Transcript Ch. 3 Radiation

Ch. 3 Radiation
Text: Wallace and Hobbs, Ch. 4,
Radiative Transfer p113-152,
Liou, An introduction to atmospheric
radiation Chs. 3, 4
3.3 Radiative transfer in planetary atmosphere:
• 3.3.1 Beer’s law:
– quantify radiation attenuation due to scattering and
absorption as it passes through a layer gases, in our
case, the atmosphere.
Based on dI l = -Il rrk l ds (4.17 in text), a beam of radiation passes the
earth's atmosphere to a altitude z, can be determined by I l ,z = Il ,¥ ò z k l rrdz,
¥
Il = Il¥e -t l secq
where Il¥ incident light at the top of the atmosphere, q : zenith angle of
incident beam
tl =
ò
¥
z
k l rrdz, optical depth or thickness, depletion of radiation that a
overhead beam (zenith = 0) would experience as it passes
through length the atmospheric layer.
Tl = e -t l secq : transimissivity of the atmospheric layer
Absorptivity : a l = 1 - Tl
dIl = -Il kl rrsec qdz
• Attenuation is
strongest near t~1.
small attenuation
when t>3, because
most of radiation has
already lost during
the transmission.
• The earth’s effective
temperature is the
temperature
approximately at t~1
in the earth’s
atmosphere.
Fig. 4.23 Vertical profiles of the monochromatic
intensity of incident radiation, the rate of absorption
of incident radiation per unit height, air density and
optical depth, for kλ and r independent of height.
• Why?
Example:
• A beam of radiation passes through a layer of 100 m thick
with zenith angle of 600 and 2 types of absorbing gases, each
contributes 50% to the total air mass. Gas A has density of 0.1
kg/m3 with mass absorption coefficient of 10-3 for wavelength
l. Gas B has density of 0.2 kg/m3 with mass absorption
coefficient of 10-2 for the same wavelength.
– A) Calculate the optical depth, transmissivity and absorption of the
layer at wavelength l.
– B) Which gas dominates the optical depth and transmissivity of the
layer?
a.
For the gas A, the mass of absorbing gas that the beam encounter along its slant path length is
u a = sec60 o × 0.1 kg/m3 ´ 0.5 ´ 100m = 2 ´ 5kg/m2 = 10kg/m2
t l ,a = kl ,a ua = 10 -3 m2 /kg ´ 10kg/m2 = 10 -2 = 0.01
For gas B :
u b = sec60 o × 0.2 kg/m3 ´ 0.5 ´ 100m = 2 ´ 10kg/m2 = 20kg/m2
t l ,b = kl ,b ub = 10 -2 m2 /kg ´ 20kg/m2 = 2 ´ 10 -1 = 0.2
The total t l = 0.21,
-(t
+t
Tl = e l ,a l ,b = 0.81
al = 1 - Tl = 0.19
)
Gas B is the main contributor for optical depth, transimissivity and absorpitivity
of the air layer
AERONET-sun photometers
• World largest observational network for
aerosols. Based on Beer’s law, optical depth
can be measured by
I l = I l¥e-t l secq , ln
t l = ln
I l ,q1
I lq2
I l ,q1
I lq2
= t l (secq1 - secq 2 )
/ (secq1 - secq 2 )
3.3.2 Absorption and emission of infrared radiation
• An beam of infrared radiation will be absorbed and emitted as
it passes through a layer of the atmosphere.
• The question that governs transfer of infrared radiation
through a gaseous medium.
Schwarzschild’s equation
dI l = -dI l (absorption) + dI l (emission) = -(I l - Bl (T ))kl rrds
s1
I l (s1 ) =
ò
I l ,oe -t l (s1 ,0)
+
k l rrBl [T (s)]e -t l (s1 ,s)ds
monochromatic intensity
0
from 0 to s account for the monochromatic intensity
1
extincition along the way of the radiation emitted by the
gas along the path from 0 to s1
account for extincition along
the way.
Derivation of the Schwarzchild’s law:
dIl = [ -Il + Bl (T)]× r × r× kl ds Þ
ò
t l (s1,s) =
-
s1
s
dIl
= -Il + Bl (T)
r × r× kl ds
rrkl ds' Þ dt l (s1,s) = - rrk l ds
dIl
= -Il + Bl (T) Þ - dIl e -t l (s1 ,s) = -Il e -t l (s1 ,s)dt l (s1,s) + Bl (T)e -t l (s1 ,s)dt l (s1,s)
dt l (s1,s)
s1
ò { - dIl e
-t l (s1 ,s)
+ Il e
-t l (s1 ,s)
dt l (s1,s)} =
0
s1
ò
Bl (T)e -t l (s1 ,s)dt l (s1,s)
0
d [ Il e
-t l (s1 ,s)
s1
] | = ò B (T)e
s1
s
l
-t l (s1 ,s)
dt l (s1,s)
0
Il e
-t l (s1 ,s1 )
= Il (0)e
-t l (s1 ,0)
+
s1
ò Bl (T)e t
-
l
dt l (s1,s)
(s1 ,s)
0
Il = Il (0)e -t l (s1 ,0) +
s1
ò Bl (T)e t
0
-
l
dt l (s1,s)
(s1 ,s)
The plane parallel approximation:
53o
•
•
Calculation of 3D radiative transfer is very
difficult and expensive.
Although radiation transmission has 3-D
structure in reality, in most of case, the
variations of atmospheric constituents and
cloud aerosols layers are primarily a function
of height. Thus, radiation transmission can be
considered primarily a function of its linear
distance.
