Equations of Continuity
Outline
1.Time Derivatives & Vector Notation
2.Differential Equations of Continuity
3.Momentum Transfer Equations
Introduction
In order to calculate forces exerted by a moving fluid as
well as the consequent transport effects, the dynamics of
flow must be described mathematically (kinematics).
FLUID
Continuous
medium
Infinitesimal
pieces of fluid
Perspectives of Fluid Motion
Eulerian Perspective – the flow as seen at fixed locations in space, or over fixed
volumes of space (the perspective of most analysis)
Lagrangian Perspective – the flow as seen by the fluid material (the perspective
of the laws of motion)
Control volume: finite fixed region
of space (Eulerian)
Coordinate: fixed point in space
(Eulerian)
Fluid system: finite piece of the fluid material
(Lagrangian)
Fluid particle: differentially small finite piece of
the fluid material (Lagrangian)
Lagrangian Perspective
The motion of a fluid particle is relative to a specific initial
position in space at an initial time.
position:
x  f1 ( x p , y p , z p , t )
y  f2 ( x p , y p,z p , t)
z  f3 ( x p , y p,z p , t )
Lagrangian Perspective
z
Lagrangian coordinate system
pathline
velocity: v x 
vy 
position vector
r  x pi  y p j  z pk
vz 
 f1 ( x p , y p , z p , t )
t
f 2 ( x p , y p , z p , t )
t
f3 ( x p , y p, z p , t )
t
y
partial (local)
time derivatives
x
Lagrangian Perspective
Consider a small fluid element with a mass concentration 
moving through Cartesian space:
t = t1
y
t = t2
y
x
x
z
z
position 1:
position 2:
x  f 1 ( x p 1 , y p 1, z p 1 , t )
x  f1 ( x p 2 , y p 2 , z p 2 , t )
y  f 2 ( x p 1 , y p 1, z p 1 , t )
y  f 2 ( x p 2 , y p 2, z p 2 , t )
z  f 3 ( x p 1 , y p 1, z p 1 , t )
z  f3 ( x p 2 , y p 2, z p 2 , t )
Lagrangian Perspective
Consider a small fluid element with a mass concentration 
moving through Cartesian space:
t = t1
y
t = t2
y
x
x
z
z
concentration 1:
 1   ( x p 1 , y p 1, z p 1 , t )
concentration 2:
 2   ( x p 2 , y p 2, z p 2 , t )
Lagrangian Perspective
Total change in the mass concentration with respect to time:
 2  1
t 2  t1
     t 2  t1       x 2  x1 






t
t

t

x
t

t

 2 1  
 2 1 
     y 2  y1       z 2  z 1 






y
t

t

z
t

t
 2 1 

 2 1  
If the timeframe is infinitesimally small:
  2  1       
 x 2  x1 
lim 


lim 




t 2  t1
t 2  t1
t

t

t

x
t

t




 2 1 
 2 1 
 y 2  y1     
 z 2  z1 
  

lim

lim
 

 t t 
 t t 
2
1
2
1

y
t

t

z
t

t




 2 1 
 2 1 
Lagrangian Perspective
Total Time Derivative
d
       x      y      z 








dt

t

x

t

y

t

z

t

 

 
 



Substantial Time Derivative
D
  
  
  
  

  vz 
  vx 
  vy 

Dt

t

x

y

z








local derivative
convective derivative
Lagrangian Perspective
D
  
  
  
  

  vz 
  vx 
  vy 

Dt
 t 
 x 
 z 
 y 
stream velocity
D
  

  v 
Dt
 t 
vector notation
 v  vxi  v y j  vzk

  

  
  
 j 
i  
k
   
 x 
 z 
 y 

 gradient
Lagrangian Perspective
D
  

  v 
Dt
 t 
Problem with the Lagrangian Perspective
The concept is pretty straightforward but very difficult to
implement (since to describe the whole fluid motion, kinematics
must be applied to ALL of the moving particles), often would
produce more information than necessary, and is not often
applicable to systems defined in fluid mechanics.
Eulerian Perspective
z
Motion of a fluid as a
continuum
flow
Fixed spatial position is
being observed rather
than the position of a
moving fluid particle
(x,y,z).
y
x
Eulerian Perspective
z
Motion of a fluid as a
continuum
flow
Velocity expressed as a
function of time t and
spatial position (x, y, z)
y
Eulerian coordinate system
x
Eulerian Perspective
Difference from the Lagrangian approach:
Eulerian
Lagrangian
Eulerian Perspective
Difference from the Lagrangian approach:
Eulerian
Lagrangian
Outline
1.Time Derivatives & Vector Notation
2.Differential Equations of Continuity
3.Momentum Transfer Equations
Equation of Continuity
differential control volume:
Differential Mass Balance
R ate of

