Download Constitutive equations

CFD5
Computer Fluid Dynamics
2181106
E181107
Transport equations,
Navier Stokes equations
Fourier Kirchhoff
equation
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
CFD5
Cauchy’s Equations
MOMENTUM transport
Newton’s law (mass times acceleration=force)
Du

 pressure_force  viscous_stress  gravity  centrifugal_forces
Dt
mas
s
acceleration
Sum of forces on fluid particle
D
V  Dt dv=  S n  Pds  V dv
Du
V  Dt dv  S n   ds  V  gdv
 p 
u
P  
flux
total stress
CFD5
Pressure forces on fluid element surface
p
 x y z
x
Resulting pressure force acting on
sides W and E in the x-direction
z
 y z ( p 
nx  1
p  x
)
x 2
 y z ( p 
y
N
y
nx  1
T
W
S
B
x
x
E
z
x
Du
V  Dt dv  S npds  V  gdv
p  x
)
x 2
CFD5
Viscous forces on fluid element surface
 xx  yx  zx
 x y z (


)
x
y
z
Resulting viscous force acting on all
sides (W,E,N,S,T,B) in the x-direction
f  n 
f x   ni ix  ni ix
i
z
 x z ( yx 
 yx  y
y 2
N ny  1
y
)
 zx  z
)
z 2
nz  1
 x y( zx 
T
 xx  x
)
x 2
nx  1 W
 xx  x
)
x 2
En  1, n  n  0
x
y
z
 y z ( xx 
 y z ( xx 
x
y
B
  z
 x y( zx  zx )
z 2
nz  1
z
x
x
 yx  y
)
y 2
ny  1 S
 x z ( yx 
CFD5
Balance of forces
Du

 x y z  pressure_force  viscous_stress  gravity
Dt
Du x
p  xx  yx  zx

 


 Sx
Dt
x x
y
z
p  xy  yy  zy

 


 Sy
Dt
y x
y
z
Du z
p  xz  yz  zz

 


 Sz
Dt
z x
y
z
Du y
Cauchy’s equation
of momentum
balances (in fact 3
equations)
Du

 p     S M
Dt
[N/m3]
CFD5
Balance of forces
D 


   (  u )
Dt
t
Do you remember?
This is a generally
valid relationship
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…)
 u
   (  uu )  p     S M
t
formulation with primitive variables,u,v,w,p.
Suitable for incompressible flows (very large
speed of sound)
Du

 p     S M
Dt
CFD5
Balance of ENERGY
TOTAL ENERGY transport

DE
 heat  mechanical _ work
Dt
1 2 2
E  i  (u  v  w2 )
2
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.
Kinetic
energy
[J/kg]
Total energy
[J/kg]
Internal energy
all form of
energies (chemical, intermolecular,
thermal) independent of coordinate
system
CFD5
Balance of ENERGY
D

   P 

Dt
Power done by total stresses ( u)
E
P  q   k T
Heat flux by
conduction
Fourier's law
CFD5
Heat conduction
q  kT
Heat transfer by conduction is described by Fourier’s law
z
 y z (qx 
nx  1
qx  x
)
x 2
 y z (qx 
y
N
y
W
S
B

qx  x
)
x 2
nx  1
T
x
x
E
T
x
T
q y  k
y
T
qz   k
z
qx   k
z
x
DE
 T
 T
 T
   q    (kT )  (k )  (k )  (k )
Dt
x x y y z z
CFD5
Mechanical work - pressure
z
 y z ( pu 
nx  1
pu  x
)
x 2
 y z ( pu 
y
N
y
nx  1
T
W
S
B
x
x
E
z
x
DE
   (k T )    ( pu ) 
Dt
 T
 T
 T
 (k
 pu )  (k
 pv)  (k
 pw)
x x
y y
z z

pu  x
)
x 2
CFD5
Mechanical work - stresses
 x y ( zy v 
 zy v  z
z
2
)
z
 x y ( zxu 
nz  1
 u  x
 y z ( xxu  xx
)
x 2
nx  1
 zxu  z
)
z 2
nz  1
y
T
N
y
 y z ( xxu 
W
DE

   (k T )    ( pu )    (  u ) 
Dt
 T
 (k
 pu   xx u   xy v   xz w) 
x x

T
 (k
 pv   yxu   yy v   yz w)
y y
 T
 (k
 pw   zx u   zy v   zz w)
z z
nx  1
S
B
 xxu  x
)
x 2
x
x
E
z
x
The situation is more complicated
because not only the work of normal
but also shear stresses must be
included.
CFD5
Total energy transport
This is scalar equation for total
energy, comprising internal energy
(temperature) and also kinetic energy.
DE

   (k T )    ( pu )    (  u )  S E
Dt

DE
   (k T  pu    u )  S E
Dt
[W/m3]
CFD5
Fourier Kirchhoff equation
Kinetic energy can be eliminated from the total energy equation
1
D(i  u  u )
2

