CFD Group
Conservation Equations
Prof. Carlo Cravero
DIME – Università di Genova
Governing equations
• The governing equations include the following conservation laws
of physics:
– Conservation of mass.
– Newton’s second law: the change of momentum equals the sum of
forces on a fluid particle.
– First law of thermodynamics (conservation of energy): rate of change
of energy equals the sum of rate of heat addition to and work done
on fluid particle.
• The fluid is treated as a continuum. For length scales of, say, 1m
and larger, the molecular structure and motions may be ignored.
2
Lagrangian vs. Eulerian description
A fluid flow field can be thought
of as being comprised of a large
number of finite sized fluid
particles which have mass,
momentum, internal energy, and
other properties. Mathematical
laws can then be written for each
fluid particle. This is the
Lagrangian description of fluid
motion.
Another view of fluid motion is
the Eulerian description. In the
Eulerian description of fluid
motion, we consider how flow
properties change at a fluid
element that is fixed in space
and time (x,y,z,t), rather than
following individual fluid
particles.
Governing equations can be derived using each
method and converted to the other form.
3
Fluid element and properties
• The behavior of the fluid is described in terms
macroscopic properties:
–
–
–
–
–
Fluid element for
ofconservation laws
Velocity u.
Pressure p.
Density r.
Temperature T.
Energy E.
dz
(x,y,z)
dy
dx
• Typically ignore (x,y,z,t) in the notation.
y
z
x
• Properties are averages of a sufficiently large
number of molecules to keep continuum
properties.
• A fluid element can be thought of as the smallest
volume for which the continuum assumption is
valid.
Faces are labeled
North, East, West,
South, Top and Bottom
4
Mass balance – CONTINUITY EQUATION
• Rate of increase of mass in fluid element equals the net rate of
flow of mass into element.
 ( rdxdydz)  r dxdydz
• Rate of increase is:
t
t
• The inflows (positive) and outflows (negative) are shown here:
( r w) 1 

. d z  d xd y
 rw 

z
2 


 ( r v) 1 
r
v

. d y  d xd z

y
2 

( ru) 1 

r
u

. d x  d yd z

x
2 

( ru ) 1 

. dx  dydz
 ru 
x 2 

y

( rv) 1 
. dy  dxdz
 rv 

y
2 

z
x  rw   ( rw) . 1 dz  dxdy

z
2

5
Continuity equation
• Summing all terms in the previous slide and dividing by the
volume dxdydz results in:
r  ( ru)  ( rv)  ( rw)  0
t
x
y
z
• In vector notation:
Change in density
r  div ( r u)  0
t
Net flow of mass across boundaries
Convective term
• For incompressible fluids r / t = 0, and the equation becomes:
div u = 0.
ui
u  v  w  0
• Alternative ways to write this: x y z
and x  0
i
6
Different forms of the continuity equation
Finite control volume
fixed in space
Finite control volume fixed
mass moving with flow

 r dV   r U  dS  0
t V
S
Integral form
D
 r dV  0
Dt V
Integral form
Conservation form
Non  conservation form
U
Infinitesimally small
element fixed in space
Infinitesimally small fluid element of fixed
mass (“fluid particle”) moving with the flow
r
   ( rU)  0
t
Differential form
Dr
 r  U  0
Dt
Differential form
Conservation form
Non  conservation form
7
Rate of change for a fluid particle (Lagrangian)
• Terminology: fluid element is a volume stationary in space, and a
fluid particle is a volume of fluid moving with the flow.
• A moving fluid particle experiences two rates of changes:
– Change due to changes in the fluid as a function of time.
– Change due to the fact that it moves to a different location in the fluid
with different conditions.
• The sum of these two rates of changes for a property per unit
mass  is called the total or substantive derivative D /Dt:
D   dx  dy  dz




Dt t x dt y dt z dt
• With dx/dt=u, dy/dt=v, dz/dt=w, this results in:
D


 u.grad 
Dt
t
8
Rate of change for a stationary fluid element
(Eulerian)
• In most cases we are interested in the changes of a flow property
for a fluid element, or fluid volume, that is stationary in space.
• However, some equations are easier derived for fluid particles.
For a moving fluid particle, the total derivative per unit volume of
this property  is given by:
(for moving fluid particle)
r
D
 

r
 u.grad  
Dt
 t

(for given location in space)
• For a fluid element, for an arbitrary conserved property  :
r  div ( r u)  0
t
Continuity equation
 ( r )
 div ( r u)  0
t
Arbitrary property
9
Fluid particle and fluid element
• We can derive the relationship between the equations for a fluid
particle (Lagrangian) and a fluid element (Eulerian) as follows:
( r )
D
 

