Navier-Stokes Equation

Newtonian Fluid
Constant Density, Viscosity
 Cartesian, Cylindrical, spherical coordinates

IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Cartesian Coordinates
DV
2

 P   V  g
Dt
  2Vx  2Vx  2Vx 
 Vx
Vx
Vx
Vx 
P
  
 
 Vx
 Vy
 Vz
   2  2  2    gx
x
y
z 
x
y
z 
 t
 x
  2Vy  2Vy  2Vy 
Vy
Vy
Vy 
 Vy
P
  
 
 Vx
 Vy
 Vz
  2  2  2    gy
x
y
z 
y
y
z 
 x
 t
  2Vz  2Vz  2Vz 
 Vz
Vz
Vz
Vz 
P
  
 
 Vx
 Vy
 Vz
   2  2  2    gz
x
y
z 
z
y
z 
 t
 x
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Cylindrical Coordinates
DV
2

 P   V  g
Dt
Centrifugal force
 Vr
Vr V Vr V2
V 
P
 
 Vr


 Vz r   
r
r 
r
z 
r
 t
2
2
  1 
 1  Vr  Vr 2 V 
rVr   2 2  2  2
  
   gr
z
r  
 r 
 r  r r
Coriolis force
V V V VrV
V
 V
 Vr   

 Vz 
r
r 
r
z
 t

1 P




r 

2
2
  1 
 1  V  V 2 Vr 
rV   2 2  2  2    g
  
z
r  
 r 
 r  r r
 1   Vz  1  2Vz  2Vz 
Vz
Vz
Vz 
P
 Vz

 Vr
 V
 Vz
 
 2    gz

r
 2
2
r

z 
z
z 
 t
 r r  r  r 
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Spherical Coordinates
DV
2

 P   V  g
Dt
2
2
 Vr
V
V

V
V

V

V

V
P


 

r
r
r



 Vr





r
r  r sin  
r
r
 t

 1 2 2
1
 
Vr 
1  2Vr 
   2 2 r Vr   2
  gr
 sin 
 2
2 
r sin   
  r sin   
 r r
2
 V
V
V
V

V

V
V

V

V

r 
 cot  




   1 P
 
 Vr





t

r
r


r
sin



r
r 


 1 2 2
1   1 
 
V sin    

 2 2 r V  2
r r
r   sin  
 


  g
2
 1  V 2 Vr 2 cot  V

 2
 2
 2

2
r
sin



r


r
sin







IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Spherical Coordinates
DV
2

 P   V  g
Dt
V V V
V V VrV  V V cot  
 V
1 P
  
 
 Vr



r
r  r sin  
r
r sin  
 t

 1   2 V  1   1 
 
  2
V sin   

 2  r
r r  r  r   sin  
 


  g
 1  2V

2 Vr 2 cot  V

 2

 2
 2
2
r sin   r sin  
 r sin  

BSL has g here instead of g
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Spherical Coordinates (3W)
DV
2

 P   V  g
Dt
2
2
 Vr
V
V

V
V

V

V

V


 

r
r
r
   P
 
 Vr





t

r
r


r
sin



r
r


 2
2
2 V 2
2 V 
   Vr  2 Vr  2
 2 V cot   2
  gr

r
r  r
r sin   

3W &R have the formula in terms of
 2However, the expression for  2 is incorrect in the book
2
1


1


1





2  2  r 2   2
 sin 
 2
r r  r  r sin   
  r sin   2
IIT-Madras, Momentum Transfer: July 2005-Dec 2005

 .V   0
t
Continuity
 


 Vx   Vy   Vz   0
t x
y
z
 1 
1 

 r Vr  
V   Vz   0

t r r
r 
z
 1 
1

1

2
V sin   
V   0
 2
 r Vr 
t r r
r sin  
r sin  


IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Newton’s law of viscosity

    V  V 
Constant Density, zero dilatational viscosity
 yx   xy
 Vy Vx 
  



x

y


Cylindrical coordinates
 r    r
   V  1 Vr 
   r   


r
r
r


 


