Aula Teórica 13
Equação da Continuidade e Equação
de Navier-Stokes
Mass conservation Principle
• Mass is conserved. This implies that the rate of
accumulation inside a volume balances the input and
output budgets:



 


dV    u.n    c.n dA  (So  Si)

t vc
surface




dV


dV

dxdydz

t 

t
t
vc
vc




u
 .n dA  dydzu x x  dydzu x x dx 
surface
 dxdz u y y  dxdz u y y  dy 
 dxdyu z z  dxdyu z z  dz

dxdydz
 dydz u x x  dydz u x x  dx 
t
 dxdz u y y  dxdz u y y  dy 
 dxdyu z z  dxdyu z z  dz
 u x   u y   u z 




t
x
y
z
Em incompressível
ui
0
xi
 u x x  u x x  dx


t
dx
u y y  u y y dy

dy
u z z  u z z  dz
dz
 ui 


t
xi
u


 ui
  i
t
xi
xi
u
d
  i
dt
xi
Interpretation
ui
d
 
dt
xi
The total derivative is proportional to the local velocity
divergence. As the fluid moves its density changes according
to the local divergence. If it is negative there is contraction
consequently the density increases.
 ui 


t
xi
The local derivative is proportional to the local mass flux
divergence. Meaning that ion a local infinitesimal volume the
rate of change of density is proportional to the incoming flux
minus the outgoing flux. If these fluxes balance there is no
mass accumulation and consequently there is no change of
local density.
u


 ui
  i
t
xi
xi
ui
0
xi
If the flow is stationary the local derivatives are null. In this
case the advection, i.e. the rate at which the property leaves
minus the rate at which it enters balances the local
production (i.e. the difference between what leaves and
enters is the production)
If the fluid is incompressible there is no accumulation and the
density entering and leaving are equal. As a consequence
there is no production.
Equation could be obtained from
dm
0
dt


d   dVol 


System

 0
dt
• The Reynolds theorem states that:

t
d
dVol 

dt
VC

t

 dVol    v .n dS
sistema
SC

 dVol     v .n dS
VC
SC
Momentum conservation


du
F m
dt
=>


F
du

Vol
dt
• One must evaluate the forces per unit of volume.
Pressure force resultant
Force resultant (including friction and weight)
 
yx y  dy
 
yx y
Weight  g dxdydz 
Summing up
p
pressure  
xi


dydz  xi x  dx   xi x  dxdz  yi y  dy   yi y dxdy  zi y  dz   zi z
Friction 


dxdydz
dxdydz
dxdydz

weight  g dxdydz 
dui
p  ji



 g i
dt
xi x j
This is the Momentum transport equation into its differential form. It states that
the fluid acceleration is the result of the 3 forces (pressure, friction and gravity).

Case of Momentum
 ji
 u i u j
  

 x
 j xi
 2 u k
   ij
 3 x
k

Replacing in the momentum evolution equation:
 ui

dui

u
p

i


 ji   gi


uj




dt
x j 
xi x j
 t
One gets:
 ui

dui

u
p

i




uj




dt
x j 
xi x j
 t
 ui

 x
j


  g i


This is the Navier-Stokes Equation, i.e., the Momentum evolution equation of a
Newtonian fluid.
Viscosidade
• Friction force (shear) along a surface appears when molecules from
one side cross to the other and velocities on both sides are
different.
• The gradient responsible for shear is the velocity gradient and not
the gradient of momentum .
• Velocities being different on both sides of the surface require the
velocity of the “migrating molecules” to change sharing momentum
with the local molecules. As a consequence local velocity changes
and momentum changes as well. The change of velocity generates a
inertia force proportional to the density . Thus the diffusive flux of
momentum is equal to the diffusive flux of velocity multiplied by
the density:
u 
u

          
s 
s

synthesis
• The shear stress is the diffusive flux of
momentum per unit of area;
• Shear stress is tangent to the velocity and
momentum flux is normal to the velocity.
• Each velocity component can change in 3
space directions and thus shear stress can
have 9 components.
• Pensemos num volume infinitesimal com a
forma de um cubo e na componente “1” da
velocidade (representada a verde).
• Esta componente pode varia na direcção “1”,
na direcção “2” e na direcção “3”, dando
origem respectivamente às tensões
 11; 21; 31;
• Estas tensões actuam nas faces do cubo, cujas
normais não na direcção “1”, ou “2”, ou “3”.
Existem duas faces para cada uma das direcções.
• A resultante das forças é a diferença entre as
tensões que actuam em faces correspondentes,
que por unidade de volume dá:

 j1 

x j
• A convenção de sinais é: o que entra é positivo.
Como determinar as tensões?
• São proporcionais ao gradiente
de velocidade,
• Não pode haver efeito de
pressão (a força é tangencial):
 11   22   33  0
• E o momento resultante sobre
um volume de controlo tem
que ser nulo (caso contrário
teria aceleração angular).
Caso geral
 x ydy
 
 
y xdx
y x
 x  y
    
O sistema de tensões segundo x criaria um binário.
Para o equilibrar tem que haver outro binário
equilibrado por tensões iguais segundo y.
y x
x y
Expressão geral da Tensão de corte
 u i u j
 ji    

 x j xi




Quando i=j esta expressão dá a divergência da velocidade,
que se o fluido for compressível tem que ser anulada. A
expressão geral fica:
 ji
 u i u j
  

 x
 j xi
 2 u k
   ij
 3 x
k