 r

 div( r u)  r   u.grad      div( ru)  r
t
Dt
 t

 t

zero because of continuity
 ( r )
D
 div( r u)  r
t
Dt
Rate of increase of
 of fluid element
Net rate of flow of
 out of fluid element
=
Rate of increase of
 for a fluid particle
10
To remember so far:
• We need to derive conservation equations that we can solve to
calculate fluid velocities and other properties.
• These equations can be derived either for a fluid particle that is
moving with the flow (Lagrangian) or for a fluid element that is
stationary in space (Eulerian).
• For CFD purposes we need them in Eulerian form, but they are
somewhat easier to derive in Lagrangian form.
• Luckily, when we derive equations for a property  in one form, we
can convert them to the other form using the relationship shown on
the bottom in the previous slide.
11
Relevant entries for Φ
x-momentum
u
Du
r
Dt
 ( ru )
 div ( ruu)
t
y-momentum
v
r
Dv
Dt
 ( rv)
 div ( rvu)
t
z-momentum
w
r
Dw
Dt
 ( rw)
 div ( rwu)
t
Energy
E
DE
r
Dt
 ( rE )
 div ( rEu)
t
12
Momentum equation
• We will first derive conservation equations for momentum and
energy for fluid particles. Next we will use the above relationships
to transform those to an Eulerian frame (for fluid elements).
• We start with deriving the momentum equations.
• Newton’s second law: rate of change of momentum equals sum
of forces.
• Rate of increase of x-, y-, and z-momentum:
Du
Dv
Dw
r
r
r
Dt
Dt
Dt
• Forces on fluid particles are:
– Surface forces such as pressure and viscous forces.
– Body forces, which act on a volume, such as gravity, centrifugal,
Coriolis, and electromagnetic forces.
13
Viscous stresses
• Stresses are forces per area. Unit is
N/m2 or Pa.
• Viscous stresses denoted by t.
• Suffix notation tij is used to indicate
direction.
• Nine stress components.
– txx, tyy, tzz are normal stresses. E.g.
tzz is the stress in the z-direction on
a z-plane.
– Other stresses are shear stresses.
E.g. tzy is the stress in the y-direction
on a z-plane.
• Forces aligned with the direction of a
coordinate axis are positive.
Opposite direction is negative.
14
Forces in the x-direction
(t yx 
t yx 1
. dy )dxdz
y 2
t zx 1
(t zx 
. dz )dydz
t yx 1
z 2
 (t yx 
. dy )dxdz
y 2
p 1
( p  . dx)dydz
x 2
p 1
 ( p  . dx)dydz
x 2
t xx 1
 (t xx 
. dx)dydz
x 2
z
y
x
(t xx 
t xx 1
. dx)dydz
x 2
t zx 1
 (t zx 
. dz )dxdy
z 2
Net force in the x-direction is the sum of all the force components in that direction.
15
Momentum equation
• Set the rate of change of x-momentum for a fluid particle Du/Dt
equal to:
– the sum of the forces due to surface stresses shown in the previous
slide, plus
– the body forces. These are usually lumped together into a source
term SM:
Du  ( p  t xx ) t yx t zx
r



 S Mx
Dt
x
y
z
– p is a compressive stress and txx is a tensile stress.
• Similarly for y- and z-momentum:
r
Dv t xy  ( p  t yy ) t zy



 S My
Dt
x
y
z
Dw t xz t yz  ( p  t zz )
r



 S Mz
Dt
x
y
z
16
Energy equation
• First law of thermodynamics: rate of change of energy of a fluid
particle is equal to the rate of heat addition plus the rate of work
done.
• Rate of increase of energy is rDE/Dt.
• Energy E = i + ½ (u2+v2+w2).
• Here, i is the internal (thermal energy).
• ½ (u2+v2+w2) is the kinetic energy.
• Potential energy (gravitation) is usually treated separately and
included as a source term.
• We will derive the energy equation by setting the total derivative
equal to the change in energy as a result of work done by viscous
stresses and the net heat conduction.
• Next we will subtract the kinetic energy equation to arrive at a
conservation equation for the internal energy.
17
Work done by surface stresses in x-direction
 (ut zx ) 1
(ut zx 
. dz )dydz
 (ut yx ) 1
 (ut yx ) 1
z 2
(ut yx 
. dy )dxdz
 (ut yx 
. dy )dxdz
y 2
y 2
 (up) 1
(up 
. dx)dydz
x 2
 (up 
 (ut xx ) 1
 (ut xx 
. dx)dydz
x 2
(ut xx 
 (up) 1
. dx)dydz
x 2
 (ut xx ) 1
. dx)dydz
x 2
z
y
x
 (ut zx 
 (ut zx ) 1
. dz )dxdy
z 2
Work done is force times velocity.
18
Work done by surface stresses
• The total rate of work done by surface stresses is calculated as
follows:
– For work done by x-components of stresses add all terms in the
previous slide.
– Do the same for the y- and z-components.
• Add all and divide by dxdydz to get the work done per unit volume
by the surface stresses:
 (ut xx )  (ut yx )  (ut zx )  (vt xy )
 div ( pu) 



