Transcript Lecture5

CEE 262A
HYDRODYNAMICS
Lecture 5
Conservation Laws Part I
1
Conservation laws
The ultimate goal is to use Newton's 2nd law (F=ma) to relate
forces on fluid parcels to their acceleration. A natural way
to do this is to compute the equations of motion following a
volume of fluid; We first begin with the governing
equations for a solid, non-deformable object with volume
V, surface area A, and density r:
V
A
Mass of object: m   rdV
V
Momentum of object: P   r u dV
~
V
~
Newton's 2nd law for the object is given by:
F 
~
d
d
P(t )   r u dV   F
 F body
~ surface
~
dt ~
dt V ~
Surface forces
e.g. Friction
Body forces
e.g. Gravity
Example: Block of mass m pushed with force F
along surface with friction coefficient b:
u
F
d
r u dV   F
  F body

~
~
~
surface
dt V
d
1
1
3
3
u  rdV   bu a  F a  N a   rg a dV
~
~
~
~
dt ~ V




V

 

Surface
du
m
~
dt
Body
 bu a  F a  N a  mg a
1
~
1
~
3
~
~
3
du

m
 bu  F

dt
 dw
m
 N  mg  0
 dt
m
N
g
The problem for fluids is that the volume is generally not fixed in
time, and so mass and momentum may leave the volume unless a
material volume is employed.
The key to understanding what to do is Leibniz's Theorem:
Consider a volume V(t) for which the bounding surface moves
(but not necessarily at the fluid velocity)
d
dt
u
n
x b ( t )
F
db
da
F ( x, t )dx  
dx 
F (b, t ) 
F ( a, t )

t
dt
dt
x a (t )
a
b
Which in 3D becomes:
d
F
F ( x, t )dV  
dV   F u  n dA

~
~A ~
dt V (t )
t
V (t )
A( t )
dA
V (t )
A(t )
velocity of boundary
Fixed volume–VF : Flow of fluid through system boundary
(control surface) is non zero, but velocity of boundary is
zero. For this case we get
d
F
F ( x, t )dV  
dV

~
dt VF
t
VF
Material Volume–VM : Consists of same fluid particles and
thus the bounding surface moves with the fluid velocity.
Thus, the second term from the Leibnitz rule is now nonzero, so
D
F
F ( x, t )dV  
dV   F u n dA

~
~ ~
Dt VM
t
VM
A
Using Gauss' theorem:
 F u n dA      F u  dV
A
~ ~
VM
~
 
D
 F

F ( x, t )dV   
   F u  dV

~
~
Dt VM
t

VM 
This is Reynolds transport theorem, where D/Dt is the same as d/dt
but implies a material volume.
Note that the Reynolds transport theorem is often written in the
more general form which does not assume that the control volume
is bounded by a material surface. Instead, the control volume is
assumed to move at some velocity u and that of the fluid is
~b
defined as relative to the control volume, such that u~  u~  u~
r
D
Dt
 F ( x, t )dV  
V
~
V
b
F
dV   F u  n dA   F u  n dA
~b ~
~r ~
t
A
A
In this case, V is not necessarily a material surface. If ur=0, then
~
ub=u and we revert to the form on the previous page.
~ ~
n
The general form of all conservation
laws that we will use is:
dA
D
F dV   D dV   C  n dA

Dt VM
VM
A
V (t )
A(t )
Rate of change of F = effects of volume sources + effects of surface sources
Quantity (F) Volume sources (D)
Surface sources (C)
Momentum1 Gravity
Stresses
(pressure/viscous)
Heat
Dissipation
Diffusion/radiation
Salt (scalar)
None
Diffusion
Algae (e.g.)
Growth
Diffusion
1
Note: momentum is a vector quantity
Conservation of Mass
For any arbitrary material volume
 
D
 r

r
dV

0




r
u
dV




~
Dt VM
t

VM 
Mass is conserved (non-relativistic fluid mechanics)
Since integral is zero for any volume, the integrand must be zero

 
r
   ru  0
~
t
Process: We have taken an integral conservation law and used it
to produce a differential balance for mass at any point
However,
r u 
~

rui   r ui  ui r
xi
xi
xi
ui
r
r

r
 ui
0
t
xi
xi
and
Dr r
r

 ui
Dt t
xi


1 Dr
  u
~
r Dt
Thus if the density of fluid particles changes, the velocity field must
be divergent. Conversely, if fluid densities remain constant,
 u  0
~
Any other fluid property (scalar, vector,.. also drop triple integral)
Let F  rf where f is an intensive property (amount/mass)


