk Aij = "isotropic" = - pt ij
6.1.1.1 Newtonian Fluid
Assume a linear relationship between the surface stress tensor, Tij , and the velocity gradient, dul dxk {
u r V
Thus, the deviatoric stress tensor is defined as follows to ensure a linear relationship u d ij = Bijkl l xk Most general form of a linear relationship Bijkl u f l xk
Note: Bijkl is a rank-4 tensor with 81 components, obviously we need to simplify! Assumptions for the constitutive relationship between the surface stress tensor, Tij , and the velocity gradient 1. linear relationship between deviatoric stress tensor and velocity gradient (Newton)
2. ij is symmetric by definition 3. Tij is symmetric via angular momentum equation 4. fluid is isotropic (no preferred direction) We can decompose deviatoric stress tensor d ij = Bijkl kl { + Bijkl
symmetric
d i j
antisymmetric
dt {
Recall assumptions #2 and #3, so Bijkl must be symmetric d ij = Bijkl kl Since the fluid is isotropic (without preferred direction (assump. #4)), Bijkl is also isotropic, so it can be represented as a "general rank-4, isotropic tensor" (see web handout), Bijkl = ' ik jl + '' il jk + ij kl , where ', '', and are thermodynamic variables. Apply symmetry (assumptions #2 and #3) d ij : i and j can switch so that ik jl can be factored from the first two terms of Bijkl . Set = ' = '' . d ij = ( ' ik jl + '' il jk + ij kl ) kl = ( 2 ik jl + ij kl ) kl For the first term to be nonzero, i = k , j = l , and for the second term to be nonzero, i = j , k = l . By applying the definition of the strain rate tensor and i = j , k = l to the second term such that 1 u u u u kl kl = kl k + l k or l = 2 xl xk x k xl Now the deviatoric stress is u d ij = 2 ij + ij k xk Now the stress tensor can be written such that u Tij = - p ij + 2 ij + ij k { x Aij 1 4 4 2 4 43k
d ij
where is the first coefficient of dynamic viscosity N s is the second coefficient of dynamic viscosity 2 . m
6.1.2 Stokes Hypothesis
Recall, Stokes Hypothesis: pt = pm
pt : thermodynamic pressure; function of the thermodynamic state 1 p pm : mechanical pressure, due to translational motion only m = - Tii 3 Recall that pt = pm when normal viscous stresses can be neglected ( ii ) . 1 1 u pm = - Tii = - pt ii + 2 ii + ii k - 3 3 xk 1 u u = - 3 pt + 2 k + 3 k - 3 xk xk 2 u = pt - + k 3 xk u r 1 D pm - pt = - + 2 = + 2 V 3 3 Dt Stokes Hypothesis: pt = pm is satisfied when
(
)
(
)
2 Coefficient of bulk viscosity is zero: = + 2 3 = 0! : = - or 3 u r Incompressible: = 0 or V if the gas always in thermodynamic equilibrium...this requires physical insight.
(
)
6.1.2.1 Physical Insight into Stokes Hypothesis
1 Recall pm = - Tii ; from kinetic theory, we know that this is a measure or time average of the 3 translational kinetic energy of the molecules only. On the other hand, pt : thermodynamic pressure, pt ( e, ) , is a measure of the total energy of the molecules (translation, rotational, and vibrational) Note: e and are well defined for non-equilibrium Schematic representation of energy modes for a typical diatomic molecule
Translational energy ( pm measures this)
Observation: monatomic molecules only have translational energy Rotational Energy
Vibrational Energy
At thermodynamic equilibrium (constant T ), the internal energy is equally distributed (polyatomic) among the translational, rotational and vibrational modes of energy. If the temperature is suddenly raised or lowered (i.e. shock, high frequency acoustics) rate of change of energy occurs in each level dE Eequilibrium - Einstantaneous = , where is the relaxation time. This dt is different (function of molecules) for each mode! The rate of energy transfer is
Example O2 -9 rot ~ trans ~ 10 sec
vib ~ 10-3 sec (very slow) , requires many collisions to equilibrate
6.1.2.2 Physics of when Stokes Hypothesis Fails
Consider an oscillatory or periodic compression/expansion of a gas containing O2 . The energy rate of excitation will also be periodic 9 Rotation and translation: will be in phase with the compression for f < 10 Hz = 1GHz Vibration: there will be a phase lag with respect to the compression for f > 1kHz In addition, during the oscillation process pt pm
The work done on the gas during compression and expansion is not fully recovered! Therefore there is dissipation of energy. Stokes Hypothesis Fails when High frequency acoustics in gases (strongly dependent on water vapor) Underwater acoustics- strong function of impurities Shocks
6.1.3 Summary
