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Appendix D.1 – Sample Midterm Lab Exam Questions:
UNBC ENSC 201
Appendix D.1: Sample Lab Midterm Exam ANSWERS
Use the equation sheet supplied with the Lab Midterm Review materials (Appendix D.1)
when you are preparing as this is what you can expect to see with the exam.
1. What is a contour line (i.e. define the concept of a contour line).
A contour line is a line of “equal value” on a map or spatial image. In meteorology,
these lines take various forms, for example: isotherms are lines of equal temperature;
isohyets are lines of equal precipitation; isobars are lines of equal pressure; etc.. (For
most people, the contour lines seen on topography maps are most familiar. They are
called “contour lines” and this can be confusing as in this context these lines represent
lines of equal elevation, not just lines of equal value. )
2. A site has the following characteristics. It is night. Note the “percentage of the site”
values given below represent both L↑ and L↓ down values.
a. Calculate L*.
b. Under these conditions, how does L* relate to Q*?
c. Answer qualitatively (calculations aren’t required).
i. How would L*change if the sky were to become clear?
ii. What would happen to the air temperature at the surface?
Type
Clear sky
Clouds
Buildings
Trees
ground
Temperature % of site
-40 0C
0%
0
-10 C
100%
5 0C
10%
0
0 C
10%
0
5 C
80%
Answer Comments
At this site these make up L↓ values and sum to
100%. The clear sky value is obviously not used.
At this site, these make up L↑ values and sum to
100%. The cloud value at 100% would be how
you could know if the buildings or trees were
considered to be L↑ or L↓ values.
ANSWER: For all calculations: MAKE SURE TO SHOW YOUR UNITS AND HOW THEY CANCEL.
a. Method: determine the energy emitted for L↓ and L↑ using StephenBoltzman’s Law (L↓ = σ T4 and L↑ = σ T4) and then compute L*
using L* = L↓ - L↑ .
To compute L↓ and L↑:
L↓ = σ T4 Where: T = 100% @ -100C  in Kelvin = 273.15 + (-100C) = 263.15 K
L↓= (5.67 x 10-8 W m-2 K-4)(283.15 K)4
L↓= 271.89 W m-2
L↑ = σ T4 Where: T = (80% @ 50C + 10% @ 50C + 10% @ 00C + 10% @ 50C)
T = (90% @ 50C + 10% @ 00C) = (0.9 x 50C + 0.1 x 00C) = 4.50C
T  in Kelvin = 273.15 + 4.50C = 277.65 K
L↑ = (5.67 x 10-8 W m-2 K-4)(277.65 K)4
L↑ = 336.96 W m-2
Then L* = L↓ - L↑ :
L* = L↓ - L↑ = ( 271.89 W m-2 - 336.96 W m-2 ) = - 65.07 W m-2
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Appendix D.1 – Sample Midterm Lab Exam Questions:
UNBC ENSC 201
b. Under these conditions (i.e. nighttime – given in the questions preamble),
Q* =L*.
c. If the sky were to become clearer:
i. L* would become a larger negative number as the value for L↓
would be smaller, but L↑ would not change as much or as quickly
so L* would be more negative.
ii. the temperature at the surface would decrease since L*=Q* will be
a larger negative number, cooling the air to a greater extent.
3. Given that the radiation temperature of the Earth-Atmosphere is about 254 K, and
assuming it is a black body:
a. Calculate the flux density of longwave radiation (Wm-2) emitted by the EarthAtmosphere.
ANSWER:
L↑E-A = σ T4
where T = 254 K
L↑ E-A = (5.67 x 10-8 W m-2 K-4) (254 K)4
L↑ E-A = 236 W m-2
b. What is the total flux emitted (W) by Earth?
ANSWER:
The flux emitted by Earth is equal to the flux density times the surface area of Earth.
L↑ E-A = 236 W m-2
Flux = (L↑ E-A) (4πrE2) = (236 W m-2)[(4)(3.14)((6.37 x106m)2)]
= (236 W m-2) (5.10 x1014m2)
Flux = 1.20 x1017 W
c. How any joules (J) does the Earth emit per year?
ANSWER:
Since a watt (W) is defined as a joule per second (J/s), you can determine this
by multiplying the flux by the number of seconds in a year.
Flux = (1.20 x1017 W) = (1.20 x1017 J s-1)
Joules = (1.20 x1017 J s-1)(365 day yr-1)(24 hr day-1)(60 min hr-1)(60 s min-1)
Joules emitted by Earth = (3.78 x 1024 J)
d. The flux density of Earth radiation reaching the moon can be viewed as that
on a spherical shell concentric about Earth whose radius is the mean distance
between Earth and the moon. Calculate the value of this flux density.
ANSWER:
The flux density of Earth radiation that reaches the moon is determined by the
flux emitted by the Earth (determined in b. above) divided by the area of a
sphere whose radius is the mean distance between the Earth and moon. This
question is analogous to the calculation made to determine the solar constant.
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Appendix D.1 – Sample Midterm Lab Exam Questions:
UNBC ENSC 201
The Earth radiation that reaches the moon is represented by the radiation that
reaches the inside of a shell that is the distance from the Earth to the moon.
Flux emitted by the Earth = 1.20 x1017 W
Distance between Earth and moon = 3.84 x 108 m
Flux density from the earth incident on the moon 