D


rf dV    ( rf ) 
( rfui ) dV

Dt VM
t
xi

VM 
 r
f

f 
dV
   f
r
f
( rui )  rui
t
t
xi
xi 
VM 
 f
r 
f 

dV

( ru i )  0   r   u i
But
t xi
t
xi 
VM 
Df
 r
dV
Dt
VM
D
Df

rf dV   r dV

Dt VM
Dt
VM
Why is this important/useful?
Rate of Change of Momentum = Net Applied Force
Because Newton’s
2nd
D
r udV   F dV
law:

~
Dt VM ~
VM
Du
D
~
But from above:
r
u
dV

r
dV


Dt VM ~
Dt
VM
F  r
~
Du
~
Dt
Net Applied Force = Mass
 Acceleration
Independent of volume type!
Some Observations
1 Dr

  u
~
r Dt
1. Incompressible
1 Dr

0
r Dt
[ No volumetric dilatation, fluid particle density conserved]

ui
 u 
0
~
xi
Differential form of “Continuity”
2. Slightly Compressible
• Typically found in stratified conditions where
r  r( x, y, z, t )  r0  r( z )  r '( x, y, z, t )
Reference density (1000
kg/m3 for water)
Background variation (typ. 110 kg/m3 for water)
Perturbation density
due to motion (typ.
0.1-10 kg/m3 for
water)
• Boussinesq Approximation
- Vertical scale of mean motion << scale height
or r  r '  1
r0
Allows us to treat fluid as if it were
slightly incompressible
Note: Sound and shock waves are not included !
Informal “Proof”
If a fluid is slightly compressible then a small disturbance
caused by a change in pressure, dP , will cause a change in
density dr . This disturbance will propagate at celerity, c.
dP
c
dr
 dP  c 2 d r
• If pressure in fluid is “hydrostatic”
dP
dr
rg
 rg 
 2
dz
dz
c
Now
d r d r dz

dt dz dt
and
dz
w
dt
[ Streamline curvature small]
dr  rgw


dt
c2
1 dr gw

 2
r dt c
Typically: g ≈10 m/s2 ; c ≈ 1500 m/s ; w ~ 0.1m/s

 u  0
~
Conservation of Momentum – Navier-Stokes
We have:
 u

 F  r ~  r  ~  u u 
 t ~ ~ 
Dt


Du
Two kinds of forces:
Two kinds of acceleration:
• Body forces
• Unsteady
• Surface forces
• Advective (convective/nonlinear)
Two kinds of surface forces:
• Those due to pressure
• Those due to viscous stresses
Divergence of Stress Tensor
Plan for derivation of the
Navier Stokes equation
1. Determine fluid accelerations from velocities etc. (done)
2. Decide on forces (done)
3. Determine how surface forces work : stress tensor
4. Split stress tensor into pressure part and viscous part
5. Convert surface forces to volume effect (Gauss' theorem)
6. Use integral theorem to get pointwise variable p.d.e.
7. Use constitutive relation to connect viscous stress tensor to strain
rate tensor
8. Compute divergence of viscous stress tensor (incompressible fluid)
9. Result = Incompressible Navier Stokes equation
Forces acting on a fluid
a) Body Forces: - distributed throughout the mass of the
fluid and are expressed either per unit
mass or per unit volume
- can be conservative & non-conservative
g   
~
Force potential
Examples:
(1) force due to gravity   gz (acts only in negative z
direction)
(2) force due to magnetic fields
We only care about gravity
b) Surface forces: - are those that are exerted on an area
element by the surroundings through
direct contact
- expressed per unit of area
- normal and tangential components
dFn
dF
~
dA
dFs
c) Interfacial forces:
- act at fluid interfaces, esp. phase discontinuities
(air/water)
- do not appear directly in equations of motion (appear as
boundary conditions only)
- e.g. surface tension – surfactants important
- very important for multiphase flows (bubbles, droplets,.
free surfaces!)
Very important deviation from text!!!
CEE262a (and most others)

Full stress

Deviatoric (viscous) stress
Kundu and Cohen


Full stress
Deviatoric (viscous) stress
Stress at a point
(From K&C – remember difference in
nomenclature,i.e. ij ← ij)
What is the force vector I need to apply at a face defined by the
unit normal vector n to equal that of the internal stresses?
~
Consider a small (differential) 2-D element
 22
11
cut away
 21
n2
12
1
11
11
dx2
2
12
 21
 22
n
12
n1
ds
dx1
 21
 22
dF
dF1  force component in x1 direction  11dx2   21dx1
dx1
dx2
dF1
  21
 11
 f1 
ds
ds
ds
 11 cos 1  21 cos 2
 11n1  21n2
[ n has magnitude of 1]
~
Defining the stress tensor to be  ij
 11 12

