Problem Set 4
1.
From the expression of the Planck function Bν (T ) , obtain the expression for Bλ (T ) .
2.
Prove that the values of ν and λ which maximize Bν (T ) and Bλ (T ) are respectively νmax = C1T and λ max = C 2 / T ,
where C1 and C2 are constants. Given C2 = 2.9×10-3 m K, find the temperatures where Bλ (T ) attains a maximum
for wavelengths 550 nm and 10.5 µm.
3.
Assuming that the Sun behaves like a blackbody at a temperature T=5800 K, calculate Bν (T ) and Bλ (T ) at 550
nm and the percentage increase in these spectral radiances if the temperature is increased by 100 K.
4.
Surface elements of area ΔS1 and ΔS2 respectively are separated by a distance r with their normal are inclined at
angles θ1 and θ2 to the line joining them. If the element ΔS1 is at a temperature T1, then the net energy flux ΔE
from ΔS1 to ΔS2 in the optical bandpass Δν is given by
B (T )
ΔE = v 2 1 Δν ΔS1 ΔS2 cos θ1 cos θ 2 .
r
State the conditions under which the above expression is obtained.
5.
Using the substitution x =
2π k
5
with σ =
3
hν
kT
∞
and
x3
∫ ex − 1
0
dx =
π4
, evaluate
15
∞
∞
0
0
4
∫ Bν dν and show that π ∫ Bν dν = σ T ,
4
15h c
2
as the Stefan-Boltzmann constant. Using the result of Problem (4) above, show that the solar
⎛ RS ⎞ , where T is the
constant Fs (i.e. the normal irradiance just outside the atmosphere) is given by Fs = σ TS
s
⎝ d⎠
effective blackbody temperature of the Sun, Rs its radius and d is the distance of the sun from earth. Evaluate Fs
2
4
for Ts = 5800 K.
6.
A horizontal disc of radius r is suspended at height h over an infinite black plane. Perform appropriate integration
and find out the power absorbed by the horizontal disc out of that emitted from the black plane, which is at
temperature T1. State any assumptions made.
7.
A meteorological satellite circles the earth at a height h above the surface. Let the radius of the earth be
Show that solid angle (ω) under which the earth is seen by the satellite sensor is 2π [1 −
ae .
(2ae h + h ) / (ae + h)].
2
Find the ω for the sensors of (i) the Indo-French satellite “Megha-Tropiques” orbiting the earth at a height of 867
km; and (ii) India’s geostationary satellite “INSAT 3D” which orbits in the space synchronously with the earth.
[Hint: h=36,000 km for a geostationary satellite].
8.
An infrared scanning radiometer aboard a meteorological satellite measures the outgoing radiation emitted from
earth’s surface at a wavelength of 10 µm. Assuming the atmosphere transparent at this length, what is the
temperature of the earth’s surface if the observed radiance is 9.8 J s-1 m-2 µm-1 sr-1?
9.
The infrared scanning sensor (channel 4) of the ScaRAB instrument aboard the meteorological satellite “MeghaTropiques” measures the outgoing longwave radiation emitted from the earth’s surface in the 11.5 µm window
region. Assuming that the effect of atmosphere between the satellite and the earth’s surface may be neglected,
−2
−1
−1
what would be temperature of the surface if the observed spectral radiance at 11.5 µm were 10.65W m µ m sr .
10.
An infrared scanning radiometer aboard a meteorological satellite measures the outgoing radiation emitted from
earth’s surface at a wavelength of 10 µm. Assuming the atmosphere transparent at this length, what is the
temperature of the earth’s surface if the observed radiance is 9.8 J s-1 m-2 µm-1 sr-1?
11.
Write a Fortran program to calculate this temperature from Bλ (T ) and verify your answer obtained by hand
calculations for Prob.8 and Prob.9 given above.
1
12.
Find the power absorbed by a sphere of radius r suspended at a height h above the black horizontal plane, which
is at temperature T1.
13.
In an isothermal atmosphere, the density of an absorbing gas varies with height z as ρ (z) = ρ exp(−z / H ) with
s
−1
H = RT / g and R = 287J kg K
−1
as the gas constant; g is the acceleration due to gravity; and ρ is the surface
s
concentration of the gas. If the mixing ratio of the absorbing gas is r (= ρ / ρ a ) , ρ a the density of air and k the
absorption coefficient ; show that the optical depth ( χ ) in such an atmosphere is given by
χ=
ps
r k exp(−z / H ) and
dχ
=−
χ
.
g
dz
H
Using the expression of the absorption rate, find the altitude where it peaks.
