Transcript Lesson14.ppt

METO 621
Lesson 14
Prototype Problem I: Differential Equation
Approach
• In this problem we will ignore the thermal
emission term
• First add and subtract the two two-stream
equations


d (I  I )



 (1  a)( I  I )
d
d (I   I  )

 (I   I  )
d
Prototype Problem I
• Differentiating the second equation with respect to
 and substituting for d(I+-I-)/d from the first
equation we get


d ( I  I ) (1  a) 


(
I

I
)
2
2
d

2
• Similarly, differentiate the first equation, and
substitute for d(I+-I-)/d
d 2 ( I   I  ) (1  a) 


(
I

I
)
2
2
d

Prototype Problem I
• We have the same differential equation to solve for both
quantities. Calling the unknown Y, we obtain a simple
second order diffusion equation.
d 2Y
2


Y where   ( 1 - a ) / 
2
d
• for which the general solution is a sum of positive and
negative exponentials
Y  A' e
 '
 B' e
  '
• A’ and B’ are arbitrary constants to be determined
Prototype Problem I
• Since the sum and differences of the two intensities are both
expressed as sums of exponentials, each intensity component
must also be expressed in the same way.
I  ( )  Ae  Be  , I  ( )  Ce  De 
• where A, B, C.and D are additional arbitrary constants.
• We now introduce boundary conditions at the top and
bottom of the medium. For prototype problem 1 these are
I  (  0)  constant
I  ( *)  0
Prototype Problem I
• The solution of this two stream problem has
the simplest analytic form.
• The equation shows four constants of
integration, but in fact only two of these are
independent.
C B
a
1  
 

 
A D 2  a  2  1   
the solutions are
I    ( * )  ( * )
I ( ) 
[e
e
]
D
I  ( * )

I ( )  [e
  2 e  ( * ) ]
D

Prototype Problem I
where D  e
 *
2   *

 e
Two stream solution for uniform
illumination (Problem 1)
Parameters I=1.0, τ*=1.0, a=0.4
μ=0.5 , p=1
Downward flux = solid line
Upward flux = dotted line
Mean intensity = dashed line
Parameters as above except
a=1.0
Prototype problem 2
• Consider the only source of radiation is thermal
emission within the slab. The two stream equations
are

dI
( )
a 
a 



 I ( )  I ( )  I ( )  (1  a) B
d
2
2

dI
( )
a 
a 



 I ( )  I ( )  I ( )  (1  a) B
d
2
2
with the boundary conditions I  (0)  I  ( *)  0
• Note the extra inhomogeneous term on the RHS
Prototype problem 2
• Solving these simultaneous equation starts by
seeking the homogeneous solution, and then a
particular equation that satisfies the whole
equation – using the boundary conditions. We
get
I  ( )  Ae   De  , I  ( )   Ae  De 
• The particular solution is obtained by guessing
that I+=B and I-=B are solutions
Prototype problem 2
• This gives for the intensities




B 2  
I ( ) 
  e  e     [e  ( * )  e  ( * ) ]  B
D
B 2  ( * )  ( * )

I ( ) 
e
e
   [ e    e   ]  B
D

Two-stream solution for an
imbedded source
Parameters B=100, τ*=1.0,
a=0.4, μ=0.5 , p=1
Downward flux = solid line
Upward flux = dotted line
Mean intensity = dashed line
Parameters as above except
a=1.0
Source function, S, divided by B
Prototype problem 3
• Assume an isotropically scattering homogeneous
atmosphere with a black lower boundary. The
appropriate two-stream equations are
dI d
a  a  a S  /  0


 Id  Id  Id 
F e
d
2
2
4

dI d
a  a  a S  /  0


 Id  Id  Id 
F e
d
2
2
4
Prototype problem 3
• As before we differentiate and substitute into the
equations and get two simultaneous equations
2


d
(
I

I
a S  / 0
2


d
d)

 (1  a)( I d  I d ) 
F e
2
d
4
2


d
(
I

I
a S  /  0
2


d
d)

 (1  a)( I d  I d ) 
F e
2
d
4
Prototype problem 3
Prototype problem 3
• Using the same solution method as for
problem 2 we consider the homogeneous
solution
I d  Ae    De   , I d    Ae  De  
• We now guess that a particular solution will be
proportional to exp(-/0). We get

I  Ae   De

d

I   Ae  De

d
 

  /  0
Z e
  /  0
Z e
• Z+ and Z- can be determined by substitution
Eddington Approximation
• Two stream approximations are used primarily to
compute fluxes and mean intensities in plane
geometry. These quantities depend only on the
azimuthally averaged radiation field. We are
interested in solutions valid for anisotropic
scattering
dI d ( , u )
a
u
 I d ( , u )   du ' p(u ' , u )I d ( , u ' )  S * ( , u )
d
2 1
1
Eddington Approximation
• Another approach is to approximate the angular
dependence of the intensity by a polynomial in u.
• We choose
I(,u)=I0()uI1()
• This approach is referred to as the Eddington
approximation. Upon substitution we get
d ( I 0  uI1 )
a
u
 I 0  uI1   du ' p (u ' , u )( I 0  u ' I1 )
d
2 1
1
aF S
 /  0

p (   0 , u )e
4
Eddington Approximation
• Remember that the phase function can be
expanded in terms of Legendre polynomials

p (u ' , u )   (2l  1)  l Pl (u ) Pl (u ' )
l 0
where the moments  are given by
1
1
 l   du ' p (u ' , u )Pl (u ' ) Pl (u )
2 1
for l  0  l  1, for l  1  l  g
Eddington Approximation
• If we only retain these first two terms then the equation
becomes
1
a
du ' p (u ' , u )( I 0  u ' I1 )  a ( I 0  3 gu u  2 I1 )

2 1
1
1
where u  2   duu 2
2 1