ATM OCN (Meteorology) 100

Answers for Homework 2

Summer 1998

Date Due: Tuesday, 14 July 1998

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

1. ATMOSPHERIC HUMIDITY

PART A. SLING PSYCHROMETER

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 difference between the dry bulb (or air) temperature and the wet bulb temperature.

 

Wet bulb depression =   Relative Humidity =   Dewpoint temperature =

3 deg C  70%  9.6 deg C (10 deg C is acceptable)

 

PART B. OTHER HUMIDITY MEASURES

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

b. If air temperature were 62°F 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 76 deg F, you can determine that the saturation vapor pressure es at that temperature is 30.64 mb from the table. Using the dewpoint temperature of 56 deg, the tabular entry would indicate an actual vapor pressure of 15.30 mb.  Hence, RH = (15.30 mb / 30.64 mb) x 100 = 49.93%.

 

b. (4 pts.)

 

Dewpoint = 56 deg F  In this case from the air temperature = 62 deg F, 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. 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 - or 55.8 deg F (using a simple interpolation), which can be rounded up to 56 deg F.

 

 

Both cases, a and b, have the same water vapor content,   since the dewpoint in each case is 56 deg; 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. NOTE: Suggestion that b is more humid because of a higher relative humidity is incorrect!

 

2. ADIABATIC PROCESSES - THE MOUNTAIN BARRIER

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 10°C and a dewpoint temperature of 5°C. 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? 

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

A.1.

 

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 deg C per 1000 meters and that the surface dewpoint remains constant. Since the parcel only needs to cool by 5 deg C (= 10 deg - 5 deg C) to become saturated, only a 500 meter lift is needed.

 

A.2

 

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

 

A.3

 

Dewpoint = 5 deg C  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 deg C per 1000 meters).

 

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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°C per 1 km]

B.2. What is the dewpoint of the parcel at the crest of the mountain? 

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

B.1.

 

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

 

B.2

 

Dewpoint = -5.5 deg C  When the saturated air is lifted, condensation takes place and the dewpoint simultaneously cools along with the air temperature, since the parcel will remain saturated.

 

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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? 

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

C.1.

 

Air temperature = 14.5 deg C  Using the dry adiabatic lapse rate, causing an air parcel to sink over 2000 meters, will heat the parcel by 20 C degrees, from -5.5 deg C to 14.5 deg C.

 

C.2

 

Dewpoint = -5.5 deg C  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.

 

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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?

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Part. D (9 pts. or 3 pts. each)

D.1.

 

Air temperature = 4.5 C deg warmer Parcel started at 10 deg C and ended at 14.5 deg C for a net warming of +4.5 C deg.

 

D.2
 

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

 

 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 = 10 deg C = 50 deg F, which yields a saturation vapor pressure of 12.3 mb and a dewpoint of 5 deg C = 41 deg F yields an actual vapor pressure of 8.7 mb)  The relative humidity at the end was 24.5% (since T = 14.5 deg C = 58.1 deg F or saturation vapor pressure of 1.5 mb and a dewpoint of -5.5 deg C = 22.1 deg F has an actual vapor pressure of 4.05 mb).

 

3. APPARENT TEMPERATURE

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

Does this situation appear reasonable? Why? Explain with personal examples.

 

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 low (low dewpoints or low relative humidities) typically would remark that they do not feel as hot as in the more humid Southeast. 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.

 

(2 pts)

How is heat lost most effectively from the human body on a hot summer day?
 

4. PRECIPITATION FORMATION THEORIES

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

 

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

 

 

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?
 

 

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.

 

5. PRECIPITATION TYPES

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 rain storm 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.

 

(7 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 or 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.

 

 

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).

 

6. PRECIPITATION EXTREMES

Last revision: 15 July 1998

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