Transcript METO 621

METO 621
Lesson 10
Upper half-range intensity
• For the upper half-range intensity we
use the integrating factor e-t/m
B (t ) t / m
d  t / m  dI 1   t / m
I e
 
 I e

e
dt
m
 dt m 


• In this case we are dealing with
upgoing beams and we integrate from
the bottom to the top.
Upper half-range intensity

0

d  t '/ m


t */ m
d
t
'
I
e

I
(
0
,
m
,

)

I
(
t
*,
m
,

)
e


t * dt ' 
t*
dt ' t '/ m
dt ' t '/ m
 
e
Bt (t ' )  
e
Bt (t ' )
0
t*
m
0
m
or
t */ m
I (0, m,  )  I (t *,m,  )e



t*

0
dt '
m
B (t ')e
t '/ m
Upper half-range intensity
• To find the intensity at an interior point t,
integrate from t* to t and obtain
I (t , m,  )  I (t *,m,  )e




(t *t )/ m

t*

t
dt '
(t 't )/ m
B (t ')e
m
• What happens when m approaches zero. This
is where the line of sight traverses an infinite
distance parallel to the slab. Here
It (t , m  0,  )  B (t )

Formal solution including Scattering and
Emission
• Note that the source is now due to
thermal emission and multiple scattering
a(t )
ˆ
ˆ
ˆ
ˆ
S(t , )  1 a(t )B(t ) 
d

'
p(
t
,

',

)I(
t
,

')

4 4 
• The independent variable is the extinction
optical depth, the sum of the absorption and
scattering optical depths. We can write
ˆ)
dI (t , 
ˆ )  S (t , 
ˆ)
m
  I (t , 
dt
Formal solution including scattering and
emission
• The method of using an integrating factor
can be applied as before

 

t (P1,P 2)
ˆ
ˆ
I t (P2 ), I t (P1), e

•
t (P P2 )
ˆ
 dt S(t,)e
t (P1 )
In slab geometry the solutions become
t / m
I (t , m,  )  I (0, m,  )e

t (P2 )


t

0
dt '
m
(t t ' )/ m
S(t ', m,  )e

Formal solution including scattering and
emission
I (t , m,  )  I (t *,m,  )e


•
where
•
and
(t *t )/ m

t*

t
dt '
S (t ')e(t 't )/ m
m
I (t,m  0,)  S (t ,m  0, )

S(t , m,  )  1 aB(t )
a

4
2
1
0
0
a

4
2
1
0
0

d

'
d
m
'
p(
m
',

';
m
,

)I
(t , m',  ')
 

d

'
d
m
'
p(
m
',

';
m
,

)I
(t , m',  ')
 
Radiative Heating Rate
• The differential change of energy over
the distance ds along a beam is
(d E)  dI dAdt d d
4

• If we divide this expression by dsdA,
(the unit volume, dV), and also ddt then
we get the time rate of change in radiative
energy per unit volume per unit frequency,
due to a change in intensity for beams
within the solid angle d.
Radiative Heating Rate
• Since there is (generally) incoming
radiation from all directions, the total change
in energy per unit frequency per unit time
per unit volume is
dI
ˆ  I )
d


d

(


4 ds 4
• The spectral heating rate H is
ˆ  I )
   d ( 


4

Radiative Heating Rate
• The net radiative heating rate H is

ˆ  I
    d  d (
)
0
4
• In a slab geometry the radiative
heating rate is written


1
F
I

H    d
 2  d  cosd(cos )
z
z
0
0
1


where F  F  F is the radiative flux in the
z direction
Separation into diffuse and direct(Solar)
components
• Two distinctly different components of the
shortwave radiation field. The solar
component:
S t / m 0
ˆ  
ˆ )
(
0
S t / m 0
(m  m0 )(  0 )
I (t ,,  )  F e

S
F e
• We have two sources to consider, the Sun
and the rest of the medium
I (t,m,)  I (t ,m, )  I (t,m,)


d

S
Diffuse and direct components
• Assume (1) the lower surface is black, (2)
no thermal radiation from the surface, the
we can write the half range intensities as
ˆ) 
dI (t , 
ˆ )  (1 a)B
m
 I (t , 
dt
a

ˆ
ˆ
ˆ

d

'
p(

',

)I
(
t
,

')

4 
a

4

ˆ
ˆ
ˆ ')
 d ' p(',)I (t ,

Diffuse and direct components
• And for the +ve direction
ˆ)
dI (t , 
ˆ )  (1 a)B
m
 I  (t , 
dt
a

ˆ
ˆ
ˆ ')

d

'
p(

',

)I
(t , 

4 
a

4

ˆ
ˆ
ˆ ')
d

'
p(

',

)I
(t , 


Diffuse and direct components
Now substitute the sum of the direct and diffuse components
ˆ)
ˆ)
dId (t , 
dIS (t , 
ˆ )  I  (t , 
ˆ )  (1  a) B
m
m
 I d (t , 
S
dt
dt
a
a

ˆ
ˆ
ˆ
ˆ ' , 
ˆ ) I  (t , 
ˆ ')

d

'
p
(


'
,


)
I
(
t
,

'
)

d

'
p
(


S
d
4 
4 
a
ˆ ' , 
ˆ ) I  (t , 
ˆ ')

d

'
p
(


d
4 
Diffuse and direct components
But I S is thedirect solar beam and dIS   I S dt / m
henceonegets
ˆ) 
dI (t , 
*
ˆ
ˆ)
m
 Id (t , )  (1 a)B  S (t ,
dt
a

ˆ
ˆ
ˆ

d

'
p(

',

)I
(
t
,

')

d
4 

d
a

4

ˆ
ˆ
ˆ
d

'
p(

'
,

)
I
(
t
,

')

d

Diffuse and direct components
• where
a
ˆ
S (t ,) 
4
*
S t / m 0
ˆ
ˆ
ˆ
ˆ
d

'
p(

',

)F
e

(




0)

a
S t / m 0
ˆ
ˆ

p(0 ,)F e
4
• One can repeat this procedure for the
upward component
Diffuse and direct components
ˆ) 
dI (t , 
*
ˆ
ˆ)
m
 Id (t , )  (1 a)B  S (t ,
dt
a

ˆ
ˆ
ˆ

d

'
p(

'
,

)
I
(
t
,

')

d
4 

d
a

4

ˆ
ˆ
ˆ
d

'
p(

'
,

)
I
(
t
,

')

d

Diffuse and direct components
a
ˆ
S (t , ) 
4
*
S t / m 0
ˆ
ˆ
ˆ
ˆ
d

'
p(

',

)F
e

(




0)

a
S t / m 0
ˆ
ˆ

p(0 ,)F e
4
Diffuse and direct components
• If we combine the half-range intensities we
get
dI(t ,u,  )
a
u
 I(t ,u,  ) 
dt
4
2
1
0
1
 d'  du', p(u'.';u,)I(t,u',')
 (1 a)B  S* (t ,u,  )
• Where u is cosand not |cos|