Download 501 09 Quiz 6 Solutions

ATMS 501
Quiz #6 Solutions
November 20, 2009
1. Multiple choice or fill in the blank (2 points each).
(a) If kλ is independent of height, –dIλ/dz is largest (well above, near, well below) the level of
unit optical depth.
(b) If kλ is independent of height, optical depth is linearly proportional to (pressure, density,
temperature). Pressure is a measure of the mass in the overlying layer.
(c) The better indicator of the heating rate in °C per day per due to the absorption of incident
radiation is (–dIλ/dz, dIλ/dp). The heating rate is proportional to the rate of absorption of
radiation –dIλ/dz divided by ρcp. and ρ is proportional to –dp/dz. It follows that the heating
rate tends to be largest well above the level of unit optical depth. For example, the heating
due to the absorption of solar UV radiation by ozone is strongest near 50 km but the level of
unit optical depth is closer to 30 km.
(d) In reality, kλ often tends to (increase, decrease) with geometric depth in a planetary
atmosphere. The increase is due to the effect of pressure broadening, which increases in
proportion to the frequency of collisions between molecules.
(e) The infrared radiation emitted by the troposphere in the upward direction tends to be (greater
than, the same as, less than) that emitted in the downward direction. The troposphere is not
an isothermal layer. It exhibits a strong positive lapse rate; i.e., a decline in temperature with
increasing height. It is also optically thick in the infrared part of the spectrum; i.e., radiation
passing through it tends to be absorbed and re emitted multiple times. Hence, the radiation
emitted in the upward direction tends to be emitted from the relatively cold upper troposphere
while the radiation emitted in the downward direction tends to be emitted from the relatively
warm lower troposphere. The emitted radiation is blackbody radiation weighted by the
emissivity and a blackbody at a higher temperature emits more radiation than a blackbody at
a lower temperature. It follows that the upward emission from the troposphere is about 50%
greater than the downward emission, as documented in Fig. 10.1.
(f) During the afternoon hours the level of unit optical depth for incident solar ultraviolet
radiation tends to (rise, drop). At the level of unit optical depth τλ sec θ = 1. During the
afternoon hours, sec θ increases. It follows that τλ must decrease. If we assume that the
density and the absorptivity remain constant, the only way for τλ to decrease is for the level of
unit optical depth to rise so that there is a decrease in the mass of air overlying the level of
unit optical depth.
(g) A layer of aerosols is more likely to contribute positively to the Earth’s albedo if its single
scattering albedo is (high, low). The higher the single scattering albedo, the greater the
scattering and the more radiation that is likely to be backscattered to space.
(h) A layer of aerosols is more likely to contribute positively to the Earth’s albedo if it overlies a
(light, dark) surface. If it overlies a dark surface it will backscatter radiation that would
otherwise be absorbed and consequently increase the fraction of incident solar radiation that
is backscattered to space.
ATMS 501
Quiz #6 Solutions
November 20, 2009
(i) As the optical thickness of a layer of a gas increases, the intensity of the radiation that the
layer emits exponentially approaches (zero, the intensity of blackbody radiation, infinity,
the intensity of the radiation incident from the other side). This is the essence of
Schwarzchild’s equation.
(j) As the optical thickness of a layer of a gas becomes very small, the intensity of the radiation
that the layer emits approaches (zero, the intensity of blackbody radiation, infinity, the
intensity of the radiation incident from the other side). Schwarzchild’s equation again. If
the layer is infinitesimally thin (in an optical sense) it will be transparent; i.e., it will have a
transmissivity approaching 1 so what goes in one side comes out the other side.
(k) Other things being equal, the level of unit optical depth tends to be higher (near the centers,
in the wings of) absorption lines. The higher the absorptivity kλ the smaller the mass of gas
required to absorb 1/e of the incident radiation and hence the higher the level of unit optical
depth.
(l) The flux absorptivity is generally (greater than, equal to, less than) the intensity absorptivity
at zero zenith angle. Intensity absorptivity increases with path length. For all path lengths
except the one with zero zenith angle, the path length is longer that for a ray incident at zero
zenith angle. For isotropic radiation it is exactly twice as long.
(m) The flux transmissivity is generally (greater than, equal to, less than) the intensity
transmissivity at zero zenith angle. Tλ = 1 – αλ so if the flux absorptivity is greater, the flux
transmissivity will be correspondingly less.
(n) Averaged over the stratosphere, radiative transfer produces (heating, cooling, almost no net
time rate of change of temperature). The stratosphere is in radiative equilibrium so the
radiative heating due to the absorption of solar radiation almost exactly balances the net
radiative cooling due to the emission of infrared radiation (mainly) to space.
