Solutions to Practice Problem Set 2
CIV300/ENV346
Fall 2015
This assignment primarily relates to the marvelous summary diagram of global energy exchanges
in a “3 layer” model by Kiehl and Trenberth (1997). The final question is a closely related
question from last term’s final exam.
In the diagram below, as is hopefully obvious yellow “blocks” (with red arrows) depict short wave
radiation exchanges, the red line blocks (with yellow arrows) long wave radiation.
Note that when an object or system receives radiation (or photons) of a given wavelength, three
things can happen: some photons can be reflected off the surface (the statistical measure
associated with this is called the albedo or reflectivity), some photons on average can be
absorbed (the absorptivity), and a fraction of them might be transmitted through the object (the
transmissivity). An opaque object will naturally not transmit photos more than a few atomic
diameters into a surface, and a perfect blackbody will only absorb them.
a. Based on these numbers, is the Earth’s energy as a whole balanced? What about the
ground surface? What about the atmospheric layer?
Ans: The energy balance for the three layers can be written as follows:
i)
For the earth:
Incoming Energy = Incoming solar radiation=342 W/m2
Out coming Energy= Outgoing long-wave radiation + Reflected solar radiation = 235 + 107 =
342 W/m2. Therefore it is balanced.
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ii)
For the ground surface:
Incoming Energy = Incoming solar radiation + Back radiation (greenhouse gases) = 198+324 =
522 W/m2
Outgoing Energy = Thermals + Evapotranspiratiopn (in terms of latent heat) + Surface radiation
+ Reflected solar by the surface = 24 + 78 + 390 + 30 = 522 W/m2. Therefore it is balanced.
iii)
For the atmospheric layer:
Incoming Energy = Incoming solar radiation + Thermals + Evapotranspiration + Surface
radiation = 67 + 24 + 78 + 350 = 519 W/m2
Outgoing Energy = Emitted by the atmosphere + back radiation by the atmosphere = (165 + 30) +
324 = 559 W/m2. Therefore it is balanced.
In all cases, the balance reflects that any energy storage term is likely to be small relative to the
fluxes (even allowing for climate change, the rate of accumulation is small overall).
b. What are the average short-wave albedo and the absorptivity of the atmosphere? Of the
ground surface? Of the Earth as a whole? What is the transmissivity of the atmosphere
both to short and to long wave radiation?
Ans:
i)
For the atmosphere:
The short-wave albedo:
77
= 0.22
342
67
= 0.20
342
€
198
and transmissivity:
= 0.58
342
€ wave transmissivity is 40/390
Long
absorptivity:
ii)
For the ground surface (transmissivity of the ground surface and the earth is zero):
€
The short-wave albedo:
and absorptivity:
iii)
30
= 0.15
198
168
= 0.85
198
€ the earth:
For
The short-wave
albedo:
€
and absorptivity:
107
= 0.31
342
168 + 67
= 0.69
342
€ terms for each component adds to 1.
Note: the set of
c.
Make some “personal observation connections” with at least five of these fluxes. For
example, the obvious shift in “apparent temperature” (coolness) you experience in
€
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summer when a cloud moves in front and blocks the sun (the short-wave transmission of
a cloud is very small). Etc.
Ans: Most of these terms are directly observable in least in terms of a local example. For
example, one can see thermals move heat from a surface like pavement in the summer; one can
feel the cooling effect of evaporation when you climb out of a lake in the summer, particularly if
the wind is blowing. You can experience the direct influence of the warmth of a sunbeam
through your skin, and the change that you experience when a cloud obscures this transfer. You
can experience the “blanketing” effect of clouds by observing how much colder (and sometimes
frostier) clear sky nights are compared to cloudy ones. The reflection of light from the ground
surface is perhaps most obvious from snow and the “snow blindness” and pain you can
experience from reflected light over a fresh snowy field (snow ages quickly, and the reflectivity
decreases as this happens). And so on….
d. The “evapotranspiration” (ET) represents the combined effect of evaporation from open
water sources and the movement of water through plants (transpiration). It is difficult to
measure but its closely dependent term, precipitation, is much easier to measure, though
global and temporal averages are still not trivial. What is the global average depth of
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precipitation that is consistent with the 78 W/m indicated here? Does this seem
reasonable?
Ans: The 78 W/m2 is corresponding to the latent heat required by the liquid water to have a phase
change into vapour (gas). If the average global depth of the precipitation is considered to be h
mm/day and the latent heat of the water 2.5 MJ/kg, we can estimate the h as follows;
Energy available for the phase change = 78 W/m2 = 78
J
86400s
J
×
= 6.739 ×10 6 2
2
s.m
day
m .day
Energy required for the h mm of water to be evaporated over the earth surface per day =
2.5
MJ
mm
kg
J
× h ×10−3
×1000 3 = 2.5€
×10 6 h 2
kg
day
m
m .day
if we balance these two energy values the h would be 2.7 mm/day.
€
It should be noted that the spatial and temporal variation of the precipitation round the world are
much different from the above-estimated value.
e. On average, 24 W/m2 is conveyed from the surface of the Earth as thermals, smaller
than the 78 W/m2 associated with ET. Describe how the relative value of these two terms
is likely very different depending on the surface features of specific places and times.
(For example, you might compare a desert to a district with many lakes, or a large city to
the wetland it replaced.)
Ans:
Generally you will get more ET where the surface has a higher moisture content i.e. lake region.
However the colder the water, the less the ET. Loss from thermals will be much higher where it’s
primarily land, and of course higher where the average albedo is lower. Lower albedo means
more energy is absorbed at the surface, therefore higher temperatures and then conduction to the
air mass, resulting in air warming and thermals. Other factor is the specific heat capacity of the
ground material. Lower SHC, higher temperature increase, more thermals.
f.
