ATM OCN (Meteorology) 100

Answers for Homework 3

Summer 2004

Date Due: Thursday, 15 July 2004

The total maximum points were 100. Point distribution for each question noted below. 



Use Tables 6.5 and 6.6 of your textbook (The Psychrometric Tables), to determine the dewpoint and relative humidity when the air temperature is 15˚C and the wet bulb temperature is 12C. [Please include appropriate units!]

Part A (6 pts. or 2 pts each)

This question involves reading (hopefully, correctly) the Psychrometric Tables. Note that the vertical columns in the Psychrometric Tables contain the Wet-bulb depression, defined as the arithmetic difference between the dry bulb (or air) temperature and the wet-bulb temperature.  Thus, you need to make this Wet-bulb depression calculation first.


Wet-bulb depression = 

Relative Humidity = 

Dewpoint temperature =

15C - 12C = 3
9.6C (10C is acceptable)


Use the table for saturation vapor pressure (see bottom of 1st homework page) to calculate the following:

a. If the air temperature were 76F and the dewpoint were 56F, what is the relative humidity?

b. If air temperature were 62F and the relative humidity were 80%, what is the dewpoint?

c. Which condition a. or b. would have more water vapor? Please explain!

Part B
a. (4 pts.)

Relative humidity = 49.93% (or 50% is acceptable) 
The relative humidity is determined from RH = e/es x 100. Since the air temperature is 76F, you can determine that the saturation vapor pressure es at that temperature to be 30.64 mb from the table. Using the dewpoint temperature of 56F, the tabular entry would indicate an actual vapor pressure of e = 15.30 mb (since cooling the air from 76F to 56F to get saturation and dew to form would reduce the value of the saturation vapor pressure, es ,  to become equal to the actual vapor pressure, e).
Hence, RH = (15.30 mb / 30.64 mb) x 100 = 49.93%. 

b. (4 pts.)

 Dewpoint = 56
In this case from the air temperature = 62F, the saturation vapor pressure es = 18.96 mb can determined from the Table. Since the relative humidity is known (80%), you can solve for the actual vapor pressure, by rearranging the equation from above to yield:
e = es x RH/100% or e = 18.96 mb x [80% / 100%] = 15.17 mb.
Now enter the table to find the dewpoint corresponding to this vapor pressure of 15.2 mb -- or 55.8F (using a simple interpolation), which can be rounded up to 56F.  (Use this latter value for the next part c.)

        c. (6 pts.)

Both cases, a and b, have essentially the same water vapor content, 
since the dewpoint in each case is approximately 56F; equivalently, both have a vapor pressure of approximately 15.3 mb. This problem is meant to illustrate that while the relative humidity in b is higher (80% as compared with 50% in a), the air temperature is lower, but the dewpoint is essentially the same.  Recall that the dewpoint is the best indicator of actual water vapor content of the air.  NOTE: Suggestion that b is more humid because of a higher relative humidity is incorrect!


PART A. The initial ascent 
(Clearly show your work for partial credit!)

An air parcel is part of a Pacific maritime air mass moving toward the West Coast. The parcel, located just above the ocean surface, has an initial air temperature of 10C and a dewpoint temperature of 5C. This air parcel is forced to ascend the Coastal Range, with a 2000 m elevation.
A.1. How far up the west (windward) slope of the mountain range would the air have to be lifted in order for it to become saturated? (Assume no change in moisture content of the air, no physical phase changes and that the dewpoint remained constant).
[Please include units!]
A.2. What is the air temperature at this level?
A.3. What is the dewpoint temperature at this level? 

Part A (6 pts. or 2 pts. each)

The height of the cloud base, where the lifted air just becomes saturated is 500 meters (1/2 km). 
Assume that the air parcel cools at a dry adiabatic lapse rate of 10 Celsius degrees per 1000 meters and that the surface dewpoint remains constant. Since the parcel only needs to cool by 5 Celsius degrees (= 10C - 5C) to become saturated, only a 500 meter lift of the air is needed to saturate the air. 


