rudolf Žitný, Ústav procesní a zpracovatelské techniky Čvut fs 201 3
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Remark: foils with „black background“ could be skipped, they are aimed to the more advanced courses
Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2013
Computer Fluid Dynamics E181107CFD5
2181106
Transport equations,Navier Stokes equationsFourier Kirchhoff equation
Cauchy’s EquationsCFD5
_ _ _Du
pressure force viscous stress gravity centrifugal forcesDt
mass
acceleration Sum of forces on fluid particle
V S Vp
Dudv n ds gdv
Dt
u P
total stress
flux
=V S V
Ddv n ds dv
Dt
P
MOMENTUM transport
Newton’s law (mass times acceleration=force)
Pressure forces on fluid element surfaceCFD5
x
x
y
z
zSW
T
E
N
B
x
y
( )2
1x
p xy z p
xn
px y z
x
Resulting pressure force acting on sides W and E in the x-direction
( )2
1x
p xy z p
xn
V S V
Dudv npds gdv
Dt
Viscous forces on fluid element surfaceCFD5
( )yxxx zxx y zx y z
Resulting viscous force acting on all sides (W,E,N,S,T,B) in the x-direction
x i ix i ixi
f n
f n n
x
x
y
z
z
x
y
( )2
1
xxxx
x
xy z
xn
( )2
1
yxyx
y
yx z
y
n
( )2
1, 0
xxxx
x y z
xy z
xn n n
( )2
1
yxyx
y
yx z
y
n
( )2
1
zxzx
z
zx y
zn
S
E
T
W
N
( )2
1
zxzx
z
zx y
zn
B
Balance of forces CFD5
_ _Du
x y z pressure force viscous stress gravityDt
yxx xx zxx
y xy yy zyy
yzxzz zzz
Du pS
Dt x x y z
Du pS
Dt y x y z
Du pS
Dt z x y z
M
Dup S
Dt
[N/m3]
Cauchy’s equation of momentum
balances (in fact 3 equations)
Balance of forces CFD5
M
Dup S
Dt
Do you remember? This is a generally valid relationship( )
Du
Dt t
( ) M
uuu p S
t
Therefore the following formulations are equivalent (mathematically but not from the point of view of numerical solution - CFD)
conservative formulation using momentum as the unknown variable (suitable for
compressible flows, shocks…)
formulation with primitive variables,u,v,w,p. Suitable for incompressible flows (very large
speed of sound)
Balance of ENERGY CFD5
2 2 21( )
2
_
E i u v w
DE heat mechanical workDt
TOTAL ENERGY transport
Total energy [J/kg]
Internal energy all form of energies (chemical, intermolecular, thermal) independent of coordinate
system
Kinetic energy [J/kg]
Heat added to FE by diffusion only (convective transport is
included in DE/Dt)
Power of mechanical forces is velocity times force. Internal
forces are pressure and viscous forces.
Balance of ENERGY CFD5
E Fourier's law
q k T
PHeat flux by conduction
Power done by total stresses ( u)
D
Dt
P
Heat conduction CFD5
x
x
y
z
zSW
T
E
N
B
x
y
( )2
1
xx
x
q xy z q
xn
( )2
1
xx
x
q xy z q
xn
Heat transfer by conduction is described by Fourier’s lawx
y
z
Tq k
xT
q ky
Tq k
z
q k T
( ) ( ) ( ) ( )DE T T T
q k T k k kDt x x y y z z
Mechanical work - pressureCFD5
x
x
y
z
zSW
T
E
N
B
x
y
( )2
1x
pu xy z pu
xn
( )2
1x
pu xy z pu
xn
( ) ( )
( ) ( ) ( )
DEk T pu
DtT T T
k pu k pv k pwx x y y z z
Mechanical work - stressesCFD5
( ) ( ) ( )
( )
( )
( )
xx xy xz
yx yy yz
zx zy zz
DEk T pu u
DtT
k pu u v wx x
Tk pv u v w
y y
Tk pw u v w
z z
x
x
y
z
zSW E
N
B
x
y
( )2
1
xxxx
x
u xy z u
xn
( )2
1
xxxx
x
u xy z u
xn
( )2
1
zxzx
z
u zx y u
zn
T
( )2
1
zyzy
z
v zx y v
zn
The situation is more complicated because not only the work of normal but also shear stresses must be included.