Newton's viscosity law implies that a fluid has the following properties: s t r T ~ The coefficients relating stress and strain-rate are thermodynamics properties which are a function of the thermodynamics state! When the fluid is stationary, Tii , is a function of thermodynamics pressure, 1 Tii = - pt 3 Fluid is isotropic Tij is symmetric pt = pm for Stokes' hypothesis, but this only is valid for specific situations (incompressible, monotonic or equilibrium gases)
6.1.4 Observations of In general = f ( T , p ) Gas: : collision dominated, collision rate o when T
,T
o
( p ) , as long as p < pcr
Liquid: intermolecular force dominate; weakened when T o , when T o
( p)
6.2 Kinetic Theory Model of Viscosity
"Hard Sphere Model" The molecules of an ideal gas are very far apart. They spend most of their time between intermolecular collisions in which their direction and speed are abruptly changed. Consider a plane separating an upper and lower region in space (intermolecular forces are negligible) x2
u r u r V = u1 ( x1 ) e1 + 2 l 3
2 l 3
-
[ plane]
x1
The shear stress at the plane u u u 21 = 1 1 + 2 21 = 1 1 x1 x2 x2 which is microscopic transport of momentum across the plane by random molecular motions. Molecules crossing the plane carry with them the bulk viscosity from their region of origin.
Basic facts from Kinetic Theory Molecules that cross the plane begin their travel on the average of 2 3 l from the plane ( l mean-free path) 1 l= 2 d n o 2 d = diameter of the molecule o n = number density (no. of molecules/unit volume) The flux of molecules across the plane from one-side to another = 1 nV 4 + : flux above the plane - : flux below the plane where o V is the average random molecular speed o "rate at which molecules cross a unit area" V =V (T ) = o o o o 8kT where m k = Boltzmann's constant T = absolute temperature m = molecular mass Consider 21
u r u d mV r "net change of momentum" F= dt
( )
t
t rate of change of momentum or unit area
net flux of momentum in a given direction
21 = ( + )
( momentum of the molecules in x1-direction ) - ( - ) ( momentum of the molecules in x1 -direction )
Apply a Taylor Series expansion Momentum above the plane u [ mu1 ] x +2 3l = m 1 +
2
u1 2 2 + l O(l ) x2 3
where u1 is evaluated at the plane ( x2 )
Similarly, below the plane u [ mu1 ] x -2 3l = m 1 +
2
u1 x2
2 2 - + l O(l ) 3 x2
Therefore, the shear stress is given by 21 = + [ mu1 ] x + 2 3l - - [ mu1 ] x -2 3l
2 2
u 2 u 2 21 = + m 1 + 1 l - 1 - 1 l u - mu x2 3 x2 3 in thermodynamic equilibrium + = - = 4 u 21 = l m 1 3 { x2
Plugging in 4 1 = l nVm 3 4 Plugging in V 2 mkT : 3d 2 2 where T 1 = lV 6 : lV , = nm
=
T
Therefore, Viscosity => diffusion of momentum
6.3 Navier Stokes Equations
Governing Equations Continuity u r + V = 0 t u r 1 D u 1 D =- k = - or V Substantial Derivative: Dt xk Dt Conservative:
( )
Conservation of Momentum:
Conservative:
u r u r t DV Substantial Derivative: = F - + p Dt
s r t u u s r ( V ) + rVr = ur + where = -p + T V F T ( ) t
u r
Then we assumed t t Linear : s t r Symmetry of T , Fluid is isotropic = ' = '' The stress tensor became Tij = - pt ij + ij Momentum uk + 2 ij xk
u r s r u r t u r DV = - + V + + F p 2 Dt Du p uk i =- + + 2 ij Fi + ij Dt xi xj xk
( (
))
(
)
Energy
{ ( e + 12 v ) } + x { u ( e + 12 u ) } = - t
2 2 i i 2
qi + ui Fi + ( u jTij ) xi xi r u u r r s u r r
q { (4e + 1 4u 3) } + 44( 2+414 u43) } = - + 1V4 +2V ) { V e 2 1 2 3 F 4( T 43 t 4 4 2 42 4 44 4 14 1 u r
2 time rate of change of energy net flux of energy net heat into work done due to body and surface forces
Navier-Stokes If = - 2 3 , the momentum equations become the Navier Stokes equations Assumptions of Navier Stokes Linear constitutive relationship between the surface stress tensor, Tij , and the strain rate tensor, ij symmetry (always because of ij , Tij ) isotropy (in gas no reason to think there would be a preferred direction) pt = pm Stoke's Hypothesis; true for one of the following
o Monatomic o Incompressible flows o = -23 u r V The equations are greatly simplified by assuming incompressible flow = 0 u r t u r DV = - + 2 + F p Dt
(
)
( )
If we also assume that
r x then the momentum equation becomes u r t u r DV = - + 2 + F p Dt
( )
From the kinematic decomposition, we know that u 1 u ij = i + j 2 x j xi The divergence of the strain-rate tensor is ui ij 1 u 1 uj 1 uj ui = + j = + = or xi 2 xi x j xi 2 x j xi xi xi 2 xi xi 1 3 4 2 4 incompressible t r 2 = V Thus, the incompressible, constant property Navier Stokes equation is u r u r u r DV 2 = - + F + V p Dt Assumptions Linear constitutive relationship symmetry (always because of ij , Tij ) isotropy (in gas no reason to think there would be a preferred direction) pt = pm Stoke's Hypothesis incompressible r r x (Always check = ( T ) and T = T x , is it important?)