Air temperature = 5
See above discussion; using the dry adiabatic lapse rate over 500 meters, the air parcel cools by 5 Celsius degrees, yielding (10C - 5C) = 5C. 


Dewpoint = 5
For this simple model, the dewpoint is assumed to remain constant for a dry adiabatic ascent (See above discussion). {In reality, the dewpoint does decrease when lifted by a rate of about 0.2 Celsius degrees per 1000 meters.} 

PART B. To the top

The air parcel continues up the 2000-meter mountain, condensing out moisture as clouds as it rises and all condensate falls to the ground as precipitation.
B.1. What is the air parcel temperature at the crest of the mountain (2000 m above sea level)?
[Assume the average moist adiabatic lapse rate of 7 Celsius degrees per 1 km]
B.2. What is the dewpoint of the parcel at the crest of the mountain? 

Part B (6 pts. or 3 pts. each)

Air temperature = -5.5
Since the air parcel is now saturated, the air parcel will cool at the saturation adiabatic lapse rate (7 Celsius degrees per 1000 meters) over (2000 m - 500 m) = 1500 meters. In other words, the parcel cools by 10.5 Celsius degrees, meaning that 5C - 10.5C = -5.5C. 


  Dewpoint = -5.5
When the saturated air is lifted, the relative humidity will remain at 100% as condensation takes place; in other words the dewpoint simultaneously cools along with the air temperature, since the parcel will remain saturated. 

PART C: The descent.

The air parcel now moves down the east slope of the mountain, to a valley that has an elevation essentially at sea level.
C.1. What is the air parcel temperature in this valley?
    [Assume no physical phase changes in any residual moisture or clouds]
C.2. What is the dewpoint temperature of the parcel in this valley? 

Part C (6 pts. or 3 pts. each)

Air temperature = 14.5
Using the dry adiabatic lapse rate, an air parcel that sinks over 2000 meters will be heated by 20 Celsius degrees, from -5.5C to 14.5C. 


Dewpoint = -5.5
For this simple model, since the dewpoint is assumed to remain constant for a dry adiabatic descent (see above discussion), the dewpoint will remain the same as air at the top of the mountain. 

PART D: Comparison.

D.1. How has the parcel temperature changed from start to the end?
D.2. How has the parcel dewpoint changed from start to the end?
D.3. How has the relative humidity changed from beginning to end?

Part. D (6 pts. or 3 pts. each)

Air temperature = 4.5 Celsius degrees higher
Parcel started at 10C and ended at 14.5C for a net warming of +4.5 Celsius degrees [or (14.5 - 10) C.] 


Dewpoint = 10.5 Celsius degrees less
Parcel dewpoint started at 5C and ended at -5.5C, for a net change of -10.5 Celsius degrees.

Note: A follow-up question (not asked) would be:
          How has the relative humidity changed from beginning to end?

Relative humidity is less at the end
Since the air temperature increased and the dewpoint decreased, the temperature-dewpoint spread increased, which means that the relative humidity decreased. (This qualitative answer was sufficient) 
For a rigorous proof, the Relative humidity of the parcel initially was approximately 71% (since T = 10C = 50F, which yields a saturation vapor pressure of 12.3 mb and a dewpoint of 5C = 41F yields an actual vapor pressure of 8.7 mb) 
The relative humidity at the end was 24.5% (since T = 14.5C = 58.1F or saturation vapor pressure of 1.5 mb and a dewpoint of -5.5˚C = 22.1˚F has an actual vapor pressure of 4.05 mb).