 Vz V 

z 
 
 z    z    r
 Vr Vz 

z2005-Dec2005
r 
July
IIT-Madras, Momentum Transfer:
 rz   z r    


2

     .V 
3

Newton’s law of viscosity

    V  V 

Spherical coordinates:
Constant Density, zero dilatational viscosity
 r    r
   V
   r 
 r  r
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
 1 Vr 


 r  

2

     .V 
3

N-S Equation: Examples
•ODE vs PDE
•Spherical and cylindrical coordinates
•Eqn for pipe flow (Hagen Poiseulle)
•Flow between rotating cylinders (not solved in class)
•Thin film flow with temp variation (not solved in class, steps were
discussed briefly. BSL ‘worked out’ example)
•Radial flow between circular plates (BSL 3B.10)
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Example problems
1. Pressure driven steady state flow of fluid
between two infinite parallel plates
inside a circular tube
2. Steady state Couvette flow of a fluid
between two infinite parallel plates with top plate moving
at a known velocity
between two circular plates of finite radius, with the top
plate rotating at a known angular velocity
between two circular cylinders with outer cylinder
rotating at a known angular velocity (end effects are negligible)
between a cone and plate (stationary plate and cone is
rotating at a known angular velocity). Angle of cone is very small
(almost a parallel plate with almost zero gap)
3. Coutte Poisseuille flow
between two parallel plates
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Equation: Examples
•PDE
• Please refer to the book “Applied Mathematical Methods for Chemical
Engineers” by Norman W Loney (CRC press), pages 330 to 342 for
“worked out” examples for Momentum Transfer problems involving PDE.
• Either multi dimensional or time dependent (however multidimensional
and time dependent cases are not discussed in detail)
• Steady state in Rectangular channel: pressure driven , coutte flow
• Plan suddenly moving with constant velocity (or stress) from time t=0
(Stokes problem)
• Sudden pressure gradient in a cylindrical tube (unsteady flow ,
converging to Hagen-Poisseuille’s flow (Bessel functions)
• Flow between two (non rotating) cylinders, caused by boundary
movement (coutte flow). Unsteady vs steady (not discussed in class or
covered in tutorial)
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Guidelines for solving PDE in Momentum
Transfer

Method:


If the problem involves finite scales, “separation of variable” method should
be tried
If the problem involves infinite (or semi-infinite) distances, “combination of
variables” method should be tried
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Guidelines for solving PDE in Momentum
Transfer

Solution forms, for finite scales:


Applying the separation of variables directly may not always give proper
results
If the equation is non-homogenous




For time dependent problems, first try to get steady state solution (and try
that as the ‘particular solution’ for the equation). Unsteady state solution
may be the ‘general solution’ for the corresponding homogenous equation
For multi dimensional problems, first try to get solution for ‘one
dimensional’ problem and try that as particular solution. The ‘correction
term’ may be the ‘general solution’ for corresponding homogenous
equation.
Even if the equation is homogenous, you can try the above methods of
obtaining ‘steady state’ or ‘one dimensional’ solution. The ‘complete
solution’ will be the sum of ‘steady state + transient’ solution OR ‘one
dimensional solution + correction for presence of plates’ (for example).
Always make sure that the ‘correction term’ goes to zero in the appropriate
limit (eg time --> infinity, or the ‘width of the channel --> infinity)
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Guidelines for solving PDE in Momentum
Transfer

Other relevant Information:



Problems in Cartesian coordinates tend to give Cosine/ Sine series solution.
In cylindrical coordinates, Bessel functions. In spherical coordinates,
Legendre functions
When you attempt a ‘complete solution’ as ‘steady state+ transient’ (OR
‘one dimensional + correction’), make sure that you also translate the
boundary conditions correctly
While solving for the ‘transient’ or ‘correction’ terms, you may encounter a
situation where you have to choose an arbitrary constant (either positive or
negative or zero). Usually the constant will not be zero. Choose the
constant as positive or negative, depending on the boundary conditions
(otherwise, you will proceed only to realize that it will not work).
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Stoke’s first problem (Please refer to
BSL for solution)
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Equation: Example: Steady
state flow in Rectangular channel
• Steady state in Rectangular channel: pressure driven flow,
incompressible fluid
2b
2h
• Vy = Vz =0
•Vx is function of y and z
• gravity has no component in x direction
b
h
2h
y
x
z
h
  2Vx  2Vx  2Vx 
 Vx
Vx
Vx
Vx 
P
  
 
 Vx
 Vy
 Vz
   2  2  2    gx
x
y
z 
x
y
z 
 t
 x
  2Vx  2Vx 
P
0
  2  2 
1
x
z 
 y
• Method employed: Find a particular solution satisfying above
equation; then find a general solution satisfying following differential
2
eqn   2V

