Download Gas Dynamics, Lecture 3

Prof. dr. A. Achterberg, Astronomical Dept. , IMAPP, Radboud Universiteit
Aim: To cast all equations in the same generic form:
Reasons:
1. Allows quick identification of conserved quantities
2. This form works best in constructing numerical
codes for Computational Fluid Dynamics
Generic Form:
Transported quantity is
a scalar S, so flux F must
be a vector!
Component form:
Generic Form:
Transported quantity is a
vector M , so the flux must
be a tensor T .
Component form:
Mass conservation: already in conservation form!
Continuity Equation:
transport of the scalar 
Excludes ‘external mass sources’ due to
processes like two-photon pair production etc.
Define mass flux F =V :

Ñ  F  0
t

 m     F  O
t
cell faces
Density
inside
a cell
Fluxes at four
cell boundaries!
Mass conservation: already in conservation form!
Continuity Equation:
transport of the scalar 
Momentum conservation: transport of a vector!
Algebraic Manipulation
Starting point: Equation of Motion
Use:
1. product rule for differentiation
2. continuity equation for density
Use divergence chain rule for dyadic tensors
Rewrite pressure gradient as a divergence
Momentum density
Stress tensor
=
momentum flux
Momentum source:
gravity
Energy density is a scalar!
Kinetic energy
density
Internal energy
density
Gravitational
potential energy
density
Irreversibly lost/gained
energy per unit volume
Internal energy per unit mass
Specific enthalpy
Irreversible gains/losses, e.g. radiation losses
“Dynamical Friction”
Summary: conservative form of the fluid equations
in an ideal fluid:
Mass
Momentum
Energy
 s

 V Ñ  s   0 , s  cv ln( P   )
 t

T 
(  s)
Ñ   Vs   0
t
ADIABATIC FLUID
Extra mathematical constraints one can put on a flow:
1. Incompressibility:
d
Ñ V  0 
0
dt
2. No vorticity (“swirl-free flow”):
Ñ V = 0  V Ñ 
3. Steady flow:

 any flow quantity   0
t

  constant , V =Ñ  , V     Uxˆ ,
 any flow qauntity   0

Ñ V  0
1   2  
1
 
 

2




r

sin



 2

0
2
r r  r  r sin   
 

V Ñ  
1   2  
1
 
 
r

sin




0
r 2 r  r  r 2 sin   
 
Solution:


B 

 (r , )  m   Ar m  m1  Pm  cos   (m  1, 2,3,....)
r
1   2  
1
 
 
r

sin




0
r 2 r  r  r 2 sin   
 
Solution:
B 
 m
 (r , )  m   Ar  m1  Pm  cos   (m  1, 2,3,....)
r 

Far away from sphere:
V  Uxˆ =Ñ Ux  Ñ Ur cos 
This suggests: m = 1 !
1   2  
1
 
 
r

sin




0
r 2 r  r  r 2 sin   
 
Trial Solution:


 (r , )   Ar 
Vr 
B
cos 
2 
r 
 
2B 
1 
B

  A  3  cos  , V 
   A  3  sin 
r 
r 
r 
r 

B

 (r , )   Ar  2  cos 
r 

Trial Solution:
Vr 
 
2B 
1 
B

  A  3  cos  , V 
   A  3  sin 
r 
r 
r 
r 

For r   : Vr  A cos   U cos  , V  A sin   U sin 
( xˆ = cos  rˆ  sin  qˆ , V     Uxˆ )
 A=U
B

 (r , )   Ur  2  cos 
r 

Trial Solution:
Vr 
 
2B 
1  
B
  U  3  cos  , V 
  U  3  sin 
r 
r 
r  
r 
2B 

On surface sphere, for r  a : Vr (r  a)   U  3  cos   0 for 
a 

Ua 3
 B
2
  a 3 
Vr 
 1  3  U cos  ,
r  r 

1 
a3 
V 
  1  3  U sin 
r 
 2r 

a3 
 (r , )   r  2 U cos 
2r 

Constant density flow:





2
2
V 
 V 
  V 2 P 
V Ñ V Ñ    V  Ñ V  Ñ    Ñ      0
2 
 2 
 2 
 t
vanishes for
potential flow



V Ñ 


P
V
 V Ñ V  Ñ  
t

Steady constant-density flow around sphere:
  V 2 P 
 V 2 P
U 2 P
Ñ 

 0 

  constant 

2 
t
2 
2

 t
P x

2
2
U

V
x

P



2
  a 3 
Vr 
 1  3  U cos  ,
r  r 

1 
a3 
V 
  1  3  U sin 
r 
 2r 
3U
Vr (r  a)  0 , V (r  a)  
sin 
2
U 2 
9 2 
P  P 
1  sin  
2  4

P    P 
PARADOX OF D’ALAMBERT
U 2 
9 2 
1

sin  

2  4

NO fore-aft symmetry,
Now there is a drag force!
Viscosity = internal friction due to molecular diffusion,
viscosity coefficient :
Viscous force density:
f visc  Ñ  Tvisc   2V
(incompressible flow!)
Equation of motion:
 V