(o) Stratospheric ozone depletion has tended to (warm cool) the stratosphere. Less ozone means
less radiative heating so that radiative equilibrium can be maintained with the greenhouse
gases in the stratosphere emitting radiation at a lower temperature..
(p) Increasing the concentrations of greenhouse gases would tend to (warm, cool) the
stratosphere. Higher greenhouse gas concentrations implies higher emissivity, so that the
lower stratospheric can emit just as much radiation (i.e., an amount equal to the absorbed
solar radiation) without being as warm.
(q) When a layer of gas is in radiative equilibrium conditions (its temperature isn’t changing with
time, it is isothermal, it emits equal fluxes of radiation in the upward and downward
directions, the net radiative fluxes into and out of the layer are in balance). Temperature
may change with time if there are energy fluxes that are non-radiative. A layer can be in
radiative equilibrium without being isothermal. In this case it would emit different amounts
of radiation in the upward and downward directions.
(r) The volume extinction coefficient kλρr has units of m-1.
ATMS 501
Quiz #6 Solutions
November 20, 2009
(s) Blue light tends to be scattered preferentially to red light when the scattering efficiency Kλ
(increases, decreases) with size parameter x = 2πr/ λ. In the Rayleigh scattering regime
increases in proportion to the fourth power of size parameter for particles of a prescribed
radius.
(t) Radiative cooling due to the emission of infrared radiation at cloud tops tends to make the
lapse rate within cloud layers (more, less) stable. Unless the cloud layer is optically very thin,
the cooling will tends to be concentrated near the top and so will tend to increase the lapse
rate. This is what drives cellular convection in stratus and stratocumulus cloud layers at the
top of the boundary layer. During the daytime this effect is opposed by heating near the cloud
top due to the absorption of solar radiation but at night it is unopposed.
(u) The level of unit optical depth marks the (top, midpoint, bottom) of the layer in which the
upward flux of energy from the surface of a planet or interior of a star is dominated by
radiative transfer. It is above the layer of unit optical depth, where upward radiation can pass
through with relatively little absorption, that radiative transfer is the dominant mechanism for
the upward transport of solar radiation absorbed at the surface of the planet or energy
generated internally by nuclear fusion or (as on Jupiter) by gravitational collapse. Below the
level of unit optical depth the radiative equilibrium lapse rate is superadiabatic and
convection is the dominant mechanism for the upward transfer of energy.
(v) Exactly (1/2, 1/e, 1 – 1/e) of the incident radiation is absorbed above the level of unit optical
depth. By definition the fraction 1/e remains unabsorbed at the level of unit optical depth, so
the fraction absorbed above that level is 1 – 1/e.
(w) The presence of N2 and O2 in the atmosphere (enhances, reduces, has no effect on) the
absorption of outgoing infrared radiation by greenhouse gases. Pressure broadening increases
with the frequency of occurrence of collisions experienced by greenhouse gas molecules,
irrespective of the chemical composition of the molecules that they collide with.
(x) The radiative flux over land surfaces is often downward (during the day, at night). Because it
radiates as a blackbody, the land surface cools rapidly after sunset so that it often becomes
cooler than the air above it. If the temperature difference becomes appreciable, the downward
radiation from the warm atmosphere will exceed the upward radiation from the colder
surface. The downward transfer of energy by the radiation cools the atmosphere at the lowest
levels, while it slows the cooling of the land surface. This is what creates nighttime
inversions over land surfaces. Inversions tend to form when when wind speeds are light so
that mixing is small and skies are clear so that the downward radiation from levels above the
boundary layer is minimized.
(y) The presence of cloud cover tends to (increase, decrease) the diurnal temperature range. ...in
two ways: it lowers daytime temperatures by backscattering to space some fraction of the
solar radiation that would otherwise be absorbed at the Earth’s surface and it raises nighttime
temperatures by blocking the emission of infrared radiation emitted by the earth’s surface
(i.e., by absorbing it and emitting some of it downward toward the Earth’s surface.
ATMS 501
Quiz #6 Solutions
November 20, 2009
Consider a planet with a one-layer, isothermal atmosphere with absorptivity a in the infrared
part of the spectrum. The atmosphere is in radiative equilibrium and it is completely transparent
to incoming solar radiation. The surface of the planet emits radiation as a blackbody. (a) Derive
expressions for the flux densities of radiation emitted by the surface of the planet F0 and by the
atmosphere F1 in terms of a and the flux density of solar radiation Fs. (b) Using the Stefan
Boltzmann law, calculate the radiative equilibrium temperatures of the surface of the planet and
the atmosphere, assuming that the flux density of solar radiation is 1368 W m-2 and the
absorptivity of the atmosphere is 0.5. (25 points)
(a) We can write simultaneous equations for the flux at the top of the atmosphere and the flux at
the Earth’s surface and solve them to obtain the values of F0 and F1 in terms of a and Fs. The
equation at the top of the atmosphere is (1) F1 + (1 – a) F0 = Fs and the equation at the
Earth’s surface is (2) F0 = Fs + F1. Substituting for F0 in (1) yields F1 = aFs / (2 – a). If this
looks a bit familiar, it’s because we’ve encountered this same situation before in Exercise 4.39.