[Hint: Write the expression of downward flux density (F) as solution of Beer Law in terms of
derivative of F with respect to χ ; set the derivative to zero to get the answer.]
χ and obtain the
14.
Consider the atmosphere as an isothermal layer with temperature 260 K and surface pressure 1025 hPa. Find the
altitude where maximum power will be absorbed when sun is overhead, by a solar radiation absorbing gas with
mixing ratio 1 g kg-1 and absorption coefficient 5 m2 kg-1.
15.
A horizontal plane is placed just above the earth and the sun is at the Zenith (i.e. sun is overhead). If both earth
and sun radiate as black bodies at temperatures 288 K and 5800 K, respectively; show that the spectral irradiance
(power per unit area, per unit wave length interval) falling on side of the horizontal plane facing the earth is
⎛ Rs ⎞ πΒ (T ) ; α is the albedo of the earth-atmosphere
λ
sun
⎝ d⎠
2
πΒλ (Tearth ) and on the side facing the sun is (1 − α )
system, Rs is the radius of the sun and d is the distance between the Sun and the Earth. Plot Βλ (Tearth ) and
Βλ (Tsun ) as a function of wavelength (µm) and estimate the maximum values of the solar and terrestrial spectral
irradiances from the graphs of irradiances with wavelength.
−2
16.
If the emittance of the sun is 6.2 × 10 W m
emittance intercepted by the earth?
17.
A low altitude earth satellite in an equatorial orbit carries below it two isolated spherical radiation sensors of
negligible heat capacity. One is painted white and may be assumed to be perfectly reflecting at all wavelengths
where there is significant solar energy (λ < 4 µm) and perfectly absorbing at longer wavelengths (λ > 4 µm). The
other spherical radiation sensor is painted black and is perfectly absorbing at all wavelengths of electromagnetic
spectrum. Assuming no input of direct or scattered energy from the spacecraft;
(i)
Calculate the temperatures of the two spheres at midnight over a thick cloud sheet at a temperature of
280 K.
(ii)
Find the radiance due to diffusely scattered sun light above a cloud layer of albedo α for an overhead
sun. [Hint: equate radiances integrated over the solid angle to the αFS, FS is the solar constant].
(iii)
With both the spherical sensors being shadowed from overhead direct sunlight by the satellite, calculate
their temperatures over a thick cloud of temperature 270 K and albedo 0.8, given that the solar constant
FS = 1366 Wm-2.
18.
If one third of the solar energy incident outside the atmosphere ( Fs ) were absorbed by the atmosphere, show that
the average rate of temperature rise (assuming no loss) of the atmosphere would be ~1 K per day. Take
7
and the radius of the earth is 6371 km, what is the fraction of sun’s
−2
Fs = 1370W m in your calculations. If the loss of energy by the troposphere is due to emission of longwave
radiation, calculate the average rate of longwave cooling of the troposphere per day. The rates of temperature
change due to radiative processes are also expressed in units of K per day.
19.
Assume that a gas with uniform absorption coefficient is uniformly mixed with height. The surface temperature
of the earth is 280 K and the tropopause at 250 hPa pressure level is at a temperature 220 K. Assuming radiative
equilibrium conditions, find φ = F↑- F↓. Also find χ* from the relation π B(T )(1 + χ *) as a function of pressure.
What would be the temperature discontinuity at the surface under such assumptions?
2
20.
The effective temperature Te of a planet’s surface is obtained by equating the solar radiation it absorbs at its
{
}
surface to the infrared radiation it emits. That is, 4 π a σ Te = π a (1 − α )FS / R with a is the radius of the
2
planet; α is the albedo of the planet, Fs = 1366W m
−2
4
2
2
is the value of the solar constant for earth, R =
d
d earth
is the
normalized distance of planet with that of earth from the sun if d is the distance of planet. R gives the distance of
a planet from the sun in astronomical units and it is expressed in terms of the distance of a fixed planet (say earth)
from a star (i.e. Sun). Taking R = 0.72 for Venus and its surface albedo as 0.77, find its effective temperature Te
of its surface.
21.
