Name:___________________________________
ATMS 111
18 November 2015
INDIVIDUAL EXAM#02
Choose the single best answer in Questions (1) – (15), each question is worth four points.
Questions (16) - (18) together are worth 40 points. EXAM#02 is worth 100 total points.
-----------------------------------------------------------------------------------------------------------An expression relating fish metabolism (dependent or “y” variable, units of Joules per
second [J s-1]) to lake temperature (independent or “x” variable, units of degrees Celsius
[oC]) is defined as
y = Bx + C
where “B” and “C” are constants that can be determined from the following observations;
{a} fish metabolism was measured to be 10.0 J s-1 at a lake temperature of 30oC and {b}
fish metabolism was measured to be 1.0 J s-1 at a lake temperature of 0oC.
Use this information to help answer Questions (1)  (3).
(1) Constant “B” has the value ________ [J s-1] / [oC].
(a) 0.1
(b) 0.2
(c) 0.3, PE#4, Q(1)
(d) 0.4
(2) Constant “C” has the value ________ [J s-1].
(a) 0.0
(b) 1.0, PE#4, Q(1)
(c) 2.0
(d) 3.0
(3) A measured lake temperature of 20oC would likely indicate a fish metabolism of
________ [J s-1].
(a) 7.0, PE#4, Q(1)
(b) 8.0
(c) 9.0
(d) 10.0
(4) A tipping bucket gauge registers a tip on June 18 at 18.90000 fractional day (FD).
This corresponds to the decimal military time (DMT) on June 18 of ________.
(a) 20.0
(b) 20.7
(c) 21.2
(d) 21.6, PE#4, Q(5)
(5) The gauge registers another tip at 21.8 DMT on June 18, showing an accumulation of
1.0 mm between this time and the DMT tip time in Problem (4). Calculate the rain rate
[mm per hour] at the tipping bucket rain gauge over the period between the two tips.
{Please show your work for complete credit.}1.0 mm/(21.8-21.6 h) = 5 mm h-1, PE#4, Q(6)
1
Name:___________________________________
ATMS 111
18 November 2015
INDIVIDUAL EXAM#02
(6) Assuming a flat earth and a straight radiation beam from a WSR88D station, a
radiation beam for an elevation angle of 0.5o above the horizon at a horizontal distance of
100 kilometers from the radar station will be found ________ kilometers above the
ground. Assume that the radar dome is sitting on the ground.
(a) 0.5236
(b) 0.8727, PE#4, Q(8)
(c) 0.9599
(d) 1.3963
(7) What might cause a problem with using the lowest radar scan data from the Greer, SC
WSR88D radar to estimate rainfall for locations in Haywood County?
{a} scan over the tops of clouds  missing precip –OR- {b} beam blocking due
to the mountains, LR#4, EQ(c)
(8) Hiking up Mt. LeConte you measure air pressure of 925 hPa at 825 meters and 850
hPa at 1555 meters. The scale height at Mt. LeConte on the day of your hike is ________
meters.
(a) 7844
(b) 7933
(c) 8544
(d) 8633, PE#5, Q(1)
(9) On the day of the hike in Problem (8), the estimated air pressure at the top of Mt.
LeConte (altitude 2010 meters) would be ________ hPa.
(a) 806, PE#5, Q(2)
(b) 816
(c) 826
(d) 836
(10) Describe if the air pressure estimate in Problem (9) involves an interpolation or an
extrapolation given the measurements made in Problem (8).
Measurements were made at Z=825 and 1555 meters, the top of Mt. LeConte is outside
of the range of measurements and so the estimate is an extrapolation
(11) If colder air moves over Mt. LeConte on Day 2 than on the day of the hike in
Problem (8), one would expect that the scale height on Day 2 between 825 and 1555
meters would be ________ on the day of the hike and that the pressure at the top of Mt.
LeConte on Day 2 would be ________ on the day of the hike.
(a) greater than, greater than
(b) greater than, less than
(c) less than, greater than
(d) less than, less than, PE#5, Q(3)
2
Name:___________________________________
ATMS 111
18 November 2015
INDIVIDUAL EXAM#02
We travel from Station A (elevation=2200 feet or 670.7 meters) where we measure a
surface temperature of 18oC and a surface air pressure of 950 hPa. We travel 100 km to
the east (compass heading 90o) and measure a surface air pressure of 993 hPa at Station B
(elevation = 900 feet or 274.4 meters). Find the illustration at Problem (17) on Page 5.
The stations are located in the Northern Hemisphere.
(12) A measured air temperature of 18oC at an elevation of 670.7 meters corresponds to a
“below ground” air temperature of ________oC at an elevation of 274.4 meters at Station
A, assuming an environmental lapse rate of 6.5oC/ km.