x
y
z
x
 (vt yy )  (vt zy )  ( wt xz )  ( wt yz )  (ut zz )





y
z
x
y
z
19
Energy flux due to heat conduction
q y 1
(q y 
. dy )dxdz
y 2
q x 1
(q x 
. dx)dydz
x 2
(q z 
q z 1
. dz )dxdy
z 2
q x 1
(q x 
. dx)dydz
x 2
z
y
x
q z 1
(q z 
. dz )dxdy
z 2
q y 1
(q y 
. dy )dxdz
y 2
The heat flux vector q has three components, qx, qy, and qz.
20
Energy flux due to heat conduction
• Summing all terms and dividing by dxdydz gives the net rate of heat
transfer to the fluid particle per unit volume:
q x q y q z



  div q
x y z
• Fourier’s law of heat conduction relates the heat flux to the local
temperature gradient:
qx   k
T
x
qy   k
T
y
qz   k
T
z
• In vector form: q   k grad T
• Thus, energy flux due to conduction:  div q  div(k grad T )
• This is the final form used in the energy equation.
21
Energy equation
• Setting the total derivative for the energy in a fluid particle equal to
the previously derived work and energy flux terms, results in the
following energy equation:
  (ut xx )  (ut yx )  (ut zx )  (vt xy )
DE
r
  div ( pu)  



Dt
y
z
x
 x

 (vt yy )
y

 (vt zy )
z
 ( wt xz )  ( wt yz )  (ut zz ) 



x
y
z 
 div ( k grad T )  S E
• Note that we also added a source term SE that includes sources
(potential energy, sources due to heat production from chemical
reactions, etc.).
22
Kinetic energy equation
• Separately, we can derive a conservation equation for the kinetic
energy of the fluid.
• In order to do this, we multiply the u-momentum equation by u,
the v-momentum equation by v, and the w-momentum equation
by w. We then add the results together.
• This results in the following equation for the kinetic energy:
D[ 12 (u 2  v 2  w 2 )]
 t xx t yx t zx 
r
  u.grad p  u



Dt
y
z 
 x
 t xy t yy t zy 
 t xz t yz t zz 
 v




  w
  u. S M
y
z 
y
z 
 x
 x
23
Internal energy equation
• Subtract the kinetic energy equation from the energy equation.
• Define a new source term for the internal energy as
Si = SE - u.SM. This results in:
Di
u
u
v
 u
r
  p div u  t xx
 t yx
 t zx
 t xy
Dt
y
z
x
 x
v
v
w
w
u 
 t yy
 t zy  t xz
 t yz
 t zz 
y
z
x
y
z 
 div (k grad T )  S i
24
Enthalpy equation
• An often used alternative form of the energy equation is the total
enthalpy equation.
– Specific enthalpy h = i + p/r.
– Total enthalpy h0 = h + ½ (u2+v2+w2) = E + p/r.
 ( r h0 )
 div( r h0u)  div(k grad T )
t
 (ut yx )
 (vt xy )
  (ut xx )
 (ut zx )
 



y
z
x
 x

 (vt yy )
y

 (vt zy )
z
 ( wt yz )
 ( wt xz )
 (ut zz ) 



x
y
z 
 Sh
25
Equations of state
• Fluid motion is described by five partial differential equations for
mass, momentum, and energy.
• Amongst the unknowns are four thermodynamic variables: r, p, i,
and T.
• We will assume thermodynamic equilibrium, i.e. that the time it
takes for a fluid particle to adjust to new conditions is short
relative to the timescale of the flow.
• We add two equations of state using the two state variables r and
T: p=p(r,T) and i=i(r,T).
• For a perfect gas, these become: p=r RT and i=CvT.
• At low speeds (e.g. Ma < 0.2), the fluids can be considered
incompressible. There is no linkage between the energy equation,
and the mass and momentum equation. We then only need to
solve for energy if the problem involves heat transfer.
5 conservation eq.
6 primitive variables (p,r,u,v,w,T)
Additional condition to close the problem
: equation of state
26
Viscous stresses
• A model for the viscous stresses tij is required.
• We will express the viscous stresses as functions of the local
deformation rate (strain rate) tensor.
• There are two types of deformation:
– Linear deformation rates due to velocity gradients.
• Elongating stress components (stretching).
• Shearing stress components.
– Volumetric deformation rates due to expansion or compression.
• All gases and most fluids are isotropic: viscosity is a scalar.
• Some fluids have anisotropic viscous stress properties, such as
certain polymers and dough. We will not discuss those here.
27
Stress Tensor – Constitutive equation
Newtonian vs. non-Newtonian
t (Pa)
Newtonian fluids: water, air.
Newtonian
(high μ)
Pseudoplastic fluids: paint,
printing ink.
Casson fluid
Dilatant fluids: dense slurries,
t0
wet cement.
Bingham fluids: toothpaste,
clay.
Bingham-plastic
Pseudo-plastic
(shear-thinning)
tc
Newtonian
(low μ)
Casson fluids: blood, yogurt.
Dilatant (shear-thickening)
Visco-elastic fluids: polymers
(not shown in graph because
viscosity is not isotropic).
Strain rate (1/s)
28
Viscous stress tensor – Newtonian Fluid
• Using an isotropic (first) dynamic viscosity  for the linear
deformations and a second viscosity l=-2/3  for the volumetric
deformations results in:
 t xx t xy