22.
Let the surface temperature of a planet in the solar system be Tg ; and Te is its effective temperature that has been
calculated using the formula given in Prob.20. Find the expression for the optical depth χ * of the planet’s
atmosphere in terms of Tg and Te using the formulae,
F
π B(Tg ) = O (2 + χ *) , FO = π B(Te )
2
The surface temperature of Venus is given as Tg = 750 K ; taking the value of Te as that has been calculated in
Problem.20, calculate the optical depth of the atmosphere of Venus using the formulae given in Prob.20 above.
−2
Based on the value of χ*, comment on the greenhouse effect on the planet Venus. Assume Fs = 1370W m ,
albedo of Venus as 0.77 and its normalized distance R = 0.72 from the Sun.
23.
The effective temperatures of Earth and Venus are 255 K and 230 K respectively. However, the measured surface
temperatures of these planets are 280 K and 750 K respectively. Compare the optical depths of Earth and Venus
atmospheres and comment on the greenhouse effect of their atmospheres.
24.
A small black body satellite is orbiting the earth at a distance far enough away from it so that the flux density of
earth radiation is negligible in comparison to solar radiation. If the satellite suddenly passes into the earth
shadow, then at what rate will it initially cool? In your calculations, take satellite as a spherical body of radius
r = 1 m , mass (M) = 1000 kg and specific heat (C ) = 103 J kg-1 K-1. Further, you may take surface temperature
uniform and albedo to be zero. [Hint: rate of cooling = MC
dT
. Find, for the satellite, the black body
dt
temperature Te and apply the radiation balance: Rate of cooling = the amount of emitted radiation].
25.
Suppose a Zebra, standing under the shadow of a tree, suddenly moves out to enjoy the Sun. What is the
difference between the temperatures of the white and black strips of the Zebra if its body temperature is 38.2°C?
In calculating the difference of temperature between the strips, take white strips perfectly reflecting and black
strips perfectly absorbing. What kind of optical effect it can create to an animal or a viewer, which only sees
object in black and white (colour blind) as temperature of the air in the immediate contact with black strip would
be higher then of the white strip that is at body temperature of zebra.
Nota Bene: The formulation of the above problem was inspired by an episode of “National Geography” showing
a lion hunting zebras. In a herd of zebras, lion fails to identify an individual animal; indeed, it perceives the herd
as a monster due to the optical effect arising from the temperature difference produced by black and white strips
of zebras’ assembly in the hot sun. In fact due to illusion, the predator perceives the assembly as a creature it has
never seen before and surprisingly the lion makes a retreat. On the contrary, lion could easily identify a lone
zebra standing in the sun despite the optical effect of its strips coupled with that of atmosphere and hence
becomes its kill. Staying in a herd protects, every zebra knows.
26.
Estimate the rate at which the earth’s surface would cool if the Sun stopped shining. State any assumption that is
made in getting the answer.
27.
Estimate the rate at which the temperature of oceans would rise if they absorbed all the energy from the Sun and
there were no cooling. What is the corresponding value if you assume that earth absorbs all the heat from the
Sun, without cooling?
28.
In a simple one dimensional “greenhouse” model, the atmosphere is represented by a single layer at an effective
radiometric temperature of the earth. This layer is completely transparent to solar radiation and completely
opaque to longwave infrared emission from earth’s surface. What is the equilibrium temperature of the surface
according to this model, assuming that the radiometric temperature of the earth is 255 K.
3
29.
30.
If TE is the earth’s equivalent blackbody temperature and FE is the density of the flux emitted from top of its
atmosphere; show that small perturbations in the earth radiation balance are governed by the relation
δ Τ E δ FE
=
TE
FE
From the above relationship, find the change in TE that would occur in response to:
(a) an increase in albedo from 0.30 to 0.35 (state clearly any assumption made);
(b) the seasonal variation in the sun-earth distance due to the eccentricity of earth’s orbit, which is approximately
3.5% from one season to other.
Carbon dioxide (CO2) is assumed well mixed in the atmosphere with its volume mixing ratio, wco2 = 360 ppm
(parts per million by volume). Compute the absorber amount u of CO2 in the atmosphere.
[Hint: u = sec θ
∫
ZT
ZB
ρ r dz ; ρ is the air density and r the mass mixing ratio of absorber gas; and mass mixing
ratio (r) of CO2 is (Mco2/Mair) wco2; M = molecular weight of the gas.]
