1

ATS 210 Spring Term 2001 NAME:_________________

Final Examination

This is a 2 hour, closed-book examination. Calculators may be used. All answers should be written on the examination paper. Use the final sheet for any notes or rough workings. Your answers must be your own work. NOTE: Academic dishonesty is a serious offence that may result in grade penalties and other disciplinary action. SIGN HERE: I have read the statement about dishonesty and understand the consequences ……………………….. Part A Answer ONLY TWELVE QUESTIONS in Part A. (Each question is 3 points) 1. (Circle the correct answer) The boundary between the troposphere and the stratosphere is called the

tropopause / stratopause/ mesopause ANSWER: tropopause 2. (Circle the correct answer) Near the surface of the earth, the atmospheric pressure decreases by about

1, 10, 30, or 100 mb for every 100m increase in elevation. ANSWER: 10 3. Changes of state involving phases of water are important for the latent heat energy associated with

them. Name three processes involving changes of state that require energy to be supplied from the environment (i.e. that result in cooling of the environment)

ANSWER: melting, evaporation, sublimation 4. (Circle the correct answers) Solar radiation energy reaching the earth’s surface is mainly in the

waveband 0.3-3 micrometers, 30-300 micrometers, or 0.3-3millimeters ; longwave radiation emitted from the earth’s surface is mainly in the waveband 0.3-3 micrometers, 3-30 micrometers, or 30-300micrometers.

ANSWER: 0.3-3 micrometers, 3-30 micrometers 5. List three greenhouse gases that have increased in concentration over the last 50 years in the global

atmosphere. ANSWER: CO2, Methane, N2O, CFCs 6. List the three resistances that control the rate of heat loss from the core of an animal to the surrounding

atmosphere. Which of the resistances are directly influenced by wind speed? ANSWER: Boundary layer, tissue, clothing or fur 7. On a cold, windy day, a weather station reported that the air temperature was -8°C and the Wind-chill

Equivalent Temperature was -25°C. Explain what is meant by ‘Wind Chill Equivalent Temperature’. ANSWER: Wind Chill Equivalent Temperature is the temperature still air would have to have to produce the same heat loss as in the windy conditions. 8. (Complete the definition) Relative humidity is …….. ANSWER: The percentage ratio of the current humidity to the saturation humidity of the current air

temperature: )(

100aTse

e× .

2

9. (Complete the statement.) The dry adiabatic lapse rate is greater than the moist adiabatic lapse rate because …….

ANSWER: Latent heat released by the condensation of water out of the air warms the air causing it to cool less quickly than unsaturated air. 10. Circle one of the following mechanisms that is not a primary way by which clouds form.

(a) lifting over topographic barriers (b) forced lifting along weather fronts (c) surface heating and convection (d) lifting by supersaturation (e) lifting by convergence

ANSWER: d 11. The figure below shows an upper air (500mb) map.

(a) Explain the meaning of the solid contour lines (b) Explain the meaning of the arrows (c) Outline the zones of flow divergence on the map

ANSWER: a) height, usually in meters, of the 500mb level; b) Wind direction c) no map L 12. Briefly explain how radiation fog is formed. ANSWER: A calm, clean night is required. Conditions need to be calm so that there is little mixing. A clear sky allows longwave radiation to escape, cooling the ground quickly. The air right next to the ground is also cooled, and if it drops to its dew point, condensation into a fog cloud occurs. 13. The satellite image below is of the Yucatan Peninsula in Mexico. Briefly explain the processes that probably produced the observed cloud pattern along the north coast (top) of the peninsula.

ANSWER: Sea breeze circulation produces lifting onshore, producing the band of clouds inland from the shore.

3

14. Name the 2 processes that cause cloud droplets to grow large enough to precipitate ANSWER: Collision-coalescence and ice-crystal formation(Bergeron process). 15. Draw a side view of a typical warm front. Include details of (a) the common cloud types you would

observe, (b) the horizontal scale of distance, and (c) the locations of warm and cold air at the surface. ANSWER:

16. List four atmospheric conditions necessary for creating a high probability for a major air pollution

incident. ANSWER: Number of pollution sources; a deep, stationary high pressure area; light surface winds; subsidence inversion; shallow mixing layer; topography(valleys); clear skies at night to form a surface inversion; sunlight to drive chemical reactions.