1.20  1017W
1.20  1017W

4rE2M
4 [(3.84  108 m) 2 ]
1.20  1017W
W
 0.065 2
18
2
1.85  10 m
m
e. How does this compare with the solar constant?
ANSWER:
The solar constant is 1367 W m-2 and the flux density from the Earth incident
on the moon is 0.065 W m-2. The solar constant is much, much larger.
4. The Sun is 20 degrees above the horizon.
a. What is the zenith angle? (Hint: go back to the diagrams in Lab 2 to see how
zenith angle is defined).
b. Calculate the flux density of radiation entering the atmosphere.
c. If the atmosphere attenuates (absorbs and reflects) 20 % of this value and Earth’s albedo
is 0.25, what is K*, the net short wave radiation?
ANSWERS:
a. Since the Sun is 20 degrees above the horizon,
and the zenith angle is defined as the angle
between the perpendicular to the horizon and the
solar beam, the situation indicates that the zenith angle (Z) must be 900 – 200 = 700.
(Also see Figure 2.1 in the lab manual; incident angle also discussed in lecture.)
b. Flux density entering the atmosphere: determined by I = I0 cos Z : where I is the flux
density entering the atmosphere, I0 = the solar constant, Z = the zenith angle.
I = I0 cos Z = (1367 W m-2)(cos 700) = 467 W m-2
c. If the atmosphere attenuates 20% and the αEarth = 0.25 what is K*? To answer this
you must remove 20% of the incoming energy to determine K↓ (this is the same
as retaining 80%); then use αEarth and K↑ = (αEarth)(K↓) to determine K↑; now you
can compute K* from K*= K↓- K↑.
So the answer is:
I = 467 W m-2 but I is attenuated so K↓ = (0.8)(467 W m-2) = 373.6 W m-2
And K↑ = (αEarth)(K↓) = (0.25)(373.6 W m-2) = 93.4 W m-2
So:
K* = K↓- K↑ = (373.6 - 93.4) W m-2
K* = 280.2 W m-2
3
Appendix D.1 – Sample Midterm Lab Exam Questions:
UNBC ENSC 201
5. What is the incoming solar radiation (I) on Feb 14 at 8 a.m. at Prince George; assume
the latitude of Prince George is 540? [Extension for studying: go back to your Lab 2
answers and compare this value to the one you determined for your lab time January.
How are they different? How much radiation are we gaining each day?]
ANSWER:
Use: I  I 0 cos Z
where:
Io = the max solar flux density = 1367 Wm-2
I = the incident radiation received by the surface
cos Z = sin sin + cos cos  cos h
where:
 = the latitude of Prince George = 540 N
h = the hour angle (150 per hour from solar noon and we assumed
that solar noon is the same as local noon). For 8:00 PST  h =
(4 hours x -150/ hour) = -600
 = the solar declination (i.e. the latitude where the sun is directly
overhead. A positive value indicates 0North; and a negative
value indicates 0South). Determine  using:
  23.4 0 cos 360TJ  10
365
where TJ = the Julian Day = (31+14) = 45 for Feb 14
  23.40 cos 36045  10  13.67
(~140 south of the equator)
365
Determining: cos Z = sin sin + cos cos  cos h
we get:
cos Z = sin(540N) sin(-13.670) + cos(540N) cos(-13.670) cos (-600)= 0.09437
Substituting cosZ into I = Io cosZ
we get:
-2
I = Io cosZ = 1367 Wm (0.09437)
I = 129 Wm-2
6. Earth-Atmosphere Radiation Budget. Using your understanding of the concept of
radiation balance complete the missing values in the following table of the annual
radiation balance of the Earth (E), the Atmosphere (A) and the Earth-Atmosphere
system (E-A). All units are in gigajoules per meter squared per year (GJ m-2 yr-1;
G = 109). You are given the average annual solar radiation to the E-A system as
338 Wm-2; the planetary albedo (αE-A) is 0.30; and the following table.
System
Atmosphere (A)
Earth (E)
Earth-Atmosphere (E-A)
Net solar (K*)
GJ m-2 yr-1
2.32
5.14
7.46
Net infra-red (L*)
GJ m-2 yr-1
-5.34
-2.12
-7.46
ANSWERS in the table are colour coded below:
4
Net all-wave (Q*)
GJ m-2 yr-1
-3.02
3.02
0
Appendix D.1 – Sample Midterm Lab Exam Questions:
UNBC ENSC 201
Like the final question in the Radiation Calculation Lab (Lab 2), when answering this
question recognize the following relationships exist:
o The first two columns add to equal the third column (i.e. net solar + net infra-red
= net all-wave radiation or in symbols: K* + L* = Q*).
o The first two rows add to equal the last row (Atmosphere + Earth = EarthAtmosphere or in symbols: A + E = E-A).
o The net all-wave radiation balance of the Earth-Atmosphere system must equal
zero if this system is in radiative balance (i.e. the system isn’t gaining or losing
energy on an annual basis).
These relationships allow determination of QE* and LE*
The value for K*E-A can be derived from understanding that:
o the average annual solar radiation to the E-A system of 338 Wm-2 = K↓E-A and
o using the planetary albedo of αE-A = 0.30, K↑ can be calculated from
K↑ E-A = 338(0.3) = 101.4 Wm-2
o From these K*E-A= K↓- K↑= (338 -101.4) Wm-2 = 236.6 Wm-2 (for those of you who are
noticing a short cut to this value it looks like: K*E-A= (338 Wm-2(1-0.30)) = 236.6 Wm-2
o To compare this value to those already in the table, K*E-A= 236.6 Wm-2 must be converted
from W m-2 to GJ m-2 yr-1 as follows:
 236.6 J
K E*  A  
2
 sm
 60 s  60 min  24 hr  365 day  1 GJ