Total energy transport CFD5
[W/m3]( ) ( ) ( ) E
DEk T pu u S
Dt
This is scalar equation for total energy, comprising internal energy
(temperature) and also kinetic energy.
( ) E
DEk T pu u S
Dt
Fourier Kirchhoff equationCFD5
[W/m3]
1( )
2 ( ) ( ) ( ) E
D i u uk T pu u S
Dt
Kinetic energy can be eliminated from the total energy equation
using Cauchy’s equation multiplied by velocity vector (scalar product, this is the way how to obtain a scalar equation from a vector equation)
12
M
D u uDuu u p u u SDt Dt
(1)
(2)
Subtracting Eq.(2) from Eq.(1) we obtain transport equation for internal energy
( ) : E M
Dik T p u u S u S
Dt
Fourier Kirchhoff equationCFD5
Interpretation using the First law of thermodynamics
di = dq - p dv
( ) : E M
Dik T p u u S u S
Dt
Heat transferred by conduction into FE Expansion cools
down working fluid
This term is zero for incompressible fluid
Dissipation of mechanical energy to heat by viscous friction
Dissipation termCFD5
:
x x xxx xy xz
y y yyx yy yz
z z zzx z
iij
i j
y zz
j
u u u
x y z
u u u
x y z
u u u
x
x
z
uu
y
Heat dissipated in unit volume [W/m3] by viscous forces
Dissipation termCFD5
U=ux(H)y
1 1 1( )
2 2 2x
xy yx x y y x
ue e u u
y
1( ( ) )
2Te u u
1: : ( ( ) ) :
2Tu u u e
This identity follows from the stress tensor symmetry Rate of deformation
tensor
: : xy yx yx xy xyu e e e
Example: Simple shear flow (flow in a gap between two plates, lubrication)
Example tutorialCFD5
Gap width H=0.1mm, U=10 m/s, oil M9ADS-II at 00C
=3.4 Pa.s, =105 1/s, =3.4.105 Pa, = 3.4.1010 W/m3
At contact surface S=0.0079 m2 the dissipated heat is 26.7 kW !!!!
U=10 m/sy
H=0.1 mm
D=5cm
L=5cm
Rotating shaft at 3820 rpm
Fourier Kirchhoff equationCFD5
[W/m3]
Internal energy can be expressed in terms of temperature as di=cpdT or di=cvdT. Especially simple form of this equation holds for liquids when cp=cv and divergence of velocity is zero (incompressibility constraint):
( ) : i
DTc k T u SDt
An alternative form of energy equation substitute internal energy by enthalpy
( ) ( ) i
DH pk T p u u S
Dt t
where total enthalpy is defined as 1
2
pH i u u
Pressure
energy
Thermal energy
Kinetic energy
Example tutorialCFD5
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 3.4.1010 W/m3, uniform inlet temperature T0=0oC and thermally insulated walls, or constant wall temperature, respectively.
Parameters: density = 800 kg/m3, cp=1.9 kJ/(kg.K), k=0.14 W/(m.K).
Approximate FK equation in 2D by finite differences. Use upwind differences in convection terms
( ) :DTc k T uDt
T0=0
x
y
SummaryCFD5
( ) : i
DTc k T u SDt
M
Dup S
Dt
( ) 0ut
Mass conservation
(continuity equation)
Momentum balance
(3 equations)
Energy balance
State equation F(p,T,)=0, e.g. =0+T
Thermodynamic equation di=cp.dT
formulation with primitive variables u,v,w,p suitable for incompressible flows
Constitutive equationsCFD5
Macke
Constitutive equationsCFD5
Constitutive equations represent description of material properties
Kinematics (rate of deformation) – stress (dynamic response to deformation)
p
Viscous stresses affected by fluid flow. Stress is in fact momentum flux due to molecular diffusion
1( ( ) )
2Te u u
rate of deformation is symmetric part of gradient of velocity)
Gradient of velocity is tensor with components
ji j
i
uu
x
2 ( )u II e
Dynamic viscosity
[Pa.s]
Second viscosity
[Pa.s]
Constitutive equationsCFD5
Rheological behaviour is quite generally expressed by viscosity function
and by the coefficient of second viscosity, that represents resistance of fluid to volumetric expansion or compression. According to Lamb’s hypothesis the second (volumetric) viscosity can be expressed in terms of dynamic viscosity
2 ( )u II e
3 3
1 1
where( ), : ij jii j
II II e e e e
2
3
3 2 0xx yy zztrace u u
This follows from the requirement that the mean normal stresses are zero (this mean value is absorbed in the pressure term)
This is second invariant of rate of deformation tensor – scalar value (magnitude of
the shear rate squared)
Constitutive equationsCFD5
2 ( )( )3
uII e
The simplest form of rheological model is NEWTONIAN fluid, characterized by viscosity independent of rate of deformation. Example is water, oils and air. Can be solved by FLUENT.