Warm Cold

500-1000km

10km Stratus

Nimbo-stratus

Cirrus

4

Part B NAME: _____________ Answer ONLY THREE QUESTIONS in Part B. (Each question is 5 points) 17. The sketch below shows evenly spaced isobars indicating a constant pressure gradient from south to

north at an altitude above the friction level. Sketch a rough path along which an air parcel starting at point A would accelerate until it reached a steady speed, and indicate the magnitude and direction of the forces acting on it at several points on its path.

North 900mb_________________________________________

904mb_________________________________________ 906mb_________________________________________ • A South ANSWER: The path will start out by going straight across the isobars, but would then start turning to the right(because of the Coriolis force) until it is going parallel to the isobars. 18. Describe three ways by which a fruit farmer might protect his trees from damage on cold nights in

spring. ANSWER: Smudge pots, fans to mix air, spraying water on the plants so that the freezing water will release latent heat to protect the plants. 19. Expla in, with the aid of a diagram, what happens to a rising parcel of air when the atmosphere is

conditionally unstable. In your example, choose values of the air temperature and dew point temperature at the ground and assume an appropriate value for the environmental lapse rate. (HINT: the dry adiabatic lapse rate is –10oC per km, and the moist adiabatic lapse rate is typically about –6.5oC per km near the surface)

ANSWER: The parcel is rising at the dry adiabatic lapse rate, which is stable with respect to the environmental lapse rate, until it becomes saturated and starts condensing out water. The condensation process releases heat into the air, causing the parcel to rise at the moist adiabatic lapse rate, which is unstable with respect to the environmental lapse rate.

20. Explain, with the aid of an example, why the surface air on the downwind side of a mountain can be drier than the surface air on the upwind side. (HINT: values for dry and moist adiabatic lapse rates are given in Question 19)

ANSWER: As the air rises up the upwind side of the mountain it cools(rising air cools). When it reaches its dew point, it begins to condense water, forming clouds, which adds heat to the rising air. The clouds can then precipitate, losing much of the water content of the clouds and making the air drier. When the air begins to descend on the downwind side, it heats up, but does not have to give up energy to evaporate much water(since it precipitated out), leaving it much warmer than it would otherwise be.

5

21. Briefly explain what causes the upwelling of cold water along the Oregon coast in summer. ANSWER: When the Pacific High moves northward in the summer, it produces winds that blows from north to south along the Washington, Oregon, and northern California coast. This has a tendency to push the top layer ocean waves in a southerly direction. The Coriolis force turns this to the right, that is, to the west away from shore. Since the upper portion is being pushed away from the shore, it has to be replaced by the colder water underneath.

22. If you were given a surface weather map showing isobars, weather fronts and current observations, and

a 500mb upper air chart, how might you use this information to predict where a storm would move and how it would develop over the next 24 hours?

ANSWER: Storm systems tend to follow the winds in the upper air chart, so predicting where the storm will go is a matter of following the upper air chart contours. Things that would contribute to the strengthening of the storm would be a strong surface low(because of friction, causing convergence) below a divergence of upper air flow, an occluded front(which provides extra lifting). Additionally, a strong jet stream over the top of the storm system can produce an effective boost to the upper air divergence, strengthening the storm. An upper air convergence can “shut down” the convergence of a surface low, or even reverse it, which has the effect of weakening the storm.

6

Part C Answer ONLY ONE QUESTION from Part C. (Each question is 9 points) 23. The net radiation measured over a dense forest in Oregon at noon on a cloudless June day was

730 W m-2. Given the following other measurements made over the forest, calculate the temperature of the top of the forest canopy, assuming that the foliage radiated like a perfect black body .