With the plane parallel assumption, the irradiance passes through
a layer of atmopshere can be expressed as:
Fn-¯(t v ) =
ò pI
2
-¯
v
(t v , cosq )cosq dw assume I-¯
v is isotropic with azimuthal angle f
we have Fn-¯(t v ) = 2p
Fn-¯ (t v ) = 2p
ò
ò
90o
0
-¯
I (t v , cosq )cosq sin q dq let m =cosq
v
1 -¯
1
0
0
I (t v , m )m dm = 2p ò I -¯
v,0 Tn (t v , m ) m dm analogous to Beer's law for flux density
v
assume Tn (t v , m ) » e
-
tl
m
» e-1.66t l where
1
m
= 1.66 is a diffusivity factor, equivalent
to radiation beam passes through a layer at 53o zenith angle.
Fn (t v ) = 2p
-¯
ò
1 -¯
0
-
tv
m
I e m dm = 2p I
v,0
-¯
v,0
-
e
tv
m
ém2 ù 1
ê ú |0 = p I
ë2û
-¯
v,0
-
e
tv
m
= pI
-¯
v
(t v , m )
When consider the diffusive
factor (3D), a bean of
radiation passes through a
layer of atmosphere is
equivalent to an beam
passes through the layer at
53o without diffusive effect
under the plane parallel
assumption.
Plane parallel approximation
is not accurate for cumuli
clouds.
Radiative transfer integrated over all wave numbers:
•
•
•
•
Radiative flux: the total radiative energy is cumulative radiative energy (irradiance) across all
wavelength (number).
The transmission of radiative flux is
However, to accurately estimate radiative flux, we need to use 10-1 to 10-3 cm-1 wave number
interval to calculate extinction for each interval. It is very expensive and slow to do so.
To improve computation efficiency, a “correlated k-distribution method” is commonly used. It
group lines with similar absorption coefficiencts into one bin and g represent the fractional
distribution of the lines with this absorption relative to total lines.
Radiative heating/cooling in the atmosphere:
• Vertical convergence or divergence causes radiative heating or cooling
• Usually we use unit of oC/day. Radiative cooling can reach a few oC/day to
50oC/day at top of dense marine cloud deck.
rC p
dT
¶F (z)
dT
1 ¶
=- R
Þ
=dt
¶z
dt
rC p ¶z
dT
1 é
=
ò
dt rC p êë 2p
=
ò
1
-1
[ò
]
ù
1 é ¶I v
m
d
w
ò
ê
úû
rC p ë 4p ¶z
I mdw =
4p v
ù
¶Iv
dfdm ú where ds = dz/m, m = cosq , dw = secqdqdf
û
¶s
2p é 1 ¶Iv ù 2p
ò dm =
rC p êë -1 ¶s úû rC p
[ò
1
-1
rrkn (Iv - Bv )dm
]
based on Schwarzchild's equation
If we let Iv » 0 i.e., assuming the incidence of longwave from the adjacent layers does not
change with height in a thin isothermal atmosphere, we have
dT
2p
=dt
Cp
[ ò rk B (z)e
1
0
n
v
-t v / m
]
dm = -
p
rkn Bv (z)e -t
mC p
v
/m
where m = 1.66 as the diffusivity factor
Longwave heating rate is mainly determined by
Emission and optical depth, thus, heading rate
Is much larger in lower troposphere.
Radative heating/cooling
T trends caused by Radative heating/cooling
Of change in greenhouse gases:
Why does CO2 warm the troposphere, but cool the stratosphere?
Summary-1:
• What is the main difference in determining
radiative energy extinction between a ray of
solar radiation and that of thermal radiation
emitted from the earth surface?
– Solar radiation experiences scattering and
absorption by particles (clouds and aerosols) and
also gases (O3 and H2O) when it passes through
the atmosphere, whereas terrestrial infrared
radiation experiences both absorption and
emission.
Summary-2:
• What physics laws are used to determine the radiative energy
absorption and emission extinction in the atmosphere
energy?
– Beer’s law: Radiative energy loss as it passes through a medium is an exponential
function of optical depth of the medium and incident angle of radiation beam. This is
used in radiative transfer calculation of the solar radiation in earth’s atmosphere.
I l = I l,¥e-t l secq
– Schwarzchild’s law: used for longwave radiative transfer, also consider
emission of the atmosphere.
s1
I l (s1 ) = I l ,oe
+ ò kl rrBl [T (s)] e-t l (s1,s) ds
absorption 0
emission
- t l (s1,0)
Summary-3:
• What are the most commonly used approximation?
– Plane parallel approximation: When consider the diffusive
factor (3D), a bean of radiation passes through a layer of
atmosphere is equivalent to an beam passes through the
layer at 53o without diffusive effect
Fn-¯ = p In-¯(t n , m )
• Correlated k-distribution: great speed up calculation by group lines with
similar absorption coefficiencts into one bin and g represent the fractional
distribution of the lines with this absorption relative to total lines.
1
T=
Dv
ò
Dv
e
-kvm
dv =
ò
1 -k (g) m
0
e
dg
Summary-4:
• How do we separate the warming due an increase of
greenhouses gases from those of solar variability, ozone
depletion and volcanic aerosol?
– Through finger prints of these radiative agents.
• Why does greenhouse gases warm the troposphere, but cool
the stratosphere?
– In troposphere, greenhouse gases absorb more radiation from the
surface than emit to space. Thus, they trap more radiation emitted
from the surface. In stratosphere, greenhouse gases absorb radiation
in the stratosphere and emit to space. Thus, they cause net loss of
radiation in stratosphere.