More complicated constitutive equations (POLYFLOW) exist for fluids exhibiting
yield stress (fluid flows only if stress exceeds a threshold, e.g. ketchup, tooth paste, many food products),
generalized newtonian fluids (viscosity depends upon the actual state of deformation rate, example are power law fluids )
thixotropic fluids (viscosity depends upon the whole deformation history, examples thixotropic paints, plasters, yoghurt)
viscoelastic fluids (exhibiting recovery of strains and relaxation of stresses). Examples are polymers.
1)2( nIIK
Constitutive equationsCFD5
Ansys POLYFLOW suggests many constitutive models of generalised newtonian fluids and viscoelastic fluids.
Differential models of viscoelastic fluids:
Upper convected Maxwell model UCM, Oldroyd-B, White Metzner, Phan-Thien-Tanner model and other models. The UCM is typical and the simplest
euuDt
D T
2))((
tensorstress of derivative timeconvectiveupper called so is this
The UCM model reduces to Newtonian liquid for zero relaxation time =0.
Integral models of viscoelastic fluids written in a general form:
1
1 2
0 fading Cauchy-Greenmemory strain
( )[ ( ) ( ) ]M s f C t s f C t s ds
Unknowns / EquationsCFD5
There are 13 unknowns:
u,v,w, (3 velocities), p, T, , i, xx, xy,…(6 components of symmetric
stress tensor)
And the same number of equations
Continuity equation
3 Cauchy’s equations
Energy equation
State equation p/=RT
Thermodynamic equation di=cp.dT
6 Constitutive equations
( ) 0ut
( ) : E M
Dik T p u u S u S
Dt
M
Dup S
Dt
2 ( )( )3
uII e
Navier Stokes equationsCFD5
22 ( ( )( )) ( ( )( ( ) )) ( ( )( ) )
3 3Tu
II e II u u II u
2 2
2( ) ( ) ( ( ) )
3
2( ) ( ) ( ( ))
3
2( ) ( )
3
( )3
ij j i kij
i i i i j i k
j i i
i i i j j i
j i i i i
i i i j j i j i j i
j i
i i i j
u u u
x x x x x x x
u u u
x x x x x x
u u u u u
x x x x x x x x x x
u u
x x x x
2 2
3i i
j i j i
u u
x x x x
( ) 2( ( ) ) ( ) ( ) ( ) ( ) ( )
3 3T II
II u II u u u II
These terms are ZERO for incompressible fluids
These terms are small and will be replaced by a parameter sm
Using constitutive equation the divergence of viscous stresses can be expressed
This is the same, but written in the index notation (you cannot make mistakes when calculating derivatives)
Navier Stokes equationsCFD5
( ( ) ) m M
Dup II u s S
Dt
General form of Navier Stokes equations valid for compressible/incompressible Non-Newtonian (with the exception of viscoelastic or thixotropic) fluids
Special case – Newtonian liquids with constant viscosity
2M
Dup u S
Dt
2 2 2
2 2 2
2 2 2
2 2 2
2 2 2
2 2 2
( ) ( )
( ) ( )
( ) ( )
x
y
z
u u u u p u u uu v w g
t x y z x x y z
v v v v p v v vu v w g
t x y z y x y z
w w w w p w w wu v w g
t x y z z x y z
Written in the cartesian coordinate
system
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