Downward solar radiation 900 W m-2

Albedo 0.10 (10%) Downward longwave radiation 400 W m-2

(HINT: Stefan’s Law L= σT4 where σ is 56.7x10-9 W m-2 K-4, and T is the temperature in degrees

Kelvin.) ANSWER:

( )( )( )

CK

RadiationT

TRadiation

Radiation

Radiation

NetRadiationRadiationRadiationRadiationNet

KmW

mW

out

out

mW

mW

mW

out

mW

mW

mW

mW

mW

mW

mW

mW

mW

in

inout

outin

o30303

107.56

480

480

7301210

1210

400810

4009009.0

4009001.01

4009001

49

4

4

42

2

2

22

2

22

22

22

22

==

×=

=

=

=

−=

=

+=

+=

+−=

+−=

−=−=

−

σ

σ

α

7

24. A student using a wet-and-dry bulb thermometer for his project breaks the dry bulb. He measures a

wet bulb temperature of 10.0 oC. A friend tells him that the vapor pressure in the atmosphere is 10.0mb. (a) What dry bulb temperature would he have recorded if the thermometer had not been broken? (b) What is the dew point temperature of the air? (c) What is the relative humidity?

[HINT: the wet bulb equation is

e = es(Tw) - γ(Ta – Tw) where e = atmospheric water vapor pressure, es(Tw) is saturation vapor pressure at wet bulb temperature Tw , Ta is dry bulb (air) temperature, and γ is the psychrometer constant 0.66 mb o C-1 . Tables of saturation vapor pressure es are provided. ]

ANSWER:

a)

C

mbmbCT

tablefrommbe

eeTT

TTee

TTee

Cmba

Cs

Ts

wa

wa

Ts

waTs

w

w

w

o

o

o

o

4.13

66.00.1027.12

0.10

)(27.12

)(

)0.10(

)(

)(

)(

=

−+=

=

−+=

−=−

−−=

γ

γ

γ

b) Use chart, table or equation to find the temperature that would saturate air that has a vapor pressure of 10.0mb. Get Td=7? C.

c) At Ta=13.4? C, mbeCs

36.15)4.13(

=o . So, %6536.150.10

100100)(

=×=×=mbmb

ee

RHaTs

8

25. At the summit of Mary’s Peak (elevation 1250m), a student measures that the atmospheric pressure

is 850mb and the air density is 1.07 kg m-3. She then drives to Waldport, on the coast. Using the pressure reading from Mary’s Peak, estimate

(i) the temperature at the top of Mary’s Peak in °C (ii) the atmospheric pressure she would measure at sea level in Waldport.

[HINTS: (a) Assume that the change of pressure with height is given by P=Po (exp –z/H), where P

is the pressure (mb) at height z (in km), Po is sea level pressure, and H is approximately 7.6 km. (b) The Gas Law is P= ρ R T where P is air pressure (N m-2), ρ is density (kg m-3), R is the gas constant of air (287 N m kg -1 K-1), and T is temperature (K) (c) 1mb = 100 N m-2, 273 K = 0°C]

ANSWER:

i)

( )( )CK

T

mbpRp

T

RTp

KkgNm

mkg

mN

mN

o4277

28707.1

85000

85000850

3

2

2

==

=

==

=

=

ρ

ρ

ii) Solve for p0 in the pressure equation:

mbe

mbe

pp

Hz

epp

kmkm

Hz

1001

850

0

66.725.1

0

=

=

=

−

=

−

−

9

26. A student, dressed only in shorts, is partying on a beach in Mexico one night during Spring Break.

The air temperature Ta is 30oC, and windspeed u is 5 m s-1. If his rate of heat loss by convection C is 100 W m-2, estimate his skin temperature Ts. Explain any assumptions you have made.

[HINT: Ignore effects of clothing. Convective heat loss is given by C = a (Ts – Ta) / rb , where the constant a is 1.2 x 103 J m-3 oC-1. Boundary layer resistance rb is given by rb = 307 (d/u)0.5 , where rb is in s m-1 when d is in m and u is in m s-1. Assume that d is 0.17m for the student.]

ANSWER: Assume rc = 0(shorts won’t stop very much). In class, we called a ?cp.

( )

( )( ) ( )( )

C

CC

Tc

CrT

r

TTcC

ud

r

CmJ

ms

smJ

CmJ

ms

mW

ap

bs

b

asp

ms

b

o

oo

oo

7.34

301200

6.5610030

1200

6.56100

6.56307

3

2

3

2

=

+=+=

+=

−=

==

ρ

ρ