( )
( )
Observations The fundamental equations have been know for more than a100 years, but they are in general difficult to solve 3 basic conservation laws
3 equations: Mass, momentum, energy u r 3 unknowns V , T , p To solve these equations, we still need relationship for 4 variables
Thermodynamic variable Transport variables
( T , p) e ( T , p)
( T , p) ( T , p)
Fortunately, there are many valid simplifications that can be made using engineering judgment.
6.4 Boundary Conditions
6.4.1 Solid Wall
u r u r V fluid = V wall No-slip V fluidtangential = Vwalltangential (viscous) u r r Solid surface normal V = 0 @ wall (inviscid) n
Tliq = Twall Permeable (Porous Wall) ut = uwall "no-slip" -u suction un = u blowing
6.4.2 Kinetic Theory of Slip Velocity
For a gas, if the mean free path is large, there may be a finite slip velocity relative to the wall (White pg 47).
Figure 1 White page 47
Specular reflection: for a perfectly smooth wall: particles reflect at molecules conserve tangential momentum => no shear on wall perfect slip case Diffuse reflection: wall is extremely rough in reference to the particles Lack of reflected tangential momentum is balanced across the plane "y" by shear force provided by finite wall velocity uw uw where l is the mean free path of the gas If only a fraction of the molecules are diffusely reflected: du 2 uw = - 1 l f dy w where f ~ 1 Knowing (from KT) that du 2 la and xy = dy 3 3 w , where a is the speed of sound 2 a u l y w
uw
Divide by the freestream velocity u to get the wall slip velocity in terms of the skin-friction coefficient ( C f )
uw u note: CF Physics: 1) l Re -1 n
3 u w 3 w u = Ma C f Cf = 1 2 , where Ma = and 4 a 1 u2 4 u a 2 2 n = 2 laminar vL where Re = n = 7 turbulent
0 "continuum" 2) High Ma, low Re (high CF ) have slip (re-entry) 3) High Re ( CF ~ 0.001 ) B-Ls have no slip
0, uw
6.4.3 Kinetic Theory for Wall-Temperature Jump
Similar physics to slip velocity, see Panton or White.
6.4.4 Liquid-vapor interface
Most general case: vapor exerts pressure, shear, heat flux and mass flux at the surface. At a point along the interface
Figure 2 White page 50
The conservation of mass, momentum, and energy must hold (Note: shear stress => momentum flux) u u r r lim [ [ Note: V 0 ] dV + ] V d S t 4V 2 4 3 1 4 2 4 3 S 1 S ~ L2 V ~ L3 u r r u r r & n n Mass flux balance: mint = v V v int = liqV liq int
Momentum flux balance: int = v Tv n
u r Vv n
= liq
int
u r V liq n
int
'' Energy flux balance: qint = kv
= kliq
int
Tliq n
int
Normal force balance:
( p - nn ) v - ( p - nn ) l =
1 1 + x y
Simplified version: neglecting surface tension and mass transfer u r r u r r u r u r n n 1. V v = V liq V v = V liq
int int n n
2. pv = pl 3. we know that v << l , kv << kl , therefore u r V liq 0 n int Tliq n Example:
int
0
What are the boundary conditions? Assumptions
Incompressible Steady Fully-developed u r V =0 x = constant
d 2u 2 = g need two boundary conditions: dy u ( y = 0 ) = u0 u y =0
y =