τ  t yx t yy
t
 zx t zy
t xz 

t yz 
t zz 
 u 2
 u v 
 u w  
 2    div u
  
  
 z x  
 y x 
 x 3

v 2
 v w  
 u v 
   
2    div u
    
y 3
 z y  
 y x 

w 2
 v w 


 u w 
  div u 
2
  
   z  x 
z 3
 z y 


Note: div u = 0 for incompressible fluids.
29
Navier-Stokes equations
• Including the viscous stress terms in the momentum balance and
rearranging, results in the Navier-Stokes equations:
x  momentum :
( ru )
p
 div( ruu)    div(  grad u )  S Mx
t
x
y  momentum :
( rv)
p
 div( rvu)    div(  grad v)  S My
t
y
z  momentum :
( rw)
p
 div( rwu)    div(  grad w)  S Mz
t
z
30
Viscous dissipation
• Similarly, substituting the stresses in the internal energy equation
and rearranging results in:
Internal energy :
( ri )
 div( riu)   p div u  div(k grad T )  F  Si
t
• Here F is the viscous dissipation term. This term is always
positive and describes the conversion of mechanical energy to
heat.
2
2
  u  2  v   w  2   u v 
F   2             
  x   y   z    y x 
 u w   v w 
    

 z x   z y 
2
2
 2
2


(
div
u
)

 3
31
Summary of equations in conservation form
r
 div( r u)  0
t
Mass :
x  momentum :
( ru )
p
 div( ruu)    div(  grad u )  S Mx
t
x
y  momentum :
 ( rv)
p
 div( rvu)    div(  grad v)  S My
t
y
( rw)
p
 div( rwu)    div(  grad w)  S Mz
t
z
z  momentum :
Internal energy :
( ri )
 div( riu)   p div u  div(k grad T )  F  Si
t
Equations of state :
p  p( r , T ) and i  i ( r , T )
e.g. for perfect gas : p  r RT and i  CvT
32
General transport equations
• The system of equations is now closed, with seven equations for seven
variables: pressure, three velocity components, enthalpy, temperature,
and density.
• There are significant commonalities between the various equations.
Using a general variable , the conservative form of all fluid flow
equations can usefully be written in the following form:
 r 
 div r u   div grad    S
t
• Or, in words:
Rate of increase
of  of fluid
+
element
Net rate of flow
of  out of
fluid element
(convection)
=
Rate of increase
of  due to
+
diffusion
Rate of increase
of  due to
sources
33
Integral form
• The key step of the finite volume method is to integrate the differential
equation shown in the previous slide, and then to apply Gauss’ divergence
theorem, which for a vector a states:
 div a dV   n  a dA
CV
A
• This then leads to the following general conservation equation in integral
form:


  r dV    n  ( r u) dA   n  ( grad  ) dA   S dV
t  CV

A
A
CV
Rate of
increase
of 
Net rate of
decrease of  due
+ to convection
=
across boundaries
Net rate of
increase of  due
+
to diffusion
across boundaries
Net rate of
creation
of 
• This is the actual form of the conservation equations solved by finite
volume based CFD programs to calculate the flow pattern and associated
scalar fields (discretization techniques)
34
Compact vector form
Conservative variables
The compact vector form contains a
set of conservation equations of the
same form (see General Transport
Equation) and it is used for software
implementation
It can be easily extended to solve
additional equations …..
35
Mathematical problem
A coupled system of non-linear second order partial
differential equations (PDE)
PDE can be:
- Hyperbolic
- Parabolic
- Elliptic
PDE are solved by setting:
- Initial conditions
- Boundary Conditions
36