  R ate of   R ate of 
mass balance: 



accum
ulation
m
ass
in
m
ass
out

 
 

 R ate of 

    vx  x yz    v y  y xz    vz  z  x y
 m ass in 
 R ate of 

    v x  x  x  y  z    v y  y  y  x  z    v z  z  z  x  y
 m ass out 
 R ate of m ass   
xyz


t
 accum ulation 
Differential Mass Balance
Substituting:

xyz    vx  x yz    v y  xz    vz  z xy 
y


t
   v x  x  x  y  z    v y 
 x  z    v z  z  z  x  y 
y  y


Rearranging:

 x  y  z     v x  x    v x  x   x   y  z
t
 xz
   v y     v y 
y
y  y 

    v z  z    v z  z   z   x  y
Differential Equation of
Continuity
Dividing everything by ΔV:
 v
   v x  x   v y  y  y    v y  y   v z  z  z    v z  z

x  x  x



t
x
y
z



Taking the limit as ∆x, ∆y and ∆z  0:
    vx     v y     vz  
 



t
y
 z 
  x





Differential Equation of
Continuity
    vx     v y     vz  
 


      v 
t
y
 z 
  x

divergence of mass velocity vector (v)
Partial differentiation:
 vx v y vz   

 
  


 vy
 vz
   vx

t
y
z  
x
y
z 
 x

Differential Equation of
Continuity
Rearranging:
 vx v y vz 
 vx
 vy
 vz
  



t
x
y
z

x

y

z


substantial time derivative




 vx v y vz 
  


      v 
Dt
y
z 
 x
D
If fluid is incompressible:   v  0
Differential Equation of
Continuity
In cylindrical coordinates:
d
dt

1    rv r 
r
r

1    v
r


w here r 

   vz 
z
0
x  y ,   tan
2
2
1
 y
 
 x
If fluid is incompressible:
vr
r

vr
r

1  v
r 

v z
z
0
Outline
1.Time Derivatives & Vector Notation
2.Differential Equations of Continuity
3.Momentum Transfer Equations
Differential Equations of
Motion
Control Volume
Fluid is flowing in 3 directions
For 1D fluid flow, momentum
transport occurs in 3 directions
Momentum transport
is fully defined by 3
equations of motion
Momentum Balance
Consider the x-component of the momentum transport:
 S um of forces 
R ate of
R ate of
R ate of











acting
in








accum
ulation
m
om
entum
in
m
om
entum
out

x 
x 
x 

 the system  x
R ate of
R ate of









 m om entum in  x  m om entum out  x

R ate of
R ate of


 
 
 
 
m
om
entum
in
m
om
entum
out
x 
 x  convective


R ate of
R ate of


 
 
 
 
m
om
entum
in
m
om
entum
out
x 
 x  m olecular

Momentum Balance
Due to convective transport:

R ate of
R ate of


 

 
 
m
om
entum
in
m
om
entum
out
x 
 x  convective

    v x v x  x    v x v x  x   x   y  z
 xz
   v y vx     v y vx 
y
y  y 

    v z v x  z    v z v x  z   z   x  y
Momentum Balance
Due to molecular transport:

R ate of
R ate of


 

 
 
  m om entum in  x  m om entum out  x  m olecular
    xx  x    xx  x   x   y  z
 xz
    yx     yx 
y
y  y 

    zx  z    zx  z   z   x  y
Momentum Balance
Consider the x-component of the momentum transport:
 S um of forces 
R ate of
R ate of
R ate of











acting
in








accum
ulation
m
om
entum
in
m
om
entum
out

x 
x 
x 

 the system  x
 S um of forces 


acting in



 the system 

x
 px
 p x  x   y  z   g x  x  y  z
Momentum Balance
Consider the x-component of the momentum transport:
 S um of forces 
R ate of
R ate of
R ate of