31.
Carbon dioxide (CO2) is assumed well mixed in the atmosphere. It is a dominant absorbent of the infrared (LW)
radiation at wavelength 15 µm. Find the change in transmittance of terrestrial radiation through the atmosphere if
mixing ratio of CO2 increases from 360 ppm to 400 ppm using the expression
−Su ⎛
Su ⎞
τ ν (u) = exp
1+
.
δ ⎝
πα ⎠
The values of the parameter on the right may be read from Table 11.2 of Module 11.
{
32.
Water vapour (H2O) is the dominant absorbent in the solar and near-infrared wavelength band 0.7–4 µm. The
path length of H2O in an atmospheric layer the column height of precipitable water (cm) which can be calculated
in two ways:
u1 =
33.
}
∫
ZT
ZB
ρ r dz and
u2 =
∫
ZT
ZB
ρ r ( p / p0 ) x dz ; x = 0.68 , p0 = 1013.25 hPa.
Consider the 1000 – 500 hPa layer in the atmosphere with water vapour mixing ratio 12 g/kg and calculate the
path lengths u1 and u2 and compare their magnitudes in this layer.
Ozone is the dominant absorber in the stratosphere. Using the absorption cross-section of O3 from Table 11.1,
find the effective transmittance of solar beam through the atmospheric layers 10 – 1 hPa with O3 number
concentration 1016 molec/m3 and the layer 100 – 10 hPa with O3 number concentration 5×1017 molec/m3 by
appropriately evaluating the transmission function integrals. Find the absorption of solar radiation in both the
layers.
34.
Write the computer program for problems (29) – (31) and check the result with your hand calculations.
35.
A body is emitting radiation in the following spectrum of monochromatic flux density.
λ < 0.35 µ m Fλ = 0
−2
−1
−2
−1
−2
−1
0.35 µ m < λ < 0.50 µ m
Fλ = 1.0W m µ m
0.50 µ m < λ < 0.70 µ m
Fλ = 0.5W m µ m
0.70 µ m < λ < 1.00 µ m
Fλ = 0.2 W m µ m
λ > 1.00 µ m Fλ = 0
Calculate the flux density of radiation. If this flux density of radiation falls on an opaque surface, which absorbs
all radiation for wavelength λ < 0.70 µ m but reflects all radiation with λ > 0.70 µ m . Calculate the amount of
radiation absorbed and reflected.
31.
During the Kuwait oil fires in February 1991, the smoke plumes fanned out to form a layer, which was 40 km
wide, 160 km long and 3 km thick. The plumes were generated from 650 oil wells (of 800 wells that were
detonated), which burnt for several months. Take the 2-month period to calculate the daily mean soot
concentration and optical depth of the smoke layer. Answer the following questions.
(i)
The total oil burned was 2,408,750 barrels/day of which 86% constituted gasoline. If the heating value
−1
of gasoline is 4.7 × 10 J kg (1 barrel of Kuwait oil weighs = 138.2 kg), calculate the height to which
7
4
the hot air bubbles will rise in the atmosphere if average surface air density is 1.20 kg m
−3
and surface

temperature 40 C . State any assumption made.
(ii)
Smoke with soot (elemental carbon) is produced when gasoline burns. Given that 0.11kg of smoke
mass is produced if 1.0 kg of gasoline is burned and this smoke mass contains 30% of elemental
carbon particles on ageing. During this 2-month period, deposition process also removes the soot mass
from the atmospheric column. Calculate the daily average density (kg/m3) and the mixing ratio (kg/kg)
of the soot particles in the smoke layer from Kuwait oil fires that developed during 60 days by taking
the deposition rate of soot particles as 25 g/m2/month to which the entire depth of the smoke layer
responds. The smoke layer may be considered as well mixed. (During ageing, different particles mix
internally/externally and their radiative properties change; the simplest being the external mixing)
(iii)
Consider a vertical section of smoke layer (assumed as an isothermal layer) over a region far away (say
100 km) from the oil fires and answer the following questions:
(a) Using the daily average density or the mixing ratio of soot particles in the smoke layer from part
(ii) of this question, find the optical depth of the smoke column extending from surface to top of the
2
−1
smoke layer if the absorption coefficient of the aged soot particles is 1.8 m g (fresh soot particle
−1
have absorption coefficient varying between 4.5 − 9.82 m g ).