 9


 min  hr  day  yr  10 J

GJ
  7.46 2
m
yr

7. To answer this question, make reference to the figure below.
a. Explain the reason(s) for the dip in the QE curve (and the related rise in QH) in
July and August.
b. How is it possible for the heat used in evapotranspiration (QE) to exceed the
available net radiation (Q*) in the September-February period?
c. Discuss the behaviour of QG throughout the year (i.e. how does it vary, at
what times and why; what is the net QG for the year, why?).
d. Discuss and explain the relationship of Q* to the other energy flux density
values shown in this graph?
ANSWER:
a. The QE curve dips in July and August because water for evapotranspiration
becomes limited and there is not enough water available to meet the
evapotranspiration demand. But, because Q* is still high, the dip in QE causes
a rise in QH as the energy goes into heating the air.
b. In Sept-Feb heat from the ground (QG) supplements the energy needed for
evapotranspiration. The negative value of QG indicates that energy is being
conducted from the ground to the surface. (Also, as QH is negative, it is losing
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UNBC ENSC 201
Appendix D.1 – Sample Midterm Lab Exam Questions:
heat to the surface and energy is being made available for evaporation from
QH. Often this happens when warmer air masses moves into an area.)
c. Throughout the year QG is balanced in terms of negative and positive energy
flux densities; this makes intuitive sense if on an annual basis the ground is
neither heating or cooling. It is positive during the summer months when Q*
is highest and the deeper ground is cooler than the surface so heat flows into
the deeper ground. QG is negative during the winter when the deeper ground
is warmer than the surface and it releases energy to the surface.
d. Q* drives the other fluxes for most of the year. During the summer months,
large positive values of Q* result in a surplus of energy at the surface which
goes into the air (as convective sensible and latent heat fluxes, QH and QE) and
into the ground as the conductive heat flux QG. During the winter, negative
values of Q* result in a deficit in radiant energy at the surface causing the
ground and the air to cool and provide heat to the surface.
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