acting
in








accum
ulation
m
om
entum
in
m
om
entum
out

x 
x 
x 

 the system  x
R ate of
   vx 


xyz

 
t
 accum ulation  x
Differential Momentum Balance
Substituting:
   vx 
t
 x  y  z     v x v x  x    v x v x  x   x   y  z
 xz
   v yvx     v yvx 
y
y  y 

    v z v x  z    v z v x  z   z   x  y
    xx  x    xx  x   x   y  z
  x z
    yx     yx 
y
y  y 

    z x  z    zx  z   z   x  y
  p x  p x  x   y  z   g x  x  y  z
Differential Momentum Balance
Dividing everything by ΔV:
   vx 
t
  v v     v v 

  vxvx     vxvx 

y x
y x
y
y  y 
x
x  x 




x
y

  vz vx     vzvx 
    xx     xx 
x
x  x 

z
z  z 



z
x
     



yx
y  y 
 yx y
   zx  z    zx  z   z 


y
z

 px
 p x  x 
x
  gx
Differential Equation of Motion
Taking the limit as ∆x, ∆y and ∆z  0:
   vx 
t


   vxvx 
x
   xx 
x


   v yvx 
y
   yx 
y


   vzvx 
  zx 
z
z

p
x
  gx
Rearranging:
   vx 
t

   vxvx 
x

   v yvx 
y

   vzvx 
z
        yx        p
xx
zx
 


  gx

y
z  x
  x

Differential Momentum Balance
For the convective terms:
   vxvx 
x

   v yvx 
y

   vzvx 
z
    vx     v y     vz  
 vx
vx
vx 
   vx
 vy
 vz



  vx 
x
y
z 
y
 z 

  x
For the accumulation term:
   vx 
vx

t
 
t
 vx
t
  vx v y vz   

  
 
 vx   


 vy
 vz
   vx

t
y
z  
x
y
z  
  x
vx
Differential Equation of Motion
Substituting:
  vx v y vz   

  

 vx   


 vy
 vz
   vx

t
y
z  
x
y
z  
  x
vx
 vx
vx
vx 
   vx
 vy
 vz


x

y

z


    vx     v y     vz  
 vx 



y
 z 
  x
        yx        p
xx
zx
 


  gx

y
z  x
  x

Differential Equation of Motion
Substituting:
 vx
vx
vx 

   vx
 vy
 vz

t
x
y
z 

vx
        yx        p
xx
zx
 


  gx

y
z  x
  x

EQUATION OF MOTION FOR THE x-COMPONENT
Differential Equation of Motion
Substituting:
v y
v y
v y 
 v y

   vx
 vy
 vz

t
x
y
z 

    xy     yy     zy    p
 


 gy

y
z  y
  x

EQUATION OF MOTION FOR THE y-COMPONENT
Differential Equation of Motion
Substituting:
 vz
vz
vz 

   vx
 vy
 vz

t
x
y
z 

vz
        yz        p
xz
zz
 


  gz

y
z  z
  x

EQUATION OF MOTION FOR THE z-COMPONENT
Differential Equation of Motion
Substantial time derivatives:
        yx        p
xx
zx

 


  gx

Dt
y
z  x
  x

D vx
    xy     yy     zy    p

 


 gy

Dt
y
z  y
  x

Dvy
        yz        p
xz
zz

 


  gz

Dt
y
z  z
  x

D vz
Differential Equation of Motion
In vector-matrix notation:
   
xx

 x
vx 

D  
    xy 
vy  



 x
Dt
 v z 

    xz 

  x

Dv
Dt
   yx 
y
   yy 
y
   yz 
y
   zx  
  p 
z  

x

gx 
 
   zy     p 
 
  gy

 
z   y 

 g z 
 
   zz     p 
   z 
z 

        p   g
Differential Equation of Motion

Dv
Dt
    τ   p   g
Cauchy momentum equation
• Equation of motion for a pure fluid
• Valid for any continuous medium (Eulerian)
• In order to determine velocity distributions, shear
stress must be expressed in terms of velocity
gradients and fluid properties (e.g. Newton’s law)
Cauchy Stress Tensor
Stress distribution:
 xx 

 yy  norm al stresses

 zz 
 xy   yx 
 xz   zx
 yz   zy

 shear stresses


Cauchy Stress Tensor
Stokes relations (based on Stokes’ hypothesis)
 xx   2 
 yy   2 
 zz =  2 
vx
x
v y
y
vz
z