(a) 13.4
(b) 15.6
(c) 18.4
(d) 20.6, PE#6, Q(1A)
(13) The layer average air temperature between 274.4 and 670.7 meters at Station A is
________ Kelvin.
(a) 288.7
(b) 289.8
(c) 291.2
(d) 292.3, PE#6, Q(1B)
(14) The calculated scale height using Eq. (6.2) between 274.4 and 670.7 meters at
Station A is ________ meters.
(a) 459
(b) 532
(c) 8551, PE#6, Q(1C)
(d) 8651
(15) Reducing the surface air pressure measured at Station A to 900 feet elevation using
Eq. (6.3) results in an estimated air pressure of ________ hPa.
(a) 985.1
(b) 995.1, PE#6, Q(1D)
(c) 1005.1
(d) 1015.1
(BONUS) The average air pressure at 900 feet elevation between Stations A and B is
________ Pascals.
(a) 98905
(b) 99405, PE#6, Q(1E)
(c) 99905
(d) 100405
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Name:___________________________________
ATMS 111
18 November 2015
INDIVIDUAL EXAM#02
(16) 15 points
Calculate the surface wind speed [m s-1] estimate between Station A and Station B
given the conditions in Problems (12) – (15) using Eq. (6.4), being careful to use the proper
units in the equation.
V = [(287 J K-1 kg-1)(292.3 K)/(99405 Pa)(8x10-5 s-1)] * (210 Pa)/(100x103m)
= 22.15 m s-1
4
Name:___________________________________
ATMS 111
18 November 2015
INDIVIDUAL EXAM#02
(17) 15 points
Let us assume that the air pressure changes most rapidly from Station A to Station
B in the horizontal on a direct line between Station A and Station B (see figure above)
given the conditions in Problems (12) – (15). Determine the most likely wind direction
between Station A and Station B given the calculated and measured air pressures at an
elevation of 900 feet, assuming that the winds are geostrophic and that air flow is without
the influence of friction. How would the wind direction change from this estimate if the
influence of friction suddenly became important (explain how the numerical estimate of
geostrophic wind direction would increase or decrease based on the influence of
friction)?
995.1 hPa at Station A and 993 hPa at Station B at an elevation of 900 feet
indicates that lower pressure is toward Station B. Given the assumption that air pressure
changes most rapidly from Station A to Station B in the horizontal on a direct line
between Station A and Station B, the isobars are directed with a north-south orientation so
that the (geostrophic) wind direction must be blowing from the north or from the south.
Since low pressure is found toward Station B and we are in the Northern Hemisphere, the
winds must be blowing so that low pressure is toward the left of the direction of the
winds, hence, the winds must be blowing from the north (360o).
If friction becomes important, the winds change to being cross-isobar flow,
blowing toward low pressure. The winds would shift ~30o so that the winds were blowing
from the northwest (360o - 30o = 330o).
5
Name:___________________________________
ATMS 111
18 November 2015
INDIVIDUAL EXAM#02
(17) 10 points
Weather station observations are plotted on the map below. The number to the upper
right of the station circle is related to the sea level pressure (mb). Draw the 1012 mb
isobar (with proper labeling) over the lower 48 states of the U.S. and write the words
“HIGH” and “LOW” to indicate areas of high and low sea level pressure, respectively.
6
7
Name:_____________________________________
Work to be included for Problem _____ of Exam _____ in class ATMS _____
8
Formula Sheet 1 for Exam#2
ATMS 111
(𝑧−𝑧 )
𝐵
𝐻 = {− ln[𝑝(𝑧)]−ln[𝑝
(𝑧
𝐵 )]
} (6.0)
𝑇(𝑧) = 𝑇(𝑧𝐵 ) − Γ ∗ [𝑧 − 𝑧𝐵 ] (6.1)
𝐻=
𝑅𝑇̅
𝑔
(6.2)
𝑝(𝑧) = 𝑝(𝑧𝐵 )exp {−
𝑉=
𝑅𝑇̅ Δ𝑝
=
𝑝̅ 𝑓 Δ𝑛
(𝑧−𝑧𝐵 )
𝐻
} (6.3)
𝑅𝑇̅ |pS − pN |
(6.4)
𝑝̅ 𝑓
Δ𝑛
where the terms in Eq. (6.1) are air temperature at some altitude [oC], air temperature at
some lower level [oC], environmental lapse rate (= 6.5 oC/ km), altitude [m], and altitude
at some lower level [km]. The terms in Eq. (6.2) are scale height [m], dry air gas constant
[= 287 J kg-1 K-1], layer average air temperature [K], and the acceleration due to gravity
[= 9.81 m s-2]. Eq. (6.3) is identical to the pressure equation introduced in Preparatory
and Laboratory Exercises #5. The terms in Eq. (6.4) are the wind speed [m s-1], dry air
gas constant [= 287 J kg-1 K-1], layer average air temperature [K], average air pressure
[Pascals] between a north and south location, Coriolis parameter [=8.0x10-5 s-1], air
pressure difference [Pascals] between a north and south location, and horizontal distance
between the north and south locations [meters].
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