 V Ñ V   Ñ P   2V
 t

coll


3
2
 coll 

3
coll
 V

 V Ñ V   Ñ P   2V
 t


Inertial force  V Ñ V
Re 

viscous force
 2V
(   /   specific viscosity)
V 2 / L VL

2
V / L

Inertial force  V Ñ V
Re 

viscous force
 2V
V 2 / L VL

2
V / L

(   /   specific viscosity)
Very viscous flow:
>> VL, Re << 1
Friction-free flow:
<< VL, Re >> 1
Because of viscosity: no slip, velocity vanishes on sphere!
Ñ V 
1  2
1

r
V

 sin  V   0 (spherical coordinates!)

r
2
r r
r sin  
Automatically satisfied by writing:
  r ,  ˆ
1 
1 
V Ñ  A with A 
f  Vr  2
, V  
r sin 
r sin  
r sin  r
 V Ñ V  Ñ P 2V
Steady flow equation
Slow flow approximation of this equation:
Ñ P   V
2
From:
P
0   V  Ñ  w Ñ  
 
2
2V  Ñ Ñ V  Ñ  Ñ V  , w  Ñ V 
vanishes:
incompressible flow!
P
 V  Ñ  w Ñ  
 
Steady slow flow equation
2
Take divergence of slow flow equation:
P
Ñ  Ñ  w   0       2 P  0
 
2
 P0
2
 2 P(r ,  ) 
1   2 P 
1
 
P 
r

sin


 2

0
2
r r  r  r sin   
 
General solution with constant pressure at infinity:
2
3
1  a 
a
a
3
P  r ,   P0  b1    b2 cos     b3  cos 2      ...
2  r 
r
r
2
monopole term
dipole term
quadrupole term
For this particular case:
a
P  r ,    P0  b cos   
r
Components of
pressure gradient:
2
P
2ba 2 cos 
1 P
ba 2 sin 

,

3
r
r
r 
r3
P
 V  Ñ  w Ñ  
 
Steady slow flow equation
2
1 Vr
1 
w Ñ V  
 rV  
r 
 r r
ˆ
 1
  ˆ
f



 f


 r sin 


Vorticity:
 2 sin 
  2  2
r
r

   1   
   sin    



P
 V  Ñ  w Ñ  
 
Steady slow flow equation
2
1 Vr
1 
w Ñ V  
 rV  
r 
 r r
- Ñ w 
1
r sin 
ˆ
 1
  ˆ
f



 f


 r sin 


  ˆ
1 

ˆ


r

  q 




r
 r 

P
 V  Ñ  w Ñ  
 
Steady slow flow equation
2

2


P
2
ba
cos 






  r
r 2 sin  
r3

 
1 P
ba 2 sin 



r sin  r
r 
r3



ba 2

sin 2 
  
r



2
ba
  
sin 2 
r
 2 sin   
 2 
2
r
r  
ba 2
 1   
2


sin

 sin    
r

Trial solution:
 d2 f 2 f  2
ba 2
  r ,   f (r )sin    2  2  sin   
sin 2 
r 
r
 dr
2
2
ba
  
sin 2 
r
 d2 f 2 f  2
ba 2
d2 f 2 f
ba 2
2
sin   2  2  
 2  2  sin   
r 
r
dr
r
r
 dr
Solution:
f (r ) 
B
Ar 
r
2
solution homogeneous
equation
ba 2 r

2
particular
solution
 2 B ba 2 r  2
 (r , )   Ar  
 sin 
r
2 

Conditions at infinity:
1 


U
cos

as
r



r 2 sin  

U

A


2

1 
V  
 U sin  as r   
r sin  r

Vr  
 U 2 B ba 2 r  2
 (r ,  )   r  
 sin 
r
2 
2
Conditions at surface sphere:
1 


0
as
r

a

r 2 sin  

U 3
3 U

B

a
,
b



4
2 a

1 
V  
 0 as r  a 
r sin  r

Vr  
Ua 2 sin 2 
  r ,  
4
  r 2  a   r  
2       3  
  a   r   a  
All flow quantities can now be determined:
 3  a  1  a 3 
 3  a  1  a 3 
Vr  U cos  1        , V  U sin  1       
 2  r  2  r  
 4  r  4  r  
3 U cos 
a
P  r ,   P0  b cos     P0 
2
a
r
2
a
 
r
2
t
force
 T  nˆ =  T11 xˆ1  T21 xˆ 2  T31 xˆ 3 
area
Tij  T ji  momentum flux tensor
 Vi V j
 P ij   

 x
 j xi

 (Cartesian coordinates!)

T  P I  T visc

= P I -  Ñ V + Ñ V 
FD    dO Trr cos   Tr sin  



=   2 a 2 d sin   P  T visc rr  cos   T visc r sin  |r  a
0
surface element
†

For this particular flow at r=a:
Trr  P  2
Tr   r
ur
3U
=P =P0 
cos 
r
2a
  u

r  r
  ur 3U

sin 

r


2
a


FD    2 a 2 d sin  Trr cos   Tr sin  
0

3U

   2 a 2 d sin   P0 cos  
2a

0

  6 aU