Using the tables of Apparent Temperature Index and Heat Stress Hazards in your text book (pages 132133), determine the apparent temperatures for the following set of air temperatures and relative humidities.  Indicate what human hazards are possible for these cases. (8 pts)

Air Temperature


Apparent temperature

Hazard to humans 




IV Fatigue possible 
III Sunstroke, heat cramps 
and heat exhaustion possible 
II Sunstroke, heat cramps 
and heat exhaustion likely 
No adverse effects 

a.    Under what condition(s) would the apparent temperature be less than the observed air temperature? (2 pts)

The Apparent temperature would be lower than the observed ambient air temperature for low relative humidity values, typically less than 50%. 

b.     Does this situation appear reasonable? Why? Explain with personal examples. (3 pts)

Yes, this situation is very reasonable since greater evaporation of perspiration (sweat) occurs when the relative humidity is low, making a person feel cooler due to the latent heat of evaporation. Any one who has traveled to the Desert Southwest or other places in the West where the moisture content of the air is usually low (low dewpoints or low relative humidities) typically would remark that they do not feel as hot as in the more humid Southeast (such as Florida), where dewpoints generally are higher. Even in Madison, days where a hot conditions associated with a west wind with lower humidities do not feel as uncomfortable as those days when a south wind would carry humid tropical air masses from the Gulf of Mexico. 

c.     How is heat lost most effectively from the human body on a hot summer day? (2 pts)

Perspiration (sweat) is evaporated from the skin and the latent heat of evaporation aids to cool the human body since approximately 590 calories per gram of water are utilized. 


(17 pts., 4 pts each for a and b, 3 pts for c, 6 pts for d)

a. Name two (2) requirements for the collision-coalescence theory:

1. Warm clouds with temperatures greater than 0˚C;
2. Non uniform sized droplets. 

b. Name two (2) requirements for the ice crystal (Bergeron) theory:

1. Cold clouds with temperatures less than 0˚C;
2. A mixture of ice crystals and supercooled water droplets. 

c. Why is the simple diffusion-condensation process not considered a plausible precipitation formation process?

The simple diffusion-condensation process would take as much as 36 to 48 hours to produce a sufficiently large raindrop from a cloud droplet - too long a time. 

d. By how many times does the volume of a typical cloud droplet (with a diameter of 20 micrometers) have to increase to form a typical raindrop (diameter of 2 millimeter)? (Please show your work for partial credit!) (Hint: 1 millimeter = 1000 micrometers)

Recall that the volume of a sphere is proportional to the cube of the sphere's radius. So, you can form a proportion between the radii (or one half the diameter) after you have converted the radii to common units: 
Radius of the cloud droplet is 20 micrometers/2 = 0.01 millimeter 
Radius of the droplet is 200 millimeters/2 = 1 millimeter 
Thus, the ratio between the radii is 1/100, but the ratio of the volumes is the ratio of the cubes of the radii or 1/1,000,000. 
In other words, the volume of the raindrop in this example is million (1,000,000) times the volume of the droplet. 


(6 pts, 3 pts each)

Distinguish between freezing rain and ice pellets:

Freezing rain

Freezing rain contains raindrops that freeze upon contact with a cold surface (temperatures at or below the nominal freezing point of 0˚C or 32˚F). These droplets do not bounce, but form a glaze. Large accumulations of ice or glaze in a freezing rainstorm is also known as an ice storm. Freezing rain occurs when only a relatively shallow layer of cold air remains near the surface, while a deep layer of warm air (temperatures above freezing) occur aloft. 

Ice pellets:

Ice pellets (also known as sleet) are frozen raindrops that have frozen in a relatively deep (as compared with the above freezing rain situation) layer of cold air that lies above the surface, but below a deep warm air layer. Ice pellets bounce when they hit the ground. 

(11 pts)

During the winter of 1995-96, Madison received four major snowstorms with total snowfall amounts for each storm totaling 4 inches or more. On 27 Nov. 1995, 7.6 inches of snow fell, with a liquid water equivalent of 1.33 inch. On 26 to 27 Jan 1996, 13.4 inches of snow fell, with a liquid water equivalent of 1.00 inch. Calculate the snow to liquid ratio (the number of inches of snow to every inch of liquid water equivalent) for each storm.