Vx  general 
Vx  Vx  general  Vx  particular
x  general
0


2
2
2

y

z
 Transfer: July 2005-Dec 2005 
IIT-Madras, Momentum
3
N-S Equation: Example;
Rectangular channel
• Hint: To obtain a physically meaningful format, we can take
particular solution to resemble one dimensional flow (when b goes to
infinity)
2b  b
h
2h
2h
h
  P  h  y 
1  2 
Vx  particular  

4
 x  2  h 
• Note: Check that the above solution is a valid particular solution
• Before trying to get general solution, write down the boundary
conditions for the over all solution Vx
2
2
Vx  y  h, z   0
Vx  y, z  b  0
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Vx
y

y 0
Vx
z
0
z 0
5
y
x
z
N-S Equation: Examples;
rectangular channel
2
2
  P  h  y 
1  2 
Vx  particular  

 x  2  h 
• Translate that to get the boundary conditions for Vx-general
• We know
Vx  general  Vx  Vx  particular
•Hence, from equation 5,
Vx  general  y   h, z   0
3
Vx  general
y
Vx  general  y, z  b    Vx  particular

y 0
Vx  general
z
6
0
z 0
• Use separation of variables method
Assume Vx  general  f  y  g z 
Eqn
2
  2Vx  general  2Vx  general  implies
0


2
2

y

z


IIT-Madras, Momentum Transfer: July 2005-Dec 2005
7a
f  g  g  f  0
7b
N-S Equation: Examples;
rectangular channel
f  g  g  f  0
implies
f   g 

f
g
• Since LHS is only a function of y and RHS is fn of z, both must be
equal to a constant
•We say
f   g 
8

 2
f
g
2
• Note: Why do we say
, why not   ? What will
 2
happen if you try that? Or if we say
 ? 0
• In any case, the chosen constant leads to
g  z   C3 e  z  C 4 e   z
f  y   C1 cos(y)  C2 sin( y)
From
6
Vx  general
y
0
implies C2  0
y 0
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
and
Vx  general
z
0
z 0
implies
9
C3  C4
N-S Equation: Examples;
rectangular channel
Hence, substituting
9in
2
2
  P  h  y 
1  2 
Vx  particular  

 x  2  h 
7a

Vx general  f  y  g z   C1C3 cos(y) ez  ez
Now, from
6
Vx  general  y   h, z   0
Using superposition principle
implies

  2n  1
Vx general   Cn cos(y) ez  ez
Again, from


2h

6
Vx  general  y, z  b    Vx  particular
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
implies
2
2
  P  h  y 

   Cn eb  e b cos(y)

1


2 

 x  2  h 


N-S Equation: Examples;
rectangular channel
2
2
  P  h  y 
1  2 
Vx  particular  

 x  2  h 
• Using Fourier cosine expansion for an even function
2
2
  P  h  y 
1  2    K n cos(y)


 x  2  h 
•Equating the co-efficients, we get

we can find Kn
Kn
C n  b   b
e e


C n e b  e  b  K n
Hence, general solution part is
Vx  general  

Kn
z
 z
cos(

y
)
e

e
e b  e   b

IIT-Madras, Momentum Transfer: July 2005-Dec 2005



N-S Equation: Examples;
rectangular channel
2
2
  P  h  y 
1  2 
Vx  particular  

 x  2  h 
“Complete” solution for the original problem is given by
2
2
K
  P  h  y 
1  2    b n b cos(y) ez  e z
Vx  

e e
 x  2  h 




• Note: When “b” goes to infinity, the ‘correction’ part goes to zero
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Equation: Other examples
• To determine the velocity profile in a rectangular channel, where the
top plate is moving at a constant velocity of V-zero, under steady state
conditions
• Try out a solution of the form “V-parallel-plate + V-correction”
• Use separation of variable techniques, to determine V-correction
• What happens if you try separation of variable in the first place?
• To determine the unsteady state solution for a flow in a cylindrical
pipe, caused by sudden application of pressure
• Try a solution of the form ‘Steady state + Transient’, just like the
one we saw for flow between parallel plates
• You will get Bessel Equations. Just like we represented
functions in rectangular coordinates by sine and cosine functions,
we can represent functions in cylindrical co ordinates by Bessel
functions, because they are orthogonal.
IIT-Madras, Momentum Transfer: July 2005-Dec 2005