2
3

2
3

2
3
   v 
   v 
   v 
w h ere   v 
vx
x
 xy   yx
 vx v y 
  



y

x


 xz   zx
 vx vz 
  



z

x


 yz   zy
 v y vz 
  


y 
 z

v y
y

vz
z
Navier-Stokes Equations
Assumptions
1. Newtonian fluid
2. Obeys Stokes’ hypothesis
3. Continuum
4. Isotropic viscosity
5. Constant density
Divergence of the stream velocity is zero
Navier-Stokes Equations
Applying the Stokes relations per component:
   xx 
x
   xy 
x
   xz 
x



   yx 
y
   yy 
y
   yz 
y



   zx 
z
   zy 
z
   zz 
z
  2vx  2vx  2vx 
  


2
2
2 
y
z 
 x
  2v y  2v y  2v y
  


2
2
 x 2

y

z




  2vz  2vz  2vz 
  


2
2
2 
y
z 
 x
Navier-Stokes Equations
Navier-Stokes equations in rectangular coordinates
  2vx  2vx  2vx  p

 



  gx
2
2
2 
Dt
y
z  x
 x
D vx
  2v y  2v y  2v y

 


2
2
2
 x
Dt

y

z

Dvy
 p
 gy
 
 y
  2vz  2vz  2vz  p

 



  gz
2
2
2 
Dt
y
z  z
 x
D vz

Dv
Dt
  p   g    v
2
Cylindrical Coordinates
2
 vr
 v r v  v r v
vr 

 vr


 vz


t

r
r


r

z


2
   1   rv r   1  2 v

v

vr
2


r
r

  gr   
 2 

 2

2
2
r

r
r

r
r


r



z





p



2
 vr
 v r v  v r v
vr 

 vr


 vz


t

r
r


r

z


2
   1   rv   1  2 v
2   v   v

  g   
 2 

 2

2
2
r 
r  r 
r    z
  r  r
1 p
2
 vr
 v r v  v r v
vr 

 vr


 vz


t

r
r


r

z


 1   vz  1  2vz  2vz 

  gz   

r
 2
2
2 
z
r

r

r
r



z




p



Applications of Navier-Stokes
Equations
Application
The Navier-Stokes equations may be reduced using the
following simplifying assumptions:
1. Steady state flow

t
v
t
0
0
 all com ponents 
Application
The Navier-Stokes equations may be reduced using the
following simplifying assumptions:
2. Unidirectional flow
  vx  vx  vx 
 vx



 
2
2
2 
2
y
z 
x
 x
2
 flow
2
2
2
along x -direction only 
Application
The Navier-Stokes equations may be reduced using the
following simplifying assumptions:
2. Unidirectional flow
2
   1   rv   1  2 v
2   v   v
 
 2 
 2

2
2
r  r 
r    z
  r  r
 v
2
 
z
2
 flow



along z -direction only 
Application
The Navier-Stokes equations may be reduced using the
following simplifying assumptions:
3. Constant fluid properties
• Isotropy (independent of position/direction)
• Independent with temperature and pressure
incom pressible fluid:
v  0
 from
the equation of continuity 
Application
The Navier-Stokes equations may be reduced using the
following simplifying assumptions:
4. No viscous dissipation (INVISCID FLOW)
 0  τ0

Dv
Dt
  p   g
Euler’s Equation
Application
The Navier-Stokes equations may be reduced using the
following simplifying assumptions:
5. No external forces acting on the system
p  0
 no
external pressure gradient 
g  0
 no
gravity effects 
Application
The Navier-Stokes equations may be reduced using the
following simplifying assumptions:
6. Laminar flow
 v x
v x
v x
v x 

 
 vx
 vy
 vz

Dt

t

x

y

z


Dvx

Dvx
Dt
 
v x
t
Example
Derive the equation giving the velocity distribution at
steady state for laminar flow of a constant-density
fluid with constant viscosity which is flowing between
two flat and parallel plates. The velocity profile
desired is at a point far from the inlet or outlet of the
channel. The two plates will be considered to be fixed
and of infinite width, with flow driven by the pressure
gradient in the x-direction.
Example
Derive the equation giving the velocity distribution at
steady state for laminar, downward flow in a vertical
pipe of a constant-density fluid with constant viscosity
which is flowing between two flat and parallel plates.