27 Nov. 1995:  The snow to liquid ratio is 7.6 inches of snow to 1.33 inches, or forming a proportion so as to state the ratio with one inch in the denominator requires dividing 7.6 to 1.33 which equals 5.71 to 1, 
              or approximately 6 to 1. 
26 to 27 Jan. 1996:  The snow to liquid ratio is 13.4 inches of snow  to 1.00 inches of rain, or
             approximately, 13 to 1.

Which storm had the "fluffy stuff" (as compared with the heavy, wet snow with a high liquid content)?

The 26 to 27 Jan 1996 storm had the more fluffy snow, since the 13.4 to 1 ratio is larger than the 5.7 to 1 ratio for the storm of 27 Nov. 1995. In other words, the January storm (which had thunder and lightning) had a higher amount to each inch of snow, meaning more air space in between the ice crystals and snowflakes. 

How do these ratios for the two storms compare with the typical 10 to 1 ratio often used?

The 10 to 1 ratio is only a rough rule of thumb - since in the first case, the snow was heavier and gave a ratio that was about half the typical ratio (meaning that the standard ratio would underestimate the liquid content from the snow depth), while in the second case the snow was more fluffy and would pile to a 30% greater depth than would be estimated from the liquid equivalent (or stated differently, the liquid equivalent would be overestimated from the snow depth). 


(9 pts)
             An explanation: This exercise was originally used with an older book by  J. M. Moran that had an appendix with precipitation extremes.  While the newer book does not have such a table, the decision was made to use the tables from the links from the precipitation lecture:

A different version of the homework was created, since some of the questions could not be answered easily.  However, the older version was inadvertently posted.  This section will be graded with that in mind. EJH

Where is the greatest annual total precipitation in the United States? 

Kuki, Maui HI  (1982)
NOTE: Several other answers are also accepted:
a. Mt. Waialeale, Kauai, HI (a multi-year average) 
b. Hilo, HI  (a multi-year average of large cities)

How much precipitation was observed at this locale? 

1788 cm (704 inches) for Kuki, Maui HI 
Other locations:
a.  1199 cm (460 inches) for Mt. Waialeale, Kauai, HI (a 30-year average)
b.  325.5 cm (128 inches) for Hilo, HI

What was the greatest amount of rain that has been recorded in one day in the United States? 

1092 mm (43 inches) 

Where was this daily record rainfall recorded? 

Alvin, TX 

Approximately how long is the world record dry spell? 

14 years and 2 months Oct 1903 to Dec 1917  at Arca, Chile
In U.S., 767 days at Bagdad, CA (3 Oct 1912 - 8 Nov 1914).

What was the record amount of snow that has fallen in one day in the United States? 

157.5 cm =  62.0 inches -- in one calendar day (midnight to midnight)
192.5 cm = 75.8 inches  -- in 24-hour span

Where was this daily record snowfall recorded? 

Thompson Pass, Alaska  (29 Dec 1955) -- (calendar day record)
Silver Lake, Boulder County, Colorado (14-15 Apr 1921)  (24-hr record)

Where is the greatest annual total snowfall in the United States? 

Paradise Ranger Station, Mt. Rainier, WA (as found in the text for the 1971-72 snow year).
Since then a new "unofficial" world record has been established at Mt. Baker, WA in Washington State for the 1998-1999 snow year.

How much snow fell at this station? 

2850 cm = 1122 inches (as found in the text for Paradise RS, Mt. Rainier).
(The "unofficial" world record at Mt. Baker, WA in 1998-1999 was 1140 inches.)

Latest revision: 21 July 2004 (1205 UTC)

Produced by Edward J. Hopkins, Ph.D.
Department of Atmospheric and Oceanic Sciences
University of Wisconsin-Madison, Madison, WI 53706

URL Address: aos100/homework/s04hmk03a.html

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