   21 22

 31 32
 f1   j1n j and
And in general
13 

23 
33 
f2   j 2n j
f i   ji n j
But  ji   ij [see Kundu p90]
dFi
f i  ij n j 
ds
“ Surface force
or
per unit area”
dF
(note this is a 2D area)
f  n 
ds
or (3D)
dF
f  n 
 Ftotal     n dA
dA
CS
Total, or net, force due to surface stresses
Conservation of momentum
 31 ( x1 , x2 ) 
 21 dx2
 21 ( x1 , x2 ) 
x2 2
 31 dx3
x3 2
x2
 dx
 11 ( x1 , x2 )  11 1
x1 2
x1
x3
 21 ( x1 , x2 ) 
 31 dx3
 31 ( x1 , x2 ) 
x3 2
 11 dx1
 11 ( x1 , x2 ) 
x1 2
 21 dx2
x2 2
Dimensions:
dx1 . dx2 . dx3
Sum of surface forces in x1 direction:

 dx
 dx 
  11  11 1  11  11 1  dx2 dx3
x1 2
x1 2 


 21 dx2
21 dx2 
   21 
  21 
 dx1dx3
x2 2
x2 2 


31 dx3
31 dx3 
  31 
 31 
 dx1dx2
x3 2
x3 2 

 11  21 31 



 dx1dx2 dx3
x2
x3 
 x1
 ji

dV
x j
 Defining i component of surface force per unit volume
to be FVi
In general :

F 
ij   
x j
i
V
Force = divergence of stress tensor
For body forces we use gravity = r g  rgi
~
Du i
r
 FV  r g
Dt
~
 ij
Du i
r
 rg i 
Dt
x j
Note that usually
g  - g e3
“Cauchy’s equation
of motion”
Important Note: This can also be derived from the Integral
From of Newton’s 2nd Law for a Material
Volume VM
But
and
D
rui dV   rg i dV    ij dA j

VM
A
Dt VM
Du i
D
rui dV   r
dV

V
V
M
Dt M
Dt
 ij
A ij dAj  VM x j dV [Gauss' Theorem]
 Dui
 ij 
 r
 rgi 
 dV  0
VM 
x j 
 Dt
 ij
Dui
 r
 rg i 
Dt
x j
Constitutive relation for a Newtonian fluid
“Equation that linearly relates the stress to the rate of
strain in a Newtonian Fluid Medium”
(i) Static Fluid: - By definition cannot support a shear stress
- still feels thermodynamic pressure
(in compression)
 ij   pij
(ii) Moving Fluid: - develops additional components of stress
(due to viscosity)
 ij   pij  ij
Hypothesis
Note difference from Kundu !
 ij  Deviatoric stress tensor [Viscous stress tensor]
Assume
 ij  K ijmnemn
K ijmn  4th order tensor (81 components!) that depend
on thermodynamic state of medium
If medium is isotropic and stress tensor is symmetric
 only 2 non-zero elements of K
which gives
ij  2eij  e mm ij
or
See derivation of  in Kundu, p 100
2
 ij  ( p    uˆ ) ij  2eij
3
Special cases
(i) Incompressible    uˆ  0
  ij   p ij  2eij
(ii) Static  eij  0
  ij   p ij
In summary
 ij
Dui
(i ) r
 rg i 
Dt
x j
2
(ii )  ij  ( p    uˆ ) ij  2eij
3
Cauchy's equation
Constitutive relation for
a compressible,
Newtonian fluid.
Navier-Stokes equation
The general form of the Navier-Stokes equation is given by substitution
of the constitutive equation for a Newtonian fluid into the Cauchy
equation of motion:
Du i

r
 rg i 
Dt
xi
 

2

  p  3 ekk  ij  2eij 

 

Incompressible form (ekk=0):
Dui

  p ij  2eij 
r
 rgi 
Dt
x j
p

 2eij 
 rgi 

xi x j
Assuming
  f ( x, y , z , t )
eij
Dui
p
r

 rgi  2
Dt
xi
x j
 1  u u j
  i 
 2  x j xi
 2u j
 2ui
p

 rgi  

xi
x j x j
xi x j
p


 rgi  2
xi
x j

 
 
p

 rgi   2ui
xi
where
2
2
2
2

u

u

u

ui
2
i
i
i
 ui 



2
2
2
x j x j x1
x2
x3
Or in vector notation
r
Du
~
Dt
Inertia
~
Pressure
gradient
 0
If “Inviscid”
r
 p  r g   2 u
Du
~
Dt
 p  r g
~
~
Gravity
Divergence of
(buoyancy) viscous stress
(friction)
Euler Equation