   (k T )    ( pu )    (  u )  S E
Dt
(1)
using Cauchy’s equation multiplied by velocity vector (scalar product, this
is the way how to obtain a scalar equation from a vector equation)
1
D u u
Du
u 
 2
 u p  u    u  S M
Dt
Dt
(2)
Subtracting Eq.(2) from Eq.(1) we obtain transport equation for internal energy
Di
    (kT )  p  u   : u  S E  u  S M
Dt
[W/m3]
CFD5
Fourier Kirchhoff equation
Interpretation using the First
di
=
dq
-
law of thermodynamics
p dv
Di
    (kT )  p  u   : u  S E  u  S M
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
CFD5
Dissipation term
ui
 : u   ij

x j
i
j
u x
u x
u x
 xx
  xy
  xz

x
y
z
u y
u y
u y
 yx
  yy
  yz

x
y
z
u z
u z
u z
 zx
  zy
  zz
x
y
z
Heat dissipated in
unit volume [W/m3]
by viscous forces
CFD5
Dissipation term
1
 : u   : (u  (u )T )   : e
2
This identity follows from
the stress tensor symmetry
1
e  (u  (u )T )
2
Rate of deformation
tensor
Example: Simple shear flow (flow in a gap between two plates, lubrication)
U=ux(H)
y
1
1 ux 1
exy  eyx  ( xu y   y u x ) 
 
2
2 y 2
 : u   : e   xy eyx   yxexy   xy
CFD5
Example tutorial
Rotating shaft at
3820 rpm
D=5cm
L=5cm
U=10 m/s
H=0.1 mm
y
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 !!!!
CFD5
Fourier Kirchhoff equation
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):
DT
c
   ( k T )   :  u  S i
Dt
[W/m3]
An alternative form of energy equation substitute internal energy by enthalpy
DH
p

   (k T )  p  u     (  u )  Si
Dt
t
where total enthalpy is defined as
H i
Thermal
energy
p
1
 u u
 2
Pressur
e
energy
Kinetic
energy
Example tutorial
CFD5
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
DT
c
   (kT )   : u
Dt
y
T0=0
x
CFD5
Summary
Mass conservation
(continuity equation)

   ( u )  0
t
formulation with primitive
variables u,v,w,p suitable
for incompressible flows
Momentum balance
(3 equations)
Energy balance
Du

 p     S M
Dt
DT
c
   ( k T )   :  u  S i
Dt
State equation F(p,T,)=0, e.g.
=0+T
Thermodynamic equation
di=cp.dT
CFD5
Constitutive equations
Macke
CFD5
Constitutive equations
Constitutive equations represent description of material properties
Kinematics (rate of deformation) – stress (dynamic response to deformation)
1
e  (u  (u )T )
2

i u j 

   p  
Viscous stresses affected by fluid
flow. Stress is in fact momentum
flux due to molecular diffusion
rate of deformation is symmetric part of
gradient of velocity)
Gradient of velocity is
tensor with components

u j
xi
Second
viscosity
[Pa.s]
Dynamic
viscosity
[Pa.s]
    u  2 ( II )e
CFD5
Constitutive equations
    u  2 ( II )e
Rheological behaviour is quite generally expressed by viscosity function
3
3
This is second invariant of
 ( II ),
where
II  e : e   eij e ji
rate of deformation tensor –
scalar value (magnitude of
the shear rate squared)
i 1 j 1
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
 
3
This follows from the requirement that the mean normal stresses
are zero (this mean value is absorbed in the pressure term)
trace   xx   yy   zz  3  u  2  u  0
CFD5
Constitutive equations
 u
  2 ( II )(e 
)
3
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   K ( 2 II ) n 1 )
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.
CFD5
Constitutive equations
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


D  
 T 
  (
   u  (u )  )  2e
Dt




this is so called upper convectivetime
derivativeof stress tensor
The UCM model reduces to Newtonian liquid for zero relaxation time =0.
Integral models of viscoelastic fluids written in a general form:

   M ( s)[ f1C 1 (t  s)  f 2 C (t  s) ]ds
0 fading
memory
Cauchy-Green
strain
CFD5
Unknowns / Equations
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
Thermodynamic equation
6 Constitutive equations

   ( u )  0
t
Du

 p     S M
Dt

Di
   (k T )  p  u   : u  S E  u  S M
Dt
p/=RT
di=cp.dT
 u
  2 ( II )(e 
)
3
CFD5
Navier Stokes equations
Using constitutive equation the divergence of viscous stresses can be expressed
 u
2
T
   2  (  ( II )(e 
 ))    (  ( II )(u  (u ) ))    (  ( II )(  u ) )
3
3
This is the same, but
written in the index
notation (you cannot
make mistakes when
calculating derivatives)
 ij
xi


u j
ui
u


2 
(
)
(
)
(  ( k ) ij ) 
xi
xi
xi
x j
3 xi
xk
u j
ui
u


2 
(
)
(
)
(  ( i )) 
xi
xi
xi
x j
3 x j
xi
u j
 2ui
 2ui

 ui
2  ui

(
)

 (

)
xi
xi
xi x j
x j xi 3 x j xi
x j xi
u j

 ui
  2ui
2  ui

(
)


xi
xi
xi x j
3 x j xi 3 x j xi
     (  ( II )u )   ( II )  (u ) 
T
These terms are small and will
be replaced by a parameter sm
 ( II )
3
2
(  u )  (  u ) ( II )
3
These terms are ZERO for
incompressible fluids
CFD5
Navier Stokes equations
General form of Navier Stokes equations valid for compressible/incompressible
Non-Newtonian (with the exception of viscoelastic or thixotropic) fluids
Du

 p    (  ( II )u )  sm  S M
Dt
Special case – Newtonian liquids with constant viscosity
Du
2

 p   u  S M
Dt
Written in the
cartesian coordinate
system
u
u
u
u
p
 2u  2u  2u
(  u  v  w )    ( 2  2  2 )   gx
t
x
y
z
x
x y z
v
v
v
v
p
 2v  2v  2v
(  u  v  w )    ( 2  2  2 )   gy
t
x
y
z
y
x y z
w
w
w
w
p
2w 2w 2w
 (  u  v  w )    ( 2  2  2 )   gz
t
x
y
z
z
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