Substituting for F1 in (2) yields F0 = 2Fs/(2 – a).
Although it’s not required as a part of the solution, I offer further proof that this is, in fact, the
correct solution.
• The fluxes at the top of the atmosphere balance: the emission from the atmosphere
F1 = aFs / (2 – a) plus the emission from the ground 2Fs/(2 – a) times the factor
(1 – a) are equal to Fs.
• The fluxes through the surface balance: Fs plus the downward emission from the
atmosphere F1 = aFs / (2 – a) equal the upward emission from the surface F0 = 2Fs/(2 – a).
• The fluxes into and out of the layer balance, The energy absorbed is 2aFs/(2 – a). The
energy emitted is 2 F1 = 2aFs/(2 – a).
We can also demonstrating the validity of this solution by tracking what happens to the
incident solar radiation after it is absorbed at the surface and radiated back to space. The
surface emits Fs of which aFs/2 comes back. The surface then emits aFs/2 of which aFs/2 ×
aFs/2 comes back, etc. Hence, the total emission from the surface is equal to Fs [1 + (aFs/2) +
(aFs/2)2....] and the sum of this infinite geometric series is 2Fs/(2 – a).
(b) Substituting a = 0.5 and Fs = 1368 / 4 = 342 W m-2 [no albedo is given in the problem, so
I’m assuming here that it is 1.0] we obtain F0 = 0.2 Fs = 68.4 W m-2 and F1 = 1.333 Fs = 456 W
m-2. . Substituting these values into the Stefan Boltzmann law F = σT 4, where σ = 5.67 × 10-8
yields T0 = 299 K and T1 = 186 K. I gave full credit if you did not divide by 4 because the
statement of the problem wasn’t explicit.
ATMS 501
Quiz #6 Solutions
November 20, 2009
2. Given the following data, based on ground based measurements of solar radiation taken at the
same location at different times on a clear day
zenith angle θ
0°
60°
monochromatic intensity Iλ
0.566 units
0.400
Making use of Beer’s Law Iλ = Iλ∞ e –τλsec θ under the assumption that the optical thickness of
the atmosphere is the same in both measurements, estimate (a) estimate Iλ∞ (b) the normal
optical thickness τλ of the atmosphere at this wavelength, (c) the monochromatic intensity at a
zenith angle of 75°. [Hint: to make the equation linear, take the antilog of both sides. Then
linearly extrapolate the data to zero atmospheric mass, noting that mass is linearly
proportional to sec θ.] (25 points).
My apologies: the hint should have said “taking the log”, not “taking the antilog. Beer’s law
can be written in the form
ln Iλ∞ – ln Iλ = τλ sec θ
Substituting the data
ln Iλ∞ + 0.569 = τλ
ln Iλ∞ + 0.916 = 2 τλ
Solving, we obtain τλ = 0.347, ln Iλ∞ = –0.222 and Iλ = 0.800 units
For part (c) we can write
ln Iλ∞ – ln Iλ
–0.222 – ln Iλ
–ln Iλ
Iλ
= τλ sec 75°
= 0.347 × 3.86
= 1.34 + 0.22 = 1.56
= 0.21 units
This problem can also be solved without taking logs. Substituting the data into Beer’s Law we
can write
0.566 = Iλ∞ e –τλ
(1)
–2τ
0.400 = Iλ∞ e λ
(2)
Dividing (2) by (1) yields
0.707 = e –τλ
Taking the natural log, we obtain τλ = 0.347
Substituting into (1) yields 0.566 = Iλ∞ × 0.707; Iλ∞ = 0.800
Iλ = Iλ∞ e –τλsec θ
ATMS 501
Quiz #6 Solutions
=
=
=
=
November 20, 2009
0.800 e –0.347 sec 75°
0.800 e –0.347 × 3.86
0.800 e –1.34
0.21 units
In retrospect, I can see that the hint for this problem wasn’t helpful. Taking logs simplifies the
graphical depiction of the solution to the problem but it makes the calculations more
complicated and more abstract and more prone to errors (as I discovered for myself).