2
(b) What will be the radiative cooling or warming produced by the smoke layer at the surface in this
region, if the incident solar flux is 550W m
32
−2
at the top of the layer?
Consider the earth’s atmosphere from surface (1013.00 hPa) to the top of the stratosphere (1.00 hPa) consisting
of three layers:
Top layer thickness
1.00 hPa ≤ p ≤ 100.00 hPa
(Layer-1 stratosphere)
Middle layer thickness
100.00 hPa ≤ p ≤ 500.00 hPa
(Layer-2 upper troposphere)
Bottom layer thickness
500.00 hPa ≤ p ≤ 1013.00 hPa (surface)
(Layer-3 lower troposphere)
The absorbent gases present in these layers are as follows:
(i)
Water vapour (H2O) is a strong absorber in wave band (1200 – 2200 cm-1) of infrared radiation centered
at wavelength 6.3 µm. Assume that it is present only in the Layer-3 of thickness 513 hPa, with its
−1
average mass mixing ratio, rH 2O = 12 g kg .
(ii)
Carbon dioxide (CO2) is the dominant absorbent of the infrared (LW) radiation at wavelength 15 µm.
Assume that it is present only in Layer-2 of thickness 400 hPa and Layer-3 of thickness 513 hPa, and its
average volume mixing ratio, wco2 = 370 ppm (parts per million).
(iii)
Ozone (O3) is the dominant absorbent of the solar (SW) radiation in the wave length band 0.22–0.30
µm. Assume that it is present only in Layer-1, and its average number concentration, No3 = 5.0 E+16
molecules per cubic meter ( # m-3) in Layer-1. The absorption cross-sections (CS) of the O3 molecule in
the ultra-violet region 220–300 nm is as follows:
Band-1: 220-240 nm, Average absorption CS, k1 = 4.85763 E-22 m2/molecule
Band-2: 250-270 nm, Average absorption CS, k2 = 9.90320 E-22 m2/molecule
Band-3: 280-300 nm, Average absorption CS, k3 = 1.86136 E-22 m2/molecule
Assuming atmosphere to be plane-parallel, calculate the heating rate in these layers due to absorption of radiation
at respective wavebands by the gases present therein.
33
Using the integrating factor exp(− χ ) , integrate the Schwarzchild’s equation
∂I
34
=I−B
∂χ
Show that differentiation with respect to χ on both sides of the following expression
I ( χ , µ ) = B( χ * ) e
↑
− ( χ * − χ )/ µ
+
∫
χ
χ
*
B( χ ′ ) e
− ( χ ′ − χ )/ µ
d χ ′ , (µ > 0 upward transfer)
µ
5
gives the equation of radiative transfer in plane parallel atmosphere
↑
dI ( χ , µ )
↑
µ
= I ( χ , µ ) − B( χ )
dχ
for the upward component of the two-stream approximation.
[Hint: Use the Leibniz’s rule of differentiation of integrals under the variable limits of integration
b( y)
d ⎡
⎢∫
dy ⎣ a ( y )
35
⎤
F(x, y) dx ⎥ =
⎦
b( y)
∫
a( y)
dF
dy
dx + F(x, b)
db
dy
− F(x, a)
da
dy
Consider a beam of radiation passing through a path of length l containing an absorbing gas of density ρ . The
transmission of this path is τ ν = exp(−kν ρ l) . Set u = ρl (absorber mass), then integrated absorptance W is
given by
W =
∫
∞
0
[1 − exp(−kν u)] dν .
The Lorentz profile expression of the absorption coefficient
kν =
S
α
π (ν − ν 0 ) + α
2
2
kν
; α = (2π tc )
−1
.
Obtain the final form of the integral after substituting kν in the integral of W and derive the expression of W
under following approximations:
(i)
Weak-line approximation: Expand the exponential term and retain the first expansion term and show that
absorptance W = S ρ l .
(ii) Strong-line approximation: The absorption is saturated at the line centre under strong-line condition. Thus
ν − ν 0 >> α ; under such a condition the term α can be neglected in the denominator of the Lorentz
2
profile expression of the absorption coefficient
kν . Substitute the resulting expression of kν in the
expression for W, and by evaluating the integral show that W = 2 Sα ρ l .
6