The “surface radiation” term is larger than one might have expected as the difference
between the energy absorbed by the surface and the amount lost by thermal and ET.
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This is because a fraction of the outgoing radiation is reflected back to the surface, and a
portion of THIS is also reflected back, and so on. Assume A is the primary amount and
that a fraction f of A is re-radiated back again to the surface and then back to clouds, in
an infinite (but converging) geometric series. Estimate this fraction f.
Ans:
A is the expected value of the surface radiation calculated as follows: (168 is the absorbed surface
radiation, 78 is the energy going to latent heat, 24 is the energy carried up into the atmosphere by
thermals).
A = 168 − 78 − 24 = 66
W
m2
Infinite geometric series is given by A + Af + Af
2
+ Af 3 + Af 4 + ... which converges to
A
.
1− f
The series converges to 390
W
according to the diagram of the 3 layer model, i.e.,
m2
390 =
66
1− f
and therefore f = 0.83 . Remarkable, since this holds at each stage of the interaction, it holds for
the total as well. Thus there is a wonderful confirmation directly available: f = 324 / 390 = 0.83
g. (i) Estimate the implied average Earth temperature using the Stefan-Boltzmann equation,
and using the back radiation value, (ii) estimate the average “sky temperature” of the
cloud layer. Do these relative values seem reasonable? (iii) More radiation is radiated
by clouds back to the Earth than is lost by clouds through their upper surface. What is
the average upper cloud (outgoing radiation) temperature? Does this make sense?
Ans:
i) Assume the Earth surface temperature is T in 100 degree of Kelvin, so using Stefan-Boltzmann
equation we have:
E = εσT 4
390
W
= (1)(5.67)(T) 4
m2
T = 2.88 of 100 degree of K = 288K = 15 C
ii) The radiation from the lower layer of the cloud to the earth is 324 W/m2. So the average sky
temperature can be estimated using E = 324 W/m2 which results is T = 275 K or 2ºC.
iii) For radiation
back to the earth from the upper layer of the cloud and atmosphere E = 195
€
W/m2, so the upper cloud temperature is 242 K or -31ºC.
The results seem reasonable since we expect the temperature decreases gradually in altitude until
tropopause.
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h. Can you recognize any other processes that are likely to experience significant feedback,
where the output of one system influences the behaviour of another?
Ans:
GHG emissions: CO2 can (indirectly) melt ice that’s trapping peat bogs and, once melted,
it can release methane from the bogs to warm the planet.
Ice ages: Ice and snow increase the Earth’s albedo, making for less energy retained on
the planet and making things colder.
Stars: More gravity makes more gravity.
Hurricanes: Faster wind rotation around the centre brings currents closer to the centre
where their linear velocity is increase (I think because of the retention of rotational
velocity…I can’t remember), so it makes for even faster wind.
i.
Discuss the measurement challenges associated with obtaining a summary like this.
What do you think would be the most difficult terms to obtain with any accuracy? Why
might the evaluation of any potential climate change be particularly difficult to measure or
model in such a system?
Ans:
Difficulties would arise in obtaining accurate temperature and radiation effects, in
determining the effects of local and regional distributions of energy, and in accurately
measuring and modelling feedback effects.
We have huge variability in space and time to contend with, complicated by large local and
regional effects, large parts where data is scarce or hard to get (either over the ocean or
uninhabited regions), etc. There are instrument issues, funding issues, averaging issues,
statistical issues, and endless communication issues (timing effects). The storage component,
if any, would be the final effect of all of these, and it is very difficult to measure. The best
measure of a change in temperature would be reflected by an expansion of the sea, but this is
a slow process too with feedback too. If you like challenges, it is a delightfully complex
system. Certainly orbiting observations have hugely extended our ability to monitor the
Earth, and that has increased the confidence of other measurements.
2. (Final Exam Question from Spring 2011 version of CIV 300.) A Wikipedia article on
“insolation” discusses the “fate” (or division) of solar radiation received on a surface as follows:
“Some of the solar radiation will be absorbed while the remainder will be reflected. Most
commonly, the absorbed solar radiation causes radiant heating, however, some systems may
store or convert some portion of the absorbed radiation, as in the case of photovoltaics or plants.
The proportion of radiation reflected or absorbed depends on the object’s reflectivity or albedo,
respectively.” How would you improve this statement to be a more complete and more accurate
reflection of what takes place on Earth?
Ans: (A possible answer at least; obviously details were expected to differ.)
“Some of the solar radiation will be absorbed, some of the remainder will be reflected and
through some transparent media (ocean, atmosphere, etc.) some will be transmitted. Most
commonly, the absorbed solar radiation causes radiant heating, and resulting temperature
increases cause re-radiation in the infrared. Some systems may store or convert some portion of
the absorbed radiation, as in the case of photovoltaics or plants, but the portion of radiation
absorbed in these manners is trivial relative to the incoming radiation. The proportion of
radiation reflected or absorbed depends on the object’s albedo or absorptivity, respectively.”
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When marking we would expect to see the three following items: (i) recognition of the missing
transmission term (some photons are neither absorbed nor reflected), (ii) the recognition that
feedback occurs and make the picture more complex than this treatment implies, and (iii) a
comment that “reflectivity” and “albedo” are synonyms (mean the SAME thing), and thus cannot
logically be paired with “reflected and absorbed” through “respectively”. The point is that
Wikipedia is great if you already know most of the answer; not so good sometimes is you wish to
learn.
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